Quantum Chemistry 
Third Edition
Quantum Chemistry 
Third Edition 
John P. Lowe 
Department of Chemistry 
The Pennsylvania State University 
University Park, Pennsylvania 
Kirk A. Peterson 
Department of Chemistry 
Washington State University 
Pullman, Washington 
Amsterdam • Boston • Heidelberg • London • NewYork • Oxford 
Paris • San Diego • San Francisco • Singapore • Sydney • Tokyo
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Library of Congress Cataloging-in-Publication Data 
Lowe, John P. 
Quantum chemistry. -- 3rd ed. / John P. Lowe, Kirk A. Peterson. 
p. cm. 
Includes bibliographical references and index. 
ISBN 0-12-457551-X 
1. Quantum chemistry. I. Peterson, Kirk A. II. Title. 
QD462.L69 2005 
541'.28--dc22 
2005019099 
British Library Cataloguing in Publication Data 
A catalogue record for this book is available from the British Library 
ISBN-13: 978-0-12-457551-6 
ISBN-10: 0-12-457551-X 
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To 
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-J. L.
THE MOLECULAR CHALLENGE 
Sir Ethylene, to scientists fair prey, 
(Who dig and delve and peek and push and pry, 
And prove their findings with equations sly) 
Smoothed out his ruffled orbitals, to say: 
“I stand in symmetry. Mine is a way 
Of mystery and magic. Ancient, I 
Am also deemed immortal. Should I die, 
Pi would be in the sky, and Judgement Day 
Would be upon us. For all things must fail, 
That hold our universe together, when 
Bonds such as bind me fail, and fall asunder. 
Hence, stand I firm against the endless hail 
Of scientific blows. I yield not.” Men 
And their computers stand and stare and wonder. 
W.G. LOWE
Contents 
Preface to the Third Edition xvii 
Preface to the Second Edition xix 
Preface to the First Edition xxi 
1 ClassicalWaves and the Time-Independent Schr¨odingerWave Equation 1 
1-1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 
1-2 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 
1-3 The ClassicalWave Equation . . . . . . . . . . . . . . . . . . . . . 4 
1-4 StandingWaves in a Clamped String . . . . . . . . . . . . . . . . . 7 
1-5 Light as an ElectromagneticWave . . . . . . . . . . . . . . . . . . . 9 
1-6 The Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . 10 
1-7 TheWave Nature of Matter . . . . . . . . . . . . . . . . . . . . . . 14 
1-8 A Diffraction Experiment with Electrons . . . . . . . . . . . . . . . 16 
1-9 Schr¨odinger’s Time-IndependentWave Equation . . . . . . . . . . . 19 
1-10 Conditions on . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 
1-11 Some Insight into the Schr¨odinger Equation . . . . . . . . . . . . . 22 
1-12 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . 25 
Reference . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 
2 Quantum Mechanics of Some Simple Systems 27 
2-1 The Particle in a One-Dimensional “Box” . . . . . . . . . . . . . . . 27 
2-2 Detailed Examination of Particle-in-a-Box Solutions . . . . . . . . . 30 
2-3 The Particle in a One-Dimensional “Box” with One FiniteWall . . . 38 
2-4 The Particle in an Infinite “Box” with a Finite Central Barrier . . . . 44 
2-5 The Free Particle in One Dimension . . . . . . . . . . . . . . . . . . 47 
2-6 The Particle in a Ring of Constant Potential . . . . . . . . . . . . . . 50 
2-7 The Particle in a Three-Dimensional Box: Separation of Variables . . 53 
2-8 The Scattering of Particles in One Dimension . . . . . . . . . . . . . 56 
2-9 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . 65 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
ix
x Contents 
3 The One-Dimensional Harmonic Oscillator 69 
3-1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 
3-2 Some Characteristics of the Classical One-Dimensional Harmonic 
Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 
3-3 The Quantum-Mechanical Harmonic Oscillator . . . . . . . . . . . . 72 
3-4 Solution of the Harmonic Oscillator Schr¨odinger Equation . . . . . . 74 
3-5 Quantum-Mechanical Average Value of the Potential Energy . . . . . 83 
3-6 Vibrations of Diatomic Molecules . . . . . . . . . . . . . . . . . . . 84 
3-7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . . 88 
4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 89 
4-1 The Schr¨odinger Equation and the Nature of Its Solutions . . . . . . . 89 
4-2 Separation of Variables . . . . . . . . . . . . . . . . . . . . . . . . . 105 
4-3 Solution of the R, , and  Equations . . . . . . . . . . . . . . . . . 106 
4-4 Atomic Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 
4-5 Angular Momentum and Spherical Harmonics . . . . . . . . . . . . . 110 
4-6 Angular Momentum and Magnetic Moment . . . . . . . . . . . . . . 115 
4-7 Angular Momentum in Molecular Rotation—The Rigid Rotor . . . . 117 
4-8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . . 125 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 126 
5 Many-Electron Atoms 127 
5-1 The Independent Electron Approximation . . . . . . . . . . . . . . . 127 
5-2 Simple Products and Electron Exchange Symmetry . . . . . . . . . . 129 
5-3 Electron Spin and the Exclusion Principle . . . . . . . . . . . . . . . 132 
5-4 Slater Determinants and the Pauli Principle . . . . . . . . . . . . . . 137 
5-5 Singlet and Triplet States for the 1s2s Configuration of Helium . . . . 138 
5-6 The Self-Consistent Field, Slater-Type Orbitals, and the Aufbau 
Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 
5-7 Electron Angular Momentum in Atoms . . . . . . . . . . . . . . . . . 149 
5-8 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . . 164 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165 
6 Postulates and Theorems of Quantum Mechanics 166 
6-1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 166 
6-2 TheWavefunction Postulate . . . . . . . . . . . . . . . . . . . . . . . 166 
6-3 The Postulate for Constructing Operators . . . . . . . . . . . . . . . . 167 
6-4 The Time-Dependent Schr¨odinger Equation Postulate . . . . . . . . . 168 
6-5 The Postulate Relating Measured Values to Eigenvalues . . . . . . . . 169 
6-6 The Postulate for Average Values . . . . . . . . . . . . . . . . . . . . 171 
6-7 Hermitian Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 171
Contents xi 
6-8 Proof That Eigenvalues of Hermitian Operators Are Real . . . . . . . 172 
6-9 Proof That Nondegenerate Eigenfunctions of a Hermitian Operator 
Form an Orthogonal Set . . . . . . . . . . . . . . . . . . . . . . . . 173 
6-10 Demonstration That All Eigenfunctions of a Hermitian Operator May 
Be Expressed as an Orthonormal Set . . . . . . . . . . . . . . . . . 174 
6-11 Proof That Commuting Operators Have Simultaneous Eigenfunctions 175 
6-12 Completeness of Eigenfunctions of a Hermitian Operator . . . . . . 176 
6-13 The Variation Principle . . . . . . . . . . . . . . . . . . . . . . . . 178 
6-14 The Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . . . . 178 
6-15 Measurement, Commutators, and Uncertainty . . . . . . . . . . . . 178 
6-16 Time-Dependent States . . . . . . . . . . . . . . . . . . . . . . . . 180 
6-17 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . 189 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 
7 The Variation Method 190 
7-1 The Spirit of the Method . . . . . . . . . . . . . . . . . . . . . . . . 190 
7-2 Nonlinear Variation: The Hydrogen Atom . . . . . . . . . . . . . . 191 
7-3 Nonlinear Variation: The Helium Atom . . . . . . . . . . . . . . . . 194 
7-4 Linear Variation: The Polarizability of the Hydrogen Atom . . . . . 197 
7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion . . . 206 
7-6 Molecular Orbitals of Homonuclear Diatomic Molecules . . . . . . . 220 
7-7 Basis Set Choice and the VariationalWavefunction . . . . . . . . . . 231 
7-8 Beyond the Orbital Approximation . . . . . . . . . . . . . . . . . . 233 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . 241 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 242 
8 The Simple H¨uckel Method and Applications 244 
8-1 The Importance of Symmetry . . . . . . . . . . . . . . . . . . . . . 244 
8-2 The Assumption of s–p Separability . . . . . . . . . . . . . . . . . 244 
8-3 The Independent p-Electron Assumption . . . . . . . . . . . . . . . 246 
8-4 Setting up the H¨uckel Determinant . . . . . . . . . . . . . . . . . . 247 
8-5 Solving the HMO Determinantal Equation for Orbital Energies . . . 250 
8-6 Solving for the Molecular Orbitals . . . . . . . . . . . . . . . . . . 251 
8-7 The Cyclopropenyl System: Handling Degeneracies . . . . . . . . . 253 
8-8 Charge Distributions from HMOs . . . . . . . . . . . . . . . . . . . 256 
8-9 Some Simplifying Generalizations . . . . . . . . . . . . . . . . . . 259 
8-10 HMO Calculations on Some Simple Molecules . . . . . . . . . . . . 263 
8-11 Summary: The Simple HMO Method for Hydrocarbons . . . . . . . 268 
8-12 Relation Between Bond Order and Bond Length . . . . . . . . . . . 269 
8-13 p-Electron Densities and Electron Spin Resonance Hyperfine 
Splitting Constants . . . . . . . . . . . . . . . . . . . . . . . . . . . 271 
8-14 Orbital Energies and Oxidation-Reduction Potentials . . . . . . . . . 275 
8-15 Orbital Energies and Ionization Energies . . . . . . . . . . . . . . . 278 
8-16 p-Electron Energy and Aromaticity . . . . . . . . . . . . . . . . . . 279
xii Contents 
8-17 Extension to Heteroatomic Molecules . . . . . . . . . . . . . . . . 284 
8-18 Self-Consistent Variations of a and ß . . . . . . . . . . . . . . . . 287 
8-19 HMO Reaction Indices . . . . . . . . . . . . . . . . . . . . . . . . 289 
8-20 Conclusions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 295 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 296 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . 305 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 
9 Matrix Formulation of the Linear Variation Method 308 
9-1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 308 
9-2 Matrices and Vectors . . . . . . . . . . . . . . . . . . . . . . . . . 308 
9-3 Matrix Formulation of the Linear Variation Method . . . . . . . . . 315 
9-4 Solving the Matrix Equation . . . . . . . . . . . . . . . . . . . . . 317 
9-5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323 
10 The Extended H¨uckel Method 324 
10-1 The Extended H¨uckel Method . . . . . . . . . . . . . . . . . . . . 324 
10-2 Mulliken Populations . . . . . . . . . . . . . . . . . . . . . . . . . 335 
10-3 Extended H¨uckel Energies and Mulliken Populations . . . . . . . . 338 
10-4 Extended H¨uckel Energies and Experimental Energies . . . . . . . 340 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 347 
11 The SCF-LCAO-MO Method and Extensions 348 
11-1 Ab Initio Calculations . . . . . . . . . . . . . . . . . . . . . . . . 348 
11-2 The Molecular Hamiltonian . . . . . . . . . . . . . . . . . . . . . 349 
11-3 The Form of theWavefunction . . . . . . . . . . . . . . . . . . . . 349 
11-4 The Nature of the Basis Set . . . . . . . . . . . . . . . . . . . . . 350 
11-5 The LCAO-MO-SCF Equation . . . . . . . . . . . . . . . . . . . . 350 
11-6 Interpretation of the LCAO-MO-SCF Eigenvalues . . . . . . . . . 351 
11-7 The SCF Total Electronic Energy . . . . . . . . . . . . . . . . . . 352 
11-8 Basis Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 
11-9 The Hartree–Fock Limit . . . . . . . . . . . . . . . . . . . . . . . 357 
11-10 Correlation Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 357 
11-11 Koopmans’ Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 358 
11-12 Configuration Interaction . . . . . . . . . . . . . . . . . . . . . . . 360 
11-13 Size Consistency and the Møller–Plesset and Coupled Cluster 
Treatments of Correlation . . . . . . . . . . . . . . . . . . . . . . 365 
11-14 Multideterminant Methods . . . . . . . . . . . . . . . . . . . . . . 367 
11-15 Density Functional Theory Methods . . . . . . . . . . . . . . . . . 368 
11-16 Examples of Ab Initio Calculations . . . . . . . . . . . . . . . . . 370 
11-17 Approximate SCF-MO Methods . . . . . . . . . . . . . . . . . . . 384 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 386 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 388
Contents xiii 
12 Time-Independent Rayleigh–Schr¨odinger Perturbation Theory 391 
12-1 An Introductory Example . . . . . . . . . . . . . . . . . . . . . . . 391 
12-2 Formal Development of the Theory for Nondegenerate States . . . . 391 
12-3 A Uniform Electrostatic Perturbation of an Electron in a “Wire” . . 396 
12-4 The Ground-State Energy to First-Order of Heliumlike Systems . . 403 
12-5 Perturbation at an Atom in the Simple H¨uckel MO Method . . . . . 406 
12-6 Perturbation Theory for a Degenerate State . . . . . . . . . . . . . 409 
12-7 Polarizability of the Hydrogen Atom in the n=2 States . . . . . . . 410 
12-8 Degenerate-Level Perturbation Theory by Inspection . . . . . . . . 412 
12-9 Interaction Between Two Orbitals: An Important Chemical Model . 414 
12-10 Connection Between Time-Independent Perturbation Theory and 
Spectroscopic Selection Rules . . . . . . . . . . . . . . . . . . . . 417 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 420 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . 427 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 428 
13 Group Theory 429 
13-1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 
13-2 An Elementary Example . . . . . . . . . . . . . . . . . . . . . . . 429 
13-3 Symmetry Point Groups . . . . . . . . . . . . . . . . . . . . . . . 431 
13-4 The Concept of Class . . . . . . . . . . . . . . . . . . . . . . . . . 434 
13-5 Symmetry Elements and Their Notation . . . . . . . . . . . . . . . 436 
13-6 Identifying the Point Group of a Molecule . . . . . . . . . . . . . . 441 
13-7 Representations for Groups . . . . . . . . . . . . . . . . . . . . . . 443 
13-8 Generating Representations from Basis Functions . . . . . . . . . . 446 
13-9 Labels for Representations . . . . . . . . . . . . . . . . . . . . . . 451 
13-10 Some Connections Between the Representation Table and Molecular 
Orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 452 
13-11 Representations for Cyclic and Related Groups . . . . . . . . . . . 453 
13-12 Orthogonality in Irreducible Inequivalent Representations . . . . . 456 
13-13 Characters and Character Tables . . . . . . . . . . . . . . . . . . . 458 
13-14 Using Characters to Resolve Reducible Representations . . . . . . 462 
13-15 Identifying Molecular Orbital Symmetries . . . . . . . . . . . . . . 463 
13-16 Determining in Which Molecular Orbital an Atomic Orbital Will 
Appear . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 465 
13-17 Generating Symmetry Orbitals . . . . . . . . . . . . . . . . . . . . 467 
13-18 Hybrid Orbitals and Localized Orbitals . . . . . . . . . . . . . . . 470 
13-19 Symmetry and Integration . . . . . . . . . . . . . . . . . . . . . . 472 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 476 
Multiple Choice Questions . . . . . . . . . . . . . . . . . . . . . . 481 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 483 
14 Qualitative Molecular Orbital Theory 484 
14-1 The Need for a Qualitative Theory . . . . . . . . . . . . . . . . . . 484 
14-2 Hierarchy in Molecular Structure and in Molecular Orbitals . . . . 484 
14-3 H+2
Revisited . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 485 
14-4 H2: Comparisons with H+2 . . . . . . . . . . . . . . . . . . . . . . 488
xiv Contents 
14-5 Rules for Qualitative Molecular Orbital Theory . . . . . . . . . . . 490 
14-6 Application of QMOT Rules to Homonuclear Diatomic Molecules . 490 
14-7 Shapes of Polyatomic Molecules: Walsh Diagrams . . . . . . . . . 495 
14-8 Frontier Orbitals . . . . . . . . . . . . . . . . . . . . . . . . . . . 505 
14-9 Qualitative Molecular Orbital Theory of Reactions . . . . . . . . . 508 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 521 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524 
15 Molecular Orbital Theory of Periodic Systems 526 
15-1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 526 
15-2 The Free Particle in One Dimension . . . . . . . . . . . . . . . . . 526 
15-3 The Particle in a Ring . . . . . . . . . . . . . . . . . . . . . . . . . 529 
15-4 Benzene . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 530 
15-5 General Form of One-Electron Orbitals in Periodic Potentials— 
Bloch’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 533 
15-6 A Retrospective Pause . . . . . . . . . . . . . . . . . . . . . . . . 537 
15-7 An Example: Polyacetylene with Uniform Bond Lengths . . . . . . 537 
15-8 Electrical Conductivity . . . . . . . . . . . . . . . . . . . . . . . . 546 
15-9 Polyacetylene with Alternating Bond Lengths—Peierls’ Distortion . 547 
15-10 Electronic Structure of All-Trans Polyacetylene . . . . . . . . . . . 551 
15-11 Comparison of EHMO and SCF Results on Polyacetylene . . . . . 552 
15-12 Effects of Chemical Substitution on the p Bands . . . . . . . . . . 554 
15-13 Poly-Paraphenylene—A Ring Polymer . . . . . . . . . . . . . . . 555 
15-14 Energy Calculations . . . . . . . . . . . . . . . . . . . . . . . . . 562 
15-15 Two-Dimensional Periodicity and Vectors in Reciprocal Space . . . 562 
15-16 Periodicity in Three Dimensions—Graphite . . . . . . . . . . . . . 565 
15-17 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 576 
Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 578 
References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 580 
Appendix 1 Useful Integrals 582 
Appendix 2 Determinants 584 
Appendix 3 Evaluation of the Coulomb Repulsion Integral Over 1s AOs 587 
Appendix 4 Angular Momentum Rules 591 
Appendix 5 The Pairing Theorem 601 
Appendix 6 H¨uckel Molecular Orbital Energies, Coefficients, Electron 
Densities, and Bond Orders for Some Simple Molecules 605 
Appendix 7 Derivation of the Hartree–Fock Equation 614 
Appendix 8 The Virial Theorem for Atoms and Diatomic Molecules 624
Contents xv 
Appendix 9 Bra-ket Notation 629 
Appendix 10 Values of Some Useful Constants and Conversion Factors 631 
Appendix 11 Group Theoretical Charts and Tables 636 
Appendix 12 Hints for Solving Selected Problems 651 
Appendix 13 Answers to Problems 654 
Index 691
Preface to the Third Edition 
We have attempted to improve and update this text while retaining the features that 
make it unique, namely, an emphasis on physical understanding, and the ability to 
estimate, evaluate, and predict results without blind reliance on computers, while still 
maintaining rigorous connection to the mathematical basis for quantum chemistry. We 
have inserted into most chapters examples that allowimportant points to be emphasized, 
clarified, or extended. This has enabled us to keep intact most of the conceptual 
development familiar to past users. In addition, many of the chapters now include 
multiple choice questions that students are invited to solve in their heads. This is not 
because we think that instructors will be using such questions. Rather it is because we 
find that such questions permit us to highlight some of the definitions or conclusions 
that students often find most confusing far more quickly and effectively than we can 
by using traditional problems. Of course, we have also sought to update material 
on computational methods, since these are changing rapidly as the field of quantum 
chemistry matures. 
This book is written for courses taught at the first-year graduate/senior undergraduate 
levels, which accounts for its implicit assumption that many readers will be relatively 
unfamiliar with much of the mathematics and physics underlying the subject. Our 
experience over the years has supported this assumption; many chemistry majors are 
exposed to the requisite mathematics and physics, yet arrive at our courses with poor 
understanding or recall of those subjects. That makes this course an opportunity for 
such students to experience the satisfaction of finally seeing how mathematics, physics, 
and chemistry are intertwined in quantum chemistry. It is for this reason that treatments 
of the simple and extended Hückel methods continue to appear, even though these are no 
longer the methods of choice for serious computations. These topics nevertheless form 
the basis for the way most non-theoretical chemists understand chemical processes, 
just as we tend to think about gas behavior as “ideal, with corrections.” 
xvii
Preface to the Second Edition 
The success of the first edition has warranted a second. The changes I have made reflect 
my perception that the book has mostly been used as a teaching text in introductory 
courses. Accordingly, I have removed some of the material in appendixes on mathematical 
details of solving matrix equations on a computer. Also I have removed computer 
listings for programs, since these are now commonly available through commercial 
channels. I have added a new chapter on MO theory of periodic systems—a subject 
of rapidly growing importance in theoretical chemistry and materials science and one 
for which chemists still have difficulty finding appropriate textbook treatments. I have 
augmented discussion in various chapters to give improved coverage of time-dependent 
phenomena and atomic term symbols and have provided better connection to scattering 
as well as to spectroscopy of molecular rotation and vibration. The discussion 
on degenerate-level perturbation theory is clearer, reflecting my own improved understanding 
since writing the first edition. There is also a new section on operator methods 
for treating angular momentum. Some teachers are strong adherents of this approach, 
while others prefer an approach that avoids the formalism of operator techniques. To 
permit both teaching methods, I have placed this material in an appendix. Because this 
edition is more overtly a text than a monograph, I have not attempted to replace older 
literature references with newer ones, except in cases where there was pedagogical 
benefit. 
A strength of this book has been its emphasis on physical argument and analogy (as 
opposed to pure mathematical development). I continue to be a strong proponent of 
the view that true understanding comes with being able to “see” a situation so clearly 
that one can solve problems in one’s head. There are significantly more end-of-chapter 
problems, a number of them of the “by inspection” type. There are also more questions 
inviting students to explain their answers. I believe that thinking about such questions, 
and then reading explanations from the answer section, significantly enhances learning. 
It is the fashion today to focus on state-of-the-art methods for just about everything. 
The impact of this on education has, I feel, been disastrous. Simpler examples are often 
needed to develop the insight that enables understanding the complexities of the latest 
techniques, but too often these are abandoned in the rush to get to the “cutting edge.” 
For this reason I continue to include a substantial treatment of simple H¨uckel theory. 
It permits students to recognize the connections between MOs and their energies and 
bonding properties, and it allows me to present examples and problems that have maximum 
transparency in later chapters on perturbation theory, group theory, qualitative 
MO theory, and periodic systems. I find simple H¨uckel theory to be educationally 
indispensable. 
xix
xx Preface to the Second Edition 
Much of the new material in this edition results from new insights I have developed 
in connection with research projects with graduate students. The work of all four of 
my students since the appearance of the first edition is represented, and I am delighted 
to thank Sherif Kafafi, John LaFemina, Maribel Soto, and Deb Camper for all I have 
learned from them. Special thanks are due to Professor Terry Carlton, of Oberlin 
College, who made many suggestions and corrections that have been adopted in the 
new edition. 
Doubtless, there are new errors. I would be grateful to learn of them so that future 
printings of this edition can be made error-free. Students or teachers with comments, 
questions, or corrections are more than welcome to contact me, either by mail at the 
Department of Chemistry, 152 Davey Lab, The Pennsylvania State University, University 
Park, PA 16802, or by e-mail directed to JL3 at PSUVM.PSU.EDU.
Preface to the First Edition 
My aim in this book is to present a reasonably rigorous treatment of molecular orbital 
theory, embracing subjects that are of practical interest to organic and inorganic as well 
as physical chemists. My approach here has been to rely on physical intuition as much 
as possible, first solving a number of specific problems in order to develop sufficient 
insight and familiarity to make the formal treatment of Chapter 6 more palatable. My 
own experience suggests that most chemists find this route the most natural. 
I have assumed that the reader has at some time learned calculus and elementary 
physics, but I have not assumed that this material is fresh in his or her mind. Other 
mathematics is developed as it is needed. The book could be used as a text for undergraduate 
or graduate students in a half or full year course. The level of rigor of the book 
is somewhat adjustable. For example, Chapters 3 and 4, on the harmonic oscillator and 
hydrogen atom, can be truncated if one wishes to know the nature of the solutions, but 
not the mathematical details of how they are produced. 
I have made use of appendixes for certain of the more complicated derivations or 
proofs. This is done in order to avoid having the development of major ideas in the 
text interrupted or obscured. Certain of the appendixes will interest only the more 
theoretically inclined student. Also, because I anticipate that some readers may wish 
to skip certain chapters or parts of chapters, I have occasionally repeated information 
so that a given chapter will be less dependent on its predecessors. This may seem 
inelegant at times, but most students will more readily forgive repetition of something 
they already know than an overly terse presentation. 
I have avoided early usage of bra-ket notation. I believe that simultaneous introduction 
of new concepts and unfamiliar notation is poor pedagogy. Bra-ket notation is 
used only after the ideas have had a change to jell. 
Problem solving is extremely important in acquiring an understanding of quantum 
chemistry. I have included a fair number of problems with hints for a few of them in 
Appendix 14 and answers for almost all of them in Appendix 15.1 
It is inevitable that one be selective in choosing topics for a book such as this. This 
book emphasizes ground stateMOtheory of molecules more than do most introductory 
texts, with rather less emphasis on spectroscopy than is usual. Angular momentum 
is treated at a fairly elementary level at various appropriate places in the text, but 
it is never given a full-blown formal development using operator commutation relations. 
Time-dependent phenomena are not included. Thus, scattering theory is absent, 
1In this Second Edition, these Appendices are numbered Appendix 12 and 13. 
xxi
xxii Preface to the First Edition 
although selection rules and the transition dipole are discussed in the chapter on timeindependent 
perturbation theory. Valence-bond theory is completely absent. If I have 
succeeded in my effort to provide a clear and meaningful treatment of topics relevant to 
modern molecular orbital theory, it should not be difficult for an instructor to provide 
for excursions into related topics not covered in the text. 
Over the years, many colleagues have been kind enough to read sections of the 
evolving manuscript and provide corrections and advice. I especially thank L. P. Gold 
and O. H. Crawford, who cheerfully bore the brunt of this task. 
Finally, I would like to thank my father,Wesley G. Lowe, for allowing me to include 
his sonnet, “The Molecular Challenge.”
Chapter 1 
ClassicalWaves 
and the Time-Independent 
Schr ¨ odingerWave Equation 
1-1 Introduction 
The application of quantum-mechanical principles to chemical problems has revolutionized 
the field of chemistry. Today our understanding of chemical bonding, spectral 
phenomena, molecular reactivities, and various other fundamental chemical problems 
rests heavily on our knowledge of the detailed behavior of electrons in atoms and 
molecules. In this book we shall describe in detail some of the basic principles, 
methods, and results of quantum chemistry that lead to our understanding of electron 
behavior. 
In the first fewchapters we shall discuss some simple, but important, particle systems. 
This will allow us to introduce many basic concepts and definitions in a fairly physical 
way. Thus, some background will be prepared for the more formal general development 
of Chapter 6. In this first chapter, we review briefly some of the concepts of classical 
physics as well as some early indications that classical physics is not sufficient to explain 
all phenomena. (Those readers who are already familiar with the physics of classical 
waves and with early atomic physics may prefer to jump ahead to Section 1-7.) 
1-2 Waves 
1-2.A TravelingWaves 
A very simple example of a traveling wave is provided by cracking a whip. A pulse of 
energy is imparted to the whipcord by a single oscillation of the handle. This results 
in a wave which travels down the cord, transferring the energy to the popper at the end 
of the whip. In Fig. 1-1, an idealization of the process is sketched. The shape of the 
disturbance in the whip is called the wave profile and is usually symbolized .(x). The 
wave profile for the traveling wave in Fig. 1-1 shows where the energy is located at a 
given instant. It also contains the information needed to tell how much energy is being 
transmitted, because the height and shape of the wave reflect the vigor with which the 
handle was oscillated. 
1
2 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
Figure 1-1 
 Cracking the whip. As time passes, the disturbance moves from left to right along 
the extended whip cord. Each segment of the cord oscillates up and down as the disturbance passes 
by, ultimately returning to its equilibrium position. 
The feature common to all traveling waves in classical physics is that energy is transmitted 
through a medium. The medium itself undergoes no permanent displacement; 
it merely undergoes local oscillations as the disturbance passes through. 
One of the most important kinds of wave in physics is the harmonic wave, for which 
thewave profile is a sinusoidal function. Aharmonicwave, at a particular instant in time, 
is sketched in Fig. 1-2. The maximum displacement of the wave from the rest position 
is the amplitude of the wave, and the wavelength . is the distance required to enclose 
one complete oscillation. Such a wave would result from a harmonic1 oscillation at 
one end of a taut string. Analogous waves would be produced on the surface of a quiet 
pool by a vibrating bob, or in air by a vibrating tuning fork. 
At the instant depicted in Fig. 1-2, the profile is described by the function 
.(x)=Asin(2px/.) (1-1) 
(. =0 when x =0, and the argument of the sine function goes from 0 to 2p, encompassing 
one complete oscillation as x goes from 0 to ..) Let us suppose that the situation 
in Fig. 1-2 pertains at the time t =0, and let the velocity of the disturbance through the 
medium be c. Then, after time t , the distance traveled is ct, the profile is shifted to the 
right by ct and is now given by 
(x, t)=Asin[(2p/.)(x -ct)] (1-2) 
Figure 1-2 
 A harmonic wave at a particular instant in time. A is the amplitude and . is the 
wavelength. 
1A harmonic oscillation is one whose equation of motion has a sine or cosine dependence on time.
Section 1-2 Waves 3 
A capital  is used to distinguish the time-dependent function (1-2) from the timeindependent 
function (1-1). 
The frequency . of a wave is the number of individual repeating wave units passing 
a point per unit time. For our harmonic wave, this is the distance traveled in unit time 
c divided by the length of a wave unit .. Hence, 
. =c/. (1-3) 
Note that the wave described by the formula 

(x, t)=Asin[(2p/.)(x -ct)+] (1-4) 
is similar to  of Eq. (1-2) except for being displaced. If we compare the two waves 
at the same instant in time, we find 
 to be shifted to the left of  by ./2p. If 
 =p, 3p, . . . , then 
 is shifted by ./2, 3./2, . . . and the two functions are said to be 
exactly out of phase. If  =2p, 4p, . . . , the shift is by ., 2., . . . , and the two waves 
are exactly in phase.  is the phase factor for 
 relative to . Alternatively, we can 
compare the two waves at the same point in x, in which case the phase factor causes 
the two waves to be displaced from each other in time. 
1-2.B StandingWaves 
In problems of physical interest, the medium is usually subject to constraints. For 
example, a string will have ends, and these may be clamped, as in a violin, so that 
they cannot oscillate when the disturbance reaches them. Under such circumstances, 
the energy pulse is unable to progress further. It cannot be absorbed by the clamping 
mechanism if it is perfectly rigid, and it has no choice but to travel back along the string 
in the opposite direction. The reflected wave is now moving into the face of the primary 
wave, and the motion of the string is in response to the demands placed on it by the two 
simultaneous waves: 
(x, t)=primary(x, t)+reflected(x, t) (1-5) 
When the primary and reflected waves have the same amplitude and speed, we can 
write 
(x, t) = Asin [(2p/.)(x -ct)]+Asin [(2p/.)(x +ct)] 
= 2Asin(2px/.) cos(2pct/.) (1-6) 
This formula describes a standing wave—a wave that does not appear to travel through 
the medium, but appears to vibrate “in place.” The first part of the function depends 
only on the x variable. Wherever the sine function vanishes,  will vanish, regardless 
of the value of t . This means that there are places where the medium does not ever 
vibrate. Such places are called nodes. Between the nodes, sin(2px/.) is finite. As 
time passes, the cosine function oscillates between plus and minus unity. This means 
that  oscillates between plus and minus the value of sin(2px/.). We say that the xdependent 
part of the function gives the maximum displacement of the standing wave, 
and the t-dependent part governs the motion of the medium back and forth between 
these extremes of maximum displacement. A standing wave with a central node is 
shown in Fig. 1-3.
4 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
Figure 1-3 
 A standing wave in a string clamped at x =0 and x =L. The wavelength . is equal 
to L. 
Equation (1-6) is often written as 
(x, t)=.(x) cos(.t) (1-7) 
where 
.=2pc/. (1-8) 
The profile .(x) is often called the amplitude function and . is the frequency factor. 
Let us consider how the energy is stored in the vibrating string depicted in Fig. 1-3. 
The string segments at the central node and at the clamped endpoints of the string 
do not move. Hence, their kinetic energies are zero at all times. Furthermore, since 
they are never displaced from their equilibrium positions, their potential energies are 
likewise always zero. Therefore, the total energy stored at these segments is always 
zero as long as the string continues to vibrate in the mode shown. The maximum kinetic 
and potential energies are associated with those segments located at the wave peaks 
and valleys (called the antinodes) because these segments have the greatest average 
velocity and displacement from the equilibrium position. Amore detailed mathematical 
treatment would show that the total energy of any string segment is proportional to 
.(x)2 (Problem 1-7). 
1-3 The ClassicalWave Equation 
It is one thing to draw a picture of a wave and describe its properties, and quite another 
to predict what sort of wave will result from disturbing a particular system. To make 
such predictions, we must consider the physical laws that the medium must obey. One 
condition is that the medium must obey Newton’s laws of motion. For example, any 
segment of string of massmsubjected to a force F must undergo an acceleration ofF/m 
in accord with Newton’s second law. In this regard, wave motion is perfectly consistent 
with ordinary particle motion. Another condition, however, peculiar to waves, is that 
each segment of the medium is “attached” to the neighboring segments so that, as 
it is displaced, it drags along its neighbor, which in turn drags along its neighbor,
Section 1-3 The ClassicalWave Equation 5 
Figure 1-4 
 A segment of string under tension T . The forces at each end of the segment are 
decomposed into forces perpendicular and parallel to x. 
etc. This provides the mechanism whereby the disturbance is propagated along the 
medium.2 
Let us consider a string under a tensile force T . When the string is displaced from 
its equilibrium position, this tension is responsible for exerting a restoring force. For 
example, observe the string segment associated with the region x to x +dx in Fig. 1-4. 
Note that the tension exerted at either end of this segment can be decomposed into 
components parallel and perpendicular to the x axis. The parallel component tends to 
stretch the string (which, however, we assume to be unstretchable), the perpendicular 
component acts to accelerate the segment toward or away from the rest position. At 
the right end of the segment, the perpendicular component F divided by the horizontal 
component gives the slope of T . However, for small deviations of the string from 
equilibrium (that is, for small angle a) the horizontal component is nearly equal in 
length to the vector T . This means that it is a good approximation to write 
slope of vector T =F/T at x +dx (1-9) 
But the slope is also given by the derivative of , and so we can write 
Fx+dx =T (./.x)x+dx (1-10) 
At the other end of the segment the tensile force acts in the opposite direction, and we 
have 
Fx =-T (./.x)x (1-11) 
The net perpendicular force on our string segment is the resultant of these two: 
F =T 
(./.x)x+dx -(./.x)x (1-12) 
The difference in slope at two infinitesimally separated points, divided by dx, is by 
definition the second derivative of a function. Therefore, 
F =T .2/.x2 dx (1-13) 
2Fluids are of relatively low viscosity, so the tendency of one segment to drag along its neighbor is weak. For 
this reason fluids are poor transmitters of transverse waves (waves in which the medium oscillates in a direction 
perpendicular to the direction of propagation). In compression waves, one segment displaces the next by pushing 
it. Here the requirement is that the medium possess elasticity for compression. Solids and fluids often meet this 
requirement well enough to transmit compression waves. The ability of rigid solids to transmit both wave types 
while fluids transmit only one type is the basis for using earthquake-induced waves to determine how deep the 
solid part of the earth’s mantle extends.
6 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
Equation (1-13) gives the force on our string segment. If the string has mass m per 
unit length, then the segment has mass mdx, and Newton’s equation F =ma may be 
written 
T .2/.x2 =m.2/.t2 (1-14) 
where we recall that acceleration is the second derivative of position with respect to time. 
Equation (1-14) is the wave equation for motion in a string of uniform density 
under tension T . It should be evident that its derivation involves nothing fundamental 
beyond Newton’s second law and the fact that the two ends of the segment are linked 
to each other and to a common tensile force. Generalizing this equation to waves in 
three-dimensional media gives 
 .2 
.x2 + 
.2 
.y2 + 
.2 
.z2 
 (x, y, z, t) =ß 
.2 (x, y, z, t) 
.t2 (1-15) 
where ß is a composite of physical quantities (analogous to m/T ) for the particular 
system. 
Returning to our string example, we have in Eq. (1-14) a time-dependent differential 
equation. Suppose we wish to limit our consideration to standing waves that can be 
separated into a space-dependent amplitude function and a harmonic time-dependent 
function. Then 
(x, t)=.(x) cos(.t) (1-16) 
and the differential equation becomes 
cos(.t)
d2. (x) 
dx2 = 
m
T 
.(x)
d2 cos(.t) 
dt2 =-
m
T 
.(x).2 cos(.t) (1-17) 
or, dividing by cos(.t), 
d2.(x)/dx2=-(.2m/T ).(x) (1-18) 
This is the classical time-independent wave equation for a string. 
We can see by inspection what kind of function .(x) must be to satisfy Eq. (1-18). 
. is a function that, when twice differentiated, is reproduced with a coefficient of 
-.2m/T . One solution is 
. =Asin .m/T x (1-19) 
This illustrates that Eq. (1-18) has sinusoidally varying solutions such as those discussed 
in Section 1-2. Comparing Eq. (1-19) with (1-1) indicates that 2p/. = .vm/T . 
Substituting this relation into Eq. (1-18) gives 
d2.(x)/dx2=-(2p/.)2.(x) (1-20) 
which is a more useful form for our purposes. 
For three-dimensional systems, the classical time-independent wave equation for an 
isotropic and uniform medium is 
(.2/.x2 +.2/.y2 +.2/.z2).(x, y, z)=-(2p/.)2.(x, y, z) (1-21)
Section 1-4 StandingWaves in a Clamped String 7 
where . depends on the elasticity of the medium. The combination of partial derivatives 
on the left-hand side of Eq. (1-21) is called the Laplacian, and is often given the shorthand 
symbol .2 (del squared). This would give for Eq. (1-21) 
.2.(x, y, z)=-(2p/.)2.(x, y, z) (1-22) 
1-4 StandingWaves in a Clamped String 
We now demonstrate how Eq. (1-20) can be used to predict the nature of standing waves 
in a string. Suppose that the string is clamped at x = 0 and L. This means that the 
string cannot oscillate at these points. Mathematically this means that 
.(0)=.(L)=0 (1-23) 
Conditions such as these are called boundary conditions. Our question is, “What 
functions . satisfy Eq. (1-20) and also Eq. (1-23)?” We begin by trying to find the most 
general equation that can satisfy Eq. (1-20). We have already seen that Asin(2px/.) 
is a solution, but it is easy to show that Acos(2px/.) is also a solution. More general 
than either of these is the linear combination3 
.(x)=Asin(2px/.)+B cos(2px/.) (1-24) 
By varying A and B, we can get different functions .. 
There are two remarks to be made at this point. First, some readers will have 
noticed that other functions exist that satisfy Eq. (1-20). These are Aexp(2pix/.) 
and Aexp(-2pix/.), where i = v-1. The reason we have not included these in 
the general function (1-24) is that these two exponential functions are mathematically 
equivalent to the trigonometric functions. The relationship is 
exp(±ikx)=cos(kx)±i sin(kx). (1-25) 
This means that any trigonometric function may be expressed in terms of such exponentials 
and vice versa. Hence, the set of trigonometric functions and the set of exponentials 
is redundant, and no additional flexibility would result by including exponentials in 
Eq. (1-24) (see Problem 1-1). The two sets of functions are linearly dependent.4 
The second remark is that for a given A and B the function described by Eq. (1-24) 
is a single sinusoidal wave with wavelength .. By altering the ratio of A to B, we cause 
the wave to shift to the left or right with respect to the origin. If A=1 and B =0, the 
wave has a node at x =0. If A=0 and B =1, the wave has an antinode at x =0. 
We nowproceed by letting the boundary conditions determine the constantsAandB. 
The condition at x =0 gives
.(0)=Asin(0)+B cos(0)=0 (1-26) 
3Given functions f1,f2,f3 . . . . A linear combination of these functions is c1f1 +c2f2 +c3f3 +···, where 
c1, c2, c3, . . . are numbers (which need not be real). 
4If one member of a set of functions (f1,f2,f3, . . . ) can be expressed as a linear combination of the remaining 
functions (i.e., if f1 =c2f2 +c3f3 +···), the set of functions is said to be linearly dependent. Otherwise, they 
are linearly independent.
8 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
However, since sin(0)=0 and cos(0)=1, this gives 
B =0 (1-27) 
Therefore, our first boundary condition forces B to be zero and leaves us with 
.(x)=Asin(2px/.) (1-28) 
Our second boundary condition, at x =L, gives 
.(L)=Asin(2pL/.)=0 (1-29) 
One solution is provided by settingAequal to zero. This gives. =0, which corresponds 
to no wave at all in the string. This is possible, but not very interesting. The other 
possibility is for 2pL/. to be equal to 0,±p,±2p, . . . ,±np, . . . since the sine function 
vanishes then. This gives the relation 
2pL/.=np, n=0,±1,±2, . . . (1-30) 
or 
.=2L/n, n=0,±1,±2, . . . (1-31) 
Substituting this expression for . into Eq. (1-28) gives 
.(x)=Asin(npx/L), n=0,±1,±2, . . . (1-32) 
Some of these solutions are sketched in Fig. 1-5. The solution for n=0 is again the 
uninteresting . =0 case. Furthermore, since sin(-x) equals -sin(x), it is clear that 
the set of functions produced by positive integers n is not physically different from the 
set produced by negative n, so we may arbitrarily restrict our attention to solutions with 
positive n. (The two sets are linearly dependent.) The constant A is still undetermined. 
It affects the amplitude of the wave. To determine A would require knowing how much 
energy is stored in the wave, that is, how hard the string was plucked. 
It is evident that there are an infinite number of acceptable solutions, each one 
corresponding to a different number of half-waves fitting between 0 and L. But an even 
larger infinity of waves has been excluded by the boundary conditions—namely, all 
waves having wavelengths not divisible into 2L an integral number of times. The result 
Figure 1-5 
 Solutions for the time-independent wave equation in one dimension with boundary 
conditions .(0)=.(L)=0.
Section 1-5 Light as an ElectromagneticWave 9 
of applying boundary conditions has been to restrict the allowed wavelengths to certain 
discrete values. As we shall see, this behavior is closely related to the quantization of 
energies in quantum mechanics. 
The example worked out above is an extremely simple one. Nevertheless, it demonstrates 
how a differential equation and boundary conditions are used to define the 
allowed states for a system. One could have arrived at solutions for this case by simple 
physical argument, but this is usually not possible in more complicated cases. The differential 
equation provides a systematic approach for finding solutions when physical 
intuition is not enough. 
1-5 Light as an ElectromagneticWave 
Suppose a charged particle is caused to oscillate harmonically on the z axis. If there 
is another charged particle some distance away and initially at rest in the xy plane, 
this second particle will commence oscillating harmonically too. Thus, energy is being 
transferred from the first particle to the second, which indicates that there is an oscillating 
electric field emanating from the first particle. We can plot the magnitude of 
this electric field at a given instant as it would be felt by a series of imaginary test 
charges stationed along a line emanating from the source and perpendicular to the axis 
of vibration (Fig. 1-6). 
If there are some magnetic compasses in the neighborhood of the oscillating charge, 
these will be found to swing back and forth in response to the disturbance. This means 
that an oscillating magnetic field is produced by the charge too. Varying the placement 
of the compasses will show that this field oscillates in a plane perpendicular to the 
axis of vibration of the charged particle. The combined electric and magnetic fields 
traveling along one ray in the xy plane appear in Fig. 1-7. 
The changes in electric and magnetic fields propagate outward with a characteristic 
velocity c, and are describable as a traveling wave, called an electromagnetic wave. 
Its frequency . is the same as the oscillation frequency of the vibrating charge. Its 
wavelength is . = c/.. Visible light, infrared radiation, radio waves, microwaves, 
ultraviolet radiation, X rays, and . rays are all forms of electromagnetic radiation, 
their only difference being their frequencies .. We shall continue the discussion in the 
context of light, understanding that it applies to all forms of electromagnetic radiation. 
Figure 1-6 
 A harmonic electric-field wave emanating from a vibrating electric charge. The wave 
magnitude is proportional to the force felt by the test charges. The charges are only imaginary; if 
they actually existed, they would possess mass and under acceleration would absorb energy from the 
wave, causing it to attenuate.
10 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
Figure 1-7 
 A harmonic electromagnetic field produced by an oscillating electric charge. The 
arrows without attached charges show the direction in which the north pole of a magnet would be 
attracted. The magnetic field is oriented perpendicular to the electric field. 
If a beam of light is produced so that the orientation of the electric field wave is 
always in the same plane, the light is said to be plane (or linearly) polarized. The planepolarized 
light shown in Fig. 1-7 is said to be z polarized. If the plane of orientation 
of the electric field wave rotates clockwise or counterclockwise about the axis of travel 
(i.e., if the electric field wave “corkscrews” through space), the light is said to be right 
or left circularly polarized. If the light is a composite of waves having random field 
orientations so that there is no resultant orientation, the light is unpolarized. 
Experiments with light in the nineteenth century and earlier were consistent with 
the view that light is a wave phenomenon. One of the more obvious experimental 
verifications of this is provided by the interference pattern produced when light from a 
point source is allowed to pass through a pair of slits and then to fall on a screen. The 
resulting interference patterns are understandable only in terms of the constructive and 
destructive interference of waves. The differential equations of Maxwell, which provided 
the connection between electromagnetic radiation and the basic laws of physics, 
also indicated that light is a wave. 
But there remained several problems that prevented physicists from closing the book 
on this subject. Onewas the inability of classical physical theory to explain the intensity 
andwavelength characteristics of light emitted by a glowing “blackbody.” This problem 
was studied by Planck, who was forced to conclude that the vibrating charged particles 
producing the light can exist only in certain discrete (separated) energy states. We 
shall not discuss this problem. Another problem had to do with the interpretation of a 
phenomenon discovered in the late 1800s, called the photoelectric effect. 
1-6 The Photoelectric Effect 
This phenomenon occurs when the exposure of some material to light causes it to eject 
electrons. Many metals do this quite readily. A simple apparatus that could be used to 
study this behavior is drawn schematically in Fig. 1-8. Incident light strikes the metal 
dish in the evacuated chamber. If electrons are ejected, some of them will strike the 
collecting wire, giving rise to a deflection of the galvanometer. In this apparatus, one 
can vary the potential difference between the metal dish and the collecting wire, and 
also the intensity and frequency of the incident light. 
Suppose that the potential difference is set at zero and a current is detected when 
light of a certain intensity and frequency strikes the dish. This means that electrons
Section 1-6 The Photoelectric Effect 11 
Figure 1-8 
 A phototube. 
are being emitted from the dish with finite kinetic energy, enabling them to travel to 
the wire. If a retarding potential is now applied, electrons that are emitted with only a 
small kinetic energy will have insufficient energy to overcome the retarding potential 
and will not travel to the wire. Hence, the current being detected will decrease. The 
retarding potential can be increased gradually until finally even the most energetic 
photoelectrons cannot make it to the collecting wire. This enables one to calculate the 
maximum kinetic energy for photoelectrons produced by the incident light on the metal 
in question. 
The observations from experiments of this sort can be summarized as follows: 
1. Below a certain cutoff frequency of incident light, no photoelectrons are ejected, no 
matter how intense the light. 
2. Above the cutoff frequency, the number of photoelectrons is directly proportional 
to the intensity of the light. 
3. As the frequency of the incident light is increased, the maximum kinetic energy of 
the photoelectrons increases. 
4. In cases where the radiation intensity is extremely low (but frequency is above the 
cutoff value) photoelectrons are emitted from the metal without any time lag. 
Some of these results are summarized graphically in Fig. 1-9. Apparently, the kinetic 
energy of the photoelectron is given by 
kinetic energy=h(. -.0) (1-33) 
where h is a constant. The cutoff frequency .0 depends on the metal being studied (and 
also its temperature), but the slope h is the same for all substances. 
We can also write the kinetic energy as 
kinetic energy=energy of light-energy needed to escape surface (1-34)
12 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
Figure 1-9 
 Maximum kinetic energy of photoelectrons as a function of incident light frequency, 
where .0 is the minimum frequency for which photoelectrons are ejected from the metal in the absence 
of any retarding or accelerating potential. 
The last quantity in Eq. (1-34) is often referred to as the work function W of the metal. 
Equating Eq. (1-33) with (1-34) gives 
energy of light-W =h. -h.0 (1-35) 
The material-dependent term W is identified with the material-dependent term h.0, 
yielding 
energy of light=E =h. (1-36) 
where the value of h has been determined to be 6.626176×10-34 J sec. (See Appendix 
10 for units and conversion factors.) 
Physicists found it difficult to reconcile these observations with the classical electromagnetic 
field theory of light. For example, if light of a certain frequency and intensity 
causes emission of electrons having a certain maximum kinetic energy, one would 
expect increased light intensity (corresponding classically to a greater electromagnetic 
field amplitude and hence greater energy density) to produce photoelectrons of higher 
kinetic energy. However, it only produces more photoelectrons and does not affect their 
energies. Again, if light is a wave, the energy is distributed over the entire wavefront 
and this means that a low light intensity would impart energy at a very low rate to an 
area of surface occupied by one atom. One can calculate that it would take years for an 
individual atom to collect sufficient energy to eject an electron under such conditions. 
No such induction period is observed. 
An explanation for these results was suggested in 1905 by Einstein, who proposed 
that the incident light be viewed as being comprised of discrete units of energy. Each 
such unit, or photon, would have an associated energy of h.,where . is the frequency 
of the oscillating emitter. Increasing the intensity of the light would correspond to 
increasing the number of photons, whereas increasing the frequency of the light would 
increase the energy of the photons. If we envision each emitted photoelectron as 
resulting from a photon striking the surface of the metal, it is quite easy to see that 
Einstein’s proposal accords with observation. But it creates a new problem: If we are 
to visualize light as a stream of photons, how can we explain the wave properties of 
light, such as the double-slit diffraction pattern? What is the physical meaning of the 
electromagnetic wave?
Section 1-6 The Photoelectric Effect 13 
Essentially, the problem is that, in the classical view, the square of the electromagnetic 
wave at any point in space is a measure of the energy density at that point. Now 
the square of the electromagnetic wave is a continuous and smoothly varying function, 
and if energy is continuous and infinitely divisible, there is no problem with this theory. 
But if the energy cannot be divided into amounts smaller than a photon—if it has 
a particulate rather than a continuous nature—then the classical interpretation cannot 
apply, for it is not possible to produce a smoothly varying energy distribution from 
energy particles any more than it is possible to produce, at the microscopic level, a 
smooth density distribution in gas made from atoms of matter. Einstein suggested that 
the square of the electromagnetic wave at some point (that is, the sum of the squares 
of the electric and magnetic field magnitudes) be taken as the probability density for 
finding a photon in the volume element around that point. The greater the square of 
the wave in some region, the greater is the probability for finding the photon in that 
region. Thus, the classical notion of energy having a definite and smoothly varying 
distribution is replaced by the idea of a smoothly varying probability density for finding 
an atomistic packet of energy. 
Let us explore this probabilistic interpretation within the context of the two-slit 
interference experiment. We know that the pattern of light and darkness observed on 
the screen agrees with the classical picture of interference of waves. Suppose we carry 
out the experiment in the usual way, except we use a light source (of frequency .) so 
weak that only h. units of energy per second pass through the apparatus and strike 
the screen. According to the classical picture, this tiny amount of energy should strike 
the screen in a delocalized manner, producing an extremely faint image of the entire 
diffraction pattern. Over a period of many seconds, this pattern could be accumulated 
(on a photographic plate, say) andwould become more intense. According to Einstein’s 
view, our experiment corresponds to transmission of one photon per second and each 
photon strikes the screen at a localized point. Each photon strikes a new spot (not to 
imply the same spot cannot be struck more than once) and, over a long period of time, 
they build up the observed diffraction pattern. If we wish to state in advance where the 
next photon will appear, we are unable to do so. The best we can do is to say that the 
next photon is more likely to strike in one area than in another, the relative probabilities 
being quantitatively described by the square of the electromagnetic wave. 
The interpretation of electromagnetic waves as probability waves often leaves one 
with some feelings of unreality. If the wave only tells us relative probabilities for 
finding a photon at one point or another, one is entitled to ask whether the wave has 
“physical reality,” or if it is merely a mathematical device which allows us to analyze 
photon distribution, the photons being the “physical reality.” We will defer discussion 
of this question until a later section on electron diffraction. 
EXAMPLE 1-1 Aretarding potential of 2.38 volts just suffices to stop photoelectrons 
emitted from potassium by light of frequency 1.13×1015 s-1. What is the work 
function, W, of potassium? 
SOLUTION 
 Elight =h. =W +KEelectron,W =h. -KEelectron =(4.136×10-15eV s) 
(1.13×1015 s-1)-2.38eV=4.67eV-2.38eV=2.29eV [Note convenience of using h in units 
of eV s for this problem. See Appendix 10 for data.] 
14 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
EXAMPLE 1-2 Spectroscopists often express E for a transition between states in 
wavenumbers , e.g., m-1, or cm-1, rather than in energy units like J or eV. (Usually 
cm-1 is favored, so we will proceed with that choice.) 
a) What is the physical meaning of the term wavenumber? 
b) What is the connection between wavenumber and energy? 
c) What wavenumber applies to an energy of 1.000 J? of 1.000 eV? 
SOLUTION 
 a)Wavenumber is the number of waves that fit into a unit of distance (usually of 
one centimeter). It is sometimes symbolized ˜.. ˜. =1/., where . is the wavelength in centimeters. 
b)Wavenumber characterizes the light that has photons of the designated energy. E=h. =hc/.= 
hc˜.. (where c is given in cm/s). 
c) E = 1.000 J = hc˜.; ˜. = 1.000 J/hc = 1.000 J /[(6.626×10-34 J s)(2.998×1010 cm/s)] = 5.034×1022 cm-1. Clearly, this is light of an extremely short wavelength since more than 1022 
wavelengths fit into 1 cm. For 1.000 eV, the above equation is repeated using h in eV s. This gives 
˜. =8065cm-1.  
1-7 TheWave Nature of Matter 
Evidently light haswave and particle aspects, and we can describe it in terms of photons, 
which are associated withwaves of frequency .=E/h. Nowphotons are rather peculiar 
particles in that they have zero rest mass. In fact, they can exist only when traveling 
at the speed of light. The more normal particles in our experience have nonzero rest 
masses and can exist at any velocity up to the speed-of-light limit. Are there also waves 
associated with such normal particles? 
Imagine a particle having a finite rest mass that somehow can be made lighter and 
lighter, approaching zero in a continuous way. It seems reasonable that the existence 
of a wave associated with the motion of the particle should become more and more 
apparent, rather than the wave coming into existence abruptly when m=0. De Broglie 
proposed that all material particles are associated with waves, which he called “matter 
waves,” but that the existence of these waves is likely to be observable only in the 
behaviors of extremely light particles. 
De Broglie’s relation can be reached as follows. Einstein’s relation for photons is 
E =h. (1-37) 
But a photon carrying energy E has a relativistic mass given by 
E =mc2 (1-38) 
Equating these two equations gives 
E =mc2 =h. =hc/. (1-39) 
or 
mc=h/. (1-40)
Section 1-7 TheWave Nature of Matter 15 
Anormal particle, with nonzero rest mass, travels at a velocity v. If we regard Eq. (1-40) 
as merely the high-velocity limit of a more general expression, we arrive at an equation 
relating particle momentum p and associated wavelength .: 
mv =p =h/. (1-41) 
or 
.=h/p (1-42) 
Here, m refers to the rest mass of the particle plus the relativistic correction, but the 
latter is usually negligible in comparison to the former. 
This relation, proposed by de Broglie in 1922, was demonstrated to be correct shortly 
thereafter when Davisson and Germer showed that a beam of electrons impinging on a 
nickel target produced the scattering patterns one expects from interferingwaves. These 
“electron waves” were observed to have wavelengths related to electron momentum in 
just the manner proposed by de Broglie. 
Equation (1-42) relates the de Broglie wavelength . of a matter wave to the momentum 
p of the particle. A higher momentum corresponds to a shorter wavelength. Since 
kinetic energy T =mv2 =(1/2m)(m2v2)=p2/2m (1-43) 
it follows that 
p =v2mT (1-44) 
Furthermore, Since E = T + V , where E is the total energy and V is the potential 
energy, we can rewrite the de Broglie wavelength as 
.= 
h 
v2m(E -V ) 
(1-45) 
Equation (1-45) is useful for understanding the way in which . will change for a 
particle moving with constant total energy in a varying potential. For example, if the 
particle enters a region where its potential energy increases (e.g., an electron approaches 
a negatively charged plate), E -V decreases and . increases (i.e., the particle slows 
down, so its momentum decreases and its associated wavelength increases). We shall 
see examples of this behavior in future chapters. 
Observe that if E = V,. as given by Eq. (1-45) is real. However, if E <V,. 
becomes imaginary. Classically, we never encounter such a situation, but we will find 
it is necessary to consider this possibility in quantum mechanics. 
EXAMPLE 1-3 AHe2+ ion is accelerated from rest through a voltage drop of 1.000 
kilovolts. What is its final deBroglie wavelength? Would the wavelike properties 
be very apparent? 
SOLUTION 
 Since a charge of two electronic units has passed through a voltage drop 
of 1.000×103 volts, the final kinetic energy of the ion is 2.000×103 eV. To calculate ., we first
16 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
convert from eV to joules: KE = p2/2m = (2.000 × 103 eV)(1.60219 × 10-19 J/eV) = 3.204 
× 10-16 J. mHe = (4.003 g/mol)(10-3 kg/g)(1 mol/6.022 × 1023atoms) = 6.65 × 10-27 kg; 
p=v2mHe ·KE=[2(6.65×10-27 kg)(3.204×10-16 J)]1/2 =2.1×10-21 kgm/s. .=h/p= 
(6.626×10-34 Js)/(2.1×10-21 kg m/s)=3.2×10-13 m=0.32 pm. This wavelength is on the 
order of 1% of the radius of a hydrogen atom–too short to produce observable interference results 
when interacting with atom-size scatterers. For most purposes, we can treat this ion as simply a 
high-speed particle.  
1-8 A Diffraction Experiment with Electrons 
In order to gain a better understanding of the meaning of matterwaves, we nowconsider 
a set of simple experiments. Suppose that we have a source of a beam of monoenergetic 
electrons and a pair of slits, as indicated schematically in Fig. 1-10. Any electron 
arriving at the phosphorescent screen produces a flash of light, just as in a television 
set. For the moment we ignore the light source near the slits (assume that it is turned 
off) and inquire as to the nature of the image on the phosphorescent screen when the 
electron beam is directed at the slits. The observation, consistent with the observations 
of Davisson and Germer already mentioned, is that there are alternating bands of light 
and dark, indicating that the electron beam is being diffracted by the slits. Furthermore, 
the distance separating the bands is consistent with the de Broglie wavelength 
corresponding to the energy of the electrons. The variation in light intensity observed 
on the screen is depicted in Fig. 1-11a. 
Evidently, the electrons in this experiment are displaying wave behavior. Does this 
mean that the electrons are spread out like waves when they are detected at the screen? 
We test this by reducing our beam intensity to let only one electron per second through 
the apparatus and observe that each electron gives a localized pinpoint of light, the 
entire diffraction pattern building up gradually by the accumulation of many points. 
Thus, the square of de Broglie’s matterwave has the same kind of statistical significance 
that Einstein proposed for electromagnetic waves and photons, and the electrons really 
are localized particles, at least when they are detected at the screen. 
However, if they are really particles, it is hard to see how they can be diffracted. 
Consider what happens when slit b is closed. Then all the electrons striking the screen 
must have come through slit a. We observe the result to be a single area of light on 
the screen (Fig. 1-11b). Closing slit a and opening b gives a similar (but displaced) 
Figure 1-10 
 The electron source produces a beam of electrons, some of which pass through slits 
a and/or b to be detected as flashes of light on the phosphorescent screen.
Section 1-8 A Diffraction Experiment with Electrons 17 
Figure 1-11 
 Light intensity at phosphorescent screen under various conditions: (a) a and b open, 
light off; (b) a open, b closed, light off; (c) a closed, b open, light off; (d) a and b open, light on, . 
short; (e) a and b open, light on, . longer. 
light area, as shown in Fig. 1-11c. These patterns are just what we would expect for 
particles. Now, with both slits open, we expect half the particles to pass through slit a 
and half through slit b, the resulting pattern being the sum of the results just described. 
Instead we obtain the diffraction pattern (Fig. 1-11a). How can this happen? It seems 
that, somehow, an electron passing through the apparatus can sense whether one or 
both slits are open, even though as a particle it can explore only one slit or the other. 
One might suppose that we are seeing the result of simultaneous traversal of the two 
slits by two electrons, the path of each electron being affected by the presence of an 
electron in the other slit. This would explain how an electron passing through slit a 
would “know” whether slit b was open or closed. But the fact that the pattern builds 
up even when electrons pass through at the rate of one per second indicates that this 
argument will not do. Could an electron be coming through both slits at once? 
To test this question, we need to have detailed information about the positions of the 
electrons as they pass through the slits. We can get such data by turning on the light 
source and aiming a microscope at the slits. Then photons will bounce off each electron 
as it passes the slits and will be observed through the microscope. The observer thus 
can tell through which slit each electron has passed, and also record its final position 
on the phosphorescent screen. In this experiment, it is necessary to use light having 
a wavelength short in comparison to the interslit distance; otherwise the microscope 
cannot resolve a flash well enough to tell which slit it is nearest. When this experiment 
is performed, we indeed detect each electron as coming through one slit or the other, 
and not both, but we also find that the diffraction pattern on the screen has been lost 
and that we have the broad, featureless distribution shown in Fig. 1-11d, which is 
basically the sum of the single-slit experiments. What has happened is that the photons 
from our light source, in bouncing off the electrons as they emerge from the slits, have 
affected the momenta of the electrons and changed their paths from what they were 
in the absence of light. We can try to counteract this by using photons with lower 
momentum; but this means using photons of lower E, hence longer .. As a result, 
the images of the electrons in the microscope get broader, and it becomes more and 
more ambiguous as to which slit a given electron has passed through or that it really 
passed through only one slit. As we become more and more uncertain about the path
18 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
of each electron as it moves past the slits, the accumulating diffraction pattern becomes 
more and more pronounced (Fig. 1-11e). (Since this is a “thought experiment,” we can 
ignore the inconvenient fact that our “light” source must produce X rays or . rays in 
order to have a wavelength short in comparison to the appropriate interslit distance.) 
This conceptual experiment illustrates a basic feature of microscopic systems—we 
cannot measure properties of the system without affecting the future development of the 
system in a nontrivial way. The system with the light turned off is significantly different 
from the system with the light turned on (with short .), and so the electrons arrive at the 
screen with different distributions. No matter how cleverly one devises the experiment, 
there is some minimum necessary disturbance involved in any measurement. In this 
example with the light off, the problem is that we know the momentum of each electron 
quite accurately (since the beam is monoenergetic and collimated), but we do not know 
anything about the way the electrons traverse the slits. With the light on, we obtain 
information about electron position just beyond the slits but we change the momentum 
of each electron in an unknown way. The measurement of particle position leads to 
loss of knowledge about particle momentum. This is an example of the uncertainty 
principle of Heisenberg, who stated that the product of the simultaneous uncertainties in 
“conjugate variables,” a and b, can never be smaller than the value of Planck’s constant 
h divided by 4p: 
a · b=h/4p (1-46) 
Here, a is a measure of the uncertainty in the variable a, etc. (The easiest way to 
recognize conjugate variables is to note that their dimensions must multiply to joule 
seconds. Linear momentum and linear position satisfies this requirement. Two other 
important pairs of conjugate variables are energy–time and angular momentum–angular 
position.) In this example with the light off, our uncertainty in momentum is small 
and our uncertainty in position is unacceptably large, since we cannot say which slit 
each electron traverses. With the light on, we reduce our uncertainty in position to 
an acceptable size, but subsequent to the position of each electron being observed, we 
have much greater uncertainty in momentum. 
Thus, we see that the appearance of an electron (or a photon) as a particle or a wave 
depends on our experiment. Because any observation on so small a particle involves a 
significant perturbation of its state, it is proper to think of the electron plus apparatus 
as a single system. The question, “Is the electron a particle or a wave?” becomes 
meaningful only when the apparatus is defined on which we plan a measurement. 
In some experiments, the apparatus and electrons interact in a way suggestive of the 
electron being a wave, in others, a particle. The question, “What is the electron when 
were not looking?,” cannot be answered experimentally, since an experiment is a “look” 
at the electron. In recent years experiments of this sort have been carried out using 
single atoms.5 
EXAMPLE 1-4 The lifetime of an excited state of a molecule is 2×10-9 s. What 
is the uncertainty in its energy in J? In cm-1? How would this manifest itself 
experimentally? 
5See F. Flam [1].
Section 1-9 Schr ¨ odinger’s Time-IndependentWave Equation 19 
SOLUTION 
 The Heisenberg uncertainty principle gives, for minimum uncertainty E · t = 
h/4p.E=(6.626×10-34 J s)/[(4p)(2×10-9 s)]=2.6×10-26 J (2.6×10-26J) (5.03×1022 
cm-1J-1)=0.001 cm-1 (See Appendix 10 for data.) Larger uncertainty in E shows up as greater 
line-width in emission spectra.  
1-9 Schr ¨ odinger’s Time-IndependentWave Equation 
Earlier we saw that we needed a wave equation in order to solve for the standing waves 
pertaining to a particular classical system and its set of boundary conditions. The same 
need exists for a wave equation to solve for matter waves. Schr¨odinger obtained such 
an equation by taking the classical time-independent wave equation and substituting 
de Broglie’s relation for .. Thus, if 
.2. =-(2p/.)2. (1-47) 
and 
.= 
h 
v2m(E -V ) 
(1-48) 
then 

-(h2/8p2m).2 +V (x, y, z).(x, y, z)=E.(x, y, z) (1-49) 
Equation (1-49) is Schr¨odinger’s time-independent wave equation for a single particle 
of mass m moving in the three-dimensional potential field V . 
In classical mechanics we have separate equations for wave motion and particle 
motion, whereas in quantum mechanics, in which the distinction between particles and 
waves is not clear-cut, we have a single equation—the Schr¨odinger equation. We have 
seen that the link between the Schr¨odinger equation and the classical wave equation is 
the de Broglie relation. Let us now compare Schr¨odinger’s equation with the classical 
equation for particle motion. 
Classically, for a particle moving in three dimensions, the total energy is the sum of 
kinetic and potential energies: 
(1/2m)(p2
x +p2
y +p2
z )+V =E (1-50) 
where px is the momentum in the x coordinate, etc. We have just seen that the analogous 
Schr¨odinger equation is [writing out Eq. (1-49)] 
 -h2 
8p2m 
 .2 
.x2 + 
.2 
.y2 + 
.2 
.z2 
+V (x, y, z)	.(x, y, z)=E.(x, y, z) (1-51) 
It is easily seen that Eq. (1-50) is linked to the quantity in brackets of Eq. (1-51) by a 
relation associating classical momentum with a partial differential operator: 
px ..(h/2pi)(./.x) (1-52) 
and similarly for py and pz . The relations (1-52) will be seen later to be an important 
postulate in a formal development of quantum mechanics.
20 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
The left-hand side of Eq. (1-50) is called the hamiltonian for the system. For this 
reason the operator in square brackets on the LHS of Eq. (1-51) is called the hamiltonian 
operator6 H. For a given system, we shall see that the construction ofH is not difficult. 
The difficulty comes in solving Schr¨odinger’s equation, often written as 
H. =E. (1-53) 
The classical and quantum-mechanical wave equations that we have discussed are 
members of a special class of equations called eigenvalue equations. Such equations 
have the format 
Op f =cf (1-54) 
where Op is an operator, f is a function, and c is a constant. Thus, eigenvalue equations 
have the property that operating on a function regenerates the same function times a 
constant. The function f that satisfies Eq. (1-54) is called an eigenfunction of the 
operator. The constant c is called the eigenvalue associated with the eigenfunction 
f . Often, an operator will have a large number of eigenfunctions and eigenvalues of 
interest associated with it, and so an index is necessary to keep them sorted, viz. 
Op fi =cifi (1-55) 
We have already seen an example of this sort of equation, Eq. (1-19) being an eigenfunction 
for Eq. (1-18), with eigenvalue -.2m/T . 
The solutions . for Schr¨odinger’s equation (1-53), are referred to as eigenfunctions, 
wavefunctions, or state functions. 
EXAMPLE 1-5 a) Show that sin(3.63x) is not an eigenfunction of the operator 
d/dx. 
b) Show that exp(-3.63ix) is an eigenfunction of the operator d/dx. What is its 
eigenvalue? 
c) Show that 1
p sin(3.63x) is an eigenfunction of the operator 
((-h2/8p2m)d2/dx2). What is its eigenvalue? 
SOLUTION 
 a) d 
dx sin(3.63x)=3.63 cos(3.63x) = constant times sin(3.63x). 
b) d 
dx exp(-3.63ix) = -3.63i exp(-3.63ix) =constant times exp(-3.63ix). Eigenvalue= 
-3.63i. 
c) ((-h2/8p2m)d2/dx2) 1
p sin(3.63x) = (-h2/8p2m)(1/p)(3.63) d 
dx cos(3.63x) 
= [(3.63)2h2/8p2m] · (1/p) sin(3.63x) 
= constant times (1/p) sin(3.63x). 
Eigenvalue = (3.63)2h2/8p2m.  
6An operator is a symbol telling us to carry out a certain mathematical operation. Thus, d/dx is a differential 
operator telling us to differentiate anything following it with respect to x. The function 1/x may be viewed as a 
multiplicative operator. Any function on which it operates gets multiplied by 1/x.
Section 1-10 Conditions on . 21 
1-10 Conditions on . 
We have already indicated that the square of the electromagnetic wave is interpreted as 
the probability density function for finding photons at various places in space. We now 
attribute an analogous meaning to .2 for matter waves. Thus, in a one-dimensional 
problem (for example, a particle constrained to move on a line), the probability that the 
particle will be found in the interval dx around the point x1 is taken to be .2(x1)dx. 
If . is a complex function, then the absolute square, |.|2 =.*. is used instead of 
.2.7 This makes it mathematically impossible for the average mass distribution to be 
negative in any region. 
If an eigenfunction . has been found for Eq. (1-53), it is easy to see that c. will 
also be an eigenfunction, for any constant c. This is due to the fact that a multiplicative 
constant commutes8 with the operator H, that is, 
H(c.)=cH. =cE. =E(c.) (1-56) 
The equality of the first and last terms is a statement of the fact that c. is an eigenfunction 
of H. The question of which constant to use for the wavefunction is resolved 
by appeal to the probability interpretation of |.|2. For a particle moving on the x axis, 
the probability that the particle is between x =-8 and x =+8 is unity, that is, a 
certainty. This probability is also equal to the sum of the probabilities for finding the 
particle in each and every infinitesimal interval along x, so this sum (an integral) must 
equal unity: 
c*c 
 +8 
-8 
.*(x). (x) dx =1 (1-57) 
If the selection of the constant multiplier c is made so that Eq. (1-57) is satisfied, 
the wavefunction .
 =c. is said to be normalized. For a three-dimensional function, 
c.(x, y, z), the normalization requirement is 
c*c 
 +8 
-8 

 +8 
-8 

 +8 
-8 
.*(x, y, z).(x, y, z)dx dy dz=|c|2 

all space |.|2 dv =1 
(1-58) 
As a result of our physical interpretation of |.|2 plus the fact that . must be an 
eigenfunction of the hamiltonian operator H, we can reach some general conclusions 
about what sort of mathematical properties . can or cannot have. 
First, we require that . be a single-valued function because we want |.|2 to give an 
unambiguous probability for finding a particle in a given region (see Fig. 1-12). Also, 
we reject functions that are infinite in any region of space because such an infinity 
will always be infinitely greater than any finite region, and |.|2 will be useless as a 
measure of comparative probabilities.9 In order for H. to be defined everywhere, it 
is necessary that the second derivative of . be defined everywhere. This requires that 
the first derivative of . be piecewise continuous and that . itself be continuous as in 
Fig.1d. (We shall see an example of this shortly.) 
7If f =u+iv, then f *, the complex conjugate of f , is given by u-iv, where u and v are real functions. 
8a and b are said to commute if ab=ba. 
9There are cases, particularly in relativistic treatments, where . is infinite at single points of zero measure, so 
that |.|2 dx remains finite. Normally we do not encounter such situations in quantum chemistry.
22 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
Figure 1-12 
 (a) . is triple valued at x0. (b) . is discontinuous at x0. (c) . grows without limit 
as x approaches +8 (i.e., . “blows up,” or “explodes”). (d) . is continuous and has a “cusp” at x0. 
Hence, first derivative of . is discontinuous at x0 and is only piecewise continuous. This does not 
prevent . from being acceptable. 
Functions that are single-valued, continuous, nowhere infinite, and have piecewise 
continuous first derivatives will be referred to as acceptable functions. The meanings 
of these terms are illustrated by some sample functions in Fig. 1-12. 
In most cases, there is one more general restriction we place on ., namely, that 
it be a normalizable function. This means that the integral of |.|2 over all space 
must not be equal to zero or infinity. A function satisfying this condition is said to 
be square-integrable. 
1-11 Some Insight into the Schr ¨ odinger Equation 
There is a fairly simple way to view the physical meaning of the Schr¨odinger equation 
(1-49). The equation essentially states that E in H. =E. depends on two things, V 
and the second derivatives of .. Since V is the potential energy, the second derivatives 
of . must be related to the kinetic energy. Now the second derivative of . with respect 
to a given direction is a measure of the rate of change of slope (i.e., the curvature, or 
“wiggliness”) of . in that direction. Hence, we see that a more wiggly wavefunction 
leads, through the Schr¨odinger equation, to a higher kinetic energy. This is in accord 
with the spirit of de Broglie’s relation, since a shorter wavelength function is a more 
wiggly function. But the Schr¨odinger equation is more generally applicable because we 
can take second derivatives of any acceptable function, whereas wavelength is defined
Section 1-12 Summary 23 
Figure 1-13 
 (a) Since V = 0, E = T . For higher T , . is more wiggly, which means that . is 
shorter. (Since . is periodic for a free particle, . is defined.) (b) As V increases from left to right, . 
becomes less wiggly. (c)–(d) . is most wiggly where V is lowest and T is greatest. 
only for periodic functions. Since E is a constant, the solutions of the Schr¨odinger 
equation must be more wiggly in regions where V is low and less wiggly where V is 
high. Examples for some one-dimensional cases are shown in Fig. 1-13. 
In the next chapter we use some fairly simple examples to illustrate the ideas that 
we have already introduced and to bring out some additional points. 
1-12 Summary 
In closing this chapter, we collect and summarize the major points to be used in 
future discussions. 
1. Associated with any particle is a wavefunction having wavelength related to particle 
momentum by .=h/p =h/v2m(E -V ). 
2. The wavefunction has the following physical meaning; its absolute square is proportional 
to the probability density for finding the particle. If the wavefunction is 
normalized, its square is equal to the probability density. 
3. Thewavefunctions. for time-independent states are eigenfunctions of Schr¨odinger’s 
equation, which can be constructed from the classical wave equation by requiring 
. = h/v2m(E -V ), or from the classical particle equation by replacing pk 
with (h/2pi)./.k, k =x, y, z.
24 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
4. For . to be acceptable, it must be single-valued, continuous, nowhere infinite, with 
a piecewise continuous first derivative. For most situations, we also require . to 
be square-integrable. 
5. The wavefunction for a particle in a varying potential oscillates most rapidly where 
V is low, giving a high T in this region. The low V plus high T equals E. In another 
region, where V is high, the wavefunction oscillates more slowly, giving a low T , 
which, with the high V , equals the same E as in the first region. 
1-12.A Problems10 
1-1. Express Acos(kx)+B sin(kx)+C exp(ikx)+Dexp(-ikx) purely in terms of 
cos(kx) and sin(kx). 
1-2. Repeat the standing-wave-in-a-string problem worked out in Section 1-4, but 
clamp the string at x=+L/2 and -L/2 instead of at 0 and L. 
1-3. Find the condition that must be satisfied by a and ß in order that . (x) = 
Asin(ax)+B cos(ßx) satisfy Eq. (1-20). 
1-4. The apparatus sketched in Fig. 1-8 is used with a dish plated with zinc and also 
with a dish plated with cesium. The wavelengths of the incident light and the 
corresponding retarding potentials needed to just prevent the photoelectrons from 
reaching the collecting wire are given in Table P1-4. Plot incident light frequency 
versus retarding potential for these two metals. Evaluate their work functions 
(in eV) and the proportionality constant h (in eV s). 
TABLE P1-4 
 
Retarding potential (V) 
.(Å) Cs Zn 
6000 0.167 — 
3000 2.235 0.435 
2000 4.302 2.502 
1500 6.369 1.567 
1200 8.436 6.636 
1-5. Calculate the de Broglie wavelength in nanometers for each of the following: 
a) An electron that has been accelerated from rest through a potential change of 
500V. 
b) A bullet weighing 5 gm and traveling at 400m s-1. 
1-6. Arguing from Eq. (1-7), what is the time needed for a standing wave to go through 
one complete cycle? 
10Hints for a few problems may be found in Appendix 12 and answers for almost all of them appear in 
Appendix 13.
Section 1-12 Summary 25 
1-7. The equation for a standing wave in a string has the form 
(x, t)=.(x) cos(.t) 
a) Calculate the time-averaged potential energy (PE) for this motion. [Hint: Use 
PE =- F d; F =ma; a =.2/.t2.] 
b) Calculate the time-averaged kinetic energy (KE) for this motion. [Hint: Use 
KE =1/2mv2 and v =./.t.] 
c) Show that this harmonically vibrating string stores its energy on the average 
half as kinetic and half as potential energy, and that E(x)ava.2(x). 
1-8. Indicate which of the following functions are “acceptable.” If one is not, give 
a reason. 
a) . =x 
b) . =x2 
c) . =sin x 
d) . =exp(-x) 
e) . =exp(-x2) 
1-9. An acceptable function is never infinite. Does this mean that an acceptable 
function must be square integrable? If you think these are not the same, try to 
find an example of a function (other than zero) that is never infinite but is not 
square integrable. 
1-10. Explain why the fact that sin(x) = -sin(-x) means that we can restrict 
Eq. (1-32) to nonnegative n without loss of physical content. 
1-11. Which of the following are eigenfunctions for d/dx? 
a) x2 
b) exp(-3.4x2) 
c) 37 
d) exp(x) 
e) sin(ax) 
f) cos(4x)+i sin(4x) 
1-12. Calculate the minimum de Broglie wavelength for a photoelectron that is produced 
when light of wavelength 140.0 nm strikes zinc metal. (Workfunction of 
zinc=3.63 eV.) 
Multiple Choice Questions 
(Intended to be answered without use of pencil and paper.) 
1. A particle satisfying the time-independent Schr¨odinger equation must have 
a) an eigenfunction that is normalized. 
b) a potential energy that is independent of location. 
c) a de Broglie wavelength that is independent of location.
26 Chapter 1 ClassicalWaves and the Time-Independent Schr ¨ odingerWave Equation 
d) a total energy that is independent of location. 
e) None of the above is a true statement. 
2. When one operates with d2/dx2 on the function 6 sin(4x), one finds that 
a) the function is an eigenfunction with eigenvalue -96. 
b) the function is an eigenfunction with eigenvalue 16. 
c) the function is an eigenfunction with eigenvalue -16. 
d) the function is not an eigenfunction. 
e) None of the above is a true statement. 
3. Which one of the following concepts did Einstein propose in order to explain the 
photoelectric effect? 
a) A particle of rest mass m and velocity v has an associated wavelength . given by 
.=h/mv. 
b) Doubling the intensity of light doubles the energy of each photon. 
c) Increasing the wavelength of light increases the energy of each photon. 
d) The photoelectron is a particle. 
e) None of the above is a concept proposed by Einstein to explain the photoelectric 
effect. 
4. Light of frequency . strikes a metal and causes photoelectrons to be emitted having 
maximum kinetic energy of 0.90 h.. From this we can say that 
a) light of frequency ./2 will not produce any photoelectrons. 
b) light of frequency 2. will produce photoelectrons having maximum kinetic 
energy of 1.80 h.. 
c) doubling the intensity of light of frequency . will produce photoelectrons having 
maximum kinetic energy of 1.80 h.. 
d) the work function of the metal is 0.90 h.. 
e) None of the above statements is correct. 
5. The reason for normalizing a wavefunction . is 
a) to guarantee that . is square-integrable. 
b) to make .*. equal to the probability distribution function for the particle. 
c) to make . an eigenfunction for the Hamiltonian operator. 
d) to make . satisfy the boundary conditions for the problem. 
e) to make . display the proper symmetry characteristics. 
Reference 
[1] F. Flam, Making Waves with Interfering Atoms. Physics Today, 921–922 (1991).
Chapter 2 
Quantum Mechanics of Some 
Simple Systems 
2-1 The Particle in a One-Dimensional “Box” 
Imagine that a particle of mass m is free to move along the x axis between x =0 and 
x =L, with no change in potential (set V =0 for 0<x<L). At x =0 and L and at all 
points beyond these limits the particle encounters an infinitely repulsive barrier (V =8 for x =0, x =L). The situation is illustrated in Fig. 2-1. Because of the shape of this 
potential, this problem is often referred to as a particle in a square well or a particle in 
a box problem. It is well to bear in mind, however, that the situation is really like that 
of a particle confined to movement along a finite length of wire. 
When the potential is discontinuous, as it is here, it is convenient to write a wave 
equation for each region. For the two regions beyond the ends of the box 
-h2 
8p2m 
d2. 
dx2 +8. =E., x =0, x =L (2-1) 
Within the box, . must satisfy the equation 
-h2 
8p2m 
d2. 
dx2 =E., 0<x <L (2-2) 
It should be realized that E must take on the same values for both of these equations; 
the eigenvalue E pertains to the entire range of the particle and is not influenced by 
divisions we make for mathematical convenience. 
Let us examine Eq. (2-1) first. Suppose that, at some point within the infinite barrier, 
say x =L+dx, . is finite. Then the second term on the left-hand side of Eq. (2-1) will 
be infinite. If the first term on the left-hand side is finite or zero, it follows immediately 
that E is infinite at the point L + dx (and hence everywhere in the system). Is it 
possible that a solution exists such that E is finite? One possibility is that . =0 at all 
points where V =8. The other possibility is that the first term on the left-hand side 
of Eq. (2-1) can be made to cancel the infinite second term. This might happen if the 
second derivative of the wavefunction is infinite at all points where V =8and . =0. 
For the second derivative to be infinite, the first derivative must be discontinuous, and 
so . itself must be nonsmooth (i.e., it must have a sharp corner; see Fig. 2-2). Thus, 
we see that it may be possible to obtain a finite value for both E and . at x =L+dx, 
provided that . is nonsmooth there. But what about the next point, x =L+2 dx, and 
all the other points outside the box? If we try to use the same device, we end up with 
the requirement that . be nonsmooth at every point where V =8. A function that is 
27
28 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure 2-1 
 The potential felt by a particle as a function of its x coordinate. 
continuous but which has a point-wise discontinuous first derivative is a contradiction 
in terms (i.e., a continuous f cannot be 100% corners. To have recognizable corners, 
we must have some (continuous) edges. We say that the first derivative of . must be 
piecewise continuous.) Hence, if V =8at a single point, we might find a solution . 
which is finite at that point, with finite energy. If V =8over a finite range of connected 
points, however, either E for the system is infinite, and . is finite over this region or 
E is not infinite (but is indeterminate) and . is zero over this region. 
We are not concerned with particles of infinite energy, and so we will say that the 
solution to Eq. (2-1) is . =0.1 
Turning now to Eq. (2-2) we ask what solutions . exist in the box having associated 
eigenvalues E that are finite and positive. Any function that, when twice differentiated, 
yields a negative constant times the selfsame function is a possible candidate for .. 
Such functions are sin(kx), cos(kx), and exp(±ikx). But these functions are not all 
independent since, as we noted in Chapter 1, 
exp(±ikx)=cos(kx)±i sin(kx) (2-3) 
We thus are free to express . in terms of exp(±ikx) or else in terms of sin(kx) and 
cos(kx). We choose the latter because of their greater familiarity, although the final 
answer must be independent of this choice. 
The most general form for the solution is 
. (x)=Asin(kx)+B cos(kx) (2-4) 
where A, B, and k remain to be determined. As we have already shown, . is zero at 
x =0, x =L and so we have as boundary conditions 
.(0) = 0 (2-5) 
.(L) = 0 (2-6) 
Mathematically, this is precisely the same problem we have already solved in Chapter 
1 for the standing waves in a clamped string. The solutions are 
.(x) = Asin(npx/L), n=1, 2, . . . , 0<x <L 
.(x) = 0, x=0, x =L (2-7) 
1Thus, the particle never gets into these regions. It is meaningless to talk of the energy of the particle in such 
regions, and our earlier statement that E must be identical in Eqs. (2-1) and (2-2) must be modified; E is constant 
in all regions where . is finite.
Section 2-1 The Particle in a One-Dimensional “Box” 29 
Figure 2-2 
 As the function f (x) approaches being nonsmooth, d approaches zero (the width of 
one point) and n approaches infinity. 
One difference between Eq. (2-7) and the string solutions is that we have rejected the 
n=0 solution in Eq. (2-7). For the string, this solution was for no vibration at all— 
a physically realizable circumstance. For the particle-in-a-box problem, this solution 
is rejected because it is not square-integrable. (It gives . =0, which means no particle 
on the x axis, contradicting our starting premise. One could also reject this solution 
for the classical case since it means no energy in the string, which might contradict a 
starting premise depending on how the problem is worded.) 
Let us check to be sure these functions satisfy Schr¨odinger’s equation: 
H.(x) = -h2 
8p2m 
d2[Asin(npx/L)] 
dx2 
= -h2 
8p2m 

-A
n2p2 
L2 sin npx 
L 
 
= 
n2h2 
8mL2 Asin npx 
L 
=E. (x) (2-8) 
This shows that the functions (2-7) are indeed eigenfunctions of H. We note in passing 
that these functions are acceptable in the sense of Chapter 1.
30 Chapter 2 Quantum Mechanics of Some Simple Systems 
The only remaining parameter is the constant A. We set this to make the probability 
of finding the particle in the well equal to unity: 
 L 
0 
.2(x)dx =A2  L 
0 
sin2(npx/L)dx =1 (2-9) 
This leads to (Problem 2-2) 
A=2/L (2-10) 
which completes the solving of Schr¨odinger’s time-independent equation for the problem. 
Our results are the normalized eigenfunctions 
.n(x)=(2/L) sin(npx/L), n=1, 2, 3, . . . (2-11) 
and the corresponding eigenvalues, from Eq. (2-8), 
En =n2h2/8mL2, n=1, 2, 3, . . . (2-12) 
Each different value of n corresponds to a different stationary state of this system. 
2-2 Detailed Examination of Particle-in-a-Box Solutions 
Having solved the Schr¨odinger equation for the particle in the infinitely deep squarewell 
potential, we now examine the results in more detail. 
2-2.A Energies 
The most obvious feature of the energies is that, as we move through the allowed states 
(n = 1, 2, 3, . . . ), E skips from one discrete, well-separated value to another (1, 4, 9 
in units of h2/8mL2). Thus, the particle can have only certain discrete energies—the 
energy is quantized. This situation is normally indicated by sketching the allowed 
energy levels as horizontal lines superimposed on the potential energy sketch, as in 
Fig. 2-3a. The fact that each energy level is a horizontal line emphasizes the fact thatE is 
a constant and is the same regardless of the x coordinate of the particle. For this reason, 
E is called a constant of motion. The dependence ofE on n2 is displayed in the increased 
spacing between levels with increasing n in Fig. 2-3a. The number n is called a quantum 
number. 
We note also thatE is proportional toL-2. This means that the more tightly a particle 
is confined, the greater is the spacing between the allowed energy levels. Alternatively, 
as the box is made wider, the separation between energies decreases and, in the limit 
of an infinitely wide box, disappears entirely. Thus, we associate quantized energies 
with spatial confinement. 
For some systems, the degree of confinement of a particle depends on its total energy. 
For example, a pendulum swings over a longer trajectory if it has higher energy. The 
potential energy for a pendulum is given by V = 1
2kx2 and is given in Fig. 2-3b. If 
one solves the Schr¨odinger equation for this system (see Chapter 3), one finds that the 
energies are proportional to n rather than n2. We can rationalize this by thinking of
Section 2-2 Detailed Examination of Particle-in-a-Box Solutions 31 
Figure 2-3 
 Allowed energies for a particle in various one-dimensional potentials. (a) “box” with 
infinite walls. (b) quadratic potential, V = 1
2 kx2. (c) V =-1/ |x|. Tendency for higher levels in (b) 
and (c) not to diverge as in (a) is due to larger “effective box size” for higher energies in (b) and (c). 
the particle as occupying successively bigger boxes as we go to higher energies. This 
counteracts the n2 increase in energy levels found for constant box width. For the 
potential V =-1/ |x| (which is the one-dimensional analog of a hydrogen atom) E 
varies as 1/n2 (Fig. 2-3c), and this is also consistent with the effective increase in L 
with increasing E. 
The energy is proportional to 1/m. This means that the separation between allowed 
energy levels decreases as m increases. Ultimately, for a macroscopic object, m is so 
large that the levels are too closely spaced to be distinguished from the continuum of 
levels expected in classical mechanics. This is an example of the correspondence principle, 
which, in its most general form, states that the predictions of quantum mechanics 
must pass smoothly into those of classical mechanics whenever we progress in a continuous 
way from the microscopic to the macroscopic realm. 
Notice that the lowest possible energy for this system occurs for n=1 and is E = 
h2/8mL2. This remarkable result means that a constrained particle (i.e., L not infinite) 
can never have an energy of zero. Evidently, the particle continues to move about in 
the region 0 to L, even at a temperature of absolute zero. For this reason, h2/8mL2 
is called the zero-point energy for this system. In general, a finite zero-point energy 
occurs in any system having a restriction for motion in any coordinate. (Note that finite 
here means not equal to zero.) 
It is possible to show that, for L=8, our particle in a box would have to violate the 
Heisenberg uncertainty principle to achieve an energy of zero. For, suppose the energy 
is precisely zero. Then the momentum must be precisely zero too. (In this system, 
all energy of the particle is kinetic since V = 0 in the box.) If the momentum px is 
precisely zero, however, our uncertainty in the value of the momentum px is also 
zero. If px is zero, the uncertainty principle [Eq. (2-46)] requires that the uncertainty 
in position x be infinite. But we know that the particle is between x =0 and x =L. 
Hence, our uncertainty is on the order of L, not infinity, and the uncertainty principle 
is not satisfied. However, when L=8(the particle is unconstrained), it is possible for 
the uncertainty principle to be satisfied simultaneously with having E=0, and this is in 
satisfying accord with the fact that E=h2/8mL2 goes to zero as L approaches infinity. 
Finally, we note that each separate value of n leads to a different energy. Thus, 
no two states have the same energy, and the states are said to be nondegenerate with 
respect to energy.
32 Chapter 2 Quantum Mechanics of Some Simple Systems 
EXAMPLE 2-1 Consider an electron in a one-dimensional box of length 258pm. 
a)What is the zero-point energy (ZPE) for this system? For a mole of such systems? 
b)What electronic speed classically corresponds to this ZPE? Compare to the speed 
of light. 
SOLUTION 
 a) ZPE = Elowest =En=1 
= 12h2/8mL2 = 
(1)2(6.626×10-34 J s)2 
8(9.11×10-31 kg)(258×10-12 m)2 
= 9.05×10-19 J; 
Per mole, this equals 
(9.05×10-19 J)(6.022×1023 mol-1)(1kJ/103 J)=54.5 kJ mol-1 
b) E is all kinetic energy since V =0 in the box, so E =mv2/2. Hence, 
v =
2E 
m 

1/2 
=
	2(9.05×10-19 J) 
9.11×10-31 kg 


1/2 
=1.41×106 ms-1 
Compared to the speed of light, this is 1.41×106 ms-1 
2.998×108 ms-1 = 0.0047, or about 0.5% of the speed of 
light.  
2-2.B Wavefunctions 
We turn nowto the eigenfunctions (2-11) for this problem. These are typically displayed 
by superimposing them on the energy levels as shown in Fig. 2-4 for the three lowestenergy 
wavefunctions. (It should be recognized that the energy units of the vertical 
axis do not pertain to the amplitudes of the wavefunctions.) 
It is apparent from Fig. 2-4 that the allowed wavefunctions for this system could 
have been produced merely by placing an integral number of half sine waves in the 
Figure 2-4 
 The eigenfunctions corresponding to n=1, 2, 3, plotted on the corresponding energy 
levels. The energy units of the ordinate do not refer to the wavefunctions .. Each wavefunction has 
a zero value wherever it intersects its own energy level, and a maximum value of v2/L.
Section 2-2 Detailed Examination of Particle-in-a-Box Solutions 33 
range 0–L. The resultingwavelengthswould then yield the energy of each state through 
application of de Broglie’s relation (1-42). Thus, by inspection of Fig. 2-4, the allowed 
wavelengths are 
.=2L/n, n=1, 2, 3, . . . (2-13) 
Therefore 
p =h/.=nh/2L (2-14) 
and 
E =p2/2m=n2h2/8mL2 (2-15) 
in agreement with Eq. (2-12). As pointed out in Section 1-11, the wavefunctions having 
higher kinetic energy oscillate more rapidly. (Here V =0, and E is all kinetic energy.) 
Let us now consider the physical meaning of the eigenfunctions .. According to our 
earlier discussion, .2 summarizes the results of many determinations of the position 
of the particle. Suppose that we had a particle-in-a-box system that we had somehow 
prepared in such a way that we knew it to be in the state with n=1. We can imagine 
some sort of experiment, such as flashing a powerful . -ray flashbulb and taking an 
instantaneous photograph, which tells us where the particle was at the instant of the 
flash. Now, suppose we wish to determine the position of the particle again. We want 
this second determination to be for the n=1 state also, but we cannot use our original 
system for this because we have “spoiled” it by our first measurement process. Hitting 
the particle with one or more . -ray photons has knocked it into some other state, and 
we do not even know which one. Therefore, we must either reprepare our system, or 
else use a separate system whose preparation is identical to that of the first system. In 
general we shall assume that we have an inexhaustible supply of identically prepared 
systems. Therefore, we take a second photograph (on our second system) using the 
same photographic plate. Then we photograph a third system, a fourth, etc., until we 
have amassed a large number of images of the particle on the film. The distribution 
of these images is given by .2
1 . (Since . is always a real function for this system, 
we do not need to bother with .*..) Other states, like .2, .3, will lead to different 
distributions of images. The results for the several states are depicted in Fig. 2-5. It 
is obvious that the probability for finding the particle near the center of the box is 
predicted to be much larger for the n=1 than the n=2 state. 
The probability for finding the particle at the midpoint of the “wire” in the n=2 state 
approaches zero in the limit of our measurement becoming precise enough to observe 
a single point. This troubles many students at first encounter because they worry about 
how the particle can get from one side of the box to the other in the n=2 state. In fact, 
this question can be raised for any state whose wavefunction has any nodes. However, 
our discussion in the preceding paragraph shows that this question, like the question, 
“Is an electron a particle or a wave when we are not looking?” has no meaningful 
answer because no experiment can be conceived that would answer it. To test whether 
or not the particle does travel from one side of the box to the other, we would have to 
prepare the system in the n=2 state and measure the position of the particle enough 
times so that we either (a) always find it on the same side (requires many measurements 
for confidence), or (b) find it on different sides (requires at least two measurements).
34 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure 2-5 
 .2 and observed particle distribution for the three lowest-energy and one high-energy 
state of the particle in a one-dimensional box. 
But for our question to be answered, the system must be in the n=2 state throughout 
this entire experiment, and we have seen that the process of measuring particle position 
prevents this. (If we find the particle first on the left and later on the right, we cannot 
be sure it did not travel across the midpoint while the system was perturbed by the first 
measurement.) Thus, the sketches in Fig. 2-5 are most safely regarded as a summary 
of the results of measurements on an ensemble of systems. 
Classically, since the particle has constant energy, hence constant speed, we would 
expect the particle to spend equal time in each line segment dx between 0 and L, but 
Fig. 2-5 shows that the quantum system with n = 1 predicts that the particle spends 
more time in segments near the center. It is characteristic of lower-energy states of 
quantum-mechanical systems to display “anti-classical” distributions. With higher 
quantum numbers, the distribution becomes difficult to distinguish from the distribution 
predicted by classical physics (see Fig. 2-5). This is another example of the tendency 
of quantum-mechanical predictions to approach classical predictions when one goes 
toward the macroscopic realm (here large n and therefore large E). 
EXAMPLE 2-2 For a particle in the n=2 state in a one-dimensional box of lengthL, 
a) estimate the probability, ., for finding the particle between x =0 and x =0.20L. 
b) calculate the probability that you estimated in part a. 
c) what probability for finding the particle between x=0 and x=0.20Lis predicted 
by classical physics?
Section 2-2 Detailed Examination of Particle-in-a-Box Solutions 35 
SOLUTION 
 a) The sketch of .2
2 in Fig. 2-5 makes it clear that the probability for finding the 
particle in the range 0=x=L/4 is 0.25 (equal to the area under the curve). The range 0=x=0.20L 
is 20% shorter, and the missing 20% of the range is associated with a relatively large probability— 
almost double the average in the range. That means we are missing nearly 40% of the probablity, so 
slightly more than 60% remains. 60% of 0.25 is 0.15, so the probability, ., in the range 0-0.20L 
is slightly larger than 0.15. 
b) 
. =  0.2L 
0 
.2
2 dx = 
2
L 
 0.2L 
0 
sin2 2px 
L 
dx 
=  2
L
 L 
2p 
 0.2L 
0 
sin2 2px 
L 
 d 2px 
L 
 
= 
1
p 
 0.4p 
0 
sin2ydy 
= (from Appendix 1) 
1
p 

y
2 - 
1
4 
sin 2y|0.4L 
0 
 
= 
1
p 

0.2p - 
1
4 
sin 0.8p 
= 0.20- 
1 
4p 
sin 0.8p =0.153 
c)The classical particle travels with constant speed, hence has a constant probability function. Therefore, 
the probability for finding the particle in any 20% of the box is 0.20.  
2-2.C Symmetry ofWavefunctions 
Inspection of Fig. 2-5 shows that the particle has equal probabilities for being observed 
in the left half and right half of the box, regardless of state. This seems reasonable 
because there is no physical factor discriminating between these halves. We shall now 
show that the hamiltonian operator is invariant for a reflection through the box center, 
and that a necessary consequence of this is that . has certain symmetry properties. 
First, we showthatH is invariant. Reflection through the box center is accomplished 
by replacing x by -x +L. We can define a reflection operator R such that Rf (x)= 
f (-x +L); i.e., R reflects any function through a plane normal to x at x =L/2 (see 
Fig. 2-6). 
Figure 2-6 
 A function f (x) and its mirror image reflected at x =L/2.
36 Chapter 2 Quantum Mechanics of Some Simple Systems 
The kinetic part of the hamiltonian, T , is unchanged by R: 
RT = R 
- 
h2 
8p2m 
d2 
dx2 
= -h2 
8p2m 
d2 
d (-x +L)2 
= -h2 
8p2m 
d2 
dx2 =T (2-16) 
where we have used the fact that L is constant and d/d(-x)=-d/dx. That the 
potential part of H is unchanged by reflection through L/2 is easily seen; the identical 
infinite barriers merely interchange position. Therefore, RT = T and RV = V , and 
RH =R(T +V )=RT +RV =T +V =H. 
Now let us see what this means for eigenfunctions of H. Assume we have a normalized 
eigenfunction . 
H. =E. (2-17) 
The two sides of Eq. (2-17) will still be equal if we reflect our coordinate system 
throughout the equation. (If two functions are identical in one coordinate system, say 
a right handed system, then they are identical in any coordinate system.) Therefore,2 
(RH)(R.)=(RE)(R.) (2-18) 
But E is simply a constant, and so it is immune to R. Furthermore, we have just seen 
that RH =H. Therefore, 
H(R.)=E(R.) (2-19) 
which shows that the function R. is an eigenfunction of H with the same eigenvalue 
as .. 
We have already mentioned that the eigenfunctions of this system are nondegenerate 
with respect to energy. This is equivalent to saying that no two linearly independent 
eigenfunctions having the same eigenvalue exist for this system. But we have just shown 
that . and R. are both eigenfunctions having the same eigenvalue E. Therefore, we 
are forced to conclude that . and R. are linearly dependent, that is, 
R. =c. (2-20) 
where c is a constant. A moment’s thought shows that R. must still be normalized 
(since reflecting a function does not change its area or the area under its square), and 
it also must still be real (since reflecting a real function does not introduce imaginary 
character). Therefore, 
 L 
0 
(R.)2dx =1= L 
0 
(c.)2 dx =c2  L 
0 
.2dx =c2 (2-21) 
where we have made use of the fact that . is normalized. If c2 =1, then c=±1 and 
R. =±. (2-22) 
2The parentheses in Eq. (2-18) are meant to show the restricted extent of operation of R. This is a departure 
from the usual mathematical notational convention, but it is hoped that this temporary departure results in greater 
clarity for the student.
Section 2-2 Detailed Examination of Particle-in-a-Box Solutions 37 
When R. =+., as is the case for .1 or .3 (Fig. 2-4), . is said to be symmetric, or 
even, for reflection. If R. =-., as for .2, . is said to be antisymmetric, or odd. 
(A function that is neither symmetric nor antisymmetric is said to be unsymmetric, or 
asymmetric. Be careful to avoid confusing “asymmetric” with “antisymmetric.”) 
We have proved a very important property of wavefunctions. In general, if . is the 
wavefunction for a nondegenerate state, it must be symmetric or antisymmetric under 
any transformation that leaves H unchanged. 
2-2.D Orthogonality ofWavefunctions 
It is possible to show that integration over the product of two different particle-in-a box 
eigenfunctions, .n and .m, must give zero as the result: 
 L 
0 
.n.mdx =0, n=m (2-23) 
When functions have this property—that their product gives zero when integrated over 
the entire range of coordinates—they are said to be orthogonal. (Since the “box” 
eigenfunctions vanish for x =0 or x =L, integration from 0 to L suffices.) 
We can use symmetry arguments to demonstrate orthogonality among certain pairs 
of “box” eigenfunctions, for example, .1 and .2. Figure 2-7 shows that, since .1 
is symmetric and .2 is antisymmetric for reflection, the product of these functions is 
antisymmetric. (In fact, it is not difficult to show in general that the product of two 
symmetric or of two antisymmetric functions is symmetric, and that an antisymmetric 
function times a symmetric function gives an antisymmetric product. See Problem 2-7.) 
Integration over an antisymmetric function must give zero as the result since an antisymmetric 
function has to have equal amounts of positive and negative area. Therefore, .1 
and .2 are orthogonal “by symmetry” as, indeed, are all the symmetric–antisymmetric 
pairs of wavefunctions. Since all .’s having odd quantum number n are symmetric, 
and all .’s having even n are antisymmetric, we have used symmetry to prove .n and 
Figure 2-7 
 .1 is symmetric, .2 is antisymmetric, and .1.2 is antisymmetric. The total signed 
area bounded by the odd functions is zero since complete cancellation of positive and negative components 
occurs.
38 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure 2-8 
 The potential for a one-dimensional “box” with one infinite barrier at x = 0 and a 
barrier of V =U at x =L. 
.m orthogonal for n even and m odd. To show orthogonality for n and m both even or 
both odd requires doing the integral out explicitly (Problem 2-9). 
The eigenfunctions (2-11) are orthogonal to each other and individually normalized, 
and we refer to them as orthonormal functions. Mathematically, this is summarized as 
 L 
0 
.n.mdx =
0, n=m 
1, n=m

=dn,m (2-24) 
The quantity dn,m is called the Kronecker delta function, and it is merely a convenient 
shorthand for the information in the braces. 
EXAMPLE 2-3 .1 and .3 are both symmetric functions. Therefore their product, 
.1.3, is symmetric. How, then, can the integral of this product vanish? 
SOLUTION 
 A sketch of the product of these functions shows it to be symmetric, with negative 
values in the central region and wings of positive values on each side. Since .1 and.3 are known to 
be orthogonal, the negative region must exactly cancel the sum of the two positive regions (though 
we wouldn’t know this for sure from a sketch). This shows that, whereas the integral over an 
antismmetric integrand must equal zero, an integral over a symmetric (or unsymmetric) integrand 
may or may not equal zero.  
2-3 The Particle in a One-Dimensional “Box” 
with One FiniteWall 
Let us now modify the system just discussed by lowering the potential on one side of 
the “box” to some finite value U. The resulting potential is shown in Fig. 2-8. We can 
think of a bead on a wire encountering infinite repulsion at x =0 and finite repulsion 
for x =L. As before, it is convenient to break up the problem into separate regions of 
x. For the region x = 0 where V is infinite, . must be zero for the same reasons as 
before (Section 2-1). 
When the particle is in region I of Fig. 2-8, V =0 and all is identical to our earlier 
box. Therefore, in this region we will have harmonic waves of the general form 
.I =AI sin(2px/.I)+BI cos(2px/.I) (2-25)
Section 2-3 The Particle in a One-Dimensional “Box” with One FiniteWall 39 
where we have used the form (1-24) in which the wavelength appears explicitly. As 
before, the boundary condition that . vanish at x =0 forces BI to vanish, leaving 
.I =AI sin(2px/.I) (2-26) 
For the moment we have no other boundary condition on .I because we do not know 
that . equals zero at the finite barrier. We do know, however, that the wavelength .I, 
whatever it turns out to be, will be related to the energy through 
.I =h/2m(E -VI) (2-27) 
and, since VI =0 (in region I), 
.I =h/v2mE (2-28) 
which is a real number for positive E. 
We now turn to region II. Since V is constant here also, . will again be a harmonic 
wave. As before, we have our choice of two general forms: 
.II =AII sin(2px/.II)+BII cos(2px/.II) (2-29) 
or [see Eq. (2-3)] 
.II =CII exp(+2pix/.II)+DII exp(-2pix/.II) (2-30) 
There are two possibilities for the energy of the particle: E =U andE>U. The first 
of these corresponds to the classical situation where the particle has insufficient energy 
to escape from the box and get into region II. Let us see what quantum mechanics says 
about this case in region II. 
For this case, .II is imaginary since 
.II =h/2m(E -U) (2-31) 
and E - U is negative. Because .II is imaginary, it is more convenient to use the 
general form (2-30) because then the i in the exponential argument can combine with 
the i of .II to produce a real argument. Let us assume that .II is equal to i times a 
positive number. (This will not affect our results.) 
Let us now examine the properties of the two exponential functions in Eq. (2-30). 
The first exponential has an argument that is real (because the i’s cancel) and positive 
(because of our above assumption). As x increases, this exponential increases rapidly, 
approaching infinity. Since acceptable functions do not blow up like this, we set CII 
equal to zero to prevent it. The second exponential has a negative, real argument, so it 
decays exponentially toward zero as x approaches infinity. This is acceptable behavior, 
and we are left with 
.II =DII exp(-2pix/.II) (2-32) 
We nowhave formulas describing fragments of thewavefunction for the two regions. 
All that remains is to join these together at x = L in such a way that the resulting 
wavefunction is continuous at x =L and has a continuous first derivative there. (Recall
40 Chapter 2 Quantum Mechanics of Some Simple Systems 
from Section 2-1 that this second requirement results from the fact that the potential is 
finite at x =L. Hence, . must be smooth at x =L.) 
The continuity requirement gives 
AI sin(2pL/.I)=DII exp(-2piL/.II) (2-33) 
Taking the derivatives of .I and .II and setting these equal at x =L (to force smoothness) 
gives 
(2p/.I)AI cos(2pL/.I)=(-2pi/.II)DII exp(-2piL/.II) (2-34) 
The exponential term is common to both Eqs. (2-33) and (2-34), providing the basis 
for another equality:
AI sin(2pL/.I)=(-AI.II/i.I) cos(2pL/.I) (2-35) 
or 
tan(2pL/.I)=i.II/.I (2-36) 
Substituting for .I and .II as indicated by Eqs. (2-28) and (2-31) gives 
tan(2pLv2mE/h)=-vE/vU -E (2-37) 
The only unknown in Eq. (2-37) is the total energy E. For given values of L, m, and 
U, only certain values of E<U will satisfy Eq. (2-37). Thus, the particle can have 
only certain energies when it is trapped in the “box.” These allowed energies can be 
found by graphing the left-hand side and right-hand side of Eq. (2-37) as functions of 
E. The values of E where the plots intersect satisfy Eq. (2-37). Figure 2-9 illustrates 
the graphical solution of Eq. (2-37) for a particular set of values for L, m, and U. 
Once a value ofE is selected, .I and .II are known [from Eqs. (2-28) and (2-31)] and 
it remains only to find AI andDII. The ratio AI/DII may be found from Eq. (2-33). The 
numerical values of AI and DII will then be obtainable if we require that the wavefunction 
be normalized. A set of such solutions is shown in Fig. 2-10. 
Before solving for the case where E >U, let us discuss in detail the results just 
obtained. 
In the first place, the energies are quantized, much as they were in the infinitely deep 
square well. There is some difference, however. In the infinitely deep well or box, the 
energy levels increased with the square of the quantum number n. Here they increase 
less rapidly (the dashed lines in Fig. 2-10 show the allowed energy levels which result 
when U=8) because the barrier becomes effectively less restrictive for particles with 
higher energies (see the following). For the lowest solution, for example, slightly less 
than one-half a sine wave is needed in one box width of distance. Thus, the wavelength 
here is slightly longer than in an infinitely deep well of equal width, and so, by de 
Broglie’s relation, the energy is slightly lower. Notice that the effect of lowering the 
height of one wall is least for the levels lying deepest in the well. 
The solutions sketched in Fig. 2-10 indicate that there is a finite probability for 
finding the particle in the region x >L even though it must have a negative kinetic 
energy there. Thus, quantum mechanics allows the particle to penetrate into regions 
where classical mechanics claims it cannot go. Notice that the penetration becomes 
more appreciable as the energy of the particle approaches that of the barrier. This results
Section 2-3 The Particle in a One-Dimensional “Box” with One FiniteWall 41 
Figure 2-9 
 Graphical solution of the equation -tan(2pLv2mE/h) = vE/vU -E. Here 
L = 2.50 nm, m = 9.11 × 10-31 kg, U = 1 eV = 16.02 × 10-20 J. Intersections occur at E = 0.828×10-20 J, 3.30×10-20 J, 7.36×10-20 J and 12.8×10-20 J. 
from the fact that E - U determines the rate at which the exponential in .II decays 
[see Eqs. (2-31) and (2-32)]. In the limit that U.8, the wavefunction vanishes at 
the barrier, in agreement with the results of the infinite square well of Section 2-1. 
If the barrier in Fig. 2-8 has finite thickness (V becomes zero again at, say, x =2L), 
then there is a finite probability that a particle in the well will penetrate through the 
barrier and appear on the other side. This phenomenon is called quantum-mechanical 
tunneling, and this is the way, for example, an a particle escapes from a nucleus even 
though it classically lacks sufficient energy to overcome the attractive nuclear forces. 
We emphasize that the tunneling referred to in this example is really not a stationary 
state phenomenon. We have an initial condition (particle in the well) and ask what the 
half-life is for the escape of the particle—a time-dependent problem. 
We saw earlier that the energy quantization for the particle in the infinitely deep well 
could be thought of as resulting from fitting integral numbers of half sine waves into 
a fixed width. Most sine waves just will not fit perfectly, and so most energies are not 
allowed. In this problem the waves are allowed to leak past one of the well walls, but 
we can still see why only certain energies are allowed. Suppose that we pick some 
arbitrary energy E for the particle. We know that . must be zero at the left wall of the 
well where V =8. Starting there, we can draw a sine wave of wavelength determined 
by E across the well to the right wall, as shown in Fig. 2-11. When the wave hits the 
right wall, it must join on smoothly to a decaying exponential, which also depends on 
E. Most of the time, it will be impossible to effect a smooth junction, and that particular 
value of E will be disallowed. 
Let us now consider the case where E>U. In region I, the considerations are the 
same as before. Then, .I is a sine wave that can be drawn from the left wall and has a
42 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure 2-10 
 Solutions for particle in well with one finite wall (see Fig. 2-9 for details). Dashed 
lines correspond to energy levels which would exist if U =8. 
wavelength determined by E(=T ) from de Broglie’s relation. This sine wave arrives 
at x =L with a certain magnitude and a certain derivative (assuming that the multiplier 
A1 has been fixed at some arbitrary value). In region II, we also have a solution of the 
usual form 
.II(x)=AII sin(2px/.II)+BII cos(2px/.II) (2-38) 
where .II is real and determined by E -U, which is now positive. The question is, can 
we always adjust .II (by changing AII and BII) so that it has the same value and slope 
at x = L that .I has? A little thought shows that such adjustment is indeed always 
possible. The two adjustments allowed in .II correspond to a change of phase for .II 
(a shift in the horizontal direction) and a change in amplitude for .II. The only thing 
about .II we cannot change is the wavelength, since this is determined by E -U. This 
is just a physical description of the mathematical circumstance in which we have two 
adjustable parameters and two requirements to fit—a soluble problem. The essential 
difference between this case and that of the trapped particle is that here we have fewer
Section 2-3 The Particle in a One-Dimensional “Box” with One FiniteWall 43 
Figure 2-11 
 An example of partial wavefunctions for an arbitrary energy E. These functions 
cannot be joined smoothly at x =L and so this value of E is not allowed. 
boundary conditions. Before, our square-integrability requirement was used to remove 
a positive exponential term. That requirement is, in effect, a boundary condition—. 
must vanish at x=8—and it led to energy quantization. Then we used the normality 
requirement to achieve unique values for AI and DII. In this case we cannot get a 
square-integrable solution. .II goes on oscillating as x.8, and so we have no 
boundary condition there. As a result, E is not quantized and . is not normalizable, 
so that only ratios of AI, AII, and BII are obtainable. 
The energy scheme for the particle in the potential well with one finite wall, then, is 
discrete when E<U, and continuous when E>U. 
Notice the way in which the wavelengths vary in Fig. 2-10. We have already seen 
that the time-independent Schr¨odinger equation states that the total energy for a particle 
in a stationary state is the same at all particle positions (i.e., a constant of motion). The 
kinetic and potential energies must vary together, then, in such a way that their sum is 
constant. This is reflected by the fact that the wavelength of an unbound solution is 
shorter in region I than it is in region II. In region I, V = 0, so that all energy of the 
particle is kinetic (T =E). In region II,V >0, so that the kinetic energy (T =E -V ) 
is less than it was in region I. Therefore, the de Broglie wavelength, which is related to 
kinetic energy, must be greater in region II. 
EXAMPLE 2-4 For the system described in the caption for Fig. 2-9, calculate the 
percentage drop of the lowest-energy state that results from barrier penetration. 
SOLUTION 
 For this, we need to solve the problem for the simple particle-in-a-box system 
(for which U =8). 
E1 = 
n2h2 
8mL2 = 
(1)2(6.626×10-34 J s)2 
8(9.11×10-31 kg)(2.500×10-10 m)2 =9.64×10-21 J 
compared to 8.28×10-21 J. Barrier penetration lowers E1 by 14%. 
44 Chapter 2 Quantum Mechanics of Some Simple Systems 
2-4 The Particle in an Infinite “Box” with 
a Finite Central Barrier 
Another example of barrier penetration in a stationary state of a system is provided by 
inserting a barrier of finite height and thickness at the midpoint of the infinite square 
well of Section 2-1 (see Fig. 2-12). 
The boundary conditions for this problem are easily obtained by obvious extensions 
of the considerations already discussed. Rather than solve this case directly, we shall 
make use of our insights from previous systems to deduce the main characteristics of 
the solutions. Let us begin by considering the case where the barrier is infinitely high. 
Figure 2-12 
 (a) Solutions for identical infinite square wells. (b) Effect of finite partition on half 
waves. (c) Symmetric combination of half waves. (d) Antisymmetric combination of half waves.
Section 2-4 The Particle in an Infinite “Box” with a Finite Central Barrier 45 
Then the problem becomes merely that of two isolated infinite square wells, each well 
having solutions as described in Sections 2-1 and 2-2. 
Now, as the height of the barrier is lowered from infinity, what happens? The levels 
lying deepest in the two sections should be least affected by the change. They must 
still vanish at the outer walls but now they can penetrate slightly into the finite barrier. 
Thus, the lowest state in, say, the left-hand section of the well will begin to look as 
given in Fig. 2-12b. The solution on the right side will do likewise, of course. As 
this happens, their energies will decrease slightly since their wavelengths increase. 
However, since the two wells are no longer separated by an infinite barrier, they are no 
longer independent. We can no longer talk about separate solutions for the two halves. 
Each solution for the Schr¨odinger equation is now a solution for the whole system from 
x=-L to +L. Furthermore, symmetry arguments state that, since the hamiltonian for 
this problem is symmetric for reflection through x =0, the solutions, if nondegenerate, 
must be either symmetric or antisymmetric through x =0. 
This requirement must be reconciled with the barrier-penetration behavior indicated 
by Fig. 2-12b, which is also occurring. One way to accomplish this is by summing the 
two halfwaves as shown in Fig. 2-12c, giving a symmetricwavefunction. Alternatively, 
subtraction gives the antisymmetric form shown in Fig. 2-12d. Both of these solutions 
will be lower in energy than their infinite-well counterparts, because the wavelengths 
in Fig. 2-12c and d continue to be longer than in 2-12a. Will their energies be equal to 
each other? Not quite. By close inspection, we can figure out which solution will have 
the lower energy. 
In Figs. 2-12b to 2-12d, the slopes of the half wave, the symmetric, and the antisymmetric 
combinations at the finite barrier are labeled respectively m, m, and m. What 
can we say about their relative values? The slope m should be less negative than m 
because the decaying exponential producing m has an increasing exponential added to 
it when producing m. Slope m should be more negative than m since the decaying 
exponential has an increasing exponential subtracted from it in case d, causing it to 
decay faster. This means that the sine curve on the left-hand side of Fig. 2-12c cannot 
be identical with that on the left side of Fig. 2-12d since they must arrive at the barrier 
with different slopes. (The same is true for the right-hand sides, of course.) How 
can we make the sine wave arrive with a less negative slope m?—by increasing the 
wavelength slightly so that not quite so much of the sine wave fits into the left well 
(see Fig. 2-13a). Increasing the wavelength slightly means, by de Broglie’s relation, 
that the energy of the particle is decreased. Similarly, the sine curve in Fig. 2-12d must 
be shortened so that it will arrive at the barrier with slope m, which corresponds to 
an energy increase. Of course, now that the energy has changed outside the barrier, it 
must change inside the barrier too. This would require going back and modifying the 
exponentials inside the barrier. But the first step is sufficient to indicate the qualitative 
results: The symmetric solution has lower energy. In Fig. 2-13a is a detailed sketch of 
the final solution for the two lowest states. 
There is a simpler way to decide that the symmetric solution has lower energy. As 
the barrier height becomes lower and lower, the two solutions become more and more 
separated in energy, but they always remain symmetric or antisymmetric with respect to 
reflection since the hamiltonian always has reflection symmetry. In the limit when the 
barrier completely disappears we have a simple square well again (but larger), the lowest 
solution of which is symmetric. (See Fig. 2-13b.) This lowest symmetric solution must
46 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure 2-13 
 (a) Detailed sketch of the two lowest solutions for the infinite square-well divided 
by a finite barrier at the midpoint. The waves are sketched from a common energy value for ease of 
comparison. Actually, the symmetric wave has a lower energy. (b) A correlation diagram relating 
energies when the barrier is infinite (left side) with those when the barrier vanishes. Letters A and S 
refer to antisymmetric and symmetric solutions, respectively. 
“come from” the symmetric combination of smaller-well wavefunctions sketched at the 
left of Fig. 2-13b; similarly, the second lowest, antisymmetric solution of the large well 
correlates with the antisymmetric small-well combination (also at left in Fig. 2-13b). 
A figure of the kind shown in Fig. 2-13b is called a correlation diagram. It shows how 
the energy eigenvalues change throughout a continuous, symmetry-conserving process.
Section 2-5 The Free Particle in One Dimension 47 
We shall see that the correlation of wavefunction symmetries in such a manner as this 
is a powerful technique in understanding and predicting chemical behavior. 
The splitting of energy levels resulting from barrier penetration is an extremely 
pervasive phenomenon in quantum chemistry. It occurs regardless of whether the barrier 
separates identical or nonidentical potential regions, i.e., regardless of whether the final 
system is symmetric or unsymmetric. When two atoms (N and N, or C and O) interact 
to form a molecule, the original atomic wavefunctions combine to form molecular 
wavefunctions in much the same way as was just described. One of these molecular 
wavefunctions may have an energy markedly lower than those in the corresponding 
atoms. Electrons having such a wavefunction will stabilize the molecule relative to the 
separated atoms. 
Another case in which energy level splitting occurs is in the vibrational spectrum 
of ammonia. Ammonia is most stable in a pyramidal configuration, but is capable 
of inverting through a higher-energy planar configuration into an equivalent “mirror 
image” pyramid. Thus, vibrations tending to flatten out the ammonia molecule occur 
in a potential similar to the double well, except that in ammonia the potential is not 
discontinuous. The lowest vibrational energy levels are not sufficiently high to allow 
classical inversion of ammonia. However, these vibrational levels are split by interaction 
through barrier penetration just as quantum mechanics predicts. The energy 
required to excite ammonia from the lowest of these sublevels to its associated sublevel 
can be accurately measured through microwave spectroscopy. Knowledge of the level 
splittings in turn allows a precise determination of the height of the barrier to inversion 
in ammonia (see Fig. 2-14). 
It is easy to anticipate the appearance of the solutions for the square well with central 
barrier for energies greater than the partition height. They will be sinusoidal waves, 
symmetric or antisymmetric in the well, and vanishing at the walls. Their wavelengths 
will be somewhat longer in the region of the partition than elsewhere because some of 
the kinetic energy of the particle is transformed to potential energy there. A sketch of 
the final results is given in Fig. 2-15. 
EXAMPLE 2-5 Fig. 2-15 shows energy levels for states when the barrier has finite 
height. When the barrier is made infinitely high, the levels at E1 and E2 merge 
into one level. Where does the energy of that one level lie—below E1, between 
E1 and E2, or above E2—and why? 
SOLUTION 
 It lies above E2. When the barrier is finite, there is always some penetration, so 
. is always at least a little larger than is the case for the infinite barrier. If . is larger, E is lower.  
2-5 The Free Particle in One Dimension 
Suppose a particle of mass m moves in one dimension in a potential that is everywhere 
zero. The Schr¨odinger equation becomes 
-h2 
8p2m 
d2. 
dx2 =E. (2-39)
48 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure 2-14 
 Sketch of potential for inversion vibrational mode in ammonia. The lowest levels 
are split by tunneling. The low energy transition E1 is visible in the microwave region whereas the 
second transition E2 is visible in the infrared. E1 =0.16×10-22 J; E2 =7.15×10-22 J. 
which has as solutions 
. =Aexp(±2piv2mEx/h) (2-40) 
or alternatively, trigonometric solutions 
. =A sin(2pv2mEx/h), . =A cos(2pv2mEx/h) (2-41) 
As is most easily seen from the exponential forms (2-40), if E is negative, . will 
blow up at either +8 or -8, and so we reject negative energies. Since there are 
no boundary conditions, it follows that E can take on any positive value; the energies 
of the free particle are not quantized. This result would be expected from our earlier 
results on constrained particles. There we saw that quantization resulted from spatial 
constraints, and here we have none. 
The constants A and A of Eqs. (2-40) and (2-41) cannot be evaluated in the usual 
way, since the solutions do not vanish at x =±8. Sometimes it is convenient to 
evaluate them to correspond to some experimental situation. For instance, suppose that 
one was working with a monoenergetic beam of electrons having an intensity of one 
electron every 10-6 m. Then we could normalize . of Eq. (2-40) so that 
 10-6 m 
0 |.|2 dx =1 
There is a surprising difference in the particle distributions predicted from expressions 
(2-40) and (2-41). The absolute square .*. of the exponentials is a constant 
(A*A), whereas the squares of the trigonometric functions are fluctuating functions
Section 2-5 The Free Particle in One Dimension 49 
Figure 2-15 
 Wavefunctions for the infinite square well with finite partition. 
of x. It seems sensible for a particle moving without restriction to have a constant 
probability distribution, and it seems absurd for it to have a varying probability distribution. 
What causes this peculiar behavior? It results from there being two independent 
solutions for each value of E (except E =0). This degeneracy with respect to energy 
means that from a degenerate pair, . and ., one can produce any number of new 
eigenfunctions, . = a. + b. (Problem 2-11). In such a situation, the symmetry 
proof of Section 2-2 does not hold. However, there will always be an independent pair 
of degenerate wavefunctions that will satisfy certain symmetry requirements. Thus, in 
the problem at hand, we have one pair of solutions, the exponentials, which do have 
the proper symmetry since their absolute squares are constant. From this pair we can 
produce any number of linear combinations [one set being given by Eq. (2-41)], but 
these need not display the symmetry properties anymore. 
The exponential solutions have another special attribute: A particle whose state is 
described by one of the exponentials has a definite linear momentum, whereas, when 
described by a trigonometric function, it does not. In Section 1-9, it was shown that 
the connection between classical and wave mechanics could be made if one related the 
classical momentum, px, with a quantum mechanical operator (h/2pi)d/dx. Now, 
for a particle to have a definite (sharp) value p for its momentum really means that, 
if we measure the momentum at some instant, there is no possibility of getting any 
value other than p. This means that the particle in the state described by . always has 
momentum p, no matter where it is in x; i.e., its momentum is a constant of motion,
50 Chapter 2 Quantum Mechanics of Some Simple Systems 
just as its energy is. This corresponds to saying that there is an eigenvalue equation for 
momentum, just as for energy. Thus 
h 
2pi 
d. 
dx =p. (2-42) 
The statement made earlier, that the exponential solutions correspond to the particle 
having sharp momentum, means that the exponentials (2-40) must be solutions to 
Eq. (2-42). This is easily verified: 
h 
2pi 
d 
dx 
	
Aexp±2piv2mEx 
h 


=±v2mE 
	
Aexp±2piv2mEx 
h 

 
Thus, the positive and negative exponential solutions correspond to momentum values 
of +v2mE and -v2mE, respectively, and are interpreted as referring to particle 
motion toward +8 and -8 respectively. Since energy is related to the square of 
the momentum, these two solutions have identical energies. (The solution for E =0 
corresponds to no momentum at all, and the directional degeneracy is removed.) A 
mixture of these states contains contributions from two different momenta but only one 
energy, so linear combinations of the exponentials fail to maintain a sharp value for 
momentum but do maintain a sharp value for energy. 
EXAMPLE 2-6 An electron is accelerated along the x axis towards x=8from rest 
through a potential drop of 1.000 kV. 
a) What is its final momentum? 
b) What is its final de Broglie wavelength? 
c) What is its final wavefunction? 
SOLUTION 
 a) px = v2mE = [2(9.105 × 10-31 kg)(1.602 × 10-16 J)]1/2 = 1.708 × 10-23 kgms-1 
b) .= h
p = 6.626×10-34 J s 
1.708×10-23 kgms-1 =3.879×10-11 m 
c) . =Aexp(+2piv2mEx/h)(choose + because moving towards x=+8)=Aexp(2pip/h)x 
=Aexp(2pix/.)=Aexp[2pi(2.578×1010 m-1)x].  
2-6 The Particle in a Ring of Constant Potential 
Suppose that a particle of mass m is free to move around a ring of radius r and zero 
potential, but that it requires infinite energy to get off the ring. This system has only 
one variable coordinate—the angle f. In classical mechanics, the useful quantities and 
relationships for describing such circular motion are those given in Table 2-1. 
Comparing formulas for linear momentum and angular momentum reveals that the 
variables mass and linear velocity are analogous to moment of inertia and angular velocity 
in circular motion, where the coordinate f replaces x. The Schr¨odinger equation 
for circular motion, then, is 
-h2 
8p2I 
d2. (f) 
df2 =E.(f) (2-43)
Section 2-6 The Particle in a Ring of Constant Potential 51 
TABLE 2-1 
 
Quantity Formula Units 
Moment of inertia I =mr2 g cm2 or kgm2 
Angular velocity .=f/t =v/r s-1 
Angular momentum (linear 
momentum times orbit radius) 
mvr =I. g cm2/s or erg s or J s 
which has, as solutions 
Aexp(±ikf) (2-44) 
or alternatively 
A sin(kf) (2-45) 
and 
A cos(kf) (2-46) 
where [substituting Eq. (2-45) or (2-46) into (2-43) and operating] 
k =2pv2IE/h (2-47) 
Let us solve the problem first with the trigonometric functions. Starting at some 
arbitrary point on the ring and moving around the circumference with a sinusoidal function, 
we shall eventually reencounter the initial point. In order that our wavefunction 
be single valued, it is necessary that . repeat itself every time f changes by 2p radians. 
Thus, for f given by Eq. (2-45), 
sin(kf)=sin[k(f +2p)] (2-48) 
Similarly, for . given by Eq. (2-49) 
cos(kf)=cos(kf +2kp) (2-49) 
Either of these relations is satisfied only if k is an integer. The case in which k =0 is 
not allowed for the sine function since it then vanishes everywhere and is unsuitable. 
However, k =0 is allowed for the cosine form. The normalized solutions are, then, 
. = (1/vp)sin(kf), k =1, 2, 3, . . . 
. = (1/vp)cos(kf), k =1, 2, 3, . . . 
. = (1/v2p) (from the k =0 case for the cosine) (2-50) 
Now let us examine the exponential form of . (Eq. 2-44). The requirement that . 
repeat itself for f.f +2p gives 
Aexp(±ikf)=Aexp[±ik(f +2p)]=Aexp(±ikf) exp(±2pik)
52 Chapter 2 Quantum Mechanics of Some Simple Systems 
or 
exp(±2pik)=1 
Taking the positive case and utilizing Eq. (2-3), we obtain 
cos(2pk)+i sin(2pk)=1 or cos(2pk)=1 and sin(2pk)=0 
Again, k must be an integer. (The same result arises by requiring that d./df repeat 
for f.f +2p.) Thus, an alternative set of normalized solutions is 
. =1/v2p exp(ikf) k =0,±1,±2,±3, . . . (2-51) 
The energies for the particle in the ring are easily obtained from Eq. (2-47): 
E =k2h2/8p2I, k =0,±1,±2,±3, . . . (2-52) 
The energies increase with the square of k, just as in the case of the infinite square 
well potential. Here we have a single state with E =0, and doubly degenerate states 
above, whereas, in the square well, we had no solution at E =0, and all solutions were 
nondegenerate. The solution at E =0 means that there is no finite zero point energy 
to be associated with free rotation, and this is in accord with uncertainty principle 
arguments since there is no constraint in the coordinate f. 
The similarity between the particle in a ring and the free particle problems is striking. 
Aside from the fact that in the ring the energies are quantized and the solutions 
are normalizable, there are few differences. The exponential solutions (2-51) are 
eigenfunctions for the angular momentum operator (h/2pi)d/df. The two angular 
momenta for a pair of degenerate solutions correspond to particle motion clockwise or 
counterclockwise in the ring. (The nondegenerate solution for E =0 has no angular 
momentum, hence no ability to achieve degeneracy through directional behavior.) The 
particle density predicted by the exponentials is uniform in the ring, while that for the 
trigonometric solutions is not. Since the trigonometric functions tend to localize the 
particle into part of the ring, thereby causing f =8, it is consistent that they are 
impure momentum states ( ang. mom. =0). (Infinite uncertainty in the coordinate 
f means that all values of f in the range 0–2p are equally likely.) 
EXAMPLE 2-7 Demonstrate that any two degenerate exponential eigenfunctions 
for a particle in a ring are orthogonal. 
SOLUTION 
 Such a pair of degenerate wavefunctions can be written as .+ = 1 v2p 
exp(ikf), 
.- = 1 v2p 
exp(-ikf). These are orthogonal if  2p 
0 .*+
.- df =0 where we must use the complex 
conjugate of either one of the wavefunctions since . is complex. But .*+
=.-, so 
 2p 
0 
.-.- df = 
1 
2p 
 2p 
0 
exp(-2if)df = 
1 
2p 
 2p 
0 [cos(2f)-i sin(2f)]df 
= 
1 
2p 
sin(2f)|2p 
0 -i(-cos(2f)|2p 
0  
= 
1 
2p [sin(4p)-sin(0)+i cos(4p)-i cos(0)] 
= 
1 
2p [0-0+i -i]=0. 

Section 2-7 The Particle in a Three-Dimensional Box: Separation of Variables 53 
2-7 The Particle in a Three-Dimensional 
Box: Separation of Variables 
Let us now consider the three-dimensional analog of the square well of Section 2-1. 
This would be a three-dimensional box with zero potential inside and infinite potential 
outside. As before, the particle has no probability for penetrating beyond the box. 
Therefore, the Schr¨odinger equation is just 
-h2 
8p2m 
 .2 
.x2 + 
.2 
.y2 + 
.2 
.z2 
. =E. (2-53) 
and . vanishes at the box edges. 
The hamiltonian operator on the left side of Eq. (2-53) can be written as a sum of 
operators, one in each variable (e.g., Hx = (-h2/8p2m).2/.x2). Let us assume for 
the moment that . can be written as a product of three functions, each one being a 
function of a different variable, x,y, or z (i.e., . =X(x)Y(y)Z(z)). If we can show 
that such a . satisfies Eq. (2-53), we will have a much simpler problem to solve. Using 
this assumption, Eq. (2-53) becomes 
(Hx +Hy +Hz)XYZ =EXYZ (2-54) 
This can be expanded and then divided through by XYZ to obtain 
HxXYZ 
XYZ + 
HyXYZ 
XYZ + 
HzXYZ 
XYZ =E (a constant) (2-55) 
Now, since Hx, for example, operates only on functions of x, but not y or z, we 
can carry out some limited cancellation. Those functions that are not operated on 
in a numerator can be canceled against the denominator. Those that are operated on 
cannot be canceled since these are differential operators [e.g., in (1/x)dx2/dx it is not 
permissible to cancel 1/x against x2 before differentiating: (1/x)dx2/dx =dx/dx]. 
Such cancellation gives
HxX 
X + 
HyY 
Y + 
HzZ 
Z =E (a constant) (2-56) 
Now, suppose the particle is moving in the box parallel to the x axis so that the variables 
y and z are not changing. Then, of course, the functions Y and Z are also not changing, 
so HyY/Y and HzZ/Z are both constant. Only HxX/X can vary—but does it vary? 
Not according to Eq. (2-56), which reduces under these conditions to 
HxX 
X +constant+constant=E (a constant) (2-57) 
Therefore, even though the particle is moving in the x direction, HxX/X must also be a 
constant, which we shall callEx. Similar reasoning leads to analogous constantsEy and 
Ez . Furthermore, the behavior of HxX/X must really be independent of whether the 
particle is moving parallel to the y and z axes. Even if y and z do change, they do not 
appear in the quantity HxX/X anyway. Thus we may write, without restriction, 
HxX 
X =Ex, 
HyY 
Y =Ey, 
HzZ 
Z =Ez (2-58)
54 Chapter 2 Quantum Mechanics of Some Simple Systems 
and, from Eq. (2-56), 
Ex +Ey +Ez =E (2-59) 
Our original equation in three variables has been separated into three equations, one in 
each variable. The first of these equations may be rewritten 
HxX =ExX (2-60) 
which is just the Schr¨odinger equation for the particle in the one-dimensional square 
well, which we have already solved. For a rectangular box with Lx =Ly =Lz we have 
the general solution 
. =XYZ =2/Lx sin (nxpx/Lx)2/Ly sin nypy/Ly2/Lz sin (nzpz/Lz) 
(2-61) 
and 
E =Ex +Ey +Ez =h2/8m n2x
/L2x 
+n2y
/L2y 
+n2z
/L2z
 (2-62) 
For a cubical box, Lx =Ly =Lz =L, and the energy expression simplifies to 
E =h2/8mL2 n2x 
+n2y 
+n2z
 (2-63) 
The lowest energy occurs when nx =ny =nz =1, and so 
E1(1)=3h2/8mL2 (2-64) 
Thus, the cubical box has three times the zero point energy of the corresponding onedimensional 
well, one-third coming from each independent coordinate for motion (i.e., 
“degree of freedom”). The one in parentheses indicates that this level is nondegenerate. 
The next level is produced when one of the quantum numbers n has a value of two 
while the others have values of one. There are three independent ways of doing this; 
therefore, the second level is triply degenerate, and E2(3)=6h2/8mL2. Proceeding, 
E3(3)=9h2/8mL2, E4(3)=11h2/8mL2, E5(1)=12h2/8mL2, E6(6)=14h2/8mL2, 
etc. Apparently, the energy level scheme and degeneracies of these levels do not proceed 
in the regular manner which is found in the one-dimensional cases we have studied. 
EXAMPLE 2-8 Verify that E6 is six-fold degenerate. 
SOLUTION 
 E6 = 14h2/8mL2, so n2x 
+ n2y 
+ n2z 
= 14. There is only one combination of 
integers that satisfies this relation, namely 1, 2, and 3. So we simply need to deduce how many 
unique ways we can assign these integers to nx,ny , and nz . There are three ways to assign 1. For 
each of these three choices, there remain but two ways to assign 2, and then there is only one way 
to assign 3. So the number of unique ways is 3×2×1=6. (Or one can simply write down all of 
the possibilities and observe that there are six of them.)  
We shall now briefly consider what probability distributions for the particle are 
predicted by these solutions. The lowest-energy solution has its largest value at the 
box center where all three sine functions are simultaneously largest. The particle
Section 2-7 The Particle in a Three-Dimensional Box: Separation of Variables 55 
Figure 2-16 
 Sketches of particle probability distributions for a particle in a cubical box. 
(a) nx =ny =nz =1. (b) nx =2, ny =nz =1. (c) nx =ny =nz =2. 
distribution is sketched in Fig. 2-16a. The second level may be produced by setting 
nx = 2, and ny = nz = 1. Then there will be a nodal plane running through the box 
perpendicular to the x axis, producing the split distribution shown in Fig. 2-16b. Since 
there are three ways this node can be oriented to produce distinct but energetically 
equal distributions, this energy level is triply degenerate. The particle distribution for 
the state where nx =ny =nz =2 is sketched in Fig. 2-16c. It is apparent that, in the 
high energy limit, the particle distribution becomes spread out uniformly throughout 
the box in accord with the classical prediction. 
The separation of variables technique which we have used to convert our threedimensional 
problem into three independent one-dimensional problems will recur in 
other quantum-chemical applications. Reviewing the procedure makes it apparent that 
this technique will work whenever the hamiltonian operator can be cleanly broken into 
parts dependent on completely different coordinates. This is always possible for the 
kinetic energy operator in cartesian coordinates. However, the potential energy operator 
often prevents separation of variables in physical systems of interest. 
It is useful to state the general results of separation of variables. Suppose we have 
a hamiltonian operator, with associated eigenfunctions and eigenvalues: 
H.i =Ei.i (2-65) 
Suppose this hamiltonian can be separated, for example, 
H (a,ß)=Ha (a)+Hß (ß) (2-66)
56 Chapter 2 Quantum Mechanics of Some Simple Systems 
where a and ß stand for two different coordinates or groups of coordinates. Then it 
follows that 
.j,k =fj (a) gk(ß) (2-67) 
where 
Hafj =ajfj (2-68) 
and 
Hßgk =bkgk (2-69) 
Furthermore, 
Ej,k =aj +bk (2-70) 
In other words, if a hamiltonian is separable, then the eigenfunctions will be products 
of eigenfunctions of the subhamiltonians, and the eigenvalues will be sums of the 
subeigenvalues. 
2-8 The Scattering of Particles in One Dimension 
Consider the potential shown in Fig. 2-17a. We imagine that a beam of particles, each 
having energy E, originates from the left and travels toward x=8, experiencing a 
constant potential everywhere except at the potential step at x =0. We are interested in 
what becomes of these particles—what fraction makes it all the way to the “end” (some 
kind of particle trap to the right of the step) and what fraction is reflected back toward 
x=-8. Problems of this type are related to scattering experiments where electrons, 
for example, travel through potential jumps produced by electronic devices or through 
a dilute gas where potential changes occur in the neighborhood of atoms. 
This kind of problem differs from most of those discussed earlier because the particle 
is not trapped (classically), so all nonnegative energy values are possible. We already 
know what the form of the eigenfunctions is for the constant potentials to the left and 
right of the step for any choice of E. On the left they are linear combinations of 
exp(±iv2mEx/¯h), where ¯h = h/2p, and to the right they are linear combinations 
of exp(±iv2m(E -U)x/¯h). The only thing we do not yet know is which linear 
combinations to take. That is, we need to find A,B,C, and D in 
.left = Aexp(ikx/¯h)+B exp(-ikx/¯h), x <0, k=v2mE (2-71) 
.right = C exp(ikx/¯h)+Dexp(-ikx/¯h), x >0, k =2m(E -U) (2-72) 
We have seen earlier that the exponentials having positive arguments correspond to 
particles traveling from left to right, etc. We signify this with arrows in Fig. 2-17a. 
The nature of .right is qualitatively different depending on whether E is larger or 
smaller than the step height U. If E <U, the exponential arguments become real. 
One of the exponentials decays and the other explodes as x increases, just as we saw 
in Section 2-3. We discard the exploding exponential. The decaying exponential on 
the right must now be made to join smoothly onto .left at the step. That is, .left and
Section 2-8 The Scattering of Particles in One Dimension 57 
Figure 2-17 
 (a) A potential step of height U at x = 0. (b) A wavefunction having E = U/2. 
(c) Fraction of particles reflected as a function of E. (d) Fraction of particles transmitted as a function 
of E. (Note: The vertical line at left of parts (a) and (b) is not a barrier. It is merely an energy 
ordinate.) 
.right must have the same value and slope at x = 0 (Fig. 2-17b). This means that 
(fch2:eqn2-72 must have real value and slope at the step, which forces it to be a 
trigonometric wave. Because there is no additional boundary condition farther left, this 
trigonometric wave can always be shifted in phase and amplitude to join smoothly onto 
the decaying exponential at the right. (Compare Fig. 2-17b to Fig. 2-11.) The final 
values of A and B are simply those that give the appropriate phase and amplitude. 
The ratio A*A/B*B is the relative fluxes of particles traveling toward the right or 
left in the region to the left of the step. If |A|=|B|, the fluxes are equal, corresponding 
to total reflection of the beam from the step potential. 
It is not difficult to show (Problem 2-24) that |A| = |B| whenever .Ieft has real 
value and slope at any point, i.e., for any trigonometric wave, and so the potential of 
Fig. 2-17a gives total reflection if E<U. (The fact that some particle density exists 
at x >0 due to barrier penetration does not affect this conclusion. The evaluation of
58 Chapter 2 Quantum Mechanics of Some Simple Systems 
extent of reflection assumes that a time-independent (steady-state) situation has been 
achieved, so the penetration population remains constant and none of the new particles 
entering from the left are “lost” due to barrier penetration.) 
The situation changes when we consider cases for E>U. For any such E, we now 
have two acceptable exponential functions on both the right and the left. We proceed 
by realizing that the function with coefficient D in .right should be rejected since it 
corresponds to particles traveling from right to left, i.e., to particles that have been 
reflected from the trap. But we assume the trap to be 100% effective, so once again we 
have only one term in .right As before, we set about forcing the values and slopes of 
Eqs. (2-71) and (2-72) (with D =0) to be equal at x =0. This time, however, there is 
no requirement that these values be real. We arrive at the relations (Problem 2-25) 
B
A = 
k -k 
k +k 
and 
C
A = 
2k 
k +k 
(2-73) 
The extent of reflection is equal to 
|B|2 
|A|2 = k -k
2 
(k +k)2 (2-74) 
This can be seen to range from zero, when k=k, to one, when k=0. k=k when E= 
E-U, i.e., when U is negligible compared toE. So zero reflection (total transmission) 
is approached in the high-energy limit. Only whenE=U does k=0, so total reflection 
occurs only when E equals the barrier height (or, as we saw previously, is lower). 
A plot of the fraction of particles reflected versus E/U appears in Fig. 2-17c. The 
transmission, equal to 1.0–reflection, is plotted in Fig. 2-17d. 
The approach represented by this scattering problem is to identify the two terms 
that can contribute to the wavefunction in each region; then to recognize that one of 
the terms in one region is lost, either because it explodes or because it corresponds 
to reflection from the particle trap; then to force a smooth junction at the position of 
the discontinuity in the potential; and finally to draw conclusions about reflection and 
transmission from the values of the absolute squares of the coefficients. Notice that, 
for E >U, we could just as well have postulated the beam to be coming from the 
right, with the trap at the left. This would lead us to set A=0 in Eq. (2-71). Even in 
cases like this, where the particles are passing over the edge of a potential cliff, there 
is backscatter (Problem 2-26). 
An additional feature appears when we consider potentials that change at two points 
in x, as in Fig. 2-18a. The solution now involves three regions and two places (x=±a) 
where . and d./dx must be made equal. As before, we decide where the particle 
source and trap are and set one coefficient equal to zero (say G). Detailed solution 
of this problem is tedious.3 For our purposes, it is the nature of the result that is 
important. The extent of transmission as a function of E/U is plotted in Fig. 2-18b. 
There are two obvious ways this differs from the step-potential transmission function 
of Fig. 2-17d: First, some of the particles are transmitted even when their energy is less 
than U. This is the result of barrier penetration leading to finite particle density at the 
right-hand side of the barrier transmission due to tunneling. (The extent of tunneling 
transmission depends on the thickness of the barrier.) Second, there are oscillations 
3See Merzbacher [2, Chapter 6].
Section 2-9 Summary 59 
Figure 2-18 
 (a) A rectangular-hill potential of height U and width 2a. (b) Fraction of beam 
transmitted as a function of E/U, where U =h2/2p2ma2. 
in the transmission function for E>U, with 100% transmission occurring at intervals 
in E instead of only in the infinite limit. These come about because of interference 
between waves reflecting off the front and back edges of the barrier. This is most easily 
understood by recognizing that 100% transmission corresponds to no reflection, so then 
B =0. This occurs when the wave reflecting back from x =-a is of opposite phase 
to that reflecting back from x =+a, and this happens whenever there is an integral 
number of de Broglie half-wavelengths between x=-a and a. At energies where this 
happens, the beam behaves as though the potential barrier is not there. 
The variation of reflection from thin films (e.g., soap bubbles) of light of different 
wavelengths results in the perception of colors and is a familiar example of scattering 
interference. Less familiar is the variation in reflection of a particle beam, outlined 
above. However, once we recognize thewave nature of matter, we must expect particles 
to manifest the same sort of wave properties we associate with light. 
2-9 Summary 
In this chapter we have discussed the following points: 
1. A particle constrained in the classical sense (i.e., lacking the energy to overcome 
barriers preventing its motion over the entire coordinate range) will have quantized 
energy levels and a finite zero-point energy. In the mathematical analysis, this arises 
from requirements on . at boundaries.
60 Chapter 2 Quantum Mechanics of Some Simple Systems 
2. . can be nonsmooth, or cusped, where V is infinite at a point. If V is infinite over 
a finite range, . must be zero there. 
3. Nondegenerate eigenfunctions of H must be symmetric or antisymmetric for any 
operation that leaves H unchanged. 
4. |.|2 may be regarded as a statistical measure—a summary of many measurements 
of position on independent, but identically prepared, systems. 
5. Quantum-mechanical predictions approach classical predictions in the limits of 
large E, or large mass, or very high quantum number values. 
6. Integrals with antisymmetric integrands must vanish. 
7. |.|2 does not vanish in regions whereV >E if V is finite. This is called “barrier 
penetration.” 
8. One-dimensional motion of a free particle has a continuum of energy levels. Except 
for E = 0, the states are doubly degenerate. Therefore, any mixture of such a 
pair of states is still an eigenfunction of H. But only two eigenfunctions (for a 
given E = 0) are also eigenfunctions for the momentum operator. These are the 
exponential functions. Since they correspond to different momenta, mixing them 
produces functions that are not eigenfunctions for the momentum operator. 
9. Motion of a particle on a ring has quantum-mechanical solutions very similar to 
those for free-particle motion in one dimension. In both cases, there is no zero-point 
energy. Both are doubly degenerate forE>0 because two directional possibilities 
are present. Both have a set of exponential solutions that are eigenfunctions for 
momentum. The main difference is that the particle-in-a-ring energies are quantized, 
due to head-to-tail “joining conditions” on .. 
10. Increasing the dimensionality of a particle’s range of motion increases the number 
of quantum numbers needed to define the wavefunctions. In cases where the 
hamiltonian operator can be written as a sum of operators for different coordinates 
(i.e., is “separable”), the problem greatly simplifies; the wavefunctions become 
products, and the energies become sums. 
11. Scattering problems are treated by selecting an energy of interest from the continuum 
of possibilities, removing functions that describe nonphysical processes 
such as backscatter from the trap, and matching wave values and slopes at region 
boundaries. Resulting wavefunctions show wave interference effects similar to 
those observed for light. 
2-9.A Problems 
2-1. Ascertain that the expression (2-12) for energy has the proper dimensions. 
2-2. Solve Eq. (2-9) for A. 
2-3. There is a simple way to show that A in Eq. (2-9) must equal v2/L. It involves 
sketching .2, recognizing that sin2 x +cos2 x =1, and asking what A must equal
Section 2-9 Summary 61 
in order to make the area under .2 equal 1. Show this for n=1, and argue why 
it must give the same result for all n. 
2-4. Evaluate the probability for finding a particle in the middle third of a onedimensional 
box in states with n=1, 2, 3, 104. Compare your answers with the 
sketches in Fig. 2-5 to see if they are reasonable. 
2-5. a) Estimate the probability for finding a particle in the n = 1 state in the line 
elementx centered at the midpoint of a one-dimensional box ifx=0.01L. 
How does this compare to the classical probability? 
b) Repeat the problem, but with x centered one third of the way from a box 
edge. 
2-6. a) Use common sense to evaluate the following integral for the particle in a 
one-dimensional box, assuming that . is normalized. 
 L/5 
0 
.2
5dx 
b) How does this value compare to that for the integral over the same range, but 
using .1 instead of .5? (Larger, smaller, or equal?) Use a sketch to defend 
your answer. 
2-7. Let S and A be respectively symmetric and antisymmetric functions for the 
operator R. Evaluate the following, where R operates on every function to its 
right: (a) RS (b)RA(c) RSS (d)RAA(e)RAS(f)RAASASSA(g) RAASASAA. 
Can you think of a simple general rule for telling when a product of symmetric 
and antisymmetric functions will be antisymmetric? 
2-8. Using the concept of odd and even functions, ascertain by inspection of sketches 
whether the following need be identically zero: 
a)  p 
0 sin . cos. d. 
b)  p 
-p sin . cos. d. 
c)  1 
-1 x cos x dx 
d)  a 
-a cos y sin2 y dy 
e)  p 
0 sin3 . cos2 . d. 
f)  p 
0 sin2 . cos3 . d. 
g)  1 
-1  1 
-1 x2y dx dy 
h)  p 
-p x sin x cosx dx 
i)  p 
0 sin x d 
dx sin2 x dx 
j)  p 
-p sin2 x d 
dx sinx dx 
2-9. Verify Eq. (2-23) for the general case n =m by explicit integration. 
2-10. For the potential of Fig. 2-8, when E<U the energies are discrete, and when 
E >U, they are continuous. Is there a solution with E = U? What special 
requirements are there, if any, for such a solution to exist?
62 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure P2-12 
 
2-11. Prove the following statement: any linear combination of degenerate eigenfunctions 
of H is also an eigenfunction of H. 
2-12. In a few words, indicate what is wrong with the wavefunctions sketched in the 
potentials shown in Fig. P2-12. If the solution appears to be acceptable, indicate 
this fact. 
2-13. A double-well potential ranges from x = 0 to x = 2L and has a thin (width = 0.01L) rectangular barrier of finite height centered at L. 
a) Sketch the wavefunction that goes with the fourth energy level in this system, 
assuming that its energy is less than the height of the barrier. 
b) Estimate the energy of this level for a particle of mass m. 
2-14. Use the simple approach presented in Problem 2-3 to demonstrate that A = 1/vp for the trigonometric particle-in-a-ring eigenfunctions and 1/v2p for the 
exponential eigenfunctions. 
2-15. Explain why (2p)-1/2 exp(iv2f) is unacceptable as a wavefunction for the 
particle in a ring. 
2-16. For a particle in a ring, an eigenfunction is . =(1/vp)cos(3f). 
a) Write down H. 
b) Evaluate H. and identify the energy. 
c) Is this a state for which angular momentum is a constant of motion? Demonstrate 
that your answer is correct.
Section 2-9 Summary 63 
2-17. Consider two related systems—a particle in a ring of constant potential and 
another just like it except for a very thin, infinitely high barrier inserted at f =0. 
When this barrier is inserted, 
a) are any energies added or lost? 
b) do any degeneracies change? 
c) are exponential and sine–cosine forms both still acceptable for eigenfunctions? 
d) is angular momentum still a constant of motion? 
2-18. Consider a particle of mass m in a two-dimensional box having side lengths Lx 
and Ly with Lx =2Ly and V =0 in the box,8outside. 
a) Write an expression for the allowed energy levels of this system. 
b) What is the zero-point energy? 
c) Calculate the energies and degeneracies for the lowest eight energy levels. 
d) Sketch the wavefunction for the fourth level. 
e) Suppose V = 10 J in the box. What effect has this on (i) the eigenvalues? 
(ii) the eigenfunctions? 
2-19. Consider the particle in a three-dimensional rectangular box with Lx = Ly = 
Lz/2. What would be the energy when nx = 1,ny = 2,nz = 2? For nx = 1, 
ny =1, nz =4? Can you guess the meaning of the term “accidental degeneracy?” 
2-20. Consider a particle of mass m in a cubical box with V =0 at 0<x, y, z<L. 
a) Is 1/v2 .5,1,1 -.3,3,3 an eigenfunction for this system? Explain your 
reasoning. 
b) Estimate the probability for finding the particle in a volume element V = 0.001V at the box center when the system is in its lowest-energy state. What 
is the classical value? 
2-21. Kuhn [1] has suggested that the mobile p electrons in polymethine dyes can be 
modeled after the one-dimensional box. Consider the symmetric carbocyanine 
dyes (I) where the positive charge “resonates” between the two nitrogen atoms. 
The zigzag polymethine path along which the p electrons are relatively free to 
move extends along the conjugated system between the two nitrogens. Kuhn 
assumed a box length L equal to this path length plus one extra bond length 
on each end (so that the nitrogens would not be at the very edge of the box 
where they would be prevented from having any p-electron charge). This 
gives L = (2n + 10)l where l is 1.39 Å, the bond length of an intermediate 
(i.e., between single and double) C–C bond. The number of p electrons in the 
polymethine region is 2n + 10. Assume that each energy level in the box is 
capable of holding no more than two electrons and that the electronic transition 
responsible for the dye color corresponds to the promotion of an electron from 
the highest filled to the lowest empty level, the levels having initially been 
filled starting with the lowest, as shown in Fig. P2-13. Calculate E and . 
for the cases n=0, 1, 2, 3 and compare with the observed values of maximum 
absorption of about 5750, 7150, 8180, and 9250 Å, respectively.
64 Chapter 2 Quantum Mechanics of Some Simple Systems 
Figure P2-13 
 
2-22. Showwhether momentum in the x direction is a constant of motion for a free particle 
of massmin states described by the following functions. In cases where it is 
a constant of motion, give its value. In all cases, evaluate the kinetic energy of the 
particle. 
a) . =sin 3x 
b) . =exp(3ix) 
c) . =cos 3x 
d) . =exp(-3ix) 
2-23. Show whether angular momentum perpendicular to the plane of rotation is a 
constant of motion for a particle of massm moving in a ring of constant potential 
in states described by the following functions. In cases where it is a constant of 
motion, give its value. In all cases, evaluate the kinetic energy of the particle. 
a) . =(1/vp)cos 3f 
b) . =(1/v2p)exp(-3if) 
c) . =(1/vp)sin 3f 
d) . =(1/v2p)exp(3if) 
2-24. Demonstrate that the requirement that . =Aexp(ikt)+B exp(-ikt) have real 
value and slope at a point in x suffices to make |A|=|B|. 
2-25. Derive relations (2-73) and (2-74) by matchingwave values and slopes at the step.
Section 2-9 Summary 65 
2-26. Solve the problem for the step potential shown in Fig. 2-17a, but with the beam 
traveling from a source at right toward a trap at left. 
2-27. The reflection coefficient is defined as |B|2 / |A|2 for a beam originating from the 
left of Fig. 2-17a. The transmission coefficient is defined as (k/k) |C|2 / |A|2. 
a) Why is the factor k/k needed? 
b) Show that the sum of these coefficients equals unity, consistent with a 
steady-state situation. 
2-28. For scattering from a potential such as that in Fig. 2-18a, 100% transmission 
occurs at various finite particle energies. Find the lowest two values of E/U 
for which this occurs for particles of mass m, barrier width d, and barrier height 
U =2h2/p2md2. 
2-29. Calculate “frequencies” in cm-1 needed to accomplish the transitions E1 and 
E2 in Fig. 2-14. 
Multiple Choice Questions 
(Try to answer these by inspection.) 
1. The integral  a 
-a cos(x) sin(x) dx 
a) equals zero for any value of a, and cos(x) is antisymmetric in the range of the 
integral. 
b) is unequal to zero except for certain values of a, and cos(x) is symmetric in the 
range of the integral. 
c) equals zero for any value of a and cos(x) is symmetric in the range of the integral. 
d) is unequal to zero except for certain values of a, and sin(x) is antisymmetric in 
the range of the integral. 
e) equals zero for any value of a, and sin(x) is symmetric in the range of the integral. 
2.  2p 
0 x sin(x) cos(x) dx 
Which one of the following statements is true about the above integral and the three 
functions in its integrand? 
a) All three functions are antisymmetric in the range and the integral equals zero. 
b) Two functions are antisymmetric and one is symmmetric in the range, and the 
integral is unequal to zero. 
c) Two functions are symmetric and one is antisymmetric in the range, and the 
integral is equal to zero. 
d) One function is symmetric, one is antisymmetric, and one is unsymmetric in the 
range, and the integral is unequal to zero. 
e) None of the above is a true statement.
66 Chapter 2 Quantum Mechanics of Some Simple Systems 
3. In solving the particle in a one-dimensional box problem with infinite repulsive 
walls at x =0 and L, we started with the function Asin(kx)+B cos(kx). Which 
one of the following is a true statement? 
a) The value of k is found by requiring that the solution be normalized. 
b) Adding C exp(ikx) to the above function would prevent it from being an eigenfunction 
of the hamiltonian operator. 
c) It is necessary that this function equal L when x =0. 
d) The boundary condition at x =L is used to show that B =0. 
e) None of the above is a true statement. 
4. It is found that a particle in a one-dimensional box of length L can be excited from 
the n=1 to the n=2 state by light of frequency .. If the box length is doubled, the 
frequency needed to produce the n=1 to n=2 transition becomes 
a) ./4 
b) ./2 
c) 2. 
d) 4. 
e) None of the above is correct. 
5. For a particle in a one-dimensional box with infinite walls at x =0 and L, and in the 
n=3 state, the probability for finding the particle in the range 0=x =L/4 is 
a) greater than 1/3. 
b) exactly 1/6. 
c) exactly 1/3. 
d) less than 1/6. 
e) None of the above is correct. 
6. A student calculates the probability for finding a particle in the left-most 10% of a 
one-dimensional box for the n=1 state. Which one of the following answers could 
be correct? 
a) 1.742 
b) 0.024 
c) 0.243 
d) 0.100 
e) None of the above is reasonable. 
7. Which one of the following statements is true about the particle in a one-dimensional 
box with infinite walls? (All integrals range over the full length of the box.) 
a)  .2
3 dx =0 because these wavefunctions are orthogonal. 
b)  .1.3 dx =0 because both of these wavefunctions are symmetric. 
c)  .1.2 dx =0 because these wavefunctions are normalized. 
d)  .1.3 dx =0 because these wavefunctions are orthogonal. 
e) None of the above is a true statement.
Section 2-9 Summary 67 
8. Which one of the following is the correct formula for the lowest-energy eigenfunction 
for a particle in a one-dimensional box having infinite barriers at x =-L/2 
and L/2? 
a) 2
L sin(px/L) 
b) 2
L cos(px/L) 
c) 2
L exp(ipx/L) 
d) 2
L exp(-ipx/L) 
e) None of the above is correct. 
9. A particle is free to move in the x dimension without constraint (i.e., under the 
influence of a constant potential, which we assume to be zero). For this system, 
the wavefunction .(x)=exp(3.4x) is not acceptable because 
a) it is not an eigenfunction of the hamiltonian operator. 
b) it is multi-valued. 
c) it is discontinuous. 
d) it approaches infinity as x approaches infinity. 
e) it goes to zero as x approaches minus infinity. 
10. Consider two identical one-dimensional square wells connected by a finite barrier. 
Which one of the following statements about the quantum-mechanical timeindependent 
solutions for this system is true when two equivalent “half-solutions” 
in the two wells are joined together to produce two overall solutions? 
a) The combination that is symmetric for reflection through the central barrier 
always has the lower energy of the two. 
b) The combination that places a node in the barrier always has the lower energy 
of the two. 
c) The sum of the two half-solutions always has the lower energy of the two. 
d) The difference of the two half-solutions always has the lower energy of the two. 
e) None of the above is a true statement. 
11. For a single particle-in-a-ring system having energy 9h2/8p2I we can say that the 
angular momentum, when measured, will equal 
a) 3¯h 
b) v12¯h 
c) either 3¯h or -3¯h 
d) zero 
e) None of the above is a true statement. 
12. A particle in a ring has wavefunctions that 
a) result from placing an integral number of half-waves in the circumference of 
the ring. 
b) must be eigenfunctions for (h/2pi)d/df. 
c) are all doubly degenerate, due to two rotational directions.
68 Chapter 2 Quantum Mechanics of Some Simple Systems 
d) correspond to energies that increase with the square of the quantum number. 
e) None of the above is a true statement. 
13. The hamiltonian operator for a system is 
H =-(h2/8p2m).2 +x2 +y2 +z2 
For this system we should expect 
a) two quantum numbers at most. 
b) eigenfunctions that are sums of functions, each depending on only one of the 
variables. 
c) eigenvalues that are products of eigenvalues of separated equations. 
d) eigenvalues that are sums of eigenvalues of separated equations. 
e) None of the above is a correct statement. 
14. For a particle in a one-dimensional box with one infinite barrier and one finite 
barrier of height U, 
a) . =0 at both barriers if E is less than U. 
b) barrier penetration is smallest for the lowest-energy state. 
c) no more than one quantized state can exist, and its energy is less than U. 
d) only states having E greater than U can exist. 
e) None of the above statements is correct. 
15. For a particle of mass m in a cubical box having edge length L, 
a) the zero point energy is 3h2/8mL2. 
b) the probability density has its maximum value at the box center for .211. 
c) the degeneracy of the ground state is 3. 
d) .111 has three nodal planes. 
e) None of the above is a true statement. 
References 
[1] H. Kuhn, J. Chem. Phys. 17, 1198 (1949). 
[2] B. Merzbacher, Quantum Mechanics, 3rd ed.Wiley, NewYork, 1998.
Chapter 3 
The One-Dimensional Harmonic 
Oscillator 
3-1 Introduction 
In Chapter 2 we examined several systems with discontinuous potential energies. In this 
chapter we consider the simple harmonic oscillator—asystem with a continuously varying 
potential. There are several reasons for studying this problem in detail. First, the 
quantum-mechanical harmonic oscillator plays an essential role in our understanding 
of molecular vibrations, their spectra, and their influence on thermodynamic properties. 
Second, the qualitative results of the problem exemplify the concepts we have presented 
in Chapters 1 and 2. Finally, the problem provides a good demonstration of mathematical 
techniques that are important in quantum chemistry. Since many chemists are not 
overly familiar with some of these mathematical concepts, we shall deal with them in 
detail in the context of this problem. 
3-2 Some Characteristics of the Classical 
One-Dimensional Harmonic Oscillator 
A pendulum consisting of a large mass hanging by an almost weightless wire, and 
swinging through a very small angle, is a close approximation to a classical harmonic 
oscillator. It is an oscillator since its motion is back and forth over the same path. 
It is harmonic to the extent that the restoring force on the mass is proportional to the 
horizontal component of the displacement of the mass from its rest position. This force 
law, known as Hooke’s law, is the common first approximation made in the analysis of 
a system vibrating about an equilibrium position. If we let the x axis be the coordinate 
of displacement of the mass, with x = 0 as the rest position, then we may write the 
restoring force as 
F =-kx, (3-1) 
where k is the force constant. The minus sign assures that the force on the displaced 
mass is always directed toward the rest position. 
We can use this force expression to determine an equation of motion for the mass, 
that is, an equation relating its location in space, x, to its location in time, t : 
F =-kx(t)=ma =m
d2x(t) 
dt2 (3-2) 
69
70 Chapter 3 The One-Dimensional Harmonic Oscillator 
or 
d2x(t) 
dt2 =(-k/m)x(t) (3-3) 
According to this equation, x(t) is a function that, when differentiated twice, is regenerated 
with the multiplier -k/m. A general solution is 
x(t)=a sin
 k
m 
t

+b cos
 k
m 
t
 (3-4) 
If we require that x(t) be at its maximum value L at t = 0 (as though the pendulum 
is held at its position of maximum displacement and then released at t = 0), then it 
follows that b=L. Since the pendulum is also motionless at t =0, (dx(t)/dt)t=0 =0, 
and so a =0. Hence, 
x (t)=Lcos
 k
m 
t
 (3-5) 
Thus, the equation of motion [Eq. (3-3)] leads to a function, x(t), that describes the 
trajectory of the oscillator. This function has the trigonometric time dependence characteristic 
of harmonic motion. From this expression, we see that x(t) repeats itself 
whenever the argument of the cosine increases by 2p. This will require a certain time 
interval t 
. Thus, the pendulum makes one complete back and forth motion in a time t 
 
given by 
 k
m 
t 
 =2p t
 =2pm
k 
(3-6) 
so the frequency of the oscillation . is 
. =1/t
 = 
1 
2p 
 k
m 
(3-7) 
Suppose that one were to take a multiflash photograph of a swinging pendulum from 
above. The result would look as shown in Fig. 3-1a, the number of images being much 
greater near the termini of the swing (called the “turning points”) than at the middle 
because the pendulum is moving fastest as it crosses the middle. This, in turn, results 
from the fact that all the potential energy of the mass has been converted to kinetic 
energy at the middle point. We thus arrive at a classical prediction for the time-averaged 
distribution of the projection of the harmonic oscillator in the displacement coordinate: 
This distribution function is greatest in regions where the potential energy is highest 
(Fig. 3-1b) (Problem 3-1). 
Let us calculate and compare the time-averaged potential and kinetic energies for the 
classical harmonic oscillator. When the particle is at some instantaneous displacement 
x
, its potential energy is 
V (x
) = (applied force times distance to return to x =0) 
=  x
 
0 
kx dx = 
1
2
kx
2 (3-8)
Section 3-2 Some Characteristics of the Classical One-Dimensional Harmonic Oscillator 71 
Figure 3-1 
 (a) Results of a uniform multiflash photograph of a swinging pendulum as photographed 
from above. (b) Distribution function corresponding to the continuous limit of discrete 
distribution shown in (a). 
The cumulative value of the potential energy over one complete oscillation, Vc, is given 
by the integral 
Vc t 
 -0= t 
 
0 
V (t) dt = 
1
2
k  t 
 
0 
x (t)2 dt (3-9) 
Substituting for x(t) as indicated in Eq. (3-5) 
Vc t 
 -0 = 
1
2
kL2  t 
 
0 
cos2 
 k
m 
t
 
dt 
= 
1
2
kL2m/k  t 
 
0 
cos2 
 k
m 
t
 
d 

 k
m 
t
 (3-10) 
When t =t 
, vk/mt =2p, and we may rewrite Eq. (3-10) as 
Vc t 
 -0= 
1
2
kL2m/k  2p 
0 
cos2 y dy =(p/2)kL2m/k (3-11) 
If we now divide by t 
 to get the average potential energy per unit time, we find 
V¯ = 
Vc t 
 -0 
t 
 = 
(p/2)kL2vm/k 
2pvm/k = 
kL2 
4 
(3-12) 
Thus, we have the average potential energy. If we knew the total energy, which is a 
constant of motion, we could get the average kinetic energy by taking the difference.
72 Chapter 3 The One-Dimensional Harmonic Oscillator 
It is easy to evaluate the total energy by taking advantage of its constancy over time. 
Since we can choose any point in time to evaluate it, we select the moment of release 
(t =0 and x =L). The mass is motionless so that the total energy is identical to the 
potential energy: 
E = 
1
2
kL2 (3-13) 
Comparing Eqs. (3-12) and (3-13) we see that, T¯ =kL2/4. On the average, then, the 
classical harmonic oscillator stores half of its total energy as potential energy and half 
as kinetic energy. 
EXAMPLE 3-1 A 10.00 g mass on a Hooke’s law spring with force constant k = 0.0246 N m-1 is pulled from rest at x =0 to x =0.400m and released. 
a) What is its total energy? 
b) What is its frequency of oscillation? 
c) How do these answers change if the mass weighs 40.00 g? 
SOLUTION 
 a) E =T +V. At x =0.400m,T =0,E =V =kx2/2=0.50(0.0246N m-1) 
(0.400m)2 =0.0020 J 
b) . = 1 
2p  k
m	
1/2 
= 1 
2p 0.0246 N m-1 
0.0100 kg 	
1/2 
=0.25 s-1 
c) E is unchanged, and . is halved to 0.125 s-1. The energy depends on the force constant and 
the displacement of the oscillator, but not on the mass of the oscillator. The frequency depends not 
only on the force constant, but also on the mass because the frequency is affected by the inertia of 
the oscillator. A greater mass has a lower frequency of oscillation for the same force constant.  
3-3 The Quantum-Mechanical Harmonic Oscillator 
We have already seen [Eq. (3-8)] that the potential energy of a harmonic oscillator is 
given by 
V (x)= 
1
2
kx2 (3-14) 
so we can immediately write down the one-dimensional Schr¨odinger equation for the 
harmonic oscillator: 

(-h2/8p2m)(d2/dx2)+ 
1
2
kx2.(x)=E.(x) (3-15) 
The detailed solution of this differential equation is taken up in the next section. 
Here we show that we can understand a great deal about the nature of the solutions to 
this equation by analogy with the systems studied in Chapter 2. 
In Fig. 3-2a are shown the potential, some eigenvalues, and some eigenfunctions for 
the harmonic oscillator. The potential function is a parabola [Eq. (3-14)] centered at 
x=0 and having a value of zero at its lowest point. For comparison, similar information
Section 3-3 The Quantum-Mechanical Harmonic Oscillator 73 
Figure 3-2 
 The potential function, energy levels, and wavefunctions for (a) the harmonic oscillator, 
(b) the particle in the infinitely deep square well. (c) .2 for the harmonic oscillator in the state 
n=5. 
is graphed in Fig. 3-2b for the particle in a box with infinitely high walls. Some 
important features of the harmonic oscillator are: 
1. The energy-level spacing for the harmonic oscillator is constant. As mentioned 
in Chapter 2, we expect the energy levels for the harmonic oscillator to diverge 
less rapidly than those for the square well because the higher energy states in the 
harmonic oscillator have effectively larger “boxes” than do the lower states (that is, 
the more energetic the oscillator, the more widely separated are its classical turning 
points). That the spacing for the oscillator grows less rapidly than that for the box 
is therefore reasonable. That it is actually constant is something we will show 
mathematically in the next section. 
2. The wavefunctions for the harmonic oscillator are either symmetric or antisymmetric 
under reflection through x=0. This is necessary because the hamiltonian is invariant
74 Chapter 3 The One-Dimensional Harmonic Oscillator 
to reflection through x =0 and because the eigenfunctions are nondegenerate. Both 
of these conditions apply also to the box problem. If we imagine moving from the 
box potential to the parabolic potential by a process of continuous deformation, the 
symmetry is not altered and there is no reason to expect nondegenerate box levels 
to come together and become degenerate. Therefore, the oscillator wavefunction 
symmetries are not surprising. As a consequence, we have at once that the symmetric 
wavefunctions are automatically orthogonal to antisymmetric wavefunctions, as 
discussed earlier. (Actually, all the wavefunctions are orthogonal to each other. This 
is proved in Section 3-4.) 
3. The harmonic oscillator has finite zero-point energy. (The evidence for this in 
Fig. 3-2a is the observation that the line for the lowest (n = 0) energy level lies 
above the lowest point of the parabola, where V = 0.) This is expected since the 
change from square well to parabolic well does not remove the restrictions on particle 
position; it merely changes them. 
4. The particle has a finite probability of being found beyond the classical turning 
points; it penetrates the barrier. This is to be expected on the basis of earlier 
considerations since the barrier is not infinite at the classical turning point. (The 
potential becomes infinite only at x=±8.) 
5. In the lowest-energy state the probability distribution favors the particle being in 
the low-potential central region of the well, while at higher energies the distribution 
approaches more nearly the classical result of favoring the higher potential regions 
(Fig. 3-2c). 
3-4 Solution of the Harmonic Oscillator 
Schr ¨ odinger Equation 
3-4.A Simplifying the Schr ¨ odinger Equation 
Equation (3-15) is simplified by substituting in the following relations: 
a = 8p2mE/h2 (3-16) 
ß2 = 4p2mk/h2 (3-17) 
The quantities a and ß have units of m-2. We will assume that ß is the positive root 
of ß2. The quantity a is necessarily positive. The Schr¨odinger equation now can be 
written 
d2.(x) 
dx2 +a -ß2x2. (x)=0 (3-18) 
3-4.B Establishing the Correct Asymptotic Behavior 
At very large values of |x|, the quantity a (which is a constant since E is a constant) 
becomes negligible compared to ß2x2. That is, the Schr¨odinger equation (3-18) 
approaches more and more closely the asymptotic form 
d2.(x)/dx2 =ß2x2.(x), |x|.8 (3-19)
Section 3-4 Solution of the Harmonic Oscillator Schr ¨ odinger Equation 75 
What we need, then, are solutions .(x) that approach the solutions of Eq. (3-19) at 
large values of |x|. The solutions for Eq. (3-19) can be figured out from the general 
rule for differentiating exponentials: 
(d/dx) exp(u(x))=exp(u(x))du(x)/dx (3-20) 
Then 
(d2/dx2) exp(u(x))=[(du(x)/dx)2 +d2u(x)/dx2]exp(u(x)) (3-21) 
We want the term in square brackets to become equal to ß2x2 in the limit of large |x|. 
We can arrange for this to happen by setting 
u(x)=±ßx2/2 (3-22) 
for then 
(d2/dx2) exp(u(x))=(ß2x2 ±ß) exp(±ßx2/2) (3-23) 
At large values of |x|, ß is negligible compared to ß2x2, and so exp(±ßx2/2) are 
asymptotic solutions for Eq. (3-19). As |x| increases, the positive exponential increases 
rapidly whereas the negative exponential dies away. We have seen that, for the wavefunction 
to be physically meaningful, we must reject the solution that blows up at large 
|x|. On the basis of these considerations, we can say that, if . contains exp(-ßx2/2), 
it will have the correct asymptotic behavior if no other term is present that dominates 
at large |x|. Therefore, 
.(x)=q(x) exp(-ßx2/2) (3-24) 
and it remains to find the function q(x). 
3-4.C The Differential Equation for q(x) 
Substituting Eq. (3-24) into the Schr¨odinger equation (3-18) gives 
exp(-ßx2/2) 
-ßq (x)-2ßx 
dq(x) 
dx + 
d2q(x) 
dx2 +aq (x)=0 (3-25) 
This equation is satisfied only if the term in brackets is zero: 
d2q(x) 
dx2 -2ßx 
dq (x) 
dx +(a -ß) q (x)=0 (3-26) 
Thus, we now have a differential equation for q(x). 
At this point it is convenient to transform variables to put the equation into a simpler 
form. Let 
y =ßx (3-27) 
Then 
d/dy =d/d ßx	=1/ß	d/dx (3-28)
76 Chapter 3 The One-Dimensional Harmonic Oscillator 
so that 
d/dx =ßd/dy (3-29) 
Similarly 
d2 
dx2 = 
ßd2 
dy2 (3-30) 
and 
x = 
y 
vß 
(3-31) 
Substituting Eqs. (3-29)–(3-31) into (3-26), and defining f (y) as 
f (y)=f ßx	=q(x) (3-32) 
we obtain (after dividing by ß) 
d2f (y) 
dy2 -2y 
df (y) 
dy +[(a/ß)-1] f (y)=0 (3-33) 
3-4.D Representing f as a Power Series 
Now f (y) is some function of y that must be single valued, continuous, and smooth 
(i.e., have a continuous first derivative), if . is to be properly behaved. Can we think 
of any functions that satisfy these properties? Of course, we can think of a limitless 
number of them. For example, 1, y, y2, y3, y4, etc., are all single valued, continuous, 
and have continuous derivatives, and so is any linear combination of such functions (e.g., 
4y3-y +3). Other examples are sin(y) and exp(y). These functions can be expressed 
as infinite sums of powers of y, however, and so they are included, in principle, in the 
first example. Thus 
sin(y)=y -y3/3!+y5/5!-y7/7!+··· (3-34) 
and 
exp(y)= 
8 
n=0 
yn/n! (3-35) 
that is, sin(y) and the set of all positive powers of y are linearly dependent. Because the 
powers of y can be combined linearly to reproduce certain other functions, the powers 
of y are called a complete set of functions. However, we must exercise some care with 
the concept of completeness. The positive powers of y cannot be used to reproduce a 
discontinuous function, or a function with discontinuous derivatives. Hence, there are 
certain restrictions on the nature of functions g(y), that satisfy the relation 
g(y)= 
8 
n=0 
cnyn (3-36)
Section 3-4 Solution of the Harmonic Oscillator Schr ¨ odinger Equation 77 
These restrictions define a class of functions, and the powers of y are a complete set 
only within this class. The positive powers of y, then, form a complete set, but if we 
remove one of the members of the set, say 1 (the zero power of y), then the set is no 
longer complete. This means that the remaining members of the set cannot compensate 
for the role played by the missing member. In other words, the missing member cannot 
be expressed as a linear combination of the remaining members. In this example 
1 = 
8 
n=1 
cnyn (3-37) 
This is easily demonstrated to be true since the left-hand side of Eq. (3-37) is unity 
whereas the right-hand side must be zero when y = 0 for any choice of coefficients. 
Thus, the powers of y are linearly independent functions (no one of them can be 
expressed as a linear combination of all the others). 
The function f (y) involved in . should be a member of the class of functions for 
which the powers of y form a complete set. Therefore, we may write 
f (y)= 
8 
n=0 
cnyn (3-38) 
and seek an expression for the unknown multipliers cn. 
3-4.E Establishing a Recursion Relation for f 
Notice that if 
f (y)=c0 +c1y +c2y2 +c3y3 +c4y4+··· (3-39) 
then 
df (y)/dy =c1 +2c2y +3c3y2 +4c4y3+··· (3-40) 
and 
d2f (y)/dy2 =2c2 +2 · 3c3y +3 · 4c4y2+··· (3-41) 
Thus, substituting Eq. (3-38) into (3-33) gives 
1 · 2c2 +2 · 3c3y +3 · 4c4y2+···-2c1y -2 · 2c2y2 -2 · 3c3y3-··· 
+[(a/ß)-1] c0 +[(a/ß)-1] c1y +[(a/ß)-1] c2y2+··· 
=0 (3-42) 
Now we will take advantage of the fact that the various powers of y form a linearly 
independent set. Equation (3-42) states that the expression on the LHS equals zero 
for all values of y. There are two ways this might happen. One of these is that minus 
the constant part of the expression is always exactly equal to the y-dependent part, no 
matter what the value of y. This would require a relationship like Eq. (3-37) (except 
with an equality), which we have seen is not possible for independent functions. The 
remaining possibility is that the various independent parts of Eq. (3-42) are individually
78 Chapter 3 The One-Dimensional Harmonic Oscillator 
equal to zero—the constant is zero, the coefficient for y is zero, etc. This gives us a 
whole set of equations. Setting the constant term equal to zero gives 
2c2 +[(a/ß)-1]c0 =0 ··· m=0 (3-43) 
Setting the coefficient for the first power of y to zero gives 
2 · 3c3 -2c1 +[(a/ß)-1]c1 =0 ··· m=1 (3-44) 
The y2 term gives 
3 · 4c4 -2 · 2c2 +[(a/ß)-1]c2 =0 ··· m=2 (3-45) 
By inspecting this series, we can arrive at the general result 
(m+1)(m+2)cm+2 +[(a/ß)-1-2m]cm =0 (3-46) 
or 
cm+2 = -[(a/ß)-2m-1] 
(m+1)(m+2) 
cm (3-47) 
Equation (3-47) is called a recursion relation. If we knew c0, we could produce 
c2, c4, c6, etc., by continued application of Eq. (3-47). Similarly, knowledge of c1 
would lead to c3, c5, etc. Thus, it appears that the coefficients for even powers of y 
and those for odd powers of y form separate sets. Choosing c0 determines one set, 
and choosing c1 determines the other, and the choices for c0 and c1 seem independent. 
This separation into two sets is reasonable when we recall that our final solutions 
must be symmetric or antisymmetric in x, hence also in y. The asymptotic part of ., 
exp(-ßx2/2), is symmetric about x =0, and so we expect the remainder of ., f (y), to 
be either symmetric (even powers of y =vßx) or antisymmetric (odd powers). Thus, 
we can anticipate that some of our solutions will have c0 =0, c2 =0, c4 =0, . . . and 
c1=c3=c5 =···=0. This will produce symmetric solutions. The remaining solutions 
will have c0 =c2 =c4=···=0 and c1 =0, c3 =0, . . . , and be antisymmetric. 
EXAMPLE 3-2 Evaluate c3 and c5 if c1 =1, for arbitrary a and ß. What ratio a/ß 
will make c3 =0? What ratio a/ß will make c5 =0 but c3 =0? 
SOLUTION 
 c1 =1, For c3, we use m=1, c3 = -[a/ß-2-1] (2)(3) ×1= 3-a/ß 
6 . 
For c5 we use m=3, c5 = -[a/ß-6-1] (4)(5) × [3-a/ß] 6 = (7-a/ß)(3-a/ß) 
120 . 
a/ß =3 makes c3 =0 (and also c5). a/ß =7 makes c5 =0, but not c3.  
3-4.F Preventing f(y) from Dominating the Asymptotic Behavior 
We now examine the asymptotic behavior of f (y). Recall that, at very large values 
of |y|, f (y) must become insignificant compared to exp(-ßx2/2)=exp(-y2/2). We 
will show that f (y) fails to have this behavior if its power series expression is infinitely 
long. That is, we will show that f (y) behaves asymptotically like exp(y2), which 
dominates exp(-y2/2).
Section 3-4 Solution of the Harmonic Oscillator Schr ¨ odinger Equation 79 
We know that the series expression for exp(y2) is 
exp(y2)=1+y2 + 
y4 
2! + 
y6 
3! +···+ 
yn 
(n/2)! + 
yn+2 
(n/2+1)! +··· (3-48) 
The series for f (y) has terms 
···+cnyn +cn+2yn+2 +cn+4yn+4+··· (3-49) 
The ratio between coefficients for two adjacent terms high up in the series (large n) for 
exp(y2) is, from Eq. (3-48) 
coeff for yn+2 
coeff for yn = 
(n/2)! 
[(n/2)+1]! = 
1 
(n/2)+1 
large n ---.
2
n 
(3-50) 
and for f (y) it is, from Eqs. (3-49) and (3-47), 
coeff for yn+2 
coeff for yn = 
cn+2 
cn = -(a/ß)+1+2n 
n2 +3n+2 
large n ---. 
2
n 
(3-51) 
This means that, at large values of y, when the higher-order terms in the series dominate, 
f (y) behaves like exp(y2). Then 
lim 
y.8
.(y)= lim 
y.8
f (y) exp(-y2/2)=exp(y2/2)-.8 (3-52) 
The asymptotic behavior of . is ruined. We can overcome this problem by requiring 
the series for f (y) to terminate at some finite power. In other words, f (y) must be 
a polynomial. This condition is automatically fulfilled if any one of the coefficients 
in a given series (odd or even) is zero since Eq. (3-47) guarantees that all the higher 
coefficients in that series will then vanish. Therefore, we require that some coefficient 
vanish: 
cn+2 =0 (3-53) 
Assuming that this is the lowest zero coefficient (i.e., cn =0), Eq. (3-47) gives 
(a/ß)-2n-1=0 (3-54) 
or 
a =ß(2n+1) (3-55) 
3-4.G The Nature of the Energy Spectrum 
Now we have a recipe for producing acceptable solutions for the Schr¨odinger equation 
for the harmonic oscillator. If we desire a symmetric solution, we set c1=0 and c0=1. 
If we want the polynomial to terminate at yn, we require that a and ß be related as in 
Eq. (3-55). In this way we can generate an unlimited number of symmetric solutions, 
one for each even value of n that can be chosen for the terminal value. Similarly, an 
unlimited set of antisymmetric solutions results from setting c0 = 0 and c1 = 1 and 
allowing the highest contributing value of n to take on various odd integer values.
80 Chapter 3 The One-Dimensional Harmonic Oscillator 
(Solved Problem 3-2 provides an example by showing that c5 becomes the first oddpower-
coefficient equal to zero if a/ß =7.) 
Since we now know that an acceptable solution satisfies Eq. (3-55), we can examine 
the energy spectrum. Substituting into Eq. (3-55) the expressions for a and ß 
[Eqs. (3-16) and (3-17)], we obtain 
8p2mE/h2 =(2pvmk/h)(2n+1) (3-56) 
or 
E =h
n+ 
1
2

 1 
2p 
k/m=
n+ 
1
2
h. (3-57) 
where the classical definition of . [Eq. (3-7)] has been used. 
This result shows that, whenever n increases by unity, the energy increases by h., so 
the energy levels are evenly spaced as shown in Fig. 3-2. At absolute zero, the system 
will lose its energy to its surroundings insofar as possible. However, since n=0 in the 
lowest permissible state for the system, there will remain a zero-point energy of 1
2h.. 
3-4.H Nature of theWavefunctions 
The lowest energy solution corresponds to n = 0. This means that c0 is the highest 
nonzero coefficient in the power series expansion for f (y). Hence, we must set c2 =0. 
(The odd-powered series coefficients are all zero for this case.) Thus, for n=0, we can 
write the unnormalized wavefunction as [from Eq. (3-24)] 
.0 =c0 exp(-y2/2)=c0 exp(-ßx2/2) (3-58) 
This is just a constant times a Gauss error function or simple “gaussian-type” function. 
This wavefunction is sketched in Fig. 3-2 and is obviously symmetric. 
The next solution has n=1, c1 =0 but c3 =c5=···=0. (The even-powered series 
coefficients are all zero for this case.) The unnormalized wavefunction for n=1 is then 
.1(y)=c1y exp(-y2/2) (3-59) 
The exponential is symmetric and y is antisymmetric, and so their product, .1(y), is 
antisymmetric (Fig. 3-2). 
To get .2 we need to use the recursion relation (3-47). We know that odd-index 
coefficients are all zero and that only c0 and c2 of the even-index coefficients are 
nonzero. Assuming c0 =1, we have (using (Eq. 3-47) with m=0) 
c2 = -[(a/ß)-2 · 0-1] 
(1)(2) · 1=-
(a/ß)-1 
2 
(3-60) 
But the ratio a/ß is determined by the requirement that c4=0. Referring to Eq. (3-55), 
this gives (for n=2) a/ß =5, and so 
c2=-4/2=-2 (3-61) 
and the unnormalized wavefunction is 
.2(y)=(1-2y2) exp(-y2/2) (3-62)
Section 3-4 Solution of the Harmonic Oscillator Schr ¨ odinger Equation 81 
Polynomials like (1-2y2), which are solutions to the differential equation (3-33), 
are known as Hermite (her·meet) polynomials, Hn(y). In addition to the recursion 
relation (3-47), which we have derived, other definitions are known. One of these 
involves successive differentiation: 
Hn(y)=(-1)n exp(y2)
dn exp(-y2) 
dyn 
(3-63) 
Thus, if we want H2(y), we just set n=2 in Eq. (3-63) and evaluate that expression to 
get 
H2(y)=4y2 -2 (3-64) 
which differs from our earlier result by a factor of -2. Yet another means of producing 
Hermite polynomials is by using the generating function 
G(y, u)=exp[y2 -(u-y)2]= 
8 
n=0
(Hn(y)/n!)un (3-65) 
We use this expression as follows: 
1. Express the exponential in terms of its power series, writing downa fewof the leading 
terms. There will exist, then, various powers of u and y and factorial coefficients. 
2. Collect together all the terms containing u2. 
3. The coefficient for this term will be equal to H2(y)/2. 
This is a fairly clumsy procedure for producing polynomials, but Eq. (3-65) is 
useful in determining general mathematical properties of these polynomials. For 
instance, Eq. (3-65) will be used in showing that the harmonic oscillatorwavefunctions 
are orthogonal. 
3-4.I Orthogonality and Normalization 
We will now show that the harmonic oscillator wavefunctions are orthogonal, i.e., that 
 +8 
-8 
.n(y).m(y) dy = +8 
-8 
Hn(y)Hm(y) exp(-y2)dy =0 ··· n =m (3-66) 
Consider the integral involving two generating functions and the exponential of y2: 
 +8 
-8 
G(y, u)G(y, v) exp(-y2)dy 
=
n 

m 
unvm  +8 
-8 
Hn(y)Hm(y) 
n!m! 
exp(-y2)dy 
 cnm  
(3-67) 
where we label the integral cnm for convenience. The left-hand side of Eq. (3-67) may 
also be written as [using Eq. (3-65)] 
 +8 
-8 
exp[-(y -u-v)2]exp(2uv)dy =exp(2uv)  +8 
-8 
exp[-(y -u-v)2]dy 
(3-68)
82 Chapter 3 The One-Dimensional Harmonic Oscillator 
However, we can add constants to the differential element without affecting the integral 
value, and u and v are constants when only y varies. Therefore, (3-68) becomes (see 
Appendix 1 for a table of useful integrals) 
exp(2uv)  +8 
-8 
exp[-(y -u-v)2]d(y -u-v)=exp(2uv)vp 
=vp{1+2uv +4u2v2/2!+8u3v3/3!+···+2nunvn/n!+···} (3-69) 
This expression is equal to the right-hand side of Eq. (3-67). Comparing Eq. (3-67) 
with (3-69), we see that c11=2vp since the term u1v1 is multiplied by 2p in Eq. (3-69) 
and by c11 in Eq. (3-67). Similarly c22 =4vp/2! But c12 =0. Hence, we arrive at the 
result 
cnm = +8 
-8 
Hn (y)Hm (y) 
n!m! 
exp -y2 dy =vp 2n/n! dn,m (3-70) 
(dn,m is the “Kronecker” delta. It is a discontinuous function having a value of unity 
when n=m but zero when n =m.) So 
 +8 
-8 
.n (y).m (y) dy =vpm!2ndn,m (3-71) 
This proves the wavefunctions to be orthogonal and also provides us with a normalizing 
factor. Normality refers to integration in x, rather than in y =vßx, so we must change 
the differential element in Eq. (3-71): 
 +8 
-8 
.2
n (y) dy =ß  +8 
-8 
.2
n (y) dx =vpn!2n (3-72) 
Requiring that  +8 
-8 .2
n (y) dx = 1 leads to the expression for the normalized wavefunctions: 
.n (y)=

ß
p 
1 
2nn!

1/2
Hn (y) exp -y2/2, n=0, 1, 2, . . . (3-73) 
The first members of the set of Hermite polynomials are 
H0(y) = 1, H1(y)=2y, H2(y)=4y2 -2, H3(y)=8y3 -12y 
H4(y) = 16y4 -48y2 +12, H5(y)=32y5 -160y3 +120y (3-74)
Section 3-5 Quantum-Mechanical Average Value of the Potential Energy 83 
3-4.J Summary of Solution of Harmonic-Oscillator 
Schr ¨ odinger Equation 
The detailed solution is so long that the reader may have lost the broad outline. 
The basic steps were: 
1. Determine the asymptotic behavior of the Schr¨odinger equation and of .. This 
produces a gaussian factor exp(-y2/2) times a function of y, f (y). 
2. Obtain a differential equation for the rest of the wavefunction, f (y). 
3. Representf (y) as a power series in y, and find a recursion relation for the coefficients 
in the series. The symmetries of the wavefunctions are linked to the symmetries of 
the series. 
4. Force the power series to be finite (i.e., polynomials) so as not to spoil the asymptotic 
behavior of thewavefunctions. This leads to a relation between a and ß that produces 
uniformly spaced, quantized energy levels. 
5. Recognize the polynomials as being Hermite polynomials, and utilize some of the 
known properties of these functions to establish orthogonality and normalization 
constants for the wavefunctions. 
EXAMPLE 3-3 Which of the following expressions are, by inspection, unacceptable 
eigenfunctions for the Schr¨odinger equation for the one-dimensional harmonic 
oscillator? 
a) (64y6 -480y4 +720y2 -120) exp(y2/2) 
b) (64y6 -480y5 +720y3 -120) exp(-y2/2) 
c) (64y6 -480y4 +720y2 -120) exp(-y2/2) 
SOLUTION 
 a) is unacceptable because exp(y2/2) blows up at large |y|. 
b) is unacceptable because the polynomial contains terms of both even and odd powers. 
c) is acceptable.  
3-5 Quantum-Mechanical Average Value of 
the Potential Energy 
We showed in Section 3-2 that the classical harmonic oscillator stores, on the average, 
half of its energy as kinetic energy, and half as potential. We now make the analogous 
comparison in the quantum-mechanical system for the ground (n=0) state. 
The wavefunction is 
.0(x)=(ß/p)1/4 exp(-ßx2/2) (3-75) 
and the probability distribution of the particle along the x coordinate is given by .2
0 (x). 
The total energy is constant and equal to 
E0 = 
1
2
h. =(h/4p)k/m (3-76)
84 Chapter 3 The One-Dimensional Harmonic Oscillator 
and the potential energy as a function of x is 
V (x)= 
1
2
kx2 (3-77) 
The probability for finding the oscillating particle in the line element dx around some 
point x1 is .2
0 (x1)dx, since .0 of Eq. (3-75) is normalized. Hence, the average value 
for the potential energy is simply the sum of all the potential energies due to all the 
elements dx, each weighted by the probability for finding the particle there: 
V¯ = 8 
-8[prob. to be in dx][V at dx]dx = +8 
-8 
.2
0 (x)V (x) dx (3-78) 
This is
V¯ =(ß/p)1/2 · 
1
2
k  +8 
-8 
exp -ßx2 x2 dx =ß/p · 
1
2
k · 
1
2
p/ß3 (3-79) 
where we have referred to Appendix 1 to evaluate the integral. Using the definition of 
ß2 (Eq. 3-17) we have 
V¯ =k/4ß =(k/4) · h/(2pvmk)=(h/8p)k/m (3-80) 
which is just one half of the total energy [Eq. (3-76)]. This means that the average 
value of the kinetic energy must also equal half of the total energy, since V¯ +T¯ =E. 
We thus arrive at the important result that the ratio of average potential and kinetic 
energies is the same in the classical harmonic oscillator and the ground state of the 
quantum-mechanical system. This result is also true for the higher states. For other 
kinds of potential, the storage need not be half and half, but whatever it is, it will be the 
same for the classical and quantum-mechanical treatments of the system. We discuss 
this point in more detail later when we examine the virial theorem (Chapter 11 and 
Appendix 8). 
3-6 Vibrations of Diatomic Molecules 
Two atoms bonded together vibrate back and forth along the internuclear axis. The 
standard first approximation is to treat the system as two nuclear masses, m1 and m2 
oscillating harmonically with respect to the center of mass. The force constant k is 
determined by the “tightness” of the bond, with stronger bonds usually having larger k. 
The two-mass problem can be transformed to motion of one reduced mass, µ, vibrating 
harmonically with respect to the center of mass. µ is equal to m1m2/(m1 +m2). 
The force constant for the vibration of the reduced mass remains identical to the force 
constant for the two masses, and the distance of the reduced mass from the center 
of mass remains identical to the distance between m1 and m2. Thus we have a very 
convenient simplification: We can use the harmonic oscillator solutions for a single 
oscillating mass µ as solutions for the two-mass problem. All of the wavefunctions and 
energy formulas are just what we have already seen except thatm is replaced by µ. The 
practical consequence of this is that we can use the spectroscopically measured energy 
spacings between molecular vibrational levels to obtain the value of k for a molecule.
Section 3-7 Summary 85 
EXAMPLE 3-4 There is a strong absorption in the infrared spectrum of H35Cl at 
2992cm-1, which corresponds to an energy of 5.941×10-20 J. This light energy, 
E, is absorbed in order to excite HCl from the n=0 to the n=1 vibrational state. 
What is the value of k, the force constant, in HCl? 
SOLUTION 
 The vibrational spacing h. must be equal to 5.941 × 10-20 J. We know that, 
since µ replaces m, . = (1/2p)vk/µ, which means that k = 4p2E2µ/h2. The formula for µ 
is mHmCl/(mH +mCl)=1.614×10-27 kg. It follows that k is equal to 4p2(5.941×10-20J)2 
(1.614×10-27kg)/(6.626×10-34 J s)2 =512N m-1.  
3-7 Summary 
In this chapter we have discussed the following points: 
1. The energies for the quantum-mechanical harmonic oscillator are given by the formula 
En = (n + 1/2)h., n = 0, 1, 2, . . . , where . = (1/2p)vk/m. This gives 
nondegenerate energy levels separated by equal intervals (h.) and a zero-point 
energy of h./2. 
2. The wavefunctions for this system are symmetric or antisymmetric for reflection 
through x =0. This symmetry alternates as n increases and is related to the presence 
of even or odd powers of y in the Hermite polynomial in .. 
3. Each wavefunction is orthogonal to all of the others, even in cases where the symmetries 
are the same. 
4. The harmonic oscillator wavefunctions differ from particle-in-a-box wavefunctions 
in two important ways: They penetrate past the classical turning points (i.e., past 
the values of x where E =V ), and they have larger distances available to them as a 
result of the opening out of the parabolic potential function at higher energies. This 
gives them more room in which to accomplish their increasing number of wiggles 
as n increases, and so the energy does not rise as quickly as it otherwise would. 
5. The manner in which the total energy is partitioned into average potential and kinetic 
parts is the same for classical and quantum-mechanical harmonic oscillators, namely, 
half and half. 
6. Vibrations in molecules are usually approximately harmonic. Mass is replaced by 
reduced mass in the energy formula. Measuring the energy needed to excite a 
molecular vibration allows one to calculate the harmonic force constant for that 
particular vibrational mode. 
3-7.A Problems 
3-1. From the equation of motion (3-5) show that the classical distribution function is 
proportional to (1-x2/L2)-1/2.
86 Chapter 3 The One-Dimensional Harmonic Oscillator 
3-2. A classical harmonic oscillator with mass of 1.00 kg and operating with a force 
constant of 2.00 kg s-2 = 2.00 J m-2 is released from rest at t = 0 and x = 0.100 m. 
a) What is the function x(t) describing the trajectory of the oscillator? 
b) Where is the oscillating mass when t =3 seconds? 
c) What is the total energy of the oscillator? 
d) What is the potential energy when t =3 seconds? 
e) What is the time-averaged potential energy? 
f) What is the time-averaged kinetic energy? 
g) How fast is the oscillator moving when t =3 seconds? 
h) Where are the turning points for the oscillator? 
i) What is the frequency of the oscillator? 
3-3. Find the expression for the classical turning points for a one-dimensional harmonic 
oscillator in terms of n, m, h, and k. 
3-4. a) Equation (3-73) for .n(y) is a rather formidable expression. It can be broken 
down into three portions, each with a certain purpose. Identify the three parts 
and state the role that each plays in meeting the mathematical requirements 
on .. 
b) Produce expressions for the normalized harmonic oscillator with n=0, 1, 2. 
3-5. Operate explicitly on .0 with H and show that .0 is an eigenfunction having 
eigenvalue h./2. 
3-6. a) At what values of y does .2 have a node? 
b) At what values of y does .2
2 have its maximum value? 
3-7. Give a simple reason why (2 + y - 3y2)exp(-y2/2) cannot be a satisfactory 
wavefunction for the harmonic oscillator. What about 2y exp(+y2/2)? You 
should be able to answer by inspection, without calculation and without reference 
to tabulations. 
3-8. Consider the function (32x5 -160x3 +120x) exp(-x2/2). 
a) How does this function behave at large values of ±x? Explain your answer. 
b) What can you say about the symmetry of this function? 
c) What are the value and slope of this function at x =0? 
3-9. Let f (x)=3 cos x +4. f (x) is expressed as a power series in x: f (x)=cnxn, 
with n=0, 1, 2, . . . . 
a) What is the value of c0? 
b) What is the value of c1? 
3-10. Only one of the following is H5(y), a Hermite polynomial. Which ones are not, 
and why? 
a) 16y5 +130y 
b) 24y5 -110y3 +90y -18 
c) 32y5 -160y3 +120y
Section 3-7 Summary 87 
3-11. Sketch the function (1-2x2) exp(-x2) versus x. 
3-12. Given that  8 0 x exp(-x2)dx =1/2, and 8 0 x2 exp(-x2)dx =vp/4, evaluate 
a)  8 
-8 x exp(-x2)dx 
b) 8 
-8 x2 exp(-x2)dx 
3-13. Evaluate  8 
-8(x +4x3) exp(-5x2)dx. 
3-14. For the n=1 state of the harmonic oscillator: 
a) Calculate the values of the classical turning points. 
b) Calculate the values of the positions of maximum probability density. 
c) What is the probability for finding the oscillator between x =0 and x=8? 
d) Estimate the probability for finding the oscillator in a line increment x equal 
to 1% of the distance between classical turning points and centered on one of 
the positions of maximum probability density. 
3-15. Calculate the probability for finding the ground state harmonic oscillator beyond 
its classical turning points. 
3-16. Use the differential expression (3-63) for Hermite polynomials to produceH2(y). 
3-17. Use the generating function (3-65) to produce H2(y). 
3-18. Write down the Schr¨odinger equation for a three-dimensional (isotropic) harmonic 
oscillator. Separate variables. What will be the zero-point energy for 
this system? What will be the degeneracy of the energy level having a value of 
(9/2)h.? (5/2)h.? Sketch (roughly) each of the solutions for the latter case 
and note their similarity to case (b) in Fig. 2-16. 
3-19. Suppose V (x)=(1/2)kx2 for x =0, and8for x <0. What can you say about 
the eigenfunctions and eigenvalues for this system? 
3-20. a) Evaluate H3(x) at x =2. 
b) Evaluate H2(sin .) at . =30.. 
3-21. What is the average potential energy for a harmonic oscillator when n=5? What 
is the average kinetic energy? 
3-22. Each degree of translational or rotational freedom can contribute up to 1
2R to the 
molar heat capacity of an ideal diatomic gas, whereas the vibrational degree of 
freedom can contribute up to R. Explain. 
3-23. Calculate the force constants for vibration in H19F, H35Cl, H81Br, and H127I, 
given that the infrared absorptions for the n = 0 to n = 1 transitions are seen, 
respectively, at 4138, 2991, 2649, and 2308 cm-1. What do these force constants 
imply about the relative bond strengths in these molecules?
88 Chapter 3 The One-Dimensional Harmonic Oscillator 
Multiple Choice Questions 
(Try to answer these without referring to the text.) 
1. Which one of the following statements conflicts with the quantum mechanical results 
for a one-dimensional harmonic oscillator? 
a) The smaller the mass of the oscillating particle, the greater will be its zero-point 
energy, for a fixed force constant. 
b) The frequency is the same as that of a classical oscillator with the same mass and 
force constant. 
c) Increasing the force constant increases the spacing between adjacent energy 
levels. 
d) The spacing between adjacent energy levels is unaffected as the vibrational quantum 
number increases. 
e) The vibrational potential energy is a constant of motion. 
2. A quantum-mechanical harmonic oscillator 
a) spends most of its time near its classical turning points in its lowest-energy state. 
b) has . =0 at its classical turning points. 
c) has doubly degenerate energy levels. 
d) has energy levels that increase with the square of the quantum number. 
e) None of the above is a correct statement. 
3. Light ofwavelength 4.33×10-6 mexcites a quantum-mechanical harmonic oscillator 
from its ground to its first excited state. Which one of the following wavelengths 
would accomplish this same transition if i) the force constant only was doubled? ii) 
the mass only was doubled? 
a) 4.33×10-6 m 
b) 2.16×10-6 m 
c) 3.06×10-6 m 
d) 6.12×10-6 m 
e) 8.66×10-6 m 
4. For a classical harmonic oscillator, the probability for finding the oscillator in the 
middle 2% of the oscillation range is 
a) greater than 0.02. 
b) equal to 0.02. 
c) less than 0.02. 
d) unknown since it depends on the force constant. 
e) unknown since it depends on the amplitude.
Chapter 4 
The Hydrogenlike Ion, Angular 
Momentum, and the Rigid Rotor 
4-1 The Schr ¨ odinger Equation and the Nature 
of Its Solutions 
4-1.A The Schr ¨ odinger Equation 
Consider the two-particle system composed of an electron (charge -e) and a nucleus 
having atomic number Z and charge Ze. (See Appendix 12 for values of physical 
constants, such as e.) Let x1, y1, z1 be the coordinates of the nucleus and x2, y2, z2 
be those for the electron. The distance between the particles is, then, [(x1 - x2)2 + 
(y1 -y2)2 +(z1 -z2)2]1/2. The potential energy is given by the product of the charges 
divided by the distance between them. If e is expressed in coulombs, C, the potential 
energy in joules is 
V = -Ze2 
4p e0 
(x1 -x2)2 +(y1 -y2)2 +(z1 -z2)2
1/2 (4-1) 
where e0 is the vacuum permittivity (8.8542 × 10-12 J-1 C2 m-1). The timeindependent 
Schr¨odinger equation for this system is 
 -h2 
8p2M 
 .2 
.x2
1 + 
.2 
.y2
1 + 
.2 
.z2
1 

- 
h2 
8p2me 
 .2 
.x2
2 + 
.2 
.y2
2 + 
.2 
.z2
2 
 
- 
Ze2 
4pe0 
(x1 -x2)2 +(y1 -y2)2 +(z1 -z2)2
1/2 

.(x1, y1, z1, x2, y2, z2) 
=E.(x1, y1, z1, x2, y2, z2) (4-2) 
where M and me are the masses of the nucleus and electron, respectively. The 
hamiltonian operator in brackets in Eq. (4-2) has three terms, corresponding to a 
kinetic energy operator for the nucleus, a kinetic energy operator for the electron, and 
a potential term for the pair of particles. 
Equation (4-2) has eigenfunctions . that are dependent on the positions of both the 
electron and the nucleus. It is possible to convert to center-of-mass coordinates and 
then to separate Eq. (4-2) into two equations, one for the motion of the center of mass 
and one for a particle of reduced mass moving around a fixed center to which it is 
attracted in exactly the same way the electron is attracted to the nucleus. Because this 
89
90 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
conversion is rather tedious, we will not perform it in this book,1 but merely discuss the 
results. The first of the two resulting equations treats the center of mass as a free particle 
moving through field-free space; its eigenvalues are simply translational energies of 
the ion. For us, the interesting equation is the second one, which is 
 -h2 
8p2µ 
 .2 
.x2 + 
.2 
.y2 + 
.2 
.z2 
	- 
Ze2 
4pe0 
x2 +y2 +z2
1/2

.(x, y, z) 
=E.(x, y, z) (4-3) 
The quantity µ is the reduced mass for the particle in our center-of-mass system, and 
is given by 
µ=meM/(me +M) (4-4) 
The coordinates x, y, and z are the coordinates of the reduced-mass particle with respect 
to the center of mass of the system. 
Even without going through the detailed procedure of converting to center-of-mass 
coordinates, we can show that Eq. (4-3) makes sense. In the idealized case in whichM 
is infinitely greater than me, µ equals me, and Eq. (4-3) becomes just the Schr¨odinger 
equation for the motion of an electron about a fixed nucleus at the coordinate origin. 
For real atoms or ions this would not be a bad approximation because, even in the 
case of the lightest nucleus (i.e., the hydrogen atom), M is nearly 2000 times me, and 
so µ is very close to me, and the center of mass is very near the nucleus. Therefore, 
the result of using center-of-mass coordinates to separate the Schr¨odinger equation is 
almost identical to making the assumption at the outset that the nucleus is fixed, and 
simply writing down the one-particle Schr¨odinger equation: 
 -h2 
8p2me 
 .2 
.x2 + 
.2 
.y2 + 
.2 
.z2 
	- 
Ze2 
4pe0 
x2 +y2 +z2
1/2 

.(x, y, z)=E.(x, y, z) (4-5) 
The use of me instead of µ [i.e., Eq. (4-5) instead of (4-3)] has no effect on the qualitative 
nature of the solutions. However, it does produce small errors in eigenvalues— 
errors that are significant in the very precise measurements and calculations of atomic 
spectroscopy (Problem 4-1). In what follows we shall use µ, but for purposes of 
discussion we will pretend that the nucleus and center of mass coincide. 
Equation (4-3) can be transformed into spherical polar coordinates. (Some important 
relationships between spherical polar and Cartesian coordinates are given in Fig. 4-1). 
The result is 
[(-h2/8p2µ)
2 -(Ze2/4pe0r)].(r, .,f)=E.(r, .,f) (4-6) 
Here 
2 is understood to be in spherical polar coordinates. In these coordinates it 
looks quite complicated:2 

2 = 
1 
r2 
. 
.r 
r2 . 
.r 
	+ 
1 
r2 sin . 
. 
.. 
sin . 
. 
.. 
	+ 
1 
r2 sin2 . 
.2 
.f2 (4-7) 
1See, for example, Eyring et al. [1, Chapter VI] or Levine [2, pp. 127–130]. 
2See, e.g., Eyring et al. [1, Appendix III].
Section 4-1 The Schr ¨ odinger Equation and the Nature of Its Solutions 91 
Figure 4-1 
 The spherical polar coordinate system. The angle f is called the azimuthal angle. 
Notice that the differential volume element is not equal to dr d. df and that the ranges of values for 
r, ., f are not -8 to +8. 
However, this coordinate system is the natural one for this system and leads to the 
easiest solution despite this rather formidable looking operator. 
Notice that the potential term,-Ze2/4pe0r, has no . or f dependence. The potential 
is spherically symmetric. However, . and f dependence does enter the hamiltonian 
through 
2, so the eigenfunctions . may be expected to show angular dependence. 
Next we describe the solutions of the Schr¨odinger equation (4-6), relegating to later 
sections the mathematical details of how the solutions are obtained. 
EXAMPLE 4-1 Using the spherical polar coordinate system of Fig. 4-1, calculate 
the volume occupied by the skin of a spherical shell, where the inside radius of the 
skin is 100.0mm and the thickness of the skin is 1.000 mm. 
SOLUTION 
 One way to solve this problem is to calculate the volume inside the entire sphere, 
including the skin, and then to subtract the volume of the sphere occupying the space inside the 
skin. The formula for the volume of a sphere of radius r can be calculated from dv by integrating 
r from 0 to r, . from 0 to p, and f from 0 to 2p: 
V = 
 r 
0 
r2dr 
 p 
0 
sin .d. 
 2p 
0 
df = 
r3 
3 |r
0 ·-cos .|p0 
· f|2p 
0 
= 
r3 
3 
(-(-1-1)) · 2p = 
4
3
pr3 
(You presumably already knew this formula, but it is useful to review how it comes out of this 
integration.) Proceeding, 
Vskin = V (r =101 mm)-V (r =100 mm) 
= 
4
3
p[(101 mm)3 -(100 mm)3]=1.269×105mm3 
Another way (slightly less exact) to approximate this volume is to calculate the area of the spherical 
shell (4pr2) and multiply by its thickness: 
V ~4pr2r = 4p(100 mm)2 · 1.00 mm 
= 1.257×105 mm3 
For a skin whose thickness is small compared to its radius, we see that this approximation is 
very good. 
92 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
4-1.B The Nature of the Eigenvalues 
The potential energy, -Ze2/4pe0r, becomes negatively infinite when r = 0 and 
approaches zero as r becomes very large. This potential is sketched in Fig. 4-2 for the 
case in which Z =1. We expect the energy levels to diverge less rapidly here than was 
the case for the harmonic oscillator since the “effective box size” increases more rapidly 
with increasing energy in this case than in the harmonic oscillator case. (See Fig. 2-3 
for the one-dimensional analogs.) Since the harmonic oscillator levels are separated by 
a constant (h., for one- or three-dimensional cases), the hydrogenlike ion levels should 
converge at higher energies. Figure 4-2 shows that this is indeed the case. Furthermore, 
by analogy with the case of the particle in a box with one finite wall, we expect the 
allowed energies to form a discrete set for the classically trapped electron (E <0) 
and a continuum for the unbound cases (E >0). Thus, the spectrum of eigenvalues 
sketched in Fig. 4-2 is in qualitative accord with understandings developed earlier. 
The lowest allowed energy for the system is far above the low-energy limit (-8) of 
the potential well. This corresponds to the finite zero point energy which we have seen 
Figure 4-2 
 The potential function V = -e2/4pe0r with eigenvalues superimposed (dashed 
lines). Degeneracies for the first few levels are noted on the right.
Section 4-1 The Schr ¨ odinger Equation and the Nature of Its Solutions 93 
in other systems where particle motion is constrained. Here it means that, at absolute 
zero, the electron does not come to rest on the nucleus (which would give T = 0, 
V =-8, E=-8), but rather continues to move about with a finite total energy. 
All of the energy levels of Fig. 4-2 are degenerate except for the lowest one. The 
order of the degeneracy is listed next to each of the lowest few levels in Fig. 4-2. 
This degeneracy is not surprising since we are dealing here with a three- dimensional 
system, and we have earlier seen that, in such cases (e.g., the cubic box), the physical 
equivalence of different directions in space can produce degeneracies (called “spatial 
degeneracies”). We shall see later that some of the degeneracies in this system do 
indeed result from directional equivalence (here, spherical symmetry), whereas others 
do not. 
The discrete, negative eigenvalues are given by the formula 
En = -µZ2e4 
8e2
0h2n2 =(-13.6058eV)
Z2 
n2 , n=1, 2, 3, . . . (4-8) 
EXAMPLE 4-2 Calculate the ionization energy (IE) of C5+ in its ground state, in 
electron volts. 
SOLUTION 
 The ionization energy equals the negative of the ground state energy. Z =6 and 
n=1, so IE =(13.6058eV) 36 
1 =489.808eV.  
4-1.C The Lowest-EnergyWavefunction 
We will now discuss the lowest-energy eigenfunction of Eq. (4-6) in some detail, since 
an understanding of atomic wavefunctions is crucial in quantum chemistry. The derivation 
of formulas for this and other wavefunctions will be discussed in later sections, 
but it is not necessary to labor through the mathematical details of the exact solution of 
Eq. (4-6) to be able to understand most of the essential features of the eigenfunctions. 
The formula for the normalized, lowest-energy solution of Eq. (4-6) is 
.(r)=(1/vp)(Z/a0)3/2 exp(-Zr/a0) (4-9) 
where a0=5.2917706×10-11 m(called the Bohr radius) and Ze is the nuclear charge. 
A sketch of . versus r for Z=1 is superimposed on the potential function in Fig. 4-3a. 
It is apparent that the electron penetrates the potential barrier (Problem 4-3). 
The square of the wavefunction (4-9) tells us how the electron is distributed about 
the nucleus. In Fig. 4-3b is plotted .2(r) as a function of r. We refer to .2 as the 
electron probability density function. In this case, the probability density is greatest at 
the nucleus (r =0) and decays to zero as r becomes infinite. 
It is important for the chemist to be able to visualize the electron distributions, or 
charge “clouds,” in atoms and molecules, and various methods of depicting electron 
distributions have been devised. In Fig. 4-3 a few of these are presented for the lowestenergy 
wavefunction. The dot picture (Fig. 4-3c) represents what one would expect 
if one took a multiflash photograph of a magnified, slowed-down hydrogenlike ion 
(assuming no disturbance of the ion by the photographing process). Each dot represents 
an instantaneous electron position, and the density of these dots is greatest at the nucleus.
94 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
0 
0 
0 
Cusp 
Penetrates 
barrier 
Classical 
turning point 
0 
0 
Z = 1 
r Z = 2 
r 
0 r 
. 
.H 
v 
(a) (b) (c) 
(d) (e) (f) 
2 
.He+ 
2 
Figure 4-3 
 (a) H-atom wavefunction superposed on -e2/r potential curve. (b) Wavefunction 
squared for H and He+. (c) Dot picture of electron distribution. (d) Contour diagram of electron 
distribution. (e) Computer-generated graphic version of (d). (f) Single-contour representation of 
electron distribution. 
An alternative way of picturing the charge is to draw a contour diagram, each contour 
indicating that the density has increased or decreased by a certain amount (Fig. 4-3d). 
A more striking version of the contour plot for .2
1s is shown in Fig. 4-3e. Perhaps the 
simplest representation is a sketch of the single contour that encloses a certain amount 
(say 90%) of the electronic charge (Fig. 4-3f). (We have been describing the electron 
as a point charge moving rapidly about the nucleus. However, for most purposes it is 
just as convenient to picture the electron as being smeared out into a charge cloud like 
those sketched in Fig. 4-3. Thus, the statements the electron spends 90% of its time 
inside this surface, and 90% of the electronic charge is contained within this surface, 
are equivalent.) 
The multiflash “photograph” sketched in Fig. 4-3c shows the electron probability 
density to be greatest at the nucleus. Suppose that we were just to take a single flash 
photograph. Then the electron would appear as a single dot. At what distance from the 
nucleus would this dot be most likely to occur? Surprisingly, the answer is not zero. 
Despite the fact that the probability density is a maximum at r =0, the probability for 
finding the electron in a volume element at the nucleus approaches zero. This is because 
the probability density, .2(r), is the measure of the probability per unit volume for the 
electron being at various distances from the nucleus. When we compare a tiny volume
Section 4-1 The Schr ¨ odinger Equation and the Nature of Its Solutions 95 
Figure 4-4 
 The volume-weighted probability density for the lowest-energy eigenfunction of the 
hydrogenlike ion. The most probable value of r occurs at rmp. 
element near the nucleus with an identical one farther out, we see from Fig. 4-3 that 
there is indeed more likelihood for the electron being in the volume element nearer the 
nucleus. But there are more volume elements associated with the larger distance. (The 
number of identical volume elements varies as the area of the surface of the sphere, 
4pr2.) Hence, the probability for the electron being in a radial element dr at a given 
distance r from the nucleus is given by the number of volume elements at r times the 
probability density per unit volume element. The reason for the near-zero probability 
for finding the electron in a volume element at the nucleus is that the number of volume 
elements associated with r =0 is vanishingly small compared to the number associated 
with larger r values. Figure 4-4 is a graph of 4pr2.2, the volume-weighted probability 
density. It is apparent that the most probable value of r, rmp, occurs at a nonzero 
distance from the nucleus. [The reader is familiar with analogous distinctions. Rhode 
Island has a higher population density than Texas, but the population of Texas (density 
times area) is greater. Again, matter in the universe has a much higher mass density 
in stars and planets than in intergalactic gas or dust, but the total mass of the latter far 
exceeds that of the former due to the much greater volume of “empty” space.] 
EXAMPLE 4-3 Estimate, for the hydrogen atom in the 1s state, the amount of 
electronic charge located in a spherical shell that is 1.00 pm thick and which has a 
radius of 60.0 pm. 
SOLUTION 
 Recall from Example 4-1 that, for relatively thin shells like this, we can 
estimate the volume of the shell by taking its area times its thickness. Also, we expect 
the charge density, .2, to change very little over the 1.00 pm thickness of the shell, so 
we can take its value at 60.0 pm as constant. Then charge density = .2(r = 60.0pm) = 
(pa3
0)-1 exp(-2(60.0pm)/a0). Recalling that a0 = 52.9 pm, this gives 0.0329a-3 
0 . This gives 
density per cubic bohr. To proceed, we can either convert this to density per cubic picometer by 
dividing by (52.9pm/bohr)3 or by converting r to bohr, by dividing by 52.9pm/bohr.We choose the
96 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
latter. Then r =60.0pm/52.9pm/bohr =1.13a0, and r = 1.00pm/52.9pm/bohr = 0.0189a0, 
so volume of shell = 4pr2r = 4p(1.13a0)2(0.0189a0) = 0.3028a3
0. Volume times density 
=(0.3028a3
0)(0.0329a-3 
0 )=0.010. So 1% of the electronic charge resides in this shell.  
We can calculate the value of rmp by finding which r value gives the maximum value 
of 4pr2.2. Recall that we can do this by finding the value of r that causes the first 
derivative of 4pr2.2 to vanish, that is, we require 
(d/dr)[4pr2p-1(Z/a0)3 exp(-2Zr/a0)]=0 (4-10) 
This gives 
constants · [2r -(2Zr2/a0)]exp(-2Zr/a0)=0 (4-11) 
The term in brackets vanishes when r = a0/Z, so this is the value of rmp. For 
Z =1, rmp =a0; a0 is the most probable distance of the electron from the nucleus in 
the hydrogen atom. For the He+ ion (Z =2), the most probable distance is only half 
as great, consistent with a more contracted charge cloud. 
Of more interest, often, is the average value of the distance of the electron from 
the nucleus. If we could sample the instantaneous distance of the electron from the 
nucleus a large number of times and calculate the average value, what sort of result 
would we obtain? The probability for finding the electron at any given distance r is 
given by the volume-weighted probability density of Fig. 4-4. Inspection of that figure 
suggests that the average value of the position of the electron ¯r is greater than rmp, the 
most probable value. But exactly how much bigger is ¯r than rmp? How should we 
compute the average value? The reader is familiar with the way an average test score 
is calculated from a collection of scores. For example, suppose the series of scores to 
be averaged is 0, 2, 6, 6, 7, 7, 7, 10, and that 0 and 10 are the minimum and maximum 
possible scores. The average is given by 
average = 
sum of scores 
number of scores 
= 
0+2+6+6+7+7+7+10 
8 = 
45 
8 
(4-12) 
Another way to write this is 
average = 
frequency of score×score 
sum of frequencies 
= 
1 · 0+0 · 1+1 · 2+0 · 3+0 · 4+0 · 5+2 · 6+3 · 7+0 · 8+0 · 9+1 · 10 
1+0+1+0+0+0+2+3+0+0+1 
or 
average= 10 
i=0 (frequency of i) · i 
10 
i=0 (frequency of i) 
(4-13) 
The same idea is used to compute a quantum-mechanical average. For the average value 
of r we take each possible value of r times its frequency (given by .2 dv) and sum
Section 4-1 The Schr ¨ odinger Equation and the Nature of Its Solutions 97 
over all these values.3 For a continuous variable like r, we must resort to integration to 
accomplish this. We divide by the “sum” of frequencies by dividing by  .2 dv. Thus 
¯r =  all space 
r.2dv 
 all space 
.2dv =  2p 
0 df  p 
0 sin. d. 8 0 r.2r2dr 
 all space 
.2dv 
(4-14) 
The denominator is unity since . is normalized. The integrals over . and f involve 
parts of the volume element dv, and not .2, because this wavefunction (4-9) does not 
depend on . or f. Continuing, 
¯r = f|2p 
0 ·-cos.|p0 
· p-1 (Z/a0)3 
 8 
0 
r3 exp (-2Zr/a0) dr (4-15) 
Utilizing the information in Appendix 1 for the integral over r, this becomes 
¯r = 
2p[-(-1)+1](1/p)(Z/a0)33! 
(2Z/a0)4 (4-16) 
= 4p · (1/p)(Z/a0)3 · 6a4
0/16Z4 = 
3a0 
2Z 
(4-17) 
(It is useful to remember that integration over the f and . parts of dv gives 4p as the 
result if no other angle-dependent functions occur in the integral.) Comparing (4-17) 
with our expression for rmp indicates that ¯r is 1.5 times greater than rmp. 
Notice that the lowest-energy eigenfunction is finite at r =0 even though V is infinite 
there. This is allowed by our arguments in Chapter 2 because the infinity in V occurs 
at only one point, so it can be cancelled by a discontinuity in the derivative of .. This 
is possible only if . has a corner or cusp at r =0 (see Fig. 4-3a and e). 
4-1.D Quantum Numbers and Nomenclature 
There are three quantum numbers, n, l, andm(all integers), characterizing each solution 
of the Schr¨odinger equation (4-6). Of these, only n enters the energy formula (4-8), 
so all solutions having the same values of n but different values of l and m will be 
energetically degenerate. As is shown in following sections, these quantum numbers 
are related in their allowed values. The l quantum number must be nonnegative and 
smaller than n. The m number may be positive, negative or zero, but its absolute value 
cannot exceed l. Thus, 
l =0, 1, 2, . . . ,n-1 (4-18) 
|m|=l (4-19) 
For the lowest-energy wavefunction we have already described, n=1, so l =m=0. 
No other choices are possible, so this level is nondegenerate. The convention (from 
atomic spectroscopy) is to refer to an l =0 solution as an “s function,” or “s orbital.” 
(For l =0, 1, 2, 3, 4, 5, the spectroscopic designation goes s, p, d, f, g, h.) Because n 
equals unity, the wavefunction is labeled 1s. 
3The 4pr2 part of 4pr2.2 in Fig. 4-4 is included in dv, as will be seen shortly.
98 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
When n=2, there are four sets of l and m quantum numbers satisfying rules (4-18) 
and (4-19). They are listed below with their spectroscopic labels: 
l =0,m=0 2s l =1,m=0 2p0 
l =1,m=-1 2p-
1 l =1,m=+1 2p+
1 
(4-20) 
Extending these rules to the n=3 energy level produces nine functions designated 
3s, 3p-1, 3p0, 3p+1, 3d-2, 3d-1, 3d0, 3d+1, 3d+2. In general, the degeneracy of the 
energy level characterized by n is n2. 
EXAMPLE 4-4 Explain how it comes about from the quantum number rules that 
the degeneracy equals n2. 
SOLUTION 
 The rules indicate that, for each value of n, there are n values of l (e.g., 
n = 1, l = 0;n = 2, l = 0 and 1, . . . ), and each value of l is associated with 2l +1 ml values (e.g., 
l = 2, ml =-2,-1, 0, 1, 2, which is five values). Note that 2l + 1 must be an odd number of 
member states, and the odd number within each set keeps increasing as l increases. Thus, for 
n=4, l =0, 1, 2, 3, leading to degenerate sets of states, respectively having 1, 3, 5, and 7 members. 
But a sequence of n odd numbers, starting from 1, is always equal to n2. QED  
4-1.E Nature of the Higher-Energy Solutions 
The second energy level is associated with the 2s, 2p-1, 2p0 and 2p+1 orbitals. The 
wavefunction for the 2s state is 
.2s = 
1 
4v2p 
 Z
a0
	
3/2 
2- 
Zr 
a0 
	exp-Zr 
2a0 
	 (4-21) 
Since this is a function of r only, it is a spherically symmetric function. (In fact, all 
s orbitals are spherically symmetric.) The 2s orbital is more “spread out” than the 1s 
orbital because the exponential in .2s decays more slowly and because the exponential 
is multiplied by Zr/a0 (the 2 becomes negligible compared to Zr/a0 at large r). As a 
result, the charge cloud associated with the 2s orbital is more diffuse. (For this reason, 
when we approximate a polyelectronic atom like beryllium by putting electrons in 
1s and 2s orbitals the 2s electrons are referred to as “outer” and the 1s electrons are 
called “inner.”) 
At small values of r, (2-Zr/a0) is positive, and at large distances it is negative, 
so .2s has a spherical radial node (a zero in the r coordinate). In Fig. 4-5 the first 
three s orbitals are plotted. We see that the nth s orbital has (n-1) spherical nodal 
surfaces dividing regions where the wavefunctions have different sign. The appearance 
of more and more nodes in the radial coordinate as the energy increases is certainly 
familiar from previous examples. Notice how the wavefunctions oscillate most rapidly 
and nodes are most closely spaced in the regions near the nucleus where the electron 
classically would have its greatest kinetic energy. 
The 2s orbital is orthogonal to the 1s orbital, and also to all higher s orbitals. This 
would not be possible if there were no radial nodes. The product .1s.2s will vanish 
upon integration only if it either vanishes everywhere or else has positive and negative 
regions that cancel on integration. Since .1s and .2s are almost everywhere finite, the
Section 4-1 The Schr ¨ odinger Equation and the Nature of Its Solutions 99 
former condition does not occur. Since .1s has the same sign everywhere, their product 
can have positive and negative regions only if .2s has positive and negative regions, 
and hence, a node. 
Let us now consider the 2p functions. They are4 
.2p0 = 
1 
4v2p 
 Z
a0
	
3/2 
Zr 
a0 
exp-Zr 
2a0 
	cos . (4-22) 
.2p±1 = ± 
1 
8vp 
 Z
a0
	
3/2 
Zr 
a0 
exp-Zr 
2a0 
	sin . exp (±if) (4-23) 
All of these functions have the same radial exponential decay as the 2s orbital, so we 
can say that the 2s and 2p orbitals are about equal in size. However, since the 2p orbitals 
contain the factor Zr/a0 where the 2s contains (2-Zr/a0), the 2p orbitals vanish at 
the nucleus and not at any intermediate r value; they have no radial nodes. The 2p 
orbitals are endowed with directional properties by their angular dependences. The 
2p0 orbital is particularly easy to understand because the factors r cos . behave exactly 
like the z Cartesian coordinate. Hence, we can rewrite 2p0 (also called 2pz ) in mixed 
coordinates as 
.2pz = 
1 
4v2p 
 Z
a0
	
5/2 
z exp-Zr 
2a0 
	 (4-24) 
0 0
0
0
0 
r r 
0 r 
0 r 
r
r 
0
0 
. 
. 4pr 
2 
1s 1s 
.2s 
.3s 
2 
. 4pr 
2 
2s 
2 
. 4pr 
2 
3s 
2 
Figure 4-5 
 s wavefunctions versus r and volume-weighted electron densities versus r for the 
hydrogenlike ion. 
4The ± factor in .2p±1 results from a phase factor that is omitted from many textbooks. It has no effect on 
our discussion here, but is consistent with an implicit choice of sign for certain integrals appearing in Appendix 4. 
See, e.g., Zare [4, Chapter 1].
100 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
(Be careful not to confuse the atomic number Z with the coordinate z.) The exponential 
term in Eq. (4-24) is spherically symmetric, resembling a diffuse 1s orbital. The 
function z vanishes in the xy plane and becomes increasingly positive or negative as 
we move away from the plane in either direction. As a result, .2pz looks as sketched in 
Fig. 4-6. It has nearly spherical lobes, one with positive phase and one with negative 
phase. When .2pz is squared, the contour lines remain unchanged in relative position, 
but the function becomes everywhere positive in sign. 
The 2p±1 orbitals are more difficult to visualize since they are complex functions. 
The charge distributions associated with these orbitals must be real, however. These 
are given by .*., where .* is the complex conjugate of .. (Recall that, for complex 
wavefunctions, .*. must be used for probability distributions, rather than .2.) One 
obtains the complex conjugate of a function by merely reversing the signs of all the i’s 
in the function. It is evident from Eq. (4-23) that .*2p+1 =-.2p-1 and .*2p-1 =-.2p+1 . 
Hence .*2p+1
.2p+1 =.*2p-1
.2p-1 =-.2p+1.2p-1 : both the 2p+1 and the 2p-1 orbitals 
give the same charge distribution. This distribution is the same as that for the 2pz 
orbital except that the angle dependence is 1
2 sin2 . instead of cos2 .. However, since 
sin2 . +cos2 . =1, it follows that the sum of 2p+1 and 2p-1 charge clouds must be such 
Figure 4-6 
 (a) Drawing of the 2pz orbital. (b) Drawing of the square of the 2pz orbital. (c) Drawing 
of .2
2pz +.*2p-1
.2p-1 +.*2p+1
.2p+1 = spherically symmetric distribution. The curved lines in 
(c) are a visualization aid and are not mathematically significant.
Section 4-1 The Schr ¨ odinger Equation and the Nature of Its Solutions 101 
that, when added to the charge cloud for 2p0 a spherical charge cloud results (since the 
angular dependence is removed). We already know that the 2pz distribution is dumbbell 
shaped, so it follows that 2p+1 and 2p-1 produce doughnut-shaped distributions. 
(A sphere minus a dumbbell equals a doughnut. See Fig. 4-6.) The shape can also be 
inferred from the sin2 .-dependence, which is a maximum in the xy plane. 
When solving the particle-in-a-ring problem, we saw that we could arrive at either a 
set of real trigonometric solutions or a set of complex exponential solutions. Since the f 
dependence of the 2p+1 orbitals is identical to that for m=±1 solutions of the particle 
in the ring, the same situation holds here. Because .2p+1 and .2p-1 are energetically 
degenerate eigenfunctions, any linear combination of them is also an eigenfunction of 
the hamiltonian (Problem 2-11). Therefore, let us find linear combinations that are 
entirely real. The complex part of .2p±1 , exp(±if), satisfies the relation 
exp(±if)=cosf ±i sin f (4-25) 
so that 
exp(+if)+exp(-if)=2 cosf (4-26) 
and 
i-1[exp(+if)-exp(-if)]=2 sin f (4-27) 
Thus, we have two linear combinations of exp(±if) that are real. It follows immediately 
that 
.2px = 
1 
v2 
.2p-1 -.2p+1= 
1 
4v2p 
 Z
a0
	
3/2 
Zr 
a0 
exp-Zr 
2a0 
	sin . cosf (4-28) 
.2py = 
iv
2 
.2p-1 +.2p+1= 
1 
4v2p 
 Z
a0
	
3/2 
Zr 
a0 
exp-Zr 
2a0 
	sin . sin f (4-29) 
where the factor 2-1/2 is used to maintain normality. Since r sin . cos f and r sin . sin f 
are equivalent to the Cartesian coordinates x and y, respectively, Eqs. (4-28) and (4-29) 
are commonly referred to as the 2px and 2py orbitals. They are exactly like the 2pz 
orbital except that they are oriented along the x and y axes (merely replace the z in 
Eq. (4-24) with x or y). 
The 2s, 2px, 2py, and 2pz orbitals are all orthogonal to one another. This is easily 
shown from symmetry considerations. Each 2p orbital is antisymmetric for reflection 
in its nodal plane, whereas 2s is symmetric for all reflections. Hence, the product 
.2s.2p is always antisymmetric with respect to some reflection so its integral vanishes. 
The 2p functions are mutually orthogonal because, if one 2p orbital is antisymmetric 
for some reflection, the others are always symmetric for that reflection. Hence, the 
product is antisymmetric for that reflection. Another way to show that the 2p orbitals 
are mutually orthogonal is to note that they behave like x, y, and z vectors and that these
102 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
vectors are orthogonal (i.e., perpendicular; orthogonality in functions is equivalent to 
perpendicularity in vectors). Sometimes the orthogonality of functions is most clearly 
seen if we sketch out the product and note whether the positive and negative regions 
cancel by symmetry (Fig. 4-7). 
The n = 3 level has nine solutions associated with it. The 3s orbital, plotted in 
Fig. 4-5, has one more node than the 2s orbital and is more diffuse. The 3p orbitals have 
the same angular terms as did the 2p orbitals so they can be written as real functions 
having the same directional properties as x, y and z vectors. The 3p orbitals differ from 
the 2p orbitals in that they possess a radial node and are more diffuse (see Fig. 4-8). 
The remaining five levels, 3d levels, may also be written in either complex or real form. 
The real orbitals are given by the formulas 
3dz2 = 
3dx2-y2 = 
3dxy = 
3dxz = 
3dyz =
....... 
...... 
2 
v2592p 
 Z
a0
	
3/2 
2Zr 
3a0 
	
2 
exp-Zr 
3a0 
	
........ 
....... 
1/v3 
3 cos2 . -1 
sin2 . cos 2f 
sin2 . sin 2f 
sin 2. cosf 
sin 2. sinf (4-30) 
These angular factors, times r2, have directional properties identical to the Cartesian 
subscripts on the left, except that 3dz2 is a shorthand for 3d3z2-r2 . These orbitals are 
sketched in Fig. 4-8. It is obvious from these figures that 3dx2-y2 has the same symmetry 
and orientation as the sum of the two vectors x2 and -y2, and that the other 3d orbitals 
have a similar connection with the notation (except for 3dz2 ). The 3d functions are 
Figure 4-7 
 Drawings of orbitals and their products to demonstrate orthogonality.
Section 4-1 The Schr ¨ odinger Equation and the Nature of Its Solutions 103 
Figure 4-8 
 Some hydrogenlike orbitals at the n=3 level. 
about the same size as the 3p and 3s functions, but have no radial nodes at intermediate 
r values. 
A general pattern emerges when we examine the nodal properties of the orbitals at 
various energies. At the lowest energy we have no nodes and the level is nondegenerate. 
At the next level, we find that each function possesses a single node. There is oneway to 
put in a radial node and so we get one 2s orbital. Or we can put in a planar node. But we
104 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
have three choices for orthogonal orientations of this plane leading to three independent 
p orbitals. At the n=3 level we find orbitals containing two nodes. The possibilities 
are: two radial nodes (3s), a radial node and a planar node (3px, 3py, 3pz ), two planar 
nodes (3dxy, 3dxz, 3dyz, 3dx2-y2 , 3dz2-x2 , 3dz2-y2 ). (But, since z2-y2-(z2-x2)= 
x2 - y2, the last three 3d orbitals are not linearly independent. Hence the last two 
are combined to form 3dz2 : z2 - x2 + z2 - y2 = 3z2 - (x2 + y2 + z2) = 3z2 - r2. 
This function can be seen to correspond to a positive dumbbell encircled by a small 
negative doughnut, or “belly band.” The two nodes in this orbital arising from nodes in 
. are conical surfaces rather than planes.) It is apparent that the degeneracies between 
various 2p orbitals, or 3d orbitals, are spatial degeneracies, due only to the physical 
equivalence of various directions in space. The degeneracy between 2s and 2p, or 3s, 
3p, and 3d is not due to spatial symmetry. The fact that an angular node is energetically 
equivalent to a radial node is a peculiarity of the particular potential (-Ze2/4pe0r) for 
this problem. This degeneracy is removed for noncoulombic central-field potentials. 
The eigenfunctions corresponding to states in the energy continuum, like the bound 
states, can be separated into radial and angular parts. The radial parts of the spherically 
symmetric eigenfunctions at two nonnegative energies are given in Fig. 4-9. Note that 
the rate of oscillation of these functions is greatest at the nucleus, where the local kinetic 
energy is largest, in accord with the ideas presented in Chapters 1 and 2. Unbound-state 
Figure 4-9 
 Radial part of unbound H-atom states (times r) versus r at two energies: (a) E = 13.6eV; (b) E =0.
Section 4-2 Separation of Variables 105 
wavefunctions are not used in most quantum chemical applications, so we will not 
discuss them further in this book. 
EXAMPLE 4-5 Anodal plane is one through which awavefunction is antisymmetric 
for reflection. Consider the 3dxy and 3dyz orbitals of Fig. 4-8. Through which 
planes do these two orbitals have different reflection symmetries? 
SOLUTION 
 3dxy is antisymmetric for reflection only through the x, z and y, z planes. 3dyz 
is antisymmetric for reflection only through the x, z and x,y planes. Hence, these orbitals differ 
in their symmetries for reflection through the y, z and x,y planes.  
4-2 Separation of Variables 
We shall indicate in some detail the way in which the Schr¨odinger equation (4-6) is 
solved. Recall the strategy of separating variables which we used in Section 2-7: 
1. Express . as a product of functions, each depending on only one variable. 
2. Substitute this product into the Schr¨odinger equation and try to manipulate it so that 
the equation becomes a sum of terms, each depending on a single variable. These 
terms must sum to a constant. 
3. Since terms for different variables are independent of each other, the terms for each 
variable must equal a constant. This enables one to set up an equation in each 
variable. If this can be done, the initial assumption (1) is justified. 
In this case we begin by assuming that 
.(r, .,f)=R(r)	(.)
(f) (4-31) 
Substituting into Eq. (4-6) gives 
-h2 
8p2µr2 
	
 
d 
dr 
r2dR 
dr 
	+R
 
1 
sin . 
d 
d. 
sin . 
d	 
d. 
	+R	 
1 
sin2 . 
d2
 
df2 
 
- 
Ze2 
4pe0r
R	
=ER	
 (4-32) 
Since each derivative operator now acts on a function of a single coordinate, we use 
total, rather than partial, derivative notation. 
Let us first see if we can isolate the f dependence. Multiplying Eq. (4-32) by 
(-8pµr2 sin2 ./h2R	
) and rearranging gives 
sin2 . 
R 
d 
dr 
r2dR 
dr 
	+ 
8p2µr2 sin2 . 
h2 
E + 
Ze2 
4pe0r 
	 
+ 
sin . 
	 
d 
d. 
sin . 
d	 
d. 
	+ 
1

 
d2
 
df2 =0 (4-33) 
The r and . dependence is still mixed in the first two terms, but we now have a rather 
simple term in the coordinate f. Now we can argue, as in Section 2-7, that, as f alone
106 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
changes, the first three terms in Eq. (4-33) do not change. That is, if only f changes, 
Eq. (4-33) may be written 
constant+constant+constant+(1/
)(d2
/df2)=0 (4-34) 
so that 
(1/
)(d2
/df2)=-m2 (a constant) (4-35) 
We call the constant -m2 for future mathematical convenience. We can rearrange 
Eq. (4-35) into the more familiar form for an eigenvalue equation: 
d2
/df2=-m2
 (4-36) 
We arrived at Eq. (4-36) by assuming that only f changes while r and . are constant. 
However, it should be obvious that the behavior of the term in f is uninfluenced by 
changes in r and . since it has no dependence on these coordinates. Thus, by establishing 
that this term is constant under certain circumstances, we have actually shown 
that it must be constant under all circumstances, and we have produced an eigenvalue 
equation for 
. 
We can now proceed with further separation of variables. Since we know that the 
last term in Eq. (4-33) is a constant, we can write 
1
R 
d 
dr 
r2dR 
dr 
	+ 
8p2µr2 
h2 
E + 
Ze2 
4pe0r 
	+ 
1 
	sin . 
d 
d. 
sin . 
d	 
d. 
	- 
m2 
sin2 . =0 
(4-37) 
Note that we have separated the . and r dependences by dividing through by sin2 .. 
We now have two terms wholly dependent on r and two wholly dependent on ., their 
sum being zero. Hence, as before, the sum of the two r-dependent terms must equal a 
constant, ß, and the sum of the .-dependent terms must equal -ß. Thus 
d 
dr 
r2dR 
dr 
	+ 
8p2µr2 
h2 
E + 
Ze2 
4pe0r 
	R = ßR (4-38) 
1 
sin . 
d 
d. 
sin . 
d	 
d. 
	- 
m2	 
sin2 . = -ß	 (4-39) 
where we have multiplied through by R in the first equation and by 	 in the second. 
The assumption that . =R	
 has led to separate equations for R, 	, and 
. This 
indicates that the assumption of separability was valid. However, there is some linkage 
between R and 	 via ß, and between 	 and 
 via m. 
4-3 Solution of the R, 
, and  Equations 
4-3.A The  Equation 
The solution of Eq. (4-36) is similar to that of the particle in a ring problem of 
Section 2-6. The normalized solutions are 

=(1/v2p)exp(imf), m=0,±1,±2, . . . (4-40) 
As shown in Section 2-6, the constant m must be an integer if 
 is to be a single-valued 
function.
Section 4-3 Solution of the R, 
, and  Equations 107 
4-3.B The 
 Equation 
There is great similarity between the mathematical techniques used in solving the R and 
	 equations and those used to solve the one-dimensional harmonic oscillator problem 
of Chapter 3. Hence, we will only summarize the steps involved in these solutions and 
make a few remarks about the results. More detailed treatments are presented in many 
texts.5 
The 	 equation can be solved as follows: 
1. Change the variable to obtain a more convenient form for the differential equation. 
2. Express the solution as a power series and obtain a recursion relation. 
3. Observe that the series diverges for certain values of the variables, producing 
nonsquare-integrable wavefunctions. Correct this by requiring that the series terminate. 
This requires that the truncated series be either symmetric or antisymmetric 
in the variable and also that ß of Eq. (4-38) and (4-39) be equal to l(l +1) with l 
an integer. 
4. Recognize these truncated series as being the associated Legendre functions. 
5. Return to the original variable to obtain an expression for 	 in terms of the starting 
coordinate. 
Reference to the end of Section 3-4 will illustrate the similarity between this and the 
harmonic oscillator case. 
The final result for m0 is 
	l,m(.)=(-1)m (2l +1) 
2 
(l -|m|)! 
(l +|m|)!

1/2 
P|m| l (cos .) (4-41) 
For m<0 the phase factor (-1)m should be omitted.6 The term in square brackets 
is a normalizing function, and P|m| l (cos .) represents some member of the series of 
associated Legendre functions. When m = 0, these become the ordinary Legendre 
polynomials. The first few ordinary Legendre polynomials are 
P0(x) = 1, P1(x)=x, P2(x)= 
1
2
(3x2 -1) 
P3(x) = 
1
2
(5x3 -3x) (4-42) 
The first few associated Legendre functions are 
P1 
1 (x)=(1-x2)1/2, P1 
2 (x)=3(1-x2)1/2x, 
P2 
2 (x)=3(1-x2), P1 
3 (x)= 3
2 (1-x2)1/2(5x2 -1), 
P2 
3 (x)=15(1-x2)x, P3 
3 (x)=15(1-x2)3/2 
(4-43) 
It is also true that 
P|m| l (x)=0 if |m|>l (4-44) 
5See, e.g., Pauling andWilson [3, Chapter 5]. 
6This is the same phase factor that we saw earlier for .2p±1 .
108 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
Thus, 	(.), and hence .(r, .,f), vanishes unless |m|=l, giving us one of our quantum 
number rules [Eq. (4-19)]. 
The associated Legendre functions satisfy an orthogonality relation: 

 +1 
-1 
P|m| l (x)P |m| l (x) dx = 
2 
(2l +1) 
(l +|m|)! 
(l -|m|)!
dll (4-45) 
For a further discussion of these functions, the reader should consult a more advanced 
text on quantum mechanics. 
4-3.C The R Equation 
The R equation can be solved as follows: 
1. Assume thatE is negative (this restricts us to bound states), and note that ß =l(l +1) 
from the previous solving of the 	 equation. 
2. Change variables for mathematical convenience. 
3. Find the asymptotic solution pertaining to the large r limit, where the R equation 
becomes simplified. 
4. Express the wavefunction as a product of the asymptotic solution and an unknown 
function. Express this unknown function as a power series and (after dealing with 
some singularities) obtain a recursion relation. 
5. Note that the power series overpowers the asymptotic part of the solution unless the 
series is truncated. This leads to the requirement that n be an integer and hence that 
E be quantized. It also requires that n>l. 
6. Recognize the truncated series to be associated Laguerre polynomials times .l , 
where . is defined below. 
The resulting solution is, if µ=me, 
Rnl =-
 2Z 
na0
	
3 
(n-l -1)! 
2n [(n+l)!]3 

1/2 
exp (-./2) .lL
2l+1 
n+l (.) (4-46) 
where .=2Zr/na0 and a0=0h2/pmee2=5.2917706×10-11 m. The term in brackets 
is a normalizing function. The exponential term is the asymptotic solution and it 
guarantees that R(r) will approach zero as r approaches infinity. The third term, .l, is 
produced in the course of removing singularities (i.e., places where parts of a differential 
equation become infinite). The last term, L(.), symbolizes the various members
Section 4-4 Atomic Units 109 
of the set of associated Laguerre polynomials. Like the Legendre functions, these 
are mathematically well characterized. A few of the low-index associated Laguerre 
polynomials are 
L11
(.)=1, L12
(.)=2. -4, 
L13
(.)=-3.2 +18. -18, L33
(.)=-6 
(4-47) 
4-4 Atomic Units 
It is convenient to define a system of units that is more natural for working with atoms 
and molecules. The commonly accepted system of atomic units for some important 
quantities is summarized in Table 4-1. [Note: the symbol ¯h (“h-cross or h-bar”) is 
often used in place of h/2p.] Additional data on values of physical quantities, units, 
and conversion factors can be found in Appendix 10. 
In terms of these units, Schr¨odinger’s equation and its resulting eigenfunctions and 
eigenvalues for the hydrogenlike ion become much simpler to write down. Thus, the 
TABLE 4-1 
 Atomic Units 
Quantity Atomic unit in cgs or other units 
Values of some atomic 
properties in atomic units 
(a.u.) 
Mass me=9.109534×10-28 g Mass of electron = 1 a.u. 
Length a0=4pe0 ¯h2/mee2 Most probable distance of 1s 
=0.52917706×10-10m electron from nucleus ofH 
(=1 bohr) atom = 1 a.u 
Time t0=a0 ¯h/e2 Time for 1s electron in H 
=2.4189×10-17 s atom to travel one bohr 
=1 a.u. 
Charge e=4.803242×10-10 esu Charge of electron=-1 a.u. 
=1.6021892 
×10-19 coulomb 
Energy e2/4pe0a0=4.359814×10-18 J Total energy of 1s electron in 
(=27.21161eV=1 hartree) H atom =-1/2 a.u. 
Angular ¯h=h/2p Angular momentum for 
momentum =1.0545887×10-34 J s particle in ring = 0, 1, 
2, . . . a.u. 
Electric field 
strength 
e/a2
0=5.1423×109 V/cm Electric field strength at 
distance of 1 bohr from 
proton = 1 a.u.
110 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
TABLE 4-2 
 Eigenfunctions for the Hydrogenlike Ion in Atomic Units 
Spectroscopic symbol Formula 
1s (1/vp)Z3/2 exp(-Zr) 
2s (1/4v2p)Z3/2(2-Zr) exp(-Zr/2) 
2px (1/4v2p)Z5/2r exp(-Zr/2) sin . cosf 
2py (1/4v2p)Z5/2r exp(-Zr/2) sin . sin f 
2pz (1/4v2p)Z5/2r exp(-Zr/2) cos . 
3s (1/81v3p)Z3/2(27-18Zr +2Z2r2) exp(-Zr/3) 
3px (v2/81vp)Z5/2r(6-Zr) exp(-Zr/3) sin . cosf 
3py (v2/81vp)Z5/2r(6-Zr) exp(-Zr/3) sin . sin f 
3pz (v2/81vp)Z5/2r(6-Zr) exp(-Zr/3) cos . 
3dz2 (=3d3z2-r2) (1/81v6p)Z7/2r2 exp(-Zr/3)(3 cos2 . -1) 
3dx2-y2 (1/81v2p)Z7/2r2 exp(-Zr/3) sin2 . cos 2f 
3dxy (1/81v2p)Z7/2r2 exp(-Zr/3) sin2 . sin 2f 
3dxz (1/81v2p)Z7/2r2 exp(-Zr/3) sin 2. cosf 
3dyz (1/81v2p)Z7/2r2 exp(-Zr/3) sin 2. sin f 
Schr¨odinger equation in atomic units is (assuming infinite nuclear mass, so that µ=me) 
-
1
2.2 - 
Z
r 
	. =E. (4-48) 
The energies are 
En=- 
Z2 
2n2 (4-49) 
The lowest-energy solution is 
.1s =Z3/p exp(-Zr) (4-50) 
The formulas for the hydrogenlike ion solutions (in atomic units) of most interest 
in quantum chemistry are listed in Table 4-2. The tabulated functions are all in real, 
rather than complex, form. Problems involving atomic orbitals are generally far easier 
to solve in atomic units. 
4-5 Angular Momentum and Spherical Harmonics 
We have now discussed three problems in which a particle is free to move over the 
entire range of one or more coordinates with no change in potential. The first case 
was the free particle in one dimension. Here we found the eigenfunctions to be simple 
trigonometric or exponential functions of x. The trigonometric form is identical to the 
harmonic amplitude function of a standing wave in an infinitely long string. We might 
refer to such functions as “linear harmonics.” The second case was the particle-in-aring 
problem, which again has solutions that may be expressed either as sine-cosine or
Section 4-5 Angular Momentum and Spherical Harmonics 111 
exponential functions of the angle f. By analogy with linear motion, we could refer 
to these as “circular harmonics.” Finally, we have described the hydrogenlike ion, 
where the particle can move over the full ranges of . and f (i.e., over the surface of a 
sphere) with no change in potential. The solutions we have just described—the products 
	l,m(.)
m(f)—are called spherical harmonics and are commonly symbolized 
Yl,m(.,f). Thus for m0 
Yl,m(.,f)=(-1)m (2l +1) 
4p 
(l -|m|)! 
(l +|m|)!

1/2 
P|m| l (cos .) exp (imf) (4-51) 
and for m<0 the factor (-1)m is omitted. Because so many physical systems have 
spherical symmetry, spherical harmonics are very important in classical and quantum 
mechanics. 
Closely linked with spherical harmonics is angular momentum. Angular momentum 
is an important physical property because it is conserved in an isolated dynamical 
system; it is a constant of motion for the system. Angular momentum is described by 
magnitude and direction, so it is a vector quantity.7 The classical system, in the absence 
of external forces, is constrained to move in such a way as to preserve both the direction 
and the magnitude of this vector. For a mass of m kg moving in a circular orbit of 
radius r m with an angular velocity of . radians per second, the angular momentum 
has magnitude mr2. kg m2/s (or, alternatively, joule seconds). The direction of the 
vector is given by the right-hand rule: the index finger of the right hand points along the 
particle trajectory and the extended thumb points along the angular momentum vector 
(see Fig. 4-10). (Alternatively, in a right-handed coordinate system, motion of a mass 
in the xy plane from +x toward +y produces angular momentum in the +z direction.) 
In vector notation, L=r×p, where L is angular momentum, r is the position vector, 
and p is the linear momentum. 
Some of the more interesting properties of angular momentum relate to the situation 
where circular motion occurs in the presence of an external field. A familiar example 
is a gyroscope mounted on a pivot and experiencing the gravitational field of the earth. 
The gyroscope flywheel is usually started with the gyroscope in an almost vertical 
position. After release, the gyroscope precesses about the axis of field direction. As 
time passes, the tilt of the gyroscope away from the field direction (which we take to be 
the z direction) increases (see Fig. 4-11). If there were no friction in the bearings, the 
angle of tilt would not change, and the gyroscope would precess about z indefinitely, 
maintaining whatever angle of tilt it found itself with initially. Notice that, in such 
Figure 4-10 
 The angular momentum vector L for a particle of mass m moving with angular 
velocity . about a circular orbit of radius r in the direction indicated. 
7Strictly speaking, angular momentum is a pseudovector—it is dual to a second order antisymmetric tensor. 
However, for the remainder of this book, we can and shall ignore this distinction.
112 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
a case, the angular momentum due to the flywheel, Lf, is conserved in magnitude 
only. Its direction is constantly changing. Thus, Lf is not a constant of motion in the 
presence of a z-directed field. Neither are the components Lfx and Lfy , which change 
in magnitude as the gyroscope precesses. However, Lfz is a constant of motion if the 
angle of tilt does not change. If we add on to Lf the angular momentum Lg due to the 
precession of the gyroscope as a whole (including the center of mass of the flywheel 
but ignoring its rotation), we find that the total angular momentum for the gyroscope 
(including flywheel motion), L and its components Lx, Ly, and Lz behave similarly 
to Lf and its components (see Fig. 4-12). We may summarize these observations from 
classical physics as follows: a rotating rigid body conserves L (hence, Lx, Ly, Lz ) 
in the absence of external forces. In the presence of a z-directed, time-independent 
external force, Lz and |L|, the magnitude of L (but not its direction) are conserved. 
Furthermore, in a system comprising several moving parts, the total angular momentum 
is the sum of the individual angular momenta, and the z component is the sum of the 
individual z components: 
L =
i 
Li (4-52) 
Lz =
i 
Lzi (4-53) 
Many characteristics of the classical situation are maintained in quantum mechanics. 
In particular, it can be shown that a hydrogenlike ion eigenfunction can always 
be associated with “sharp” values for Lz , but not for Lx or Ly, and that the magnitude 
of L is sharp, but not its direction. We have indicated several times in this 
book that a sharp value (i.e., a constant of motion) exists when a state function is an 
eigenfunction for an operator associated with the property. For example, all of our 
hydrogenlike ion wavefunctions are eigenfunctions for the hamiltonian operator, so all 
are associated with sharp energies. We introduced the operator for the z-component of 
angular momentum, in Section 2-6, as (h/2pi)d/df. Generalizing to situations with 
several variables requires switching to partial derivative notation. Then our operator, 
Figure 4-11 
 A gyroscope with the angular momentum of the flywheel, Lf, together with x, y, 
and z components of Lf, at a given instant.
Section 4-5 Angular Momentum and Spherical Harmonics 113 
Figure 4-12 
 The total angular momentum of the gyroscope L is shown as the sum of Lf, the 
angular momentum of the flywheel, and Lg, the angular momentum of the gyroscope. L precesses, 
so only Lz and the magnitude of L are constants of motion. 
symbolized Lˆz , is (h¯/i)./.f. In atomic units, Lˆz is (1/i)./.f. This operator was 
introduced in Section 2-6, where it was given the symbol pf. (A general discussion on 
operators will be given in Chapter 6. The carat symbol is frequently used to denote an 
operator.) Our statement that hydrogenlike eigenfunctions have sharp Lz means that 
we expect Lˆz.n,l,m(r, .,f)= constant · .n,l,m(r, .,f). Since all these eigenfunctions 
have exp(imf) as their only f-dependent term, it follows immediately that 
Lˆz.n,l,m =mh¯.n,l,m (4-54) 
or, equivalently, 
LˆzYl,m (.,f)=mh¯Yl,m (.,f) (4-55) 
Hence, the quantum number m is equal to the z component of angular momentum in 
units of ¯h for the state in question. This means that the angular momentum associated 
with an s state (l = 0, so m = 0) has a zero z component, while a p state (l = 1, so 
m=-1, 0,+1) can have a z component of -¯h, 0, or ¯h. 
The other quantity that we have stated is conserved in these systems is the magnitude 
of L. In quantum mechanics, it is convenient to deal with the square of this magnitude 
L2. The quantum-mechanical operator associated with this quantity is 
Lˆ2 = -¯h2 
(.2/..2)+cot .(./..)+(1/ sin2 .)(.2/.f2) 
= -¯h2 
(1/ sin .)(./..) sin .(./..)+(1/ sin2 .)(.2/.f2) (4-56) 
The result of operating on Yl,m(.,f) with this operator is 
Lˆ2Yl,m (.,f)=l(l +1) ¯h2Yl,m (.,f) (4-57) 
This means that the square of the magnitude of the total angular momentum equals 
l(l+1) ¯h2. Hence, for an s state it is zero, for a p state it is 2¯h2, for a d state it is 6¯h2, etc. 
One can construct vector diagrams to parallel these relationships. A few of these are 
sketched in Fig. 4-13. 
Operators for Lˆx and Lˆy can also be constructed. They are 
Lˆx = ih¯[sin f(./..)+cot . cosf(./.f)] (4-58) 
Lˆy = -ih¯[cosf(./..)-cot . sin f(./.f)] (4-59)
114 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
Figure 4-13 
 Vector relationships that satisfy the following rules: L2 = l(l + 1),Lz = m, 
m=-l,-l +1, . . . , 0, . . . , l -1, l. The quantum rules correspond to a classical analog where the 
gyroscope can have only certain discrete angles of tilt. (a) s; l = 0; l(l + 1) = 0; m = 0. (b) p; 
l =1; l(l +1)=2; m=-1, 0,+1. (c) d; l =2;l(l +1)=6; m=-2,-1, 0,+1,+2. (all in atomic 
units) 
The hydrogenlike eigenfunctions are not necessarily eigenfunctions for either of these 
operators (Problem 4-23). 
It is interesting to consider the physical meaning of these results. If a quantity has a 
sharp value, it means that we will always get that value no matter when we measure that 
property for systems in the state being considered. Thus, the z component of angular 
momentum for hydrogen atoms in the 2p+1 state will always be measured to be +1¯h, 
or one atomic unit. For the x or y component, however, repeated measurements (on an 
ensemble of 2p+1 atoms) will yield a spread of values. We can measure (or compute) 
an average value of Lx or Ly but not a sharp value. In terms of our mental model (a 
gyroscope) this seems sensible enough except for one thing. Our hydrogenlike eigenfunctions 
are solutions for a central field potential with no external field. Under such 
conditions, L, Lx, Ly, and Lz are classically all constants of motion. Why, then, are 
they not all sharp quantum mechanically? The answer is that quantum-mechanical state 
functions never contain more information than is, in principle, extractable by measurement. 
To measure a component of angular momentum in a system always means, in 
practice, subjecting the system to some sort of external force. Furthermore, this system 
must obey the limitations decreed by the uncertainty principle. The hydrogenlike ion 
wavefunctions cannot simultaneously be eigenfunctions for Lˆx, Lˆy, and Lˆz because 
that would give simultaneous sharp values (i.e., no uncertainty) for the conjugate variables 
angular momentum vector length and angular momentum vector orientation. This 
would violate the uncertainty principle, which is in turn a reflection of limitations on 
our ability to measure one variable without affecting another (see Section 1-8). 
It is possible, working only with the quantum-mechanical operators, to generate 
the eigenvalues of Lˆz and Lˆ2. This approach is a deviation from our main line of
Section 4-6 Angular Momentum and Magnetic Moment 115 
development and is contained in Appendix 4. (It is recommended that Chapter 6 be 
completed before reading Appendix 4.) We give only the results. They are 
Lˆzfl,m = mh¯fl,m, m=-l,-l +1, . . . , l -1, l (4-60) 
Lˆ2fl,m = l (l +1) ¯h2fl,m (4-61) 
These look like the results already given in Eqs. (4-55) and (4-57). There is a difference, 
however. Here there is no indication that m is an integer, whereas in Eq. (4-55) m 
must be an integer, as indicated by the presence of zero in its value list. There are two 
ways in which we can have a sequence of the form -l,-l +1, . . . , l -1, l. One way 
is to have an integer series, for example, -2,-1, 0,+1,+2, which must contain zero. 
The other way is to have a half-integer series, for example, -3
2 ,-1
2 ,+1
2 ,+3
2 , which 
skips zero. If we work only with the properties of the operators, we find that either 
possibility is allowed. But if we assume that the as-yet-unspecified eigenfunctions fl,m 
are separable into .- and f-dependent parts, we find ourselves restricted to the integer 
series. For orbital angular momentum (due to motion of the electron in the atomic 
orbital), the z component must be (in atomic units) an integer, for we have seen that 
the state functions . contain the spherical harmonics Yl,m, which are indeed separable. 
Spin angular momentum for an electron (to be discussed in more detail in the next 
chapter), has half-integer z components of angular momentum, and the eigenfunctions 
corresponding to spin cannot be expressed with spherical harmonics. 
4-6 Angular Momentum and Magnetic Moment 
If a charged particle is accelerated, a magnetic field is produced. Since circular motion 
of constant velocity requires acceleration (classically) it follows that a charged particle 
having angular momentum will also have a magnetic moment. The magnetic moment 
is proportional to the angular momentum, colinear with it, and oriented in the same 
direction if the charge is positive. For an electron, the magnetic moment is given by 
µ=-ßeL (4-62) 
where ße, the Bohr magneton, has a value of 9.274078×10-24 J T-1 (equal to 1
2 a.u.), 
where T is magnetic field strength in Tesla. (ße is defined to contain the ¯h that belongs 
to L, so it is only the vl(l +1) part of L that is used in the calculation.) 
EXAMPLE 4-6 What is the magnitude of the orbital magnetic moment for an electron 
in a 3d state of a hydrogen atom? In a 4d state of He+? 
SOLUTION 
 For any d state, l = 2, so, in a.u., |µ| = ßeL = ßevl(l +1) = v6ße = 2.27×10-23JT-1. (Since we want magnitude, we can ignore the minus sign.) The value does 
not depend on the quantum number n nor on atomic number Z, so it is the same for He+.  
Applying a magnetic field of strength B defines a z-direction about which the magnetic 
moment vector precesses. The z-component, µz , of the precessing vector interacts 
with the applied field B. The interaction energy is 
E=-µzB =ßeLzB =ßemB (4-63)
116 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
This means that some of the degeneracies among the energy levels of the hydrogenlike 
ion will be removed by imposing an external magnetic field. For instance, the 2p+1 
and 2p-1 energy levels will be raised and lowered in energy, respectively, while 2s and 
2p0 will be unaffected (see Fig. 4-14). This, in turn, will affect the atomic spectrum 
for absorption or emission. The splitting of spectral lines due to the imposition of an 
external magnetic field is known as Zeeman splitting. Because the splitting of levels 
depicted in Fig. 4-14 is proportional to the z component of orbital angular momentum, 
given by m¯h, it is conventional to refer to m as the magnetic quantum number. 
In the absence of external fields, eigenfunctions having the same n but different l 
and m are degenerate. We have seen that this allows us to take linear combinations 
of eigenfunctions, thereby arriving at completely real eigenfunctions like 2px and 2py, 
instead of 2p+1 and 2p-1. When a magnetic field is imposed, the degeneracy no longer 
exists, and we are unable to perform such mixing. Under these conditions, 2px, 2py, 
3dxy, etc. are not eigenfunctions, and we are restricted to the pure m=0,±1,±2 . . . 
type solutions. 
Thus far we have indicated that the stationary state functions for the hydrogenlike 
ions are eigenfunctions for Lˆ2 and Lˆz , and we have compared this to the fact that |L| and Lz are constants of motion for a frictionless gyroscope precessing about an external 
field axis. But howabout atoms with several electrons? And howabout molecules? Are 
their stationary state functions also eigenfunctions for Lˆ2 and Lˆz? A general approach 
to this kind of question is discussed in Chapter 6. For now we simply note that the 
spherical harmonics are eigenfunctions of Lˆ2 and Lˆz [Eqs. (4-55) and (4-57)] and 
that any state function of the form .(r, .,f)=R(r)Yl,m(.,f) will necessarily be an 
eigenfunction of these operators. But the spherical harmonics are solutions associated 
with spherically symmetric potentials. Therefore, it turns out that eigenfunctions of 
the time-independent hamiltonian operator are also eigenfunctions for Lˆ2 and Lˆz only 
if the potential is spherically symmetric. In the more restricted case in which . has 
Figure 4-14 
 Energy levels of a hydrogenlike ion in the absence and presence of a z-directed 
magnetic field.
Section 4-7 Angular Momentum in Molecular Rotation—The Rigid Rotor 117 
the form .(r, .,f)=f (r, .) exp(imf), . will still be an eigenfunction of Lˆz but not 
of Lˆ2. This situation applies in systems having cylindrically symmetric potentials, 
dependent on r and . but not f (e.g., H+2
, N2). We discuss such cases in more detail in 
Chapter 7. 
4-7 Angular Momentum in Molecular 
Rotation—The Rigid Rotor 
We have seen that the two-particle system of an electron and a nucleus rotating about a 
center of mass (COM) can be transformed to the one-particle system of a reduced mass 
rotating about a fixed point. However, this transformation can be made for any two-mass 
system, and so it applies also to the case of the nuclei of a rotating diatomic molecule. 
As we now show, the mathematical outcome for the rotating diatomic molecule is 
strikingly similar to that for the hydrogenlike ion. 
The simplest treatment of molecular rotation ignores vibrational motion by assuming 
that the distance between the nuclei is fixed. The resulting model is therefore called 
the rigid-rotor model. Let there be two nuclear masses, m1 and m2, separated from the 
COM by distances r1 and r2, respectively. Then, because of the way that the COM is 
defined, we have that 
m1r1 =m2r2 (4-64) 
The moment of inertia, I, is 
I =m1r2
1 +m2r2
2 (4-65) 
It is not difficult to show (Problem 4-35) that the same moment of inertia results from 
a reduced mass µ rotating about a fixed point at a distance r =r1 +r2. That is, if 
µ= 
m1m2 
m1 +m2 
(4-66) 
then 
I =µr2 (4-67) 
Therefore, solving the problem of a reduced mass µ rotating about a fixed point at the 
fixed distance r =r1 +r2 is equivalent to solving the two-mass rigid-rotor problem. In 
effect, the rotating-diatomic problem is transformed to a particle-on-the-surface-of-asphere 
problem. 
As usual, we write the Schr¨odinger equation by starting with the general prescription 
[(-¯h2/2µ).2 +V ]. =E.. Then we recognize that V is constant over the spherical 
surface (corresponding to the diatomic molecule having no preferred orientation), so 
we can set V =0. Since r is a constant, the first term in .2 [Eq. (4-7)] vanishes due to 
the ./.r operators. The resulting Schr¨odinger equation is [using Eq. (4-67)] 
[(1/ sin .)(./..) sin .(./..)+(1/ sin2 .)(.2/.f2)].(.,f)=(-2IE/¯h2).(.,f) 
(4-68)
118 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
Equation (4-68) is the same as the 	, 
 equations seen earlier [(4-36) and (4-39)] 
with ß = 2IE/¯h, and so we already know the eigenvalues and eigenfunctions. We 
noted earlier that ß =l(l +1). For molecular rotation it is conventional to symbolize 
the 	 quantum number as J , rather than l. This leads to 
J(J +1)=2IE/¯h2 (4-69) 
or 
E =J(J +1) ¯h2/2I J=0, 1, 2, . . . (4-70) 
Since V =0, E is entirely kinetic energy. Because r is not a variable, there is no analog 
to the principal quantum number n, and J is not limited in its highest value. (Note also 
that atomic units, which are designed to simplify electronic problems, are not normally 
used for molecular rotation or vibration.) 
The eigenfunctions are, as before, the spherical harmonics YJ,mJ (.,f), with mJ = 0,±1,±2, . . . ,±J . Thus we have an s-type solution (J =0,mJ =0) that has constant 
value over the spherical surface, three p-type solutions (J =1,mJ =+1, 0,-1), five 
d-type solutions, etc. For each value of J , there are 2J +1 eigenfunctions. 
The s-type solution has zero energy. One can imagine that the reduced mass is 
motionless on the surface of the sphere and has equal probability for being found 
anywhere. This transforms back to a picture where the diatomic molecule is not 
rotating and where there is no preferred orientation. Since E = 0 when J = 0, we 
conclude that there is no zero-point energy for free rotation. (However, if rotation is 
restricted so that some orientations become preferred, the zero-point energy becomes 
finite.) 
The three p-type solutions are degenerate, hence can be mixed to give real functions 
analogous to px, py, pz . We could represent the pz function by taking a globe and 
marking a circular region of positive phase around the northern polar region, with a 
matching region of negative phase around the southern pole. Clearly, these rigid-rotor 
wavefunctions have the same symmetries as their hydrogenlike counterparts. A pz 
rotational state corresponds to a molecule (or ensemble of molecules) rotating with 
kinetic energy 2¯h2/2I (since J =1) with little likelihood of finding the reduced mass 
near the equator (i.e., with little probability of finding the diatomic molecule oriented 
nearly perpendicular to the z axis.) 
Angular momentum for the rigid rotor also follows the hydrogenlike system rules. 
The square of the total angular momentum equals J(J +1) ¯h2, and the z component 
equals mJ ¯h. 
Transitions between rotational energy levels can be detected spectroscopically. Once 
such energy differences are identified with specific changes in J , it becomes a simple 
matter to solve for r, the internuclear distance in the rigid rotor. 
For example, suppose an absorption peak for H81Br seen at 101.58 cm-1 is 
assigned to the J = 6 . 5 transition. Since the energies of these levels are 
42¯h2/2I and 30¯h2/2I , respectively, their difference is 12¯h2/2I . This energy is 
equal to that of the photons of light at 101.58 cm-1. Solving this relation gives 
I = 3.3069×10-47 kg m2. We know that this equals µr2, and we know how to get 
µ from mH and mBr [Eq. (4-46)]. So we can solve for r, finding r = 141.44 pm 
(Problem 4-37).
Section 4-8 Summary 119 
4-8 Summary 
1. The motion of two masses moving about a center of mass can be transformed to 
motion of a single reduced mass moving about a fixed point. The radius of rotation 
for the reduced mass is identical to the distance of separation of the original two 
masses. For hydrogenlike ions, the nuclear mass is so much greater than the electron 
mass that the reduced mass is almost identical to the electron mass. 
2. The bound-state energies for time-independent states of the hydrogenlike ion depend 
on only the atomic number Z and the quantum number n (a positive integer) and 
vary as -Z2/2n2. This means that the energies get closer together as n increases 
and that there is an infinite number of such negative energy levels. Each such energy 
level has degeneracy n2. A continuum of energies exists for unbound (E>0) states. 
3. Each stationary state wavefunction is characterized by three quantum numbers, n, 
l, and m, all integers, with l ranging from 0 to n - 1 and m ranging from -l to 
+l. If l =0, we have an s state and . is spherically symmetric with a cusp at the 
nucleus. If l =1, 2, . . . we have a p, d, . . . state, and . vanishes at the nucleus and 
is not spherically symmetric. In all states there is a finite probability for finding the 
electron beyond the classical turning point. 
4. Eigenfunctions Rn,l(r)	l,m(.)
m(f) with m=0 are complex but can be mixed to 
form real eigenfunctions. However, if an external field causes states of different m 
to be nondegenerate, such mixing will produce noneigenfunctions. 
5. All the stationary state eigenfunctions are orthogonal to each other, and radial and/or 
angular (usually planar) nodes are instrumental in this. The effect of a radial node 
on energy is the same as that of an angular node, so that all eigenfunctions with, 
say, three nodes (all radial, all angular, or a combination) are degenerate. This is 
peculiar to the -r-1 potential. 
6. Separation of variables is not “perfectly clean” since the differential equations for R 
and	(Eqs. (4-38) and (4-39)), are linked through ß and those for	and
(Eq. 4-40) 
are linked through m. This leads to interdependencies in the values of n, l, and m. 
7. Spherical harmonics are the angular parts of solutions to Schr¨odinger equations for 
systems having spherically symmetric potentials. These functions are eigenfunctions 
of Lˆz and Lˆ2 as well as Hˆ , so such states have sharp values of Lz , L2, and E. 
The value of Lz is m¯h, and for L2 it is l(l +1) ¯h2, where l and m must be integers. 
In atomic units the quantity ¯h does not appear in these formulas. 
8. The z component of the magnetic moment due to orbital motion of a charged particle 
is proportional to m¯h, and so m is called the magnetic quantum number. This 
magnetic moment contributes to the Zeeman splitting seen in spectra of hydrogenlike 
ions in magnetic fields. 
9. Eigenfunctions other than spherical harmonics exist for Lˆ2 and Lˆz , but these are not 
separable into .- and f-dependent functions. In these cases, l and m can be halfintegers. 
These cases do not arise in orbital motion, but do arise in spin problems.
120 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
10. If V is cylindrically symmetric [i.e., V =V (r,.)], the eigenfunctions of the hamiltonian 
are still eigenfunctions for Lˆz but not for Lˆ2. Hence, mh¯ still equals the z 
component of angular momentum for such a system. Here, z is the direction of the 
axis of rotational symmetry, i.e., the internuclear axis of the molecule. 
11. The rules for allowed angular momentum magnitudes and orientations are the 
same for the rigid rotor as for the hydrogenlike ion. This leads to the following 
relationships in terms of the rotational quantum numbers J = 0, 1, 2, . . . , mJ = 0,±1, . . . ,±J : 
Length of angular momentum vector =vJ(J +1) ¯h. 
Component of angular momentum perpendicular to internuclear axis = mJ ¯h. 
Kinetic energy of rotation: TJ =J(J +1) ¯h2/2I. 
Degeneracy of level: gJ =2J +1. 
4-8.A Problems 
4-1. An observed spectroscopic transition in the hydrogen atom involves the 2p.1s 
transition. Using Eq. (4-8), evaluate this energy difference in units of hertz (Hz). 
(1 Hz=1 s-1.) Do the calculation using both me and µ. (See Appendix 10 for 
constants and conversion factors.) How much error in this calculation, in parts 
per million, is introduced by ignoring the finite mass of the nucleus (i.e., using 
me instead of µ)? 
4-2. Sketch qualitatively, on the same r axis, .2(r) for a 1s state and 4pr2 for the 
variation of dv with r. Sketch the radial distribution function, which is the product 
of these two functions, and explain why it vanishes at r =0,8. 
4-3. For a hydrogen atom in the 1s state, . =(1/vp)exp(-r), in atomic units. 
a) Calculate the value of r (in a.u.) at the classical turning point. 
b) Calculate the percentage of the electronic charge that is predicted to be beyond 
the classical turning point. (See Appendix 1 for useful integrals.) 
4-4. Using atomic units, compute for a 1s electron of the hydrogenlike ion [. = 
(Z3/p) exp(-Zr)]. (See Appendix 1 for useful integrals.) 
a) the most probable distance of the electron from the nucleus, 
b) the average distance of the electron from the nucleus, 
c) the distance from the nucleus of maximum probability density, 
d) the average value of the potential energy (V =-Z/r). Note how these quantities 
depend on atomic number Z. Also, why do you think that, when Z=1, (d) 
is not minus the reciprocal of (b)? Why is (d) lower than minus the reciprocal 
of (b)? 
4-5. Demonstrate by integration that the 1s and 2s orbitals of the hydrogen atom are 
orthogonal. (See Appendix 1 for useful integrals.)
Section 4-8 Summary 121 
4-6. Normalize the function r exp(-r) cos .. 
4-7. Obtain the average value of position ¯x, for a particle moving in a one-dimensional 
harmonic oscillator potential in a state with the normalized wavefunction 
. =(ß/482p)1/4 (2ßx)3 -12ßx exp(-ßx2/2) 
[There is an easy way to do this problem.] 
4-8. For a particle in a one-dimensional box with boundaries at x =0 and L and for 
any quantum number n: 
a) Show how you would set up the calculation for the mean square deviation of 
the particle from its average position. 
b) Explain qualitatively how you would expect the value of (a) to vary with 
quantum number n. 
c) Evaluate the expression from part (a) in terms of n and L. Calculate the value 
for n=1, 2. Discuss the relative values of these numbers for reasonableness. 
4-9. Sketch the 2pz and the 3dxy wavefunctions. Demonstrate, without explicitly 
integrating, that these are orthogonal. 
4-10. Repeat Problem 4-4, but for the 2p0 wavefunction. 
4-11. Find an expression for the classical turning radius for a hydrogenlike ion in terms 
of n, l,m, and Z. 
4-12. Show that the sum of the charge distributions of all five 3d orbitals is spherically 
symmetric. 
4-13. Try to answer the following questions without looking up formulas or using 
pencil and paper. Use atomic units. 
a) What is the energy of the hydrogen atom in the 1s state? 
b) What is the energy of He+ when n=1? n=2? 
c) What is the degeneracy of the n=5 energy level of hydrogen? 
d) How many planar nodes does the 4dxz orbital have? How many radial nodes? 
e) What is the potential energy in a hydrogen atom when the electron is 0.50 a.u. 
from the nucleus? 
4-14. You should be able to answer the following questions (use a.u.) in your head or 
with trivial calculations. 
An unnormalized eigenfunction for the hydrogen atom is 
. =(27-18r +2r2) exp(-r/3) 
a) What are the l and m quantum numbers for this state? 
b) How many radial nodes does this function possess? 
c) What is the energy of this state? 
d) What is the classical turning radius for this state?
122 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
4-15. . =N(6r -r2) exp(-r/3) sin . sin f is an eigenfunction, in a.u., for the hydrogen 
atom hamiltonian. N is the normalizing constant. Without looking at 
formulas in the text, answer the following questions by inspection: 
a) Is there a node in the r coordinate? If so, where? 
b) Which state is this? (Give orbital symbol—e.g., 1s, 2pz , etc.) 
4-16. Without comparing to tabulated formulas, state whether each of the following 
could reasonably be expected to be an eigenfunction (unnormalized) of the hamiltonian 
for a hydrogenlike ion. Explain why. 
a) (27-18Zr +2Z2r2) exp(-2Zr/3) 
b) r exp(Zr/2) sin . cosf 
c) r sin . exp(-if). 
4-17. Calculate the average value of x for the 1s state of the hydrogen atom. Explain 
why your result is physically reasonable. 
4-18. a) Calculate the most probable value of . in the 2pz state of the hydrogen atom. 
b) Calculate the values of . corresponding to nodal cones in the 3dz2 orbital. 
4-19. Demonstrate that Eq. (4-30) for 3dxy has the same angular dependence as the 
function xy. 
4-20. Verify Eq. (4-45) using P1(x) with P1(x) and with P3(x) [from Eq. (4-42)]. 
What does Eq. (4-45) imply for the integral over all space of .3p0.3d-1? 
4-21. Using Eqs. (4-51) and (4-43), write the normalized spherical harmonic function 
Y3,-2(.,f). For which type of hydrogenlike AO does this function give the 
angular dependence? 
4-22. Verify Eq. (4-55). 
4-23. Test the 2p0 eigenfunction to see if it is an eigenfunction for Lˆx or Lˆy, [Eqs. 
(4-58) and (4-59)]. Show that the 1s function is an eigenfunction of Lˆx, Lˆy, 
Lˆz , and Lˆ2. Explain, in terms of the vector model, this seeming violation of the 
discussion in the text. 
4-24. Work out the value of Lˆ2.2p0 by brute force and show that the result agrees with 
Eq. (4-57). 
4-25. Sketch the vector diagram (as in Fig. 4-13) for the 4f orbitals of hydrogen. How 
does this compare to the diagram for the 6f orbitals of He+? 
4-26. Sometimes eigenfunctions for an operator can be mixed together to produce 
new functions that are still eigenfunctions. Listed below are some operators 
with pairs of their eigenfunctions. Indicate in each case whether mixtures of 
these pairs will or will not continue to be eigenfunctions for the operator shown. 
(You should be able to do this by inspection, using your knowledge of these 
systems.)
Section 4-8 Summary 123 
Operator Eigenfunctions 
a) Hˆ (1 dim. box) .1 .3 
b) Hˆ (ring) sin 3f cos 3f 
c) Lˆz (ang. mom.) exp(3if) exp(-3if) 
d) Hˆ (cubical box) .1,2,3 .2,2,2 
e) Hˆ (H atom) .2s .2p0 
f) Hˆ (H atom) .3s .4s 
4-27. A hydrogenlike ion is in a state having a z-component of angular momentum 
equal to -2 a.u. 
a) What is the smallest possible value of the length of the angular momentum 
vector for this state? 
b) What symbol describes the state corresponding to your answer to part (a)? 
4-28. Calculate the average angular momentum, L¯z , for a particle in a ring of constant 
potential having wavefunction 
a) (1/vp)sin 3f 
b) (1/v2p)exp(-3if) 
4-29. Evaluate each of the following integrals. Look for labor-saving approaches. 
Integrals are over all space unless otherwise indicated. 
a)  .2px Lˆz.2pxdv 
b)  .2px Lˆ2.2pxdv 
c)  .2px Lˆx.2pxdv 
d)  .3dx2-y2.2pxdv 
e)  2p 
0 exp (2if) exp (-3if) df 
4-30. Evaluate each of the following in a.u. (You should be able to answer these by 
inspection.) Note: .n,l,m stands for a hydrogen atom eigenfunction of Hˆ . 
a) Lˆ2.3,2,1 
b) Lˆ2.2px 
c) Hˆ .3px 
d) (1/i) (./.f).2p-1 
4-31. For each of the following operators, indicate by “yes” or “no” whether .3px 
(with Z =1) is an eigenfunction. If it is, then also give the eigenvalue in a.u. 
a) -1
2.2 -1/r 
b) -1
2.2 -3/r 
c) Lˆz 
d) -1
2.2 
e) Lˆx 
f) Lˆ2
124 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
g) r 
h) 1/r 
4-32. For the 3s state of the hydrogen atom, estimate the amount of electronic charge 
between 211 pm and 213 pm. 
4-33. Calculate in hertz the splitting between 2p0 and 2p+1 of a hydrogen atom by a 
magnetic field of 2 tesla. Compare this with the 2p.-1s transition energy (in 
parts per million). 
4-34. Try to answer (by inspection) the following questions about the n=4 states of 
the hydrogen atom. 
a) What is the energy of the level, in a.u.? 
b) What is the degeneracy of the level? 
c) What values for the length of the orbital angular momentum vector (in a.u.) 
are possible? 
d) Into how many sublevels (of energy) is the n=4 level split by imposition of 
a magnetic field? 
e) What is the degeneracy of the unshifted portion of the sublevels referred to 
in (d)? 
4-35. Show that the moment of inertia for two masses, m1 and m2, moving on a rigid 
massless bar about the center of mass at distances r1 and r2, respectively, is 
identical to that of a reduced mass µ=m1m2/(m1 +m2) moving about a point 
at a distance of r =r1 +r2. 
4-36. Show that Eq. (4-68) can be written, using Lˆ2, in a form that is analogous to the 
classical relation for a freely rotating mass, L2/2I =T . 
4-37. Using thatmH=1.0078 a.m.u. andmBr = 80.9163 a.m.u., calculate µ and verify 
the value of r given at the end of Section 4-7. 
4-38. Assuming an internuclear distance of 127.5 pm in D35Cl, compute the expected 
positions in cm-1 for the absorption peaks corresponding to J =1.0, 2.1, 
3.2 (D = 2.0141 a.m.u., 35Cl = 34.9688 a.m.u.). 
4-39. The J =1.0 transition in 12C16O occurs at 3.86 cm-1. Calculate the internuclear 
distance. (12C = 12 a.m.u. by definition, 16O = 15.9949 a.m.u.) 
4-40. HCl has a permanent electric dipole moment, which means that the reduced 
mass has a partial electric charge in the transformed version of the system. 
This in turn means that there is a magnetic moment vector parallel to the total 
angular momentum vector. Describe qualitatively what happens to the energies 
of the J =3 rotational states when HCl is subjected to a uniform magnetic 
field. 
4-41. Uncertainty in position in one dimension, x, is defined as x2 - ¯x2
1/2 
That is, 
it is the square root of the difference between the average squared position and 
the square of the average position. Calculate r for the 1s state of the hydrogen 
atom.
Section 4-8 Summary 125 
Multiple Choice Questions 
(Try to answer these without referring to the text or using pencil and paper.) 
1. A particle on the surface of a sphere has quantum number J =7. The energy level 
to which this state belongs has a degeneracy of 
a) 56 
b) 49 
c) 42 
d) 14 
e) None of these. 
2. A particle on the surface of a sphere in the state having J =4, mJ =4 
a) has E =16¯h2/2I. 
b) has a z-component of angular momentum of 4¯h. 
c) doesn’t exist because this state violates quantum number rules. 
d) has a degeneracy of 20. 
e) None of the above is a true statement. 
3. HI and DI are made to undergo the same transition (say J =11.-J =10, but the 
particular J values are not important). The light frequency inducing the transition 
for HI is equal to .. Approximately which frequency would you expect to induce 
the same transition for DI? 
a) 2. 
b) ./2 
c) v2. 
d) ./v2 
e) None of these. 
4. The electronic energy of Li2+ in the 2s state is 
a) the same as that of H in the 1s state. 
b) nine times that of H in the 1s state. 
c) one-fourth that of H in the 1s state. 
d) four-ninths that of H in the 1s state. 
e) nine-fourths that of H in the 1s state. 
5. Consider the following expressions, where . is a hydrogen-atom wavefunction. 
1)  allspace 
.*r. dv 2) d 
dr.*.r2 sin . =0 
3) d 
dr.*. =0 4)d 
dr. =0 
Which one of the following is a true statement? 
a) Expression 1 is equal to unity if . is normalized. 
b) Expression 4 is true when r is the position of a radial node. 
c) Expression 3 is true everywhere because .*. is a constant. 
d) Expression 2 is true when r is at its most probable value. 
e) Expression 3 is true when r is at its most probable value.
126 Chapter 4 The Hydrogenlike Ion, Angular Momentum, and the Rigid Rotor 
6. A 5dxy atomic orbital has 
a) 2 planar and 1 radial nodes. 
b) 2 planar and 3 radial nodes. 
c) 3 planar and 1 radial nodes. 
d) 5 nodes all together. 
e) None of the above is correct. 
7. For a hydrogen atom in an n = 4 state, the maximum possible z-component of orbital 
angular momentum is 
a) 2¯h 
b) 3¯h 
c) v12¯h 
d) v6¯h 
e) None of the above is correct. 
8. The following is an eigenfunction for the hydrogen atom: 
. =1/32pa3
0
 (r/a0) exp(-r/2a0) cos . 
Which one of the following statements about . is true? 
a) The term on the left, up to the exponential, is the normalizing constant. 
b) This . has a nonzero value at the nucleus. 
c) For this state, l =1 and ml =0. 
d) This state has a spherical electron cloud distribution. 
e) None of the above statements is true. 
9. The radial distribution function for a 1s state, 4pr2.2
1s , indicates that 
a) the most probable value of the distance from the nucleus is zero. 
b) the average value of r is zero. 
c) the average value of r is greater than the most probable value. 
d) the average value of r is equal to the most probable value. 
e) the electron cloud density per cubic picometer is greatest at a radius other 
than zero. 
References 
[1] H. Eyring, J. Walter, and E. D. Kimball, Quantum Chemistry, Chapter VI. Wiley, 
NewYork, 1944. 
[2] I. N. Levine, Quantum Chemistry, 5th ed. Prentice Hall, Upper Saddle River, New 
Jersey, 2000. 
[3] L. Pauling and E. B. Wilson, Jr., Introduction to Quantum Mechanics. McGraw- 
Hill, NewYork, 1935. 
[4] R. N. Zare, Angular Momentum.Wiley, NewYork, 1988.
Chapter 5 
Many-Electron Atoms 
5-1 The Independent Electron Approximation 
In previous chapters we have dealt with the motion of a single particle in various 
potential fields. When we deal with more than one particle, new problems arise and 
new techniques are needed. Some of these are discussed in this chapter. 
In constructing the hamiltonian operator for a many electron atom, we shall assume 
a fixed nucleus and ignore the minor error introduced by using electron mass rather 
than reduced mass. There will be a kinetic energy operator for each electron and 
potential terms for the various electrostatic attractions and repulsions in the system. 
Assuming n electrons and an atomic number ofZ, the hamiltonian operator is (in atomic 
units) 
H(1, 2, 3, . . . ,n)=-
1
2 
n 


i=1 
.2
i - 
n 


i=1
(Z/ri)+
n-1 


i=1 
n 

 
j=i+1 
1 
rij 
(5-1) 
The numbers in parentheses on the left-hand side of Eq. (5-1) symbolize the spatial 
coordinates of each of the n electrons. Thus, 1 stands for x1, y1, z1, or r1, .1, f1, 
etc. We shall use this notation frequently throughout this book. Since we are not here 
concerned with the quantum-mechanical description of the translational motion of the 
atom, there is no kinetic energy operator for the nucleus in Eq. (5-1). The index i refers 
to the electrons, so we see that Eq. (5-1) provides us with the desired kinetic energy 
operator for each electron, a nuclear electronic attraction term for each electron, and an 
interelectronic repulsion term for each distinct electron pair. (The summation indices 
guarantee that 1/r12 and 1/r21 will not both appear in H. This prevents counting the 
same physical interaction twice. The indices also prevent nonphysical self-repulsion 
terms, such as 1/r22, from occurring.) Frequently used alternative notations for the 
double summation in Eq. (5-1) are 1
2n
i=j 1/rij , which counts each interaction twice 
and divides by two, and n
i<j or 
i,j which is merely a shorthand symbol for the 
expression in Eq. (5-1). In each of these alternative notations, the summation is still 
over two indices, but the secondsymbol is “understood.” 
For the helium atom, Eq. (5-1) becomes (see Figure 5-1) 
H(1, 2)=-
1
2.2
1 - 
1
2.2
2 -(2/r1)-(2/r2)+(1/r12) (5-2) 
The helium hamiltonian (5-2) can be rewritten as 
H(1, 2)=h(1)+h(2)+1/r12 (5-3) 
127
128 Chapter 5 Many-Electron Atoms 
Figure 5-1 
 Interparticle coordinates for a three-particle system consisting of two electrons and 
a nucleus. 
where 
h(i)=-
1
2.2
i -2/ri (5-4) 
In Eq. (5-3) we have merely grouped H into two one-electron operators and one twoelectron 
operator. There is no way to separate this hamiltonian completely into a sum 
of one-electron operators without loss of rigor. However, if we wish to approximate the 
hamiltonian for helium in such a way that it becomes separable, we might try simply 
ignoring the interelectronic repulsion term: 
Happrox =h(1)+h(2) (5-5) 
If we do this, our approximate hamiltonian Happrox treats the kinetic and potential 
energies of each electron completely independently of the motion or position of the 
other. For this reason, such a treatment falls within the category of “independent 
electron approximations.” 
Notice that each individual one-electron hamiltonian (5-4) is just the hamiltonian for 
a hydrogenlike ion, so it has as eigenfunctions the 1s, 2s, 2p, etc., functions of Chapter 4 
with Z=2. Such one-electron functions are referred to as atomic orbitals.1 Representing 
them with the symbol fi (e.g., f1 = 1s, f2 = 2s, f3 = 2px, f4= 2py, etc.) we have, 
then, 
h(1)fi(1)=ifi(1) (5-6) 
where i is referred to as the orbital energy, or one-electron energy for atomic orbital 
fi . As we saw in Chapter 4, i is given in atomic units by 
i =-
1
2
Z2/n2 (5-7) 
where n is the principal quantum number for fi , and Z is the nuclear charge. The “1” in 
Eq. (5-6) indicates that fi(1) is a function whose variable is the position of electron 1. 
We will now show that products of the atomic orbitals f are eigenfunctions of 
Happrox. Let the general product of atomic orbitals for helium be written fi(1)fj (2). 
Then 
Happroxfi(1)fj (2) = (h(1)+h(2))fi(1)fj (2) (5-8) 
= h(1)fi(1)fj (2)+h(2)fi(1)fj (2) (5-9) 
1The term “atomic orbital” is used for any one-electron function used to describe the electronic distribution 
about an atom.
Section 5-2 Simple Products and Electron Exchange Symmetry 129 
But h(1) does not contain any of the variables in fj (2), and so they commute. Similarly, 
h(2) and fi(1) commute, and 
Happroxfi(1)fj (2) = fj (2)h (1)fi(1)+fi(1)h (2)fj (2) 
= fj (2)ifi(1)+fi(1)jfj (2) [from Eq. (5-6)] 
= i +j fi(1)fj (2)=Efi(1)fj (2). (5-10) 
Thus, fi(1)fj (2) is an eigenfunction of Happrox, and the eigenvalue E is equal to 
the sum of the orbital energies. These results are yet another example of the general 
rules stated in Section 2-7 for separable hamiltonians. Indeed, once we recognized that 
Happrox is separable, we could have written these results down at once. 
Since the above terminology and results are so important for understanding many 
quantum-chemical calculations, we will summarize them here: 
1. The hamiltonian for a multielectron system cannot be separated into one-electron 
parts without making some approximation. 
2. Ignoring interelectron repulsion operators is one way to allow separability. 
3. The one-electron operators in the resulting approximate hamiltonian for an atom are 
hydrogenlike ion hamiltonians. Their eigenfunctions are called atomic orbitals. 
4. Simple products of atomic orbitals are eigenfunctions for the approximate hamiltonian. 
5. In this approximation the total energy is equal to the sum of the one-electron energies. 
EXAMPLE 5-1 What electronic energy is predicted by the above approximation for 
the lithium atom in its ground state? What is the experimental value for the total 
electronic energy, given that the first and second ionization energies are 0.198 a.u. 
and 2.778 a.u.? 
SOLUTION 
 The ground state configuration for lithium is 1s22s, so Eapprox =2e1s +e2s = 
2(-1
2 · 32 
12 a.u.)+(-1
2 · 32 
22 a.u.)=-10.125 a.u. The experimental value of E equals minus the 
sum of all three ionization energies. The first two values are given, and the third can be calculated 
using the formula for one-electron ions: IE3 =-ELi2+ =-(-1
2 · 32 
12 ) = 4.500 a.u. Therefore, 
Eexp=-(0.198+2.778+4.500) a.u.=-7.476 a.u. Clearly, the approximate hamiltonian predicts 
an electronic energy that is much lower than the experimental value.  
5-2 Simple Products and Electron Exchange Symmetry 
In the independent particle model just described, thewavefunction for the lowest-energy 
state for helium is 1s(1)1s(2) since this has the lowest possible sum of one-electron 
energies. The electronic configuration for this state is symbolized 1s2, the superscript 
telling us howmany electrons are in 1s orbitals. What might we expect for the electronic 
configuration of the lowest excited state? The answer is 1s2s (superscript “ones” are 
implicit). (At this point there is no reason for preferring this configuration to, say,
130 Chapter 5 Many-Electron Atoms 
1s2px, but we shall show later that, in multielectronic systems, the 2s orbital has a 
lower energy than does a 2p orbital, even though they have the same principal quantum 
number.) Thus, we might write 
.(1, 2)=1s(1)2s(2)=8/p exp(-2r1) 
 He+1s 	
1/p(1-r2) exp(-r2) 
 He+2s 	 
(5-11) 
If one were to calculate ¯r1, the average distance from the nucleus for electron 1, using this 
wavefunction, onewould obtain a value of 3
4 a.u., consistent with the 1s state of a helium 
ion. For electron 2 one would find an average value, ¯r2, of 3 a.u., characteristic of the 2s 
state (Problem 5-2). How does this correspond to what we would find experimentally? 
Before answering this question, we must recall that there are special problems associated 
with measuring the properties of an atomic system. The process of “seeing” 
electrons in atoms well enough to pinpoint their positions perturbs an atom so strongly 
that it cannot be assumed to be in the same state after the measurement. To get around 
this problem, we can assume that we have a very large number of identically prepared 
helium atoms, and that a single measurement of electronic positions will be made on 
each atom. It is assumed that the average of the instantaneous r values for a billion 
systems is identical to the average r value for a billion instants in a single undisturbed 
system. 
When we consider the measurement of average values for r1 and r2 in helium, we 
immediately encounter another problem. Say we can effect a simultaneous measurement 
of the two electronic distances in the first He atom. We call these r1 and r2 and 
tabulate them for future averaging. Then we move on to a new helium atom and measure 
r1 and r2 for it. But we clearly have no way of identifying a particular one of these 
electrons with a particular one of the earlier pair. There is no connection between r1 
for one atom and r1 for the next since all electrons are identical. If we want to know ¯r, 
we can only average them all together and leave it at that. 
Thus, the wavefunction (5-11) does not seem to be entirely satisfactory since it 
enables us to calculate ¯r1 = ¯r2, something that is in principle impossible to measure. 
We need to modify the wavefunction so that it yields an average value for r1 and r2 
(or for any quantity) that is independent of our choice of electron labels. This means 
that the electron density itself, given by .(1, 2)2, must be independent of our electron 
labeling scheme. 
In a two-electron system like helium, there are only two ways to arrange the labels 
“1” and “2” in a single product function. For example, the product 1s2s can be written 
1s(1)2s(2) or 2s(1)1s(2) (5-12) 
Squaring these gives two different functions, namely, 
1s2(1)2s2(2)=(8/p) exp(-4r1)(1/p)(1-2r2 +r2
2 ) exp(-2r2) 
2s2(1)1s2(2)=(8/p) exp(-4r2)(1/p)(1-2r1 +r2
1 ) exp(-2r1) 
(5-13) 
These are different since they predict, for instance, different distributions for electron 1. 
The functions (5-12) are said to differ by an interchange of electron indices, or coordinates. 
(Since electron labels denote position coordinates, interchange of labels in 
the mathematical formula corresponds to interchanging positions of electrons in the 
physical model.) For .2 to be invariant under such an interchange, it is necessary that
Section 5-2 Simple Products and Electron Exchange Symmetry 131 
. itself be either symmetric or antisymmetric under the interchange. That is, if P is 
an interchange operator such that Pf (1, 2)=f (2, 1) then we need a . such that 
P. =±. (5-14) 
since then 
P(.2)=(P.)2 =(±.)2 =.2 
One such wavefunction is given by the sum of eigenfunctions (5-12), 
.s =(1/v2)[1s(1)2s(2)+2s(1)1s(2)] (5-15) 
since 
P.s =(1/v2)[1s(2)2s(1)+2s(2)1s(1)]=.s 
(the factor 1/v2 keeps the wavefunction normalized). Wavefunction (5-15) is thus 
symmetric under electron interchange. Is Eq. (5-15) still an eigenfunction for Happrox? 
Yes, because the eigenfunctions (5-12) are degenerate (both have E = 1s +2s) and 
can therefore be mixed together in any way we choose and still be eigenfunctions. The 
antisymmetric combination is 
.a =(1/v2)[1s(1)2s(2)-2s(1)1s(2)] (5-16) 
Thus far we have shown that simple products of atomic orbitals give us two degenerate 
eigenfunctions of Happrox for the configuration 1s2s and that these eigenfunctions 
fail to have the required symmetry properties for interchange of electron coordinates. 
But we have shown that, by taking the sum and difference of these simple products, we 
can form new eigenfunctions of Happrox that are respectively symmetric and antisymmetric 
with respect to the interchange of electron coordinates, so that .2 is invariant 
to electron interchange. 
There is another way we can demonstrate that the helium atom eigenfunctions ought 
to be symmetric or antisymmetric for electron exchange: We can examine the hamiltonian 
operator. We have shown in Chapter 2 that nondegenerate eigenfunctions must be 
symmetric or antisymmetric for any operation that leaves the hamiltonian unchanged 
and that degenerate eigenfunctions may always be mixed together in some combination 
so that they too are symmetric or antisymmetric. This suggests that, for the case 
under discussion (the helium atom), the hamiltonian operator might be unchanged by 
an exchange of electrons. First we examine Happrox: 
PHapprox =P[h(1)+h(2)]=h(2)+h(1)=Happrox (5-17) 
Our approximate hamiltonian is invariant to electron exchange, so any nondegenerate 
eigenfunctions must be symmetric or antisymmetric for interchange of electron labels 
(or positions). Only because the 1s2s configuration leads to degenerate eigenfunctions 
were we able to find unsymmetric2 eigenfunctions like Eq. (5-12). This situation is 
reminiscent of the particle-in-a-ring system discussed in Chapter 2, where degenerate, 
2A function is unsymmetric for any operation that produces neither plus nor minus that function; i.e., if Pf =y 
and y =±f, then f is unsymmetric under the operation P.
132 Chapter 5 Many-Electron Atoms 
symmetric exponential eigenfunctions could be mixed to form degenerate unsymmetric 
trigonometric eigenfunctions. Let us now examine the full hamiltonian H: 
PH(1, 2)=P[h(1)+h(2)+1/r12]=h(2)+h(1)+1/r21 =H(1, 2) (5-18) 
Since r12 and r21 are the same distance, it is evident that the exactH is likewise invariant 
to interchange of electron labels. Thus, we see that appeal either to physical argument 
or to the invariance of H and of Happrox to exchange of electrons leads us to recognize 
the need to impose symmetry conditions on the wavefunctions. 
We now summarize the points we have tried to convey in this section. 
1. A simple product function of the type 1s(1)2s(2) enables one to calculate different 
values of ¯r for electrons 1 and 2. This makes no physical sense since the electrons 
are identical particles and hence are not physically distinguishable. 
2. Wavefunctions that overcome this difficulty must be either symmetric or antisymmetric 
with respect to exchange of electron labels (coordinates). 
3. The fact that this kind of “exchange symmetry” must be present is also (or alternatively) 
seen from the fact that H (and also Happrox) is invariant under such an 
exchange operation. 
EXAMPLE 5-2 Given the functions f (x1)=x2
1 and g(x2)=exp(x2), show that, for 
x1 = 1, x2 = 2, f (x1)g(x2) is unsymmetric for exchange of the two x positions, 
f (x1)g(x2)+g(x1)f (x2) is symmetric, and f (x1)g(x2)-g(x1)f (x2) is antisymmetric. 
SOLUTION 
 For fg, we are examining what happens when x2
1 exp(x2) turns into x2
2 exp(x1). 
That is, we are comparing 12 exp(2) to 22 exp(1). The resulting values are 7.389 and 10.873— 
obviously neither plus or minus times each other. fg + gf equals 12 exp(2) + 22 exp(1), or 
7.389+10.873. After switching positions, we get 10.873+7.389, which is obviously the same 
thing. fg-gf gives 7.389-10.873. After switching, it gives 10.873-7.389, which is obviously 
minus one times the first value.  
5-3 Electron Spin and the Exclusion Principle 
Chemical and spectral evidence indicates that metals in Groups IA and IB of the periodic 
table are reasonably well represented by an electron configuration wherein one loosely 
held “valence” electron occupies an s orbital and all other electrons occur in pairs in 
orbitals of lower principal quantum number. Thus, lithium has a ground-state electronic 
structure approximated by the configuration 1s22s, sodium by 1s22s22p63s, copper by 
1s22s22p63s23p63d104s, etc. (A configuration indicating that all orbitals of given n 
and l are doubly occupied, leaving no other electrons, is often called a closed shell. Thus, 
the above-cited examples each consist of a closed shell plus one s valence electron.) 
The observation that each atomic orbital in such configurations is occupied by no more 
than two electrons was without a theoretical explanation for some time.
Section 5-3 Electron Spin and the Exclusion Principle 133 
When an atom like sodium is placed in an external magnetic field, what should 
be the magnetic moment of the atom due to orbital motion of the electrons? The 
s electrons should contribute nothing since, by definition of s, l = 0 and hence the 
magnetic quantum number m = 0 for such electrons. An electron in a p orbital may 
have an orbital magnetic moment, but if all p levels (l =1,m=+1, 0,-1) are equally 
occupied, the net magnetic moment should be zero. It is clear, then, that we might 
expect atoms in Groups IA and IB to possess no magnetic moment due to electron 
orbital motion. Nevertheless, Gerlach and Stern [1, 2] found that, when a beam of 
unexcited silver atoms is passed through an inhomogeneous magnetic field, it splits 
into two components as though each silver atom possesses a small magnetic moment 
capable of taking on either of two orientations in the applied field. (In a homogeneous 
magnetic field, the north and south poles of a magnetic dipole experience equal but 
oppositely directed forces, causing the dipole to become oriented. An example is a 
magnetic compass in the magnetic field of the earth. In an inhomogeneous magnetic 
field the poles experience opposite but unequal forces, causing the entire dipole to be 
accelerated through space in addition to being oriented.) Uhlenbeck and Goudsmit 
[3] and Bichowsky and Urey [4] independently suggested that the electron behaves 
as though it were a particle of finite radius spinning about its center of mass. Such a 
spinning particle would classically have angular momentum and, since it is charged, 
an accompanying magnetic moment.3 
If we accept the model of electron spin, then we can rationalize our experimental 
facts if we assume each electron is capable of being in one of but two possible states of 
opposite spin. This is done in the following way. If we attribute opposite spins to the 
two 1s electrons in, say, silver, their spin moments should cancel. Similarly, all other 
orbital-sharing electrons would contribute nothing if their spins were opposed. Only 
the outermost (5s) electron would have an uncanceled spin moment. Its two possible 
orientations would cause the beam to split into two components as is observed.4 
The evident need for the introduction of the concept of electron spin means that our 
wavefunctions of earlier sections are incomplete. We need a wavefunction that tells us 
not only the probability that an electron will be found at given r, ., f coordinates in 
three-dimensional space, but also the probability that it will be in one or the other spin 
state. Rather than seeking detailed mathematical descriptions of spin state functions, 
we will simply symbolize them a and ß. Then the symbol f(1)a(1) will mean that 
electron number 1 is in a spatial distribution corresponding to space orbital f, and that 
it has spin a. In the independent electron scheme, then, we could write the spin orbital 
(includes space and spin parts) for the valence electron of silver either as 5s(1)a(1) or 
5s(1)ß(1). These two possibilities both occur in the atomic beam and interact differently 
with the inhomogeneous magnetic field. 
We now focus on the manner in which spin considerations affect wavefunction 
symmetry. The electrons are still identical particles, so our particle distribution must be 
3This classical model, developed in the 1920s, is pedagogically useful and is responsible for the term spin, which 
is still employed to describe the fourth quantum number. However, itwas not until 1948 and 1967 that mathematical 
studies of the properties of linearized equivalents of the Sch¨odinger equation revealed the mathematical connection 
to this quantum number. For an entry to the literature, see Roman [10]. 
4Actually, other experimental evidence, such as splitting of atomic spectral lines due to applied magnetic fields, 
was also available. Furthermore, experience with the quantum theory of orbital angular momentum played a role 
in the treatment of electron spin. The reader should not think that the historical development of quantum theory 
of spin was as naive or simple as we make it appear here.
134 Chapter 5 Many-Electron Atoms 
insensitive to our choice of labels. This last statement is equivalent to saying that. must 
be symmetric or antisymmetric for interchange of electron space and spin coordinates. 
Let us examine this situation in the case of ground state helium and lithium atoms. 
In the independent electron approximation, the lowest-energy configuration for 
helium is 1s2. Let us write the various conceivable spin combinations for this configuration. 
They are 
1s(1)a(1)1s(2)a(2).... 
... 
.... 
... 
a(1)a(2) (5-19) 
1s(1)a(1)1s(2)ß(2) =1s(1)1s(2) 
a(1)ß(2) (5-20) 
1s(1)ß(1)1s(2)a(2) ß(1)a(2) (5-21) 
1s(1)ß(1)1s(2)ß(2) ß(1)ß(2) (5-22) 
It is easy to see that the common space term 1s(1)1s(2) is symmetric for electron interchange. 
Likewise, a(1)a(2) and ß(1)ß(2) are each symmetric, so Eqs. (5-19) and 
(5-22) are totally symmetricwavefunctions. The spin parts of Eqs. (5-20) and (5-21) are 
unsymmetric (not antisymmetric) for interchange, so these wavefunctions are not satisfactory. 
However, we can take the sum and difference of Eqs. (5-20) and (5-21) to obtain 
1s(1)1s(2) (1/v2)[a(1)ß(2)+ß(1)a(2)] (5-23) 
(1/v2)[a(1)ß(2)-ß(1)a(2)] (5-24) 
The 2-1/2 serves to maintain normality if we assume a and ß to be orthonormal: 
 a*(1)a(1)d.(1) =  ß*(1)ß(1)d.(1)=1 (5-25) 
 a*(1)ß(1)d.(1) =  ß*(1)a(1)d.(1)=0 (5-26) 
Here we use integrals and a differential element d. in a “spin coordinate ..” This is 
notationally convenient but not, for our purposes, worth delving into. We can interpret 
integration over . to be in effect equivalent to summing over the possible electron 
indices. If, for a particular electron index, the spins agree, then the integral equals unity. 
If they disagree, the integral vanishes. Wavefunction (5-23) consists of symmetric space 
and spin parts, so it is overall symmetric. Wavefunction (5-24) contains a symmetric 
space part times an antisymmetric spin part, so it is overall antisymmetric. We have 
succeeded, then, in writing down four wavefunctions for the configuration 1s2 having 
proper symmetry for electron interchange. Three of these, Eqs. (5-19), (5-22), (5-23), 
are symmetric and one, Eq. (5-24), is antisymmetric. Experimentally, we know that 
the ground state of helium is a singlet, that is, there is but one such state. This suggests 
that the wavefunction must be antisymmetric for exchange of electron space and spin 
coordinates. 
EXAMPLE 5-3 We have just shown four wavefunctions resulting from four spin 
functions times a symmetric space part (1s2). Can we manipulate the 1s2 configuration 
to obtain an antisymmetric space part, as we did for the 1s2s configuration?
Section 5-3 Electron Spin and the Exclusion Principle 135 
SOLUTION 
 We can try to produce an antisymmetric space function by taking the product 
difference 1s(1)1s(2)-1s(2)1s(1). Since these two products are really the same, this combination 
equals zero. Thus, whereas two different orbitals can be arranged in two product combinations, one 
symmetric and the other antisymmetric, we find that a single orbital, doubly occupied, can appear 
only in one simple product, which must be symmetric for electron exchange.  
Now let us try lithium. The lowest-energy configuration should be 1s3, and we can 
write eight unique space-spin orbital products: 
1s(1)1s(2)1s(3)
.............. 
............. 
a(1)a(2)a(3) (5-27) 
a(1)a(2)ß(3) (5-28) 
a(1)ß(2)a(3) (5-29) 
ß(1)a(2)a(3) (5-30) 
a(1)ß(2)ß(3) (5-31) 
ß(1)a(2)ß(3) (5-32) 
ß(1)ß(2)a(3) (5-33) 
ß(1)ß(2)ß(3) (5-34) 
Of these, the first and last are totally symmetric for all electron interchanges. The 
remaining six are unsymmetric for two out of three possible interchanges. Can we 
make appropriate linear combinations of these as we did for helium? Let us try. The 
problem is simplified by recognizing that, if we start with, say, two a’s and one ß, 
we still have that number of a’s and ß’s after interchange of electron labels. Hence, 
we mix together only functions that agree in total numbers of a’s and ß’s, i.e., (5-28), 
(5-29), (5-30) with each other, or (5-31), (5-32), (5-33) with each other. Let us try the 
sum of (5-28), (5-29), and (5-30). Ignoring normalization, this gives the spin function 
a(1)a(2)ß(3)+a(1)ß(2)a(3)+ß(1)a(2)a(3) (5-35) 
Interchanging electron spin coordinates 1 and 2 gives 
a(2)a(1)ß(3)+a(2)ß(1)a(3)+ß(2)a(1)a(3) 
which, upon reordering each product, is easily seen to be identical to (5-35). The same 
result arises from interchanging 1 and 3 or 2 and 3, and so (5-35) is symmetric for all 
interchanges. The sum of (5-31), (5-32), and (5-33) is likewise symmetric. Can we find 
any combinations that are totally antisymmetric? A few attempts with pencil and paper 
should convince one that it is impossible to find a combination that is antisymmetric 
for all interchanges. Experimentally, we know that no state of lithium corresponds to 
a 1s3 configuration. 
To summarize, we have found that for the configuration 1s2 we can write three wavefunctions 
that are symmetric and one that is antisymmetric under exchange of electron 
space and spin coordinates, while for the configuration 1s3 we can construct symmetric 
or unsymmetric wavefunctions, but no antisymmetric ones. The physical observation
136 Chapter 5 Many-Electron Atoms 
is that atoms exist in only one state having an electronic structure approximately represented 
by the configuration 1s2, but that there are no atoms having any state represented 
by 1s3. This and other physical evidence has led to the recognition of the exclusion principle: 
Wavefunctions must be antisymmetric with respect to simultaneous interchange 
of space and spin coordinates of electrons.5 In invoking the exclusion principle, we 
exclude all of the 1s3 wavefunctions and three out of the four wavefunctions we were 
able to construct for the ground state of helium, leaving (5-24) as the only acceptable 
wave-function. 
We have seen that the ground state configuration of lithium cannot be 1s3. Can we 
satisfy the exclusion principle with the next-lowest energy configuration 1s22s? We 
will try to find a satisfactory solution, but our manipulations will be simplified if we 
streamline our notation. We will write a function such as 1s(1)1s(2)2s(3)a(1)ß(2)a(3) 
as 1s1s2saßa, allowing position in the sequence to stand for the electron label. Interchanging 
electrons 1 and 2 is then represented by switching the order of space functions 
in positions 1 and 2 and spin functions in positions 1 and 2 thus 
1s1s2saßa 
12 -.1s1s2sßaa (5-36) 
This interchange produced a new function rather than merely reversing the sign 
of our starting function. But if we take the difference between the two products 
in Eq. (5-36), we will have a function that is antisymmetric to 1, 2 interchange: 
1s1s2s(aßa -ßaa). Now we subject this to a 1, 3 interchange and the new products 
produced are subtracted to give a function that is antisymmetric to 1, 3 interchange: 
1s1s2s(aßa -ßaa)-2s1s1s(aßa -aaß). The first pair of terms is still not antisymmetric 
to 2, 3 interchange, and the second pair is not antisymmetric to 1, 2 interchange. 
We can use either one of these interchanges to produce two new terms to subtract. 
Either way, the resulting wavefunction, antisymmetric for all interchanges, is 
1 
v6[1s1s2s(aßa -ßaa)+1s2s1s(ßaa -aaß)+2s1s1s(aaß -aßa)] (5-37) 
The factor 6-1/2 normalizes (5-37) since all of the six space-spin products are normalized 
and orthogonal to each other product by virtue of either space-orbital or spin-orbital 
disagreement, or both. Note that, whereas the two-electron wavefunction for helium 
was separable into a single space function times a spin function, the lithium wavefunction 
must be written as a linear combination of such products. This is usually true when 
we deal with more than two electrons. 
Since 1s22s is the lowest-energy configuration for which we can write an antisymmetrized 
wavefunction, this is the ground state configuration for lithium in this 
independent-electron approximation. 
In summary, phenomenological evidence suggests that an electron can exist in either 
of two “spin states” in the presence of a magnetic field. Writingwavefunctions including 
spin functions and comparing these with experimental facts indicates that states exist 
only for wavefunctions that satisfy the exclusion principle. 
5A broader statement is: Wavefunctions must be antisymmetric (symmetric) with respect to simultaneous 
interchange of space and spin coordinates of fermions (bosons). A fermion is characterized by half-integral 
spin quantum number; a boson is characterized by integral spin quantum number. Electrons have spin quantum 
number 1
2 and are therefore fermions.
Section 5-4 Slater Determinants and the Pauli Principle 137 
5-4 Slater Determinants and the Pauli Principle 
It was pointed out by Slater [5] that there is a simple way to write wavefunctions 
guaranteeing that they will be antisymmetric for interchange of electronic space and spin 
coordinates: one writes the wavefunction as a determinant. For the 1s22s configuration 
of lithium, one would write 
. = 
1 
v6
 
1s(1)a(1) 1s(2)a(2) 1s(3)a(3) 
1s(1)ß(1) 1s(2)ß(2) 1s(3)ß(3) 
2s(1)a(1) 2s(2)a(2) 2s(3)a(3)
 
(5-38) 
Expanding this according to the usual rules governing determinants (seeAppendix 2) 
gives 
. = 
1 
v6[1s(1)a(1)1s(2)ß(2)2s(3)a(3)+2s(1)a(1)1s(2)a(2)1s(3)ß(3) 
+1s(1)ß(1)2s(2)a(2)1s(3)a(3)-2s(1)a(1)1s(2)ß(2)1s(3)a(3) 
-1s(1)ß(1)1s(2)a(2)2s(3)a(3)-1s(1)a(1)2s(2)a(2)1s(3)ß(3)] (5-39) 
This can be factored and shown to be identical to wavefunction (5-37) of the preceding 
section. 
A simplifying notation in common usage is to delete the a, ß symbols of the spinorbitals 
and to let a bar over the space orbital signify ß spin, absence of a bar being 
understood to signify a spin. In this notation, Eq. (5-38) would be written 
. = 
1 
v6
 
1s(1) 1s(2) 1s(3) 
1¯s(1) 1¯s(2) 1¯s(3) 
2s(1) 2s(2) 2s(3)
 
(5-40) 
The general prescription to follow in writing a Slater determinantal wavefunction is 
very simple: 
1. Choose the configuration to be represented. 1s1¯s2s was used above. (Here we write 
1s1¯s2s rather than 1s22s to emphasize that the two 1s electrons occupy different spinorbitals.) 
For our general example, we will let Ui stand for a general spin-orbital 
and take a four-electron example of configuration U1U2U3U4. 
2. For n electrons, set up an n×n determinant with (n!)-1/2 as normalizing factor. 
Every position in the first row should be occupied by the first spin-orbital of the 
configuration; every position in the second row by the second spin-orbital, etc. Now 
put in electron indices so that all positions in column 1 are occupied by electron 1, 
column 2 by electron 2, etc. 
In the case of our four-electron configuration, the recipe gives 
. = 
1 
v4!
 
U1(1) U1(2) U1(3) U1(4) 
U2(1) U2(2) U2(3) U2(4) 
U3(1) U3(2) U3(3) U3(4) 
U4(1) U4(2) U4(3) U4(4)
 
(5-41)
138 Chapter 5 Many-Electron Atoms 
Notice that the principal diagonal (top left to bottom right) contains our original configuration 
U1U2U3U4. Often, the Slater determinant is represented in a space-saving 
way by simply writing the principal diagonal between short vertical bars. The normalizing 
factor is deleted. Thus, Eq. (5-41) would be symbolized as |U1(1)U2(2) 
U3(3)U4(4)|. 
We have indicated the general recipe for writing down a Slater determinant, and we 
have seen that, for the configuration 1s1¯s2s, this gives an antisymmetric wavefunction. 
Now we will give a general proof of the antisymmetry of such wavefunctions 
for exchange of electrons. We have already seen that interchanging the space and spin 
coordinates of electrons 1 and 2 corresponds to going through the wavefunction and 
changing all the 1s to 2s and vice versa; i.e., electron labels denote coordinates. In 
a Slater determinant, interchanging electron labels 1 and 2 is the same thing as interchanging 
columns 1 and 2 of the determinant. [See Eq. (5-41) and note that columns 1 
and 2 differ only in electron index.] But a determinant reverses sign upon interchange 
of two rows or columns. (See Appendix 2 for a summary of the properties of determinants.) 
Hence, any Slater determinant reverses sign (i.e., is antisymmetric) upon the 
interchange of space and spin coordinates of any two electrons. 
Suppose we tried to put two electrons into the same space-orbital with the same spin. 
This would require that the same spin-orbital be written twice in the configuration, 
causing two rows of the Slater determinant to be identical. [If both 1s electrons in 
Eq. (5-40) had a spin, the bars would be absent from row 2.] We just stated that 
the determinant must reverse sign upon interchange of two rows. If we interchange 
two identical rows, we change nothing yet the sign must reverse: the determinant 
must be equal to zero. Thus, the determinantal wavefunction vanishes when we try 
to put more than one electron into the same spin-orbital, indicating that this is not a 
physically allowed situation. This is a generalization of our earlier discovery that no 
1s3 configuration is allowed by the exclusion principle, such a configuration requiring 
at least two electrons to have the same space and spin functions. 
This restriction on allowable electronic configurations is more familiar to chemists as 
the Pauli principle: In assigning electrons to atomic orbitals in the independent electron 
scheme, no two electrons are allowed to have all four quantum numbers (n, l, m, spin) 
the same. The Pauli principle is a restatement of the exclusion principle as it applies in 
the special case of an orbital approximation to the wavefunction. 
5-5 Singlet and Triplet States for the 1s2s Configuration 
of Helium 
We showed in Section 5-2 that two space functions having proper space symmetry 
could be written for the configuration 1s2s. One was symmetric (Eq. 5-15) and one 
was antisymmetric (Eq. 5-16). Now we find that spin functions must be included in 
our wavefunctions, and in a way that makes the final result antisymmetric when space 
and spin coordinates are interchanged. We can accomplish this by multiplying the 
symmetric space function by an antisymmetric spin function, calling the result .s,a. 
Thus, 
.s,a(1, 2)=(1/v2)1s(1)2s(2)+2s(1)1s(2)(1/v2)a(1)ß(2)-ß(1)a(2) (5-42)
Section 5-5 Singlet and Triplet States for the 1s2s Configuration of Helium 139 
Alternatively, we can multiply the antisymmetric space term by any one of the three 
possible symmetric spin terms: 
.a,s(1, 2)=(1/v2)1s(1)2s(2)-2s(1)1s(2)
... 
.. 
a(1)a(2) (5-43a) 
(1/v2)a(1)ß(2)+ß(1)a(2) (5-43b) 
ß(1)ß(2) (5-43c) 
All four of these wavefunctions satisfy the exclusion principle and each is linearly 
independent of the others, indicating that four distinct physical states arise from the 
configuration 1s2s. 
There are a number of important points that can be illustrated using these wavefunctions. 
The first has to do with Slater determinants. Let us write down a Slater 
determinantal expression corresponding to wavefunction (5-43a). The configuration is 
1s(1)a(1)2s(2)a(2), giving the Slater determinant (where absence of a bar indicates 
a spin) 
.a,s(1, 2)= 
1 
v2
 
1s(1) 1s(2) 
2s(1) 2s(2)
 
(5-44) 
which, upon expansion, gives us Eq. (5-43a). If we attempt the same process to 
obtain Eq. (5-43b), we encounter a difficulty. The configuration 1s(1)a(1)2s(2)ß(2) 
leads to a 2 × 2 determinant, which, upon expansion, gives two product terms, 
whereas Eq. (5-43b) involves four product terms. The Slater determinantal functions 
corresponding to Eqs. (5-42) and (5-43b) are, in fact, 
.s,a 
a,s
(1, 2)= 
1 
v2 
 1 
v2
 
1s(1) 1s(2) 
2¯s(1) 2¯s(2)

± 
1 
v2
 
1¯s(1) 1¯s(2) 
2s(1) 2s(2)
 
 (5-45) 
The lesson to be gained from this is that a single Slater determinant does not always 
display all of the symmetry possessed by the correct wavefunction. (In this particular 
case, a single determinant restricts one of the AOs to a spin and the other to ß, which 
is an artificial limitation.) 
Next we will investigate the energies of the states as they are described by these 
wavefunctions. We have already pointed out that they are degenerate eigenfunctions 
of Happrox, but we will now examine their interactions with the full hamiltonian (5-2). 
Since ourwavefunctions are not eigenfunctions of this hamiltonian, we cannot compare 
eigenvalues. Instead we must calculate the average values of the energy for each 
wavefunction, using the formula 
E¯ =  .*H. dt 
 .*. dt 
(5-46) 
The symbol “dt” stands for integration over space and spin coordinates of the electrons: 
dt = dv d.. Since both space and spin parts of our wavefunctions are normalized 
[cf. Eqs. (5-25) and (5-26)], the denominator of Eq. (5-46) is unity and may be ignored. 
The energy thus is given by the expression 
E¯ = .* -
1
2.2
1 - 
1
2.2
2 -(2/r1)-(2/r2)+(1/r12). dt (5-47)
140 Chapter 5 Many-Electron Atoms 
Notice that the energy operator H contains no terms that would interact with spin 
functions a and ß. (Such terms do arise at higher levels of refinement, but we ignore 
them for now.) Hence, the spin terms of . can be integrated separately, and, since all 
spin factors in Eqs. (5-42) and (5-43) are normalized, this gives a factor of unity in all 
four cases. This means that the average energies will be entirely determined by the 
space parts of the wavefunctions. This, in turn, means that all three states (5-43), which 
have the same space term, will have the same energy but that the state approximated by 
the function (5-42) may have a different energy. If our approximate representation of 
the exact eigenfunctions is physically realistic, we expect helium to display two excited 
state energies in the energy range consistent with a 1s2s configuration. Furthermore, 
we expect one of these state energies to be triply degenerate. 
Which of these two state energies should be higher? To determine this requires that 
we expand our energy expression (5-47) for each of the two space functions (5-42) 
and (5-43). 
¯ E13
= 
1
2 
 [1s*(1)2s*(2)±2s*(1)1s*(2)]-
1
2.2
1 - 
1
2.2
2 -(2/r1) 
-(2/r2)+(1/r12) [1s(1)2s(2)±2s(1)1s(2)]dv(1)dv(2) (5-48) 
(The subscript on E¯ refers to the degeneracy of whichever energy level we are considering.) 
This expands into a large number of terms. Integrals over one-electron operators 
may be written as products of two integrals, each over a different electron.6 Thus, the 
expansion over the kinetic energy operators gives 
(5-49) 
6We have already shown that, if the 1/r12 term is absent, the energy is equal to E1s +E2s, for He+, which is 
equal to -2.5 a.u. Therefore, the detailed breakdown leading to Eqs. (5-49)–(5-51) is not necessary. However, we 
will present it in detail in the belief that some students will benefit from another specific example of integration 
of two-electron products over one-electron operators.
Section 5-5 Singlet and Triplet States for the 1s2s Configuration of Helium 141 
The orthogonality of the 1s and 2s orbitals causes the terms preceded by ± to vanish. 
Furthermore, integrals that differ only in the variable label [such as those in the second 
and third terms of (5-49)] are equal, so that this expansion becomes 
 1s*(1) -
1
2.2
1
 1s(1)dv(1)+ 2s*(1) -
1
2.2
1
 2s(1)dv(1) (5-50) 
Expansion of Eq. (5-48) over (-2/r1 -2/r2) proceeds analogously to give 
 1s*(1)(-2/r1)1s(1)dv(1)+ 2s*(1) (-2/r1) 2s(1)dv(1) (5-51) 
The final term in the hamiltonian, 1/r12, occurs in four two-electron integrals: 
1
2 
 1s*(1)2s*(2)(1/r12)1s(1)2s(2)dv(1)dv(2) 
+  2s*(1)1s*(2)(1/r12)2s(1)1s(2)dv(1)dv(2) 
±  1s*(1)2s*(2)(1/r12)2s(1)1s(2)dv(1)dv(2) 
±  2s*(1)1s*(2)(1/r12)1s(1)2s(2)dv(1)dv(2) (5-52) 
The first two integrals of (5-52) differ only by an interchange of labels “1” and “2 ,” and 
so they are equal to each other. The same is true of the second pair. Thus, the average 
value of the energy is 
¯ E13
=  1s*(1) -
1
2.2
1
 1s(1)dv(1)+ 1s*(1) [-2/r1] 1s(1)dv(1) 
+ 2s*(1) -
1
2.2
1
 2s(1)dv(1) + 2s*(1) [-2/r1] 2s(1)dv(1) 
+ 1s*(1)2s*(2)(1/r12)1s(1)2s(2)dv(1)dv(2) 
±  1s*(1)2s*(2)(1/r12)2s(1)1s(2)dv(1)dv(2) (5-53) 
Notice that, since -1
2.2 -2/r is the hamiltonian for He+, the first two integrals of 
Eq. (5-33) combine to give the average energy of He+ in its 1s state. The second pair 
gives the energy for He+ in the 2s state. Thus, Eq. (5-53) can be written 
¯ E13
=E1s +E2s +J ±K (5-54) 
where J and K represent the last two integrals in Eq. (5-53). No bars appear on E1s 
or E2s because these “average energies” are identical to the eigenvalues for the He+ 
hamiltonian (Problem 5-15). 
The integral J denotes electrons 1 and 2 as being in “charge clouds” described by 
1s*1s and 2s*2s, respectively. The operator 1/r12 gives the electrostatic repulsion 
energy between these two charge clouds. Since these charge clouds are everywhere negatively 
charged, all the interactions are repulsive, and it is necessary that this “coulomb
142 Chapter 5 Many-Electron Atoms 
Figure 5-2 
 The function produced by multiplying together hydrogenlike 1s and 2s orbitals. R 
is the radius of the spherical nodal surface. 
integral” J be positive. Alternatively, we can argue that the functions 1s*1s, 2s*2s, 
and 1/r12 are everywhere positive, so the integrand of J is everywhere positive and J 
must be positive. 
The integral K is called an “exchange integral” because the two product functions in 
the integrand differ by an exchange of electrons. This integral gives the net interaction 
between an electron “distribution” described by 1s*2s, and another electron in the same 
distribution. (These distributions are mathematical functions, not physically realizable 
electron distributions.) The 1s2s function is sketched in Fig. 5-2. Because the 2s orbital 
has a radial node, the 1s2s function (which is the same as 1s*2s since the 1s function 
is real) also has a radial node. Now the function 1s(1)2s(1)1s(2)2s(2) will be positive 
whenever r1 and r2 are either both smaller or both larger than the radial node distance 
(R in Figure 5-2). But when one r value is smaller than R and the other is greater, 
corresponding to the electrons being on opposite sides of the nodal surface, the product 
1s(1)2s(1)1s(2)2s(2) is negative. These positive and negative contributions to K are 
weighted by the function 1/r12, which is always positive and hence unable to affect 
the sign of the integrand. But 1/r12 is smallest when the electrons are far apart. This 
means that 1/r12 tends to reduce the contributions where the electrons are on opposite 
sides of the node (i.e., the negative contributions), and so the value of K turns out to be 
positive (although not as large in magnitude as J , which has no negative contribution 
at all). 
Since the integralK is positive, we can see from Eq. (5-54) that the triply degenerate 
energy level lies below the singly degenerate one, the separation between them being 
2K. (We note in passing that these independent-electron wavefunction energies are not 
simply the sum of one-electron energies as was the case when we used Happrox, thereby 
ignoring interelectronic repulsion.) 
The experimental observation agrees qualitatively with these results. There are two 
state energies associated with the 1s2s configuration. When the atom is placed in an 
external magnetic field, the lower-state-energy-level splits into three levels. The state 
having the higher energy has a “multiplicity” of one and is called a singlet. The lowerenergy 
with multiplicity three is called a triplet. (The reference to a “triplet state” 
should not be construed to mean that this is one state. It is a triplet of states.) 
It is possible to use vector arguments similar to those presented in Chapter 4 to understand 
why the triply degenerate level splits into three different levels in the presence of 
a homogeneous magnetic field. Let us first consider the case of a single electron. We 
have already indicated that two spin states are possible, which we have labeled a and ß.
Section 5-5 Singlet and Triplet States for the 1s2s Configuration of Helium 143 
In a magnetic field the angular momentum vectors precess about the field axis z, as 
depicted in Figure 5-3. The z components of the angular momentum vectors are constant 
but the x and y components are not. Because the allowed z components must 
in general differ by one atomic unit (stated but not proved in Chapter 4), and because 
there are but two allowed values (inferred from observations such as the splitting of 
a beam of silver atoms into two components), and because the two kinds of state are 
oppositely affected by magnetic fields, it is concluded that the z components of angular 
momentum (labeled ms) are equal to +1
2 and -1
2 a.u. for a and ß, respectively. Following 
through using orbital angular momentum relations as a model, we postulate an 
electron spin quantum number s equal to the maximum z-component of spin angular 
momentum in a.u., 1
2 , and a spin angular momentum vector s having length vs(s +1) 
a.u. The degeneracy, gs =2s +1, equals 2, in agreement with the two orientations in 
Fig. 5-3. 
As noted in Chapter 4, half-integer quantum numbers correspond to eigenfunctions 
that cannot be expressed as spherical harmonics. We will not pursue the question 
of detailed expressions for a and ß here. (However, see the problems at the end of 
Chapter 9.) 
Now let us turn to the two-electron system. We will assume that the magnetic 
moments of the two electrons interact independently with the external field. This 
ignores the fact that each electron senses a small contribution to the magnetic field 
resulting from the magnetic moment of the other electron. 
Another factor that could affect the magnetic field sensed by the spin moment is the 
magnetic moment resulting from the orbital motions of the electrons, although this is 
not present if both electrons are taken to be in s atomic orbitals (AOs). For two electrons, 
we can imagine four situations: aa, aß, ßa, and ßß. We pointed out earlier (Section 
4-5) that, for a system composed of several moving parts, the total angular momentum 
is the sum of the individual angular momenta, and the z component is the sum of the 
individual z components. For the four spin situations listed above, this means that the 
net z components of spin angular momentum (labeled Ms) are +1, 0, 0, and -1 a.u., 
respectively. The spin combinations aß + ßa [from the triplet (5-43b)] and aß -ßa 
Figure 5-3 
 The angular momentum vectors for a and ß precess around the magnetic field axis 
z. The z components of these vectors are constant and have values of +1
2 and -1
2 a.u. respectively.
144 Chapter 5 Many-Electron Atoms 
Figure 5-4 
 Energy levels for singlet and triplet levels of 1s2s helium in (a) absence, and 
(b) presence of an external magnetic field. 
[from the singlet (5-42)] are linear combinations of aß and ßa. However, since these 
two functions have the same value for the z component of angular momentum (zero), 
their linear combinations will also have that value. It follows that the z components 
of the spin angular momenta of the triplet of states (5-43) are +1, 0, and -1 a.u., and 
for the singlet (5-42) it is zero. Because the electrons are charged, these spin angular 
momenta correspond to spin magnetic moments, which interact differently with the 
applied magnetic field to give splitting of the triplet (see Figure 5-4). It is customary 
to refer to all three spin states in (5-43) as having parallel spins even though the vector 
diagram for the (5-43b) state is not particularly in accord with this terminology. For 
the singlet state, the spins are said to be opposed, or antiparallel. 
5-6 The Self-Consistent Field, Slater-Type Orbitals, 
and the Aufbau Principle 
Up to now we have used wavefunctions that, while not being eigenfunctions of the 
hamiltonian, are eigenfunctions of an “effective hamiltonian” obtained by ignoring the 
interelectronic repulsion operator 1/rij . That is, these wavefunctions would be exactly 
correct if the electrons in helium were attracted by the nucleus, but did not repel each 
other. For this reason, we have referred to this as an independent electron approximation. 
Because we have neglected interelectronic repulsion, we cannot expect such 
a wavefunction to give very good numerical predictions of charge density or energy. 
We can compare the energy of He in the 1s2 (ground) state as predicted by our independent 
electron wavefunction and Happrox (-108.84 eV) with the experimental value 
(-79.0143 eV). (See Table 5-1.) This shows that the predicted energy is much too low, 
which is understandable since we have neglected an important repulsive (hence positive) 
interaction energy. But we can account for much of this neglected energy by calculating 
the average value of the energy using H (with 1/r12 included) instead of Happrox . This 
gives a value of-74.83 eV—much better, though nowtoo high by more than 4 eV. Even 
though we have now accounted for interelectronic repulsion, there is still a problem: 
Because we ignored interelectronic repulsion in arriving at these wavefunctions, they 
predict electron densities that are too large near the nucleus. In reality, interelectronic 
repulsion prevents so much build-up of charge density. Methods have been devised that 
partially overcome this problem by retaining the convenient form of orbital products 
but modifying the formulas for the orbitals themselves to make them more diffuse.
Section 5-6 The Self-Consistent Field, Slater-Type Orbitals, and the Aufbau Principle 145 
TABLE 5-1 
 Average Values for Energy Calculated from Helium Atom Ground State 
ApproximateWavefunctionsa 
Wavefunction description E¯(eV) 
1) Product of He+ orbitals -74.83 
2) Product of hyrdrogenlike orbitals with . fixed by SCF 
method -77.48 
3) Best product-type wavefunction -77.870917 
4) Nonorbital wavefunction of Pekeris [9]. This wavefunction 
uses functions of r1, r2 and r12 as coordinates and has 
the form of an exponential times a linear combination of 
1078 terms -79.00946912 
aE¯ = .*H.dt/ .*.dt , where H is given by Eq. (5-2). 
Let the ground state of helium be our example. We take the ordinary independentelectron 
wavefunction as our initial approximation: 
1s(1)1s(2)=8/p exp(-2r1)8/p exp(-2r2) (5-55) 
These atomic orbitals are correct only if electrons 1 and 2 do not “see” each other via 
a repulsive interaction. They really do repel each other, and we can approximate this 
repulsion by saying that electron 2 “sees” electron 1 as a smeared out, time-averaged 
charge cloud rather than the rapidly moving point charge that is actually present. The 
initial description for this charge cloud is just the absolute square of the initial atomic 
orbital occupied by electron 2: [1s(2)]2. Our approximation now has electron 1 moving 
in the field of a positive nucleus embedded in a spherical cloud of negative charge. 
Thus, for electron 1, the positive nuclear charge is “shielded” or “screened” by electron 
2. Hence, electron 1 should occupy an orbital that is less contracted about the nucleus. 
Let us write this new orbital in the form 
1s(1)=. 3/p exp(-.r1) (5-56) 
where . is related to the screened nuclear charge seen by electron 1. The mathematical 
methods used to evaluate . will be described later in this book. 
Next we turn to electron 2, which we now take to be moving in the field of the 
nucleus shielded by the charge cloud due to electron 1, now in its expanded orbital. 
Just as before, we find a new orbital of form (5-56) for electron 2. The value of . 
that we calculate for electron 2, however, will be different from what we found for 
electron 1 because the shielding of the nucleus by electron 1 is different from what it 
was by electron 2 in our previous step. We now have a new distribution for electron 
2, but this means that we must recalculate the orbital for electron 1 since this orbital 
was appropriate for the screening due to electron 2 in its old orbital. After revising 
the orbital for electron 1, we must revise the orbital for electron 2. This procedure 
is continued back and forth between electrons 1 and 2 until the value of . converges 
to an unchanging value (under the constraint that electrons 1 and 2 ultimately occupy 
orbitals having the same value of . ). Then the orbital for each electron is consistent
146 Chapter 5 Many-Electron Atoms 
with the potential due to the nucleus and the charge cloud for the other electron: the 
electrons move in a “self-consistent field” (SCF). 
The result of such a calculation is a wavefunction in much closer accord with the 
actual charge density distributions of atoms than that given by the complete neglect of 
interelectron repulsion.7 A plot of the electron density distribution in helium as given 
by wavefunction (5-55) and by a similar wavefunction with optimized . is given in 
Fig. 5-5. Because each electron senses only the time-averaged charge cloud of the 
other in this approximation, it is still an independent-electron treatment. The hallmark 
of the independent electron treatment is a wavefunction containing only a product of 
one-electron functions. It will not contain functions of, say, r12, which would make . 
depend on the instantaneous distance between electrons 1 and 2. 
Atomic orbitals that are eigenfunctions for the one-electron hydrogenlike ion (for 
integral or nonintegral Z) are called hydrogenlike orbitals. In Chapter 4 we noted that 
many hydrogenlike orbitals have radial nodes. In actual practice, this mathematical 
aspect causes increased complexity in solving integrals in quantum chemical calculations. 
Much more convenient are a class of modified orbitals called Slater-type orbitals 
(STOs). These differ from their hydrogenlike counterparts in that they have no radial 
nodes. Angular terms are identical in the two types of orbital. The unnormalized radial 
term for an STO is 
R(n,Z, s)=r(n-1) exp[-(Z -s)r/n] (5-57) 
Figure 5-5 
 Electron distribution in helium as given by SCF and unshielded independent electron 
approximations. 
7In practice, mathematical techniques have been found that lead to a self-consistent solution without explicit 
iteration between evolving AOs that converge to some final optimized . . Examples are described in Chapters 7 
and 12. A thorough discussion of the SCF and related methods is given in Chapter 11.
Section 5-6 The Self-Consistent Field, Slater-Type Orbitals, and the Aufbau Principle 147 
where Z is the nuclear charge in atomic units, n is the principal quantum number, and 
s is a “screening constant” which has the function of reducing the nuclear charge Z 
“seen” by an electron. Slater [7] constructed rules for determining the values of s 
that will produce STOs in close agreement with those one would obtain by an SCF 
calculation. These rules, appropriate for electrons up to the 3d level, are 
1. The electrons in the atom are divided up into the following groups: 1s|2s,2p|3s, 
3p|3d. 
2. The shielding constant s for an orbital associated with any of the above groups is 
the sum of the following contributions: 
a) Nothing from any electrons in groups to the right (in the above list) of the group 
under consideration 
b) 0.35 from each other electron in the group under consideration (except 0.30 in 
the 1s group) 
c) If the orbital under consideration is an s or p orbital, 0.85 for each electron with 
principal quantum number less by 1, and 1.00 for each electron still “farther in”; 
for a d orbital, 1.00 for all electrons farther in 
For example, nitrogen, with ground state configuration 1s22s2 2p3, would have the 
same radial part for the 2s and 2p STOs. This would be given by the formula (n=2, 
Z =7, s =4×0.35+2×0.85=3.1) 
R2s,2p(2, 7, 3.1)=r(2-1) exp[-(7-3.1)r/2]=r exp(-1.95r) 
For the 1s level, n=1,Z =7, s =0.30, and 
R1s =exp(-6.7r) 
Comparing orbital exponents, we see that the 1s charge cloud is compressed much 
more tightly around the nucleus than are the 2s and 2p “valence orbital” charge clouds. 
Slater-type orbitals are very frequently used in quantum chemistry because they provide 
us with very good approximations to self-consistent field atomic orbitals (SCF–AOs) 
with almost no effort. 
Clementi and Raimondi [8] have published a refined list of rules for the shielding 
constant, which extends to the 4p level. Their rules include contributions to shielding 
due to the presence of electrons in shells outside the orbital under consideration. Such 
contributions are not large, and, up to the 3d level, there is reasonably good agreement 
between these two sets of rules. 
The fact thatSTOs have no radial nodes results in some loss of orthogonality. Angular 
terms still give orthogonality between orbitals having different l ormquantum numbers, 
but STOs differing only in their n quantum number are nonorthogonal. Thus, 1s, 2s, 
3s, . . . are nonorthogonal. Similarly 2pz, 3pz , . . . or 3dxz, 4dxy, . . . are nonorthogonal. 
In practice, this feature is handled easily. The only real problem arises if one forgets 
about this nonorthogonality when making certain calculations. 
When carrying out SCF calculations on multielectronic atoms, one finds that the 
orbital energies for 2s and 2p functions are not the same. Similarly, 3s 3p, and 3d orbitals 
are nondegenerate. Yet these orbitals were degenerate in the one-electron hydrogenlike
148 Chapter 5 Many-Electron Atoms 
system in which energy was a function of n but not of l or m. Why are these orbital 
energies nondegenerate in the many-electron calculation? Areasonable explanation can 
be found by considering the comparative effectiveness with which a pair of 1s electrons 
screen the nucleus from a 3s or a 3p electron. Comparing the 3s, 3p, and 3d hydrogenlike 
orbital formulas in Table 4-2 shows that the 3s orbital is finite at the nucleus, decreasing 
proportionally to r for small r. The 3p orbitals vanish at the nucleus but grow as r for 
small r. The 3d orbitals vanish at the nucleus but grow as r2 for small r. The result of 
all this is that an s electron spends a larger amount of its time near the nucleus than a p 
electron of the same principal quantum number, the p electron spending more time near 
the nucleus than the d, etc. Hence, the s electron penetrates the “underlying” charge 
clouds more effectively and is therefore less effectively shielded from the nucleus. 
Since the s electron sees a greater effective nuclear charge, its energy is lower than 
that of the p electron. (This effect is not obvious in STOs since the 3s and 3p STOs 
have the same radial function which vanishes at the nucleus. However, the STO for 3d 
does reflect the nondegeneracy since Slater’s rules give it a different screening constant 
from 3s or 3p.) 
The tendency for higher l values to be associated with higher orbital energies leads 
to the following orbital ordering: 
1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 5d 4f 6p 7s 6d 5f . . . (5-58) 
When we get to principal quantum numbers of 3 and higher, the energy differences 
between different l values for the same n become comparable to the differences between 
different n levels. Thus, in some atoms, the 4s level is almost the same as the 3d level, 
etc.
In compiling data on ground states of atoms, Hund noticed that greatest stability 
results if the AOs in a degenerate set are half-filled with electrons before any of them 
are filled. This generalization, called Hund’s rule, is sometimes stated in an alternative 
form: Of the states associated with the ground state configuration of an atom or ion, 
those with greatest spin multiplicity lie deepest in energy. Chemists generally find the 
former version to be more convenient, spectroscopists the latter. The reason for the 
equivalence of these statements will emerge later in this chapter. 
EXAMPLE 5-4 An unexcited Fe atom has an electronic configuration of 
1s22s22p63s23p64s23d6. What is its spin multiplicity? 
SOLUTION 
 All electrons below 3d6 are spin-paired in orbitals, hence contribute nothing to 
MS. In 3d6, we have 4 electrons that can each occupy a 3d AO alone. If we follow Hund’s 
rule and seek maximum spin multiplicity, we make all their spins the same (a). Then maximum, 
MS =4 · 1
2 =2, so S =2, and spin multiplicity is 2S +1=5. The atom has a quintet ground “state” 
(really five states).  
The energy-ordering scheme (5-58) coupled with the Pauli or exclusion principle 
and Hund’s rule leads us to a simple prescription for “building up” the electronic configurations 
of atoms. This aufbau principle is familiar to chemists and leads naturally 
to a correlation between electronic structure and the periodic table. The procedure is to 
place all the electrons of the atom into atomic orbitals, two to an orbital, starting at the
Section 5-7 Electron Angular Momentum in Atoms 149 
low-energy end of the list (5-58) and working up in energy. In addition, when filling a 
set of degenerate levels like the five 3d levels, one half-fills all the levels with electrons 
of parallel spin before filling any of them. This prescription enables one to guess the 
electronic configuration of any atom, once its atomic number is known, unless it happens 
to put us into a region of ambiguity, where different levels have almost the same 
energy. (Electronic configurations for such atoms are deduced from experimentally 
determined chemical, spectral, and physical properties.) The configuration for carbon 
(atomic number 6) would be 1s22s22p2, with the understanding that p electrons occupy 
different p orbitals and have parallel spins. (Recall that we expect the most stable of all 
the states arising from the configuration 1s22s22p2 to be the one of highest multiplicity. 
The 2p electrons can produce either a singlet or a triplet state just as could the two 
electrons in the 1s2s configuration of helium. The triplet should be the ground state and 
this corresponds to parallel spins, which requires different p orbitals by the exclusion 
principle.) 
It is important to realize that the orbital ordering (5-58) used in the aufbau process 
is not fixed, but depends on the atomic number Z. The ordering in (5-58) cannot be 
blindly followed in all cases. For instance, the ordering shows that 5s fills before 4d. 
It is true that element 38, strontium, has a ···4p65s24d0 configuration. But a later 
element, palladium, number 46, has ···4p64d105s0 as its ground state configuration. 
The effect of adding more protons and electrons has been to depress the 4d level more 
than the 5s level. 
5-7 Electron Angular Momentum in Atoms 
Most of our attention thus far has been with wavefunction symmetry and energy. 
However, understanding atomic spectroscopy or interatomic interactions (in reactions 
or scattering) requires close attention to angular momentum due to electronic orbital 
motion and “spin.” In this section we will see what possibilities exist for the total 
electronic angular momenta of atoms and how these various states are distinguished 
symbolically. 
We encountered earlier (Section 4-5) the notion that the total angular momentum for 
a classical system is the vector sum of the angular momenta of its parts. If the system 
interacts with a z-directed field, the total angular momentum vector precesses about 
the z axis, so the z component continues to be conserved and continues to be equal to 
the sum of z components of the system’s parts. Since quantum hydrogenlike systems 
obey angular momentum relations analogous to a precessing classical system, it is this 
z-axis behavior that we focus on as we seek to construct the nature of the total angular 
momentum from the orbital and spin parts we already understand. 
Because it is the total angular momentum that is conserved in a multicomponent 
classical system, it is the total angular momentum that obeys the quantum rules we have 
previously described for separate spin and orbital components. If we consider a oneelectron 
system, the combined spin-orbital angular momentum can be associated with 
a quantum number symbolized by j (analogous to s and l). Then we can immediately 
say that the allowed z components of total angular momentum are, in a.u., mj =±j, 
±(j -1), . . . and that the length of the vector is vj (j +1) a.u. 
The implication of accepting total angular momentum as the fundamental quantized 
quantity is that the spin and orbital angular momenta do not individually obey the
150 Chapter 5 Many-Electron Atoms 
quantum rules we have so far applied to them—s and l are not “good” quantum numbers. 
However, for atoms of low atomic number they are in fact quite good, especially for 
low-energy states, and we can continue to refer to the s and l quantum numbers in such 
cases with some confidence. (Classically, this corresponds to cases where there is little 
transfer of angular momentum between modes.) 
5-7.A Combined Spin-Orbital Angular Momentum for 
One-Electron Ions 
The key to understanding the following discussion is to remember that a quantum 
number l, s, or j really tells us three things: 
1. It equals the maximum value of ml , ms, or mj. If l =2, the maximum allowed value 
of ml is 2, and the maximum z component of orbital angular momentum is 2 a.u. 
2. It allows us to know the length of the related angular momentum vector, l, s, or j, 
in a.u. For j , this is given by vj (j +1). If j =2, the length of the total angular 
momentum vector j is v6 a.u. 
3. It allows us to know the degeneracy, g, of the energy level due to states having this 
angular momentum. For s, this is 2s +1. If s =1/2, gs =2. The corresponding l 
degeneracies produce the s, p, d, f degeneracies of 1, 3, 5, 7. 
Using the hydrogen atom as our example, let us consider what the total electronic 
angular momentum is in the ground (1s) state. For an s AO, l =0, and so there is no 
orbital angular momentum. This means that the total angular momentum is the same 
as the spin angular momentum, so j =s =1/2, mj =±1/2. The diagram for the vector 
j, then, looks just like that for s (Fig. 5-3). 
Figure 5-6 
 (a) Maximum z components of orbital and spin angular momenta for a p electron 
leading to a total z component of 3/2. (b) The four states corresponding to the j =3/2 vector assuming 
its possible z intercepts (3/2, 1/2, -1/2,-3/2).
Section 5-7 Electron Angular Momentum in Atoms 151 
Figure 5-7 
 (a) j =1/2, resulting when l is oriented with its maximum z intercept and s is oriented 
in its other orientation (other than as in Fig. 5-6). (b) The two states corresponding to the J =1/2 
vector assuming its possible z intercepts (1/2,-1/2). 
More interesting is an excited state, say 2p. Now l =1 and s =1/2. From l =1 we 
can say that the maximum orbital z component of angular momentum is 1 a.u. s =1/2 
tells us that there is an additional maximum spin z component of 1/2. The maximum 
sum, then, is 3/2 for the z component of j. But, if this is the maximum mj , then j 
itself must be 3/2 and the length of j must be v(3/2)(5/2)=1.94 a.u. There must be 
2j +1=4 allowed orientations of j, with z intercepts at 3/2, 1/2, -1/2, -3/2 in a.u. 
(Fig. 5-6). 
We are not yet finished with the 2p possibilities. The total angular momentum is 
the sum of its orbital and spin parts, and we have so far found the way they combine 
to give the maximum z component. But this is not the only way they can combine. It 
is possible to have ml =1 and ms =-1/2. Then the maximum mj =1/2, so j =1/2, 
giving us a vector j of length v(1/2)(3/2) a.u. and two orientations (Fig. 5-7). 
So far we have identified six states, four with j =3/2 and two with j =1/2. This 
is all we should expect since we have three 2p AOs and two spins, giving a total 
of six combinations. It seems, though, that we could generate some more states by 
now letting ml = 0 or -1 and combining these with ms =±1/2. However, these 
possibilities are already implicitly accounted for in the multiplicity of states we recognize 
to be associated with the j = 3/2, 1/2 cases already found. This illustrates 
the general approach to be taken when combining two vectors: Orient the larger 
vector to give maximum z projection, and combine this projection with each of the 
allowed z components of the smaller vector. This gives all of the possible mj (max) 
values, hence all of the j values. In other words, it gives us all of the allowed vectors, 
j, each oriented with maximum z component, and it remains only to recognize 
that these can have certain other orientations corresponding to z intercepts of mj -1, 
mj -2, . . . ,-mj . 
States can be labeled to reflect all of the angular momentum parts they possess. The 
main symbol is simply s, p, d, f, g, etc. depending on the l value as usual. A superscript 
at left gives spin multiplicity (2s + 1) for the states. A subscript at right tells the j 
quantum number for the states. If an individual member of the group of states having 
the same j value is to be cited, it is identified by placing its mj value at upper right.
152 Chapter 5 Many-Electron Atoms 
Thus, all six of the states discussed above can be referred to as 2p states. The two 
groups having different total angular momentum are distinguished as 2p3/2 and 2p1/2. 
One of the four states in the former group is the 2p-1/2 
3/2 state. 
The general form of the symbol is 2s+1l
mj 
j . Such symbols are normally called term 
symbols, and the collection of states they refer to is called a term (except when an 
individual state is denoted by inclusion of the mj value). 
The reason for distinguishing between the 2p3/2 and 2p1/2 terms is that they occur 
at slightly different energies. This results from the different energies of interaction 
between the magnetic moments due to spin and orbital motions. For instance, if l and 
s are coupled so as to give the maximum j , their associated magnetic moments are 
oriented like two bar magnets side by side with north poles adjacent. This is a higherenergy 
arrangement than the other extreme, where l and s couple to give minimum j , 
acting as a pair of parallel bar magnets with the north pole of each next to the south 
pole of the other. So 2p1/2 should be lower in energy than 2p3/2 for hydrogen. 
EXAMPLE 5-5 For a hydrogen atom having n = 3, l = 2, what are the possible 
j values, and how many states are possible? Indicate the lengths of the j vectors 
in a.u. What term symbols apply? 
SOLUTION 
 If l =2 (d states), there are five AOs and two possible spins, so we expect a total 
of ten possible states. The maximum possible values of the z-component of angular moment for 
orbital and spin respectively are 2 and 1/2. So the maximum value is 5/2 giving a vector length 
of v35/4 a.u. and six possible z projections, hence six states. The term symbol is 2d5/2. The 
remaining possible j value is 2-1/2=3/2, accounting for four more states and giving a vector 
length of v15/4 a.u. and a term symbol of 2d3/2.  
5-7.B Spin-Orbital Angular Momentum for Many-Electron Atoms 
Much of what we have seen for one-electron ions continues to hold for many-electron 
atoms or ions. All the symbolism is the same, except that capital letters replace lowercase: 
The quantum numbers are L, S, and J , and the main symbol becomes S, P, D, 
F, G, etc. There is a total orbital angular momentum vector L with quantum number 
L that equals the maximum value of ML. The length of L is vL(L+1) a.u., and it 
has 2L+1 orientations. Vectors S and J behave analogously. When constructing the 
vectors J, we continue to place the larger of L and S to give maximum z intercept, 
and add to this the possible z intercepts of the smaller vector. The situation, then, is 
just as before except that we need to figure out the possible values for ML and MS by 
combining the allowed values of ml(1),ml(2), . . . and ms(1),ms(2), . . . .8 
8This procedure of first combining individual orbital contributions to find L and spin contributions to find S 
and then combining these to get J is referred to as “L–S coupling,” or “Russell–Saunders coupling.” The other 
extreme is to first combine l and s for the first electron to give j(1), l and s for the second electron to give j(2), . . . 
and then combine these individual-electron j’s to give J. This is more appropriate for atoms having high atomic 
number (in which electrons move at relativistic speeds in the vicinity of the nucleus), and is referred to as “j–j 
coupling.” We will not describe j–j coupling in this text. The reader should consult Herzberg [6] for a fuller 
treatment.
Section 5-7 Electron Angular Momentum in Atoms 153 
For example, if we have found that ML(max) = 2 (which means L = 2) and 
Ms(max)=1 (which means S = 1), we have that MJ (max) can be 2 +1, 2 +0, and 
2+(-1), or 3, 2, and 1. This means that the possible values of J are 3, 2, 1, giving 
three different J vectors. Since L=2 and S =1, the term symbols for these three J 
cases are 3D3, 3D2, and 3D1. Notice that the multiplicities of these three terms—7, 
5, and 3, respectively, obtained from 2J + 1—total 15 states, which is just what we 
should expect for the 3D symbol (spin multiplicity of 3, orbital multiplicity of 5). The 
15 triplet-D states are found in three closely spaced levels, differing in energy because 
of different spin-orbital magnetic interactions. 
The problem remains, howdo we find theML andMs values that allowus to construct 
term symbols? There are two situations to distinguish in this context, and a different 
approach is taken for each. 
1. Nonequivalent Electrons. The first situation is exemplified by carbon in its 
1s22s22p3p configuration. It is not difficult to show that the electrons in the 1s and 
2s AOs contribute no net angular momentum and can be ignored: The spins of paired 
electrons are opposed, hence cancel, and the s-type AOs have no angular momentum, 
hence cannot contribute. However, even p, d, etc. sets ofAOs cannot contribute if they 
are filled because then any orbital momentum having z intercept ml is canceled by one 
with -ml . The important result is that filled subshells do not contribute to orbital or 
spin angular momentum. The remaining 2p and 3p electrons occupy different sets of 
AOs, hence are called nonequivalent electrons. 
Since these electrons are never in the sameAO, they are not restricted to have opposite 
spins at any time—theirAO and spin assignments are independent. There are threeAO 
choices (p1, p0, p-1) and two spin choices—six possibilities—for each electron, hence 
36 unique possibilities. We should expect, therefore, 36 states to be included in our 
final set of terms. 
We first find the possible L values. ml for each electron is 1, 0, or -1. We orient the 
larger of the l vectors to give the maximum ml(1)=+1 and orient the second l in all 
possible ways, giving ml(2)=+1, 0,-1. (Since the vectors have equal length in this 
case there is no “larger–smaller” choice to make.) The net ML values are +2,+1, 0, 
and this tells us that the possible L values are 2, 1, 0. 
Treating ms values similarly gives Ms =1, 0, so S =1, 0. 
Thus, we have three L vectors and two S vectors. We now combine every one of the 
L, S pairs. In each case, we again take the longer in its position of greatest z overlap 
and combine it with the shorter in all of its orientations. This gives the J values shown 
in Table 5-2. The appropriate term symbols follow from L, S, and J in each case. 
Thus, our term symbols are 3D3, 3D2, 3D1, 1D2, 3P2, 3P1, 3P0,1P1, 3S1 and 1S0 for a 
total of 7+5+3+5+5+3+1+3+3+1=36 states. 
In the absence of external fields, these 36 states occur in 10 energy levels, one for 
each term. These lie at different energies for several reasons. We have already seen, 
in our discussion of 1s2s helium states, that different spin multiplicities are associated 
with different symmetries of the spin wavefunction, meaning that the space part of 
the wavefunctions also differ in symmetry. This has a significant effect on energy, so 
1S and 3S, for example, have rather different energies. Different L values amount to 
different occupancies ofAOs, which also has an effect on the spatial wavefunctions, so 
3P and 3S have different energies. Finally, we have already seen that different J values 
correspond to different relative orientations of orbital and spin angular momentum
154 Chapter 5 Many-Electron Atoms 
TABLE 5-2 
 L and S Values for Two 
Nonequivalent Electrons and Resulting J Values 
and Term Symbols 
L S J Term 
2 1 
... 
.. 
3
2
1 
3D3 
3D2 
3D1 
2 0 2 1D2 
1 1 
... 
.. 
2
1
0 
3P2 
3P1 
3P0 
1 0 1 1P1 
0 1 1 3S1 
0 0 0 1S0 
vectors, hence of magnetic moments. For light atoms, this is a relatively small effect, 
so 3P2, 3P1, and 3P0 have only slightly different energies. The resulting energies for 
states of carbon in 1s22s22p2, 1s22s22p3p, and 1s22s22p4p configurations are shown 
in Fig. 5-8. Only the major term-energy differences are distinguishable on the scale of 
the figure. The line for 3D is really three very closely spaced lines corresponding to 
3D3, 3D2, and 3D1 terms. 
2. Zeeman Effect. It was pointed out in Section 4-6 that the orbital energies of 
a hydrogen atom corresponding to the same n but different ml undergo splitting when 
a magnetic field is imposed. Now we have seen that spin angular momentum is also 
present. Therefore, a proper treatment of the Zeeman effect requires that we focus on 
total angular momentum, not just the orbital component. Since there are 2J +1 states 
with different MJ values in a given term, we expect each term to split into 2J + 1 
evenly separated energies in the presence of a magnetic field, and this is indeed what 
is seen to happen (through its effects on lines in the spectrum). For example, a 3P2 
term splits into five closely spaced energies, corresponding to MJ = 2, 1, 0,-1,-2, 
and a1P1 term splits into three energies. 
A surprising feature of this phenomenon is that the amount of splitting is not the 
same for all terms, despite the fact that adjacent members of any term always differ by 
±1 unit of angular momentum on the z axis. For instance, the spacing between adjacent 
members of the 3P2 term mentioned above is 1.50 times greater than that in the 1P1 
term. It was recognized that terms wherein S =0, so that J is entirely due to orbital 
angular momentum (J =L), undergo “normal splitting”—i.e., equal to what classical 
physics would predict for the amount of angular momentum and charge involved. On 
the other hand, terms wherein J is entirely due to spin (L = 0, so J = S) undergo 
twice the splitting predicted from classical considerations. [This extra factor of two 
(actually 2.0023)was without theoretical explanation until Dirac’s relativistic treatment 
of quantum mechanics.] 
Terms wherein J contains contributions from both L and S have Zeeman splittings 
other than one or two times the normal value, depending on the details of the way L and
Section 5-7 Electron Angular Momentum in Atoms 155 
Figure 5-8 
 Energy levels for carbon atom terms resulting from configurations 1s22s22p2, 
1s22s22p3p, and 1s22s22p4p. 
S are combined. The extent to which a term member’s energy is shifted by a magnetic 
field of strength B is 
E =gßeMjB (5-59) 
where ße is the Bohr magneton (Appendix 10) and g is the Land´e g factor, which 
accounts for the different effects of L and S on magnetic moment that we have been 
discussing: 
g =1+ 
J(J +1)+S(S +1)-L(L+1) 
2J(J +1) 
(5-60) 
It is not difficult to see that this formula equals one when S =0, J =L, and equals two 
when L=0, J =S. For the 3P2 term, S =1, L=1, J =2, and g equals 1.5, indicating 
that, in this state, half of the z-component of angular momentum is due to orbital
156 Chapter 5 Many-Electron Atoms 
motion, and half is due to spin (which is double-weighted in its effect on magnetic 
moment). 
3. Equivalent Electrons. Observe that the energy-level diagram for carbon 
(Fig. 5-8) shows the 10 expected terms for the excited 2p3p and 2p4p configurations, 
but not for the ground 2p2 configuration. There are no new terms for the latter case, 
but some of the terms present for 2p3p or 2p4p are gone, namely 3D, 3S, and 1P. The 
remaining terms account for 15 states. Evidently 21 states that are possible for a pair 
of nonequivalent p electrons are not allowed for a pair of equivalent electrons in a p2 
configuration. We will see that some of the states that are different for nonequivalent 
electrons become one and the same for equivalent electrons and must be excluded. 
Others are excluded by the Pauli exclusion principle because they would require two 
electrons to be in the same AO with the same spin. 
We now demonstrate the method for discovering the terms that exist for equivalent 
electrons. This is more difficult than for nonequivalent electrons, even though there are 
fewer terms. We first list all the orbital-spin combinations (called microstates), strike 
out those that are redundant or that violate the Pauli exclusion principle, and then infer 
from the remaining microstates what terms exist. 
Taking the p2 case for illustration, we begin with the 36 microstates listed in 
Table 5-3. Some of these microstates are equivalent to others. For instance, 
2p1(1)2p1(2)a(1)ß(2) is not a different state from 2p1(1)2p1(2)ß(1)a(2). These both 
correspond to a pair of electrons in the same pair of spin orbitals, 2p1 and 2p1. Since 
electrons are indistinguishable, we cannot expect wavefunctions differing only in the 
order of electron labels to correspond to different physical states. [The single state that 
does exist would be accurately represented by 2p1(1)2p1(2)(a(1)ß(2) - ß(1)a(2)), 
which is a linear combination of the microstates. But we do not need to go to this level 
of detail when finding terms. We only need to recognize that there is but one state here 
and omit one of the microstates as superfluous.] Accordingly, we strike out rows 3, 19, 
and 35 from Table 5-3, labeling them “R” for redundant. 
Another way to recognize this equivalence is to observe that the microstates deemed 
redundant differ only by an interchange of a pair of electrons. This reveals that the 
set of four microstates with 2p1(1)2p0(2) is equivalent to the set with 2p0(1)2p1(2). 
Therefore, we can strike out rows 13–16. A similar argument removes rows 25–32. 
Already we have removed 15 microstates. 
Next we look for violations of the Pauli exclusion principle. This leads us to strike 
out rows 1, 4, 17, 20, 33, and 36, labeling them “P” for Pauli. Our remaining microstates 
number 15 and are reassembled in Table 5-4, along with values of the quantum numbers 
for z components of the relevant angular momentum vectors for individual electrons as 
well as for their sum. 
At this stage of the argument, the final column of Table 5-4 (the term symbols) is not 
yet known. We are about to fill out this column by making use of a simple rule that is 
based on the diagrammatic device described earlier—placing the larger vector so that 
it has the maximum z extension and then placing the shorter vector in all its allowed 
orientations. It is not difficult to see that the maximum resultant z component (MJ ) can 
be achieved in one and only one way, namely when both vectors give their maximum 
z projection. This means that the maximum-MJ -member of a given set of states in 
the same term should be recognized as corresponding to one and only one microstate, 
because there is only one way to achieve this orientation. So we look for this maximum
Section 5-7 Electron Angular Momentum in Atoms 157 
TABLE 5-3 
 Unrestricted List of Space–Spin Combinations for a 
Pair of Electrons (Same Subshell). R = “Redundant,” P = “Pauli” 
Electron number 
Row number 1 2 1 2 Comment 
1 p1 p1 a a P 
2 p1 p1 a ß 
3 p1 p1 ß a R 
4 p1 p1 ß ß P 
5 p1 p0 a a 
6 p1 p0 a ß 
7 p1 p0 ß a 
8 p1 p0 ß ß 
9 p1 p-1 a a 
10 p1 p-1 a ß 
11 p1 p-1 ß a 
12 p1 p-1 ß ß 
13 p0 p1 a a R 
14 p0 p1 a ß R 
15 p0 p1 ß a R 
16 p0 p1 ß ß R 
17 p0 p0 a a P 
18 p0 p0 a ß 
19 p0 p0 ß a R 
20 p0 p0 ß ß P 
21 p0 p-1 a a 
22 p0 p-1 a ß 
23 p0 p-1 ß a 
24 p0 p-1 ß ß 
25 p-1 p1 a a R 
26 p-1 p1 a ß R 
27 p-1 p1 ß a R 
28 p-1 p1 ß ß R 
29 p-1 p0 a a R 
30 p-1 p0 a ß R 
31 p-1 p0 ß a R 
32 p-1 p0 ß ß R 
33 p-1 p-1 a a P 
34 p-1 p-1 a ß 
35 p-1 p-1 ß a R 
36 p-1 p-1 ß ß P
158 Chapter 5 Many-Electron Atoms 
TABLE 5-4 
 Allowed Space-Spin Combinations and M Quantum Numbers for a Pair of 
p Electrons (Same Subshell) 
Microstate ml(1) ml(2) ms(1) ms(2) ML MS MJ State term 
p1p1aß 1 1 1/2 -1/2 2 0 2 1D2
2 
p1p0aa 1 0 1/2 1/2 1 1 2 3P2
2 
p1p0aß 1 0 1/2 -1/2 1 0 1 (1 
D2
1) 
p1p0ßa 1 0 -1/2 1/2 1 0 1 (3 
P2
1) 
p1p0ßß 1 0 -1/2 -1/2 1 -1 0 (3 
P2
0) 
p1p-1aa 1 -1 1/2 1/2 0 1 1 (3 
P1
1) 
p1p-1aß 1 -1 1/2 -1/2 0 0 0 (1 
D2
0) 
p1p-1ßa 1 -1 -1/2 1/2 0 0 0 (3 
P1
0) 
p1p-1ßß 1 -1 -1/2 -1/2 0 -1 -1 (3P2-1) 
p0p0aß 0 0 1/2 -1/2 0 0 0 1S0
0 
p0p-1aa 0 -1 1/2 1/2 -1 1 0 (3 
P0
0) 
p0p-1aß 0 -1 1/2 -1/2 -1 0 -1 (1D2-1) 
p0p-1ßa 0 -1 -1/2 1/2 -1 0 -1 (3P1-1) 
p0p-1ßß 0 -1 -1/2 -1/2 -1 -1 -2 (3P2-2) 
p-1p-1aß -1 -1 1/2 -1/2 -2 0 -2 (1D2-2) 
MJ and, from its microstate, get the L and S values that go with it. That gives us the 
information we need to establish the term symbol. 
We start, then, by seeking the maximumMJ value in Table 5-4. This isMJ =2, and 
it occurs twice (in the first two rows). The first of these goes with ML =2, MS =0. 
Since these result when L and S are giving their maximum z component, we conclude 
that L=2, S =0. This, then, is a member of the 1D2 term. (It is the 1D22
member of 
that term, since MJ =2.)We label this row 1D22
and proceed to select microstates that 
can account for the other four members of this term. Our choice is controlled by the 
requirements that (1) the MJ values for the other members must be 1, 0,-1,-2, and 
(2) we cannot have an |Ms | value larger than zero or an |ML| value larger than 2. (That 
would be impossible for states resulting from vectors having L = 2 and S = 0.) Our 
selections are indicated in Table 5-4, with parentheses to indicate that these assignments 
follow from recognition of the leading member 1D22
. (All are symbolized as 1D2.) 
There issomearbitrariness in selecting the “inner”members, forwhich|MJ |<J: The 
parenthetical term 1D22
could just as easily be assigned to p1p0ßa as p1p0aß. (Actually, 
neither of these microstates is a correct wavefunction for 1D22
. A linear combination of 
them is. But, if we only wish to designate term symbols, we need not worry about this.) 
We have accomplished already the identification of a term 1D2 and the elimination of 
five microstates from our list. The other microstate having MJ =2 has ML =MJ =1, 
so we know this goes with L = 1, S = 1 and has the symbol 3P22
. Again, four other 
members exist down the table, and we select them, being careful that |ML| and |Ms | do not exceed 1, while MJ =1, 0,-1,-2. 
At this point, we must recognize that we are not through with the 3P family. The 
existence of the 3P part of the symbol implies the existence of nine states, but 3P2
Section 5-8 Overview 159 
accounts for only five of them. The others come from 3P1 and 3P0, resulting from 
ML =1 with Ms =0,-1. (We did not worry about this for the 1D2 term because only 
five states are implied by 1D.) So we seek the microstates associated with these terms 
and label them as shown in the table. 
Only one microstate remains. For this ML =Ms =0, so this is a state labeled 1S00
. 
The term symbols for the p2 configuration, then, are 1D2, 3P2, 3P1, 3P0, 1S0, for a 
total of 15 states. The energies for these terms are shown in Fig. 5-8. The five terms 
fall into three energy groups, since the 3P terms are found to be very close in energy. 
High resolution spectra can be used to see the slight energy differences between terms 
that appear to be at the same level at the energy scale used in Fig. 5-8. Delving further, 
the degenerate energies of microstates in the same term can be made to separate by 
imposing a magnetic field (Zeeman effect). 
Based on spectroscopic assignments of energy levels for large numbers of atoms, 
Hund proposed a set of rules enabling one to predict the energy ordering for terms 
associated with equivalent electrons. These rules are, in order of decreasing influence: 
1. Terms having greater spin multiplicity lie lower in energy. 
2. Within each spin multiplicity, terms having greater L lie lower. 
3. Within the same L and S, levels of different J behave oppositely, according to 
whether the subshell is more or less than half-filled: If less than half-filled, terms 
with lower J lie lower. 
According to these rules, the five levels for carbon in its 1s22s22p2 configuration 
should be, in order of increasing energy, 3P0, 3P1, 3P2 (closely spaced) followed by 1D0, 
followed by 1S0. The actual energies (in cm-1) are, respectively, 0, 16.4, 43.5, 10194.8, 
21647.7 (Fig. 5-8). The order of states for the excited 2p3p and 2p4p configurations 
is different. This is not a breakdown of Hund’s rules because these are not equivalentelectron 
cases. 
Hund’s first rule is the source of the aufbau rule, cited earlier, that each AO of a 
subshell becomes half-filled before any of them become filled with electrons. The 
equivalence of these statements is easily demonstrated (Problem 5-30). 
One can use Hund’s rules to find the lowest-energy state term symbol without going 
through the tedious microstate process just described. For an atom having an outer 
subshell configuration of p2 we would first recognize that we seek maximum S, so the 
electrons must have parallel spin, giving S =1. (We use Hund’s most influential rule 
first.) Subject to this constraint, we seek maximum L. Since the electrons cannot both 
be in p1 with the same spin, p1p0 is next best, giving maximum ML = 1, so L = 1. 
S =1,L=1 gives J =2, 1, 0, so we know the corresponding terms are 3P2, 3P1, 3P0. 
Since the 2p subshell is less than half- filled, 3P0 is the ground term. 
5-8 Overview 
This chapter describes the new features that appear when we deal with systems having 
more than one electron. One of these features, interelectron repulsion, is easy to 
understand in its manifestation as operators in the hamiltonian and as coulomb repulsion 
integrals, J , in the average energy expression. Another feature, antisymmetry for
160 Chapter 5 Many-Electron Atoms 
electron exchange and the resulting existence of exchange integrals, K, is unfamiliar 
and unintuitive, without a classical counterpart, yet is enormously important in its effect 
on electronic structure. 
In addition to these features, we have noted the importance of recognizing that atomic 
states conserve magnitude and z component of total angular momentum. Using this 
permits us to characterize states in terms of J , L, and S (even though the latter two are 
not “good” quantum numbers) for ground and excited configurations. This is essential 
in atomic spectroscopy (a topic we do not pursue in this book) and also allows one, 
with the assistance of Hund’s rules, to predict the energy order for states associated 
with the ground configuration of any atom. 
In closing this chapter, we should emphasize again a point frequently forgotten by 
chemist. In the orbital approach to many-electron systems we have a convenient approximation. 
This is an imperfect but useful way to describe atomic structure. There are 
more accurate ways to approximate eigenfunctions of many-electron hamiltonians, but 
this usually involves more difficulty in interpretation. The orbital representation of . 
appears to be the best compromise between accuracy and convenience for most chemical 
purposes. 
5-8.A Problems 
5-1. Write down the hamiltonian operator for the lithium atom, in a.u. 
5-2. Calculate the values of ¯r1and ¯r2 consistent with the He wavefunction .(1, 2)= 1s(1)2s(2) . . . (Eq. 5-11). 
5-3. Calculate the energy in electron volts of a photon with associated wavelength 
0.1 a.u. Compare this result with the ionization energy in electron volts of the 
hydrogen atom in its ground state. Why is this comparison relevant? 
5-4. Show that the wavefunction (5-15) is normalized if the 1s and 2s orbitals are 
orthonormal. 
5-5. Show that the wavefunction (5-16) is antisymmetric with respect to exchange of 
electron coordinates. 
5-6. Show that the wavefunction (5-37) would vanish if 2s were replaced throughout 
by 1s, giving a 1s3configuration. 
5-7. Produce a totally antisymmetric wavefunction starting from the configuration 
1s(1)a(1)2p(2)ß(2)1s(3)ß(3). Use the method described for Eq. (5-37) and use 
a determinantal function as a check. 
5-8. Set up the integral of the product between (1s1s2saßa)* and 2s1s1saaß. (Use 
symbols rather than explicit atomic orbital formulas.) Factor the integral into 
a product of integrals over one-electron space functions and one-electron spin 
functions. Indicate the value of each of the resulting six integrals and of their 
product. 
5-9. a) Write down the Slater determinantal wavefunction for the configuration 
1s1¯s2pz .
Section 5-8 Overview 161 
b) Expand this determinant into a linear combination of products. 
c) Write down the nonzero part of expansion (b) when r3 =0, r1 =1 a.u., r2 =2 
a.u. [Do not evaluate the expression; just use symbols like 1s (r =1).] Also 
write down the nonzero part of (b) when r2 = 0, r1 = 1, r3 = 2, and when 
r1=0, r2=2, r3=1. Is there any physical difference between saying “electron 
3 is at the nucleus” and saying “an electron is at the nucleus?” Explain. 
5-10. Wavefunction (5-38) describes a member of a doublet. Write the wavefunction 
for the other member. 
5-11. A particle is capable of being in any one of three spin states. Call them a, ß, 
and . . Suppose you have two such particles in a molecule. 
a) Write down all the spin functions you can that are symmetric for exchange 
of these two particles. (Do not worry about normalization.) 
b) Write down all the antisymmetric cases. 
5-12. The following wavefunction is proposed for an excited state of the lithium atom 
. = 
1 
v6
 
1s(1) 1s(2) 1s(3) 
2s(1) 2s(2) 2s(3) 
3s(1) 3s(2) 3s(3)
 
Here 1s, 2s, and 3s are eigenfunctions for the Li2+ hamiltonian. 
a) Does this wavefunction satisfy the Pauli exclusion principle? Explain. 
b) Write the exact H for the lithium atom in atomic units. 
c) Is . an eigenfunction for the exact hamiltonian? 
d) If interelectronic repulsion terms are neglected in H, what energy, in a.u., is 
associated with .? 
e) What z component of spin and orbital angular momentum (in atomic units) 
would you expect for the atom in this state, ignoring any nuclear contribution? 
5-13. Write the normalized Slater determinantal wavefunction for beryllium in the 
1s22s2 configuration. Do not expand the determinant. 
5-14. Write down the ground state configuration for the fluorine atom. Use Slater’s 
rules to find the orbital exponents . =(Z -s)/n for 1s and 2s, 2p orbitals. 
5-15. Show that the average value of an operator for a state described by an eigenfunction 
for that operator is identical to the eigenvalue associated with that eigenfunction. 
5-16. Explain briefly the observation that the energy difference between the 
1s22s1(2S1/2) state and the 1s22p1(2P1/2) state for Li is 14,904 cm-1, whereas 
for Li2+ the 2s1(2S1/2) and 2p1(2P1/2) states are essentially degenerate. (They 
differ by only 2.4 cm-1.) 
5-17. In Chapter 4 it was stated that the magnitude of the square of the angular momentum 
is given by l(l + 1) a.u., and that z components can be any of the values 
-l,-l +1, . . . l -1, l a.u. Similar relations hold for spin. From this fact plus
162 Chapter 5 Many-Electron Atoms 
the knowledge that the possible z components of spin angular momentum are 
±1
2 a.u., calculate the length of the spin angular momentum vector. 
5-18. It has been shown that, for a single spinning electron, two spin states are possible 
having z components of spin angular momentum of+1/2 and-1/2 a.u. For two 
unpaired electrons, the state of greatest multiplicity is a triplet (Ms =+1, 0,-1). 
Show that, in general, the maximum spin multiplicity resulting from n unpaired 
electrons equals n+1. 
5-19. You have been shown symbolically that 1s(1)2s(2)±2s(1)1s(2) and a(1)ß(2)± 
ß(1)a(2) are symmetric or antisymmetric for exchange of electron labels (electron 
coordinates). For a more concrete and familiar example, take two functions: 
f (x) = exp(x) and g(y) = y3. Construct combinations of these functions that 
are symmetric and antisymmetric for exchange of x and y coordinates. Set x =1 
and y = 2 and evaluate each function. Now set x = 2 and y = 1 and evaluate 
again. Compare results. 
5-20. Give all the allowed term symbols for a hydrogen atom (a) in the n = 1 level, 
(b) in the n=2 level. In each case, total up the states to see whether you have 
the expected number. 
5-21. Consider the following helium atom wavefunction: 
. =1s(1)3d+2(2)a(1)a(2) 
a) Is this a satisfactory wavefunction in the sense of meeting general symmetry 
conditions resulting from particle indistinguishability and the exclusion 
principle? If not, how would you modify it to make it satisfactory? 
b) Identify the term to which this state (modified if necessary) belongs. 
5-22. How many states exist for the configuration spd? 
5-23. A group of related terms has the common symbol 2P. (This is called a term 
multiplet.) 
a) What are the full term symbols for this multiplet? 
b) How many energy levels exist (in the absence of a magnetic field) for this 
multiplet? 
c) Indicate into how many levels each member of the multiplet splits in the 
presence of an external magnetic field. 
5-24. Given the following space part of an approximate wavefunction for a Li+ ion: 
(1/v2)[1s(1)2p1(2) + 2p1(1)1s(2)], 
a) Write a physically possible spin part for this wavefunction. 
b) What energy would this state have (in a.u.) if the 1/r12 term in H did not 
exist? 
c) What average energy (expressed in terms of symbols like J ) would this state 
have using the correct H (including 1/r12)? 
d) You have not been shown the rules for operating with S2, the operator for the 
square of total spin angular momentum, but you can nevertheless guess what 
the result would be if S2 operates on this state function. What is your guess?
Section 5-8 Overview 163 
5-25. A state in the term 3D3 is described by the wavefunction .. What is the value of 
x in each of the expressions Op . =x., where Op is as given below? (Assume 
L–S coupling to be valid. If more than one x is possible, list them all.) (a) L2 
(b) S2 (c) J 2 (d) Lz (e) Sz (f) Jz . 
5-26. Carbon (1s22s22p2) and oxygen (1s22s22p4) have a “symmetrical” relation in 
their 2p occupancy: C has one electron less than a half-filled subshell, O has one 
electron more. Another way of stating this is to note that C has 2 electrons and 
4 holes in its 2p shell, while O has 2 holes and 4 electrons. 
a) Show that this leads to the same lowest-energy family (or “multiplet”) of term 
symbols, 3P2,0,1. 
b) How do these atoms differ in the energy-ordering of these three terms? 
c) Show that this agreement in lowest-energy multiplet terms holds in general 
for atoms having this symmetrical occupation relation. 
5-27. Predict the ground state term symbol for each of the following atoms. 
a) Na (1s22s22p63s) 
b) P (1s22s22p63s23p3) 
c) Ne (1s22s22p6) 
d) Ti (1s22s22p63s23p64s23d2) 
5-28. Calcium atoms are excited to the [Ar]4s4p configuration. 
a) How many states are there? 
b) What are the term symbols? 
5-29. a) Find all the terms for boron in its ground configuration, 1s22s22p, and order 
these terms according to energy. 
b) Repeat for phosphorus, [Ne]3s23p3. 
5-30. Explain howHund’s first rule is equivalent to the aufbau rule that degenerateAOs 
half-fill with electrons before any are filled, when forming the lowest-energy 
state(s). 
5-31. How many states exist for each of the following term multiplets? 
a) 3D 
b) 5F 
5-32. For a given electron configuration, are all of the following terms possible? 
Explain your reasoning. 2P3/2, 2P1/2, 1S0. 
5-33. How many states exist for each of the following configurations? (a) sd (b) sp 
(c) s2p (d) pd (e) dd (nonequivalent) 
5-34. a) How many states are associated with the 4F term multiplet? 
b) Write down the term symbols included in this multiplet. 
5-35. By inspection, what is the term symbol with the maximum J value we can 
have for the configuration sd? What other terms would be included in the same 
multiplet?
164 Chapter 5 Many-Electron Atoms 
5-36. Derive a formula for the number of states that exist for two equivalent electrons 
in a subshell having degeneracy g. How many states does this predict for p2? 
for d2? 
5-37. Evaluate the splitting between adjacent lines in Zeeman-split terms 3D3, 3D2, 
3D1, 1D2, when B equals 1 tesla. 
Multiple Choice Questions 
(Try to answer these without referring to the text.) 
1. Which one of the following is an acceptable (unnormalized) approximate wavefunction 
for a state of the helium atom? 
a) [1s(1)1s(2)-1s(1)1s(2)]a(1)a(2) 
b) 1s(1)1s(2)[a(1)ß(2)+ß(1)a(2)] c) [1s(1)2s(2)+2s(1)1s(2)]a(1)a(2) 
d) [1s(1)2s(2)+2s(1)1s(2)][a(1)ß(2)-ß(1)a(2)] e) None of the above is acceptable. 
2. The spin multiplicity of an atom in its ground state and having the outer-shell configuration 
4s23d7 is 
a) 19 
b) 15 
c) 7 
d) 5 
e) None of the above. 
3. Which one of the following statements isNOT true for the ground state of the helium 
atom? 
a) The atom’s size (measured by rav) is larger than the size of He+ in its 1s state. 
b) The ground state is a singlet. 
c) The energy of the 2p0 orbital is above that of the 2s orbital. 
d) The effective nuclear charge seen by both electrons is less than 2. 
e) The atom’s electronic energy is equal to -108.8eV. 
4. The crudest orbital model for the ground state of He uses the 1s atomic orbitals for 
He+, for which Z =2. Which of the following statements describes correctly the 
situation that pertains to a change to a more appropriate value? 
a) The improved Z value is larger than 2, and the orbitals become more contracted. 
b) The improved Z value is larger than 2, and the orbitals become more expanded. 
c) The improvedZ value is smaller than 2, and the orbitals become more contracted. 
d) The improved Z value is smaller than 2, and the orbitals become more expanded. 
e) The improvedZ value is smaller than 2, but this only affects the computed energy, 
and not orbital size.
Section 5-8 Overview 165 
References 
[1] O. Stern, Z. Physik 7, 249 (1921). 
[2] W. Gerlach and O. Stern, Z. Physik 8, 110 (1922). 
[3] E. G. Uhlenbeck and S. Goudsmit, Naturwissenschaften 13, 953 (1925); Nature 
117, 264 (1926). 
[4] R. Bichowsky and H. C. Urey, Proc. Natl. Acad. Sci. 12, 80 (1926). 
[5] J. C. Slater, Phys. Rev. 34, 1293 (1929). 
[6] G. Herzberg, Atomic Spectra and Atomic Structure. Dover, NewYork, 1944. 
[7] J. C. Slater, Phys. Rev. 36, 57 (1930). 
[8] E. Clementi and D. L. Raimondi, J. Chem. Phys. 38, 2686 (1963). 
[9] C. L. Pekeris, Phys. Rev. 115, 1216 (1959). 
[10] P. Roman, Origins of nonrelativistic spin, Physics Today, Jan. 1985, p.126.
Chapter 6 
Postulates and Theorems 
of Quantum Mechanics 
6-1 Introduction 
The first part of this book has treated a number of systems from a fairly physical viewpoint, 
using intuition as much as possible. Now, armed with the concepts already 
developed, the reader should be in a better position to understand the more formal foundation 
to be described in this chapter. This foundation is presented as a set of postulates. 
From these follow proofs of various theorems. The ultimate test of the validity of the 
postulates comes in comparing the theoretical predictions with experimental data. The 
extra effort required to master the postulates and theorems is repaid many times over 
when we seek to solve problems of chemical interest. 
6-2 TheWavefunction Postulate 
We have already described most of the requirements that awavefunction must satisfy: . 
must be acceptable (i.e., single-valued, nowhere infinite, continuous, with a piecewise 
continuous first derivative). For bound states (i.e., states in which the particles lack the 
energy to achieve infinite separation classically) we require that . be square integrable. 
So far we have considered only cases where the state of the system does not vary with 
time. For much of quantum chemistry, these are the cases of interest, but, in general, a 
state may change with time, and . will be a function of t in order to followthe evolution 
of the system. 
Gathering all this together, we arrive at 
Postulate I Any bound state of a dynamical system of n particles is described as 
completely as possible by an acceptable, square-integrable function 
(q1, q2, . . . q3n,.1,.2, . . . ,.n, t), where the q’s are spatial coordinates, 
.’s are spin coordinates, and t is the time coordinate. *dt is 
the probability that the space-spin coordinates lie in the volume element 
dt(=dt1dt2 ···dtn) at time t , if  is normalized. 
For example, suppose we have a two-electron system in a time-dependent state 
described by thewavefunction (x1, y1, z1,.1, x2, y2, z2,.2, t). The spin coordinates 
. would each be some combination of spin functions a and ß. If we integrate * 
over the spin coordinates of both electrons, we are left with a spin-free density function. 
Call it .(x1, y1, z1, x2, y2, z2, t) = .(v1, v2, t). We interpret .(v1, v2, t)dv1 dv2 as 
166
Section 6-3 The Postulate for Constructing Operators 167 
the probability that electron 1 is in dv1 (i.e., between x1 and x1 + dx, y1 and y1 + 
dy, and z1 and z1 + dz) and electron two is in dv2 at time t . If we now integrate 
over the coordinates of electron 2, we obtain a new density function, .(v1, t), which 
describes the probability of finding electron 1 in various volume elements at various 
times regardless of the position of electron 2. 
6-3 The Postulate for Constructing Operators 
Much of the substance of the second postulate is already familiar. We earlier used 
arguments based on de Broglie waves to construct hamiltonian operators. We then 
noted that the kinetic energy part of the operators can be identified with a classical 
term like p2
x/2m through the relation px .( ¯h/i)./.x. The potential energy terms 
in the hamiltonian operators are completely classical, however. Thus, we could have 
constructed the quantum mechanical hamiltonians by writing down the classical energy 
expressions in terms of momenta and position, and then replacing everymomentumterm 
by the appropriate partial differential operator. This is an example of the use of part c of: 
Postulate II To every observable dynamical variableM there can be assigned a linear 
hermitian operator Mˆ . One begins by writing the classical expression, 
as fully as possible in terms of momenta and positions. Then: 
a) If M is q or t , Mˆ is q or t . (q and t are space and time coordinates.) 
b) If M is a momentum, pj , for the j th particle, the operator is 
( ¯h/i)./.qj , where qj is conjugate to pj (e.g., xj is conjugate to pxj ). 
c) If M is expressible in terms of the q’s, p’s and t , Mˆ is found by 
substituting the above operators in the expression forM in such a way 
that Mˆ is hermitian. 
The reason for specifying that Mˆ must be hermitian is that the eigenvalues of a 
hermitian operator must be real numbers.1 We shall discuss this and other aspects of 
hermiticity (including its definition) later in this chapter. 
As an explicit example of this procedure, we reconsider the hydrogen atom. Assuming 
a fixed nucleus (infinite inertia), the classical expression for the total energy of the 
system is 
Eclassical =(1/2me)(p2
x +p2
y +p2
z )-e2/ 
4p
0(x2 +y2 +z2)1/2 
where the first term is just the kinetic energy of the electron and the second term is the 
electrostatic potential energy. The coordinate origin is on the nucleus. Application of 
postulate II retains the position variables x, y, and z of the potential term unchanged, 
but replaces px by ( ¯h/i) ./.x, etc.: 
1 
2me 
p2
x +p2
y +p2
z  . 
1 
2me 
 h 
2pi 
. 
.x 

2 
+ h 
2pi 
. 
.y 

2 
+ h 
2pi 
. 
.z

2 
= -h2 
8p2me.2 
1In this text (and in quantum chemistry in general) a caret indicates an operator and not a unit vector quantity 
as in classical physics.
168 Chapter 6 Postulates and Theorems of Quantum Mechanics 
Thus, we arrive at 
Hˆ = -h2 
8p2me.2 - 
e2 
4p
0(x2 +y2 +z2)1/2 
and we are now free to transform Hˆ to other coordinate systems (such as r, .,f) if 
we wish. 
6-4 The Time-Dependent Schr ¨ odinger Equation Postulate 
We have discussed only caseswhere neither the hamiltonianHˆ nor. is time dependent. 
In those cases we required that . be an eigenfunction of Hˆ . In the more general case 
in which  and Hˆ are time dependent,2 a different requirement is imposed by 
Postulate III The state functions (or wavefunctions) satisfy the equation 
Hˆ  (q, t)= -¯h 
i 
. 
.t 
 (q, t) (6-1) 
where Hˆ is the hamiltonian operator for the system. 
We should check to see if this is consistent with the time-independent Schr¨odinger 
equation we have been using. Suppose that the hamiltonian is time independent. Let 
us see if a solution to Eq. (6-1) exists when (q, t) is separated into a product of spaceand 
time-dependent functions: (q, t)=.(q)f (t). Inserting this into Eq. (6-1) gives 
Hˆ .(q)f (t)= -h¯ 
i 
. 
.t 
.(q)f (t) (6-2) 
Dividing by .(q)f (t) gives 
Hˆ .(q) 
.(q) = 
(-¯h/i)(./.t)f (t) 
f (t) 
(6-3) 
Since each side of Eq. (6-3) depends on a different variable, the two sides must equal 
the same constant, which we call E. This gives 
Hˆ .(q)=E.(q) (6-4) 
and 
-¯h 
i 
d 
dt 
f (t)=Ef (t) (6-5) 
The first of these equations is just the time-independent Schr¨odinger equation we have 
been using. The second equation has the solution f (t)=Aexp(-iEt/¯h). Hence, f *f 
equals a constant, and so * =.*.f *f ..*.. Since f has no effect on energy or 
particle distribution, we can ignore it in dealing with stationary states. The situation is 
analogous to the case of standing waves discussed in Chapter 1. 
2Hˆ and  symbolize time dependence; Hˆ and . symbolize time independence.
Section 6-5 The Postulate Relating Measured Values to Eigenvalues 169 
Note that while we have shown that solutions may exist in which  is separable, this 
does notmean that every solution of Eq. (6-1)with Hˆ=Hˆ is separable (i.e., stationary). 
We can imagine a situation where a system in a stationary state is suddenly perturbed 
to produce a new time-independent hamiltonian.  will change as the system adjusts 
to this new situation, giving us a case where the hamiltonian is time-independent (after 
the perturbation, at least) but  is not a stationary state function. The way in which  
evolves in time is governed by Eq. (6-1). 
EXAMPLE 6-1 Show that the average energy for a nonstationary state of the hydrogen 
atom is conserved as the system evolves, if Hˆ is not time-dependent. 
SOLUTION 
 We choose a simple example: 
. = 
1 
v2 {.1s exp(it/2¯h)+.2s exp(it/8¯h)} 
where each exponential equals exp(-iEt/¯h). Then 
	E
=	 .*Hˆ . dv dt = 
1
2 
	 .*1s exp(-it/2h¯)Hˆ .1s exp(it/2h¯) dv dt 
+ the analogous 2s2s term +1s2s and 2s1s cross terms. Since Hˆ does not operate on functions 
of t, the exponentials in each of the first two integrals can join together, giving exp(0)=1. So timedependence 
disappears from the first two integrals, and they become respectively equal to -1
2 a.u., 
and -18
a.u. Dependence on time does not disappear from the cross-term integrals, but that doesn’t 
matter because the integration over space gives zero in each case, due to orbital orthogonality. 
Thus, <E>= 1
2 (-1
2 a.u.)+ 1
2 (-18
a.u.)=- 5 
16 a.u., which has no time-dependence.  
6-5 The Postulate Relating Measured Values 
to Eigenvalues 
The second postulate indicated that every observable variable of a system (such as 
position, momentum, velocity, energy, dipole moment) was associated with a hermitian 
operator. The connection between the observed value of a variable and the operator is 
given by 
Postulate IV Any result of a measurement of a dynamical variable is one of the eigenvalues 
of the corresponding operator. 
Any measurement always gives a real number, and so this postulate requires that 
eigenvalues of the appropriate operators be real. We will prove later that hermitian 
operators satisfy this requirement. 
If we measured the electronic energy of a hydrogen atom (the negative of its ionization 
energy), we could get any of the allowed eigenvalues (-1/2n2 a.u.) but no 
intermediate value. What if, instead, we measured the distance of the electron from the 
nucleus. By postulate II, the operator for this property is just the variable r itself. That 
is, ˆr =r. Hence, we need to consider the eigenvalues of r in the equation 
r d(r, .,f)=.d(r, .,f) (6-6)
170 Chapter 6 Postulates and Theorems of Quantum Mechanics 
where d is an eigenfunction and . is a real number (corresponding to the distance of 
the electron from the nucleus). We can rewrite this equation as 
(r -.)d(r, .,f)=0 (6-7) 
This form makes it more apparent that the function d must vanish at all points in space 
except those where r = .. But . is an eigenvalue of r and hence is a possible result 
of a measurement. Thus, we see that postulate IV implies some connection between 
a measurement of, say, r = 2 a.u. and an eigenfunction of r that is finite only at 
r =2 a.u. We symbolize this eigenfunction d(r -2 a.u.), this “delta function” being 
zero whenever the argument is not zero. If we measured the electron’s position to be 
at r =5.3 a.u., the corresponding eigenfunction would be d(r -5.3 a.u.)—a function 
that is zero everywhere except in a shell of infinitesimal thickness at r = 5.3 a.u. If 
instead we measured the point in space of the electron, rather than just the distance 
from the nucleus, and found it to be r0, .0, f0, then the corresponding eigenfunction 
of the position operator would be d(r -r0)d(. -.0)d(f -f0). This function vanishes 
everywhere except at r0, .0, f0. 
It is evident that any value of . from zero to infinity in Eq. (6-7) may be chosen 
without spoiling the ability of d to serve as an eigenfunction of r. This means that, 
unlike the energy measurement, the measurement of the distance of the electron from 
the nucleus can have any value. 
The eigenfunctions of the position operator are called Dirac delta functions. They 
are “spike” functions having infinitesimal width. They are normalized through the 
equation 
	 d (x -x0) dx =1 (6-8) 
where the integration range includes x0.3 On first acquaintance, these functions seem 
mathematically peculiar, but they make physical sense in the following way. One can 
interpret the actual measurement of position as a process that forces the particle to 
acquire a certain position at some instant. At that instant, .2 for the system (now 
perturbed by the measuring process) ought to give unit probability for finding the 
particle at that point (where it definitely is) and zero probability elsewhere, and this is 
just what the Dirac delta function does.4 
Postulate IV, then, is in accord with a picture wherein the process of measurement 
forces the measured system into an eigenstate for the appropriate operator, giving the 
corresponding eigenvalue as the measurement. This definition of “measurement” is 
somewhat restrictive and can be deceptive. Often scientists refer to measurements 
that are really measurements of average values rather than eigenvalues. This point is 
discussed further below. 
3The reader should avoid confusing the Dirac delta function d(x -x0) with the Kronecker delta di,j encountered 
earlier. They are similar in that both vanish unless x = x0 in the former and i = j in the latter. But they differ 
in that the value of di,j is definite (unity) while the value of d(x0 -x0) is not defined. The Dirac delta function 
has definite value only in integrated expressions like Eq. (6-8). The spin functions a and ß may be thought of as 
Dirac delta functions in the spin “coordinate” .. The Dirac delta function is admittedly unusual, and one tends to 
be uneasy with it at first. This function is important and useful in quantum mechanics. However, since we will 
make almost no use of it in this text, we will not develop the topic further. 
4Notice that Eq. (6-8) does not involve d*d, but merely d. Because d is nonzero only at one point, d*d is likewise 
nonzero only at the same point. d and d*d are therefore not independent functions. It is convenient to view the 
d function as both the eigenfunction for the position operator and also as the probability distribution function for 
the particle.
Section 6-7 Hermitian Operators 171 
6-6 The Postulate for Average Values 
Suppose that we had somehow prepared a large number of hydrogen atoms so that they 
were all in the same, known, stationary state. Then we could measure the distance of 
the electron from the nucleus once in each atom and average these measurements to 
obtain an average value. We have already indicated that this average would be given 
by the sum of all the r values, each multiplied by its frequency of occurrence, which is 
given by .2 dv if . is normalized. Since r is a continuous variable, the sum becomes 
an integral. This is the content of 
Postulate V When a large number of identical systems have the same state function 
., the expected average of measurements on the variable M (one measurement 
per system) is given by 
Mav =	 .*Mˆ . dt
	 .*. dt (6-9) 
The denominator is unity if . is normalized. 
It is important to understand the distinction between average value and eigenvalue 
as they relate to measurements. A good example is the dipole moment. The dipole 
moment operator for a system of n charged particles is µ=n
i=1 ziri where zi is the 
charge on the ith particle and ri is its position vector with respect to an arbitrary origin. 
(We get this by writing the classical formula and observing that momentum terms 
do not appear. Hence, the quantum-mechanical operator is the same as the classical 
expression.) What will the eigenfunctions and eigenvalues of µˆ be like? 
The charge zi is only a number, while ri is a position operator, which has Dirac 
delta functions as eigenfunctions. For a hydrogen atom, one eigenfunction of ri would 
be a delta function at r = 1 a.u., . = 0, f = 0. The corresponding eigenvalue for µˆ 
would be the dipole moment obtained when a proton and an electron are separated by 
1 a.u., clearly a finite number. But “everybody knows” that an unperturbed atom in a 
stationary state has zero dipole moment. The difficulty is resolved when we recognize 
that measurement of a variable in postulates IV and V means measuring the value of 
a variable at a given instant. Hence, we must distinguish between the instantaneous 
dipole moment of an atom, which can have any value from among the eigenvalues of 
µˆ and the average dipole moment, which is zero for the atom. In everyday scientific 
discussion, the term “dipole moment” is usually understood to refer to the average 
dipole moment. Indeed, the usual measurements of dipole moment are measurements 
that effectively average over many molecules or long times (in atomic terms) or both. 
6-7 Hermitian Operators 
Let f and . be any square-integrable functions and Aˆ be an operator, all having the 
same domain. Aˆ is defined to be hermitian if 
	 .*Aˆfdv =	 fAˆ*.*dv (6-10) 
The integration is over the entire range of each spatial coordinate. Recall that the 
asterisk signifies reversal of the sign of i in a complex or imaginary term. The hermitian 
property has important consequences in quantum chemistry.
172 Chapter 6 Postulates and Theorems of Quantum Mechanics 
As an example of a test of an operator by Eq. (6-10), let us take the . and f to 
be square-integrable functions of x and Aˆ to be i(d/dx). Then the left-hand side of 
Eq. (6-10) becomes, upon integration by parts, 
(6-11) 
Since . and f are square integrable, they (and their product) must vanish at infinity, 
giving the zero term in Eq. (6-11). We now write out the right-hand side of Eq. (6-10): 
	 +8 
-8 
f(i d/dx)*.*dx=-i 	 +8 
-8 
f(d.*/dx) dx (6-12) 
where the minus sign comes from carrying out the operation indicated by the asterisk. 
Equation (6-12) is equal to Eq. (6-11), and so the operator i(d/dx) is hermitian. 
Since the effect of i was to introduce a necessary sign reversal, it is apparent that the 
equality would not result for Aˆ = d/dx. Clearly, any hermitian operator involving a 
first derivative in any Cartesian coordinate must contain the factor i. The operators for 
linear momenta (Chapter 2) are examples of this. 
It is important to realize that Eq. (6-10) does not imply that .*Aˆf = fAˆ*.*. 
A simple example will make this clearer. Let Aˆ be the hydrogen atom hamiltonian, 
Hˆ =-1
2.2 -1/r, and let f be the 1s eigenfunction: f =(1/vp)exp(-r). Also, let 
. =v8/p exp(-2r) which is not an eigenfunction of Hˆ . Then, since Hˆ f =-1
2f, 
.*Hˆ f =-
1
2
.*f (6-13) 
But 
fH*.* = f -
1
2
(1/r2)(d/dr)r2(d/dr)-1/r
8/p exp(-2r) (6-14) 
= f [(1/r)-2]8/p exp(-2r)=[(1/r)-2].*f (6-15) 
(Since . has no . or f dependence, the parts of Hˆ * that include ./.. and ./.f have 
been omitted in Eq. (6-14).) Here we have two functions, -1
2.*f and [(1/r)-2].*f. 
They are obviously different. However, by Eq. (6-10), their integrals are equal since 
Hˆ is hermitian. 
6-8 Proof That Eigenvalues of Hermitian Operators 
Are Real 
Let Aˆ be a hermitian operator with a square-integrable eigenfunction .. Then 
Aˆ. =a. (6-16) 
Each side of Eq. (6-16) must be expressible as a real and an imaginary part. The real 
parts must be equal to each other and so must the imaginary parts. Taking the complex
Section 6-9 Eigenfunctions of a Hermitian Operator Form an Orthogonal Set 173 
conjugate of Eq. (6-16) causes the imaginary parts to reverse sign, but they remain 
equal. Therefore, we may write 
Aˆ*.* =a*.* (6-17) 
We multiply Eq. (6-16) from the left by .* and integrate over all spatial variables: 
	 .*Aˆ. dv =a 	 .*. dv (6-18) 
Similarly, we multiply Eq. (6-17) from the left by . and integrate: 
	 .Aˆ*.*dv =a* 	 ..* dv (6-19) 
SinceAˆ is hermitian, the left-hand sides of Eqs. (6-18) and (6-19) are equal by definition 
(Eq. 6-10). Therefore, the right-hand sides are equal, and their difference is zero: 
(a -a*) 	 .*. dv =0 (6-20) 
Since . is square integrable the integral cannot be zero. Therefore, a - a* is zero, 
which requires that a be real. 
6-9 Proof That Nondegenerate Eigenfunctions of a 
Hermitian Operator Form an Orthogonal Set 
Let . and f be two square-integrable eigenfunctions of the hermitian operator Aˆ: 
Aˆ. = a1. (6-21) 
Aˆ*f* = a2f* (6-22) 
Multiplying Eq. (6-21) from the left by f* and Eq. (6-22) from the left by ., and 
integrating gives 
	 f*Aˆ. dv = a1 	 f*. dv (6-23) 
	 .Aˆ*f* dv = a2 	 .f* dv (6-24) 
The left sides of Eqs. (6-23) and (6-24) are equal by (6-10), and 
(a1 -a2) 	 f*. dv =0. (6-25) 
If a1 = a2, the integral vanishes. This proves that nondegenerate eigenfunctions are 
orthogonal.
174 Chapter 6 Postulates and Theorems of Quantum Mechanics 
EXAMPLE 6-2 It has been shown (Section 6-7) that i(d/dx) is a hermitian operator. 
We know that it has eigenfunctions exp(±ikx) with eigenvalues ±k, which are 
real. So far, so good. However, this operator also has eigenfunctions exp(±kx), 
with eigenvalues ±ik, which are imaginary. This appears to violate the proof 
that eigenvalues of hermitian operators are real. Explain why neither of these 
eigenfunction sets is covered by the proof of section 6-8, and how one of them 
manages to obey the rule anyway. 
SOLUTION 
 The test for Hermiticity requires that i.*f|+-88 = 0. If f is ., and if . is 
square-integrable, this condition is satisfied, because .*. vanishes at ±8, giving 0-0=0. But 
neither of the exponential functions given above is square-integrable: They are both unequal to 
zero at ±8, so they both fall outside of the proof as given. Despite this, exp(±ikx) does have 
real eigenvalues, leading us to look more closely. Is it the case that i.*.|+-88 =0 for this set of 
functions, even though they do not vanish at infinity? It is indeed, since .*. =1, giving i -i =0 
for this term. Thus we see that our requirement that . be square integrable is more restrictive than 
what is necessary, namely that i.*.|+-88 =0. Note that the other set of exponentials, exp(±kx), 
leads to i.*. =i exp(±2kx), which does not produce a value of zero when values at x=8and 
x=-8are subtracted. Note also that the functions exp(±ikx) are orthogonal for different values 
of k, whereas the functions exp(±kx) are not.  
The point of the above example is that all of our proofs about eigenvalues or 
eigenfunctions of hermitian operators refer to cases where the eigenfunctions satisfy 
the requirement that i.* .|+-88 = 0. Square-integrability guarantees this, but some 
nonsquare-integrable sets of functions can satisfy it too. A hermitian operator can have 
eigenfunctions that are associated with complex or imaginary eigenvalues, but these 
must result from eigenfunctions that do not satisfy the requirement. 
6-10 Demonstration That All Eigenfunctions 
of a Hermitian Operator May Be Expressed 
as an Orthonormal Set 
If a1=a2, Eq. (6-25) is satisfied even when the integral is finite. Therefore, degenerate 
eigenfunctions need not be orthogonal. But they must be linearly independent or 
else they are the self-same function (to within a multiplicative constant), and if they 
are linearly independent, they can be converted to an orthogonal pair. Hence, it is 
always possible to express the degenerate eigenfunctions of a hermitian operator as 
an orthogonal set (and, as we have just proved, it is necessary that nondegenerate 
eigenfunctions be orthogonal). Furthermore, the functions must be square integrable, 
hence normalizable. In general, then, we are able to assume that all of the eigenfunctions 
of a hermitian operator can be expressed as an orthonormal set. 
Oneway to orthogonalize two nonorthogonal, linearly independent functions (which 
may or may not be eigenfunctions) will now be demonstrated. Let the functions be . 
and f (assumed normalized) and the integral of their product have the value S: 
	 .*f dv =S (6-26)
Section 6-11 Proof That Commuting Operators Have Simultaneous Eigenfunctions 175 
We keep one of the functions unchanged, say ., and let f=f-S. be our new second 
function. . and f are orthogonal since 
	 .*f dv =	 .*(f -S.)dv =	 .*f dv 
 S 
-S	 .*. dv 
 1 
=0 (6-27) 
(The new function f needs to be renormalized.) This process, known as Schmidt 
orthogonalization, may be generalized and applied sequentially to any number of linearly 
independent functions. 
EXAMPLE 6-3 Two normalized 1s AOs are located on nearby nuclei, A and B, and 
overlap each other enough so that  1sA1sB dv=0.500. Construct a function from 
these two that is orthogonal to 1sA and is normalized. 
SOLUTION 
 1sB 
=1sB -0.5 · 1sA is orthogonal to 1sA. It is not yet normalized because 
	 (1sB
)2 dv = 	 (1s2B 
+0.25 · 1s2A 
-2 · 0.5 · 1sA1sB)dv 
= 1+0.25-2 · 0.5 · 0.5=0.75= 
3
4 
. 
So the normalized function we seek is 2 v3 
(1sB -0.5 · 1sA).  
6-11 Proof That Commuting Operators Have 
Simultaneous Eigenfunctions 
Aˆ and Bˆ are commuting operators if, for the general square-integrable function f, 
AˆBˆf =BˆAˆf . This can bewritten (AˆBˆ -BˆAˆ)f =0,which requires thatAˆBˆ -BˆAˆ =0ˆ. 
(ˆ0 is called the null operator. It satisfies the equation, ˆ0f = 0.) This difference of 
operator products is called the commutator of Aˆ and Bˆ and is usually symbolized5 by 
[Aˆ,Bˆ]. If the commutator [Aˆ,Bˆ] vanishes, then Aˆ and Bˆ commute. 
We will now prove an important property of commuting operators, namely, that they 
have “simultaneous” eigenfunctions (i.e., that a set of eigenfunctions can be found for 
one of the operators that is also an eigenfunction set for the other operator). Let ßi be 
the eigenfunctions for Bˆ :Bˆßi =bißi . For the moment, assume all the numbers bi are 
different (i.e., the eigenfunctions ßi are nondegenerate). Let [Aˆ,Bˆ]=0ˆ. Then 
Bˆ(Aˆßi)=AˆBˆßi =Aˆbißi =bi(Aˆßi) (6-28) 
The parentheses emphasize that the function obtained by operating on ßi with Aˆ is 
an eigenfunction of Bˆ with eigenvalue bi . But that function can only be a constant 
times ßi itself. Hence, for nondegenerate ßi we have that Aˆßi =cßi , and so ßi is an 
eigenfunction of Aˆ. This proves that the nondegenerate eigenfunctions for one operator 
will also be eigenfunctions for any other operators that commute with it. 
5Other less common conventions are [Aˆ,Bˆ]- and (Aˆ,Bˆ).
176 Chapter 6 Postulates and Theorems of Quantum Mechanics 
If ßi is degenerate with other functions ßi,k , then we can only go so far as to say that 
Aˆßi =k ckßi,k, for this general linear combination is an eigenfunction of Bˆ having 
eigenvalue bi . But if this is so, then ßi is evidently not necessarily an eigenfunction of 
Aˆ. We shall not prove it here, but it is possible to show that one can always find some 
linear combinations of ßi,k to produce a set of newfunctions, ßi , that are eigenfunctions 
of Aˆ (and remain eigenfunctions of Bˆ as well). Therefore we can state that, if Aˆ 
and Bˆ commute, there exists a set of functions that are eigenfunctions for Aˆ and Bˆ 
simultaneously. 
An example of this property occurred in the particle-in-a-ring system described 
in Chapter 2. The hamiltonian and angular momentum operators commute for that 
system. There we found one set of functions, the trigonometric functions, that are 
eigenfunctions for Hˆ but not for Lˆz . But by mixing the energy-degenerate sines and 
cosines we produced exponential functions that are eigenfunctions for both of these 
operators. 
Another example concerns the familiar symmetry operations for reflection, rotation, 
etc. If one of these operations, symbolized Rˆ, commutes with the hamiltonian, then 
we should expect there to be a set of eigenfunctions for Hˆ that are simultaneously 
eigenfunctions for Rˆ. It was proved in Chapter 2 that this means that nondegenerate 
eigenfunctions must be symmetric or antisymmetric with respect to Rˆ. 
A symmetry operator that leaves Hˆ unchanged can be shown to commute with Hˆ . 
That is, if RˆHˆ =Hˆ , then RˆHˆ f =Hˆ Rˆf , where f is any function. To show this, let Rˆ 
be, say, a reflection operator. Then Rˆ operates on functions and operators to its right 
by reflecting the appropriate coordinates: Rˆf (q) = f (Rq). If Hˆ is invariant under 
reflection Rˆ, then H(q)=H(Rq), and it follows that RˆHˆ (q)f (q)=Hˆ (Rq)f (Rq)= 
Hˆ (q)f (Rq) = Hˆ (q)Rˆf (q), and so RˆHˆ f = Hˆ Rˆf . We shall formally develop the 
ramifications of symmetry in quantum chemistry in Chapter 13. 
The existence of simultaneous eigenfunctions for various operators has important 
ramifications for the measurement of a system’s properties. This is discussed in 
Section 6-15. 
6-12 Completeness of Eigenfunctions of a 
Hermitian Operator 
In Chapter 3 we discussed the concept of completeness in connection with the power 
series expansion of a function. Briefly, a series of functions6 {f} having certain restrictions 
(e.g., all derivatives vary smoothly) is said to be complete if an arbitrary function 
f having the same restrictions can be expressed in terms of the series7 
f =
i 
cifi (6-29) 
Proofs exist that certain hermitian operators corresponding to observable properties 
have eigenfunctions forming a complete set in the space of well-behaved (continuous, 
6A symbol in braces is frequently employed to represent an entire set of functions. 
7Equation (6-29) is overly restrictive in that it requires that the function and the series have identical values at 
every point, whereas it is possible for them to disagree at points of zero measure. However, at the level of this 
book, we can ignore this distinction and use Eq. (6-29) without encountering difficulty.
Section 6-12 Completeness of Eigenfunctions of a Hermitian Operator 177 
single-valued, square-integrable) functions. These proofs are difficult and will not be 
given here.8 Instead we shall introduce 
Postulate VI The eigenfunctions for any quantum mechanical operator corresponding 
to an observable variable constitute a complete set. (Furthermore, we 
have seen in Section 6-10 that we can assume that this set has been made 
orthonormal.) 
We will now use this property to investigate further the nature of the average value of 
an operator. Let the systembe in some state . (normalized), not an eigenfunction ofMˆ . 
However, Mˆ possesses eigenfunctions {µ} that must form a complete set. Therefore, 
we can express . in terms of µ’s: 
. =
i 
ciµi (6-30) 
Now we calculate the average value of M for the state .: 
Mav = 	 .*Mˆ .dv =	 
i 
c*i µ*i 
Mˆ 
j 
cjµj dv 
=
i 

j 
c*i cj 	 µ*i 
Mˆ µj dv (6-31) 
But Mˆ µi =miµi , and so 
Mav =
i 

j 
c*i cj 	 µ*i
mjµj dv =
i 

j 
c*i cjmj 	 µ*i
µj dv (6-32) 
But we are assuming that {µ} is an orthonormal set, and so 
Mav =
i 

j 
c*i cjmj dij =
i 
c*i cimi (6-33) 
What does this expression mean? Each measurement of the property corresponding 
to Mˆ must give one of the eigenvalues mi (postulate IV) and the average of many 
such measurements must be Mav. Equation (6-33) states how the individual measurements 
must be weighted to give the average, so it follows that each ci*ci is a measure 
of the relative frequency for observing the corresponding mi . Putting it another 
way, the absolute squares of the mixing coefficients in Eq. (6-30) give the probabilities 
that a measurement of the variable M will give the corresponding eigenvalue. 
For example, if . happens to be equal to (1/v2)µ1 + (1/v2)µ3, it follows that 
Mav =(1/2)m1 +(1/2)m3. 
EXAMPLE 6-4 What is the average value for the z-component of orbital angular 
momentum for the normalized function f =(1/v5)(.2s +2 ·.2p+1 )? 
SOLUTION 
 Since we know that the 2s and 2p+1 eigenfunctions have z-components of angular 
momentum of 0 and +1 respectively, we can say at once that p¯z = 15 
· 0 + 4
5 · 1 = 0.8 a.u. 
 
8See, e.g., Kemble [1, Section 25].
178 Chapter 6 Postulates and Theorems of Quantum Mechanics 
6-13 The Variation Principle 
Many of the calculations of quantum chemistry are based on the Rayleigh-Ritz variation 
principle which states: For any normalized, acceptable function f, 
Hav =	 f*Hˆ f dt =E0 (6-34) 
where E0 is the lowest eigenvalue of Hˆ . 
This statement is easily proved. We expand f in terms of {.i }, the complete, 
orthonormal set of eigenfunctions of Hˆ : 
f =
i 
ci.i (6-35) 
As in the preceding section, this leads to 
	 f*Hˆ fdt =
i 
c*i ciEi (6-36) 
Now c*i ci is never negative, and so Eq. (6-36) is merely a weighted average of the 
eigenvalues Ei . Such an average can never be lower than the lowest contributing 
member and the principle is proved. 
The variation principle is sometimes stated in an equivalent way by saying that the 
average value ofHˆ over f is an upper bound for the lowest eigenvalue ofHˆ . Following 
the approach of the example at the end of the previous section, if f for a hydrogen atom 
happens to be a function equal to (1/v2).1s +(1/v2).2s , the average energy for f 
is (1/2)E1s +(1/2)E2s , which obviously lies above the lowest eigenvalue E1s . 
6-14 The Pauli Exclusion Principle 
We have already discussed the Pauli exclusion principle in Chapter 5. In its most 
general form, this is: 
Postulate VII . must be antisymmetric (symmetric) for the exchange of identical 
fermions (bosons). 
6-15 Measurement, Commutators, and Uncertainty 
If we measure the exact position of the electron in a hydrogen atom, we force it into 
a state having a Dirac delta function as its wavefunction. Since this function is also 
an eigenfunction for the dipole moment operator, it follows that we also know the 
(instantaneous) dipole moment for the atom at that instant. In effect, measuring position 
measures dipole moment too. But the delta function is not an eigenfunction for the 
hamiltonian operator of the atom, and so we have not simultaneously measured the 
electronic energy of the atom. 
We have earlier seen that an eigenfunction for one operator can serve also as eigenfunction 
for another operator when the operators commute. In the above example, the
Section 6-15 Measurement, Commutators, and Uncertainty 179 
operators for position and dipole moment commute with each other but not with the 
hamiltonian operator. This leads us to recognize that we can simultaneously measure 
values for two variables only if their operators commute. 
Let us consider this situation more deeply by imagining two successive measurements 
on a hydrogen atom, one immediately following the other. If we first measure 
position and find r =2.0 a.u., and then measure dipole moment, we will get the value 
(µ=2.0 a.u.) corresponding to the electron being at r =2.0 a.u. That is where we found 
it in the first measurement, and it has not had time to move elsewhere before the second 
measurement. If we immediately follow with yet another position measurement, the 
electron will still be found at r =2 a.u. (We are imagining that no time elapses between 
measurements, which is a limit we cannot actually achieve. In the present case, though, 
since measurement of r is also a measure of µ, both measurements are done at once, so 
this is really not a problem.) Hence, it makes sense to say that we knowthese two values 
“simultaneously.” However, if we first measure position and then measure energy, we 
find something very different. Suppose that we find r =2 a.u. and then, in a subsequent 
measurement, E = -1/2 a.u. (E must, after all, be an eigenvalue of Hˆ , according 
to postulate IV.) We know that the eigenfunction during the first measurement was 
d(r -2 a.u.), and that during the second measurement was a 1sAO. If we immediately 
do yet another position measurement, we can find any value of r (with probabilities 
given by 4pr2.2
1s dr). The processes of measuring position and energy are incompatible 
in the sense that there is no single function that can describe the situation that exists 
during both measurements. The energy-measuring process can be pictured as forcing a 
reconstruction of the wavefunction in such a manner that it no longer corresponds to a 
particular position, while measurement of position forces a state function that does not 
correspond to a particular energy. (In this case, separate measurements would really 
be necessary, so the impossibility of doing a second measurement truly immediately 
after the first must be recognized. Indeed, one has to allow for the fact that finding an 
electron in one place and then at some other place must imply a lapse of time permitting 
the electron to travel.) 
The reader may suspect that there is some connection between commutators and the 
uncertainty principle, and this is indeed the case. It can be shown9 that the product 
of widths of simultaneous measurements (i.e., the uncertainty in their values) of two 
variables satisfies the relation 
a · b= 
1
2 

	 .* Aˆ,Bˆ.dt 
(6-37) 
where . is normalized, and the absolute value |X| is defined as the positive square 
root of X*X. If A and B are conjugate variables, such as position and momentum, 
Eq. (6-37) becomes a ·b = h¯/2, which is Heisenberg’s uncertainty relation. If Aˆ 
and Bˆ commute, the right-hand side of Eq. (6-37) vanishes, and the values of both 
variables may, in theory, be simultaneously known exactly. 
Among the properties of greatest interest in molecular quantum mechanics are 
energy, symmetry, and electron orbital angular momentum because, for many 
molecules, some of these operators commute. Thus, if we know that an oxygen 
molecule is in a nondegenerate stationary electronic state, we know that it is possible 
to characterize that state by a definite value of the orbital angular momentum along 
9See Merzbacher [2, Section 10-5].
180 Chapter 6 Postulates and Theorems of Quantum Mechanics 
the internuclear axis. Also, we know that the wavefunction must be symmetric or 
antisymmetric for inversion through the molecular midpoint. 
6-16 Time-Dependent States 
Much of quantum chemistry is concerned with stationary states, for which  is a 
product of a space term . (an eigenfunction of Hˆ ) and a time-dependent factor 
exp(-iEt/¯h), which we usually ignore because it has no effect on particle probability 
distribution. Sometimes, however, it becomes necessary to consider time-dependent 
states. In this section we illustrate how some of these may be treated. 
There are two types of situation to distinguish. One is situations where the potential 
is changing as a function of time, and hence the hamiltonian operator is time dependent. 
An example is a molecule or atom in a time-varying electromagnetic field. The other 
is situations where the potential and hamiltonian operator do not change with time, 
but the particle is nonetheless in a nonstationary state. An example is a particle that is 
known to have been forced into a nonstationary state by a measurement of its position. 
We deal here with the second category. 
As our first example, consider a particle in a one-dimensional box with infinite walls. 
Suppose that we measure the particle’s position and find it in the left side of the box 
(i.e., between x = 0 and L/2; we will be more specific shortly) at some instant that 
we take to be t = 0. We are interested in knowing what this implies about a future 
measurement of the particle’s position. 
Our knowing that the particle is on the left side at t =0 means that the wavefunction 
for this state is not one of the time-independent box eigenfunctions we sawin Chapter 2, 
because those all predict equal probabilities for finding the particle on the two sides 
of the box. If the state function is not stationary, it must be time dependent, and it 
must satisfy Schr¨odinger’s time-dependent equation (6-1). We have, then, that the state 
function is time dependent, and that * =||2 is zero everywhere on the right side 
of the box when t =0. (We have not yet been specific enough to describe ||2 in detail 
on the left side of the box.) 
Schr¨odinger’s time-dependent equation (6-1) is not an eigenvalue equation. However, 
Eq. (6-2) shows that Schr¨odinger’s time-dependent equation is satisfied by timeindependent 
eigenfunctions of Hˆ if they are multiplied by their time-dependent factors 
f (t) = exp(-iEt/¯h). Furthermore, Eq. (6-2) continues to be satisfied if the term 
.(q)f (t) is replaced by a sum of such terms. (See Problem 6-9.) This means that 
we can seek to express the time-dependent state function, (x, t), as a sum of timeindependent 
box eigenfunctions as long as each of these is accompanied by its time 
factor f (t). When t = 0, all the factors f (t) equal unity, so at that point in time  
becomes the same as the sum of box eigenfunctions without their time factors. 
Our strategy, then, is to find a linear combination of time-independent box eigenfunctions, 
.n, that describe  when t =0. This is easy to do because the time factors 
are all equal to unity. Once we have found the proper mixture of .n, we multiply each 
by its time factor and then observe the behavior of ||2 as t increases. 
In the case of our particle-in-a-box example, we can start with a very simple approximation 
to . at t =0 by taking a 50–50 mixture of .1 and .2: 
(x, t)=(1/v2) 
.1 exp(-iE1t/¯h)+.2 exp(-iE2t/¯h) (6-38)
Section 6-16 Time-Dependent States 181 
.1 0 
.2 
.(x, 0) 
0
0 
2/L 
2/L 
2/L 
Figure 6-1 
 Stationary eigenfunctions (n=1, 2) for the particle in a box and their normalized sum. 
We have included the functions f (t), even though they equal unity when t =0, because 
they are needed to make (x, t) a solution to Schr¨odinger’s equation (6-1) and because 
they will inform us of the nature of  at later times. 
We choose this pair of functions because, when t =0, both are positive on the left, 
but they differ in sign on the right, giving us some cancellation there. (See Fig. 6-1.) 
Obviously, we have not succeeded in describing a function that has no probability 
density on the right, but we already have a definite imbalance in that direction. (It is 
not difficult to see that some .3 with a positive coefficient should help remove much 
of the remaining probability density on the right.) 
Nowwe are in a position to examine this ||2 as time progresses—the time evolution 
of the square of a wavepacket that describes the probability distribution for a particle 
that is known to have been in the left half of the box at t =0. This is mathematically 
straightforward (Problem 6-20) and leads to the probability distributions sketched in 
Fig. 6-2 after time steps of t. The figure shows a changing distribution suggestive of 
the particle bouncing back and forth in the box with a cycle time of 8t. It is not difficult 
to see why this happens. 1 and 2 have different “frequency factors” exp(-iEnt/¯h), 
so they behave like two waves oscillating at different frequencies. Since E2 = 4E1 
(recall E .n2 in the box), 2 oscillates four times faster than 1. This means that, by 
the time 1 has made half a cycle (and is equal to -1 times its starting coordinates), 
2 has made two cycles and is just as it was at t =0. It is easy to see from Fig. 6-1 
that this will give a  that is skewed to the right, leading to the distribution shown in 
Fig. 6-2(e). (This allows us to conclude that 4t equals 1/2 of the cycle time of 1. 
See Problem 6-21.) 
If we want a more accurate starting representation for the localized particle, we must 
mix together a larger number of stationary-state wavefunctions. In order to decide 
how much of each is needed, we must have a better-defined description of  at t =0.
182 Chapter 6 Postulates and Theorems of Quantum Mechanics 
3 /L 
2 /L 
1 /L
0
0 0 L 0 L 
t = 0 
8.t 
.t 
7.t 6.t 5.t 
2.t 3.t 4.t 
(a) (b) (c) (d) (e) 
L 0 L 0 L 
Figure 6-2 
 |(x, t)|2 from Eq. (6-38) as it appears at various times. 
Figure 6-3 
 A normalized half sine wave in the left half of a “box.” The numbers at left are values 
of , not of E. 
Suppose, for instance, we choose to describe the starting wavefunction (x, 0) as a 
normalized half sine wave in the left side of the box and zero at the right (Fig. 6-3). 
Then we can calculate the amount (cn) of each of the stationary-state functions .n 
present in this function as follows: 
cn =	 L 
0 
.n(x, 0)dx (6-39) 
This follows from the completeness10 and orthonormality of {.n}. (See Problem 6-4.) 
Evaluation of Eq. (6-39) for the first few terms gives (Problem 6-22) 
(x, t)=0.600.1 +0.707.2 +0.360.3 +0.000.4 +0.086.5+··· (6-40) 
This modifies slightly our earlier commonsense combination and also verifies our prediction 
that .3 times a positive coefficient would be beneficial. 
10Because (x, 0) has a discontinuous derivative at the midpoint of the box, it falls outside the class for which 
{.n} is complete. However, because this problem is restricted to dx around one point, it should have little effect.
Section 6-16 Time-Dependent States 183 
This example illustrates the basic approach to such problems: 
1. Find a function that represents the initial particle distribution (x, 0). 
2. Expand that function as a series of eigenfunctions for the hamiltonian, and include 
the time-dependent factor for each term. 
3. Evaluate the probability distribution at other times t by examining | (x, t)|2. 
As a second example, suppose one were considering the behavior of the electronic 
state immediately after a tritium atom emits a beta particle to become a helium ion: 
31
H.32
He+0
-1 e. A crude analysis could be attempted by imagining that the nuclear 
charge suddenly changes from 1 to 2 a.u., and the orbiting electron (not the beta particle) 
suddenly finds itself in a state (the original 1s state) that is not an eigenfunction for the 
new hamiltonian. We would accordingly set (t =0) to be the 1sAO of hydrogen and 
expand this in terms of He+ eigenfunctions. Only s-typeAOs could contribute because 
of symmetry. The coefficients are given by 
cn =	 allspace 
.1s(Z =1).ns(Z =2)dv (6-41) 
and the time-dependent wavefunction is (in a.u.) 
(r, .,f, t) = c1.1s(Z =2) exp(-2it)+c2.2s(Z =2) exp(-it/2) 
+c3.3s(Z =2) exp(-2it/9)+··· (6-42) 
This function could be evaluated at various times t and would be found to give an 
oscillating spherical distribution, as though the electron cloud were shrinking, then 
rebounding to its original distance, then shrinking again, etc. 
Our next example is perhaps the most important: It is a particle initially localized in 
some region of space, say by measurement of its position, and free to move anywhere 
thereafter. Taking the one-dimensional case, we imagine that the particle has been 
detected around x =0 at t =0 (the measurement caused it to be “unfree” for an instant). 
We assume that the average momentum of the particle is zero. We seek to know how 
the probability distribution function for the particle will evolve in time. 
As before, we need a functional description of the wavefunction at t =0,(x, 0). 
We will then expand that in terms of eigenfunctions of the free-particle hamiltonian. 
As we have seen in Section 2-5, the free-particle eigenfunctions may be written 
exp(±iv2mEx/¯h), whereE is any nonnegative number. These are also eigenfunctions 
for the momentum operator, with eigenvalues v2mE¯h. 
The function usually selected to describe (x, 0) is a gaussian function: 
(x, 0)= 4 2a/p exp(-ax2) (6-43) 
The constant a affects the width of the gaussian and reflects our degree of certainty 
in our knowledge of position. Large a gives a tight function and small uncertainty 
x. The relationship between the gaussian function in x and the coefficients of the 
eigenfunctions as a function of v2mE/¯h is depicted in Fig. 6-4. Remarkably, the 
coefficient values are also described by a gaussian function (inv2mE/¯h). Furthermore, 
the tighter the gaussian function is in x, the broader the corresponding gaussian function
184 Chapter 6 Postulates and Theorems of Quantum Mechanics 
Figure 6-4 
 (a and c) Gaussianwave packets describing particles found to be at x=0 with differing 
degrees of certainty. (b and d)Values of ck (where k=v2mE/¯h) for momentum eigenfunctions that 
combine to express the gaussian wave packets to their left. (a, b) corresponds to relatively certain 
position and relatively uncertain momentum, whereas (c, d) corresponds to the opposite situation. 
is inv2mE/¯h (Problem 6-24). That is, we need to combine free-particle eigenfunction 
contributions from a wider range of momenta to create a tighter position function. This 
means that greater certainty in position goes with greater uncertainty in momentum, in 
accord with the uncertainty principle. 
Once we have the appropriate mixture of momentum eigenfunctions, each with its 
time-dependent term, we can follow the time evolution of the particle wave packet. 
We find that the packet spreads out more and more about x =0 as time passes, which 
means that our knowledge of position is decreasing as time passes. Even though the 
average position is not changing, the probability for finding the particle at a distance 
from x =0 is increasing. 
We can interpret this by remembering that the square of the wavefunction predicts 
the results of many experiments. In each of many position measurements finding the 
particle near x = 0, we impart some degree of momentum to the particle. Then, in 
a second measurement, we find that some of the particles have moved away from 
x = 0. The longer we wait before taking the second measurement, the greater this 
spread in x values. (Our assumption of zero average momentum amounts to saying that 
large deviations from zero momentum are equally likely for motion toward x =+8
Section 6-17 Summary 185 
and-8.) The more precise our first position measurement is, the greater the likelihood 
of introducing momenta quite different from zero and the more rapidly the wave packet 
spreads out as time passes. 
In the first example, a packet located in half of a one-dimensional box, we saw ||2 
oscillate back and forth, changing shape in the process so that the motion cannot be 
described with a single frequency. A related important case is that of an oscillator 
moving in a harmonic potential. Let us assume the oscillator’s position at t =0 to be 
described by a gaussian wavefunction. If this function is not centered at the oscillator’s 
equilibrium position, we have a time dependent situation analogous to the above 
particle-in-a-box case. For a harmonic potential, it can be shown11 that ||2 remains 
a gaussian function as time passes (i.e., does not change shape), and that the center 
of this gaussian oscillates back-and-forth about the equilibrium position with the classical 
frequency. This is a situation of interest to spectroscopists because it bears on 
the process of electronically exciting a sample of diatomic molecules. Suppose the 
molecules are initially in their ground electronic and ground vibrational states. Then 
their vibrational wavefunction is a simple gaussian function (the lowest-energy harmonic 
oscillator wavefunction) centered at Re (i.e., not off-center). If the molecules 
are excited by a laser pulse to a new electronic state having an equilibrium internuclear 
distance of Re
, the vibrational wavefunction at t =0 is still the simple gaussian centered 
at Re, which means that it is now off-center. As time passes, the center of this 
function oscillates back and forth about Re
in a coherent manner (i.e., describable with 
a single frequency). Thus, we have gone from an initial state describing an ensemble 
of molecules vibrating about Re with zero-point energy hv/2 and with random 
phases (i.e., a time-independent state wherein such measurable properties as average 
molecular dipole moment appear to be constant) to a final state where the molecules are 
vibrating in phase about Re
(a time-dependent state, wherein one might expect to see 
time variation of such properties). If Re were quite a bit smaller than Re
, for instance, 
then almost all the molecules would find themselves to be “too short” at t = 0, “too 
long” a short time later, etc., as they swing in phase about Re
. As a result of this simple 
behavior, it is possible to take advantage of it and time a second laser pulse to strike the 
molecules when they are almost all at their shortest, or all at their longest, extension. 
Of course, in real molecules the potential is not exactly harmonic. Furthermore, phase 
coherence is eventually lost due to collisions. So the second pulse must come very 
soon after the first one. 
6-17 Summary 
Some of the postulates and proofs described in this chapter are most important for what 
follows in this book, and we list these points here. 
1. . describes a state as completely as possible and must meet certain mathematical 
requirements (single-valued, etc.). .*. is the probability density distribution 
function for the system. 
2. For any observable variable, there is an operator (hermitian) which is constructed 
from the classical expression according to a simple recipe. (Operators related to 
11See Schiff [3, pp. 67, 68], and Tanner [4].
186 Chapter 6 Postulates and Theorems of Quantum Mechanics 
“spin” are the exception because the classical analog does not exist.) The eigenvalues 
for such an operator are the possible values we can measure for that quantity. The act 
of measuring the quantity forces the system into a state described by an eigenfunction 
of the operator. Once in that state, we may know exact values for other quantities 
only if their operators commute with the operator associated with our measurement. 
3. If the hamiltonian operator for a system is time independent, stationary eigenfunctions 
exist of the form .(q,.) exp(-iEt/¯h). The time-dependent exponential does 
not affect the measurable properties of a system in this state and is almost always 
completely ignored in any time-independent problem. 
4. The formula for the quantum-mechanical average value [Eq. (6-9)] is equivalent to 
the arithmetic average of all the possible measured values of a property times their 
frequency of occurrence [Eq. (6-33)]. This means that it is impossible to devise a 
function that satisfies the general conditions on . and leads to an average energy 
lower than the lowest eigenvalue of Hˆ . 
5. The square-integrable eigenfunctions for an operator corresponding to an observable 
quantity form a complete set, which may be assumed orthonormal. The eigenvalues 
are all real. 
6. Any operation that leaves Hˆ unchanged also commutes with Hˆ . 
7. Wavefunctions describing time-dependent states are solutions to Schr¨odinger’s timedependent 
equation. The absolute square of such a wavefunction gives a particle 
distribution function that depends on time. The time evolution of this particle distribution 
function is the quantum-mechanical equivalent of the classical concept of a 
trajectory. It is often convenient to express the time-dependent wave packet as a linear 
combination of eigenfunctions of the time-independent hamiltonian multiplied 
by their time-dependent phase factors. 
6-17.A Problems 
6-1. Prove that d2/dx2 is hermitian. 
6-2. Integrate the expressions in Eqs. (6-13) and (6-15) to show that their integrals are 
equal. 
6-3. Prove that, if a normalized function is expanded in terms of an orthonormal set of 
functions, the sum of the absolute squares of the expansion coefficients is unity. 
6-4. Show that a particular coefficient ck in Eq. (6-30) is given by ck = µ*k
. dv. 
6-5. A particle in a ring is in a state with wavefunction . =1/vp cos(2f). 
a) Calculate the average value for the angular momentum by evaluating 
 .*Lˆz. df, where Lˆz =(h¯/i) d/df. (Use symmetry arguments to evaluate 
the integral.) 
b) Express . as a linear combination of exponentials and evaluate the average 
value of the angular momentum using the formula Lz,av =i c*i ciLzi where 
Lzi is the eigenvalue for the ith exponential function.
Section 6-17 Summary 187 
6-6. Using Eq. (6-37), show that x · px = ¯h/2. 
6-7. What condition must the function f satisfy for the equality part of = to hold in 
Eq. (6-34)? 
6-8. Suppose you had an operator and a set of eigenfunctions for it that were associated 
with real eigenvalues. Does it necessarily follow that the operator is hermitian 
as defined by Eq. (6-10)? [Hint: Consider d/dr and the set of all functions 
exp(-ar), where a is real and positive definite.] 
6-9. a) Show that the nonstationary state having wavefunction 
 =(1/v2).1s exp(it/2)+(1/v2).2p0 exp(it/8) 
is a solution to Schr¨odinger’s time-dependent equation when Hˆ is the timeindependent 
Hˆ of the hydrogen atom. Use atomic units (i.e., h¯ =1). 
b) This time-dependent state is dipolar and oscillates with a characteristic frequency 
.. Show that . satisfies the relation E2 - E1 = E = 2p. in a.u. 
(The dipole oscillates at the same frequency as that of light required to drive 
the 1s .. 2p transition. This is central to the subject of spectroscopy.) 
6-10. From the definition that f =f -S. [see the discussion following Eq. (6-26)], 
evaluate the normalizing constant for f, assuming that f and . are normalized. 
6-11. Given the two normalized nonorthogonal functions (1/vp)exp(-r) and v1/3pr exp(-r), construct a new function f that is orthogonal to the first 
function and lies within the function space spanned by these two functions, and 
is normalized. 
6-12. If the hydrogen atom 1s AO is expanded in terms of the He+ AOs, what is the 
coefficient for (a) the He+ 1s AO? (b) the He+ 2p0 AO? 
6-13. The lowest-energy eigenfunction for the one-dimensional harmonic oscillator is 
.n=0 =(ß/p)1/4 exp(-ßx2/2). 
a) Demonstrate whether or not momentum is a constant of motion (i.e., is a 
“sharp” quantity) for this state. 
b) Calculate the average momentum for this state. 
6-14. Demonstrate whether x2d/dx and xd2/dx2 commute. What about xd/dx and 
x2d2/dx2? 
6-15. Evaluate the following integrals over all space. In neither case should you need 
to do this by brute force. 
a)  (3dxy)Lˆ2(3dxy) dv 
b)  3dxyLˆz(3dxy) dv
188 Chapter 6 Postulates and Theorems of Quantum Mechanics 
6-16. The operators for energy and angular momentum for an electron constrained to 
move in a ring of constant potential are, respectively, in a.u. -(1/2)d2/df2 and 
(1/i)d/df. 
a) Discuss whether or not there should be a set of functions that are simultaneously 
eigenfunctions for both operators. 
b) Discuss whether or not there is a set of functions that are eigenfunctions for 
one of these operators but not the other. 
c) Discuss whether it is reasonable to expect these two physical quantities to be 
exactly knowable simultaneously or whether the uncertainty principle makes 
this impossible. 
6-17. Suppose a hydrogen atom state was approximated by the function f=(1/v3)1s 
+(1/v3)2s+(1/v3)3s, where 1s, 2s, and 3s are normalized eigenfunctions for 
the hydrogen atom hamiltonian. What would be the average value of energy 
associated with this function, in a.u.? 
6-18. A function f is defined as follows: f = 0.1 · 1s + 0.2 · 2p1 + 0.3 · 3d2, where 
1s is the normalized eigenfunction for the 1s state of the hydrogen atom, etc. 
Evaluate the average value of the z component of angular momentum in a.u. for 
this function. 
6-19. Without looking back at the text, prove that (a) eigenvalues of hermitian operators 
are real, (b) nondegenerate eigenfunctions of hermitian operators are orthogonal, 
(c) nondegenerate eigenfunctions of Aˆ must be eigenfunctions of Bˆ if Aˆ and Bˆ 
commute. 
6-20. Using Eq. (6-38), obtain an expression for ||2 as a function of x and t . 
6-21. Evaluate t of Fig. 6-2 in terms of m, h, and L. 
6-22. Verify the values of the coefficients in Eq. (6-40). How can you tell from simple 
inspection that (a) c2 will be largest and positive, (b) c4 will be zero, (c) ci will 
tend toward small values at large i? 
6-23. a) Evaluate the first two coefficients in Eq. (6-42). 
b) What qualitative difference would you expect between  of Eq. (6-42) and 
one that takes explicit account of the changing potential resulting as the beta 
particle travels away from the nucleus? 
6-24. Showby qualitative arguments based on mathematical functions why coefficients 
ck should drop off more rapidly with k if the position wave packet is broader. 
ck =  v4 2a/p exp(-ax2) exp(ikx) dx. 
6-25. a) Prove that, ifV is real and(x, y, z, t) satisfies Schr¨odinger’s time-dependent 
equation, then (x, y, z,-t)* is also a solution. (This is called “invariance 
under time reversal.”) 
b) Show that, for stationary states, invariance under time reversal means that 
Hˆ .* =E.* if Hˆ . =E. and if V is real. 
c) Show from (b) that nondegenerate eigenfunctions of Hˆ (with real V ) must be 
real.
Section 6-17 Summary 189 
d) What becomes of the 2p-1AO(with time dependence included) upon complex 
conjugation and time reversal? the 2p0 AO? 
e) Can the statement in (c) be reworded to say that all degenerate eigenfunctions 
of Hˆ (with real V ) must be complex? 
Multiple Choice Questions 
(Try to answer these without referring to the text.) 
1. Which one of the following statements about the eigenfunctions of a timeindependent 
Hamiltonian operator is true? 
a) Any linear combination of these eigenfunctions is also an eigenfunction for Hˆ . 
b) The state function for this system must be one of these eigenfunctions. 
c) The eigenvalues associated with these eigenfunctions must all be real. 
d) These eigenfunctions must all be orthogonal to one another. 
e) These eigenfunctions have no time dependence. 
2. A hydrogen atom is in a nonstationary state having the wavefunction 1
2 {.1s 
exp(it/2¯h)+ .2s exp(it/8¯h)+.2p0 exp(it/8¯h)+.2p+1 exp(it/8¯h)} 
Which statement is true at any time t? 
a) A measurement of the energy has a 25% chance of giving -0.5 a.u. 
b) The average value of the z component of angular momentum is 0.5 a.u. 
c) The average energy is -78
a.u. 
d) A measurement of the total angular momentum has a 75% chance of giving v2 a.u. 
e) None of the above statements is true. 
3. The function rexp(-0.3r2) cos . is expanded in terms of hydrogen atom wavefunctions. 
This series may have finite contributions from bound-state eigenfunctions 
a) of all types: s,px,py,pz, dxy, dyz, etc. 
b) of all types except s. 
c) of types px,py,pz only. 
d) of type pz only. 
e) of types pz and dz2 only. 
References 
[1] E. C. Kemble, The Fundamental Principles of Quantum Mechanics with Elementary 
Applications. Dover, NewYork, 1958. 
[2] E. Merzbacher, Quantum Mechanics, 3rd ed.Wiley, NewYork, 1998. 
[3] L. I. Schiff, Quantum Mechanics, 3rd ed. McGraw-Hill, NewYork, 1968. 
[4] J. J. Tanner, J. Chem. Ed., 67, 917 (1990).
Chapter 7 
The Variation Method 
7-1 The Spirit of the Method 
The proof of the Rayleigh-Ritz variation principle (Section 6-12) involves essentially 
two ideas. The first is that any function can be expanded into a linear combination of 
other functions that span the same function space. Thus, for example, exp(ikx) can 
be expressed as cos(kx) + i sin(kx). An exponential can also be written as a linear 
combination of powers of the argument: 
exp(x)=1+x +x2/2!+x3/3!+···+xn/n!+··· (7-1) 
The second idea is that, if a function is expressed as a linear combination of eigenfunctions 
for the energy operator, then the average energy associated with the function is a 
weighted average of the energy eigenvalues. For example, if 
f =
 1 
v2
.1 +
 1 
v2
.2 (7-2) 
where 
ˆ H.1 =E1.1, ˆ H.2 =E2.2, E1 =E2 (7-3) 
then measuring the energy of many systems in states described by f would give the 
result E1 half of the time and E2 the other half. The average value, 1
2E1 + 1
2E2 must 
lie between E1 and E2. Alternatively, if 
f =1
3
.1 +2
3
.2 (7-4) 
measurements would give E1 one-third of the time, and E2 the rest of the time, for an 
average that still must lie between E1 and E2. It should be evident that, even when f 
is a linear combination of many eigenfunctions .i , the average value of E can never 
lie below the lowest or above the highest eigenvalue. 
The variation method is based on the idea that, by varying a function to give the lowest 
average energy, we tend to maximize the amount of the lowest-energy eigenfunction 
.0 present in the linear combination already discussed. Thus, if we minimize 
E¯ =  f*Hˆ f dv 
 f*f dv 
(7-5) 
the resulting f should tend to resemble .0 since we have maximized (in a sense) the 
amount of .0 in f by this procedure. 
190
Section 7-2 Nonlinear Variation: The Hydrogen Atom 191 
7-2 Nonlinear Variation: The Hydrogen Atom 
We have already seen (Chapter 4) that the lowest-energy eigenfunction for the hydrogen 
atom is (in atomic units) 
.1s =
 1 
vp 
exp(-r) (7-6) 
Suppose we did not knowthis and used the variation method to optimize the normalized 
trial function 
f =
.
.
. 3 
p 
.
.
exp(-.r) (7-7) 
In this example, when . =1, f becomes identical to .1s , but in more complicated systems 
the trial function never becomes identical to an eigenfunction of the hamiltonian. 
Nevertheless, this is a good example to start with since there are few mathematical 
complexities to obscure the philosophy of the approach. 
The variation method requires that we minimize 
E¯ =
 f*Hˆ f dv (7-8) 
by varying .. [f is normalized, so no denominator is required in Eq. (7-8).] Since the 
trial function f has no .- or f-dependent terms for .2 to operate on, only the radial 
part of .2 is needed in Hˆ . Thus [from Eq. (4-7)] 
Hˆ =-
1
2 
1 
r2 
d 
dr 
r2 d 
dr - 
1
r 
(7-9) 
According to Eq. (7-8), we need first to evaluate the quantity Hˆ f: 
Hˆ f = -
1
2 
1 
r2 
d 
dr 
r2 d 
dr - 
1
r 
. 3 
p 
exp (-.r) (7-10) 
... 
= (. -1) 
r - 
. 2 
2 
. 3 
p 
exp (-.r) (7-11) 
Incorporating this into Eq. (7-8) gives (after integrating . and f in dv to give 4p) 
E¯ = 4p 
. 3 
p 

 8 
0 
(. -1) 
r - 
. 2 
2 
 exp (-2.r) r2 dr (7-12) 
= 4. 3 
(. -1) 
 8 
0 
r exp (-2.r) dr -
. 2 
2 

 8 
0 
r2 exp (-2.r) dr (7-13) 
Using the integral table in Appendix 1, we obtain 
E¯ =
. 2 
2 
-. (7-14)
192 Chapter 7 The Variation Method 
Now we have a simple expression for E¯ as a function of . . To obtain the minimum, 
we set the derivative of E¯ to zero: 
dE¯ 
d. =0=. -1 (7-15) 
As we expected, . =1. Inserting this value for . into Eq. (7-14) gives E¯ =-1
2 a.u., 
which is identical with the lowest eigenvalue for the hydrogen atom. 
This example demonstrates thatminimizingE¯ for a trial function causes the function 
to become like the lowest eigenfunction for the system. But it is more realistic to 
examine a case where the trial function is incapable of becoming exactly identical with 
the lowest eigenfunction. Suppose we assumed a trial form (normalized) of 
f =
 . 5 
3p 
r exp(-.r) (7-16) 
Proceeding as before, we first evaluate Hˆ f: 
Hˆ f =
- 
1 
r2
+ 
2. -1 
r - 
. 2 
2 
f (7-17) 
This leads to 
E¯ (. )= 
4
3 
. 2 
8 - 
3. 
8 
 (7-18) 
so that 
dE¯ 
d. =0= 
4
3 
.
4 - 
3
8
 (7-19) 
and E¯ is a minimum when . = 3
2 . Thus, our energy-optimized f is 
f =
 35 
96p 
r exp
-
3r 
2 
 (7-20) 
This is obviously not identical to the eigenfunction given by Eq. (7-6), but it must be 
expressible as a linear combination of hydrogen atom eigenfunctions, and the amount 
of .1s present should be quite large unless the trial form was unwisely chosen. Since 
f also must contain contributions from higher energy eigenfunctions, it follows that E¯ 
must be higher in energy than -1
2 a.u. We test this by inserting . = 3
2 into Eq. (7-18), 
obtaining an E¯ of -38
a.u., (-0.375 a.u.). This value is above the lowest eigenvalue, 
but it is well below the second-lowest eigenvalue (-18
a.u.) associated with 2s, 2p 
eigenfunctions, and so we know that f does indeed contain much 1s character. We can 
find out exactly how much 1s eigenfunction is contained in f by calculating the overlap 
between f and the 1s eigenfunction. That is, since the 1s function is orthogonal to all 
the other hydrogen atom eigenfunctions; 

 .1s.j dv =
0 j =1s 
1 j =1s 
(7-21)
Section 7-2 Nonlinear Variation: The Hydrogen Atom 193 
and since 
f =c1s.1s +
j =1s 
cj.j (7-22) 
it follows that 
(7-23) 
Integrating  .1sf dv, where f is given by Eq. (7-20), gives c1s=0.9775, so f does 
indeed “contain” a large amount of .1s, (If c1s =1, then .1s and f are identical.) 
This suggests another way of trying to get a “best” approximate wavefunction. We 
could find the value of . that maximized the overlap between f and.1s. This maximizes 
c1s in Eq. (7-22). If one does this (Problem 7-6), one obtains . = 53
, which corresponds 
to an overlap of 0.9826 and an E¯ of -0.370 a.u. At first sight, this seems puzzling. 
f. = 53
 has a larger amount of .1s in it, but f. = 3
2  has a lower average energy. 
But it is really not so unreasonable. Maximizing the value of c1s causes . to take on 
a certain value that may cause the coefficient for some high-energy state (say, 6s) to 
become relatively larger, producing a tendency to raise the average energy. On the 
other hand, minimizing E¯ is a process that is implicitly concerned with what all the 
coefficients are doing. This allows for a different sort of compromise wherein c1s may 
be allowed to be a bit smaller if the associated energy loss is more than compensated 
for by a favorable shifting in values of higher coefficients (e.g., if c2s increases and c6s 
decreases). The purpose of this discussion is to emphasize that the variation method 
optimizes the trial function in a certain sense (best energy), but that other kinds of 
optimization are conceivable. The optimization that gives best overlap is not generally 
useful because it requires that we know the exact solution to begin with. If we vary 
. to obtain optimal agreement for r¯, or V¯ , or r2, we would find a different value of 
. appropriate for each property. In each case, however, we would need to know the 
correct value of ¯r, etc., before starting. A great virtue of the energy variation method 
is that it does not require foreknowledge of the exact eigenvalue or eigenfunction. 
However, the function that gives the lowest value for E¯ might not be especially good in 
describing other properties. This is demonstrated in Fig. 7-1 and Table 7-1. The figure 
indicates that the trial function differs from the exact function chiefly near the nucleus, 
where r <1. This discrepancy shows up when we compare average values for various 
powers of r. The operators r, r2, and r3 become large when r is large. Thus, these 
operators magnify .2 dv at large r. Since the two functions are fairly similar at large 
r, the average values show fair agreement. But r-1 and r-2 become large when r is 
small. Thus, the fact that the approximate function is too small at small r shows up 
as a marked disagreement in the average value of r-2, this average being much larger 
for the exact function (see Table 7-1). Any trial function that is known to be especially 
inaccurate in some region of space (e.g., at small r) will give unreliable average values 
for operators that are largest in that region of space (e.g., r-2). 
Choosing a trial form such as Eq. (7-16), which must vanish at r =0, might seem 
foolish since we know that .1s does not vanish at r =0. And if we were interested in 
electron density at the nucleus for, say, calculating the Fermi contact interaction, this 
would indeed be a self-defeating choice. But if our interest is in energies or other properties 
having operators that are large in regions where the trial function is not too deficient,
194 Chapter 7 The Variation Method 
Figure 7-1 
 Plots of .1s and f [Eq. (7-20)] versus r. 
TABLE 7-1 
 Comparison between Exact Values for Some Properties of (1s) Hydrogen and 
Values Calculated from the Function Eq. (7-20) 
Quantity Exact (1s) value (a.u.) Trial function value (a.u.) 
E¯ -1
2 -38 
Electron density 
at nucleus 1
p 0 
¯r 1.5 1.67 
r2 3.0 3.33 
r3 7.5 7.78 
r-1 1.0 0.75 
r-2 2.0 0.75 
this choice would serve. One often settles for a mathematically convenient form even 
though it is known to be inadequate in someway. Care must then be exercised, however, 
to avoid using that trial wavefunction in ways which emphasize its inadequacies. 
7-3 Nonlinear Variation: The Helium Atom 
We mentioned in Chapter 5 that the ground-state wavefunction 1s(1)1s(2) for helium 
was much too contracted if the 1s functions were taken from the He+ ion without 
modification. Physically, this arises because, in He+, the single electron sees only a
Section 7-3 Nonlinear Variation: The Helium Atom 195 
doubly positive nucleus, whereas in He each electron sees a doubly positive nucleus 
and another electron, so that in He the repulsion between electrons prevents them from 
spending as much time near the nucleus as in He+. Somehow, the 1s functions should 
be modified to reflect this behavior. We will now show how the variation method may 
be used to accomplish this. 
The form of the hydrogenlike ion 1s solution is 
1s=
Z3 
p 
exp(-Zr) (7-24) 
For He+, Z =2, but we have just seen that this gives a function that is too contracted. 
Smaller values ofZ would cause the function to die away more slowly with r. Therefore, 
it is reasonable to replace the atomic number Z with a variable parameter . and find 
the value of . that gives the lowest average energy. Hence, we let 
1s (1)=
. 3 
p 
exp(-.r1) (7-25) 
and our trial wavefunction is1 
f(1, 2)=1s(1)1s(2) 1/v2a(1)ß(2)-ß(1)a(2) (7-26) 
The average energy is [since f(1, 2) is normalized] 
E¯ =
 
 f*(1, 2)Hˆ (1, 2)f(1, 2)dt (1)dt (2) (7-27) 
Since Hˆ (1, 2) contains no spin operators at our level of approximation, the integral 
separates into an integral over the space coordinates of both electrons and an integral 
over the spin coordinates of both electrons. The integration over spins gives a factor of 
unity. There remains 
E¯ =
 
 1s(1)1s(2)Hˆ (1, 2)1s(1)1s(2)dv(1)dv(2) (7-28) 
where 
Hˆ (1, 2)=-
1
2.2
1 - 
1
2.2
2 -
 2 
r1
-
 2 
r2
+
 1 
r12
 (7-29) 
and where the . and f parts of .2 can be ignored since f(1, 2) is independent of these 
variables. The calculation is easier if we recognize that 
Hˆ (1, 2)=HˆHe+ (1)+HˆHe+ (2)+ 
1 
r12 
(7-30) 
1In this trial function, . has the same value in each atomic orbital. This is not a necessary restriction. There is no 
physical reason for not choosing the more general trial function where orbitals with different . are used. Symmetry 
requires that such a function be written 2-1/2[1s(1)1s(2)+1s(1)1s(2)]2-1/2[a(1)ß(2)-ß(1)a(2)]. This 
type of function is called a “split shell” wavefunction. It gives a lower energy for He than does the function (7-26). 
However, for most quantum-chemical calculations split shells are not used, the gain in accuracy usually not being 
commensurate with the increased computational effort.
196 Chapter 7 The Variation Method 
This allows us to express Eq. (7-28) as the sum of three integrals, the first of these being 


 1s(1)1s(2)HˆHe+ (1) 1s(1)1s(2)dv(1)dv(2) (7-31) 
Since the operator in the integrand operates only on coordinates of electron 1, we can 
separate this into a product of two integrals: 

 1s(2)1s(2)dv(2) 
 1s(1)HˆHe+ (1) 1s(1)dv(1) (7-32) 
The 1s functions are normalized, and so the first integral is unity. The second 
integral is almost identical to the integral in Eq. (7-8) and has the value . 2/2-2. . 
Therefore, the first of the three integrals mentioned above equals . 2/2 - 2. . The 
second of the three integrals is identical with Eq. (7-31) except that the operator acts 
on electron 2 instead of 1. This integral is evaluated in the same manner and gives 
the same result. The third integral, in which the operator is 1/r12, is more difficult to 
evaluate. This interesting and instructive problem constitutes a detour from the main 
sequence of ideas in this chapter and is therefore discussed in Appendix 3. We here 
simply take the result, 58
. , and proceed with the variation calculation. 
We now have an expression for E¯ as a function of . : 
E¯ =2 . 2/2-2. + 
5
8
. =. 2 - 
27 
8 
. (7-33) 
Minimizing E¯ with respect to . gives 
dE¯ 
d. =0=2. - 
27 
8 
(7-34) 
so that 
. = 
27 
16 
(7-35) 
This value of . is smaller than the unmodified He+ value of 2, as we anticipated. Let us 
see how much the average energy has been improved. According to Eq. (7-33), when 
. is 2, E¯ is equal to -2.75 a.u. When . = 27 
16 ,E¯ equals -2.848 a.u., so the average 
energy has been lowered by approximately 0.1 a.u., or 2.7 eV, or 62 kcal/mole. (The 
exact nonrelativistic energy for He is-2.903724377 a.u.) Further analysis would show 
that, by decreasing . , we have decreased the average kinetic energy (the less compressed 
wavefunction changes slope less rapidly), raised the nuclear-electron attraction energy 
from a large negative to a smaller negative value (the decreased attraction resulting 
from the electrons being farther from the nucleus, on the average), and decreased the 
interelectronic repulsion energy from a higher positive value to a lower one. The 
variational procedure has allowed the wavefunction to adjust to the best compromise it 
can achieve among these three factors. If . becomes less than 27 
16 the loss of nuclear– 
electron attraction is too great to be offset by the loss of interelectronic repulsion and 
kinetic energy. 
Varying a parameter in the argument of an exponential produces a nonlinear change 
in the function, and so calculations of the type described above are referred to as 
nonlinear variation calculations. Such calculations tend to become mathematically 
complicated and are not frequently used except for fairly simple systems. The fact
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom 197 
that the hamiltonian operator is a linear operator [i.e., Hˆ (c1f1 +c2f2) = (c1Hˆ f1 + 
c2Hˆ f2)] makes a linear variation procedure more convenient for most purposes. 
7-4 Linear Variation: The Polarizability 
of the Hydrogen Atom 
Suppose we wish to express a wavefunction . (which may be approximate) as a linear 
combination of two known functions f1 and f2: 
. (c1, c2)=c1f1 +c2f2 (7-36) 
The question is, what values of c1 and c2 give a . that best approximates the exact 
wavefunction for a particular system? The usual approach is to determine which values 
of c1 and c2 give the . associated with the minimum average energy attainable. The 
technique for achieving this, called the linear variation method, is by far the most 
common type of quantum chemical calculation performed. 
An example of a problem that can be treated by this method is the polarizability of 
the hydrogen atom. The wavefunction for the unperturbed hydrogen atom in its ground 
state is spherically symmetrical. But, when a uniform z-directed external electric field 
of strength F is imposed, the positive nucleus and the negative electron are attracted 
in opposite directions, which leads to an electronic distribution that is skewed with 
respect to the nucleus. The wavefunction describing this skewed distribution may be 
approximated by mixing with the unperturbed 1s function some 2pz function: . = 
c11s+c22pz . As indicated in Fig. 7-2, this produces a skewed wavefunction because 
the 2pz function is of the same sign as the 1s function on one side of the nucleus and 
of the opposite sign on the other. We will work out the details of this example after 
developing the method for the general case. 
Let the generalized trial function . be a linear combination of known functions 
f1,f2, . . . ,fn. (This set of functions is called the basis set for the calculation.) 
. =c1f1 +c2f2+···+cnfn (7-37) 
where the coefficients c are to be determined so that 
 .*Hˆ . dt 
 .*. dt =E¯ (7-38) 
is minimized. Substituting Eq. (7-37) into Eq. (7-38) gives 
E¯ =  c*1f*1 +c*2f*2 +···+c*nf*nHˆ (c1f1 +c2f2 +···+cnfn) dt 
 c*1f*1 +c*2f*2 +···+c*nf*n (c1f1 +c2f2+···+cnfn) dt 
= 
num 
denom 
(7-39) 
Since we will be dealing with cases in which the c’s and f’s are real, we will temporarily 
omit the complex conjugate notation to simplify the derivation. At the minimum 
value of E¯, 
.E¯ 
.c1 = 
.E¯ 
.c2 =···= 
.E¯ 
.cn =0 (7-40)
198 Chapter 7 The Variation Method 
Figure 7-2 
 Values of . versus z for 1s state of H atom (—-) and for approximate wavefunction 
given by 0.982 1s-0.188 2pz (- - -). The nucleus is at z=0 for each case. 
The partial derivative of Eq. (7-39) with respect to c1 is 
.E¯ 
.c1 =  f1Hˆ (c1f1 +···+cnfn) dt 
denom +  (c1f1 +···+cnfn)Hˆ f1 dt 
denom 
-(num) (denom)-2 
 f1 (c1f1+···+cnfn) dt 
+ 
 (c1f1+···+cnfn)f1dt 
= 0 (7-41)
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom 199 
Multiplying through by denom, recalling that num/denom equals E¯, and rearranging, 
gives 
c1 
 f1Hˆ f1 dt -E¯ 
 f1f1 dt 
+c2 
 f1Hˆ f2 dt -E¯ 
 f1f2 dt 
+···+cn 
 f1Hˆ fn dt -E¯ 
 f1fn dt 
=0 (7-42) 
At this point, it is convenient to switch to an abbreviated notation: 

 fiHˆ fj dt = Hij (7-43) 

 fifj dt = Sij (7-44) 
The integral Sij is normally called an overlap integral since its value is, in certain 
cases, an indication of the extent to which the two functions fi and fj occupy the same 
space. Use of this abbreviated notation produces, for Eq. (7-42), 
c1 H11 -E¯S11+c2 H12 -E¯S12+···+cn H1n -E¯S1n=0 (7-45) 
A similar treatment for .E¯/.ci gives a similar equation: 
c1 Hi1 -E¯Si1+c2 Hi2 -E¯Si2+···+cn Hin -E¯Sin=0 (7-46) 
Thus, requiring that .E¯/.ci vanish for all coefficients produces n homogeneous linear 
equations (homogeneous, all equal zero; linear, all ci ’s to first power). If one chooses 
a value for E¯, there remain n unknowns—the coefficients ci . (The integrals Hij and 
Sij are presumably knowable since Hˆ and the functions fi are known.) Of course, one 
trivial solution for Eqs. (7-46) is always possible, namely, c1 =c2=···=cn =0. But 
this corresponds to . =0, a case of no physical interest. Are there nontrivial solutions 
as well? In quantum chemical calculations, nontrivial solutions usually exist only for 
certain discrete values of E¯. This provides the approach for solving the problem. 
First, find those values of E¯ for which nontrivial coefficients exist. Second, substitute 
into Eqs. (7-46) whichever of these values of E¯ one is interested in and solve for the 
coefficients. (Each value of E¯ has its own associated set of coefficients.) But how do 
we find these particular values of E¯ that yield nontrivial solutions to Eqs. (7-46)? The 
answer is given in Appendix 2, where it is shown that the condition which must be met 
by the coefficients of a set of linear homogeneous equations in order that nontrivial 
solutions exist is that their determinant vanish. Notice that, in the standard treatment 
given in Appendix 2, the coefficients are known and x,y, and z are unknown. Here, 
however, the coefficients ci are unknown, and Hij and Sij are known. Therefore, it is 
the determinant of the H’s, S’s and E¯ in Eqs. (7-46) that must equal zero: 
 
H11 -E¯S11 H12 -E¯S12 ··· H1n -E¯S1n 
H21 -E¯S21 H22 -E¯S22 ··· H2n -E¯S2n 
...
...
... 
Hn1 -E¯Sn1 Hn2 -E¯Sn2 ··· Hnn -E¯Snn
 
=0 (7-47)
200 Chapter 7 The Variation Method 
Expansion of this determinant gives a single equation containing the unknown E¯. Any 
value of E¯ satisfying this equation is associated with a nontrivial set of coefficients. 
The lowest of these values of E¯ is the minimum average energy achievable by variation 
of the coefficients. Substitution of this value of E¯ back into Eqs. (7-46) produces 
n simultaneous equations for the n coefficients. Equation (7-47) is referred to as 
the secular equation, and the determinant on the left-hand side is called the secular 
determinant. 
This method is best illustrated by example, and we will now proceed with the problem 
of a hydrogen atom in a z-directed uniform electric field of strength F a.u. As 
mentioned earlier, a suitable choice of functions to mix together to approximate the 
accurate wavefunction is the 1s and 2p, hydrogenlike functions. The choice of two 
basis functions leads to a 2×2 secular determinantal equation: 
 
H11 -E¯S11 H12 -E¯S12 
H21 -E¯S21 H22 -E¯S22

=0 (7-48) 
If we arbitrarily associate the 1s function with the index 1 and the 2pz function with 
index 2 (consistent with . = c11s + c22pz ), then the terms in the determinant are 
(returning to general complex conjugate notation): 
H11 =
 1s*Hˆ 1s dt H12 =
 1s*Hˆ 2pz dt 
H21 =
 2p*z 
ˆ H1s dt H22 =
 2p*z 
Hˆ 2pz dt (7-49) 
(The electron label has been omitted since there is only one electron.) The corresponding 
S integrals are obtained from these if Hˆ is omitted in each case. The hamiltonian 
operator is just that for a hydrogen atom with an additional term to account for the 
z-directed field: 
Hˆ =-
1
2.2 -(1/r)-Fr cos . (7-50) 
[The energy of a charge -e in a uniform electric field of strength F and direction z is 
-eF z. In atomic units, one unit of field strength is e/a2
0 =5.142×1011 V/m. The unit 
of charge in atomic units is e, so this symbol does not appear explicitly in Eq. (7-50). 
Also, the identity z=r cos . has been used.] This may also be written 
Hˆ =Hˆhyd -Fr cos . (7-51) 
where Hˆhyd is the hamiltonian for the unperturbed hydrogen atom. 
The secular determinant contains four H-type and four S-type terms. However, 
evaluating these eight integrals turns out to be much easier than one might expect. 
In the first place, S12 = S21 since these integrals differ only in the order of the two 
functions in the integrand, and the functions commute. Also, because Hˆ is hermitian, 
it follows immediately that H21 =H* 12. This leaves us with three S terms and three H 
terms to evaluate. The three S terms are simple. Because the hydrogenlike functions 
are orthonormal, S11 and S22 equal unity, and S12 vanishes. These points have already 
reduced the secular determinantal equation to 
 
H11 -E¯ H12 
H1*2 H22 -E¯ 

=0 (7-52)
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom 201 
Consider next the term H11. This may be written as 
H11 =
 1s*Hˆhyd1s dt -
 1s* (Fr cos .) 1s dt (7-53) 
But the 1s function is an eigenfunction of Hˆhyd with eigenvalue -1
2 a.u. Therefore, the 
first integral on the right-hand side of Eq. (7-53) is 

 1s*Hˆhyd1s dt =-
1
2 

 1s*1s dt =-
1
2
a.u. (7-54) 
The second term on the right-hand side of Eq. (7-53) is zero by symmetry since 1s*1s 
is symmetric for reflection in the xy plane while r cos .(=z) is antisymmetric. Thus, 
H11=-1
2 a.u. Similarly, H22 =-18
a.u. [Recall that the 8 eigenvalues of hydrogen 
are equal to -1/(2n2), and here n=2.] All that remains is H12: 
H12 =
 1s*Hˆhyd2pz dt -F 
 1s* (r cos .) 2pz dt (7-55) 
The first term on the right-hand side is easily shown to be zero: 

 1s*Hˆhyd2pz dt =
 1s* 
-
1
8
2pz dt =0 (7-56) 
Here we employ the fact that 2pz is an eigenfunction of Hˆhyd and then the fact that 1s 
and 2pz AOs are orthogonal. The second term on the right-hand side must be written 
out in full and integrated by “brute force.” It is remarkable that, of the eight terms 
originally considered, only one needs to be done by detailed integration. Proceeding, 
we substitute formulas for 1s and 2pz in this last integral to obtain 
-F 


 (p)-1/2 exp(-r)r cos .(32p)-1/2r exp(-r/2) cos .(r2 sin .) dr d. df 
(7-57) 
We consider the integration over spin to have been carried out already, giving a factor 
of unity. Integrating over f to obtain 2p, and regrouping terms gives 
-2pF/(4v2p)
 8 
0 
r4 exp(-3r/2)dr 
 p 
0 
cos2 . sin. d. (7-58) 
=-F/2v2 4! 
3
2 
5 
2
3
=-
215/2F 
35 a.u. (7-59) 
This completes the task of evaluating the terms in the secular determinant.2 The final 
result is (in atomic units)
 
-1
2 -E¯ -215/2(F/35) 
-215/2(F/35) -18 
-E¯ 

=0 (7-60) 
2Since H12 is real, it is clear that H21 =H12.
202 Chapter 7 The Variation Method 
which expands to 
1 
16 +(5E¯/8)+E¯2 -215F2/310 =0 (7-61) 
This is quadratic in E¯ having roots 
E¯ =- 
5 
16 ±  9 
64 +217F2/310
1/2 
2 
(7-62) 
When no external field is present, F = 0 and the roots are just -1
2 and -18
a.u., the 
1s and 2pz eigenvalues for the unperturbed hydrogen atom. As F increases from zero, 
the roots change, as indicated in Fig. 7-3. We see that, for a given field strength, there 
are only two values of E¯ that will cause the determinant to vanish. If either of these 
two values of E¯ is substituted into the simultaneous equations related to Eq. (7-52), 
then nontrivial values for c1 and c2 can be found. Thus, at F =0.1 a.u., E¯ =-0.51425 
a.u., and E¯ = -0.1107 a.u. are values of E¯ for which .E¯/.c1, and .E¯/.c2 both 
vanish. The former is the minimum, the latter the maximum in the curve of E¯ versus 
c1 (Fig. 7-4). (The normality requirement results in the two c’s being dependent, and 
so E¯ may be plotted against either of them.) Since the variation principle tells us that 
E¯ = Elowest exact, we can say immediately that the energy of the hydrogen atom in a 
uniform electric field of 0.1 a.u. is -0.51425 a.u. or lower. That is, -0.51425 a.u. is 
an upper bound to the true energy. 
Now that we have the value of the lowest E¯1 we can solve for c1 and c2 and obtain 
the approximate ground state wavefunction. The homogeneous equations related to the 
determinant in Eq. (7-52) are 
c1(H11 -E¯)+c2H12 = 0 (7-63) 
c1H12 +c2 H22 -E¯ = 0 (7-64) 
Substituting -0.51425 for E¯, and inserting the values for H11,H22, and H12 found 
earlier gives (when F =0.1 a.u.) 
0.01425c1 -0.074493c2 = 0 (7-65) 
-0.074493c1 +0.38925c2 = 0 (7-66) 
Equation (7-65) gives 
c1 =5.2275c2 (7-67) 
If we substitute this expression for c1 into Eq. (7-66) we get 
-0.3892c2 +0.3892c2 =0 (7-68) 
This is useless for evaluating c2. It is one of the properties of such a set of homogeneous 
equations that the last unused equation is useless for determining coefficients. This 
arises because an equation like (7-63) still equals zero when c1 and c2 are both multiplied 
by the same arbitrary constant. Therefore, these equations are inherently capable of 
telling us the ratio of c1 and c2 only, and not their absolute values. We shall determine
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom 203 
Figure 7-3 
 Average energies for a hydrogen atom in a uniform electric field of strength F as 
given by a linear variation calculation using a 1s, 2pz basis. (- - -) Results from accurate calculations. 
absolute values by invoking the requirement that . be normalized. In this case, this 
means that 
c2
1 +c2
2 =1 (7-69) 
or 
(5.2275c2)2 +c2
2 =1 (7-70) 
which gives 
c2=±0.18789 (7-71) 
If we arbitrarily choose the positive root for c2, it follows from Eq. (7-67) that 
c1 =0.98219.
204 Chapter 7 The Variation Method 
Figure 7-4 
 E¯ versus c1 for a hydrogen atom in a uniform electric field of strength 0.1 a.u. 
Thus, when F =0.1 a.u., the linear variation method using a 1s, 2pz basis set gives 
an upper bound to the energy of -0.51425 a.u. and a corresponding approximate 
wavefunction of 
. =0.98219 1s+0.18789 2pz (7-72) 
As mentioned earlier, the admixture of 2pz with 1s produces the skewed charge distribution 
shown in Fig. 7-2. 
It is important to note that the extent of mixing between 1s and 2pz depends partly on 
the size of the off-diagonal determinantal element H12. When H12 is zero (no external 
field), no mixing occurs. As H12 increases, mixing increases. Generally speaking, 
the larger the size of the off-diagonal element connecting two basis functions in the 
secular determinant, the greater the degree of mixing of these basis functions in the 
final solution, other factors being equal. Hij is sometimes referred to as the interaction 
element between basis functions i and j . 
If we carried through the same procedure using the maximum E¯ of -0.1107 a.u., 
we would obtain the approximate wavefunction 
. =0.98219 2pz -0.18789 1s (7-73) 
This wavefunction is orthogonal to .. It may be proved that the second root, 
E¯ = -0.1107 is an upper bound for the energy of the second-lowest state of the
Section 7-4 Linear Variation: The Polarizability of the Hydrogen Atom 205 
hydrogen atom in the field. However, . is probably not too good an approximation 
to the exact wavefunction for that state. This is partly because the wavefunction . 
is one that maximizes E¯. Hence, there is no particular tendency for the procedure to 
isolate the second-lowest state from the infinite manifold of states. Also, our basis set 
was chosen with an eye toward its appropriateness for approximating the lowest state. 
The true second-lowest state might be expected to contain significant amounts of the 
2s AO, which is not included in this basis. 
The values of E¯ versus c1 are plotted in Fig. 7-4. The values of E¯ that we obtained 
by the variation procedure correspond to the extrema in this figure. The low-energy 
extreme corresponds to a wavefunction that shifts negative charge in the direction it 
is attracted by the field. The high-energy extreme corresponds to a wavefunction that 
shifts charge in the opposite direction. (When a calculation is performed over a more 
extensive basis to produce more than two roots, the highest and lowest roots correspond, 
respectively, to the maximum and minimum, the other roots to saddle points, on the 
energy hypersurface.) 
The detailed treatment just completed is rather involved, and so we now summarize 
the main points. Step 1 involved selection of a basis set of functions which is capable 
of approximating the exact solution. Step 2 was the construction of the secular 
determinant, including evaluation of all the Hij - and Sij -type integrals. Step 3 was 
the conversion of the determinantal equation into its equivalent equation in powers of 
E¯ and solution for the roots E¯. Step 4 was the substitution of an E¯ of interest into 
the simultaneous equations that are related to the secular determinant and solution for 
c1/c2. Finally, we used the normality requirement to arrive at convenient values for 
c1 and c2. 
There are many ways one could increase the flexibility of the trial function in an 
effort to increase the accuracy of the calculation. By adding additional basis functions, 
one would stay within the linear variation framework, merely increasing the size of the 
secular determinant. If these additional basis functions are of appropriate symmetries, 
they will cause the minimum energy root to be lowered further and will mix into the 
corresponding wavefunction to make it a better approximation to the lowest-energy 
eigenfunction for the system. Also, the additional functions will increase the number 
of roots E¯, thereby providing upper bounds for the energies of the third-, fourth-, etc. 
lowest states of the system. Another possibility is to allow nonlinear variation of the 
1s and 2pz orbital exponents, in combination with linear variation. This would be more 
involved than the calculation we have shown here, but could easily be accomplished 
with the aid of a computer. 
EXAMPLE 7-1 Suppose the variational process just described were performed 
using.(c1, c2, c3)=c11s+c22p0+c33d0. Would all three of theseAOs be present 
in the lowest-energy solution? 
SOLUTION 
 We already know that 1s and 2p0 will be mixed. They differ in reflection symmetry 
in just the right manner to provide a wavefunction skewed in the z direction, and this is 
manifested as a nonzero interaction element H12. 3d0, however, is, like 1s, symmetric for reflection 
through the x, y plane. Therefore, it cannot skew 1s in the z direction, and its interaction element 
with 1s, H13, equals zero (by symmetry). Therefore, it is tempting to think that that 3d0 will not 
contribute. However, because 3d0 and 2p0 have opposite reflection symmetries through the x,y
206 Chapter 7 The Variation Method 
plane, they do interact, H23 is not zero, and so all three basis functions show up in the lowest-energy 
solution. 1s and 3d0 are indirectly linked because they are each directly linked to 2p0. Physically, 
one can argue that a 2p0 AO that has been polarized by admixture with 3d0 can better polarize the 
1s AO than can the pure 2p0 AO. (One could say that 3d0 is brought in “on the coat-tails” of 2p0.) 
Note also that 1s has zero angular momentum along the z axis before the electric field is turned 
on. The z-directed field distorts the ground state, but has no effect on the z component of angular 
momentum. Therefore, only eigenfunctions having ml =0 can contribute to the polarized ground 
state.  
7-5 Linear Combination of Atomic Orbitals: 
The H+2
Molecule–Ion 
We are now ready to consider using the linear variation method on molecular systems. 
We begin with the simplest case, H+2 . This molecule–ion has enough symmetry so that 
we could guess many important features of the solution without calculation. However, 
to demonstrate the method, we shall first simply plunge ahead mathematically, and 
discuss symmetry later. 
The H+2 system consists of two protons separated by a variable distance R, and a 
single electron (see Fig. 7-5). The hamiltonian for this molecule is, in atomic units 
Hˆ (rA, rB, r1)=-
1
2 

.2
1 + .2
A 
1836 + .2
B 
1836

-
 1 
rA1
-
 1 
rB1
+
 1
R
 (7-74) 
Since all the particles in the system are capable of motion, the exact eigenfunction 
of Hˆ will be a function of the coordinates of the electron and the protons. However, 
the protons are each 1836 times as heavy as the electron, and in states of chemical 
interest their velocity is much smaller than that of the electron. This means that, to a 
very good degree of approximation, the electron can respond instantly to changes in 
internuclear separation. In other words, whenever the nuclei are separated by a given 
distance R, no matter how they got there, the motion of the electron will always be 
described in the same way by .. This means that we can separate the electronic and 
nuclear motions with little loss of accuracy.3 Given an internuclear separation, we can 
solve for the electronic wavefunction by ignoring nuclear motion [i.e., omitting .2
A and 
Figure 7-5 
 The H+2 molecule–ion. A and B are protons. The electron is numbered “1.” 
3However, there are times when coupling of electronic and nuclear motions becomes important.
Section 7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion 207 
.2
B in Eq. (7-74)]. This gives us a hamiltonian for the electronic energy and nuclear 
repulsion energy of the system: 
Hˆ (rA, rB, r1)=-
1
2.2
1 -
 1 
rA1
-
 1 
rB1
+
 1
R
 (7-75) 
For a given internuclear separation R, the internuclear repulsion 1/R is a constant, and 
we can omit it and merely add it on again after we have found the electronic energy. If 
we let Hˆelec stand for the first three terms on the right-hand side of Eq. (7-75), we can 
write 
Hˆelec.elec (r1)=Eelec.elec (r1) (7-76) 
and 
Eelec +Enuc rep =Eelec +1/R (7-77) 
Solving Eq. (7-76) for Eelec for every value of R allows us to plot the electronic and 
also the total energy of the system as a function of R. But Eelec(R)+1/R is just the 
potential energy for nuclear motion. Therefore, this quantity can be inserted as the 
potential in the hamiltonian operator for nuclear motion: 
Hˆnuc(rA, rB)=-
1
2 
 .2
A 
1836 + .2
B 
1836

+Eelec(R)+1/R (7-78) 
Hˆnuc(rA, rB).nuc(rA, rB)=Enuc.nuc (rA, rB) (7-79) 
The eigenfunctions of Eq. (7-79) describe the translational, vibrational, and rotational 
states of the molecule. Note that the eigenvalues of the hamiltonian for nuclear motion 
are total energies for the system because they contain the electronic energy in their 
potential parts. Hence, 
Etot =Enuc (7-80) 
But 
.tot(rA, rB, r1)=.elec(r1).nuc (rA, rB) (7-81) 
This approximation—that the electronicwavefunction depends only on the positions 
of nuclei and not on their momenta—is called the Born–Oppenheimer approximation.4 
Only to the extent that this approximation holds true is it valid, for example, to separate 
electronic and vibrational wavefunctions and treat various vibrational states as 
a subset existing in conjunction with a given electronic state. We will assume the 
Born–Oppenheimer approximation to be valid in all cases treated in this book. 
Making the Born–Oppenheimer approximation for H+2 , we seek to solve for the 
electronic eigenfunctions and eigenvalues with the nuclei fixed at various separation 
distances R. We already know these solutions for the two extremes of R. When the two 
4It is analogous to the concept of reversibility in thermodynamics: The piston moves so slowly in the cylinder 
that the gas can always maintain equilibrium, so pressure depends only on piston position; the nuclei move so 
slowly in a molecule that the electrons can always maintain their optimum motion at each R, so electronic energy 
depends only on nuclear position.
208 Chapter 7 The Variation Method 
nuclei are widely separated, the lowest-energy state is a 1s hydrogen atom and a distant 
proton. (Since there is a choice about which nucleus is “the distant proton,” there are 
really two degenerate lowest-energy states.) When R = 0, the system becomes He+ 
These two extremes are commonly referred to as the separated-atom and united-atom 
limits, respectively. 
In carrying out a linear variation calculation on H+2 , our first problem is choice of 
basis. In the separated-atom limit, the appropriate basis for the ground state would be a 
1s atomic orbital (AO) on each proton. Then, regardless of which nucleus the electron 
resided at, the basis could describe the wavefunction correctly. The appropriate basis 
at the united-atom limit is a hydrogenlike 1s wavefunction with Z=2. At intermediate 
values of R, choice of an appropriate basis is less obvious. One possible choice is 
a large number of hydrogenlike orbitals or, alternatively, Stater-type orbitals (STOs), 
all centered at the molecular midpoint. Such a basis is capable of approximating the 
exact wavefunction to a high degree of accuracy, provided a sufficiently large number 
of basis functions is used.5 Calculations using such a basis are called single-center 
expansions. A different basis, and one that is much more popular among chemists, 
is the separated-atom basis—a 1s hydrogen AO centered on each nucleus. At finite 
values of R, this basis can produce only an approximation to the true wavefunction. 
One way to improve this approximation is to allow additionalAOs on each nucleus, 2s, 
2p, 3s, etc., thereby increasing the mathematical flexibility of the basis. If we restrict 
our basis toAOs that are occupied in the separated-atom limit ground state (the 1sAOs 
in this case), then we are performing what is called a minimal basis set calculation. For 
now, we will use a minimal basis set. 
The wavefunction that we produce by linear variation will extend over the whole 
H+2 molecule, and its square will tell us how the electron density is distributed in the 
molecule. Hence, the one-electron molecularwavefunction is referred to as a molecular 
orbital (MO) just as the one-electron atomic wavefunction is referred to as an atomic 
orbital (AO). With a basis set of the type we have selected, the MOs are expressed as 
linear combinations of AOs. For this reason, this kind of calculation is referred to as 
a minimal basis set linear combination of atomic orbitals–molecular orbital (LCAO– 
MO) calculation. 
Our second problem, now that we have selected a basis, is construction of the secular 
determinant. Since we have only two basis functions (1sA, 1sB), we expect a 2 × 2 
determinant: 
 
HAA -E¯SAA HAB -E¯SAB 
HBA -E¯SBA HBB -E¯SBB

=0 (7-82) 
Here we have used the notation developed earlier, where 
HAB =
 1s*A (1)Hˆelec (1) 1sB (1) dt (1) (7-83) 
etc. E¯ of Eq. (7-82) is the average value of the energy. Henceforth, the bar is omitted. 
If we take our basis functions to be normalized, SAA = SBB = 1. Since our basis 
functions and hamiltonian are all real, their integrals will be real. Therefore, SAB =SBA 
5The hydrogenlike orbitals are a complete set if the continuum functions are included. Hence, this set can 
allow one to approach arbitrarily close to the exact eigenfunction and eigenvalue. The Slater-type orbitals do not 
constitute a complete set.
Section 7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion 209 
and HAB = H*BA = HBA. Since the hamiltonian is invariant to an interchange of the 
labels A and B, it follows that HAA =HBB. (HAA is the energy of an electron when 
it is in a 1s AO on one side of the molecule, HBB when it is on the other side.) This 
leaves us with 
 
HAA -E HAB -ESAB 
HAB -ESAB HAA -E 

=0 (7-84) 
For Eq. (7-84) to be satisfied, the product of the diagonal terms must equal that of the 
off-diagonal terms, which means that 
HAA -E=±(HAB -ESAB) (7-85) 
This gives two values for E: 
E± = 
HAA ±HAB 
1±SAB 
(7-86) 
To arrive at numerical values for E+ and E- requires that we choose a value for R 
and explicitly evaluate HAA, HAB, and SAB. We know in advance that SAB increases 
monotonically from zero at R=8 to unity at R = 0 because 1sA and 1sB are each 
normalized and everywhere positive. HAA is the average energy of an electron in a 1s 
AO on nucleus A, subject also to an attraction by nucleus B. Hence, HAA should be 
lower than the energy of the isolated H atom (-1
2 a.u.) whenever R is finite. HAB is 
easily expanded to 
HAB =
 1sA 
-
1
2.2 -1/rB1sBdv +
 1sA(-1/rA)1sB dv (7-87) 
The operator in the first integrand is simply the hamiltonian operator for a hydrogen 
atom centered at nucleus B. This operator operates on 1sB to give -1
2 1sB. Hence, 
the first integral becomes simply -1
2SAB. The second integral in Eq. (7-87) gives the 
attraction between a nucleus and the “overlap charge.” Thus, HAB is zero at R=8and 
negative for finite R. The formulas for these terms are (after nontrivial mathematical 
evaluation) 
SAB = 
 1sA1sB dv =exp(-R)1+R +R2/3 (7-88) 
HAA = 
 1sAHˆelec1sA dv =-
1
2 -(1/R) 1-e-2R (1+R) (7-89) 
HAB = 
 1sAHˆelec1sB dv =-SAB/2-e-R (1+R) (7-90) 
When R =2 a.u., SAB =0.586, HAA=-0.972 a.u. and HAB =-0.699 a.u. Inserting 
these values into Eq. (7-86) gives E+ =-1.054 a.u. and E- =-0.661 a.u. These 
are electronic energies. Internuclear repulsion energy (+1
2 a.u.) can be added to these 
values to give -0.554 a.u. and -0.161 a.u., respectively. 
The ways in which HAA, HAB, SAB, and 1/R contribute to the energy are illustrated 
for R=2 a.u. in Fig. 7-6. HAA is lower than the energy of an isolated H atom because
210 Chapter 7 The Variation Method 
Figure 7-6 
 Contributions to the energy of H+2 at R =2 a.u. in a minimal basis LCAO–MO calculation. 
the electron experiences additional nuclear attraction at R =2. The effect of the HAB 
interaction element is to split the energy into two levels equally spaced above and below 
HAA. The SAB term has the effect of partially negating this splitting. The internuclear 
repulsion energy 1/R merely raises each level by 1
2 a.u. The lower energy, E+ +1/R, 
has a final value that is lower than the separated-atom energy of -1
2 a.u. Since the exact 
energy at R=2 must be as low or lower than our value of-0.554 a.u., we can conclude 
that the H+2 molecule–ion has a state that is stable, with respect to dissociation into 
H +H+, by at least 0.054 a.u., or 1.47 eV, or 33.9 kcal/mole, neglecting vibrational 
energy effects. 
The data depicted in Fig. 7-6 are sometimes presented in the abbreviated form of 
Fig. 7-7. The energy levels for the pertinentAOs of the separated atoms are indicated on 
the left and right, and the final energies (either electronic or electronic plus internuclear) 
are shown in the center.
Section 7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion 211 
Figure 7-7 
 Separated atom energies and energies at an intermediate R for H+2 . 
Figure 7-8 
 E± + 1/R versus R for H+2 . (—) Calculation described in text. (- - -) Exact 
calculation. 
The behavior of these energies as a function of R is plotted in Fig. 7-8. Included 
for comparison are the exact energies for the two lowest-energy states of H+2 . Only 
the lower of these shows stability with respect to molecular dissociation. Both energy 
levels approach infinity asymptotically as R approaches zero because of internuclear 
repulsion. (The zero of energy corresponds to complete separation of the protons and 
electron.) 
Having found the roots E± for the secular determinant, we can now solve for the 
coefficients which describe the approximate wavefunctions in terms of our basis set. 
Let us first find the approximatewavefunction corresponding to the lower energy. To do 
this, we substitute the expression for E+ [Eq. (7-86)] into the simultaneous equations 
associated with the secular determinant (7-84): 
cA(HAA -E+)+cB(HAB -E+SAB) = 0 (7-91) 
cA(HAB -E+SAB)+cB(HAA -E+) = 0 (7-92)
212 Chapter 7 The Variation Method 
Equation (7-91) leads to 
cA HAA -(HAA +HAB)/ 1+SAB =-cB HAB -(HAA +HAB)SAB/ 1+SAB  
(7-93) 
which ultimately gives 
cA =cB (7-94) 
The same procedure for E- produces the result 
cA=-cB (7-95) 
The normality requirement is 
(7-96) 
so that 
c2
A +c2
B +2cAcBSAB =1 (7-97) 
For cA =cB, this gives 
cA =1/ [2 (1+SAB)]1/2 =cB (7-98) 
For cA=-cB, 
cA =1/ [2 (1-SAB)]1/2=-cB (7-99) 
Therefore, the LCAO–MO wavefunction corresponding to the lower energy E+ is 
.+ = 
1 
v2(1+SAB) 
(1sA +1sB) (7-100) 
The higher-energy solution is 
.- = 
1 
v2(1-SAB) 
(1sA -1sB) (7-101) 
Just aswas true forAOs, there are a number ofways to display these MOs pictorially. 
One possibility is to plot the value of .± or .2±
along a ray passing through both nuclei. 
(Other rays could also be chosen if we were especially interested in other regions.) 
Another approach is to plot contours of . or .2 on a plane containing the internuclear 
axis, Still another way is to sketch a three dimensional view of a surface of constant 
value of . or .2 containing about 90–95% of the wavefunction or the electron charge. 
Finally, one can plot the value of . as distance above or below the plane containing 
the internuclear axis. All of these schemes are shown in Fig. 7-9 for .+ and .-.
Section 7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion 213 
Figure 7-9 
 (a) Plot of .+, along the z axis. [Eq. (7-100)]. (b) Plot of .- along the z axis 
[Eq. (7-101)]. 
The wavefunctions .+ and .- may be seen from Fig. 7-9 to be, respectively, symmetric 
and antisymmetric for inversion through the molecular midpoint. [They are 
commonly called gerade (German for even) and ungerade, respectively.] This would 
be expected for nondegenerate eigenfunctions of the hamiltonian, since it is invariant 
to inversion. But .+ and .- are not eigenfunctions. They are only approximations 
to eigenfunctions. How is it that they show the proper symmetry characteristics of 
eigenfunctions? The reason is that the symmetry of the H+2 molecule is manifested as 
a symmetry in our secular determinant of Eq. (7-84). Note that the determinant is symmetric 
for reflection across either diagonal. The symmetry for reflection through the 
principal diagonal (which runs from upper left to lower right) is due to the hermiticity 
of Hˆ , and is always present in the secular determinant for any molecule regardless of 
symmetry.6 Symmetry for reflection through the other diagonal is due to the fact that 
the hamiltonian is invariant to inversion and also to the fact that the basis functions at 
6Hermiticity requires that Hij =Hji *. If Hij is imaginary, the determinant will be antisymmetric for reflection 
through the principal diagonal.
214 Chapter 7 The Variation Method 
Figure 7-9 
 (Continued) (c) Contour diagram of .+. (d) Contour diagram of .-. 
the two ends of the molecule are identical. When theAOs in a basis set are interchanged 
(times±1) by a symmetry operation of the molecule, the basis set is said to be balanced 
for that operation. Thus, a 1sA and 1sB basis is balanced for inversion in H+2 , but a 
1sA and 2sB basis is not. Whenever a symmetry-balanced basis is used for a molecule, 
the symmetry of the molecule is manifested in the secular determinant and ultimately 
leads to approximate MOs that show the proper symmetry characteristics. 
Since the eigenfunctions for H+2 must be gerade or ungerade, and since we started 
with the simple balanced basis 1sA and 1sB, it should be evident that it is unnecessary 
to go through the linear variation procedure for this case. With such a simple basis 
set, there is only one possible gerade linear combination of AOs, namely 1sA +1sB. 
Similarly, 1sA -1sB is the only possible ungerade combination. Therefore, we could 
have used symmetry to guess our solutions at the outset. Usually, however, we are not so 
limited in our basis. We shall see that, while symmetry is useful in such circumstances, 
it does not suffice to produce the variationally best solution.
Section 7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion 215 
Figure 7-9 
 (Continued) (e) Three-dimensional sketch of contour envelope for .+ and 
(f) for .-. 
According to the theorem mentioned earlier (but not proved), the nth lowest root 
of a linear variation calculation for a state function must lie above the nth lowest 
exact eigenvalue for the system. However, the two states we are dealing with have 
different symmetries. In such a case, a more powerful boundedness theorem holds— 
one that holds even if we are not using a linear variation procedure. To prove this, 
we first recognize that every H+2 eigenfunction is either gerade (i.e., symmetric for 
inversion) or ungerade. Since the lowest-energy approximate wavefunction .+ is 
gerade, it must be expressible as a linear combination of these gerade eigenfunctions. 
Hence, its average energy E+ cannot be lower than the lowest eigenvalue for the 
gerade eigenfunctions. Similarly, E- cannot be lower than the lowest eigenvalue for 
the ungerade eigenfunctions, and so we have a separate lower bound for the average 
energy of trial functions of each symmetry type. For this reason, our approximate 
energies in Fig. 7-8 must lie above the exact energies for both states. If we were to 
make further efforts to lower the average energy of the ungerade function, even by 
going outside the linear variation procedure, we could never fall below the exact energy 
for the lowest-energy ungerade state unless we somehow allowed our trial function to 
change symmetry. This means that a lowest average energy criterion can be used in 
attempting to find the lowest-energy state of each symmetry type for a system, by either 
linear or nonlinear variation methods.
216 Chapter 7 The Variation Method 
Figure 7-9 
 (Continued) (g)Value of .+ and (h) of .- versus position in plane containing the 
nuclei at points a and b. 
EXAMPLE 7-2 Suppose one were to do a variational calculation on the hydrogen 
atom using the unnormalized trialwavefunction f=exp(-ar2) cos ., with a being 
varied. What lower bound could we expect for the average energy? 
SOLUTION 
 This function has a node in the x, y plane (where . =p/2), and this nodal plane 
exists regardless of the value of a. Therefore, our trial function is like the 2p0 AO in symmetry. It 
will be represented by a linear combination of p0 AOs–a combination that changes as a is varied. 
The average value of energy cannot be lower than the lowest eigenvalue in the set, which is-1/8 a.u. 
(for the 2p0 AO).  
Inspection of Fig. 7-9 shows that the charge distribution for the state described by 
.+ is augmented at the molecular midpoint compared to the charge due to unperturbed
Section 7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion 217 
atoms. This state is also the one that gives H+2 stability at finite R. Because this MO 
puts electronic charge into the bond region and stabilizes the molecule, it is commonly 
called a bonding MO. In the state described by .-, charge is shifted out of the bond 
region and the molecule is unstable at finite R, and so .- is called an antibonding MO. 
Because the potential in H+2 (or in any linear molecule) is independent of f, the 
angle about the internuclear axis, the f dependence of the wavefunctions is always of 
the form 

(f)=(1/v2p)exp(imf), m=0,±1,±2, . . . (7-102) 
This fact may be arrived at in two ways. One way is to write down the Schr¨odinger 
equation for H+2 using spherical polar coordinates or elliptical coordinates. (f is a 
coordinate in each of these coordinate systems.) Then one attempts to separate coordinates 
and finds that the f coordinate is indeed separable from the others and yields 
the equation 
d2
(f) 
df2 =-m2
(f) (7-103) 
The acceptable solutions of this are the functions (7-102). The other approach is to note 
that, since the potential in Hˆ has no f dependence, Hˆ commutes with Lˆz , the angular 
momentumoperator, so that the eigenfunctions ofHˆ are simultaneously eigenfunctions 
of Lˆz .We know the eigenfunctions of Lˆz are the functions (7-102), and thus, we know 
that these functions must also give the f dependence of the eigenfunctions of Hˆ . 
Two important conclusions emerge. First, each nondegenerate H+2 wavefunction 
must have a f dependence given by one of the functions (7-102). This tells us something 
about the shapes of the wavefunctions. Second, the nondegenerate H+2 wavefunctions 
are eigenfunctions of Lˆz , which means that an electron in any one of these states 
has a definite, unvarying (i.e., sharp) component of angular momentum of value m¯h 
mks units (m a.u.) along the internuclear axis. It is thus useful to know the m value 
associated with a given one-electron wavefunction. A standard notation is used, which 
is analogous to the atomic orbital notation wherein s, p, d, f, correspond to l values of 
0, 1, 2, 3, respectively. The corresponding Greek lower-case letters s, p, d, f indicate 
values of |m| of 0, 1, 2, 3, respectively, in one-electron orbitals of linear molecules. 
Thus, for the case at hand, .+ and .- are both s MOs because they are cylindrically 
symmetrical (i.e., no f dependence), which requires that m be zero. Because .+ is a 
gerade function, it is symbolized sg. .-, then, is a su MO. (A simple way to determine 
whether anMOis s, p, or d is to imagine viewing it (in its real form) end-on (i.e., along 
a projection of the internuclear axis.) If the MO from this view looks like an s AO, 
it is a s MO. If it looks like a p AO, it is a p MO. A d MO has the four-leaf-clover 
appearance of a d AO.) 
Let us nowexamine the dependence of our LCAO–MOresults on our original choice 
of basis. The 1s AOs we have used are capable of giving the exact energy when 
R=8, but become increasingly inadequate as R decreases. As a result, Fig. 7-8 
shows that, for both states, the approximate energy deviates more and more from the 
exact energy as R decreases. At R =0, our sg function becomes a single 1s H atom 
(Z = 1) AO centered on a doubly positive nucleus. Yet we know that the lowestenergy 
eigenfunction for that situation is a single 1s He+(Z = 2) AO. An obvious 
way to improve our wavefunction, then, is to allow the 1s basis functions to change
218 Chapter 7 The Variation Method 
their orbital exponents as R changes. This adds a nonlinear variation, and the calculation 
is more complicated. It is very easily performed with the aid of a computer, 
however, and we summarize the results in Figs. 7-10 and 7-11. The internuclear repulsion 
has been omitted from the energies in Fig. 7-10. The lowest approximate energy 
curve is now in perfect agreement with the exact electronic energy both at R =0 and 
R=8, and shows improved, though still not perfect, agreement at intermediate R. 
The R dependence of the orbital exponent for this wavefunction (Fig. 7-11) varies 
smoothly from 1 at R=8 to 2 at R = 0, as expected. In contrast, the su function 
fails to reach a well-defined energy at R = 0 because the function becomes indeterminate 
at that point. [At R = 0, 1sA - 1sB becomes (1sA - 1sA) = 0.] However, 
Figs. 7-10 and 7-11 indicate that the exact energy E- is -1
2 a.u. at R = 0, corresponding 
to the n = 2 level of He+, and that the orbital exponent in our 1s basis 
functions approaches 0.4 in the effort to approximate this state function at small R. 
To understand this behavior, we must once again consider the symmetries of these 
functions. 
We have already seen that the nondegenerate eigenfunctions of H+2 must be either 
gerade or ungerade, and we note in Fig. 7-10 that the energy curve for the gerade state 
is continuous as is the one for the ungerade state. In other words, as we move along 
a given curve, we are always referring to a wavefunction of the same symmetry. This 
continuity of symmetry along an energy curve is central to many applications of quantum 
chemistry. The reason for continuity of symmetry can be seen by considering a 
molecule having some element of symmetry, and having a nondegeneratewavefunction 
or molecular orbital (which must be symmetric or antisymmetric with respect to the 
symmetry operations of the molecule). If we change the molecule infinitesimally (without 
destroying its symmetry), we expect thewavefunction to change infinitesimally also. 
Figure 7-10 
 E+ and E- for H+2 from (- - -) exact, (···) variable . , and (—) fixed . (. =1) 
treatments.
Section 7-5 Linear Combination of Atomic Orbitals: The H+2
Molecule–Ion 219 
Figure 7-11 
 Values of . minimizing E+ and E- as a function of R. 
In particular it should not change symmetry because this is generally not an infinitesimal 
change. (To change symmetry requires adding or removing nodes, changing signs 
in parts of the function. Such changes have more than infinitesimal effects on . and 
on kinetic and potential parts of the energy.) The entire curve can be traversed by an 
infinite number of such infinitesimal but symmetry conserving steps. 
The continuity of symmetry enables the sg state of H+2 to correlate with an s-type 
AO of He+ as R goes to zero. This correlation is symmetry allowed because the 
s-type AOs of He+ have the proper symmetry characteristics—they are gerade and 
have no dependence on angle about the axis that is the internuclear axis when R>0 
(see Fig. 7-12). In contrast, the su MO cannot correlate with an s-type AO. It must 
correlate with an AO that is cylindrically symmetrical about the old internuclear axis 
but is antisymmetric for inversion. A p-type AO pointing along the old internuclear 
axis (called a ps AO) satisfies these requirements. 
Figure 7-12 
 Sketches demonstrating howseparated atom functions can be related to united-atom 
functions through symmetry invariance. These functions are not drawn to a common scale.
220 Chapter 7 The Variation Method 
These symmetry requirements help us understand the su curves of Figs. 7-10 and 
7-11. The exact energy for E- goes to -1
2 a.u. at R =0 because the su state of H+2 
correlates with a 2p AO of He+. Our basis set is incapable of reproducing a 2p AO at 
R =0, and so the calculated energy curve fails to rejoin the exact curve at R =0. At 
small R, our basis set is attempting to approximate the two lobes of an evolving 2p 
AO. Apparently the orbital exponent that best enables the basis to accomplish this is 
around 0.4. 
7-6 Molecular Orbitals of Homonuclear 
Diatomic Molecules 
We have already seen how one produces the ground configurations for manyelectron 
atoms by placing pairs of electrons of opposite spin into AOs, starting 
with the lowest-energy AO and working up. Subsequent manipulation of this 
product function to produce proper space and spin symmetry yields the approximate 
wavefunction. Precisely the same procedure is used for molecules. Thus, 
the electronic configurations for H+2 , H2, and H-2 are 1sg, (1sg)2, and (1sg)21su, 
respectively, and the approximate wavefunction for H2 is provided by the Slater 
determinant 
1sg (1) 1 ¯sg (2)
. When we come to consider heavier homonuclear 
diatomic molecules, such as O2, we must place electrons in higher energy MOs. 
Such MOs are still provided by the minimal basis set, which now includes 1s, 2s, 
and three 2p AOs on each atom since these AOs are occupied in the separated 
atoms. We now consider the natures of the additional MOs produced by this larger 
basis. 
We begin by making a change to a basis set that is mathematically equivalent to the 
starting set but is more convenient for discussing and analyzing the problem. This new 
set is the set of symmetry orbitals (SOs) (1sA ±1sB), (2sA ±2sB), etc. From our original 
ten AOs, we thus produce ten SOs. These may be normalized, if desired. Each of 
these SOs has definite symmetry. The SOs built from 2sAOs must be of sg and su symmetry 
since the 2sAOs act like 1sAOs for all the symmetry operations of the molecule. 
The 2p AOs pointing along the internuclear axis have cylindrical symmetry and hence 
also give rise to a sg and a su SO. (We will take the internuclear axis to be coincident 
with the z axis, and so these SOs are constructed from 2p0 (or 2pz )AOs.) The functions 
(2p+1A ±2p+1B ) are p SOs because |m|=1. Since it is difficult to visualize complex 
functions, however, the usual practice is to take linear combinations of the complex 
p functions to produce a corresponding set of real functions. (This is completely analogous 
to forming real px and py AOs from complex p+1 and p-1 AOs.) Thus, we obtain 
(2pxA ± 2pxB ) and (2pyA ± 2pyB ), which are not eigenfunctions for the Lˆz operator 
anymore, but are still given the symbol p. From Fig. 7-13, we see that the positive 
combinations give ungerade p SOs. This is just the opposite of the case for s-type SOs. 
We conclude from Fig. 7-13 that sg and pu SOs will tend to place charge into the bond 
and hence contribute to bonding, whereas su and pg SOs will contribute antibonding 
character. 
Even though these SOs are only a basis set, the reader may nevertheless recognize 
that conversion to this symmetrized basis goes a longway toward producing our ultimate 
MOs. Indeed, our (1sA±1sB) SOs, if normalized, are the same as the MOs we obtained
Section 7-6 Molecular Orbitals of Homonuclear Diatomic Molecules 221 
Figure 7-13 
 Symmetry orbitals constructed from s- and p-type AOs. (a) Sketches according to 
an idealized convention that ignores overlap between AOs on A and B. (b) Effects of overlap. Note 
that the psA -psB combination is bonding. This depends on our having chosen a common z axis for 
both atoms. Sometimes the z axes are chosen to point from each atom toward the other. In that case, 
psA -psB becomes antibonding. 
for H+2 . The essential advantage of a symmetrized basis set is that it simplifies the 
secular determinant and makes it easier to understand and describe the mixing of the 
basis functions by the hamiltonian. For instance, since our MOs must have pure s, 
p, d, . . . and also g or u symmetry, and since our SOs are already of pure symmetry, 
we expect no further mixing to occur between SOs of different symmetry in forming 
MOs. This suggests that the interaction elementHij between SOs fi and fj of different 
symmetry should vanish. This is easily proved by noting that Hˆ is symmetric for 
all symmetry operations of the molecule, and, if fi and fj differ in symmetry for 
some operation, their product is antisymmetric for that operation, and therefore f*i Hfj 
is antisymmetric and its integral vanishes. Similarly, Sij vanishes, and Hij - ESij 
vanishes except in positions connecting basis functions of identical symmetry. As a 
result, our secular determinant over SOs has the form (7-104), where the notation sg[1s] indicates a sg SO made from 1sA, 1sB AOs, etc. By placing basis functions of like 
symmetry together, we have emphasized the block-diagonal form of our determinant, 
all elements in the nonshaded areas being zero by symmetry. (The pg[2px] and pg[2py] do not interact because of symmetry disagreements for reflection in the xz and yz 
planes.) Each of the nonzero blocks of (7-104) is a separate determinant (which is 
just a number), and the value of determinant (7-104) is simply the product of these six
222 Chapter 7 The Variation Method 
smaller determinants. Hence, if any one of these small determinants is zero, the large 
determinant is zero, thereby satisfying our determinantal equation. Therefore, each of 
these small determinants may be employed in a separate determinantal equation, and 
the problem is said to be partitioned into six smaller problems. It follows immediately 
that we can get mixing among the three sg SOs to produce three sg MOs and likewise 
for the su set, and that the p SOs can undergo no further mixing and hence are already 
MOs. (Notice that SOs in one block are not linked by off-diagonal elements either 
directly or indirectly, unlike the case described in Example (7-1).) 
(7-104) 
What will be the nature of the lowest-energy sg MO? It will not be pure sg[1s] because admixture of sg[2s] and sg[2p] SOs can produce charge shifts that will lower 
the energy. But the sg[2s] and sg[2p] SO energies are much higher than the sg[1s] (mainly because the 2s and 2p AOs are higher in energy in the atoms, and the atomic 
contributions still dominate in the molecule). Any energy decrease to be gained by 
charge shifting must be weighed against the energy increase due to the mixing in of 
such high energy components. The former very quickly become overbalanced by the 
latter, so the lowest energy sg MO is almost pure sg[1s] SO, the sg[2s] and sg[2p] SOs 
coming in only very slightly. This exemplifies an important general feature of quantum 
chemical calculations: mixing between basis orbitals tends to be small if they have 
widely different energies in the system. Thus, we now have two factors governing the 
extent of mixing of functions fi and fj—the size of the interaction element Hij , and 
the difference in energy between them in the system (Hii -Hjj ). 
A label we can use for this lowest-energyMOthat avoids implying that it is identical 
to the sg[1s] SO is 1sg. This stands for “the lowest-energy sg MO.” Because of the low 
energy of the 1s AOs, the next-lowest MO is almost pure su[1s], and we label it 1su. 
When considering the remaining two sg MOs, we can expect substantial mixing 
between sg[2s] and sg[2p] SOs because these functions are not very different in energy. 
In the hydrogen atom, the 2s and 2p AOs are degenerate, and, as we move along the 
periodic table, they become split farther and farther apart in energy. Therefore, we 
might expect to find the greatest mixing for B2 and C2, and to find less mixing for O2 
and F2. Figure 7-14 is a schematic diagram of the MO energy levels we should expect 
for F2. Here the 2sg MOis primarily the sg[2s] SO and the 3sg MOis mainly the sg[2p]
Section 7-6 Molecular Orbitals of Homonuclear Diatomic Molecules 223 
Figure 7-14 
 Schematic of MO energy level order for F2. The vertical axis is not an accurate 
energy scale. For instance, the 1s levels are very much lower in energy relative to the other levels 
than is suggested by the drawing. (- - -) indicates that the orbital is unoccupied in the ground state. 
SO. Similarly 2su and 3su are mainly su[2s] and su[2p], respectively. The 1p and 3s 
MOs are degenerate at the separated atom limit, where they are all 2p AOs. As the 
atoms come together and interact, the p levels split apart less than the s levels because 
the 2px and 2py AOs approach each other side to side, whereas the 2ps AOs approach 
end to end. The latter mode produces larger overlap and leads to larger interaction 
elements and greater splitting. (Because of the symmetry of the molecule, the 1pu MOs 
are always degenerate, and they may be mixed together in any way. In particular, we 
can regard them as being 1pux and 1puy or 1pu+1 and 1pu-1 with equal validity. The 
same situation holds for the 1pg pair.) From the ordering of energy levels in Fig. 7-14 
we obtain for F2 the configuration 
F2 : (1sg)2(1su)2(2sg)2(2su)2(3sg)2(1pu)4(1pg)4 (7-105) 
Now we will consider what happens for lighter molecules. Recall that here the 2s 
and 2pAOs are closer together in energy. This allows greater mixing between the SOs 
containing these AOs and produces increased energy level splitting. Thus, the sg[2s] and sg[2p] SOs mix together more, and the resulting splitting causes an additional 
lowering in energy of the 2sg level and an increase for the 3sg level, compared to the 
F2 case. In a similar way the 2su and 3su levels are lowered and raised, respectively, by 
increased mixing between su[2s] and su[2p] SOs. The resulting energy level pattern 
for C2 is shown in Fig. 7-15. Note that the extra splitting has pushed the 3sg level above 
the 1pu level. No other change in the orbital energy ordering has occurred. As a result, 
C2 has, the configuration
224 Chapter 7 The Variation Method 
Figure 7-15 
 Aschematic diagram of theMOenergy level order forC2. The 2s and 2ps symmetry 
functions mix more strongly in producing 2s and 3s MOs than is the case in F2. The level shifts 
discussed in the text are indicated by dashed lines and arrows. (-------) represent F2 levels. Note 
the inversion of the order of the 1pu and 3pg levels. The orbital ordering is deduced from spectra. 
(See Mulliken [1].) (---) represent energies of orbitals not occupied by electrons. Note that there 
is no well-defined energy ordinate for this figure, and no accurate relationship between absolute values 
of orbital energies within a molecule or between molecules is implied. 
C2 : (1sg)2(1su)2(2sg)2(2su)2(1pu)4 (7-106) 
The above discussion is an effort to rationalize orbital energies obtained from calculations 
or deduced from molecular spectra. We have not yet described the details of 
how one goes about carrying out MO calculations on these molecules, nor will we in 
this chapter. However, even in the absence of precise numerical results, it is possible for 
such qualitative arguments to be very useful in understanding and predicting features 
of a wide range of chemical reactions. 
The molecular configurations obtained from the patterns of Figs. 7-14 and 7-15 
predict molecular properties that are in strikingly good qualitative agreement with 
experimental observations. The data in Table 7-2 show that, when we go from H+2 
to H2, adding a second electron to a bonding MO, the bond length decreases and the 
dissociation energy increases. Adding a third electron (He+2
) causes partial occupation 
of the antibonding 1su MO and causes the bond length to increase and the dissociation 
energy to decrease. The four electron molecule He2 is not observed as is consistent 
with its configuration. The relative inertness of N2 as a chemical reactant becomes 
understandable from the fact that it has six more “bonding electrons” than antibonding 
electrons, giving a net of three bonds—one s and two p bonds. For N2 to react, it
Section 7-6 Molecular Orbitals of Homonuclear Diatomic Molecules 225 
TABLE 7-2 
 Some Properties of Homonuclear Diatomic Molecules and Ions in their Ground 
Electronic States 
Molecule MO configuration 
Net number 
of bonding 
electrons 
Binding 
energy, 
De (eV) 
Equilibrium 
internuclear 
separation, 
Re(Å) Termc 
H + 2 1sg 1 2.7928 1.06 2
g 
H2 1sg
2 2 4.747745 0.7414 1
g+ 
H - 2 1sg
21su 1 1.7a 0.8 2
u + 
He2+ 1sg
21su 1 2.5 1.08 2
u + 
He2 1sg
21su
2 0 0.001b 2.88 (1
g+) 
He2- [He2]2sg 1 No data 2
g+ 
Li2+ [He2]2sg 1 1.29 3.14 2
g+ 
Li2 [He2]2sg
2 2 1.05 2.673 1
g+ 
Li2- [He2]2sg
22su 1 ~1.3(?) 3.2 2
u + 
Be2+ [He2]2sg
22su 1 No definitive 
data 
2
u + 
Be2 [He2]2sg
22su
2 0 0.1 2.49 1
g+ 
Be2- [Be2]1pu 1 ~0.3 2.4 2u 
B2+ [Be2]1pu 1 1.8 — 2u 
B2 [Be2]1pu
2(?) 2 ~3 1.589 3
g- 
B2- [Be2]1pu
3 3 No data 2u 
C2+ [Be2]1pu
3 3 5.3 1.301 2u 
C2 [Be2]1pu
4 4 6.36 1.2425 1
g+ 
C2- [Be2]1pu
43sg 5 8.6 — 2
g+ 
N2+ [Be2]1pu
43sg 5 8.86 1.116 2
g+ 
N2 [Be2]1pu
43sg
2 6 9.90 1.098 1
g+ 
N2- [Be2]1pu
43sg
21pg 5 ~8.3 — 2g 
O2+ [Be2]1pu
43sg
21pg 5 6.7796 1.1171 2g 
O2 [Be2]3sg
21pu
41pg
2 4 5.2132 1.2075 3
g- 
O2- [Be2]3sg
21pu
41pg
3 3 4.14 1.32 2g 
F2+ [Be2]3sg
21pu
41pg
3 3 3.39 1.32 2g 
F2 [Be2]3sg
21pu
41pg
4 2 1.65 1.42 1
g+ 
F2- [Be2]3sg
21pu
41pg
43su 1 ~1.3 1.9 2
u + 
Ne2+ [Be2]3sg
21pu
41pg
43su 1 ~1.1 1.7 2
u + 
Ne2 [Be2]3sg
21pu
41pg
43su
2 0 0.003b 3.09 (1
g+) 
aThis state is unstable with respect to loss of an electron, but is stable with respect to dissociation into an atom 
and a negative ion. 
bFrom Hirschfelder et al. [2]. It may be shown that any two neutral atoms will have some range of R where 
the attractive part of the van der Waals’ interaction dominates. For He2, this minimum is so shallow and the 
nuclei so light that a stable state (including vibrations) probably cannot exist. For Ne2, a stable state should 
exist. The data for He2 and Ne2 are calculated from considerations of intermolecular forces. 
cThe term symbol corresponds to the configuration of column 2.
226 Chapter 7 The Variation Method 
is necessary to supply enough energy to at least partially break these bonds prior to 
forming new bonds. The reversal of the orbital energy order between 1pu and 3sg can 
be seen to occur between N2 and O2. 
The configurations of Table 7-2 indicate that some molecules will have closed shells, 
whereas others will have one or more unpaired electrons. For example, O2 has a 
configuration in which the degenerate 1pg level “contains” two electrons. Several states 
can be produced from such a configuration, just as several states could be produced from 
the 1s2s configuration for He. Hund’s first rule has been found to hold for molecules, 
and so the state of highest multiplicity is lowest in energy. For n unpaired electrons, 
the highest multiplicity achievable is n+1. For O2, this is three, and we expect the 
ground state of the O2 molecule to be a triplet. This is conveniently demonstrated to be 
so by observing that liquid oxygen is attracted into the gap between poles of a magnet, 
consistent with the existence of uncanceled electron spin magnetic moments. 
Close perusal of Table 7-2 indicates that some of the data are not in accord with 
the simple qualitative ideas just presented. For example, Li+2
is more strongly bonded 
than is Li2, even though the former has fewer bonding electrons. However, the Li+2 
ion-molecule is longer than Li2. H-2 is less strongly bound than isoelectronic He+2
, yet 
it has a shorter equilibrium internuclear separation. Irregularities such as these require 
more detailed treatment. However, one of the useful characteristics of a qualitative 
approach is that it enables us to recognize cases that deviate from our expectations and 
therefore warrant further study. 
In Figs. 7-14 and 7-15, we saw how MOs are related to SOs for the separated 
atoms. Let us now consider how the separated atom SOs correlate with the united 
atomAOs. Recall that these orbitals are correlated by requiring them to be of identical 
symmetry. In Fig. 7-16 some of the possible SOs and united-atom AOs, together with 
their symmetry labels, are shown. Note that the sg SOs can correlate with s or ds AOs, 
su SOs with ps AOs, pu SOs with pp AOs and pg SOs with dp AOs. This gives us all 
the information we need except for resolving the ambiguities within a given symmetry 
type. For instance, which of the 1s, 2s, 3s, 3ds , . . . in theAOs correlates with which of 
the sg[1s], sg[2s], sg[2p], sg[3s], . . . in the SOs? This question is resolved by use of the 
noncrossing rule, which states that, in correlation diagrams, energy levels associated 
with orbitals or states of the same symmetry will not cross. This requires that we match 
up the lowest-energy united-atom AO of a given symmetry with the lowest energy SO 
of that symmetry, and so on up the ladder. This leads to the diagram in Fig. 7-17. The 
line interconnecting the 1s AO of the united atom with the 1s AOs of the separated 
atoms refers, at intermediate R, to the 1sg MO.We have already seen that thisMOmay 
contain contributions from sg[2s] and sg[2p] SOs. Thus, the correlation diagram tells 
us what orbitals the 1sg MO “turns into” at the limits of R, but does not imply that, at 
other R values, this MO is comprised totally of 1s AOs. 
Study of this correlation diagram reveals that the antibondingMOs(su andpg) are the 
MOs that correlate with higher energy united-atom AOs and hence favor the separated 
atoms in terms of energy. This results from the fact that antibonding MOs have a nodal 
plane bisecting the bond which is preserved as we proceed to the united atom, yielding 
a united atomAO having one more node than the separatedAOs we started with. Thus, 
1ssu goes to p, 2ppg goes to d, etc. The MO is said to be “promoted.” 
We have now seen several ways to explain the effects of an orbital. We may focus 
on energies, and note that bonding and antibonding MOs correlate with low-energy 
and high-energy united-atom orbitals, respectively. Or, as we saw earlier, we can focus
Section 7-6 Molecular Orbitals of Homonuclear Diatomic Molecules 227 
Figure 7-16 
 (a) Symmetry orbitals for homonuclear diatomic molecules. (b) United-atom AOs 
characterized by symmetry with respect to z axis. 
on charge distributions and their attractions for nuclei, and note that bonding MOs 
concentrate charge in the bond region, attracting the nuclei together, whereas antibonding 
MOs shift charge outside the bond, attracting the nuclei apart. Alternatively, 
we can recognize that bonding MOs result from AOs on each atom coming together 
in phase, resulting in positive overlap, while antibonding MOs result from opposite 
phases coming together to give negative overlap.
228 Chapter 7 The Variation Method 
Figure 7-17 
 Correlation diagram between separated-atom orbitals and united-atom orbitals for 
homonuclear diatomic molecules. Energy ordinate and internuclear separation abscissa are only 
suggestive. No absolute values are implied by the sketch. [Note: For H+2 sg2p correlates with 
3dsg, sg3s with 3ssg. This arises because the separability of the H+2 hamiltonian (in the Born– 
Oppenheimer approximation) leads to an additional quantum number for this molecule. In essence, 
the H+2 wavefunction in elliptical coordinates may be written . =L(.)M(µ) eimf. The function L 
may have nodal surfaces of elliptical shape. M may have nodes of hyperbolic shape. In the correlation 
diagram for H+2 , it is necessary that ellipsoidal nodes correlate with spherical nodes in the united 
atom, while hyperboloid nodes correlate with hyperboloid nodes (which may be planar). Sketching 
sg2p and sg3s and comparing them with 3ssg and 3dsg (i.e., 3d3z2-r2 ) makes clear how this “nodal 
control” results in what, at first sight, appears to be a violation of the noncrossing rule. For H+2 , 
modifications for higher-energy states will also be required. For example, 4ssg will correlate with 
sg4s, not sg3p.]
Section 7-6 Molecular Orbitals of Homonuclear Diatomic Molecules 229 
Three common labeling conventions are used in Fig. 7-17. A level may be labeled 
with reference to the separated atom AOs to which it correlates. The separated atom 
AO symbol is placed to the right of the MO symmetry symbol (e.g., sg2s). Note the 
absence of square brackets, which we used to symbolize the SO (2sA+2sB). The symbol 
sg2s means “the MO of sg symmetry that correlates with 2s AOs at R=8.” An 
alternative label indicates the united-atom orbital with which the MO correlates. Here 
theAO label is placed to the left of the symmetry symbol (e.g., 3psu). The u and g subscripts 
in the united-atom notation are redundant and are often omitted. However, they 
are helpful in drawing correlation diagrams. Finally, the MOs may be simply numbered 
in their energy order within each symmetry type, as mentioned earlier (e.g., 2sg). 
EXAMPLE 7-3 A symmetry orbital is produced by taking 3dyz, a -3dyz, b, where 
a and b are points on the z axis. Sketch the situation. What is the sign of the overlap 
between these AOs? Is this SO bonding or antibonding? What is the symmetry 
symbol for this SO? 
SOLUTION 
 The sketch would show two four-leafed AOs, each like the 3dyz AO shown in 
Fig. 7-16, except that the phase signs of one would be minus those in the matching lobes of the 
other. TheseAOs are positioned side by side, and their phases are such that the inner lobes pointing 
towards each other are in phase. Hence the overlap is positive and the SO is bonding. Looking at 
this SO along the z axis, it appears like a py AO, so it is p. Inversion causes interchange of lobes 
of opposite sign, so it has u symmetry. Its symbol, then, is pu.  
Term symbols for electronic states of homonuclear diatomic molecules are much like 
those in atoms. The main symbol gives information about the component of electronic 
angular momentum along the z axis. If ML =0,±1,±2, etc., then the main symbol is 

, , , etc. ML is the sum of ml values for the electrons. When identifying ml, we 
assume that the p MOs are complex, with ml =+1 and -1 rather than being mixed 
to give real px and py MOs. The main symbol is decorated with a superscript at left, 
giving spin multiplicity, and a g or u subscript at right for overall inversion symmetry. 
As an example, C+2
, with configuration 1s2 
g 1s2 
u 2s2 
g 2s2 
u 1p3 
u has ML=±1 (zero for all 
s electrons and either +1,+1,-1 or +1,-1,-1 for the three p electrons) so its main 
symbol is . The inversion symmetry is u because there is an odd number of electrons 
in ungerade MOs. There is one unpaired electron, so the multiplicity is 2. The term, 
symbol is 2u (“doublet-pi-you”). 

 terms are given, in addition, a+or-superscript at the right, indicating whether the 
wavefunction is symmetric or antisymmetric for reflection through a plane containing 
the nuclei. Such a reflection has the effect of reversing the direction of f: Clockwise 
motion around the internuclear axis becomes counterclockwise when viewed in a mirror 
containing that axis. This transforms exp(if) and exp(-if) into each other, so that p+ and p- turn into (minus) each other. To achieve a 
 MO in the first place requires that 
mz values sum to 0. This happens for occupied equivalent p MOs only if p+ and p- are equally occupied, which means that both have one electron or both are full. The 
full case must yield 1
+g , i.e., must be symmetric for every operation, because each 
MO occurs twice in the electron list, and even antisymmetric functions give symmetric 
results when multiplied by themselves. For the half-filled case there are only two 
possibilities: p+(1)p-(2)±p-(1)p+(2). The “plus” case is symmetric for electron
230 Chapter 7 The Variation Method 
exchange, hence goes with the antisymmetric spin function aß -ßa, hence is a singlet 
state. Reflection causes p+.p-, but this returns the same function, so this case goes 
with a 1
+ term. Similar reasoning shows that the “minus” case goes with a 3
- term. 
Hund’s rule predicts the latter term to lie below the former in energy. 
EXAMPLE 7-4 What term symbols represent possible excited states ofN2 produced 
by promoting an electron from the highest occupied MO to the lowest unoccupied 
MO? 
SOLUTION 
 When an electron is promoted, the possible resulting 1pu, 1pg configurations can 
be pictured as follows: 
1pg 
1pu 
+1 –1 +1 –1 +1 –1 +1 –1 
The excited molecule has two half-filled MOs, for which the electron spins can be opposite (singlet) 
or the same (triplet). The sum of electron orbital momenta will remain zero (giving
 if the electron 
is excited fromml =-1 to-1 or from+1 to+1, but will change to+2 or-2 (giving) if changed 
from -1 to +1 or from +1 to -1. In all cases, the excited state has an odd number of electrons in 
the MOs of ungerade symmetry, so all terms will have a u subscript. The term symbols, then, are, 
respectively 1
u, 3
u, 3u, 1u.  
These conventions for term symbols apply for any linear molecule having inversion 
symmetry (e.g., HCCH, CO2). For linear molecules lacking inversion symmetry (CO, 
HCN) all is the same except that there is no g–u symbol. 
The noncrossing rule mentioned above is an important aid in constructing correlation 
diagrams for many processes. It is called a rule rather than a law because it can only be 
shown to be highly improbable, not impossible, for two levels of the same symmetry to 
cross. Thus, imagine that we have a molecule with some variable parameter . and also 
with a symmetry operation R which is not lost as . varies. For example, . might be the 
H–O–Hangle inwater, andR could be reflection through the plane bisecting theH–O–H 
angle. Suppose that we have a complete set of basis functions and that, at each value 
of ., we manage to express exactly all but two of the eigenfunctions for the molecule. 
This uses up all but two dimensions of our function space, leaving us, at each value of ., 
with two eigenfunctions to determine and two functions in terms of which to express 
them. (These functions change with ., but the above argument has nevertheless served 
to reduce our problem to two dimensions.) Now let the two functions remaining from 
our original basis be mixed to become orthonormal and also individually either symmetric 
or antisymmetric for R. We label these symmetrized basis functions .1 and .2. 
Because we began with a complete basis, it must be possible to express the as yet 
undetermined wavefunctions .1 and .2 exactly as linear combinations of .1 and .2 
at each value of .. Furthermore, if .1 and .2 are, say, both antisymmetric for R, it 
is necessary that .1 and .2 also both be antisymmetric. If .1 and .2 have opposite 
symmetries, however, .1 and .2 also have opposite symmetries. (In the latter case, .1 
and .2 can only mix to produce unsymmetric functions, and so we know that .1 and .2
Section 7-7 Basis Set Choice and the VariationalWavefunction 231 
are already identical with .1 and .2.) To determine the mixing coefficients and state 
energies for .1 and .2, we solve the 2×2 secular equation over the basis .1, .2: 
 
H11 -E H12 
H12 H22 -E

=0 (7-107) 
The roots are 
E± = 
1
2 
H11 +H22 ±4H2 
12 +(H11 -H22)2
1/2 (7-108) 
The crossing of energy levels for .1 and .2 requires that, at some value of ., E+ equals 
E-. From Eq. (7-108), we see that this requires that the term in square brackets vanish, 
which requires that H12 and H11 - H22 vanish. Now, if .1 and .2 (and hence .1 
and .2) have opposite symmetries for R, H12 vanishes for all values of ., and the 
curves will cross whenever H11 equals H22. But if .1 and .2 (and hence .1 and .2) 
have the same symmetry for R, H12 is not generally zero. In this situation, the curve 
crossing requires that both H12 and H11-H22 happen to pass through zero at the same 
value of .. This simultaneous occurrence of two functions passing through zero is so 
unlikely that it is safe to assume it will not happen. 
If the molecule possesses several elements of symmetry, H12 will vanish at all . if 
.1 and .2 disagree in symmetry for any one of them, so the noncrossing rule applies 
only to states having wavefunctions of identical symmetry for all symmetry operations 
of the molecule. 
A similar treatment for orbitals and orbital energy levels is possible, and the noncrossing 
rule applies for orbital energies as well as for state energies. 
7-7 Basis Set Choice and the VariationalWavefunction 
One of the places where human decision can effect the outcome of a variational calculation 
is in the choice of basis. Some insight into the ways this choice effects the 
ultimate results is necessary if one is to make a wise choice of basis, or recognize which 
calculated results are “physically real” and which are artifacts of basis choice. 
One question we can ask is this: Is a minimal basis set equally appropriate for 
calculating an MO wavefunction for, say, B2 as F2? In each case we use 10 AOs 
and 2 spin functions producing a total of 20 spin MOs. With B2, however, we have 
10 electrons to go into these spin MOs, and in F2 we have 18 electrons. In all but 
the crudest MO calculations, the total energy is minimized in a manner that depends 
on the natures of only the occupied MOs. In effect, then, the calculation forB2 produces 
the 10 “best” spin MOs from a basis set of 20 spin-AOs, whereas that for F2 produces 
the 18 best MOs from a different basis set of 20 spin-AOs. In a sense, then, the basis 
for F2 is less flexible than that for B2. Of course, the use of separated atom orbitals is a 
conscious effort to choose that basis that best spans the same function space as the best 
MOs. To the extent that this strategy is successful, the above problem is obviated (i.e., if 
both sets are perfect, additional flexibility is useless). The strategy is not completely 
successful, however, and comparison of results of minimal basis set calculations down 
a series of molecules such as B2,C2,N2,O2, and F2 may be partially hampered by this
232 Chapter 7 The Variation Method 
ill-defined inequivalence in basis set adequacy. In contrast, comparison of calculated 
results in a series of molecules such as CnH2n+2 is much less likely to suffer from this 
particular problem because the minimal basis set grows with increasing n in a way to 
keep pace with the number of electrons. 
Let us now briefly consider how basis sets might vary in adequacy for different 
states of a given molecule. We will compare the wavefunction for the ground state of 
a molecule with the wavefunction for a Rydberg state. Rydberg states are so named 
because their spectral lines progress toward the ionization limit in a manner similar to the 
spectral pattern for hydrogenlike ions (called a Rydberg series).7 Hence, the Rydberg 
states of molecules are in some way like excited states of the hydrogen atom. This can 
be understood by visualizing an excited state for, say, N2 wherein one electron is, on 
the average, very far away from the rest of the molecule, which is now an N+2 “core.” 
As the excited electron moves to orbitals farther and farther out, the N+2 core becomes 
effectively almost like a point positive charge. As a result, the coupling between the 
angular momentum of this orbital and the internuclear axis grows progressively weaker, 
so that the motion of the Rydberg electron becomes more and more independent of 
orientation of the core. It is not surprising that a hydrogen-likeAO centered in the bond 
becomes more and more appropriate as a basis for describing this orbital. In contrast, 
such a “single-center” basis normally requires many terms to accurately describe MOs 
in ground or non-Rydberg excited states. Thus, for a Rydberg state of N2, one would 
do well to choose a basis set of AOs located on the nuclei to describe the MOs of the 
N+2 core, and to use an AO (or several AOs) centered between the nuclei to describe 
the orbital for the Rydberg electron. 
Thus far we have kept the discussion within the framework of homonuclear diatomic 
molecules. When we come to heteronuclear diatomics, for example, CO, we lose 
inversion symmetry and we can no longer symmetry balance our basis. This means 
that a given basis may be more inadequate for representing the wavefunction on one 
end of the molecule than on the other. As a result, the electronic charge will be shifted 
toward the end where the basis set is best able to minimize the energy. This charge 
shift is an artifact of basis set imbalance, but, since we have no way to evaluate this 
imbalance, it is difficult to tell how much it affects our results. Mulliken has published 
some calculations on the HF molecule that illustrate this problem in a striking way. 
Table 7-3 is a list of total energies and dipole moments calculated for HF using a variety 
of basis sets. 
The first column of data arises from a minimal basis set of STOs (1sH, 1sF, 2sF, 2psF , 
2ppxF , 2ppyF ) with orbital exponents evaluated from Slater’s rules for atoms. The 
second column results if the orbital exponents are allowed to vary independently 
to minimize the molecular energy. The basis set for the third column is obtained 
by augmenting the previous basis with additional STOs centered on the H nucleus 
(2sH, 2psH, 2ppxH, 2ppyH). Finally, the fourth column results from use of a basis set 
that has been augmented (over the minimal basis) at both nuclei in a way thought to be 
appropriately balanced. As the basis set grows increasingly flexible, the average energy 
becomes lower, but the expectation value for the dipole moment does not converge uniformly 
toward the observed value. In particular, by augmenting the basis on hydrogen 
only, we create a very unbalanced basis, which causes charge to shift too much toward 
the hydrogen end of the molecule. 
7See A. B. F. Duncan [3].
Section 7-8 Beyond the Orbital Approximation 233 
TABLE 7-3 
 Energies and Dipole Moments for Hydrogen Fluoride Calculated by the Variation 
Method Using Different Basis Setsa 
Min STO 
Slater . 
Min STO 
best . 
Min STO F; Aug. 
STO H (very 
unbalanced) 
Aug. STO 
F and H 
(balanced) Exp 
E (a.u.) -99.4785 -99.5361 -99.6576 -100.0580 -100.527 
µ(H+F-) 0.85D 1.44D 0.92D 1.98D 1.82Db 
aSee Mulliken [4]. 
bData fromWeiss [5]. 
These problems with basis set adequacy are difficult to overcome completely. Fortunately, 
with a certain amount of experience, insight, and caution, it is nevertheless 
possible to carry out variational calculations and interpret their results to obtain reliable 
and useful information. 
7-8 Beyond the Orbital Approximation 
Most of our discussion of the variation method has been restricted to calculations within 
the orbital approximation. To avoid leaving an inaccurate impression of the capabilities 
of the variation method, we shall briefly describe some calculations on some small 
(two-electron) systems where the method can be employed to its fullest capabilities. 
These calculations are listed in Table 7-4. 
The calculation on He by Kinoshita expresses the spatial part of the wavefunction as 
.(ks, kt, ku)=e-ks/2 8 
l,m,n=0 
n,even 
cl,m,n(ks)l-m(ku)m-n(kt)n (7-109) 
where 
s =r1 +r2, u=r12, t=-r1 +r2 (7-110) 
and k and cl,m,n are variable parameters. The exponential term causes the wavefunction 
to vanish as either electron goes to infinite r, and the terms in the sum build up 
a polynomial in one- and two-electron coordinates, reminiscent of the form of eigenfunctions 
for the harmonic oscillator and the hydrogenlike ion. Kinoshita carried out 
his calculation to as many as 39 terms, obtaining an energy that he estimated to differ 
from the exact result by no more than 1.2 × 10-6 a.u. A subsequent calculation by 
Pekeris, using a related approach, required solving a secular determinant of order 1078 
and yielded an energy estimated to be accurate to 1.0×10-9 a.u. Applying corrections 
for coupling between electronic and nuclear motions, and also for relativistic effects, 
Pekeris arrived at a theoretical value for the ionization energy of He of 198310.687cm-1 
compared to the experimental value of 198310.82 ±0.15cm-1.
234 Chapter 7 The Variation Method 
TABLE 7-4 
 Results of Some Very Accurate Variational Calculations on 
Two-Electron Systems 
System 
Minimized 
energya (a.u.) 
Estimated maximum 
energy error 
E¯ -Eexact (a.u.) 
He (ground singlet)b -2.9037225 0.0000012 
He (ground singlet)c -2.903724375 0.000000001 
He (lowest triplet)c -2.17522937822 0.00000000001 
H2 (ground singlet)d -1.17447498302e 0.000000001 
aUncorrected for nuclear motion and relativistic effects. 
bFrom Kinoshita [6]. 
cFrom Pekeris [7]. 
dFrom Kolos andWolniewicz [8]. 
eAt R =1.401078 a.u. 
In the 35 years since Pekeris work was reported, even more extensive calculations 
have been reported. For example, Drake et al. [9] have calculated an upper bound for 
the ground state of helium having 22 significant figures using a wavefunction having 
2358 terms. 
Extremely accurate variational calculations have been carried out onH2 byKolos and 
Wolniewicz. They used elliptic coordinates and an r12 coordinate and expressed their 
wavefunction as an expansion in powers of these coordinates, analogous in spirit to the 
Kinoshita wavefunction described above. Their most accurate wavefunctions contain 
100 terms and are calculated for a range of R values. After including corrections for 
relativistic effects and nuclear motion, Kolos and Wolniewicz arrived at a theoretical 
value for the dissociation energy in H2 of 36117.4cm-1 compared to what was then 
the best experimental value 36113.6±0.5cm-1. Subsequent redetermination of the 
experimental value gave 36117.3±1.0cm-1.8 
The dissociation energy, D0, is the energy required to separate a molecule into its 
constituent atoms, starting with a molecule in its lowest vibrational state. The binding 
energy, De, is the energy for the corresponding process if we omit the vibrational 
energy of the molecule (see Fig. 7-18). These quantities are often much more sensitive 
measures of the accuracies of calculations than are total energies. The reason for this is 
easily understood when we recognize that the binding energy is a fairly small difference 
between two large numbers—the total energy of the molecule and the total energy of 
the separated atoms. Unless our errors in these two large energies are equal, the residual 
error is magnified (in terms of percentage) when we take the difference. Thus, the best 
total energy forH2 in a certain orbital approximation is-1.133629 a.u., which is 96.7% 
of the total energy. However, the corresponding binding energy is -0.133629 a.u., 
which is 76.6% of the correct value. The need to calculate accurate binding energies is 
sometimes referred to as the need to achieve “chemical accuracy.” 
The variational calculations cited above are among the most accurate performed, 
and they give an indication of the capabilities of the method. Properties other than 
8See Herzberg [10].
Section 7-8 Beyond the Orbital Approximation 235 
Figure 7-18 
 Schematic drawing showing the distinction between D0, the dissociation energy 
from the lowest vibrational level, and De, the binding energy, which does not take vibrational energy 
into account. The zero of energy is the energy of the separated atoms. 
energy predicted from suchwavefunctions are also very accurate. For example, Pekeris’ 
best wavefunction for the first triplet state of helium gives an electron density at the 
nucleus of 33.18416 electrons per cubic bohr compared with the experimental value 
33.18388±0.00023 deduced from hyperfine splitting. For most systems of chemical 
interest, calculations of this sort become much too impractical to be considered. For 
this reason the orbital approximation, with all its limitations, is used in most quantumchemical 
calculations on systems having more than two electrons. 
7-8.A Problems 
7-1. Given the following two functions, f (x) and g(x), for the range 0=x =L: 
Figure P7-1 

236 Chapter 7 The Variation Method 
Can these functions be expressed accurately as linear combinations of particlein-
a-box eigenfunctions (box walls at x =0,L)? Indicate your reasoning. If yes, 
what is the expression for the first two coefficients in the expansion? Can you 
evaluate any of these by inspection? 
7-2. Consider a particle in a box with a biased potential that is higher at x =L than 
at x =0. An approximate solution for the ground state could be f =v v 0.9.1 + 0.1.2, where .1 and .2 are the first and second eigenfunctions for the unbiased 
box. (a) Make a rough sketch of f, showing how it skews the particle distribution. 
(b) What is the average kinetic energy for f, in terms of h,m, and L? 
7-3. The normalized function f =(2/45p)1/2r2 exp(-r) can be expanded in terms of 
hydrogen atom eigenfunctions: 
f =c1.1s +c2.2s +c3.2p0 +··· 
where .1s=(1/vp)exp(-r) and .2p0 =(1/v32p)r exp(-r/2) cos .. Evaluate 
c1 and c3. 
7-4. Given the approximate wavefunction for the lowest state of a particle in a onedimensional 
box (Fig. P7-4): 
Figure P7-4 
 
f =3/L(2x/L), 0=x =L/2 
f =3/L[2(L-x)/L], L/2=x =L 
f =0, x<0, x>L 
a) Resolve f into the box eigenfunctions. That is, evaluate cn in the expression 
f = 
8 
n=1 
cn.n, 
where 
.n =2/Lsin (npx/L) , 0=x =L, .n =0, x<0, x>L 
b) Using the coefficients from part (a) compare the value of f at x =L/2 with 
the values one obtains from the 
fapprox = 
m 

n=1 
cn.n, with m=1, 3, 5, 7, and 9
Section 7-8 Beyond the Orbital Approximation 237 
c) Use the coefficients from part (a) to obtain an expression for E¯ appropriate 
for f. Estimate the value of the infinite series and thereby estimate E¯. Compare 
this value to Eexact. 
7-5. Let f =exp(-ar2) be a trial function (not normalized) for the ground state of 
the hydrogen atom. Use the variation method to determine the minimum energy 
attainable from this form by variation of a. Find the average value of r and the 
most probable value of r for this wavefunction. Compare these r values and the 
average energy with the exact values. 
7-6. Let f(a)=(a5/3p)1/2r exp(-ar) be a trial function for the ground state of the 
hydrogen atom: 
a) Verify that the variation method gives a = 3
2 ,E¯ =-38
a.u. 
b) Verify that f 3
2  has an overlap of 0.9775 with the 1s function. 
c) Find the value of a that maximizes the overlap of f with the 1s function and 
determine the average energy of this new f. 
7-7. A normalized approximate wavefunction f for the hydrogen atom is projected 
into its components and found to be f = (1/v2).1s + (1/v4).2s + 
(1/v8).3s +c4s.4s. There are no higher terms. 
a) What is the value of c4s? 
b) What is the value of the average energy? 
7-8. A normalized, spherically symmetric variationally optimized function for the 
hydrogen atom is analyzed by projecting out coefficients c1s, c2s, c2p0 , etc. It is 
found that c1s is equal to v0.80 and c2s =v0.15. 
a) What is the maximum possible value for c2p0? Explain your reasoning. 
b) What is the minimum possible value for E¯ that could correspond to this 
function, based on the above data? Explain your reasoning. 
7-9. A normalized trial wavefunction of the form (in a.u.) 
f =(2. )7/(4p6!)
1/2 
r2 exp (-.r) 
is variationally optimized to give the lowest possible energy for the hydrogen 
atom. The results part way along this process are  fHˆ f dv =. 2/10-./3. 
a) Complete the variational process to obtain the optimum . and the minimized 
average energy. 
b) Calculate the value of c1 in the expression fopt =c1.1s +c2.2s+···. 
c) Produce a new, normalized function . that is orthogonal to .1s but is otherwise 
as similar as possible to fopt. (Use the unexpanded symbolic terms .1s 
and f in your answer.) 
d) Suppose you are told that the average energy associated with your new function 
. is -0.133 a.u. Do you find this reasonable? Explain your answer. 
7-10. f=(a5/3p)1/2r exp(-ar) is a normalized trial function for the hydrogen atom. 
The energy is minimized when a = 3/2. Calculate the average value of the 
potential energy predicted by this function if a =3/2.
238 Chapter 7 The Variation Method 
7-11. Prove that optimized trial function (7-20) must contain contributions from continuum 
wavefunctions. 
7-12. Compare the orbital exponent for a 1sAO in He as found by the variation method 
[Eq. (7-35)] with that given by Slater’s rules (Chapter 5). 
7-13. A different trial function for calculating the polarizability of the hydrogen atom 
in a uniform electric field of strength F is 
.trial =.1s(c1 +c2z) 
This is somewhat similar to the example in the text, since z.1s gives a p-like 
function, but not exactly the 2pz eigenfunction. 
a) Use this form to find an expression for the minimum E¯ as a function of F. 
What value of E¯ does this give for F =0.1 a.u.? Can you suggest why this 
trial function is superior to the one used in the text? 
b) The polarizability is defined to be a in the expression 
E=-
1
2 - 
1
2
aF2 
What value of a do you obtain? [Exact a =4.5 a.u.] 1 a.u. of field strength is 
equal to e/a2
0 . Deduce the value of 1 a.u. of polarizability. 
7-14. Which hydrogen atom state should be more polarizable, the 1s or 2s? [Consider 
the factors that determine the extent of mixing between basis functions.] Explain 
your reasoning. 
7-15. fa and fb are chosen to be a normalized set of basis functions for an LCAO 
wavefunction for a one-electron homonuclear diatomic system. It is found that 
the values for the integrals involving these functions are 

 f*aHˆ fa dv = -2 a.u., 
 f*bHˆ fb dv =-2 a.u., 

 f*aHˆ fb dv = -1 a.u., 
 f*afb dv = 
1
4
. 
Find an upper bound for the exact lowest electronic energy for this system. Find 
the corresponding LCAO normalized approximate wavefunction. 
7-16. fa and fb are chosen as the normalized basis functions for an LCAO wavefunction 
for a one-electron, heteronuclear, diatomic molecule. It is found that the 
values for some integrals involving these functions are 

 faHˆ fa dv = -2 a.u., 
 fbHˆ fb dv =-1 a.u., 

 faHˆ fb dv = -
1
2 
a.u., 
 fafb dv = 
1
3
. 
whereHˆ is themolecular hamiltonian. Set up the secular determinantal equation 
and find the lowest electronic energy that can be computed from an LCAO 
wavefunction cafa + cbfb. Find ca and cb such that E¯ is minimized and the 
wavefunction is normalized.
Section 7-8 Beyond the Orbital Approximation 239 
7-17. Apossible basis function for representing the 1sg wavefunction ofH+2 is a 1s-like 
AO . 3/p
1/2 exp(-.r) located at the bond center. Assuming an internuclear 
separation of 2 a.u., find the . value that minimizes E¯. Is this basis function 
adequate to predict a bound H+2 molecule? [Use Appendix 3 to help you develop 
your formulas.] 
7-18. Without referring to the text, and by inspection, what is the united atom limit for 
the 1su molecular orbital’s energy (in a.u.) for H+2 ? 
7-19. Show that, at R=8, the .+ and .- wavefunctions for H+2 are capable of 
describing a state wherein the electron is in a 1s orbital on atom A. 
7-20. Evaluate Eqs. (7-89) and (7-90) at R =0 to show that HAA =HAB at this point. 
7-21. Examining Eq. (7-86), and letting HAB = kHAA what relationship between k 
and SAB is necessary if the sg MO is to be lower in energy than the su MO? 
[Assume that HAA is negative, and that k and SAB are positive.] 
7-22. Consider the one-electron molecule–ion HeH2+: 
a) Write down the hamiltonian operator (nonrelativistic, Born–Oppenheimer 
approximation) for the electronic energy in atomic units for this system. 
b) Calculate the electronic energies for the lowest energy state of this system in 
the separated atom and united atom limits. 
7-23. For a homonuclear diatomic molecule aligned as shown in Fig. P7-23, characterize 
each of the following MOs as s, p, d, and g or u, and bonding or antibonding. 
a) 2pya +2pyb 
b) 2pza +2pzb 
c) 3dz2
a +3dz2
b 
d) 3dxya +3dxyb 
e) 3dxza -3dxzb 
Figure P7-23 
 
7-24. Characterize each of the following atomic orbitals with the symbols s, p, d, and 
also g or u. Let the z axis be the reference axis for angular momentum. 
1s 2pz 3py 3dxy 
2s 2px 3dz2 3dxz 
7-25. Indicate whether you expect each of the following homonuclear diatomic MOs 
to be bonding or antibonding. Sketch the MO in each case: (a) su (b) pu (c) dg
240 Chapter 7 The Variation Method 
7-26. A homonuclear diatomic molecule MO of pu symmetry is to be expressed as a 
linear combination ofAOs centered on the nuclei, which lie on the z axis. Which 
of the AOs in the following list can contribute to the MO? 
1s 2s 2px 2py 2pz 3s 3px 
3py 3pz 3dxy 3dxz 3dyz 3dz2 3dx2-y2 
7-27. Use sketches and symmetry arguments to decide which of the following integrals 
vanish for diatomic molecules (the x, y, and z axes are shown in Fig. P7-23): 
a)  2pza 1sb dv 
b)  2pya1sb dv 
c)  2pza2pyb dv 
d)  2pya3dyzb dv 
e)  2pza3dyzb dv 
f)  1saHˆ 2pxa dv 
g)  1saHˆ 2pza dv 
7-28. Show that, for reflection through a plane containing the nuclei, if the MO dx2-y2 
is symmetric, then the MO dxy is antisymmetric. Show that the same is true for 
p MOs constructed from dxz and dyz AOs. (The internuclear axis is assumed to 
lie along the z coordinate.) 
7-29. Assuming the internuclear axis to lie along the z coordinate, what are the possible 
ML quantum numbers for an MO constructed from 3dz2
a -3z2
b 
? 
7-30. In a homonuclear diatomic correlation diagram, what MO symmetry symbols 
(s, p, d, g, u) could correlate with each of the united atom AOs listed below? 
Assume z to be the “old” internuclear axis. Indicate for each case whether this 
united atom orbital is the terminus for a bonding or an antibonding MO. (a) 2pz 
(b) 2px (c) 3dxz (d) 3dxy 
7-31. A homonuclear diatomic system has the ground-state MO configuration 
1s2 
g 1s2 
u 2s2 
g 2s2 
u 3s2 
g 1p4 
u 1p2 
g : 
a) What is the net number of bonding electrons? 
b) What spin multiplicity would you expect for the ground state? 
c) What would you expect the effect to be on the dissociation energy of this 
molecule of ionization (1) from the 1pg MO? (2) from the 3sg MO? 
d) Upon ionization (one-electron) from the 1pg level, what would be the spin 
multiplicity of the resulting ion? 
e) To what type of united atom AO does the pu MO correlate? 
7-32. a) Without referring to the text, write out the ground state configuration for O+2 
using MO symmetry symbols (1s2 
g etc.) 
b) What is the net number of bonding electrons? 
c) How does the dissociation energy for this ion compare to that for O2? 
d) What is the ground state term symbol for this ion? 
e) Which occupied MOs may contain contributions from 2pz AOs, assuming z 
to be the internuclear axis?
Section 7-8 Beyond the Orbital Approximation 241 
7-33. The reduced symmetry of heteronuclear (compared to homonuclear) diatomic 
molecules results in their having a different correlation diagram. Set up a correlation 
diagram for heteronuclear diatomics. Be sure to indicate that the energy 
levels of each type of AO are not identical for the separated atoms. Comparing 
correlation diagrams for homonuclear and heteronuclear molecules, does it seem 
reasonable that He2 is unstable, whereas the isoelectronic LiH and LiHe+ are 
stable molecules? 
7-34. Following are some Slater orbital coefficients for some MO’s of F2 calculated 
by Ransil [11] (the 2ps STOs are defined according to z axes pointing from each 
atom toward the other): 
1sg c1s,A =c1s,B =0.70483 2sg c1s,A =c1s,B =0.17327 
c2s,A =c2s,B =0.00912 c2s,A =c2s,B =-0.67160 
c2ps ,A =c2ps ,B =-0.00022 c2ps ,A =c2ps ,B =-0.08540 
We see that the 1sg MO is almost entirely made from 1s AOs on A and B. 
However, the 2sg MO contains what appears to be an anomalously large amount 
of 1s AO. This turns out to be an artifact of the fact that Slater-type 2s orbitals 
are not orthogonal to 1sAOs on the same center. For F2, the STO 1s, 2s overlap 
is 0.2377. Use this fact to construct a new orbital, 2s, that is orthogonal to 1s. 
Express the 2sg MO of Ransil in terms of the basis functions 1s, 2s, and 2ps on 
centers A and B. You should find the 1s coefficients much reduced. 
7-35. Re for H+2 equals 2.00 a.u. At this distance, Eelec =-1.1026 a.u. What is the 
value of De for H+2 ? 
Multiple Choice Questions 
(Try to answer these without referring to the text.) 
1. A homonuclear diatomic MO is given by f =2pz,a +2pz,b, where the z axis is the 
same as the internuclear axis. Which one of the following statements about f is 
correct? 
a) f is an antibonding MO, symbolized su. 
b) f is a bonding MO, symbolized pu. 
c) f is an antibonding MO, symbolized pg. 
d) f is a bonding MO, symbolized sg. 
e) f is an antibonding MO, symbolized pu. 
2. For the three species N2, N+2 , N-2 , which one of the following orders for the bond 
energy (i.e., bond strength) is most reasonable? 
a) N2 >N+2 >N-2 
b) N+2 >N2 >N-2 
c) N-2 >N2 >N+2 
d) N-2 >N+2 >N2 
e) Only N2 forms a bond; N+2 and N-2 do not.
242 Chapter 7 The Variation Method 
3. According to the LCAO-MO model, which one of the following second period 
diatomic molecules has a double bond in the ground electronic state? 
a) Li2 
b) Be2 
c) B2 
d) C2 
e) N2 
4. Which one of the following statements concerning H+2 is false? 
a) The nondegenerate LCAO-MOs (without spin) must be either symmetric or antisymmetric 
for inversion. 
b) The lowest energy MO (without spin) of the molecule is antisymmetric for 
inversion. 
c) The MOs transform into the AOs of the helium ion as the two nuclei are fused 
together. 
d) The ground state has a multiplicity of two. 
e) The Born-Oppenheimer approximation permits finding the solution for the purely 
electronic wave function at each value of the internuclear distance. 
5. Which one of the following statements is false for bonding MOs formed from linear 
combinations of AOs on atoms a and b? 
a) Only AOs that have nonzero overlap can form bonding MOs. 
b) Only AOs that have similar energies can form strongly bonding MOs. 
c) Bonding MOs cannot have a nodal plane perpendicular to the internuclear axis 
and midway between a and b. 
d) A p AO on b can combine with a p AO on a to form s,p, or d MOs. 
e) A maximum of three bonding MOs can be formed from 2p AOs on a and b. 
References 
[1] R. S. Mulliken, Rev. Mod. Phys. 2, 60, 506 (1930); 3, 90 (1931); 4, 1 (1932). 
[2] J. O. Hirschfelder, C. F. Curtiss, and R. B. Bird, Intermolecular Forces.Wiley, New 
York, 1964. 
[3] A. B. F. Duncan, Rydberg Series in Atoms and Molecules. Academic Press, 
NewYork, 1971. 
[4] R. S. Mulliken, J. Chem Phys. 36, 3428 (1962). 
[5] R.Weiss, Phys. Rev. 131, 659 (1963). 
[6] T. Kinoshita, Phys. Rev. 150, 1490 (1957). 
[7] C. L. Pekeris, Phys. Rev. 115, 1216 (1959). 
[8] W. Kolos and L.Wolniewicz, J. Chem. Phys. 49, 404 (1968).
Section 7-8 Beyond the Orbital Approximation 243 
[9] G.W. F. Drake, M. M. Cassar, and R. A. Nistor, Phys. Rev. A 65, 54501 (2002). 
[10] G. Herzberg, J. Mol. Spectrosc. 33, 147 (1970). 
[11] B. J. Ransil, Rev. Mod. Phys. 32, 245 (1960).
Chapter 8 
The Simple H¨uckel Method 
and Applications 
8-1 The Importance of Symmetry 
Our discussions of the particle in a box, the harmonic oscillator, the hydrogen atom, and 
homonuclear diatomic molecules have all included emphasis on the role that symmetry 
plays in determining the qualitative nature of the eigenfunctions. When we encounter 
larger systems, detailed and accurate solutions become much more difficult to perform 
and interpret, but symmetry continues to exert strong control over the solutions. 
In this chapter, we will describe a rather simple quantum chemical method that was 
formulated in the early 1930s by E. H¨uckel. One of the strengths of this method is 
that, by virtue of its crudeness and simplicity, the effects of symmetry and topology on 
molecular characteristics are easily seen. Also, the simplicity of the model makes it an 
excellent pedagogical tool for illustrating many quantum chemical concepts, such as 
bond order, electron densities, and orbital energies. Finally, the method and some of 
its variants continue to be useful for certain research applications. Indeed, it is difficult 
to argue against the proposition that every graduate student of organic and inorganic 
chemistry should be acquainted with the H¨uckel molecular orbital (HMO) method. 
8-2 The Assumption of s–p Separability 
The simple H¨uckel method was devised to treat electrons in unsaturated molecules like 
ethylene and benzene. By 1930 it was recognized that unsaturated hydrocarbons are 
chemically more reactive than are alkanes, and that their spectroscopic and thermodynamic 
properties are different too. The available evidence suggested the existence of 
loosely held electrons in unsaturated molecules. 
We have already seen that, when atoms combine to form a linear molecule, we 
can distinguish between MOs of type s,p, d, . . . depending on whether the MOs are 
associated with anmquantum number of 0,±1,±2, . . . Thus, in acetylene (C2H2), the 
minimal basis set ofAOs on carbon and hydrogen lead to s and p MOs. Let us imagine 
that our acetylene molecule is aligned along the z Cartesian axis. Then the px p-type 
AOs on the carbons are antisymmetric for reflection through a plane containing the 
molecular axis and the y axis. Similarly, the py p-type AOs are antisymmetric for 
reflection through a plane containing the molecular axis and the x axis. The pz AOs, 
which are s-type functions, are symmetric for reflection through any plane containing 
244
Section 8-2 The Assumption of s–p Separability 245 
the molecular axis. It has become standard practice to carry over the s-p terminology 
to planar (but nonlinear) molecules, where m is no longer a “good” quantum number. 
In this expanded usage, a p orbital is one that is antisymmetric for reflection through 
the plane of the molecule, a s orbital being symmetric for that reflection. 
H¨uckel found that, by treating only the p electrons explicitly, it is possible to reproduce 
theoretically many of the observed properties of unsaturated molecules such as 
the uniform C–C bond lengths of benzene, the high-energy barrier to internal rotation 
about double bonds, and the unusual chemical stability of benzene. Subsequent work 
by a large number of investigators has revealed many other useful correlations between 
experiment and this simple HMO method for p electrons. 
Treating only the p electrons explicitly and ignoring the s electrons is clearly an 
approximation, yet it appears to work surprisingly well. Physically, H¨uckel’s approximation 
may be viewed as one that has the p electrons moving in a potential field due 
to the nuclei and a “s core,” which is assumed to be frozen as the p electrons move 
about. Mathematically, the s-p separability approximation is 
Etot =Es +Ep (8-1) 
where Etot is taken to be the electronic energy Eel plus the internuclear repulsion 
energy Vnn. 
Let us consider the implications of Eq. (8-1). We have already seen (Chapter 5), 
that a sum of energies is consistent with a sum of hamiltonians and a product-type 
wavefunction. This means that, if Eq. (8-1) is true, the wavefunction of our planar 
molecule should be of the form (see Problem 8-1) 
.(1, . . . ,n)=.p (1, . . . , k).s (k +1, . . . ,n) (8-2) 
and our hamiltonian should be separable into p and s parts: 
Hˆ(1, 2, . . . ,n)=Hˆp (1, 2, . . . , k)+Hˆs (k +1, . . . ,n) (8-3) 
Equations (8-2) and (8-3) lead immediately to Eq. (8-1): 
E¯ = 

 .*p.*s Hˆp +Hˆs .p.s dt(1, . . . ,n) 

 .*p.*s.p.s dt(1, . . . ,n) 
= 
 .*pHˆp.pdt(1, . . . , k) 

 .*p.pdt(1, . . . , k) + 
 .*sHˆs.s dt(k +1, . . . ,n) 

 .*s.s dt(k +1, . . . ,n) 
= Ep +Es (8-4) 
If these equations were valid, one could ignore .s and legitimately minimize Ep by 
varying .p , But the equations are not valid because it is impossible to rigorously satisfy 
Eq. (8-3). We cannot define Hˆp and Hˆs so that they individually depend completely 
on separate groups of electrons and still sum to the correct total hamiltonian. Writing 
these operators explicitly gives 
Hˆp (1, . . . , k) = -
1
2 
k 

i=1 
.2
i + 
k 

i=1 
Vne (i)+ 
1
2 
k 

i=1 
k 
 
j=1,j =i 
1 
rij 
(8-5) 
Hˆs (k +1, . . . ,n) = -
1
2 
n 
 
i=k+1
.2
i + 
n 
 
i=k+1
Vne (i)+ 
1
2 
n 
 
i=k+1 
n 
 
j=k+1,j =i 
1 
rij +Vnn (8-6)
246 Chapter 8 The Simple H¨uckel Method and Applications 
where Vne (i) represents the attraction between electron i and all the nuclei. These 
hamiltonians do indeed depend on the separate groups of electrons, but they leave out 
the operators for repulsion between s and p electrons: 
Hˆ-Hˆp -Hˆs = 
k 

i=1 
n 
 
j=k+1 
1 
rij 
(8-7) 
In short, the s and p electrons really do interact with each other, and the fact that the 
HMO method does not explicitly include such interactions must be kept in mind when 
we consider the applicability of the method to certain problems. Some account of s-p 
interactions is included implicitly in the method, as we shall see shortly. 
8-3 The Independent p-Electron Assumption 
TheHMOmethod assumes further that thewavefunction.p is a product of one-electron 
functions and that the hamiltonian Hˆp is a sum of one-electron operators. Thus, for 
np electrons, 
.p (1, 2, . . . ,n) = fi(1)fj (2) . . . fl(n) (8-8) 
Hˆp (1, 2, . . . , n) = Hˆp (1)+Hˆp (2)+···+Hˆp (n) (8-9) 
and 

 f*i(1)Hˆp (1)fi(1)dt (1) 

 f*i(1)fi(1)dt (1) =Ei (8-10) 
It follows that the total p energy Ep is a sum of one-electron energies: 
Ep =Ei +Ej +···+El (8-11) 
This means that the p electrons are being treated as though they are independent of each 
other, since Ei depends only on fi and is not influenced by the presence or absence of 
an electron in fj . However, this cannot be correct because p electrons in fact interact 
strongly with each other. Once again, such interactions will be roughly accounted for 
in an implicit way by the HMO method. 
The implicit inclusion of interelectronic interactions is possible because we never 
actually write down a detailed expression for the p one-electron hamiltonian operator 
Hˆp (i). (We cannot write it down because it results froma p–s separability assumption 
and an independent p-electron assumption, and both assumptions are incorrect.) Hˆp (i) 
is considered to be an “effective” one-electron operator—an operator that somehow 
includes the important physical interactions of the problem so that it can lead to a reasonably 
correct energy value Ei . A key point is that the HMO method ultimately evaluates 
Ei via parameters that are evaluated by appeal to experiment. Hence, it is a semiempirical 
method. Since the experimental numbers must include effects resulting from
Section 8-4 Setting up the H¨uckel Determinant 247 
all the interelectronic interactions, it follows that these effects are implicitly included 
to some extent in the HMO method through its parameters. 
It was pointed out in Chapter 5 that, when the independent electron approximation 
[Eqs. (8-8)–(8-11)] is taken, all states belonging to the same configuration become 
degenerate. In other words, considerations of space-spin symmetry do not affect the 
energy in that approximation. Therefore, the HMO method can make no explicit use of 
spin orbitals or Slater determinants, and so .p is normally taken to be a single product 
function as in Eq. (8-8). The Pauli principle is provided for by assigning no more than 
two electrons to a single MO. 
EXAMPLE 8-1 If O2 were treated by the HMO method, what would be the form of 
the wavefunction and energy for the ground state? 
SOLUTION 
 The ground state configuration for O2 is 1s2 
g 1s2 
u 2s2 
g 2s2 
u 3s2 
g 1p2
u,x1p2
u,y × 1pg,x1pg,y, where we have shown the degenerate members of p levels explicitly and in their 
real forms. The HMO wavefunction is simply a product of the pi MOs, one for each of the 
six pi electrons: 1pu,x (1)1pu,x (2)1pu,y (3)1pu,y (4)1pg,x(5)1pg,y(6). The HMO energy is 
2Ep,u,x +2Ep,u,y +Ep,g,x +Ep,g,y, which reduces to 4Ep,u +2Ep,g. Note that, becauseO2 is 
linear, there is no unique molecular plane containing the internuclear axis. Therefore this molecule 
has two sets of p MOs, one pair pointing in the x direction, the other pair pointing along y. For a 
planar molecule, only one of these pairs would qualify as p MOs, as will be seen in the next section. 
 
8-4 Setting up the H¨uckel Determinant 
8-4.A Identifying the Basis Atomic Orbitals and Constructing 
a Determinant 
The allyl radical, C3H5, is a planar molecule1 with three unsaturated carbon centers 
(see Fig. 8-1). The minimal basis set of AOs for this molecule consists of a 1s AO 
on each hydrogen and 1s, 2s, 2px, 2py, and 2pz AOs on each carbon. Of all these 
AOs only the 2pz AOs at the three carbons are antisymmetric for reflection through the 
molecular plane. 
Figure 8-1 
 Sketch of the nuclear framework for the allyl radical. All the nuclei are coplanar. 
The z axis is taken to be perpendicular to the plane containing the nuclei. 
1The minimum energy conformation of the allyl system is planar. We will ignore the deviations from planarity 
resulting from vibrational bending of the system.
248 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-2 
 The three p-type AOs in the minimal basis set of the allyl radical. 
Following H¨uckel, we ignore all the s-type AOs and take the three 2pz AOs as our 
set of basis functions. Notice that this restricts us to the carbon atoms: the hydrogens 
are not treated explicitly in the simpleHMOmethod. We label our three basis functions 
.1,.2,.3 as indicated in Fig. 8-2. We will assume these AOs to be normalized. 
Suppose that we now perform a linear variation calculation using this basis set. We 
know this will lead to a 3×3 determinant having roots that are MO energies which can 
be used to obtain MO coefficients. The determinantal equation is 
 
H11 -ES11 H12 -ES12 H13 -ES13 
H21 -ES21 H22 -ES22 H23 -ES23 
H31 -ES31 H32 -ES32 H33 -ES33

=0 (8-12) 
where 
Hij =  .iHˆp.j dv (8-13) 
Sij =  .i.j dv (8-14) 
Since Hij and Sij are integrals over the space coordinates of a single electron, the 
electron index is suppressed in Eqs. (8-13) and (8-14). 
8-4.B The Quantity a 
We have already indicated that there is no way to write an explicit expression for Hˆp 
that is both consistent with our separability assumptions and physically correct. But, 
without an expression for Hˆp , how can we evaluate the integrals Hij ? The HMO 
method sidesteps this problem by carrying certain of theHij integrals along as symbols 
until they can be evaluated empirically by matching theory with experiment. 
Let us first consider the integrals H11, H22 and H33. The interpretation consistent 
with these integrals is that H11, for instance, is the average energy of an electron inAO 
.1 experiencing a potential field due to the entire molecule. Symmetry requires that 
H11 =H33. H22 should be different since an electron in AO .2 experiences a different
Section 8-4 Setting up the H¨uckel Determinant 249 
environment than it does when in .1 or .3. It seems likely, however, that H22 is not 
very different from H11. In each case, we expect the dominant part of the potential to 
arise from interactions with the local carbon atom, with more distant atoms playing a 
secondary role. Hence, one of the approximations made in the HMO method is that all 
Hii are identical if .i is on a carbon atom. The symbol a is used for such integrals. 
Thus, for the example at hand, H11 =H22 =H33 =a. The quantity a is often called 
the coulomb integral.2 
8-4.C The Quantity ß 
Next, we consider the resonance integrals or bond integrals H12, H23, and H13. (The 
requirement that Hˆp be hermitian plus the fact that the .’s and Hˆp are real suffices 
to make these equal to H21, H32, and H31, respectively.) The interpretation consistent 
with these integrals is thatH12, for instance, is the energy of the overlap charge between 
.1 and .2. Symmetry requires thatH12=H23 in the allyl system. However, even when 
symmetry does not require it, the assumption is made that all Hij are equal to the same 
quantity (called ß) when i and j refer to “neighbors” (i.e., atoms connected by a s 
bond). It is further assumed that Hij =0 when i and j are not neighbors. Therefore, 
in the allyl case, 
H12 =H23 =ß,H13 =0. 
8-4.D Overlap Integrals 
Since the .’s are normalized, Sii =1. The overlaps between neighbors are typically 
around 0.3. Nevertheless, in the HMO method, all Sij (i = j ) are taken to be zero. 
Although this seems a fairly drastic approximation, it has been shown to have little 
effect on the qualitative nature of the solutions. 
8-4.E Further Manipulation of the Determinant 
Our determinantal equation for the allyl system is now much simplified. It is 
 
a -E ß 0 
ß a-E ß 
0 ß a-E

=0 (8-15) 
Dividing each row of the determinant by ß corresponds to dividing the whole determinant 
by ß3. This will not affect the equality. Letting (a -E)/ß =x, we obtain the 
result 
 
x 1 0 
1 x 1 
0 1 x

=0 (8-16) 
2The term “coulomb integral” for a is unfortunate since the same name is used for repulsion integrals of the 
form 
 .1(1).2(2)(1/r12).1(1).2(2)dv. The quantity a also contains kinetic energy and nuclear–electronic 
attraction energy.
250 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-3 
 HMO determinants for some small systems. 
which is the form we will refer to as the HMO determinantal equation. Notice that x 
occurs on the principal diagonal, 1 appears in positions where the indices correspond 
to a bond, 0 appears in positions (e.g., 1,3) corresponding to no bond. This gives us a 
simple prescription for writing the HMO determinant for any unsaturated hydrocarbon 
system directly from a sketch of the molecular structure. The rules are (1) sketch 
the framework defined by the n unsaturated carbons; (2) number the atoms 1, . . . ,n 
(the ordering of numbers is arbitrary); (3) fill in the n×n determinant with x’s on the 
diagonal, 1’s in positions where rowcolumn indices correspond to bonds, 0’s elsewhere. 
See Fig. (8-3) for examples. As a check, it is useful to be sure that the determinant is 
symmetric for reflection through the diagonal of x’s. This is necessary since, if atoms 
i and j are neighbors, 1’s must appear in positions i, j and j , i of the determinant. 
Since the H¨uckel determinant contains only information about the number of unsaturated 
carbons and how they are connected together, it is sometimes referred to as a 
topological determinant. (Topology refers to properties that are due to the connectedness 
of a figure, but are unaffected by twisting, bending, etc.) 
8-5 Solving the HMO Determinantal Equation 
for Orbital Energies 
The HMO determinantal equation for the allyl system (8-16) can be expanded to give 
x3 -2x =0 (8-17) 
or 
x(x2 -2)=0 (8-18) 
Thus, the roots are x =0, x =v2, and x=-v2. Recalling the definition of x, these 
roots correspond respectively to the energies E =a, E =a -v2ß, E =a +v2ß. 
How should we interpret these results? Since a is supposed to be the energy of 
a pi electron in a carbon 2p AO in the molecule, we expect this quantity to be negative 
(corresponding to a bound electron). Since ß refers to an electron in a bond 
region, it too should be negative. Therefore, the lowest-energy root should be E1 = 
a + v2ß, followed by E2 = a, with E3 = a - v2ß being the highest-energy root.
Section 8-6 Solving for the Molecular Orbitals 251 
Figure 8-4 
 p-Electron configurations and total energies for the ground states of the allyl cation, 
radical, and anion. 
(It is convenient to number the orbital energies sequentially, starting with the lowest, 
as we have done here.) 
We have just seen that bringing three 2pp AOs together in a linear arrangement 
causes a splitting into three MO energy levels. This is similar to the splitting into two 
energy levels produced when two 1sAOs interact, discussed in connection with H+2 . In 
general, n linearly independent separatedAOs will lead to n linearly independent MOs. 
The ground-state p-electron configuration of the allyl system is built up by putting 
electrons in pairs into the MOs, starting with those of lowest energy. Thus far, we have 
been describing our system as the allyl radical. However, since we have as yet made 
no use of the number of p electrons in the system, our results so far apply equally well 
for the allyl cation, radical, or anion. 
Configurations and total p energies for these systems in their ground states are 
depicted in Fig. 8-4. The total p-electron energies are obtained by summing the oneelectron 
energies, as indicated earlier. 
EXAMPLE 8-2 For a planar, unsaturated hydrocarbon having formula CxHy, where 
all the carbons are part of the unsaturated framework, how many pi MOs are there? 
SOLUTION 
 Each carbon atom brings one 2pp AO into the basis set, so there are x basisAOs. 
These x independent AOs mix to form x independent MOs.  
8-6 Solving for the Molecular Orbitals 
We still have to find the coefficients that describe the MOs as linear combinations 
of AOs. Recall from Chapter 7 that this is done by substituting energy roots of the 
secular determinant back into the simultaneous equations. For the allyl system, the 
simultaneous equations corresponding to the secular determinant (8-16) are 
c1x +c2 =0 (8-19) 
c1 +c2x +c3 =0 (8-20) 
c2 +c3x =0 (8-21) 
(Compare these equations with the secular determinant in Eq. (8-16) and note the 
obvious relation.) As we noted in Chapter 7, homogeneous equations like these can give 
us only ratios between c1, c2, and c3, not their absolute values. So we anticipate using 
only two of these equations and obtaining absolute values by satisfying the normality
252 Chapter 8 The Simple H¨uckel Method and Applications 
condition. Because we are neglecting overlap betweenAOs, the latter step corresponds 
to requiring 
c2
1 +c2
2 +c2
3 =1 (8-22) 
The roots x are, in order of increasing energy, -v2, 0,+v2. Let us take x =-v2 
first. Then 
-v2c1+ c2 =0 (8-23a) 
c1-v2c2+ c3 =0 (8-23b) 
c2-v2c3 =0 (8-23c) 
Comparing Eqs. (8-23a) and (8-23c) gives c1 =c3. Equation (8-23a) gives c2 =v2c1. 
Inserting these relations into the normality equation (8-22) gives 
c2
1 +v2c1
2
+c2
1 =1 (8-24) 
4c2
1 =1, c1=±
1
2 
(8-25) 
It makes no difference which sign we choose for c1 since anywavefunction is equivalent 
to its negative. (Both give the same .2.) Choosing c1=+1
2 gives 
c1 = 
1
2
, c2 = 
1 
v2
, c3 = 
1
2 
(8-26) 
These coefficients define our lowest-energy MO, f1: 
f1 = 
1
2
.1 + 
1 
v2
.2 + 
1
2
.3 (8-27) 
A similar approach may be taken for x =0 and x=+v2. The results are 
(x =0) : f2 = 
1 
v2
.1 - 
1 
v2
.3 (8-28) 
x=+v2 : f3 = 
1
2
.1 - 
1 
v2
.2 + 
1
2
.3 (8-29) 
The allyl system MOs are sketched in Fig. 8-5. 
The lowest-energy MO, f1, has no nodes (other than the molecular-plane node 
common to all p MOs) and is said to be bonding in the C1-C2 and C2 -C3 regions. It 
is reasonable that such a bonding MO should have an energy wherein the bond-related 
term ß acts to lower the energy, as is true here. The second-lowest energy MO, f2, 
has a nodal plane at the central carbon. Because there are no p AOs on neighboring 
carbons in this MO, there are no interactions at all, and ß is absent from the energy 
expression. This MO is said to be nonbonding. The high-energy MO, f3 has nodal 
planes intersecting both bonds. Because the p AOs showsign disagreement across both 
bonds, this MO is everywhere antibonding and ß terms act to raise the orbital energy 
above a. 
EXAMPLE 8-3 According to HMO theory, do the p electrons favor a linear, or a 
bent allyl radical?
Section 8-7 The Cyclopropenyl System: Handling Degeneracies 253 
Figure 8-5 
 Sketches of the allyl system MOs. (a) emphasizes AO signs and magnitudes. 
(b) resembles more closely the actual contours of the MOs. 
SOLUTION 
 HMO theory favors neither. The difference between linear and bent allyl shows 
up as a difference in C1 - C2 - C3 angle and in C1 to C3 distance. The HMO method has no 
angular-dependent features and explicitly omits interactions between non-neighbor carbons, like 
C1 and C3.  
8-7 The Cyclopropenyl System: Handling Degeneracies 
The allyl system results when three p AOs interact in a linear arrangement wherein 
H12 = H23 = ß, but H13 = 0. We can also treat the situation where the three p AOs 
approach each other on vertices of an ever-shrinking equilateral triangle. In this case, 
each AO interacts equally with the other two. This triangular system is the cyclopropenyl 
system C3H3 shown in Fig. 8-6. 
The HMO determinantal equation for this system is 
 x 1 1 
1 x 1 
1 1 x

=0, x3 +2-3x =0 (8-30) 
Figure 8-6 
 The cyclopropenyl system (all nuclei are coplanar).
254 Chapter 8 The Simple H¨uckel Method and Applications 
This equation can be factored as 
(x +2)(x -1)(x -1)=0 (8-31) 
Therefore, the roots are x=-2,+1,+1. 
Since the root x =1 occurs twice, we can expect there to be two independent HMOs 
having the same energy–a doubly degenerate level. The energy scheme and ground 
state electron configuration for the cyclopropenyl radical (three p electrons) (I) gives 
a total Ep of 3a +3ß. We can surmise from these orbital energies that f1 is a bonding 
MO, whereas f2 and f3 are predominantly antibonding. To see if this is reflected in the 
nodal properties of the MOs, let us solve for the coefficients. The equations consistent 
with the HMO determinant and with orbital normality are 
c1x +c2 +c3 =0 
c1 +c2x +c3 =0 (8-32) 
c1 +c2 +c3x =0 
c2
1 +c2
2 +c2
3 =1 
Setting x=-2 and solving gives 
f1 = 
1 
v3
.1 + 
1 
v3
.2 + 
1 
v3
.3 (8-33) 
For this MO, the coefficients are all of the same sign, and so the AOs show phase 
agreement across all bonds and all interactions are bonding. 
To find f2 and f3 is trickier. We begin by inserting x =±1 into our simultaneous 
equations. This gives 
c1 +c2 +c3 = 0 (three times) (8-34) 
c2
1 +c2
2 +c2
3 = 1 (8-35) 
With three unknowns and two equations, an infinite number of solutions is possible. 
Let us pick a convenient one: c1=-c2, c3 =0. The normalization requirement then 
gives c1 =1/v2, c2=-1/v2, c3 =0. Let us call this solution f2: 
f2 = 
1 
v2
.1 - 
1 
v2
.2 (8-36) 
We still need to find f3. There remain an infinite number of possibilities, so let us pick 
one: c1 =1/v2, c2 =0, c3=-1/v2. We have used our experience with f2 to choose 
c’s that guarantee a normalized f3. Also, it is clear that f3 is linearly independent of
Section 8-7 The Cyclopropenyl System: Handling Degeneracies 255 
f2 since they contain different AOs. But it is desirable to have f3 orthogonal to f2. 
Let us test f2 and f3 to see if they are orthogonal: 
Since S = 0, f2 and f3 are nonorthogonal. We can project out that part of f3 that 
is orthogonal to f2 by using the Schmidt orthogonalization procedure described in 
Section 6-10. We seek a new function f3
given by 
f3 
=f3 -Sf2 (8-38) 
where 
S = f2f3dv = 
1
2 
(8-39) 
Therefore, 
f3 
=f3 - 
1
2
f2 = 
1 
2v2 
(.1 +.2 -2.3) (8-40) 
This function is orthogonal to f2 but is not normalized. Renormalizing gives 
f3 = 
1 
v6 
(.1 +.2 -2.3) (8-41) 
In summary, to produce HMO coefficients for degenerate MOs, pick any two independent 
solutions from the infinite choice available, and orthogonalize one of them to the 
other using the Schmidt (or any other) orthogonalization procedure. 
The MOs for the cyclopropenyl system as seen from above the molecular plane are 
sketched in Fig. 8-7. The MO f2 can be seen to have both antibonding (C1–C2) and 
nonbonding (C1–C3,C2–C3) interactions. f3 has antibonding (C1–C3,C2–C3) and 
bonding (C1–C2) interactions. The interactions are of such size and number as to give 
an equal net energy value (a-ß) in each case. Since nodal planes produce antibonding 
or nonbonding situations, it is not surprising that higher and higher-energy HMOs in a 
Figure 8-7 
 The HMOs for the cyclopropenyl system: (a) f1 =(1/v3)(.1 +.2 +.3); (b) f2 = 
(1/v2)(.1 -.2) (c) f3 =(1/v6)(.1 +.2 -2.3). The nodal planes intersect the molecular plane 
at the dashed lines.
256 Chapter 8 The Simple H¨uckel Method and Applications 
system display more and more nodal planes. Notice that the MOs f2 and f3 have the 
same number of nodal planes (one, not counting the one in the molecular plane) but 
that these planes are perpendicular to each other. This is a common feature of some 
degenerate, orthogonal MOs in cyclic molecules. 
It is important to notice the symmetry characteristics of these MOs. f1 is either 
symmetric or antisymmetric for every symmetry operation of the molecule. (It is 
antisymmetric for reflection through the molecular plane, symmetric for rotation about 
the threefold axis, etc.) This must be so for any nondegenerate MO. But the degenerate 
MOs f2 and f3 are neither symmetric nor antisymmetric for certain operations. (f2 is 
antisymmetric for reflection through the plane indicated by the dashed line in Fig. 8-7, 
but is neither symmetric nor antisymmetric for rotation about the threefold axis by 
120..) In fact, one can easily show that, given a cycle with an odd number of centers, 
each with oneAO of a common type, there is but one way to combine theAOs (to form a 
realMO)so that the result is symmetric or antisymmetric for all rotations and reflections 
of the cycle. Hence, an HMO calculation for a three-, five, seven-, . . . membered ring 
can give only one nondegenerate MO. However, for a cycle containing an even number 
of centers, the analogous argument shows that two nondegenerate MOs exist. 
8-8 Charge Distributions from HMOs 
Now that we have a method that provides us with orbitals and orbital energies, it should 
be possible to get information about the way the p-electron charge is distributed in the 
system by squaring the total wavefunction .p . In the case of the neutral allyl radical, 
we have (taking .p to be a simple product of MOs) 
.p =f1(1)f1(2)f2(3) (8-42) 
Hence, the probability for simultaneously finding electron 1 in dv(1), electron 2 in 
dv(2) and electron 3 in dv(3) is 
.2
p (1, 2, 3)dv(1)dv(2)dv(3)=f2
1 (1)f2
1 (2)f2
2 (3)dv(1)dv(2)dv(3) (8-43) 
For most physical properties of interest, we need to know the probability for finding an 
electron in a three-dimensional volume element dv. Since the probability for finding 
an electron in dv is the sum of the probabilities for finding each electron there, the 
one-electron density function . for the allyl radical is 
. =2f2
1 +f2
2 (8-44) 
where we have suppressed the index for the electron. If we integrate . over all space, 
we obtain a value of three. This means we are certain of finding a total p charge 
corresponding to three p electrons in the system. 
To find out how the p charge is distributed in the molecule, let us express . in terms 
of AOs. First, we write f2
1 and f2
2 separately: 
f2
1 = 
1
4
.2 
1 + 
1
2
.2 
2 + 
1
4
.2 
3 + 
1 
v2
.1.2 + 
1 
v2
.2.3 + 
1
2
.1.3 
f2
2 = 
1
2
.2 
1 + 
1
2
.2 
3 -.1.3 (8-45)
Section 8-8 Charge Distributions from HMOs 257 
If we were to integrate f2
1 we would obtain 
 f2
1 dv = 
1
4 
-.1 
 	 
 .2 
1 dv+
1
2 
-.1 
 	 
 .2 
2 dv+
1
4 
-.1 
 	 
 .2 
3 dv+ 
1 
v2 
-.0 
 	 
 .1.2 dv 
+ 
1 
v2 
-.0 
 	 
 .2.3 dv+
1
2 
-.0 
 	 
 .1.3 dv 
= 
1
4 + 
1
2 + 
1
4 =1 (8-46) 
Thus, one electron in f1 shows up, upon integration, as being “distributed” 1
4 at 
carbon 1, 1
2 at carbon 2, and 1
4 at carbon 3. We say that the atomic p-electron densities 
due to an electron in f1 are 1
4 , 1
2 , 1
4 at C1, C2, and C3, respectively, If we accumulate 
these figures for all the electrons, we arrive at a total p-electron density for each carbon. 
For the allyl radical, Table 8-1 shows that each atom has a p-electron density of unity. 
Generalizing this approach gives for the total p-electron density qi on atom i 
qi =
all MOs 

k 
nkc2
ik (8-47) 
Here k is the MO index, cik is the coefficient for anAO on atom i in MO k, and nk, the 
“occupation number,” is the number of electrons (0, 1, or 2) in MO k. (In those rare 
cases where cik is complex, c2
ik in Eq. (8-47) must be replaced by c*ikcik.) 
If we apply Eq. (8-47) to the cyclopropenyl radical, we encounter an ambiguity. If the 
unpaired electron is assumed to be inMOf2 of Fig. 8-7, we obtain q1=q2= 76
, q3= 4
6 . 
On the other hand, if the unpaired electron is taken to be in f3, q1=q2= 56
, q3= 86
. The 
HMO method resolves this ambiguity by assuming that each of the degenerate MOs is 
occupied by half an electron. This has the effect of forcing the charge distribution to 
show the overall symmetry of the molecule. In this example, it follows that q1 =q2 = 
q3 =1. The general rule is that, for purposes of calculating electron distributions, the 
electron occupation is averaged in any set of partially occupied, degenerate MOs. 
TABLE 8-1 
 HMO p Electron 
Densities in the Allyl Radical 
Carbon atom 
Electron 1 2 3 
1 in f1 
1
4 
1
2 
1
4 
2 in f1 
1
4 
1
2 
1
4 
3 in f2 
1
2 0 1
2 
– – – 
Sum 1 1 1
258 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-8 
 When the equilateral structure is distorted by decreasing R12 and increasing R13, 
R23, the energies associated with f1, f2, f3 shift as shown. 
In actuality, the equilateral triangular structure for the cyclopropenyl radical is unstable, 
and therefore the above-described averaging process is only a theoretical idealization. 
It is fairly easy to see that a distortion from equilateral to isosceles form will 
affect the MO energies E1, E2, and E3 differently. In particular, a distortion of the 
sort depicted in Fig. 8-8 would have little effect on E1 but would raise E2 (increased 
antibonding) and lowerE3 (decreased antibonding and increased bonding). Thus, there 
is good reason for the cyclopropenyl radical to be more stable in an isosceles rather 
than equilateral triangular form. This is an example of the Jahn–Teller theorem, which 
states, in effect, that a system having an odd number of electrons in degenerate MOs 
will change its nuclear configuration in a way to remove the degeneracy.3 The preference 
of the cyclopropenyl radical for a shape less symmetrical than what we might 
have anticipated is frequently called Jahn–Teller distortion.4 
Many times we are interested in comparing the p-electron distribution in the bonds 
instead of on the atoms. In the integrated expression (8-46) are cross terms that vanish 
under the HMO assumption of zero overlap. But the overlaps are not actually zero, 
especially betweenAOs on nearest neighbors. Hence, we might view the factors 1/v2 
as indicating how much overlap charge is being placed in the C1–C2 and C2–C3 bonds 
by an electron in f1. The C1–C3 bond is usually ignored because these atoms are not 
nearest neighbors and therefore have much smaller AO overlap. Since S12 =S23 =Sij 
for neighbors i and j in any p system (assuming equal bond distances), we need not 
include Sij , explicitly in our bond index. If we proceed in this manner, two electrons in 
3Linear systems are exceptions to this rule. Problems are also encountered if there is an odd number of electrons 
and spin-orbit coupling is substantial. The reader should realize that the above statement of the theorem is a little 
misleading inasmuch as it makes it sound like the molecule finds itself in a symmetric geometry that produces 
denerate MOs and then “distorts” to a lower-energy geometry. It is actually we who have guessed a geometry that 
is too symmetric. When our calculations reveal that this results in degenerate orbital energies containing an odd 
number of electrons, we are alerted that we have erred in our assumption, and that the molecule is really in a less 
symmetric, lower energy geometry. 
4See Salem [1, Chapter 8].
Section 8-9 Some Simplifying Generalizations 259 
f would then give us a “bond order” of 2/v2=1.414. It is more convenient in practice 
to divide this number in half, because then the calculated p-bond order for ethylene 
turns out to be unity rather than two. Since ethylene has one p-bond, this can be seen 
to be a more sensible index. 
As a result of these considerations, the p-bond order (sometimes called mobile bond 
order) of the allyl radical is 1/v2 = 0.707 in each bond. (Electrons in f2 make no 
contribution to bond order since c2 vanishes. This is consistent with the nonbonding 
label for f2.) 
Generalizing the argument gives, for pij , the p-bond order between nearestneighbor 
atoms i and j : 
pij =
all MOs 

k 
nkcikcjk (8-48) 
where the symbols have the same meanings as in Eq. (8-46). In cases in which partially 
filled degenerateMOsare encountered, the averaging procedure described in connection 
with electron densities must be employed for bond orders as well. 
EXAMPLE 8-4 Calculate p13 for the cyclopropenyl radical, using data in Fig. 8-7. 
SOLUTION 
 There are 2 electrons in f1 and the coefficients on atoms 1 and 3 are 1 v3
, so 
this MO contributes 2 × ( 1 v3 
)2 = 2/3. We allocate 1
2 electron to f2. Since c3 = 0 in this MO, 
the contribution to p13 is zero. The remaining 1
2 electron goes to f3, yielding a contribution of 
1
2 × 1 v6 × -2 v6 = -1 
6 . So p13 = 2
3 - 16 
= 1
2 .  
8-9 Some Simplifying Generalizations 
Thus far we have presented the bare bones of the HMO method using fairly small 
systems as examples. If we try to apply this method directly to larger molecules, it is 
very cumbersome. A ten-carbon-atom system leads to a 10 ×10 HMO determinant. 
Expanding and solving this for roots and coefficients is tedious. However, there are 
some short cuts available for certain cases. In the event that the system is too complicated 
to yield to these, one can use computer programs which are readily available. 
For straight chain and monocyclic planar, conjugated hydrocarbon systems, simple 
formulas exist for HMO energy roots and coefficients. These are derivable from the 
very simple forms of the HMO determinants for such systems.5 We state the results 
without proof. 
For a straight chain of n unsaturated carbons numbered sequentially, 
x = -2 cos[kp/(n+1)], k=1, 2, . . . ,n (8-49) 
clk = [2/(n+1)]1/2 sin[klp/(n+1)] (8-50) 
where l is the atom index and k the MO index. 
5See Coulson [2].
260 Chapter 8 The Simple H¨uckel Method and Applications 
For a cyclic polyene of n carbons, 
x = -2 cos(2xk/n), k =0, 1, . . . ,n-1 (8-51) 
clk = n-1/2 exp[2pik(l -1)/n], i=v-1 (8-52) 
The coefficients derived from Eq. (8-52) for monocyclic polyenes will be complex 
when theMOis one of a degenerate pair. In such cases one may take linear combinations 
of these degenerate MOs to produce MOs with real coefficients, if one desires. 
There is also a diagrammatic way to find the energy levels for linear and monocyclic 
systems.6 Let us consider monocycles first. One begins by drawing a circle of radius 
2 |ß|. Into this circle inscribe the cycle, point down, as shown in Fig. 8-9 for benzene. 
Project sideways the points where the polygon intersects the circle. The positions of 
these projections correspond to the HMO energy levels if the circle center is assumed 
to be at E =a (see Fig. 8-9). The number of intersections at a given energy is identical 
to the degeneracy. The numerical values for E are often obtainable from such a sketch 
by inspection or simple trigonometry. 
For straight chains, a modified version of the above method may be used: For an ncarbon 
chain, inscribe a cycle with 2n+2 carbons into the circle as before. Projecting 
out all intersections except the highest and lowest, and ignoring degeneracies gives the 
proper roots. This is exemplified for the allyl system in Fig. 8-10. 
Examination of the energy levels in Figs. 8-9 and 8-10 reveals that the orbital energies 
are symmetrically disposed about E =a. Why is this so? Consider the allyl system. 
The lowest-energy MO has two bonding interactions. The highest-energy MO differs 
only in that these interactions are now antibonding. [See Fig. 8-5 and note that the 
coefficients in f1 and f3 are identical except for sign in Eqs. (8-27) and (8-29).] The 
role of the ß terms is thus reversed and so they act to raise the orbital energy for f3 just 
as much as they lower it for f1. A similar situation holds for benzene. As we will see 
shortly, the lowest energy corresponds to an MO without nodes between atoms, so this 
is a totally bonding MO. The highest-energyMOhas nodal planes between all neighbor 
carbons, and so every interaction is antibonding. An analogous argument holds for the 
degenerate pairs of benzene MOs. These observations suggest that the energy of an 
Figure 8-9 
 HMO energy levels for benzene produced by projecting intersections of a hexagon 
with a circle of radius 2 |ß|. 
6See Frost and Musulin [3].
Section 8-9 Some Simplifying Generalizations 261 
Figure 8-10 
 HMO energy levels for the allyl system (n=3) produced by projecting the intersections 
of an octagon (n=2×3+2) with a circle of radius 2 |ß|. 
MO should be expressible as a function of the net bond order associated with it, and 
this is indeed the case. The energy of the ith MO is given by the expression 
Ei =  fiHˆpfi dv = 
k 
cki.kHˆp
l 
cli.l dv (8-53) 
=
k 

l 
ckicli  .kHˆp.l dv (8-54) 
When the atom indices k and l are identical, the integral is equal to a; when k and l are 
neighbors, it equals ß. Otherwise it vanishes. Hence, we may write 
Ei =
k 
c2
kia +
neighbors 

k,l 
ckicliß (8-55) 
However, c2
ki is qk,i , the electron density at atom k due to one electron in MO fi , 
and ckicli is pkl,i , the bond order between atoms k and l due to an electron in fi . 
Therefore, 
Ei =
k 
qk,ia +2
neighbors 

k<l 
pkl,iß (8-56) 
We have seen that the sum of electron densities must equal the total number of electrons 
present. For one electron in fi , this gives additional simplification. 
Ei =a +2ß 
bonds 

k<l 
pkl,i (8-57) 
The total p-electron energy is the sum of one-electron energies. For np electrons 
Ep =na +2ß 
bonds 

k<l 
pkl (8-58) 
where pkl is the total p-bond order between neighbors k and l. Hence, the individual 
orbital energies directly reflect the amount of bonding or antibonding described by 
the MOs, and the total energy reflects the net bonding or antibonding due to all the p 
electrons together.
262 Chapter 8 The Simple H¨uckel Method and Applications 
EXAMPLE 8-5 The total p energy of cycloheptatrienyl radical (C7H7) is 7a + 8.5429ß. What is the bond order for any bond in this molecule, assuming it to be 
heptagonal? 
SOLUTION 
 The total p-bond order must be 8.5429/2 or 4.2714. This results from seven 
identical bonds, so each bond order is 4.2714/7=0.6102.  
Does this pairing of energy levels observed for allyl and benzene always occur? It 
is easy to show that it cannot in rings with an odd number of carbon centers. Consider 
the cyclopropenyl system. The lowest-energy MO is nodeless, totally bonding and has 
an energy of 2a +2ß. [Note from Eq. (8-51) and also from the diagram method that 
every monocyclic system has a totally bonding MO at this energy.] To transform these 
three bonding interactions into antibonding interactions of equal magnitude requires 
that we cause a sign reversal across every bond. This is impossible, for, if c1 disagrees 
in sign with c2 and c3, then c2 and c3 must agree in sign and cannot yield an antibonding 
interaction. 
Not surprisingly, this has all been considered in a rigorous mathematical fashion. 
Systems containing a ring with an odd number of atoms are “nonalternant” systems. All 
other homonuclear unsaturated systems are “alternant” systems. An alternant system 
can always have asterisks placed on some of the centers so that no two neighbors are 
both asterisked or unasterisked. For nonalternants, this is not possible (see Fig. 8-11). 
It is convenient to subdivide alternant systems into even alternants or odd alternants 
according to whether the number of centers is even or odd. With this terminology 
defined, we can nowstate the pairing theorem and some of its immediate consequences. 
The theorem states that, for alternant systems, (1) energy levels are paired such that, 
for each level at E = a +kß there is a level at E = a -kß; (2) MOs that are paired 
in energy differ only in the signs of the coefficients for one of the sets (asterisked or 
unasterisked) of AOs. 
It is easy to see that an immediate result of this theorem is that an odd-alternant 
system, which must have an odd number of MOs, must have a nonbonding (E = a) 
MO that is not paired with another MO. It is also possible to show that the electron 
density is unity at every carbon for the neutral ground state of an alternant system. The 
proofs of the pairing theorem and some of its consequences are given in Appendix 5. 
Another useful short cut exists that enables one to sketch qualitatively the MOs 
for any linear polyene. The HMOs for the allyl and butadiene systems are given in 
Fig. 8-12. Notice that the envelopes of positive (or negative) phase in these MOs are 
similar in appearance to the particle in a one-dimensional “box” solutions described 
Figure 8-11 
 (a) Even and (b) odd alternants have no two neighbors identical in terms of an 
asterisk label. (c) Nonalternants have neighbors that are identical.
Section 8-10 HMO Calculations on Some Simple Molecules 263 
+ 
f 4 
f 3 
f 2 
f 1 
f 2 
f 1 
f 1 
– 
– 
– 
– 
– 
– 
– 
– 
– 
– 
–
– 
– 
– 
– – 
–
–
– 
– 
– 
+ 
+ 
+ 
+ 
+ + 
+ 
+ 
+ 
+ 
+
+ 
+ + 
+ 
+
+ 
+ 
+
+ 
–
+ 
–
+ –
+ 
Figure 8-12 
 MOs for the allyl and butadiene systems. The dashed lines emphasize the similarity 
between an envelope, or contour, of positive fp for these systems and the particle in a one-dimensional 
box solutions. 
in Chapter 2. This similarity makes it fairly easy to guess the first few MOs for 
pentadienyl, hexatriene, etc. Also, if one knows the lowest-energy half of the MOs for 
such molecules, one can generate the remaining MOs by appeal to the second part of 
the pairing theorem. (The edges of the one-dimensional box should extend one C–C 
bond length beyond the terminal atoms.) 
As we consider larger systems, brute-force solution of determinantal equations 
becomes too labor-intensive, and so such cases are always either solved on a computer7 
or else by appeal to HMO tabulations in print.8 However, the above generalizations 
continue to be useful in understanding the MO results. 
8-10 HMO Calculations on Some Simple Molecules 
Thus far, we have used the allyl and cyclopropenyl systems as examples. We will now 
describe the results of HMO calculations on some other simple but important systems. 
8-10.A Ethylene (Even Alternant) 
The H¨uckel determinantal equation is
 
x 1 
1 x

=0 
7Many types of quantum-chemical computer programs are available from: Quantum Chemistry Program 
Exchange, Chemistry Department. Indiana U¨ niversity, Bloomington, Indiana 47401. On the Internet at 
http://www.QCPE.Indiana.edu/ 
8See Coulson and Streitwieser [4], Streitwieser and Brauman [5], and Heilbronner and Straub [6]. See also 
Appendix 6 of this text.
264 Chapter 8 The Simple H¨uckel Method and Applications 
and so x2 -1=0; x=+1, -1. The resulting orbital energies and coefficients are 
E1 = a +ß,f1 = 
1 
v2
.1 + 
1 
v2
.2 
E2 = a -ß,f2 = 
1 
v2
.1 - 
1 
v2
.2 (8-59) 
These, with the ground state electronic configuration indicated, are shown in (II). 
p-Electron densities and p-bond order are indicated in the diagram beneath the MO 
sketches. Ethylene is an even alternant, so it has paired energies, unit electron densities, 
and coefficients related by a sign change. 
8-10.B Butadiene (Even Alternant) 
This problem can be solved by expansion to a polynomial in x and factoring, but it 
is simpler to use Eq. (8-49) or the decagon in a circle of radius 2 |ß|. The coefficients 
are obtainable from Eq. (8-50). The results are 
(8-60) 
The MOs are given in Fig. 8-12. f1 is bonding in all bonds, f2 is bonding in the 
outer bonds, antibonding in the central bond. As a result, butadiene has a lower p-bond 
order in the central bond than in the outer bonds. This is in pleasing accord with the
Section 8-10 HMO Calculations on Some Simple Molecules 265 
experimental observation that the central bond in butadiene is significantly longer than 
the outer bonds. 
The formal structural formula for butadiene (III), indicating two pure double bonds 
and one pure single bond, is clearly not an adequate description since we have just 
found the central bond to have some p-bonding order (0.447) and the outer bonds to 
be less p bonding than ethylene (0.894 as opposed to 1.000). The MO parlance is that 
the p electrons in butadiene are delocalized over the entire carbon system rather than 
being restricted to the formal double bonds only. 
Because the HMO method allows for no interaction between terminal carbons 
(i.e., H1,4 =0), there is no distinction between cis- and trans-butadiene in this calculation. 
However, if a weak interaction were postulated, it is not difficult to see that f1 
would give a bonding end-to-end contribution, f2 a substantially larger antibonding 
interaction, leading to a prediction (in agreement with experiment) that trans-butadiene 
is the more stable form. 
8-10.C Cyclobutadiene (Even Alternant) 
If we assume that this system is perfectly square, we can inscribe the square in a 
circle, giving us the orbital energies immediately (IV). 
Since f1 and f4 are nondegenerate, they must be symmetric or antisymmetric for 
the various rotations and reflections of the molecule. Also, f1 and f4 must have the 
same coefficients, except for sign changes. 
It follows at once that 
f1 = 
1
2
.1 + 
1
2
.2 + 
1
2
.3 + 
1
2
.4, f4 = 
1
2
.1 - 
1
2
.2 + 
1
2
.3 - 
1
2
.4 
f2 and f3 are degenerate, so there is some arbitrariness here. However, we expect each 
of these MOs to have a nodal plane, and these planes should be perpendicular to each 
other if f2 and f3 are to be orthogonal. Therefore, we choose the pair having nodal 
planes indicated by dashed lines (V). The four MOs for cyclobutadiene as seen from
266 Chapter 8 The Simple H¨uckel Method and Applications 
above are shown in (VI). It is clear that f2 and f3 are nonbonding because the nodal 
planes prevent interactions between neighbors, but it looks like f2 and f3 violate the 
pairing theorem since they cannot be interchanged by changing signs of coefficients. 
This is only an apparent violation, because it is easy to find an equivalent pair of MOs 
that follow the rule. We need only choose nodal planes that are rotated 45. (VII) from 
those we selected previously. These MOs are linear combinations of f2 and f3 and 
are equivalent to them for purposes of calculating electron densities and bond orders. 
They are still nonbonding MOs (E still equals a), but now it is not because of a nodal 
plane preventing nearest-neighbor interactions, but because bonding and antibonding 
interactions occur in equal number and magnitude. 
Notice that this is an example of an even alternant system with nonbonding MOs. 
Thus, whereas an odd alternant system must have an unpaired nonbonding MO, even 
alternants may have nonbonding MOs in pairs. 
Since this system is alternant, it must have p-electron densities of unity in its ground 
neutral state. Thiswould be necessary however, even if the molecule were not alternant, 
due to the fact that all carbons are equivalent by symmetry. Hence, they must all have 
the same electron density. Since it must sum to four electrons, the density of each atom 
must be unity. The same argument applies to the cyclopropenyl radical, a nonalternant 
that, nevertheless, has all electron densities equal to unity (if the unpaired electron is 
divided between degenerate MOs). 
Our assumed square symmetry also requires all four bonds to be identical. Since 
the total energy, 4a+4ß, is related to bond order through Eq. (8-58), it follows at once 
that the total bond order is 2, and so each bond has order 1
2 . 
EXAMPLE 8-6 We have assumed that cyclobutadiene is square. Could it be 
otherwise?
Section 8-10 HMO Calculations on Some Simple Molecules 267 
SOLUTION 
 Square cyclobutadiene leads to degenerate, nonbonding pi MOs, each of which 
we are assuming to contain one electron. (See IV.)We have further assumed that these two electrons 
have the same spin. This gives a triplet state (assumed to be the lower-energy choice) which forces 
us to put each electron into a different MO. If cyclobutadiene were distorted into a rectangle by 
shortening bonds 1–2 and 3–4, and lengthening 2–3 and 1–4, we would expect f2
of VII to rise in 
energy and f3
to drop. If we continue having one electron in each MO, these changes would tend to 
cancel but, if we put both electrons into f3
, the p energy would drop. However, this would require 
paired spins—a singlet. We have here, then, a competition between two factors: the energy rise of 
forming a singlet and the energy lowering from being distorted to a rectangle. HMO theory doesn’t 
tell us which factor wins, but it does alert us to a possible extension to the Jahn-Teller effect: An 
odd number of electrons (one or three) in a doubly-degenerate level must cause distortion. An even 
number (two) may cause distortion if the destabilization caused by going from triplet to singlet is 
more than counterbalanced by the stabilization resulting from “distortion.”9  
8-10.D Benzene (Even Alternant) 
Benzene is another molecule whose high symmetry enables one to use shortcuts. The 
orbital energies have already been found from the hexagon-in-a-circle diagram (see 
Fig. 8-9). The lowest- and highest-energy MOs must show all the symmetry of the 
molecule, as they are nondegenerate. Therefore, 
f1 
(6) = 
1 
v6 
.1 +(-)
.2 +.3 +(-)
.4 +.5 +(-)
.6 
where the carbon atoms are numbered sequentially around the ring. 
The degenerate MOs f2 and f3 should have one nodal plane each, and these should 
be perpendicular to each other.10 If we take one plane as shown in (VIII), we can 
immediately write down f2. The node for f3 is given in (IX). It is obvious that .6, 
.1, .2 have coefficients of the same sign, and that c2 = c6 =-c5 =-c3, and also 
c1 =-c4. However, c1 need not equal c2 as these atoms are differently placed with 
respect to the nodal plane. To determine these coefficients, we will use the fact that 
the neutral ground-state p densities are all unity in this system. We consider first atom 
number 1. Its electron density due to two electrons in f1 and two electrons in f2 is 
16 
+ 16 
+0+0= 13
. 
9Experiments and high-accuracy calculations indicate that the lowest-energy state for cyclobutadiene is a rectangular 
singlet. Surprisingly, calculations of high accuracy also indicate that, even in the square geometry, the 
1B1g singlet state (corresponding to each degenerate p MO containing one electron, but with spins paired) lies 
below the triplet in energy—the opposite of what Hund’s rule predicts. This is thought to result from energy 
lowering due to spin polarization. For a detailed review of studies on cyclobutadiene, see Minkin et al. [19]. 
10This statement applies to the real forms of f2 and f3. The complex forms [derivable from Eq. (8-52)] do not 
have a planar node. The situation is analogous to the 2p+1, 2p-1 versus 2px, 2py orbitals for the H atom.
268 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-13 
 f2 for benzene and f1 for ethylene have the same HMO energy. The MOs are 
sketched as seen from above. 
Therefore, two electrons in f3 must produce a contribution of 2
3 . Hence the coefficient 
for this atom in f3 must be 1/v3. Asimilar argument for atom 2 gives a coefficient 
of 1/v12 (or, one can use the normality condition for f3). As a result, 
f3 
(5) = 
1 
v3 
.1 +(-)
1
2
.2 - 
1
2
.3 -(+)
.4 - 
1
2
.5 +(-)
1
2
.6 
Appeal to the pairing theorem generates f4 and f5 as indicated. 
By symmetry, all bond orders must be identical, and their sum must be 4, since 
E =6a +8ß. Therefore, p12 =p23=···=p61 = 4
6 =0.667. 
The hexagon in a circle applies to ethylene (n=2, 2n+2=6) as well as to benzene. 
As a result, the orbital energies for f2, f3 of benzene are identical to E1 for ethylene. 
Examination of these MOs (Fig. 8-13) makes the reason for their energy agreement 
clear. Molecular orbital f2 of benzene is more revealing than is f3 in this context. 
The nodal plane produces an MO corresponding to two noninteracting ethylene MOs. 
Hence, an electron in this MO is always in a situation that is indistinguishable (under 
HMO approximations) from that in f1 of ethylene. 
In this section, we have tried to illustrate some of the properties of HMO solutions 
for simple systems and to indicate how symmetry and other relations are useful in 
producing and understanding HMO results. It is often convenient to have HMO results 
for various simple systems readily available in a condensed form. Therefore, a summary 
of results for a number of molecules is provided in Appendix 6. 
8-11 Summary: The Simple HMO Method for 
Hydrocarbons 
1. The assumption is made that the p-electron energy can be minimized independently 
of s electrons. This is an approximation. 
2. The assumption is made that each p electron sees the same field (the repulsion due to 
the other p electrons is presumably included “in effect,” in a time averaged way) so 
that the p electrons are treated as independent particles. This approximation leads 
to a total wavefunction that is a simple product of one-electron MOs and a total p 
energy that is a sum of one-electron energies. Except for use of the Pauli principle 
to build up configurations, no explicit treatment is made of electron spin. 
3. The basis set is chosen to be a 2.p AO from each carbon atom in the unsaturated 
system. Choosing a basis set of AOs means our MOs will be linear combinations 
of AOs, and so this is an LCAO–MO method.
Section 8-12 Relation Between Bond Order and Bond Length 269 
4. The H¨uckel determinant summarizes the connectedness of the unsaturated system, 
and is independent of cis–trans isomerism or bond length variation. 
5. The energy of eachMOis expressed in terms of atomic terms, a, and bond terms, ß. 
The amount of a in each MO energy is always unity because the sum of p-electron 
densities for one electron in the MO is always unity. The amount of ß present is 
related to the net bonding or antibonding character of the MO. 
6. Alternant systems display paired energy levels and corresponding MOs having coefficients 
related by simple sign reversals. For ground-state neutral alternants, the p 
electron densities are all unity. 
7. Acaveat: One can performHMOcalculations on very large systems such as pentahelicene 
(X) thereby making the implicit assumption that this is a planar molecule. But 
repulsion between protons on the terminal rings is sufficient to cause this molecule 
to deviate from planarity. Similarly, cyclooctatetraene is tub-shaped, probably due 
to considerations of angle strain that would occur in the planar molecule. Hence, 
one must recognize that, in certain cases, an HMO calculation refers to a planar 
“ideal” not actually achieved by the molecule. 
8-12 Relation Between Bond Order and Bond Length 
In this and following sections we will describe some of the relations between HMO 
theoretical quantities and experimental observations.11 
It is natural to look for a correlation between calculated p-bond orders and experimentally 
determined bond lengths. A high bond order should correspond to a large 
p charge in the bond region, which should yield a shorter, stronger bond. The bondorder–
bond-length results for certain simple systems, given as a graph in Fig. 8-14, 
do indeed show the anticipated behavior. However, as more and more data are added 
(Fig. 8-15) it becomes clear that an exact linear relation between these quantities does 
not exist at this level of refinement. 
Nevertheless, the correlation between bond order and bond length is good enough 
to make it useful for rough predictions of bond length variations. An example of this 
is given in Fig. 8-16, where calculated and observed bond lengths for phenanthrene are 
plotted. Even though the predicted absolute values of bond lengths are imperfect, the 
theoretical values show a rough parallelism with observed changes in values as we go 
from bond to bond. The relation used here to calculate the theoretical bond lengths 
from HMO bond orders is due to Coulson [8] and has the form 
R =s - 
s -d 
1+k(1-p)/p 
(8-61) 
11For a more complete discussion of these phenomena as well as many others, see Streitwieser [7].
270 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-14 
 p-Bond order versus bond length for some simple unsaturated hydrocarbons. 
Figure 8-15 
 Bond lengths versus p-bond orders for benzene, graphite, naphthalene, anthracene, 
phenanthrene, triphenylene, and pyrene. 
Here, s is the single bond length, d the double bond length, p the p-bond order, k an 
adjustable parameter, and R the predicted length. The double bond length d is taken 
to be 1.337 Å, the bond length of ethylene. The single bond length may be taken to be 
the length of the C–C bond in ethane, 1.54 Å, or it may be set by fitting to data points 
(such as those in Fig. 8-15) on the assumption that the single bond between two CH2 
groups (i.e., ethylene with its p bond “turned off”) is not necessarily the same length 
as the single bond between two CH3 groups. Both of these alternatives have been used 
in Fig. 8-16. Other mathematical forms have also been suggested. Because of the 
scatter in the data, there is little basis for preferring one formula over another. It seems 
generally true that all the proposed relationships work best for bonds in condensed ring 
systems, and most poorly for bonds in acyclic polyenes (e.g., butadiene) or between 
rings (e.g., biphenyl).
Section 8-13 ESR Hyperfine Splitting Constants 271 
Figure 8-16 
 Theoretical versus experimental bond lengths for phenanthrene in order of decreasing 
observed length: (•) experimental; (×) theoretical with s = 1.54Å, k = 0.765; (.) theoretical 
with s =1.515Å, k =1.05. 
8-13 p-Electron Densities and Electron Spin Resonance 
Hyperfine Splitting Constants12 
If a molecule with one or more unpaired electron spins is placed in an external magnetic 
field, the interaction energy of an unpaired electron’s spin dipole with the external field 
will be different depending on whether the electron’s spin is a or ß. This produces two 
slightly different energy levels for an unpaired electron. Suppose microwave radiation 
is supplied in an effort to excite such electrons from the lower to the higher of these two 
levels. If the radiation’s energy is larger than required, no absorption occurs. But, if we 
gradually increase the strength of the magnetic field, thereby increasing the energy-level 
12A number of reviews on this subject have been published. See, for example, Gerson and Hammons [9].
272 Chapter 8 The Simple H¨uckel Method and Applications 
gap, we eventually achieve a magnetic field where absorption of the microwave energy 
occurs. This would give us a spectrum (absorption vs. magnetic field) with just one 
line. 
Now suppose that our radical has a hydrogen atom attached to it. The proton nucleus 
of the hydrogen can have a spin of a or ß, and this can have a small effect on the 
energies of the electron whose spin we are trying to flip. We find now that half of the 
radicals absorb at a slightly different applied magnetic field than the other half giving us 
a two-line spectrum: The original single line now exhibits hyperfine splitting due to the 
presence of a hydrogen atom, and the distance between the two peaks on the magnetic 
axis is called the hyperfine splitting constant, usually given in millitesla (mT), or in 
gauss (G). 
If the unpaired electron is a p electron,HMOtheory can be applied to the value of the 
ESR splitting constant. The interaction between an unpaired p electron and a proton 
(or other nucleus with nonzero nuclear spin) falls off rapidly with increasing separation. 
Therefore, the hyperfine structure is generally ascribed to interactions involving protons 
directly bonded to carbons in the p system (a protons) or else separated from the p 
system by two s bonds (ß protons). The equilibrium position of an a proton is in the 
nodal plane of the p system, so it is clear that any net spin density at the proton must 
be only indirectly due to the presence of an unpaired p electron. This indirect effect 
arises because the unpaired p electron interacts slightly differently with a- and ß-spin 
s electrons on carbon, and so the spatial distributions of these become slightly different 
in the s MOs, ultimately producing net spin density at the proton; the s electrons are 
said to be spin polarized by the p electron (see Fig. 8-17). 
The extent of spin polarization at a given hydrogen should depend on the percentage 
of time the unpaired p electron spends on the carbon to which that hydrogen is 
bonded. It is therefore reasonable to look for relationships between the distribution of 
p-spin density in a radical and the hyperfine coupling constants characterizing its ESR 
spectrum. The simplest assumption one can make in this regard, called the McConnell 
relation, is that the hyperfine splitting constant, aHµ, for a proton directly bonded to 
the µth carbon, is proportional to the net p-spin density .µ on that carbon: 
aHµ =Q.µ (8-62) 
Thus, if the unpaired p electron density is twice as great on carbon 1 as on carbon 2, 
the ESR splitting constant for a proton attached to carbon 1 should be twice as great as 
that for a proton attached to carbon 2. 
Figure 8-17 
 The unpaired p-spin density at carbon repels both s electrons in the C–H bond 
region, but does not repel them equally. As a result, slight spin imbalance due to s electrons occurs 
at the proton.
Section 8-13 ESR Hyperfine Splitting Constants 273 
The HMO method predicts that the spin density at the µth carbon due to an unpaired 
electron in the mth MO is simply
cµm
2. This is only a first approximation to .µ, 
but a graph of observed splitting constants plotted against .µ calculated in this way 
shows a respectable correlation (Fig. 8-18). The proportionality factor Q, given by the 
slope of the line of best fit, varies somewhat depending on the type of system and the 
charge of the radical. Thus, in Fig. 8-18, where all the data are from aromatic fused ring 
hydrocarbon radical anions, the correlation is quite good. As data for nonalternants 
and radical cations are added (Fig. 8-19) the scatter increases. 
Some ESR spectra are interpreted to be consistent with the presence of some negative 
spin density. That is, if the extra p electron is taken to have a spin, some of the carbons 
appear to have an excess of ß-spin p density. This has the effect of giving us a total spin 
density that is greater than unity, even though we have only one electron with unpaired 
spin. (It sounds absurd, but read on.) Now the amount of splitting seen in ESR spectra 
depends only on the magnitudes of spin densities at the protons, not on whether these 
spin densities are a or ß, and the presence of negative spin density does not produce 
a qualitative change in an ESR spectrum (such as a “negative splitting,” whatever that 
might be). Rather it leads to an increase in the total amount of spin density in the 
system, and this, in turn, leads to an increase in the sum of splitting constants (over 
Figure 8-18 
 ESR splitting constants aHµ in gauss versusHMOunpaired spin densities. The systems 
are fused ring alternant hydrocarbon radical anions (naphthalene, anthracene, tetracene, pyrene). 
The underlined point is thought to result from negative spin density. (Data from Streitwieser [7].)
274 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-19 
 ESR hyperfine splitting constants in gauss versus c2 for the highest occupied MO 
of the radical. All data are from hydrocarbon radical anions or cations of alternant or nonalternant 
type. (•) Anion radical; (×) cation radical. Uncertain assignments: (.) anion; (.) cation. The two 
correlation lines are merely sight fitted to the anion and cation points separately and suggest that Q 
for cations should be larger than for anions. Points thought to result from negative spin density are 
underlined. [See Tables I, III, VIII, XII, XIII, XIV, XV, XVI, XVII, XVIII of Gerson and Hammons 
[9], and Table 6.2 of Streitwieser [7] for data plotted here.] 
what would be predicted in the absence of negative spin density). For example, the 
HMO prediction for allyl radical would give a net spin density of 1
2 at each terminal 
carbon and zero at the central carbon. [The unpaired electron is in the nonbonding MO 
1 v2
.1 - 1 v2
.3.] However, we might imagine the situation wherein there is a net spin 
of 2
3a at each terminal carbon and 13
ß at the central atom. This retains a net spin value 
of 1a yet results in a larger splitting constant aHµ for every proton in the molecule. 
It is possible to think of a physical explanation for this kind of spin distribution. The 
p electrons in lower-energy, filled MOs are being spin polarized similarly to the s 
electrons mentioned earlier. If an a-spin electron in effect repels electrons of ß spin 
more strongly than those of a spin, a buildup of ß spin on the central carbon of allyl 
radical could be expected. Because this secondary effect is expected to be fairly small,
Section 8-14 Orbital Energies and Oxidation-Reduction Potentials 275 
a net negative spin density is likely only on carbons where the primary effect (c2
µm) is 
zero or quite small. Methods for calculating this spin polarization (which will not be 
described here) indicate that certain of the splitting constants plotted in Figs. 8-18 and 
8-19 do in fact result from negative spin densities. These data points are indicated in the 
figures, and it can be seen that they do indeed occur where c2
µm and aHµ are small. This 
secondary effect is masked at higher values of c2
µm and aHµ. but presumably accounts 
for some of the scatter in the data. (Some scatter also results from solvent dependence 
of aHµ. Several solvents were used in experiments yielding the data in Fig. 8-19.) 
Because of the pairing-theorem relation between coefficients of MOs, the ESR spectra 
of the radical cation and anion of an alternant system should appear very similar. 
H¨uckel molecular orbital theory would predict them to be identical except for a slight 
change of scale due to a change in the factor Q. In practice, this similarity has been 
observed to hold fairly well. 
8-14 Orbital Energies and Oxidation-Reduction Potentials 
Many conjugated hydrocarbons can be oxidized or reduced in solution using standard 
electrochemical techniques. Since oxidation involves removing an electron from the 
highest occupied (p) MO (HOMO), it is reasonable to expect molecules with lowerlying 
HOMOs to have larger oxidation potentials. Similarly, we might expect reduction 
to be easier for compounds wherein the lowest unoccupied MO (LUMO) is lower in 
energy [see (XI)]. Thus, compoundAshould have a lower oxidation potential than compound 
B since EA
m >EBm
, but compound B should have the lower reduction potential. 
A plot of oxidation potential versus Em in units of ß for a series of aromatic hydrocarbons 
(Fig. 8-20) yields a correlation that is remarkably free of scatter. Figure 8-21 
indicates a similar correlation for reduction potential versus Em+1. 
Since oxidation-reduction potentials correlate with HOMO–LUMO energies, and 
since these energies are paired in even alternant systems, we should expect a plot of 
oxidation vs. reduction potential for such compounds to be linear also. (Problem 8-14.) 
The data in Figs. 8-20 and 8-21 provide a connection between theoretical energy 
differences in units of ß, and experimental energies. From the slope in Fig. 8-20 we
276 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-20 
 Oxidation potentials in acetonitrile solution versus energy of HOMO (in units of ß). 
(1/2 from Lund [10].) 
obtain that ß~=
-2.03 eV, or -46.8 kcal/mole. Similarly, the reduction potential data 
of Fig. 8-21 give ß =-2.44eV=-56.3 kcal/mole. 
One must be cautious in interpreting the above values of ß. The problem is that 
the experimental numbers include effects of physical processes not included in the 
theory. For example, when a neutral molecule in solvent becomes oxidized or reduced, 
solvation energy changes occur. One might argue that the fit of the data to straight lines 
in Figs. 8-20 and 8-21 implies this sort of contribution to be small, but the sensitivity 
of redox potentials to solvent nature indicates that this is not the case. However, in 
larger molecules, solvation energy change upon ionization tends to be smaller, and 
larger molecules also tend to have Em and Em+1 closer to the E = a level. In other 
words, we expect both solvation energy change and redox potential to be proportional 
to Em or Em+1. Therefore, they can be combined in a single linear relation. There
Section 8-14 Orbital Energies and Oxidation-Reduction Potentials 277 
Figure 8-21 
 First reduction potential in 2-methoxyethanol versus energy of LUMO (in units 
of ß). (1/2 from Bergman [11].) 
is, in general, no guarantee that “extra” contributing effects will be of a nature to be 
correctly assimilated into the theoretical formula, but, for this effect, it happens that 
this is at least partially the case. 
Another extra effect to consider is the p-electron repulsion energy. Because the 
energies of the HOMO and LUMO are both associated with the neutral molecule, their 
energies fail to reflect the decrease or increase of p-electron repulsion energy resulting 
from loss or gain of a p electron. Here again, however, we expect the magnitude of the 
effect to be larger for smaller molecules, where the change in p densities is greatest, 
and also where Em and Em+1 deviate most from a. Thus, as before, this extra effect 
will not necessarily upset the linear relation expected from simpler considerations. 
In view of the crudity of the HMO method together with the fact that the empirical 
value of ß includes the effects of several extra processes, the ß values cited above 
for oxidation and reduction are considered to be in fairly good agreement, especially 
considering that the electron repulsion change operates to make oxidation easier and 
reduction harder.
278 Chapter 8 The Simple H¨uckel Method and Applications 
8-15 Orbital Energies and Ionization Energies 
Suppose that monochromatic light is beamed into a gaseous sample of a compound. 
If the light is of sufficient energy, electrons will be “knocked out” of the molecules. The 
kinetic energy of such a photoelectron will be equal to the kinetic energy of the incident 
photon (h.) minus the energy needed to remove the electron from the molecule, that is, 
the ionization energy. Measurement of the kinetic energies of photoelectrons emitted 
in this manner is known as “photo-electron spectroscopy” [12]. 
In measuring the kinetic energies of photoelectrons from, say, benzene, it is 
found that a large number are near a particular energy value, another large number 
of electrons are near a different value, and so on for several kinetic energy values. 
Because the photoelectrons tend to clump near several kinetic energy values, it follows 
that we are, in effect, measuring several ionization energies. It is reasonable 
to associate these with removal of electrons from different MOs of the molecule.13 
A correlation plot (Fig. 8-22) between HMO orbital energies and experimentally 
measured ionization energies for a number of alternant and nonalternant hydrocarbons 
has been produced by Brogli and Heilbronner [13]. Their best fit was achieved using 
a=-6.553±0.340eV,ß =-2.734±0.333 eV, where the limits define a range for 
the predicted ionization energy (IE) that will include the experimental value nine 
HMO values for IE (eV) 
Experimental IE (eV) 
13
7
7 13 
12 
12 
11 
11 
10 
10 
9 
9 
8 
8 
Figure 8-22 
 Experimental ionization energies for alternant and nonalternant hydrocarbons versus 
HMO orbital energies using a=-6.553eV,ß=-2.734 eV. 
13Note that each ionization energy is associated with removal of an electron from a different MO of the neutral 
molecule. This differs from the first, second, etc. ionization energies produced by successive ionization.
Section 8-16 p-Electron Energy and Aromaticity 279 
times out of ten (a 90% confidence level). There is a fair degree of scatter in the 
correlation plot. By making some additional refinements in the theory, Brogli and 
Heilbronner succeeded in substantially improving the correlation. We will describe 
these refinements in Section 8-18. 
In this section, we have dealt only with ionization energies assigned to removal of 
a p electron. Removal of electrons from s orbitals is also observed. If the impinging 
photons are of X-radiation frequencies, the inner shell electron ionization energies are 
seen. Quantum-chemical treatments of these ionization energies are possible using 
theoretical methods described later. 
8-16 p-Electron Energy and Aromaticity 
When propene is hydrogenated to form propane, the increase in heat content 
H. 298 
(called the heat of hydrogenation) is-30.1 kcal/mole. The heat of hydrogenation for 1- 
butene is -30.3 kcal/mole. In general, the heat of hydrogenation of an isolated double 
bond is about-30 kcal/mole. Asimilar constancy holds for the contribution of a double 
bond to the heat of formation of a molecule. Therefore, isolated double bonds fit easily 
into the usual chemical device of estimating the energy of a molecule by adding together 
contributions from the substituent parts. This additivity appears to be violated when 
double bonds are conjugated. Thus, the heat of hydrogenation for trans-1,3-butadiene 
is -57.1 kcal/mole compared with the value of -60.6 kcal/mole for a pair of butene 
double bonds. Since the hydrogenation product in each case is butane, the energy difference 
of 3.5 kcal/mole must be due to the greater stability of conjugated double bonds. 
There is a theoretical parallel to this. TheHMOenergy for butadiene is 4a+4.472ß. 
For a pair of isolated double bonds, we double the energy of ethylene to obtain 4a+4ß. 
Therefore, the HMO method indicates that the conjugated double bonds are stabilized 
by 0.472ß. 
Because the p electrons in butadiene are delocalized over all three C–C bonds, this 
0.472ß has often been referred to as delocalization energy. It was common practice for 
many years to equate this theoretical delocalization energy for a molecule to its experimentally 
measured “extra” stability (e.g., for butadiene, 0.472 |ß|=3.5 kcal/mole). 
In 1969, Dewar and co-workers [14, Chapter 5; 15; 16] demonstrated that conjugated 
double bonds may be successfully included in an additivity scheme. Hess and Schaad 
[17, 18] have considered this idea in the context of the HMO method. They distinguish 
between several kinds of C–C single and double bond energy as indicated in Table 8-2. 
Using these values of bond energy, the p energy of butadiene is calculated to be 2× 2.000ß + 0.4660ß + 4a = 4a + 4.4660ß compared with the HMO result of 4a + 4.472ß. The difference is 0.006ß, less than 0.002ß per p electron. Hess and Schaad 
show that this level of agreement holds for acyclic polyenes in general, even when there 
is much branching. Thus, it seems that conjugated double bonds in acyclic molecules 
fit into an additivity scheme after all. 
There is an important difference between the energy additivity scheme for conjugated 
systems described above and the familiar additivity scheme used for C–H, C–C, and 
isolated C–C bonds. In the latter cases the energy “contributed” by the bond is generally 
thought of as being the same as the energy of that bond in the molecule (at least in 
an averaged way–some care must be exercised with definitions), and, furthermore, the 
bond is thought to be fairly independent of the identity of the molecule. For instance,
280 Chapter 8 The Simple H¨uckel Method and Applications 
TABLE 8-2 
 p-Bond Types and Effective Binding Energies for Carbon–Carbon Double 
and Single Bonds 
Bond type Effective binding energy in units of ß 
2.0000 
(cis or trans) 2.0699 
2.0000 
2.1083 
2.1716 
(cis or trans) 0.4660 
(cis or trans) 0.4362 
(cis or trans) 0.4358 
aThe eight numbers associated with these bonds are not unique. They satisfy six simultaneous equations. 
Hence, any two of them may be given arbitrary values and the remaining six found by solving the simultaneous 
equations. Different arbitrary assignments lead to different sets of numbers, all equally valid and all giving 
identical p-electron energies. These numbers can only be applied when a bond has an unambiguous formal 
identity, which means we must deal with acyclic polyenes (even number of centers and electrons). 
a C–H bond in butane is very similar to one in heptane. But this is not the case for 
conjugated molecules. Inspection of Fig. 8-23 shows that a single C–C bond between 
(formal) double bonds varies significantly in bond order from molecule to molecule. 
Since the total p-electron energy depends on bond order [Eq. (8-58)], we can tell at once 
that the actual theoretical energy contribution due to such a “single” bond varies from 
2×0.4472=0.8944ß in butadiene to 1.088ß in decapentaene. However, examination 
of Fig. 8-23 reveals that, in going from molecule to molecule, as the “single” bonds 
increase in p-bond order, the “double” bonds decrease in order. Thus, adding an 
additional C–C–C group to a chain adds a constant amount of bond energy (about 
2.54ß) to the total p energy, but this energy increment contains contributions from 
bond order changes over the whole molecule. We have, then, an additive scheme for
Section 8-16 p-Electron Energy and Aromaticity 281 
Figure 8-23 
 HMO bond orders for butadiene, hexatriene, octatetraene, and decapentaene. 
a delocalized effect. For this reason, the bond energy contributions of Table 8-2 are 
called effective bond energies. 
We are nowin a position to consider the concept of aromaticity. The term “aromatic” 
originally referred to organic molecules having pleasant odors. Later it referred to a 
class of molecules having a high degree of unsaturation. Benzene was recognized as 
the parent compound for many such molecules, and the term aromatic has come to 
mean having chemical properties peculiar to benzene and some of its relatives. The 
chemical stability of these molecules, their relatively low heats of combustion or of 
hydrogenation, and their tendency to prefer substitution rather than addition (thereby 
preserving their p systems intact) distinguish these molecules from ordinary polyenes 
and have come to be called aromatic properties, or manifestations of aromaticity.14 
These properties suggest that the p electrons in aromatic systems are unusually low in 
energy, contributing to both the thermodynamic and the kinetic stability of the systems. 
We can test whether this is the case by calculating an expected p energy for benzene 
using the bond energies of Table 8-2 for an alternating single-double bond, or Kekul´e, 
structure. The result, 6a +7.61ß, is significantly less stable (by 0.39ß) than the HMO 
energy of 6a +8ß. We shall refer to this difference as the resonance energy (RE) of 
the system. 
RE =En(HMO)-En (from Table 8-2) (8-63) 
By this definition, a positive RE (in units of ß) corresponds to extra molecular stabilization. 
If we divide the RE by the number of p electrons, we obtain the RE per electron 
(REPE). Hence, the REPE for benzene is 0.065ß. Following Dewar we shall refer to a 
system having significantly positive REPE as aromatic, significantly negative REPE as 
antiaromatic, and negligible REPE as nonaromatic, or polyolefinic. (Recall that ß is a 
negative quantity. When REPE is tabulated in eV, it is conventional to use the absolute 
value of ß in eV so as to retain positive REPE for aromatic molecules.) 
Cyclic polyenes differ from acyclic polyenes in that many of them show significant 
values for REPE. Some results of Schaad and Hess [18] are reproduced in Fig. 8-24. 
14The definition of the term “aromatic” is not generally agreed upon. Various criteria based on structural, 
magnetic, or reactivity properties have been proposed. See Minkin et al. [19] for a general review, and Schleyer 
et al. [20] for consideration of ring-current-induced nmr chemical shifts as an aromaticity probe.
282 Chapter 8 The Simple H¨uckel Method and Applications 
It is evident that a strong correlation exists between REPE and the presence or absence 
of aromatic properties. 
The change from very aromatic to very antiaromatic nature as we go from a sixmembered 
to a four-membered system is striking, If we examine the HMO energy 
levels, it is not difficult to see why these monocycles are so different. All six p electrons 
in benzene occupy bonding MOs, whereas cyclobutadiene has two bonding and two 
nonbonding electrons (XII). The situation is optimal for benzene not only because 
all of its p electrons are bonding, but also because all of the bonding levels are fully 
occupied. Because all monocycles have a nondegenerate lowest level followed by 
higher-energy pairs of degenerate levels, it is not hard to describe the conditions that 
should produce maximum stability. There should be 4n+2 electrons (2 for the lowest 
level and 4 for each of the n higher bonding levels), and there should be more than 
4n centers (either 4n+1, 4n+2, or 4n+3) to force the n doubly degenerate, occupied 
levels to all be bonding. The stability of cyclopentadienyl anion (6 electrons, 5 centers) 
and cycloheptatrienyl cation (6 electrons, 7 centers) was correctly predicted from these 
simple considerations (XIII). Extensive research has gone on in efforts to find examples 
Figure 8-24 
 Resonance energy per electron for a number of molecules. (From Schaad and 
Hess [18].)
Section 8-16 p-Electron Energy and Aromaticity 283 
where n = 1, but results are often complicated by angle strain and deviation from 
planarity in the molecule.15 
EXAMPLE 8-7 What would the above approach predict for the relative stabilities 
of cyclic ions C5H+5 and C5H-5 ? 
SOLUTION 
 C5H-5 , the cyclopentadienyl anion shown in XIII, satisfies the 4n+2 rule and 
should be unusually stable. C5H+5 has only four p electrons, hence should be less stable. (It has 
two fewer bonding electrons than C5H-5 .)  
In a molecule containing both cyclic and acyclic parts, does the RE arise only from 
the cyclic part? For certain cases, the answer is yes. Dewar has recognized that side or 
connecting chains contribute nothing to the RE of a system when the chain in question 
is the same in all formal structures for the molecule. For example, stilbene can be 
written in four equivalent formal ways, as shown below. The linking chain is identical 
in all cases, and it should not contribute to the RE. Calculations confirm that the RE 
for stilbene is just double that for benzene. 
The intimate relation between aromaticity and the possibility for more than one 
equivalent formal structure for a molecule has long been recognized. These “mobile” 
bonds tend to favor equal bond lengths in contrast to the strong alternation characteristic 
of acyclic polyenes. In benzene, the extra stabilization may be viewed as resulting from 
the fact that all six bonds are identical and have a higher bond order (0.667) than the 
average of 0.634 for double and single acyclic bonds. In antiaromatic cyclobutadiene, 
the four identical bonds have a bond order of 0.5, which is significantly below the 
15Research on aromaticity has been extended to Mobius conjugated hydrocarbons, where aromatic behavior 
should be associated with 4n electrons (see Herges [21]), and also to clusters of boron or silicon atoms 
(see Ritter [22]), as well as metal atoms (see Chen et al. [23]).
284 Chapter 8 The Simple H¨uckel Method and Applications 
average for acyclics. The very different bond orders in these two molecules is fully 
consistent with the energy argument based on the 4n+2 rule described earlier.16 
To summarize, the p bonds in acyclic polyenes exhibit delocalization in the sense 
that bond orders are not transferable from one molecule to another, additivity in effective 
bond energies, immobility in formal bonds. Cyclic molecules do not exhibit additivity 
of effective energies or immobility of formal bonds. Their energy deviations from 
energy calculated assuming additivity and immobility are good indicators of kinetic 
and thermodynamic stability. Aromatic molecules possess extra stability because their 
p electrons17 are more bonding than those in acyclic polyenes. Antiaromatics are 
unstable because their p electrons are less bonding. 
From a consideration of experimental heats of atomization, Schaad and Hess have 
evaluated ß to be -1.4199 eV. The physical processes involved in dissociating a gasphase 
molecule into constituent atoms are quite different from those involved in adding 
or removing a p electron from a molecule in a solvent. Therefore, it is not surprising 
that the ß value obtained from heats of atomization differs substantially from values 
obtained from redox experiments. Indeed, it is this variability of ß as we compareHMO 
theory with different types of experiment that compensates for many of the oversights 
and simplifications of the approach. It is remarkable that, with but one such parameter, 
HMO theory does as well as it does. 
8-17 Extension to Heteroatomic Molecules 
The range of application of the HMO method could be greatly extended if atoms other 
than carbon could be treated. Consider pyridine as an example (XV). A p electron at 
a carbon atom contributes an energy a to Ep . The contribution due to a p electron at 
nitrogen is presumably something different. Let us take it to be a = a +hß, where 
h is a parameter that will be fixed by fitting theoretical results to experiment. If the 
p electron is attracted more strongly to nitrogen than to carbon, h will be a positive 
16It has been noted that bond length equalization associated with bond mobility results in it energy lowering 
when the C–C–C angles are near 120.. When the angle is very different from this, p energies are higher than 
expected. This has led to suggestions that “strain energy” may be an important factor in aromaticity. Because 
of the present lack of a quantum mechanical quantity equivalent to strain energy, and because the HMO method 
may include effects of s electrons in an implicit but poorly understood way, it is very hard to know whether such 
suggestions are at variance with other statements or are simply equivalent to them but stated from a different 
viewpoint. 
17It is not necessarily true that all the “extra” stability of aromatic molecules is attributable to p-electron effects; 
s-electron energies also depend on bond lengths and bond angles. Hence, we may be seeing, once again, a situation 
where the p-electron treatment includes other effects implicitly. Schaad and Hess [18] indicate that s energies 
and p energies are indeed simply related over the bond-length range of interest.
Section 8-17 Extension to Heteroatomic Molecules 285 
number. In a similar spirit, we will take the energy of a p electron in a C–N bond to be 
ß =kß and evaluate k empirically. Not surprisingly, the values of h and k appropriate 
for various heteroatoms depend somewhat on which molecules and properties are used 
in the evaluation procedure. A set of values compiled and critically discussed by 
Streitwieser [7] is given in Table 8-3. Other sets have been published.18 
The dots over each symbol indicate the number of p electrons contributed by the 
atom. In pyridine, the formal bond diagram indicates a six p-electron system, implying 
that the nitrogen atom contributes one p electron. We also can argue that, of the five 
TABLE 8-3 
 Parameters for Heteroatoms in the H¨uckel Methoda 
Heteroatom h Heteroatomic bond k 
N. 0.5 C.—. .N¨ 1.0 
N¨ 1.5 C—N¨ 0.8 
+N
2.0 
. O 1.0 C—— 
O. 1.0 
O¨ 2.0 C—O¨ 0.8 
+O
2.5 N—O¨ 0.7 
¨F 3.0 C—¨F 0.7 
C¨l 2.0 C—C¨l 0.4 
B¨r 1.5 C—B¨r 0.3 
Sb 0.0 C—S 0.8 
S 0.0 C—S 0.8 
S—S 1.0 
Methyl (inductive C. a—Me)-0.5 hCa =-0.5 none — 
Methyl (heteroatom C. a—M¨e)0.2 hMe = 0.2 Ca—Me 0.7 
Methyl (conjugative C. a—C—¨ H3) hCa =-0.1 Ca—C 0.8 
hc =-0.1 C—H3 3.0 
hH3 =-0.5 
aConsistent with the philosophy of this approach is a distinction between single, double, and intermediateC—C 
bonds. Streitwieser recommends kC-C =0.9, kC... -C =1.0, kC=C =1.1. 
bSulfur is treated as a pair of AOs with a total of two p electrons, i.e., a sulfur in an aromatic ring is formally 
treated as two adjacent atoms S and S with the indicated parameters. 
18See McGlynn et al. [24, p. 87].
286 Chapter 8 The Simple H¨uckel Method and Applications 
valence electrons of nitrogen, two are involved in s covalent bonds with neighboring 
carbons, two more are in a s lone pair, leaving one for the p system. Therefore, the atom 
parameter to use for this molecule is h=0.5. The pyridine ring, like benzene, admits two 
equivalent structural formulas, and so the C–N bonds should be intermediate between 
double and single, symbolized C.–..N. in Table 8-3. Since k=1.0 in this case, ß=ß, and 
pyridine will have an HMO determinant differing from the benzene determinant only 
in the diagonal position corresponding to the nitrogen atom–the 1, 1 position according 
to our (arbitrary) numbering scheme. For this position, instead of x, we will have 
x =(a -E)/ß =(a +0.5ß -E)/ß =(a -E)/ß +0.5ß/ß =x +0.5 (8-64) 
The pyrrole molecule has a nitrogen atom of the type N¨ (XVI). Since three valence 
electrons of nitrogen are in covalent s bonds, two remain for inclusion in the p system. 
Therefore, pyrrole has a total of six p electrons. The unique structural formula indicates 
that theC–N¨ bond is formally single, and k=0.8, h=1.5 are the appropriate parameters 
here. Also, the carbon–carbon bonds are nowformally single or double. If we choose to 
distinguish among these bonds using the parameters in note a of Table 8-3, the resulting 
HMO determinant is 
 
x +1.5 0.8 0 0 0.8 
0.8 x 1.1 0 0 
0 1.1 x 0.9 0 
0 0 0.9 x 1.1 
0.8 0 0 1.1 x 
 
The methyl group can also be incorporated into the HMO method. Several 
approaches have been suggested. One is simply to modify the coulomb integral a for 
the carbon to which the methyl group is attached. A methyl group is thought to release 
sigma electrons to the rest of the molecule as compared to a substituent hydrogen. This 
suggests that an atom having a methyl group attached to it will be a bit electron rich 
and hence will be less attractive to p electrons. Use of a negative h parameter for this 
carbon is appropriate. This method is called the inductive model. Use of the inductive 
model does not add any new centers or any more p electrons to the conjugated system 
to which the methyl group is attached: The carbon to which the methyl is bonded is 
merely treated as a less attractive atom. A second approach is to treat the methyl group 
itself as a heteroatom. As we shall see shortly, the methyl group has two electrons that
Section 8-18 Self-Consistent Variations of a and ß 287 
Figure 8-25 
 Three MO symmetry solutions for a methyl group attached to a benzene ring. 
can participate (to a slight extent) in the p system, and so use of this heteroatom model 
adds one more center and two more p electrons for each methyl group included in this 
way. A third approach is the conjugative model. Because the methyl group has local 
threefold symmetry, one of the s MOs for the methyl group resembles the p MOs 
of cyclopropenyl in symmetry characteristics. The three types of symmetry solutions 
are given in Fig. 8-25 for a methyl group on a benzene ring (compare with Fig. 8-7). 
Notice that the MO at the right of the figure has its phases arranged so that it has 
nonzero overlap with a p AO on the benzene ring. This means that the two electrons in 
this “methyl group MO” can participate in the p system of the molecule. To emulate 
this picture in our HMO determinant, we must add two p electrons and two more 
centers to our system (one for C and one for H3) and find h and k values for the two 
new centers and bonds. The inductive effect of the methyl group on the neighboring 
ring carbon is often included in this model. In most cases, it is probably safe to say 
that the conjugative model is superior to the heteroatom model, which is, in turn, better 
than the inductive model, but no extensive critical comparison of these three models 
has been made. Parameters for all three approaches are included in Table 8-3. (Ca is 
the ring carbon.) 
8-18 Self-Consistent Variations of a and ß 
Efforts have been made to improve the HMO method by taking account of molecular 
p charge distribution. Suppose that we carry out anHMOcalculation on a nonalternant 
molecule and find an electron density of 1.2 at one carbon and 0.8 at another. It is 
reasonable to argue that a p electron at the latter carbon is more strongly bound because 
it experiences less repulsion from other p electrons there. We can try to account for 
this by making a at that atom more negative. Thus, we could take 
ai =ai +.(1-qi)ß (8-65) 
where qi is the p-electron density at atom i and . is a parameter (assumed positive) 
to be fixed empirically. If q <1, then a is more negative than a. If q >1, a is 
less negative. Having now modified a (using a trial value for .), we must set up 
our new HMO determinant and solve it again. This yields new MOs, new values 
of qi , and therefore new values of a. We repeat this process over and over until 
electron densities remain essentially unchanged for two successive iterations. At this 
point, the electron densities leading to the HMO determinant are the same as those 
produced by the determinant, and the solution is said to be self-consistent with respect 
to electron densities. This procedure, often referred to as the “. technique,” discourages 
extreme deviations of electronic densities from the “norm” of unity at each carbon and 
thereby helps to compensate for the lack of explicit inclusion of p electron repulsion
288 Chapter 8 The Simple H¨uckel Method and Applications 
in the HMO method. Streitwieser’s calculations have led him to favor a value for 
. of 1.4. 
A similar idea has been applied to variations of the bond integral ß. Suppose that 
we carry out an HMO calculation and find a p-bond order of 0.5 in one bond and 
0.9 in another. We expect that the latter bond is in fact shorter than the former. We 
could roughly predict how much shorter it is by using the bond-order–bond-length 
relation described earlier. It is reasonable to modify ß in these bonds on the basis 
of predicted length differences, set up a new HMO determinant, solve again, find 
new bond orders, and iterate until self-consistency is achieved with respect to bond 
orders. 
These modifications to the simple HMO method improve predictions of some properties 
but not others. For example, Brogli and Heilbronner [13] have found that orbital 
energy correlation with ionization energy, determined by photoelectron spectroscopy, 
is significantly improved through inclusion of the effects of bond length variations in 
the neutral molecule and the cation. This improved correlation, shown in Fig. 8-26, 
showed no additional improvement upon subsequent variation of a as a function of electron 
density. This is reasonable, since ß variation affects primarily bond order, hence 
MOenergy, and that is the property measured by photoelectron spectroscopy. Variation 
of a shifts charge from atom to atom, but has smaller energy effects. On the other 
hand, p-electron contributions to dipole moments, calculated from electronic excess 
or deficiency at each center, are very sensitive to variation of a, and quite insensitive 
13 
12 
11 
10 
HMO values for IE (eV) 
9
8
7 
7 8 9 10 
Experimental IE (eV) 
11 12 13 
Figure 8-26 
 Experimental ionization energy for alternant and nonalternant hydrocarbons versus 
orbital energy using a modified HMO technique which includes provisions for bond length variation 
in molecule and cation. (Compare with Fig. 8-22.)
Section 8-19 HMO Reaction Indices 289 
to variation of ß. Again, this is reasonable because dipole moments are sensitive to 
electronic distribution rather than MO energy. 
8-19 HMO Reaction Indices 
In this section, we discuss some applications of the HMO method to reactivities of 
conjugated molecules. The reactions of conjugated molecules that have received most 
of this theoretical treatment are: 
1. electrophilic aromatic substitution (XVII) 
2. nucleophilic aromatic substitution (XVIII) 
3. radical addition (XIX) 
For any of these reactions, we imagine there to be a path of least energy connecting 
reactants with products. For the two distinct reaction positions, 1 and 2 on naphthalene, 
the activation energies  may differ, as indicated in Fig. 8-27. 
The problem is somehow to relate the differences in  (inferred from relative rate 
data) to a number based on quantum chemical calculations. To do this in a sensible way 
requires that we have some idea of the detailed way in which the reaction proceeds– 
we have to know what the reaction coordinate is. In some cases, this is fairly well 
known. For electrophilic aromatic substitution reactions, evidence suggests that a
290 Chapter 8 The Simple H¨uckel Method and Applications 
Figure 8-27 
 Generalized energy versus reaction coordinate for reaction at two positions in 
naphthalene. 
positive electrophile (e.g., Cl+) approaches the substrate (say, naphthalene). As it 
draws closer, it causes a significant polarization of the p-electron charge distribution, 
drawing it toward the site of attack. Ultimately, it forms a partial bond with the carbon. 
At this stage, the carbon has already begun to loosen its bond to hydrogen, but it is at 
least partially bonded to four atoms (XX). This means that the ability of the carbon to 
participate in the aromatic system is temporarily hampered. At this point, the system is 
at or near the transition state. Thereafter, as H+ leaves, the potential energy decreases 
and Cl moves into the molecular plane. A similar detailed mechanism is thought to 
apply for nucleophilic reactions except that the attacking group is negative. 
For attack by a neutral radical, electrostatic attraction and charge polarization should 
not be significant factors. The radical bonds to the site of attack to produce a more-orless 
tetrahedrally bonded carbon. This again leads to an interruption of the p system, 
but now it is not temporary as it was in the substitution reactions. 
Based on these simple pictures, a number of MO quantities, often referred to as 
reaction indices, have been proposed as indicators of preferred sites for reaction. It is 
useful to divide these into two categories–those purporting to relate to early stages of 
the reaction, and those specifically related to the intermediate stage. 
Perhaps the most obvious reaction index to use for the earliest stages of electrophilic 
or nucleophilic reactions is the p-electron density. If Cl+ is attracted to p charge, 
it should be attracted most to those sites where p density is greatest. (Such an ion 
should be attracted to sites having excessive s charge density also, but our basic HMO 
assumptions ignore any variations in s density.) For an alternant hydrocarbon like 
naphthalene, all p densities are unity, so this index is of no use. For nonalternant 
molecules, however, it can be quite helpful. Azulene has varying HMO p densities 
(XXI). (More sophisticated calculations described in future chapters are in qualitative
Section 8-19 HMO Reaction Indices 291 
agreement with these p-electron density variations.) Experimentally, it is found that 
electrophilic substitution by Cl occurs almost entirely at position 1 (or 3). Nucleophilic 
substitution by CH3 (from CH3 Li) occurs at the position of least p-electron density, 
namely 4 (or 8). 
The charge density index refers to the nature of the molecule before allowance is made 
for perturbing effects due to the approaching reactant. Such a method is often called 
a “first-order” method, a terminology that is discussed more fully in Chapter 12. For 
alternant molecules, it is necessary to proceed to a high-order method, one that reflects 
the ease with which molecular charge is drawn toward some atom, or pushed away from 
it, as approach by a charged chemical reactant makes that atom more or less attractive for 
electrons. An index which measures this is called atom self-polarizability, symbolized 
pr,r . The formulas for this and related polarizabilities are derived in Chapter 12. For 
now, we simply note that the formula is 
pr,r = 
.qr 
.ar =4 
occ 

j 
unocc 

k 
c2
rj c2
rk 
Ej -Ek 
(8-66) 
A larger absolute value of prr means that a larger change in p density qr occurs 
as a result of making atom r more or less attractive for electrons. (pr,r is negative 
since Ej - Ek is negative. This makes physical sense because it means that if dar 
is negative, making atom r more attractive, dqr is positive, indicating that charge 
accumulates there.) Since the most polarizable site should most easily accommodate 
either a positive- or a negative-approaching reactant, this index should apply 
for both electrophilic and nucleophilic reactions. For naphthalene, the values are 
p11=-0.433/ |ß|,p22=-0.405/ |ß|. This agrees with the experimentally observed 
fact that the 1 position of naphthalene is more reactive for both types of reaction. 
Examination of Eq. (8-66) indicates that the MOs near the energy gap between 
filled and empty MOs will tend to contribute most heavily to pr,r because, for these, 
Ej - Ek is smallest. For this reason, the highest occupied and lowest unfilled MO 
(HOMO and LUMO) are often the determining factor in relative values of prr . Fukui19 
named these the frontier orbitals and suggested that electrophilic substitution would 
occur preferentially at the site where the HOMO had the largest squared coefficient. In 
nucleophilic substitution, the approaching reagent seeks to donate electronic charge to 
the substrate, so here the largest squared coefficient for theLUMOshould determine the 
preferred site. For even alternants like naphthalene, the pairing theorem forces these two 
MOs to have their absolute maxima at the same atom. The HOMO–LUMO coefficients 
for naphthalene are 0.425 and 0.263 for atoms 1 and 2, respectively, in accord with 
our expectations. For the nonalternant molecule azulene, discussed above, the largest 
HOMO coefficient occurs at atoms 1 and 3, which have already been mentioned to 
be the preferred sites for electrophilic attack. The largest LUMO coefficient occurs at 
19See Fujimoto and Fukui [25].
292 Chapter 8 The Simple H¨uckel Method and Applications 
atom 6, with atoms 4 and 8 having the second-largest value (see Appendix 6). Atoms 
4 and 8 are the preferred sites for nucleophilic attack. Here, then, is a case where 
the charge density and frontier MO indices are not in agreement. The results suggest 
that, when significant p-density variations occur, this factor should be favored over 
higher-order indices. However, the whole approach is so crude that no ironclad rule 
can be formulated. 
For radical attack, some other index should be used, for we do not expect electrostatic 
or polarization effects to be important in such reactions. An index called the 
free valence20 has been proposed for free radical reactions. One assumes that the free 
radical begins bonding to a carbon atom in early stages of the reaction and that the ease 
with which this occurs depends on how much residual bonding capacity the carbon has 
after accounting for its regular p bonds. Thus, free valence is taken to be the difference 
between the maximum p bonding a carbon atom is capable of and the amount of p 
bonding it actually exhibits in the unreacted substrate molecule. The extent of p bonding 
is taken as the sum of all the orders of p bonds involving the atom in question. 
A common choice of reference for a maximally bonded carbon is the central atom in 
trimethylenemethane in its planar conformation (which is not the most stable). (XXII). 
Each bond in this neutral system has a p-bond order of 1/v3, and so the total p-bond 
order associated with the central carbon is v3.21 Using this as reference, the free 
valence for some atom r in any unsaturated hydrocarbon is defined as 
Fr =v3 -
neighbors of r 

z 
prs (8-67) 
A common way of representing the situation schematically is indicated in (XXIII) for 
butadiene. Bond orders are indicated on the bonds and free valences by arrows. It is 
clear that butadiene has a good deal more “residual bonding capacity” on its terminal 
atoms, and this is consistent with the fact that free radical attack on butadiene occurs 
predominantly on the end atoms. Other examples of correlation between free valence 
and rate of free radical addition have been reported.22 A plot of rate data for methyl 
20See Coulson [26]. 
21Sometimes the three sigma bonds are included in this calculation, giving 3+v3. This has no effect on the 
question of relative values of Fr . 
22See Streitwieser [7] and Salem [1].
Section 8-19 HMO Reaction Indices 293 
Figure 8-28 
 Rate data for methyl radical addition are plotted against the maximum free valence 
found in each molecule. The original “methyl affinity” (Levy and Szwarc [27]) has been multiplied 
by 6/m, where m is the number of sites having maximum free valence. (The asterisks in the figure 
identify these sites.) 
radical addition to conjugated molecules versus the largest free valence of the molecule 
is shown in Fig. 8-28. (We assume that the kinetics is dominated by the atom(s) having 
the maximum free valence.) 
The indices described above are most appropriate for indicating the relative ease of 
reaction in the early stages. By the time the reactants have reached the transition state, 
the substrate is quite far from its starting condition, so charge densities, polarizabilities, 
free valences calculated from the wavefunction of the unperturbed molecule may no
294 Chapter 8 The Simple H¨uckel Method and Applications 
longer be very appropriate. If the energy curves being compared through our indices 
behave in the simple manner described in Fig. 8-27, so that the higher-energy curve in 
early stages is also the higher-energy curve in the region of the transition state, such 
indices can be useful. Also, such simple behavior is very likely to occur when we 
compare a single type of reaction down a series of molecules of similar type, as in 
Fig. 8-28 (see also Problem 8-25). Experience, however, has indicated that indices 
more closely linked to the nature of the transition state are more generally reliable. We 
now describe one such reactivity index. 
We mentioned earlier that addition and substitution reactions are expected to interrupt 
the p system at the site of attack. For substitution reactions, this interruption is only 
temporary and is presumably most severe in the transition state. For addition reactions, 
it is permanent. The localization energy is defined as the p energy lost in this process 
of interrupting the p system. 
As an example, let us return to the naphthalene molecule. The situations resulting 
from interruption of the p system by neutral radical attack at positions 1 and 2 are 
illustrated in (XXIV). The remaining unsaturated fragment is, in each case, a neutral 
radical. [If attack were by a negative ion (nucleophilic), the fragment would be topologically 
the same but would be negatively charged. Likewise, attack by a positive 
electrophilic reagent leads to a positively charged fragment.] No matter where attack 
occurs, our p energy must go from 10a+···ß to 9a+···ß. This decrease by a is 
thus not expected to differ from case to case and hence is ignored in our localization 
energy calculation. The decrease in p energy is thus 2.299 |ß| for attack at position 1 
and 2.480 |ß| for position 2. These localization energies indicate that attack at position 
1 should be favored since the energy cost is smaller there, and this is in accord with 
observation. (Localization energies are symbolized L·r , L+r
, L-r
depending, respectively, 
on whether attack is by a free radical, an electrophilic cation, or a nucleophilic 
anion.) 
It is interesting to compare the various indices we have discussed for a single 
molecule to see how well they agree. Data for azulene are collected in Table 8-4. 
Experimentally, azulene is known to preferentially undergo electrophilic substitution 
at positions 1 (and 3), nucleophilic substitution at positions 4 (and 8), and radical addi-
Section 8-20 Conclusions 295 
TABLE 8-4 
 Reactivity Indices for Azulene 
r HOMO LUMO 
(atom no) qr -|ß|pr,r c2
r c2
r Fr Lr+(|ß|) Lr•(|ß|) Lr-(|ß|) 
1 1.173 0.425 0.2946 0.0040 0.480 1.924 2.262 2.600 
2 1.047 0.419 0.0000 0.0997 0.420 2.362 2.362 2.362 
4 0.855 0.438 0.0256 0.2208 0.482 2.551 2.240 1.929 
5 0.986 0.429 0.1126 0.0104 0.429 2.341 2.341 2.341 
6 0.870 0.424 0.0000 0.2610 0.454 2.730 2.359 1.988 
tion in positions 1 (and 3). Consider first electrophilic reaction. Examining the table 
indicates that position 1 is heavily favored by qr , HOMO distribution, and L+r
. The 
only other index relevant for this process, prr , favors position 4. For nucleophilic 
reaction, qr,prr,L-r
all favor position 4. The LUMO index favors position 6, but not 
decisively over position 4. For radical addition, LUMO favors position 6, whereas Fr 
and L·r favor position 4. The latter two indices, however, favor 4 over 1 by only a 
slight margin. Thus, for a nonalternant molecule like azulene, these numbers are not 
completely trustworthy and must be interpreted with caution. One difference between 
a molecule like anthracene and one like azulene is that all the C–C–C angles in the 
former molecule are similar (~120.) whereas in azulene they differ. One might anticipate 
that the smaller angles in the five-membered subunit, being already closer to the 
tetrahedral angle characteristic of saturated carbons, would allow easier substitution or 
addition than would be the case in the seven-membered subunit. This factor is ignored 
in our calculations of L·r and might easily tip the balance to favor position 1 over position 
4 for radical attack since L·1 and L·4 are so close in value. In short, these HMO 
reactivity index approaches are once again techniques that ignore many aspects of the 
physical processes being followed. It seems likely that many of these will cancel out of 
comparisons among similar molecules, but dissimilarities between or within molecules 
(most often encountered in nonalternant systems) will cause such cancellations to be 
less complete. For more discussion of these and otherHMOreaction indices, the reader 
is referred to more specialized discussions.23 Application of some of these indices to 
carcinogenicities of polycyclic aromatic hydrocarbons has been reviewed.24 
8-20 Conclusions 
In this chapter, we have seen how certain basic features of molecular structure manifest 
themselves in molecular properties. The connectedness, or s bond network, defines 
the bond positions where p electrons can congregate to lower the energy of the system. 
23See Streitwieser [7], Salem [1], Dewar [14], and Klopman [28]. 
24See Lowe and Silverman [29]. For a dialogue for for nonspecialists by the same authors, see [30].
296 Chapter 8 The Simple H¨uckel Method and Applications 
The extent of congregation is a useful measure of bond length and also is directly 
contributory to the total energy of the system. 
The symmetry restrictions for MOs have been emphasized. The fairly successful 
correlation of HMO results with certain experimental measurements suggests that the 
method effectively accounts for the controlling factors in some molecular properties. 
However, we have omitted discussion of other properties that correlate only poorly 
with HMO theory. Notable in this regard are spectral energies. The natural idea of 
relating a spectral transition energy to a difference between orbital energies has not 
been very successful in HMO theory, except when one restricts attention to a particular 
band in a series of related molecules. This is at least partly due to improper handling of 
electron exchange symmetry in the HMO method leading to an inability to distinguish 
between different states of a given configuration. 
The HMO method can be an extremely instructive way to approach a problem since 
it can describe the manner in which certain important factors are operating. Also, 
for some situations, its predictive power is rather good. However, the limitation to 
conjugated systems, the reliance on an increasing number of parameters as extensions 
are made, the inability to conform to some kinds of experimental measurement, and 
the conceptual slipperiness of the quantities used in the method have all contributed to 
a decline of interest in further development of HMO theory. More powerful computers 
have made it possible for more complicated but better defined methods to be used. 
8-20.A Problems 
8-1. Show that, if Hˆ = Hˆ1 + Hˆ2 + Hˆ3, and Hˆifj (i) = Ejfj (i), then .prod = 
f1(1)f2(2)f3(3) and .det = |f1(1)f2(2)f3(3)| are both eigenfunctions of Hˆ 
and have the same eigenvalue. Show also that . =(1/v2)[f1(1)f2(2)f3(3)+ 
f1(1)f2(2)f4(3)] is an eigenfunction of Hˆ if and only if E3 =E4. 
8-2. Set up the HMO determinant for each of the following molecules: 
(a) (b) 
(c) (d) 
(e) 
8-3. Set up and solve the H¨uckel determinantal equation for 2-allylmethyl (also called 
trimethylenemethane) (XXV). Display the orbital energy levels and indicate the
Section 8-20 Conclusions 297 
electron configuration for the neutral ground state. Calculate En. Find the coefficients 
for all MOs. (Be sure that degenerate MOs are orthogonal.) Calculate the 
charge densities and bond orders. 
8-4. Suppose that two MOs of a molecule are given by the formulas 
f1 = (1/v3).1 +(1/v3).2 +(1/v3).3, 
f2 = (1/v3).3 +(1/v3).4 +(1/v3).5, 
where the .’s areAOs which are assumed to be orthonormal. By inspection, what 
is the overlap between these MOs? 
8-5. Given the following two degenerate MOs for cyclobutadiene (assumed square 
planar): 
Use the Schmidt procedure to obtain a normalized MO that has the same energy 
as f2 but is orthogonal to f1. 
8-6. For cyclooctatetraene (in its idealized, but incorrect, planar, octahedral form), see 
if you can answer the following without reference to tabulations: 
a) What is the HMO energy of the highest energy p MO? 
b) Sketch this MO (from above the molecule) showing signs and magnitudes 
(actual numbers) of the MO coefficients. 
8-7. a) What is the H¨uckel orbital energy for the following MO? (Assume that all 
centers are carbons.) [Hint: Use Eq. (8-56).] 
1 2 3 
4
5 
c 1 c 2 c 3 c 4 c 5 
1 0 
3 
13
13
3 2 
b) Calculate the contributions to bond orders due to one electron in this MO. 
c) Calculate the H¨uckel energy of the following MO. (Figure out the value of a 
if you need to use it.) 
8-8. Without performing an HMO calculation, sketch the MOs for the pentadienyl 
radical. Use the particle-in-a-box solutions and the pairing theorem as a guide.
298 Chapter 8 The Simple H¨uckel Method and Applications 
8-9. Which, if any, of the following systems would you expect to exhibit Jahn–Teller 
distortion? Indicate your thinking. (You should be able to answer without 
reference to HMO data tables.) (a) Benzyl radical, C7H7, (b) Cyclopentadienyl 
radical, C5H5. (c) Cyclobutadienyl radical cation, C4H+4 . (d) Benzene 
cation, C6H+6 . 
8-10. The bond lengths given in Table P8-10 have been reported for ovalene (XXVI). 
Using a library source or a computer program, obtain HMO bond orders for 
ovalene and calculate theoretical bond lengths using a relation from Section 8-12. 
Make a comparison plot for ovalene of the type shown in Fig. 8-16. (See caption 
of Fig. 8-16 for values of k and s.) 
TABLE P8-10 
 
Bond Length (Å) Bond Length (Å) 
1–2 1.445 1-10 1.401 
2–3 1.354 6-7 1.419 
3–4 1.432 4-25 1.411 
4–5 1.429 24-25 1.366 
5–6 1.429 5-22 1.424 
6–1 1.425 7-20 1.435 
8-11. Horrocks et al. [31] report experimental bond lengths for quinoline complexed 
to nickel. They display a comparison plot that uses theoretical data from an 
MO method more refined than the simple HMO method. Using the appropriate 
heteronuclear parameters, perform on the computer an HMO calculation for 
quinoline. Calculate theoretical C–C bond distances and compare them with the 
experimental and theoretical data of Horrocks et al. 
8-12. When the molecule CH2=CH–CH=O absorbs light of a certain frequency, a 
lone-pair electron on oxygen (called “n” for nonbonding) is promoted to the 
lowest empty p MO of the molecule (called p*; hence, an n-.p* transition). 
Assuming that the p MOs of this molecule are identical to those in butadiene, 
which C–C bondwould you expect to become longer and which shorter as a result 
of this transition? Calculate the expected bond length changes using butadiene 
data, using either set of k, s values in the caption of Fig. 8-16. (Observed: 

CH2–CH~=
0.06Å,
CH–CH~=
-0.04Å.)
Section 8-20 Conclusions 299 
8-13. ESR coupling constants are shown in Table P8-13 for six hydrocarbon anion radicals. 
Use HMO tabulations in the literature (or a computer) to obtain p-electron 
MO coefficients for these systems. Construct a plot of coupling constant aHµ 
versus c2
µi, where i is the MO containing the unpaired electron (aHµ values are 
in gauss). The numbered positions in Table P8-13 refer to hydrogen atoms. 
TABLE P8-13 
 
8-14. Polarographic half-wave potentials for oxidation and reduction of aromatic 
hydrocarbons are given in Table P8-14. 
a) Make separate plots of these data against energy (in units of ß) of the highest 
occupied and lowest empty MO respectively. (Use tabulations or a computer 
program.) 
b) Now plot reduction versus oxidation half-wave potential for this series. 
Explain adherence to or deviation from linearity. 
TABLE P8-14 
 
Compound Structure 
Reduction half-wave 
potential in 
2-methyoxyethanol 
(V) 
Oxidation 
half-wave 
potential in 
acetonitrile (V) 
Tetracene 1.135 0.54 
1,2-Benzpyrene 1.36 0.76 
Anthracene 1.46 0.84 
(Continued)
300 Chapter 8 The Simple H¨uckel Method and Applications 
TABLE P8-14 
 (Continued) 
Compound Structure 
Reduction half-wave 
potential in 
2-methyoxyethanol 
(V) 
Oxidation 
half-wave 
potential in 
acetonitrile (V) 
Pyrene 1.61 0.86 
1,2-Benzanthracene 1.53 0.92 
1,2,5,6-Dibenzanthracene 1.545 1.00 
Phenanthrene 1.935 1.23 
Fluoranthene 1.345 1.18 
Naphthalene 1.98 1.31 
Biphenyl 2.075 1.48
Section 8-20 Conclusions 301 
8-15. Use tabulated or computer-generated HMO data for neutral azulene (XXVII) to 
answer the following questions (Tabulated data may be found in Appendix 6.): 
a) What values would you expect for oxidation and reduction half-wave potentials 
for this molecule under conditions described in Problem 8-14? 
b) If an electron were removed from the highest occupied MO to produce an 
ion, which bonds would you expect to lengthen, which to shorten? 
8-16. Use the effective bond energies of Table 8-2 to calculate the expected p energy 
for (XXVIII). Compare this with the HMO energy of 18a+21.906ß. 
8-17. Obtain the HMO data for naphthalene (XXIX) and perylene (XXX): 
a) For each molecule, compare E to the energy predicted by use of Table 8-2. 
Categorize each molecule as aromatic, nonaromatic or antiaromatic. 
b) Compare the RE for these two molecules. Does the central ring in perylene 
appear to be contributing? 
c) Draw formal bond structures for perylene. What can you conclude about the 
two bonds connecting naphthalene units in perylene? 
d) Use HMO bond orders to calculate a predicted length for these two bonds. 
How do they compare with the observed 1.471Å value? Is this observed 
length consistent with your conclusion of part (c)? 
8-18. The third, fourth, and fifth molecules in Fig. 8-24 have dipole moments. Assuming 
that the individual rings attract or repel charge in accordance with our expectations 
from the 4n+2 rule, predict the direction of the p-electronic contribution 
to the dipoles. (Dipoles are defined by chemists as being directed from positive
302 Chapter 8 The Simple H¨uckel Method and Applications 
toward negative ends of electric dipoles. Physicists use the opposite convention.) 
Howwould you expect the p-electron densities to vary in these molecules? 
Compare your expectations with tabulated densities. 
8-19. How many p electrons are there in each of the following neutral molecules? 
Assume that each one is planar. 
8-20. Use the parameters in Table 8-3 to construct the HMO determinant for molecule 
(XXXI). Use the conjugative model for the methyl group. 
8-21. Substitution of a nitrogen for a carbon in benzene changes the HMO energy 
levels, but not drastically. Hence the stability of the six p-electron molecule 
pyridine can still be rationalized by the 4n+2 rule. Which member of each of 
the following pairs of molecules would you expect to be stable on the basis of 
such arguments 
8-22. It is observed that many even alternant hydrocarbons tend to undergo nucleophilic 
substitution, electrophilic substitution, and radical addition at the same site(s). 
Rationalize this behavior in terms of the following indices (where appropriate): 
qr , HOMO, LUMO, L·r .
Section 8-20 Conclusions 303 
8-23. Can prr be a successful index for nucleophilic and electrophilic substitution if 
these are observed to occur at different sites? 
8-24. For the methylene cyclopropene system C4H4 (see Appendix 6 for HMO data 
and atomic numbering scheme): 
a) Calculate the free valences for the neutral molecule. 
b) Decide, using three appropriate reactivity indices, which site is most susceptible 
to electrophilic attack. 
c) Decide which protons would lead to hyperfine splitting of the ESR spectrum 
of the radical anion, according to the simple H¨uckel approach. 
d) Decide whether the second-lowest MO is net bonding or net antibonding. 
Why? 
8-25. Published data for free radical (CCI·3) addition to hydrocarbons are shown in 
Table P8-25; see Kooyman and Farenhorst [32]. Assuming planarity in each 
case, use standardHMOtabulations or computer programs to obtainFr values for 
each molecule. Choose the largest value for each molecule and plot this against 
the log of the modified rate constant. In each case, modify the rate constant by 
dividing k by the number of sites on the molecule having the maximum Fr value. 
TABLE P8-24 
 
Molecule Structure Rate constant 
Benzene <10-3 (use this limit) 
Biphenyl 2.7 × 10-3 
Triphenylene <4×10-2 (use this limit) 
Phenanthrene 1.6 × 10-2 
Naphthalene 4 × 10-2 
(Continued)
304 Chapter 8 The Simple H¨uckel Method and Applications 
TABLE P8-24 
 (Continued) 
Molecule Structure Rate constant 
Chrysene 6.7 × 10-2 
Pyrene 1.3 
Stilbene 1.0 
1,2,5,6-Dibenzanthracene 3.7 
Styrene 12.5 
Anthracene 22 
Benzanthracene 30 
3,4-Benzopyrene 70 
Naphthacene 102
Section 8-20 Conclusions 305 
(If a molecule has Fr values differing by 0.002 or less, treat them as equal.) If 
any data points deviate greatly from the general trend, try to give an explanation. 
8-26. Which ring of the fourth molecule in Fig. 8-24would you predict an electrophilic 
reagent would be more likely to attack, and why? 
8-27.* Which of the following neutral unsaturated planar hydrocarbons should experience 
greater changes in bond order when a p electron is added to form the 
anion, and why? 
8-28.* Consider an ESR experiment on the odd-alternant radical shown. Identify the 
site(s) that have attached hydrogen atoms that should give 
a) the largest coupling constant. 
b) the second-largest coupling constant. 
c) no coupling constant. 
8-29.* Which of the following two seven-p-electron systems should be easier to ionize, 
and why? 
Multiple Choice Questions 
(Try to answer the following questions from sketches that you generate without reference 
to the text.) 
1. According to simple HMO theory, which one of the following statements about 
butadiene is true? 
a) It has a nonbonding MO. 
b) Exciting an electron from the HOMO to the LUMO will not change the p bond 
orders. 
*This problem assumes prior study of Appendix 5.
306 Chapter 8 The Simple H¨uckel Method and Applications 
c) Exciting an electron from the HOMO to the LUMO will not change the p charge 
densities. 
d) Ionizing an electron from the HOMO results in all p charge densities becoming 
equal to 0.75. 
e) It is most likely to undergo electrophilic substitution at one of the two inner 
carbons. 
2. Which of the following is a prediction that would result from a simple HMO treatment 
of the butadienyl cation, C4H+6 ? 
a) The ESR coupling constant is larger for hydrogens attached to the two inner 
carbons. 
b) The central C–C bond has a higher p-bond order than it has in the neutral 
molecule. 
c) The positive charge resides mostly on the two central carbons. 
d) There are only three p energy levels. 
e) The MO coefficients on the central pair of carbons are larger in all of the p MOs 
than they are in the neutral molecule. 
References 
[1] L. Salem, The Molecular Orbital Theory of Conjugated Systems. Benjamin, 
NewYork, 1966. 
[2] C. A. Coulson, Proc. Roy. Soc. (London) A164, 383 (1938). 
[3] A. A. Frost and B. Musulin, J. Chem. Phys., 21, 572 (1953). 
[4] C. A. Coulson and A. Streitwieser, Jr., Dictionary of p-Electron Calculations. 
Freeman, San Francisco, 1965. 
[5] A. Streitwieser, Jr., and J. I. Brauman, Supplemental Tables of Molecular Orbital 
Calculations. Pergamon, Oxford, 1965. 
[6] E. Heilbronner and P. A. Straub, HMOs. Springer-Verlag, Berlin and New York, 
1966. 
[7] A. Streitwieser, Jr., Molecular Orbital Theory for Organic Chemists. Wiley, 
NewYork, 1961. 
[8] C. A. Coulson, Proc. Roy. Soc. (London) A169, 413 (1939). 
[9] F. Gerson and J. H. Hammons, in Nonbenzenoid Aromatics (P. J. Snyder, ed.), 
Vol. II. Academic Press, NewYork, 1971. 
[10] H. Lund, Acta Chem. Scand. 11, 1323 (1957). 
[11] I. Bergman, Trans. Faraday Soc. 50, 829 (1954). 
[12] D. W. Turner, Molecular Photoelectron Spectroscopy. Wiley (Interscience), 
NewYork, 1970. 
[13] F. Brogli and E. Heilbronner, Theoret. Chim. Acta 26, 289 (1972).
Section 8-20 Conclusions 307 
[14] M. J. S. Dewar, The Molecular Orbital Theory of Organic Chemistry. McGraw- 
Hill, NewYork, 1969. 
[15] M. J. S. Dewar and C. de Llano, J. Amer. Chem. Soc. 91, 789 (1969). 
[16] M. J. S. Dewar, A. J. Harget, and N. Trinajstic, J. Amer. Chem. Soc. 91, 6321 
(1969) 
[17] B. A. Hess, Jr., and L. J. Schaad, J. Amer. Chem. Soc. 93, 305, 2413 (1971). 
[18] L. J. Schaad and B. A. Hess, Jr., J. Amer. Chem. Soc. 94, 3068 (1972). 
[19] V. I. Minkin, M. N. Glukhovtsev, and B. Y. Simkin, Aromaticity and Antiaromaticity.
Wiley (Interscience), NewYork, 1994. 
[20] P. v. R. Shleyer, C. Maerker, A. Dransfeld, H. Jiao, and N. J. R. van Eikema 
Hommes, J. Amer. Chem. Soc. 118, 6317, (1996). 
[21] R. Herges, and D. Ajami, Nature, 426, 819 (2003). 
[22] S. K. Ritter, Chem. Eng. News, 82, vol. 9, p. 28 (2004). 
[23] Z. Chen, C. Corminboeuf, T. Heine, J. Bohmann, and P v. R. Schleyer, J. Amer. 
Chem. Soc., 125, 13, 930, (2003). 
[24] S. P. McGlynn, L. G. Vanquickenborne, M. Kinoshita, and D. G. Carroll, Introduction 
To Applied Quantum Chemistry. Holt, NewYork, 1972. 
[25] H. Fujimoto and K. Fukui, in Chemical Reactivity and Reaction Paths (G. Klopman, 
ed.).Wiley (Interscience), NewYork, 1974. 
[26] C. A. Coulson, Trans. Faraday Soc. 42, 265 (1946). 
[27] M. Levy and M. Szwarc, J. Chem. Phys. 22, 1621 (1954). 
[28] G. Klopman, ed., Chemical Reactivity and Reaction Paths.Wiley (Interscience), 
NewYork, 1974. 
[29] J. P. Lowe and B. D. Silverman, Accounts Chem Res., 17, 332, 1984. 
[30] J. P. Lowe and B. D. Silverman, J. Mol. Structure (Theochem.) 179, 47 (1988). 
[31] W. deW. Horrocks, D. H. Templeton, and A. Zalkin, Inorg. Chem. 7, 2303 (1968). 
[32] E. C. Kooyman and E. Farenhorst, Trans. Faraday Soc. 49, 58 (1953).
Chapter 9 
Matrix Formulation of the 
Linear Variation Method 
9-1 Introduction 
In Chapter 7 we developed a method for performing linear variational calculations. The 
method requires solving a determinantal equation for its roots, and then solving a set of 
simultaneous homogeneous equations for coefficients. This procedure is not the most 
efficient for programmed solution by computer. In this chapter we describe the matrix 
formulation for the linear variation procedure. Not only is this the basis for many 
quantum-chemical computer programs, but it also provides a convenient framework 
for formulating the various quantum-chemical methods we shall encounter in future 
chapters. 
Vectors and matrices may be defined in a formal, algebraic way, but they also 
may be given geometric interpretations. The formal definitions and rules suffice for 
quantum-chemical purposes. However, the terminology of matrix algebra is closely 
connected with the geometric ideas that influenced early development. Furthermore, 
most chemists are more comfortable if they have a physical or geometric model to 
carry along with mathematical discussion. Therefore, we append some discussion of 
geometrical interpretation to the algebraic treatment.1 
9-2 Matrices and Vectors 
9-2.A Definitions 
A matrix is an ordered array of elements satisfying certain algebraic rules. We write our 
matrices with parentheses on the left and right of the array.2 Unless otherwise stated, 
we restrict the elements to be numbers (which need not be real). In general, however, 
as long as the rules of matrix algebra can be observed, there is no restriction on what 
the elements may be. In expression (9-1), we have written a matrix in three ways: 
.
.... 
1.2 3.8 -4.0 
5.0 1.0 0.0 
9.1 0.0 -3+4i 
6.0 -1.0 -1.0 
.
....
=
.
.... 
c11 c12 c13 
c21 c22 c23 
c31 c32 c33 
c41 c42 c43
.
....
=C (9-1) 
1More thorough discussions of matrix algebra at a level suitable for the nonspecialist are given by Aitken [1] 
and Birkhoff and MacLane [2]. 
2Some authors use brackets. 
308
Section 9-2 Matrices and Vectors 309 
On the left, the numerical elements are written explicitly. In the center they are symbolized 
by a subscripted letter. On the right the entire matrix is indicated by a single 
symbol. We will use sans serif, upper-case symbols to represent matrices. 
Individual matrix elements will be symbolized either by a subscripted lower-case 
symbol (e.g., c12) or by a subscripted symbol for the matrix in parentheses (e.g., (C)12). 
It is useful to recognize rows and columns in a matrix. The sample matrix given above 
has four rows and three columns, so it is said to have dimensions 4×3. When subscripts 
are used to denote position in a matrix, the convention is that the first subscript indicates 
the row, the second indicates the column. (The order “row-column” is important to 
remember. The mnemonic “RC,” or “Roman Catholic” is helpful.) Rows are numbered 
from top to bottom, columns from left to right. There is no limit on the dimensions for 
matrices, but we usually will be concerned in a practical way with finite-dimensional 
matrices in this book. 
If a matrix has only one column or row, it is called a column vector or row vector, 
respectively. We will use sans serif, lower-case symbols to denote column vectors. 
Additional symbols, described shortly, will be used to denote row vectors. These two 
kinds of vector behave differently under the rules of matrix algebra, so it is important 
to avoid confusing them. 
If a matrix has only one row and one column, its behavior under the rules of matrix 
algebra becomes identical to the familiar behavior of ordinary scalars (i.e., numbers), 
and so a 1×1 matrix is simply a number. 
The similarity in appearance between a matrix and a determinant may be deceptive. 
A determinant is denoted by bounding with vertical straight lines, and is equal to 
a number that can be found by reducing the determinant according to a prescribed 
procedure (see Appendix 2). For this to be possible, the determinant must be square 
(i.e., have the same number of rows as columns). A matrix is not equal to a number 
and need not be square. (However, one can take the determinant of a square matrix A. 
This number is symbolized |A| and is not the same as A without the vertical bars.) 
Two matrices are equal if all elements in corresponding positions are equal. Thus, 
A=B means aij =bij for all i and j . 
9-2.B Complex Conjugate,Transpose, and Hermitian Adjoint 
of a Matrix 
We define the complex conjugate of a matrix A to be the matrix A*, formed by replacing 
every element of A by its complex conjugate. If A = A*,A is a real matrix. (Every 
element is real.) 
We define the transpose of a matrix A to be the matrix A˜ , formed by interchanging 
row 1 and column 1, row 2 and column 2, etc. The transpose of the 4×3 matrix in 
expression (9-1) is the 3×4 matrix given in 
C˜ =
.
.. 
1.2 5.0 9.1 6.0 
3.8 1.0 0.0 -1.0 
-4.0 0.0 -3+4i -1.0 
.
..
(9-2) 
If we denote some column vector as p, we can symbolize the corresponding row vector 
as ˜p. Thus, the tilde symbol is one device we can use to indicate a row vector.
310 Chapter 9 Matrix Formulation of the Linear Variation Method 
Transposing a square matrix corresponds to “reflecting” it through its principal 
diagonal (which runs from upper left to lower right) as indicated in 
A=
.
.. 
1 2 3 
4 5 6 
7 8 9
.
..
, A˜ =
.
.. 
1 4 7 
2 5 8 
3 6 9
.
..
(9-3) 
If A=A˜ ,A is a symmetric matrix. 
We define the hermitian adjoint of A to be the matrix A†, formed by taking the 
transpose of the complex conjugate of A (or the complex conjugate of the transpose. 
The order of these operations is immaterial.) Hence, A† = (A˜ )* = (A˜ *). If A = A†, 
A is a hermitian matrix. 
9-2.C Addition and Multiplication of Matrices and Vectors 
Multiplication of a matrix by a scalar is equivalent to multiplying every element in the 
matrix by the scalar. Addition of two matrices is accomplished by adding elements in 
corresponding positions in the matrices. Thus, for example, 
1 2 
3 4

+25 6 
7 8

=
1+10 2+12 
3+14 4+16

=
11 14 
17 20
 
and it is evident that the operation of matrix addition is possible only within sets of 
matrices of identical dimensions. 
Matrix multiplication is a bit more involved. We start by considering multiplication 
of vectors. Two types of vector multiplication are possible. If we multiply a row vector 
on the left times a column vector on the right, we take the product of the leading element 
of each plus the product of the second element of each, plus . . . , etc., thereby obtaining 
a scalar as a result. Hence, this is called scalar multiplication of vectors. For example, 
	1 2 3

.
.. 
4
5
6
.
..
=1×4+2×5+3×6=32 
This kind of multiplication requires an equal number of elements in the two vectors. 
It is important to retain in mind the basic operation described here: summation of 
products taken by sweeping across a row on the left and down a column on the right. 
Since there exists but one row on the left and one column on the right, we obtain one 
number as the result. 
The other possibility is to multiply a column vector on the left times a row vector on 
the right. Employing the same basic operation as above, we sweep across row 1 on the 
left and down column 1 on the right, obtaining a product that we will store in position 
(1, 1) of a matrix to keep track of its origin. The product of row 1 times column 2 gives 
us element (1, 2) and so on. In this way, we generate a whole matrix of numbers. For 
example, 
1
2
	3 4 5
=
1×3 1×4 1×5 
2×3 2×4 2×5

=
3 4 5 
6 8 10

Section 9-2 Matrices and Vectors 311 
This is an example of matrix multiplication of vectors. Just as before, the number of 
columns on the left (one) equals the number of rows on the right. Now, however, the 
number of elements in the vectors may differ, and the dimensions of the matrix reflect 
the dimensions of the original vectors. 
The two types of vector multiplication may be symbolized as follows, using our 
notation for scalars, row vectors, column vectors, and matrices: 
˜ab = c (9-4) 
a˜b = C (9-5) 
Multiplying two matrices together is most simply viewed as scalar multiplying all 
the rows in the left matrix by all the columns of the right matrix. Thus, in AB=C, the 
element cij is the (scalar) product of row i in A times column j in B. This process is 
possible only when the number of columns in A equals the number of rows in B. Thus, 
AB may exist as a matrix C, while BA may not exist due to a disagreement in number of 
rows and columns. If A and B are both square matrices and have equal dimension, then 
AB and BA both exist, but they still need not be equal. That is, matrix multiplication 
is not commutative. 
In a triple product of matrices, ABC, one can multiply AB first (call the result D) 
and then multiply DC to get the final result (call it F). Or, if one takes BC=E, then 
one always finds AE=F. Thus, the result is invariant to the choice between (AB)C or 
A(BC), and so matrix multiplication is associative. 
9-2.D Diagonal Matrices, Unit Matrices, and Inverse Matrices 
A diagonal matrix is a square matrix having zeros everywhere except on the principal 
diagonal. Diagonal matrices of equal dimension commute with each other, but a 
diagonal matrix does not, in general, commute with a nondiagonal matrix. 
A unit matrix 1 is a special diagonal matrix. Every diagonal element has a value 
of unity. A unit matrix times any matrix (of appropriate dimension) gives that same 
matrix as product. That is, 1A =A1=A. It follows immediately that the unit matrix 
commutes with any square matrix of the same dimension. 
We define the left inverse of a matrix A to be A-1, satisfying the matrix equation 
A-1A=1. The right inverse is defined to satisfy AA-1 =1. 
In most of our quantum-chemical applications of matrix algebra, we will be concerned 
only with vectors and square matrices. For square matrices, the left and right 
inverses are identical, and so we refer simply to the inverse of the matrix. 
9-2.E Complex Conjugate, Inverse, and Transpose 
of a Product of Matrices 
If AB=C, then C* =(AB)* =A*B*. In words, the complex conjugate of a product 
of matrices is equal to the product of the complex conjugate matrices. This is demonstrable 
from the observation that (C)ij =(A)i1(B)1j +(A)i2(B)2j +··· and (C)*ij = 
(A)*i
1(B)*1
j + (A)*i
2(B)*2
j +··· and so the complex conjugate is produced by taking
312 Chapter 9 Matrix Formulation of the Linear Variation Method 
the complex conjugate of every element in A and B but not changing their order of 
combination. 
If AB = C, then C-1 = (AB)-1 = B-1A-1. In words, the inverse of a product 
of matrices is equal to the product of inverses, but with the order reversed. We can 
easily show that this satisfies the rules of matrix algebra. C-1C = (AB)-1AB = B-1A-1AB=B-11B=B-1B=1. If we failed to reverse the order, we would instead 
have A-1B-1AB and, because the matrices do not commute, we would be prevented 
from carrying through the reduction to 1. 
If AB = C, then C˜ = (
AB) = B˜ A˜ . The transpose of a product is the product of 
transposes, again in reverse order. Since C˜ has cij and cj i interchanged, it follows that, 
where we had row i of A times column j of B, we must now have row j of A times 
column i of B. But this is the same as column j of A˜ times row i of B˜ . To obtain row 
on left and column on right for proper multiplication, we must have B˜ A˜ . 
9-2.F A Geometric Model 
Consider a vector in two-dimensional space emanating from the origin of a Cartesian 
system as indicated in Fig. 9-1a. We can summarize the information contained in this 
vector (magnitude and direction) by writing down the x and y components of the vector 
terminus, (3 2) in this case. It must be understood that the first number corresponds to 
the x component and not the y, and so the vector (3 2) carries its information through 
number position as well as number value. 
If we multiply both numbers in the vector by 2, the result, (6 4), corresponds to a 
vector collinear to the original but twice as long. Therefore, multiplying a vector by a 
number results in a change of scale but no change in direction. Hence, the term “scalar” 
is often used in place of “number” in vector terminology. 
Suppose that we rotated the Cartesian axes counterclockwise through an angle ., 
maintaining them orthogonal to each other and not varying the distance scales. We 
imagine our original vector to remain unrotated during this coordinate transformation. 
(Equivalently, we can imagine rotating the vector clockwise by ., keeping the axes 
fixed.) We wish to know how to express our vector in the new coordinate system. The 
situation is depicted in Fig. 9-1b. Inspection reveals that the new coordinates (xy) are 
related to the old (x y) as follows: 
x =x cos . +y sin ., y=-x sin . +y cos . (9-6) 
If we make use of matrix algebra, we can express Eqs. (9-6) as a matrix equation: 
x 
y

=
 cos . sin . 
-sin . cos . 
x
y
 (9-7) 
Multiplying the two-dimensional vector (call it v) by the 2×2 matrix (call it R) generates 
a new vector v, that gives the coordinates of our vector in the new coordinate 
system: 
v =Rv (9-8)
Section 9-2 Matrices and Vectors 313 
Figure 9-1 
 (a) The vector (3 2). (b) The same vector and its relationship to two Cartesian axis 
systems. 
We have, then, a parallel between the vectors v and v and matrix R on the one hand, 
and the two-dimensional “geometrical” vector and rotating coordinate system on the 
other. R represents the rotation and is often referred to as a rotation matrix. 
If we were to perform the rotation in the opposite direction, the rotation matrixwould 
be the same except for the sin . terms, which would reverse sign: 
rotation of coordinates 
clockwise by . 

-.
cos . -sin . 
sin . cos . 
 (9-9) 
Note that this is just R˜ , the transpose of R.
314 Chapter 9 Matrix Formulation of the Linear Variation Method 
If we were to rotate counterclockwise by . and then clockwise by ., we should end 
up with our original coordinates for the vector. Thus, we should expect 
R˜ Rv=v (9-10) 
or 
R˜ R=1 (9-11) 
(Note that the order of operations is consistent with reading from right to left, just as 
with differential operators. Thus, R˜ Rmeans that firstRis performed, thenR˜ .) Relation 
(9-11) is easily verified by explicit multiplication. Thus we see that, if we think of a 
matrix as representing some coordinate transformation in geometrical space, the inverse 
of the matrix represents the reverse transformation. In this particular example, the 
transpose of the transformation matrix turns out to be the inverse transformation matrix. 
When this is so, the matrix is said to be orthogonal. (Orthogonal transformations do not 
change the angles between coordinate axes; orthogonal axes remain orthogonal—hence 
the name “orthogonal.”) The analogous transformation for matrices having complex 
or imaginary coefficients is called a unitary transformation. A unitary matrix has its 
hermitian adjoint as inverse: A†A=1. 
While it is easy to visualize a coordinate transformation in two or three dimensions, 
it is more difficult in higher-dimensional situations. Nevertheless, the mathematics and 
terminology carry forward to any desired dimension and are very useful. One may, if 
one wishes, talk of vectors and coordinate transformations in hyperspace, or one can 
eschew such mental constructs and simply follow the mathematical rules without a 
mental model. 
One must be a bit cautious about inverses of matrices. In the rotation described 
above, we have a unique way of relating each x, y point in one coordinate system to a 
point at x, y in the other. The transformation does not entail any loss of information 
and can therefore be “undone.” 
Such transformations (and their matrices) are called nonsingular. A nonsingular 
matrix is recognizable through the fact that its determinant must be nonzero. If we had a 
transformation which, for example, caused all or some points in one coordinate system 
to coalesce into a single point in the transformed system, we would lose our ability to 
back-transform in a unique way. Such a singular transformation has no inverse, and 
the determinant of a singular matrix equals zero. 
9-2.G Similarity Transformations 
A matrix product of the form A-1HA is called a similarity transformation on H. If 
A is orthogonal, then A˜HA is a special kind of similarity transformation, called an 
orthogonal transformation. If A is unitary, then A†HA is a unitary transformation onH. 
There is a physical interpretation for a similarity transformation, which will be discussed 
in a later chapter. For the present, we are concerned only with the mathematical 
definition of such a transformation. The important feature is that the eigenvalues, or 
“latent roots,” of H are preserved in such a transformation (see Problem 9-5). 
In this section we have quickly presented the salient rules of matrix algebra and 
hinted at their connection with geometric operations. The results are summarized in 
Table 9-1 for ease of reference.
Section 9-3 Matrix Formulation of the Linear Variation Method 315 
TABLE 9-1 
 Some Matrix Rules and Definitions for a Square Matrix A of Dimension n 
A=B Matrix equality; means aij =bij, i,j=1,n 
A+B=C Matrix addition; cij =aij +bij, i,j=1,n 
cA=B Multiplication of A by scalar; bij =c ·aij, i,j=1,n 
AB=C Matrix multiplication; cij =
n
k=1 aikbkj, i,j=1,n 
|A| The determinant of the matrix A (see Appendix 2) 
A-1 The inverse of A; A-1A=AA-1 =1 
If A-1 exists, A is nonsingular and |A| =0. 
A* The complex conjugate of A; aij .a*ij, i,j=1,n 
If A* =A,A is real. 
A˜ The transpose of A; (A˜ )ij =aj i (rows and columns 
inter changed) 
If A˜ =A, symmetric; if A˜ =-A, antisymmetric; 
if A˜ =A-1, orthogonal. 
A† The hermitian adjoint of A; (A†)ij =a*ji(A† =A˜ *) 
If A† =A, hermitian. If A† =A-1, unitary. 
(ABC)* =A*B*C* Complex conjugate of product 
(A
BC)=C˜ B˜ A˜ Transpose of product 
(ABC)† =C†B†A† Hermitian adjoint of product 
(ABC)-1 =C-1B-1A-1 Inverse of product 
|ABC| = |A| · |B| · |C| Determinant of product (any order) 
T-1AT A similarity transformation 
If T-1 =T†, this is a unitary transformation. 
If T-1 = ˜T, this is an orthogonal transformation. 
9-3 Matrix Formulation of the Linear Variation Method 
We have seen that the independent-electron approximation leads to a series of MOs 
for a molecular system. If the MOs are expressed as a linear combination of n basis 
functions (which are often approximations to AOs, although this is not necessary), the 
variation method leads to a set of simultaneous equations: 
(H11 -ES11)c1 +(H12 -ES12)c2+···+(H1n -ES1n)cn = 0 
...
(9-12) 
(Hn1 -ESn1)c1 +···+(Hnn -ESnn)cn = 0 
All terms have been defined in Chapter 7. Given a value forE that satisfies the associated 
determinantal equations, we can solve this set of simultaneous equations for ratios 
between the ci ’s. Requiring MO normality establishes convenient numerical values for 
the ci ’s.
316 Chapter 9 Matrix Formulation of the Linear Variation Method 
A matrix equation equivalent to Eq. (9-12) is3 
.
.. 
H11 -ES11 H12 -ES12 ··· H1n -ES1n 
... 
Hn1 -ESn1 ··· Hnn -ESnn
.
.. 
.
..... 
c1 
c2
... 
cn
.
.....
=
.
..... 
0
0
... 
0
.
..... 
(9-13) 
The matrix in Eq. (9-13) is clearly the difference between two matrices. This enables 
us to rewrite the equation in the form 
.
.. 
H11 H12 ··· H1n 
... 
... 
Hn1 ··· Hnn
.
.. 
.
..... 
c1 
c2
... 
cn
.
.....
=E
.
.. 
S11 S12 ··· S1n 
... 
... 
Sn1 ··· Snn
.
.. 
.
..... 
c1 
c2
... 
cn
.
..... 
(9-14) 
or 
Hci =EiSci, i=1, 2, . . . ,n (9-15) 
where we have introduced the subscript i to account for the fact that there are many 
possible values forE and that each one has its owncharacteristic set of coefficients. Note 
that the “eigenvector” ci is a column vector and that each element in ci is (effectively) 
multiplied by the scalar Ei according to Eq. (9-15). 
In general, there are as many MOs as there are basis functions, and so Eq. (9-15) 
represents n separate matrix equations. We can continue to use matrix notation to 
reduce these to a single matrix equation. We do this by stacking the n c vectors together, 
side by side, to produce an n×n matrix C. The numbers E must also be combined 
into an appropriate matrix form. We must be careful to do this in such a way that the 
scalar E1 still multiplies only c1 (now column 1 of C) E2 multiplies only c2, and so 
forth. This is accomplished in the following equation 
.
.. 
H11 ··· H1n 
... 
... 
Hn1 ··· Hnn
.
.. 
.
.. 
c11 ··· c1n 
... 
... 
cn1 ··· cnn
.
.. 
=
.
.. 
S11 ··· S1n 
... 
... 
Sn1 ··· Snn
.
.. 
.
.. 
c11 ··· c1n 
... 
... 
cn1 ··· cnn
.
.. 
.
..... 
E1 0 0 ··· 0 
0 E2 0 ··· 0 
... 
... 
... 
... 
0 0 0 ··· En
.
..... 
(9-16) 
or 
HC=SCE (9-17) 
The matrix E is a diagonal matrix of orbital energies (often referred to as the matrix 
of eigenvalues). C is the matrix of coefficients (or matrix of eigenvectors), and each 
3Quantum-chemical convention is to use upper case letters for individual elements of the matrices H, S, and E. 
This differs from the usual convention.
Section 9-4 Solving the Matrix Equation 317 
column in C refers to a different MO. The first column refers to the MO having energy 
E1. In multiplying E by C from the left, each coefficient in column 1 becomes multiplied 
by E1. This would not occur if we multiplied E by C1 from the right. Therefore, 
HC=SCE is correct, whereas HC=ESC is incorrect. 
9-4 Solving the Matrix Equation 
Since we know the basis functions and the effective hamiltonian (in principle, at least), 
we are in a position to evaluate the elements in H and S. How do we then find C and E? 
Let us first treat the simplified situation where our basis set of functions is orthonormal, 
either by assumption or design. Then all the off-diagonal elements of S (which 
correspond to overlap between different basis functions) are zero, and all the diagonal 
elements are unity because of normality. In short, S = 1. Therefore, Eq. (9-17) 
becomes 
HC=CE (9-18) 
and our problem is, given H, find C and E. 
Now, we want a set of coefficients that correspond to normalized MOs. We have 
seen earlier that, for an orthonormal basis set, this requires eachMOto have coefficients 
satisfying the equation (assuming real coefficients) 
c2
1i +c2
2i +···+c2
ni =1 (9-19) 
We can write this as a vector equation 
˜cici =	ci1 ci2 ··· cin

.
..... 
c1i 
c2i 
... cni
.
.....=1 (9-20) 
Furthermore, we know that any two different MOs must be orthogonal to each other. 
That is ˜cicj =0, i =j . All this may be summarized in the matrix equation 
C˜ C=1 (9-21) 
Hence, the coefficient matrix is orthogonal. In the more general case in which coefficients 
may be complex, C is unitary; i.e., C†C=1. Our problem then is, given H, find 
a unitary matrix C such that HC=CE with E diagonal. 
We can multiply both sides of a matrix equation by the same matrix and preserve the 
equality. However, because matrices do not necessarily commute, we must be careful 
to carry out the multiplication from the left on both sides, or from the right on both 
sides. Thus, multiplying Eq. (9-18) from the left by C†, we obtain 
C†HC=C†CE=1E=E (9-22) 
where we have used the fact that C is unitary. Now our problem may be stated as, 
given H, find a unitary matrix C such that C†HC is diagonal.4 Several techniques 
4Not every matrix (not even every square matrix) can be diagonalized by a unitary transformation, but every 
hermitian matrix can be so diagonalized.
318 Chapter 9 Matrix Formulation of the Linear Variation Method 
exist for finding such a matrix C. These are generally much more suitable for machine 
computation than are determinantal manipulations. 
We can illustrate that the allyl radical energies and coefficients already found by the 
HMO method do in fact satisfy the relations C†C=1 and C†HC=E. The matrix C 
can be constructed from the HMO coefficients and is 
C=
.
... 
1
2 
1 v2 
1
2 
1 v2 
0 - 1 v2 
1
2 - 1 v2 
1
2 
.
...
(9-23) 
Therefore 
C†C = 
.
... 
1
2 
1 v2 
1
2 
1 v2 
0 - 1 v2 
1
2 - 1 v2 
1
2 
.
... 
.
... 
1
2 
1 v2 
1
2 
1 v2 
0 - 1 v2 
1
2 - 1 v2 
1
2 
.
... 
= 
.
.. 
1 0 0 
0 1 0 
0 0 1
.
..
=1 (9-24) 
The matrix H for the allyl radical is, in HMO theory, 
H=
.
.. 
a ß 0 
ß a ß 
0 ß a
.
..
(9-25) 
The reader should verify that 
C†HC=
.
.. 
a +v2ß 0 0 
0 a 0 
0 0 a -v2ß 
.
..
=E (9-26) 
The diagonal elements can be seen to correspond to the HMO energies. Note that the 
energy in the (1 1) position of E corresponds to the MO with coefficients appearing in 
column 1 of C, illustrating the positional correlation of eigenvalues and eigenvectors 
referred to earlier. 
If the basis functions are not orthogonal, S =1 and the procedure is slightly more 
complicated. Basically, one first transforms to an orthogonal basis to obtain an equation 
of the form HC = CE. One diagonalizes H as indicated above to find E and C, 
where C is the matrix of coefficients in the orthogonalized basis. Then one back 
transforms C into the original basis set to obtain C. There are many choices available 
for the orthogonalizing transformation. The Schmidt transformation, based on the 
Schmidt orthogonalization procedure described in Chapter 6, is popular because it is 
very rapidly performed by a computer. Here we will simply indicate the matrix algebra 
involved. Let the matrix that transforms a nonorthonormal basis to an orthonormal one 
be symbolized A. This matrix satisfies the relation 
A†SA=1 (9-27)
Section 9-4 Solving the Matrix Equation 319 
Furthermore, |A| = 0 and so A-1 exists. We can insert the unit matrix (in the form 
AA-1) wherever we please in the matrix equation HC = SCE without affecting the 
equality. Thus, 
HAA-1C=SAA-1CE (9-28) 
Multiplying from the left by A† gives 
A†HAA-1C=A†SAA-1CE (9-29) 
By Eq. (9-27), this reduces to
(A†HA)(A-1C)=(A-1C)E (9-30) 
where the parentheses serve only to make the following discussion clearer. If we define 
A†HA to be H, and A-1C to be C, Eq. (9-30) becomes 
HC =CE (9-31) 
Since we know the matrix H and can compute A from knowledge of S, it is possible 
to write down an explicit H matrix for a given problem. Then, knowing H (which 
is just the hamiltonian matrix for the problem in the orthonormal basis), we can seek 
the unitary matrix C such that C†HC is diagonal. These diagonal elements are our 
orbital energies. (Note that E in Eq. (9-31) is the same as E in HC=SCE.) To find 
the coefficients for the MOs in terms of the original basis (i.e., to find C), we use the 
relation 
AC =A(A-1C)=1C=C (9-32) 
One nice feature of this procedure is that, even though we use the inverse matrix A-1 in 
our formal development, we never need to actually compute it. (A and A† are used to 
find H, and A is used to find C.) This is fortunate because calculating inverse matrices 
is a relatively slow process. 
A few more words should be said about the process of diagonalizing a hermitian 
matrix H with a unitary transformation. Two methods are currently in wide use. The 
older, slower method, known as the Jacobi method, requires a series of steps on the 
starting matrix. In the first step, a matrix O1 is constructed that causes the largest 
off-diagonal pair of elements of H to vanish in the transformation H1 =O˜ 1HO1. Now 
a second transformation matrix O2 is constructed to force the largest off-diagonal pair 
of elements in H1 to vanish in the transformation O˜ 2H1O2 = O˜ 2O˜ 1HO1O2. This 
procedure is continued. However, since each transformation affects more elements in 
the matrix than just the biggest pair, we eventually “unzero” the pair that was zeroed in 
forming H1 or H2, etc. This means that many more transformations are required than 
there are off-diagonal pairs. Eventually, however, the off-diagonal elements will have 
been nibbled away (while the diagonal elements have been building up) until they are 
all smaller in magnitude than some preselected value, and so we stop the process. The 
transformation matrix C corresponds to the accumulated product O1O2O3 . . . .
320 Chapter 9 Matrix Formulation of the Linear Variation Method 
Amore recently discovered, faster procedure is the Givens– Householder–Wilkinson 
method. Here, H is first tridiagonalized, which means that all elements are made to 
vanish except those on the main diagonal as well as on the codiagonals above and below 
the main diagonal. This similarity transformation can be done in a few steps, each 
step zeroing all the necessary elements in an entire row and column. The eigenvalues 
for the tridiagonal matrix (and hence for the original matrix) may be found one at a 
time as desired. If only the third lowest eigenvalue is of interest, that one alone can be 
computed. This is a useful degree of freedom which results in substantial savings of 
time. Once an eigenvalue is found, its corresponding eigenvector may be computed. 
9-5 Summary 
The steps to be performed in a matrix solution for a linear variation calculation are: 
1. From the basis set, calculate the overlap matrix S. 
2. From the basis set and hamiltonian operator, calculate the hamiltonian matrix H. 
3. If S = 1, find an orthogonalization procedure. In the Schmidt method, A is such 
that A†SA=1. The matrix equation may now be written in the form HC =CE. 
4. Find C such that C†HC is a diagonal matrix. The diagonal elements are the 
roots E. 
5. If necessary, back transform: AC =C. The columns of C contain the MO coefficients 
appropriate for the original basis set. 
9-5.A Problems 
9-1. Evaluate the following according to the rules of matrix algebra: 
a) 	6 7 8

.
.. 
9 
10 
11
.
.. 
b) 6
7
	a b c
 
c) 4 6 
i -3

+7 3 1 
-1 3
 
d) cos . -sin . 
sin . cos . 
 cos . sin . 
-sin . cos . 
 
e) 
.
.. 
i 4 
1 7 
0 -3
.
.. 
3 2 
4 7

Section 9-5 Summary 321 
f) 3 2 
4 7
.
.. 
i 4 
1 7 
0 -3
.
.. 
g) 
 
cos . -sin . 
sin . cos .
 
9-2. If Hij = .i*Hˆ .j dt and Hˆ is hermitian, show that H is a hermitian matrix. 
9-3. Let 
A=
a11 a12 
a21 a22

, B=
b11 b12 
b21 b22
 
Show that, in general, AB=BA. 
9-4. Let
A=
.
.. 
1 0 0 
0 2 0 
0 0 3
.
..
, B=
.
.. 
4 0 0 
0 5 0 
0 0 6
.
..
, and C=
.
.. 
1 0 1 
0 1 0 
1 0 1
.
.. Show that AB=BA, but AC =CA. Compare the matrix AC with CA. Do these 
matrices show any simple relationship? Can you relate this to properties of A and 
C mathematically? 
9-5. The “latent roots” .i of A are solutions to the equation |A-.i1| = 0, 
i =1, 2, . . . ,n, where n is the dimension of A. 
a) Show that, under a similarity transformation B=T-1AT, the latent roots are 
preserved. 
b) Demonstrate that diagonalization of A via a similarity transformation produces 
the latent roots as the diagonal elements. 
9-6. Show that, if a matrix has any latent roots equal to zero, it has no inverse. 
9-7. The trace (or spur) of a matrix is the sum of the elements on the principal diagonal. 
Thus, tr A=n
i=1aii . 
a) Show that the trace of a triple product of matrices is invariant under cyclic 
permutation. That is, tr(ABC)=tr(CAB)=tr(BCA) but not tr(CBA). 
b) Show that the trace of a matrix is invariant under a similarity transformation. 
9-8. The norm of a matrix is the positive square root of the sum of the absolute squares 
of all the elements. 
For a real matrix A, 
norm A=
.
. 
n 

i,j=1 
a2
ij
.
.
1/2 
=
.
.

i 

j 
	A˜ 

i,j 
(A)j,i
.
.
1/2
322 Chapter 9 Matrix Formulation of the Linear Variation Method 
Prove that the norm of a real matrix is preserved in an orthogonal transformation 
(or, you may prefer to prove that the norm of any matrix is preserved in a unitary 
transformation). 
9-9. Use the facts that the trace, the determinant, and the norm of a matrix are invariant 
under an orthogonal transformation to find the eigenvalues of the following 
matrices: 
a) 
.
.. 
0 1 1 
1 0 1 
1 1 0
.
.. 
b) 
.
... 
1
2 
1 v2 
1
2 
1 v2 
0 - 1 v2 
1
2 - 1 v2 
1
2 
.
... 
c) 
.
.. 
0 1 0 
1 2 1 
0 1 0
.
.. 
9-10. Consider the matrix 
 cos . 0 
-sin . 0
 
What is the effect of this transformation on 3
2
? On 3
3
? Can the transformation 
be uniquely reversed? (That is, for, say, . =0, and given a transformed 
vector 3
0
, can one uniquely determine the vector this was transformed from?) 
Does the matrix have an inverse? Evaluate its determinant. 
9-11. What are the eigenvectors for the matrix 
H=
.
.. 
1 0 0 
0 -3 0 
0 0 -2
.
.. 
9-12. Show that, if A and B have “simultaneous eigenvectors” (i.e., both diagonalized 
by the same similarity transformation), then A and B commute. 
9-13. IfHC=CE, andC†C=1, thenC†HC=E, and we seek a unitary transformation 
that diagonalizes H. If HC = SCE, and C†SC = 1, then C†HC = C†SCE = 1E, and C†HC = E. Since this is the same working equation as the one we 
found above, why do we not proceed in the same way? Why do we bother 
orthogonalizing our basis first? 
9-14. We have mentioned that a matrix may be used to represent the rotation of coordinates 
by some angle .. Such a rotation is a geometric operation, so we have,
Section 9-5 Summary 323 
in effect, represented an operator with a matrix. It is possible to represent other 
operators in a similar way. Indeed, an alternative approach to quantum mechanics 
exists in which the whole formalism is based on matrices and their properties 
(matrix mechanics, as opposed to wave mechanics). A particularly interesting 
example is provided by the matrices constructed by Pauli to represent spin operators 
and functions. It was mentioned in Chapter 5 that spin functions a and ß 
satisfy rules similar to those for orbital angular momentum. Two of these are 
Sˆza = 
1
2
a, Sˆzß =-
1
2
ß 
But it was pointed out that a and ß could not be expressed in terms of spherical 
harmonics. Pauli represented this operator and functions by 
a =
1
0

, ß=
0
1

, Sˆz = 
1
2 
1 0 
0 -1
 
Using these definitions, show that 
 a†ß d. =  ߆a d.=0,  a†a d.= ß†ß d.=1, 
Sˆza = 
1
2
a, Sˆzß =-
1
2
ß 
[Note: since a and ß are essentially the Dirac delta functions in the spin coordinate., 
the process of integration reduces here to scalar multiplication of vectors.] 
References 
[1] A. C. Aitken, Determinants and Matrices, 9th ed.Wiley (Interscience), NewYork, 
1958. 
[2] G. Birkhoff and S. MacLane, A Survey of Modern Algebra, 5th ed. Macmillan, 
NewYork, 1995.
Chapter 10 
The Extended H¨uckel Method 
10-1 The Extended H¨uckel Method 
The extended H¨uckel (EH) method is much like the simple H¨uckel method in many 
of its assumptions and limitations. However, it is of more general applicability since 
it takes account of all valence electrons, s and p, and it is of more recent vintage 
because it can only be carried out on a practical basis with the aid of a computer. The 
basic methods of extended H¨uckel calculations have been proposed at several times by 
various people. We will describe the method of Hoffmann [1], which, because of its 
systematic development and application, is the EHMO method in common use. 
The method is described most easily by reference to an example. We will use 
methane (CH4) for this purpose. 
10-1.A Selecting Nuclear Coordinates 
The first choice we must make is the molecular geometry to be used. For methane, we 
will take the H–C–H angles to be tetrahedral and C–H bond distances of 1.1 Å.We can 
try altering these dimensions later. 
Cartesian coordinates for the five atoms are listed in Table 10-1, and the orientation 
of the nuclei in Cartesian space is indicated in Fig. 10-1. (Even though the eigenvalues 
and MOs one finally obtains are independent of how CH4 is oriented in Cartesian 
space,1 it is generally a good idea to choose an orientation that causes some Cartesian 
and symmetry axes to coincide. The resulting expressions for MOs in terms of AOs 
are generally much simpler to sketch and interpret.) 
10-1.B The Basis Set 
Next we must select the basis set of functions with which to express the MOs. The 
extended H¨uckel method uses the normalized valence AOs for this purpose. For CH4, 
this means a 1s AO on each hydrogen and a 2s, 2px, 2py, and 2pz AO on carbon. The 
inner-shell 1s AO on carbon is not included. The AOs are represented by Slater-type 
orbitals (STOs). Except for the 1sAOs on hydrogen, the exponential parameters of the 
STOs are determined from Slater’s rules (Chapter 5). Various values for the hydrogen 
1sAO exponent have been suggested. These have ranged from the 1.0 given by Slater’s 
1It sometimes happens that an approximation is made that causes the solution to depend on orientation. This is 
called “loss of rotational invariance.” 
324
Section 10-1 The Extended H¨uckel Method 325 
TABLE 10-1 
 Cartesian Coordinates (in Angstroms) for Atoms of 
Methane Oriented as Shown in Fig. 10-1 
Atom x y z 
C 0.0 0.0 0.0 
Ha 0.0 0.0 1.1 
Hb 1.03709 0.0 -0.366667 
Hc -0.518545 0.898146 -0.366667 
Hd -0.518545 -0.898146 -0.366667 
Figure 10-1 
 Orientation of methane in a Cartesian axis system. 
rules to a value of v2. We will use a value of 1.2, which is near the optimal value for 
H2 (see Chapter 7). The STOs for methane are listed in Table 10-2. 
10-1.C The Overlap Matrix 
Knowing the AO functions and their relative positions enables us to calculate all of 
their overlaps. This would be a tedious process with pencil and paper. However, the 
formulas have been programmed for automatic computation, so this step is included in 
any EH computer program.2 The computed overlap matrix for the methane molecule 
is shown in Table 10-3. 
This matrix is symmetric (since the overlap between twoAOs is independent of their 
numbering order) and has diagonal elements of unity since the AOs are normalized. 
The zero values in the first four rows and columns reflect the orthogonality between the 
s and all pAOs on carbon. Other zero values result when hydrogen 1sAOs are centered 
in nodal planes of carbon pAOs. The geometry of the system is clearly reflected in the 
overlap matrix. For instance, the overlap of the 2pz AO of carbon with the hydrogen 
1s AO at Ha is large and positive, while its overlaps with AOs on Hb,Hc, and Hd are 
negative, equal, and of smaller magnitude. The 2s AO of carbon, on the other hand, 
overlaps all 1s AOs equally. Also, the overlap between every pair of hydrogen 1s AOs 
2Several such programs are available from Quantum Chemistry Program Exchange, Chemistry Dept., Room204, 
Indiana University, Bloomington, Indiana 47401. http://www.QCPE.Indiana.edu
326 Chapter 10 The Extended H¨uckel Method 
TABLE 10-2 
 Basis AOs for Methane 
AO no. Atom Type na la ma exp 
1 C 2s 2 0 0 1.625 
2 C 2pz
2 1 0 1.625 
3 C 2px
2 1 (1)b 1.625 
4 C 2py
2 1 (1)b 1.625 
5 Ha 1s 1 0 0 1.200 
6 Hb 1s 1 0 0 1.200 
7 Hc 1s 1 0 0 1.200 
8 Hd 1s 1 0 0 1.200 
an, l,m are the quantum numbers described in Chapter 4. 
b2px and 2py are formed from linear combinations of m=+1 and m=-1 STOs, and 
neither of these AOs can be associated with a particular value of m. 
TABLE 10-3 
 Overlap Matrix for STOs of Table 10-2 
1 2 3 4 5 6 7 8 
1 1.0000 0.0 0.0 0.0 0.5133 0.5133 0.5133 0.5133 
2 0.0 1.0000 0.0 0.0 0.4855 -0.1618 -0.1618 -0.1618 
3 0.0 0.0 1.0000 0.0 0.0 0.4577 -0.2289 -0.2289 
4 0.0 0.0 0.0 1.0000 0.0 0.0 0.3964 -0.3964 
5 0.5133 0.4855 0.0 0.0 1.0000 0.1805 0.1805 0.1805 
6 0.5133 -0.1618 0.4577 0.0 0.1805 1.0000 0.1805 0.1805 
7 0.5133 -0.1618 -0.2289 0.3964 0.1805 0.1805 1.0000 0.1805 
8 0.5133 -0.1618 -0.2289 -0.3964 0.1805 0.1805 0.1805 1.000 
is the same. Features such as these provide a useful check on the correctness of our 
initial Cartesian coordinates. 
10-1.D The Hamiltonian Matrix 
We have the overlap matrix S. Next we must find the hamiltonian matrix H. Then we 
will be in a position to solve the equation HC=SCE for C and E. (See Chapter 9.) 
The matrix H is calculated from a very approximate but simple recipe. The basic ideas 
are similar in spirit to those described in connection with the interpretations of a and 
ß in the simple H¨uckel method. The energy integral Hii in the EH method is taken to 
be equal to the energy of an electron in the ith AO of the isolated atom in the appropriate 
state. The various ionization energies of atoms are known,3 so this presents no 
great difficulty. However, one special problem must be dealt with, namely, finding the 
3See Moore [2].
Section 10-1 The Extended H¨uckel Method 327 
appropriate state. In the isolated carbon atom, the lowest-energy states are associated 
with the configuration 1s22s22p2. In a saturated molecule such as methane, however, 
carbon shares electrons with four hydrogens, and calculations indicate that the 2s and 
all three 2p AOs are about equally involved in forming occupied MOs. That is, in the 
molecule, carbon behaves as though it were in the 2s2p3 configuration. This configuration 
(shortened to sp3) is referred to as the valence state of carbon in this molecule. 
Since this is an “open-shell” configuration (i.e., not all the electrons are spin-paired 
in filled orbitals), there are several actual physical states (corresponding to different 
spin and orbital angular momenta of an isolated carbon atom) that are associated with 
this configuration. Thus, there are several real physical states, with different ionization 
energies, associated with our mentally constructed sp3 valence state for the atom in a 
molecule. The question is, what real ionization energies should we use to evaluate our 
valence state ionization energy (VSIE)? The approach that is used is simply to average 
the real IEs for loss of a 2p or a 2s electron, the average being taken over all states 
associated with the sp3 configuration. Various authors recommend slightly different 
sets of VSIEs.4 We use here the values tabulated by Pople and Segal [7]. Because of 
the rather crude nature of the EH method, the slight variations in VSIE resulting from 
different choices are of little consequence.5 For methane, we have 
(C2s):H11 = -19.44eV=-0.7144 a.u. (10-1) 
(C2p):H22 = H33 =H44=-10.67eV=-0.3921 a.u. (10-2) 
(H1s):H55 = H66 =H77 =H88=-13.60eV=-0.50000 a.u. (10-3) 
The off-diagonal elements of H are evaluated according to6 
Hij =KSij 
Hii +Hjj 
2 
 (10-4) 
where K is an adjustable parameter. The rationalization for such an expression is that 
the energy of interaction should be greater when the overlap between AOs is greater, 
and that an overlap interaction energy between low-energy AOs should be lower than 
that produced by an equal amount of overlap between higher-energy AOs. We will 
discuss the energy versus overlap relation in more detail in a later section. The value 
of K suggested by Hoffmann [1] is 1.75. The reasons for choosing this value will be 
discussed shortly. For now, we accept this value and arrive at the hamiltonian matrix 
given in Table 10-4. 
By examiningHwe can guess in advance some of the qualitative features of the MOs 
that will be produced. For instance, the value of H25 (-0.3790 a.u.) indicates a strong 
energy-lowering interaction between the 2pz AO and the 1sAO on Ha. This interaction 
refers to AOs with positive and negative lobes as they are assigned in the basis set. 
Therefore, we expect a low-energy (bonding) MO to occur where theseAOs are mixed 
with coefficients of the same sign so as not to affect thisAOsign relation. Ahigh-energy 
MO should also exist where the mixing occurs through coefficients of opposite sign, 
producing an antibonding interaction. The values of H26,H27, and H28 are positive, 
4See Skinner and Pritchard [3], Hinze and Jaff´e [4], Basch et al. [5], Anno [6], and Pople and Segal [7]. 
5The proper valence state for carbon in methane differs from that in ethylene, which in turn differs from that 
in acetylene. Generally, this is ignored in EHMO calculations and a compromise set of VSIEs is selected for use 
over the whole range of molecules. 
6This formula is often called theWolfsberg–Helmholtz relation.
328 Chapter 10 The Extended H¨uckel Method 
TABLE 10-4 
 The Extended H¨uckel Hamiltonian Matrix for CH4a 
1 2 3 4 5 6 7 8 
1 -0.7144 0.0 0.0 0.0 -0.5454 -0.5454 -0.5454 -0.5454 
2 0.0 -0.3921 0.0 0.0 -0.3790 0.1263 0.1263 0.1263 
3 0.0 0.0 -0.3921 0.0 0.0 -0.3573 0.1787 0.1787 
4 0.0 0.0 0.0 -0.3921 0.0 0.0 -0.3094 0.3094 
5 -0.5454 -0.3790 0.0 0.0 -0.5000 -0.1579 -0.1579 -0.1579 
6 -0.5454 0.1263 -0.3573 0.0 -0.1579 -0.5000 -0.1579 -0.1579 
7 -0.5454 0.1263 0.1787 -0.3094 -0.1579 -0.1579 -0.5000 -0.1579 
8 -0.5454 0.1263 0.1787 0.3094 -0.1579 -0.1579 -0.1579 -0.5000 
aAll energies in a.u. 
due to negative overlap in corresponding positions of S. In this case, energy lowering 
will be associated with mixing 2pz with 1s AOs on hydrogens b, c, and d, but now 
the mixing coefficients will have signs that reverse the AO sign relations from those 
pertaining in the original basis. TheAOs that are orthogonal have zero interaction, and 
so mixing between such AOs will not affect MO energies. (When such AOs are mixed 
in the same MO, it is often the result of arbitrary mixing between degenerate MOs. In 
such cases, one can find an orthogonal pair of MOs such that two noninteracting AOs 
do not appear in the same MO. Methane will be seen to provide an example of this.) 
10-1.E The Eigenvalues and Eigenvectors 
Having H and S, we now can use the appropriate matrix-handling programs to solve 
HC=SCE for the matrix eigenvalues on the diagonal of E and the coefficients for the 
MOs, which are given by the columns of C. The eigenvalues for methane, together 
with their occupation numbers, are given in Table 10-5. The corresponding coefficients 
are given in Table 10-6. 
TABLE 10-5 
 Energies for Methane by the 
Extended H¨uckel Method 
MO no. Energy (a.u.) Occ. no. 
8 1.1904 0 
7 0.2068 0 
6 0.2068 0 
5 0.2068 0 
4 -0.5487 2 
3 -0.5487 2 
2 -0.5487 2 
1 -0.8519 2
Section 10-1 The Extended H¨uckel Method 329 
TABLE 10-6 
 Coefficients Defining MOs for Methane 
MO number 
1 2 3 4 5 6 7 8 
1(2s) 0.5842 0.0 0.0 0.0 0.0 0.0 0.0 1.6795 
2(2pz) 0.0 0.5313 -0.0021 -0.0007 -0.0112 -0.0137 1.1573 0.0 
3(2px) 0.0 0.0021 0.5313 -0.0021 1.1573 -0.0178 0.0110 0.0 
4(2py) 0.0 0.0007 0.0021 0.5313 0.0176 1.1572 0.0139 0.0 
5(1sa) 0.1858 0.5547 -0.0022 -0.0007 0.0105 0.0128 -1.0846 -0.6916 
6(1sb) 0.1858 -0.1828 0.5237 -0.0019 -1.0260 0.0114 0.3518 -0.6916 
7(1sc) 0.1858 -0.1853 -0.2589 0.4542 0.4943 -0.8977 0.3558 -0.6916 
8(1sd) 0.1858 -0.1865 -0.2626 -0.4516 0.5213 0.8734 0.3770 -0.6916 
Only two of the eight MOs are nondegenerate. These two MOs must be symmetric 
or antisymmetric for every symmetry operation of the molecule. This is easily 
checked by sketching the MOs, referring to the coefficients in the appropriate columns 
of Table 10-6. The lowest nondegenerate energy occurs in position 1 of our eigenvalue 
list (Table 10-5), and so the coefficients for this MO are to be found in column 1 of 
Table 10-6. This column indicates that theMO f1 is equal to 0.5842 2s+0.1858 1sa + 0.1858 1sb +0.1858 1sc +0.1858 1sd , where the symbols 2s, 1sa, etc., stand for AOs 
on carbon, Ha, etc. A sketch of thisMOappears in Fig. 10-2. It is obviously symmetric 
for all rotations and reflections of a tetrahedron. Notice that this MO is bonding in all 
four C–H bond regions since the 2s STO on carbon is in phase agreement with all the 
hydrogen 1s AOs. The higher-energy, nondegenerate MO f8 is qualitatively similar to 
f1 except that the signs are reversed on the 1s AOs (see Table 10-6). Hence, this MO 
has the same symmetry properties as f1, but is antibonding in the C–H regions. 
Figure 10-2 
 A drawing of the lowest-energy nondegenerate EHMO for methane. The AOs are 
drawn as though they do not overlap. This is done only to make the drawing simpler. Actually, the 
AOs overlap strongly.
330 Chapter 10 The Extended H¨uckel Method 
Since these two MOs resemble the 2s STO in being symmetric for all the symmetry 
operations of a tetrahedron, we will refer to them as s-type MOs. 
The remaining six MOs are grouped into two energy levels, each level being triply 
degenerate. (It is possible to predict from symmetry considerations alone that the energy 
levels resulting from this calculation will be nondegenerate and triply degenerate. This 
is discussed in a later chapter.) Because they are degenerate, the MOs cannot be 
expected to show all the symmetry of the molecule, but it should be possible for them to 
show some symmetry. Consider f2, as given by column 2 of Table 10-6. This is mainly 
constructed from the 2pz AO on carbon and 1s AOs on the four hydrogens. Small 
contributions from 2px and 2py are also present, however. It would be nice to remove 
these small contributions and “clean up” the MO.We can do this, as mentioned earlier, 
by mixing f2 with appropriate amounts of f3 and f4 since these are all degenerate. 
The equations that our cleaned-up MO, f2
, must satisfy are 
f2 
=d2f2 +d3f3 +d4f4 (10-5) 
where (in order to cause all 2px and 2py contributions to vanish) 
0.0021d2 +0.5313d3 +-0.0021d4=0.0 
0.0007d2 +0.0021d3 + 0.5313d4=0.0 (10-6) 
d2
2 + d2
3 + d2
4=1 
As a result, 
f2 
=0.9999f2 -0.0040f3 -0.0013f4 (10-7) 
A similar procedure to produce an orbital with no pz or py contribution (f3
), and one 
with no pz or px contribution (f4
) gives 
f3 
= 0.9999f3 +0.0040f2 -0.0040f4 (10-8) 
f4 
= 0.9999f4 +0.0013f2 +0.0039f3 (10-9) 
The coefficients for these MOs appear in Table 10-7. 
TABLE 10-7 
 Coefficients for MOs f2
,f3
,f4 
f2
f3f4 
2s 0.0 0.0 0.0 
2pz 0.5313 0.0 0.0 
2px 0.0 0.5313 0.0 
2py 0.0 0.0 0.5313 
1sa 0.5547 0.0 0.0 
1sb -0.1849 0.5228 0.0 
1sc -0.1849 -0.2614 0.4529 
1sd -0.1849 -0.2614 -0.4529
Section 10-1 The Extended H¨uckel Method 331 
Figure 10-3 
 The three lowest-energy degenerate MOs of methane. 
There is no fundamental change produced by intermixing degenerate MOs in this 
way. The total electronic density and the orbital energies are uninfluenced. The only 
advantage is that the cleaned-up MOs are easier to sketch and visualize. The MOs f2
, 
f3
, and f4
are sketched in Fig. 10-3. 
Each of the MOs in Fig. 10-3 is symmetric or antisymmetric for some of the operations 
that apply to a tetrahedron. f2
is symmetric for rotations about the z axis by 2p/3, 
and also for reflection through the xz plane. This same reflection plane is a symmetry 
plane for f3
and f4
, but neither of these MOs shows symmetry or antisymmetry for 
rotation about the z axis. Each MO contains one pAO and, perforce, has the symmetry 
of that AO.We shall refer to these as p-type MOs. Note that hydrogen 1s AOs lying in 
the nodal plane of a pAO do not mix with that pAO in formation of MOs. This results 
from zero interaction elements in H, which, in turn, results from zero overlap elements 
in S. Note also that the MO f2
is the MO that we anticipated earlier on the basis of 
inspection of the matrix H. Because of phase agreements between the hydrogen 1s 
AOs and the adjacent lobes of the p AOs, f1
, f2
, and f3
are C–H bonding MOs. 
A similar “cleaning up” procedure can be performed on f5, f6, and f7. These turn 
out to be the C–H antibonding mates to the MOs in Fig. 10-3. 
The broad results of this calculation are that there are four occupied C–H bonding 
MOs, one of s type and three of p type. At higher energies are four unoccupied C–H 
antibonding MOs, again one of s type and three of p type. 
Note that the s- and the three p-type MOs fall into the same energy pattern as the s and 
pAOs of isolated carbon. Because of their highly symmetric tetrahedral geometry, the 
hydrogen atoms do not lift the degeneracy of the pAOs. There are many molecules and 
complexes in which a cluster of atoms or molecules surrounds a central atom in such 
a highly symmetric way that the degeneracies among certain AOs on the central atom 
are retained. 
10-1.F The Total Energy 
The total EH energy is taken as the sum of the one-electron energies. For methane, this 
is 2×(-0.8519)+6×(-0.5487), or-4.9963 a.u. There is some ambiguity as to how 
this energy is to be interpreted. For instance, does it include any of the internuclear 
repulsion energy? Also, what problems will arise from our neglect of inner-shell
332 Chapter 10 The Extended H¨uckel Method 
electrons? By comparing EH total energy changes with experimental energy changes, 
it has been decided7 that the change in EH total energy upon change of geometry is 
approximately the same as the actual change in total electronic plus nuclear repulsion 
energy for the system. Thus, our value of -4.9963 a.u. is not a realistic value for 
the total (nonrelativistic) energy of methane (the actual value is -40.52 a.u.), but it 
is meaningful when compared to EH energies for methane at other geometries. For 
example, if we uniformly lengthen or shorten all the C–H bonds in methane and repeat 
our EH calculation several times, we can generate an energy curve versus RC–H for 
the symmetrical stretch vibrational mode of methane. The resultant plot is given in 
Fig. 10-4. The EH total energy is minimized at about RC–H =1 Å, reasonably close to 
the experimentally observed 1.1- Å distance for the minimum total energy of methane. 
The appearance of the curve in Fig. 10-4 does encourage us to equate EH total energy 
changes to changes in actual electronic-plus-nuclear-repulsion energies. As we shall 
see later, this procedure fails for some molecules (notably H2) and for methane may be 
regarded as fortuitous. 
Figure 10-4 
 Total extended H¨uckel energy for CH4 as a function of C–H bond length. 
7See Hoffmann [1].
Section 10-1 The Extended H¨uckel Method 333 
Figure 10-5 
 Staggered ethane MO energies versus K. C–H distances are 1.1 Å; C–C distance 
is 1.54 Å; all angles are tetrahedral. 
10-1.G Fixing the Parameter K 
We mentioned earlier that Hoffmann suggested a value of 1.75 for K. We will now 
indicate the considerations behind this suggestion. 
Hoffmann used the ethane molecule C2H6 to evaluate K. A plot of the orbital 
energies of staggered ethane as a function of K is shown in Fig. 10-5. The energies are 
linearly dependent on K at values of K greater than about 1.5. At lower values, the 
lines curve and some crossing occurs. Hence, in order that the MO energy order not be 
highly sensitive to K, its value should exceed 1.5. A plot of the amount of electronic 
charge in a bond or at an atom in ethane, as calculated by the EHMO method, versus K 
is shown in Fig. 10-6. We will describe the details of such calculations shortly, but for 
now, we merely note that the disposition of charge in ethane becomes rather insensitive 
to K at values of K greater than about 1.5. In Fig. 10-7 a plot of the EH total energy 
difference between staggered and eclipsed ethanes versus K is given. This energy 
difference has an experimentally determined value of 2.875 ± 0.025 kcal/mole, the 
staggered form being more stable. To give reasonable agreement with this experimental 
value, K should be about 1.75. Thus, the value K =1.75 is selected because the MO 
energy order and charge distribution are not sensitive to K in this region and because 
this value of K gives the correct total energy change for a known physical process. We 
have also seen earlier that this same value of K leads to a reasonable prediction for the 
equilibrium bond length inCH4. Notice that the evaluation ofK comes by matching the
334 Chapter 10 The Extended H¨uckel Method 
Figure 10-6 
 Mulliken gross population onH[N(H)] and Mulliken overlap populations inC–Cand 
C–H bonds [n(C–C) and n(C–H)] of ethane as calculated by the EH method with various values of K. 
Figure 10-7 
 Extended H¨uckel energy difference between staggered and eclipsed ethanes as a 
function of K.
Section 10-2 Mulliken Populations 335 
EH total energy change to the total (nuclear repulsion plus electronic) observed energy 
change for internal rotation in ethane. This is consistent with our earlier interpretation 
of EH total energy. 
10-2 Mulliken Populations 
We found that the electron densities and bond orders calculated in the simple H¨uckel 
method were extremely useful for relating theory to observable molecular properties 
such as electron spin-resonance splittings or bond lengths. Hence, it is desirable that 
we find analogous quantities to describe the distribution of electrons in an all-valenceelectron 
method like the EH method. A number of suggestions have been made. The 
one we use is due to Mulliken [8]. It is the most widely used, and, as we shall see, it 
has an especially direct and useful connection with the EH method. 
Consider a real, normalized MO, fi , made up from two normalizedAOs, .j and .k: 
fi =cji.j +cki.k (10-10) 
We square this MO to obtain information about electronic distribution: 
f2
i =c2
ji.2 
j +c2
ki.2 
k +2cjicki.j.k (10-11) 
If we integrate Eq. (10-11) over the electronic coordinates, we obtain (since fi , .j , and 
.k are normalized) 
1=c2
ji +c2
ki +2cjickiSjk (10-12) 
where Sjk is the overlap integral between .j and .k. Mulliken suggested that one 
electron in fi should be considered to contribute c2
ji to the electron net AO population 
of .j , c2
ki to the population of .k, and 2cjickiSjk to the overlap population between 
.j and .k. If there are two electrons in fi , then these populations should be doubled. 
Let qi
j symbolize the net AO population of .j due to one electron in MOfi, and 
pi
jk symbolize the overlap population between .j and .k due to this same electron. The 
above example leads to the following general definitions: 
qi
j = c2
ji (10-13) 
pi
jk = 2cjickiSjk (10-14) 
We can now sum the contributions due to all the electrons present in the model system, 
obtaining a Mulliken net AO population qj for each AO .j , and a Mulliken overlap 
population pjk, for each distinct AO pair .j and .k: 
qj = 
MOs 

i 
nic2
ji =
MOs 

i 
niqi
j (10-15) 
Pjk = 2 
MOs 

i 
nicjickiSjk =
MOs 

i 
nipi
jk (10-16) 
Notice that the sum of all the netAO and overlap populations must be equal to the total 
number of electrons in the model system. (In the EH method, this is the total number
336 Chapter 10 The Extended H¨uckel Method 
of valence electrons.) This contrasts with the situation in the simple H¨uckel method 
where the sum of electron densities alone, exclusive of bond orders, equals the total 
number of p electrons. 
The Mulliken populations are useful indices of the location of electronic charge in the 
molecule and its bonding or antibonding nature. The contributions to such populations 
from one electron in MO f4
of methane are given below (data taken from Tables 10-3 
and 10-7): 
q4 
2py = (0.5313)2=0.2823 (10-17) 
q4 
1sc = q4 
1sd=(0.4529)2=0.2051 (10-18) 
p4 
2py-1sc = p4 
2py-1sd=2(0.5313)(0.4529)(0.3964)=0.1908 (10-19) 
p4 
1sc-1sd = 2(0.4529)(-0.4529)(0.1805)=-0.0740 (10-20) 
All the other populations for f4 are zero. These numbers show that an electron in this 
MOcontributes about 28%, of an electron to the carbon 2py netAOpopulation, 20–21% 
to each of two hydrogen AOs, 19% to each of two C–H bonds, and a negative 7% to 
the region between Hc and Hd . The last corresponds to an antibonding interaction, as 
is obvious from an examination of the sketch of f4 in Fig. 10-3. (A negative overlap 
population is interpreted to mean that the amount of charge in the overlap region is less 
than what would exist if one squared the two AOs and then combined them.) 
By summing over the contributions due to all eight valence electrons, we obtain the 
Mulliken populations shown in Table 10-8. These data are part of the normal output of 
an EH computer program. The matrix is symmetric, and only the unique elements are 
tabulated. Such a matrix is named a Mulliken overlap population matrix. 
The data in Table 10-8 indicate that the hydrogens have positive overlap populations 
with the 2s and 2pAOs of carbon but have small negative overlap populations with each 
other. This corresponds to saying that the hydrogens interact with the carbon AOs in a 
bonding way and in a weakly antibonding way with each other. The symmetry equivalence 
of the four hydrogen atoms results in their having identical net AO populations. 
Often we are interested in knowing how the hydrogen atoms interact with the carbon 
in toto, rather than with the 2s and 2p AOs separately. This can be obtained simply 
TABLE 10-8 
 Mulliken Net AO and Overlap Populations for Methane as Computed by the 
Extended H¨uckel Method 
2s 2pz 2px 2py 1sa 1sb 1sc 1sd 
2s 0.6827 0.0 0.0 0.0 0.2229 0.2229 0.2229 0.2229 
2pz 0.5645 0.0 0.0 0.5723 0.0636 0.0636 0.0636 
2px 0.5645 0.0 0.0 0.5087 0.1272 0.1272 
2py 0.5645 0.0 0.0 0.3815 0.3815 
1sa 0.6844 -0.0491 -0.0491 -0.0491 
1sb 0.6844 -0.0491 -0.0491 
1sc 0.6844 -0.0491 
1sd 0.6844
Section 10-2 Mulliken Populations 337 
TABLE 10-9 
 Reduced Net AO and Overlap Population Matrix for Methane 
C Ha Hb Hc Hd 
C 2.3762 0.7952 0.7952 0.7952 0.7952 
Ha 0.6844 -0.0491 -0.0491 -0.0491 
Hb 0.6844 -0.0491 -0.0491 
Hc 0.6844 -0.0491 
Hd 0.6844 
by summing all the carbon atom AO contributions together to give a reduced overlap 
population matrix, shown in Table 10-9. The reduced population matrix makes evident 
the equivalence of the four C–H bonds, all of which have a total overlap population of 
0.7952 electrons. 
The Mulliken population scheme described above assigns some electronic charge to 
AOs, the rest to overlap regions. An alternative scheme, which assigns all the charge 
to AOs, was also proposed by Mulliken. One simply divides each overlap population 
in half, assigning half of the charge to each of the two participating AOs. When all 
the overlap populations have been reassigned in this way, the electronic charge is all 
in the AOs, and the sum of these AO charges still equals the total number of electrons. 
Mulliken called theAOpopulations resulting from this procedure grossAOpopulations. 
We will use the symbol N(X) for the gross population in X, where X can be anAO or 
an atom (i.e., the sum of all gross AO populations on one atom). Clearly 
N(X)=qx + 
1
2
j =x 
pxj (10-21) 
The grossAOpopulations, gross atomic populations, and the resultant atomic charges 
(obtained by combining electronic and nuclear charges) for methane are listed in 
Table 10-10. These data suggest that the carbon has lost a very small amount of charge 
to hydrogen upon formation of the molecule. It would be very risky, however, to place 
much faith in such an interpretation. It turns out that populations are rather sensitive to 
choice of VSIEs. For example, Hoffmann’s original choice of VSIEs differs from that 
TABLE 10-10 
 Gross AO Populations, Gross Atomic 
Populations, and Net Atomic Charges for Methane 
Gross AO Gross atom Net atomic 
population population charge 
C2s 1.128 3.966 +0.0334 
C2p
a 0.946 
Ha 1.008 1.008 -0.0083 
aAll 2pAOs and all HAOs have identical values because they are 
equivalent through symmetry.
338 Chapter 10 The Extended H¨uckel Method 
used here, and he obtained gross populations for hydrogen atoms of around 0.9, giving 
net positive charges of 0.1. Thus, absolute values are not very useful, but changes 
in gross population as we go from one hydrogen to another in the same hydrocarbon 
molecule or in closely related molecules do appear to be rather insensitive to VSIE 
choice and are often in accord with results of more accurate calculations. 
10-3 Extended H¨uckel Energies and Mulliken 
Populations 
In Chapter 8 it was shown that a simple quantitative relation exists between the energy 
of a simple H¨uckel MO and the contributions of the MO to bond orders [Eq. (8-56)]. 
The more bonding such an MO is, the lower is its energy. A similar relationship will 
now be shown to hold for extended H¨uckel energies and Mulliken populations. 
The orbital energy for the real MO fi is (ignoring the spin variable) 
Ei =  fiHˆ fi dv 
 f2
i dv 
(10-22) 
= 
AOs 
j,k cjickiHjk 
AOs 
j,k cjickiSjk 
(10-23) 
Assume that fi is normalized: The denominator is unity. The numerator of 
Eq. (10-23) contains diagonal and off-diagonal terms. If we separate these, and 
substitute the relation (10-4) for the off-diagonal terms, we obtain 
Ei = 
AOs 

j,k 
c2
jiHjj +2 
AOs 

j<k 
cjickiKSjk
Hjj +Hkk 
2 
(10-24) 
Comparison with Eqs. (10-13) and (10-14) shows that the first sum contains contributions 
to net AO populations and the second sum contains overlap population contributions. 
That is, 
Ei = 
AOs 

j 
qi
jHjj + 
AOs 

j<k 
pi
jkK
Hjj +Hkk 
2 
(10-25) 
This equation indicates that the energy of an extendedH¨uckelMOis equal to its net contributions 
to AO populations times AO energy weighting factors plus its contributions 
to overlap populations times overlap energy weighting factors. 
By summing over all MOs times occupation numbers, we arrive at a relation between 
total EH energy and net AO and overlap populations: 
E = 
AOs 

j 
qjHjj + 
AOs 

j<k 
pjkK
Hjj +Hkk 
2 
(10-26) 
Equations (10-25) and (10-26) are useful because they permit us to understand 
computed energies in terms of electron distributions. This is helpful when we seek to 
understand energy changes which occur as molecules are stretched, bent, or twisted.
Section 10-3 Extended H¨uckel Energies and Mulliken Populations 339 
There is an important difference between the extended H¨uckel formulas (10-25) and 
(10-26) and their simple H¨uckel counterparts. In a simple H¨uckel MO, the charge 
density contributions qi always add up to unity, contributing a to the MO energy 
(omitting heteroatom cases). The deviation of theMOenergy from a is thus due only to 
the bond-order contributions pi . As a result, for hydrocarbons, the simple H¨uckel MO 
that is the more bonding of a pair is always lower in energy. This is not so, however, in 
extended H¨uckel MOs. Since the sum of AO and overlap populations must equal the 
number of electrons present, we can increase the total amount of overlap population 
only at the expense of net AO populations. Therefore, energy lowering due to the 
second sum of Eq. (10-25) is purchased at the expense of that due to the first sum. This 
complicates the situation and requires that we take a more detailed look before issuing 
any blanket statements. 
The following simple example provides a convenient reference point for discussion. 
Consider two identical 1sAOs, .a and .b, on identical nuclei separated by a distance R. 
TheseAOs combine to form MOs of sg and su symmetry. The sg MOhas an associated 
positive overlap population and equal positive net AO populations for .a and .b. If R 
now is decreased slightly, the overlap population increases slightly (say by 2d), so the 
net AO populations must each decrease by d. The energy change for the MO, then, is 
Esg =-d (Haa +Hbb)+2dK 
Haa +Hbb 
2 
(10-27) 
Since the molecule is homonuclear, Haa =Hbb, and 
Esg =2dHaa (K -1) (10-28) 
IfK>1,Esg is negative (since Haa is negative). This analysis shows that the choice 
of 1.75 as the value for K has the effect of making the increase in overlap population 
dominate the energy change. If K were less than unity, the netAO population changes 
would dominate. 
The above example suggests that the EHMO method lowers the energy by maximizing 
weighted overlap populations at the expense of net AO populations. It also 
suggests that the EH energy should be lowered whenever a molecule is distorted in a 
way that enables overlap population to increase. These are useful rules of thumb, but 
some caution must be exercised since the existence of several different kinds of atom 
in a molecule leads to a more complicated relation than that in the above example. 
Our methane example illustrates the above ideas. We have already seen that the total 
EH energy for methane goes through a minimum around RC–H =1 Å. Let us see how 
the individual MO energies change as a function of RC–H and try to rationalize their 
behaviors in terms of the overlap population changes. Aplot of theMOenergies is given 
in Fig. 10-8. The lowest-energyMOis s type and C–H bonding. The overlap betweenH 
1sAOs and theC2s STOincreases asRC–H decreases. As a result, the energy of thisMO 
decreases asRC–H decreases, favoring formation of the united atom. The second-lowest 
energy level is triply degenerate and belongs to p-type C–H bonding MOs. A glance at 
Fig. 10-3 indicates that overlap between AOs will first increase in magnitude as RC–H 
decreases. But ultimately, at small RC–H this behavior must be reversed because the 1s 
and 2pAOs are orthogonal when they are isocentric. Therefore, as RC–H decreases, the 
overlap population first increases, then decreases toward zero. Consequently, the EH 
energy of this level first decreases, then increases. Since there are six electrons in this
340 Chapter 10 The Extended H¨uckel Method 
Figure 10-8 
 Extended H¨uckel MO energies for methane as a function of RC–H. 
level and only two in the lowest-energy MO, this one dominates the total energy and is 
responsible for the eventual rise in total energy at small R that creates the minimum in 
Fig. 10-4. The two remaining (unoccupied) levels belong to antibonding MOs, and the 
overlap population becomes more negative as RC–H decreases. The p-type level rises 
less rapidly and eventually passes through a maximum (not shown), again because the 
overlap population must ultimately approach zero as RC–H approaches zero. 
It is important to remember that all these remarks apply to the EH method only. The 
relationships between the EH method and other methods or with experimental energies 
is yet to be discussed. 
10-4 Extended H¨uckel Energies and Experimental 
Energies 
We have seen that a change in EH energy reflects certain changes in calculated Mulliken 
populations. We now consider the circumstances that should exist in order that such 
EH energy changes agree roughly with actual total energy changes for various systems.
Section 10-4 Extended H¨uckel Energies and Experimental Energies 341 
In essence, there are two requirements. First, the population changes calculated 
by the EH method ought to be in qualitative agreement with the charge shifts that 
actually occur in the real system. This condition is not always met. Some systems, 
especially those having unpaired electrons, are not accurately representable by a singleconfiguration 
wavefunction (that is, by a single product or a single Slater determinant). 
An example of this was seen for some 1s2s states of helium (see Chapter 5). Since the 
EHMO method is based on a single configuration, it is much less reliable for treating 
such systems. Special methods exist for handling such systems, but they normally are 
not applied at the H¨uckel level of approximation. 
The second requirement is that, as the charge shifts in the real system, the total 
energy should change in the way postulated by the EH method. For instance, if the 
charge increases in bond regions, this should tend to lower the total energy. In actuality, 
this does not always happen. We can see this by returning to our example of two 
identical 1sAOs separated byR. The associated sg MOincreases its overlap population 
monotonically asR decreases. Therefore, theEHpostulate predicts that the total energy 
should decrease monotonically. It is clear, however, that this does not happen in any 
real system. If our system is H2, we know that the experimental energy decreases 
with decreasing R until R =Re, and increases thereafter (Fig. 7-18). If our system is 
He2+ 2 , the experimental total energy increases as R decreases, due to the dominance 
of internuclear repulsion.8 Consideration of many examples such as these indicates 
that the postulate is usually qualitatively correct when we are dealing with interactions 
between neutral, fairly nonpolar systems that are separated by a distance greater than 
a bond length typical for the atoms involved. These, then, are circumstances under 
which the EHMO method might be expected to be qualitatively correct. 
In brief, we find two kinds of condition limiting the applicability of the EH method. 
The first is that the system must be reasonably well represented by a single configuration. 
Hence, closed-shell systems are safest. The second condition is that we apply the 
method to uncharged nonpolar systems where the nuclei that are undergoing relative 
motion are not too close to each other. 
These conditions are often satisfied by molecules undergoing internal rotation about 
single bonds. Thus, it is indeed appropriate that the internal rotation barrier in ethane 
was used to help calibrate the method. A comparison of other EH calculated barriers 
with experimental values (Table 10-11) gives us some idea of the capabilities and 
limitations of the method. For all the molecules listed, the most stable conformation 
predicted by the EH method agrees with experiment. This suggests that one can place 
a fair amount of reliance on the conformational predictions of the method, at least 
for threefold symmetric rotors. Also, certain gross quantitative trends, such as the 
significant reduction as we proceed down the series ethane, methylamine, methanol, are 
displayed in the EH results, but it is evident that the quantitative predictions of barrier 
height are not very good. This is not too surprising since the method was calibrated on 
hydrocarbons. Introduction of halogens, nitrogen, or oxygen produces greater polarity 
and might be expected to require a different parametrization. 
For reasons outlined earlier, we expect even less accuracy in EH calculations of 
molecular deformations involving bond-angle or bond-length changes. This expectation 
is generally reinforced by calculation. Molecular shape predictions become poorer 
8Actually, He2+ 2 has a minimum at short R in its energy curve due to an avoidance of curve crossing (discussed 
in Chapter 14), but this minimum is unstable with respect to the separated ions. See Pauling [9].
342 Chapter 10 The Extended H¨uckel Method 
TABLE 10-11 
 Energy Barriers for Internal Rotation about Single Bondsa 
Barrier (kcal/mole)b 
Molecule Calculated Experiment 
CH3–CH3 3.04 2.88 
CH3–NH2 1.66 1.98 
CH3–OH 0.45 1.07 
CH3–CH2F 2.76 3.33 
CH3–CHF2 2.39 3.18 
CH3–CF3 2.17 3.25 
CH3–CH2Cl 4.58 3.68 
CH3–CHCH2 1.20 1.99 
cis–CH3–CHCHCl 0.11 0.62 
CH3–CHO 0.32 1.16 
CH3–NCH2 0.44 1.97 
aCalculated barriers are for rigid rotation, where no bond length or angle changes occur 
except for the torsional angle change about the internal axis. 
bThe stable form for the first seven molecules has the methyl C–H bonds staggered with 
respect to bonds across the rotor axis. For the last four molecules, the stable form has 
a C–H methyl bond eclipsing the double bond. 
for more polar molecules (water is calculated to be most stable when it is linear), and 
bond-length predictions are quite poor too. 
Results such as these have tended to restrict use of the EH method to qualitative 
predictions of conformation in molecules too large to be conveniently treated by more 
accurate methods. However, just as the simple H¨uckel method underwent various 
refinements (such as the . technique) to patch up certain inadequacies, so has the EH 
method been refined. Such refinements9 have been shown to give marked improvement 
in numerical predictions of various properties. The EH method has been overtaken in 
popularity by a host of more sophisticated computational methods. (See Chapter 11.) 
However, it is still sometimes used as a first step in such methods as a way to produce a 
starting set of approximate MOs. The EHMO method also continues to be important as 
the computational equivalent of qualitative MO theory (Chapter 14), which continues 
to play an important role in theoretical treatments of inorganic and organic chemistries 
(as, for example, inWalsh’s Rules and inWoodward-Hoffmann Rules). 
10-4.A Problems 
The following output is produced by an EH calculation on the formaldehyde molecule, 
and is referred to in Problems 10-1 to 10-10. 
9See Kalman [10], Boyd [11], and the references cited in these papers. Also, see Anderson and Hoffmann [12] 
and Anderson [13].
Section 10-4 Extended H¨uckel Energies and Experimental Energies 343 
Formaldehyde (Ground State) Orbital Numbering 
Orbital Atom n l m x y z exp Hii 
1 H-1 1 0 0 -0.550000 0.952600 0.0 1.200–13.60 
2 H-2 1 0 0 -0.550000 -0.952600 0.0 1.200–13.60 
3 C-3 2 0 0 0.0 0.0 0.0 1.625–19.44 
4 C-3 2 1 0 0.0 0.0 0.0 1.625–10.67 
5 C-3 2 1 1 0.0 0.0 0.0 1.625–10.67 
6 C-3 2 1 1 0.0 0.0 0.0 1.625–10.67 
7 O-4 2 0 0 1.220000 0.0 0.0 2.275–32.38 
8 O-4 2 1 0 1.220000 0.0 0.0 2.275–15.85 
9 O-4 2 1 1 1.220000 0.0 0.0 2.275–15.85 
10 O-4 2 1 1 1.220000 0.0 0.0 2.275–15.85 
Hij =KSij (Hii +Hjj )/2, with K =1.75. 
Distance Matrix (a.u.) 
1 2 3 4 
1 0.0 3.6004 2.0787 3.7985 
2 3.6004 0.0 2.0787 3.7985 
3 2.0787 2.0787 0.0 2.3055 
4 3.7985 3.7985 2.3055 0.0 
Total effective nuclear repulsion =17.69537317 a.u. 
Formaldehyde Eigenvalues 
Eigenvalues (a.u.) Occ. no. Eigenvalues (a.u.) Occ. no. 
E(1)= 1.039011 0 E(6) =-0.587488 2 
E(2)= 0.472053 0 E(7) =-0.597185 2 
E(3)= 0.314551 0 E(8) =-0.611577 2 
E(4)=-0.342162 0 E(9) =-0.755816 2 
E(5)=-0.517925 2 E(10)=-1.242836 2 
Sum=-8.625654 a.u.
Total Overlap Matrix 
1 2 3 4 5 6 7 8 9 10 
1 1.0000 0.1534 0.5133 0.0 -0.2428 0.4204 0.0813 0.0 -0.0729 0.0392 
2 0.1534 1.0000 0.5133 0.0 -0.2428 -0.4204 0.0813 0.0 -0.0729 -0.0392 
3 0.5133 0.5133 1.0000 0.0 0.0 0.0 0.3734 0.0 -0.3070 0.0 
4 0.0 0.0 0.0 1.0000 0.0 0.0 0.0 0.2146 0.0 0.0 
5 -0.2428 -0.2428 0.0 0.0 1.0000 0.0 0.4580 0.0 -0.3056 0.0 
6 0.4204 -0.4204 0.0 0.0 0.0 1.0000 0.0 0.0 0.0 0.2146 
7 0.0813 0.0813 0.3734 0.0 0.4580 0.0 1.0000 0.0 0.0 0.0 
8 0.0 0.0 0.0 0.2146 0.0 0.0 0.0 1.0000 0.0 0.0 
9 -0.0729 -0.0729 -0.3070 0.0 -0.3056 0.0 0.0 0.0 1.0000 0.0 
10 0.0392 -0.0392 0.0 0.0 0.0 0.2146 0.0 0.0 0.0 1.0000 
Eigenvectors
1 2 3 4 5 6 7 8 9 10 
1 0.5279 0.7683 0.8924 0.0 -0.4281 -0.2016 0.0 -0.2141 -0.2721 0.0011 
2 0.5279 0.7683 -0.8924 0.0 0.4281 -0.2016 0.0 0.2141 -0.2721 0.0011 
3 -1.3964 -0.5553 0.0000 0.0 -0.0000 -0.0460 0.0 -0.0000 -0.4875 0.2550 
4 0.0 0.0 0.0 0.9940 0.0 0.0 0.2456 0.0 0.0 0.0 
5 -0.6043 1.1727 -0.0000 0.0 0.0000 0.2768 0.0 0.0000 0.2245 0.0685 
6 0.0000 -0.0000 -1.2519 0.0 -0.3813 0.0000 0.0 -0.3179 0.0000 0.0000 
7 0.8367 -0.4799 0.0000 0.0 0.0000 -0.0884 0.0 -0.0000 0.3066 0.8481 
8 0.0 0.0 0.0 -0.4532 0.0 0.0 0.9181 0.0 0.0 0.0 
9 -0.6960 0.3412 -0.0000 0.0 0.0000 -0.8317 0.0 -0.0000 0.3327 0.0252 
10 -0.0000 0.0000 0.2511 0.0 0.6475 0.0000 0.0 -0.7600 -0.0000 -0.0000 
344
Mulliken Overlap Populations for 12 Electrons 
1 2 3 4 5 6 7 8 9 10 
1 0.6876 -0.0702 0.2920 0.0 0.1134 0.3890 -0.0210 0.0 -0.0225 -0.0180 
2 -0.0702 0.6876 0.2920 0.0 0.1134 0.3890 -0.0210 0.0 -0.0225 -0.0180 
3 0.2920 0.2920 0.6097 0.0 0.0 0.0 0.1058 0.0 0.1443 0.0 
4 0.0 0.0 0.0 0.1207 0.0 0.0 0.0 0.1936 0.0 0.0 
5 0.1134 0.1134 0.0 0.0 0.2634 0.0 0.1876 0.0 0.1880 0.0 
6 0.3890 0.3890 0.0 0.0 0.0 0.4929 0.0 0.0 0.0 -0.0046 
7 -0.0210 -0.0210 0.1058 0.0 0.1876 0.0 1.6420 0.0 0.0 0.0 
8 0.0 0.0 0.0 0.1936 0.0 0.0 0.0 1.6857 0.0 0.0 
9 -0.0225 -0.0225 0.1443 0.0 0.1880 0.0 0.0 0.0 1.6061 0.0 
10 -0.0180 -0.0180 0.0 0.0 0.0 -0.0046 0.0 0.0 0.0 1.9939 
Charge Matrix for MOs with Two Electrons in Each 
1 2 3 4 5 6 7 8 9 10 
1 0.1664 0.3881 0.4266 0.0 0.4258 0.1088 0.0 0.1476 0.3363 0.0004 
2 0.1664 0.3881 0.4266 0.0 0.4258 0.1088 0.0 0.1476 0.3363 0.0004 
3 0.9171 0.0561 0.0000 0.0 0.0000 0.0028 0.0 0.0000 0.7358 0.2882 
4 0.0 0.0 0.0 1.7825 0.0 0.0 0.2175 0.0 0.0 0.0 
5 0.3198 1.1155 0.0000 0.0 0.0000 0.3257 0.0 0.0000 0.1775 0.0614 
6 0.0000 0.0000 1.1204 0.0 0.4593 0.0000 0.0 0.4203 0.0000 0.0000 
7 0.2082 0.0241 0.0000 0.0 0.0000 0.0021 0.0 0.0000 0.1123 1.6534 
8 0.0 0.0 0.0 0.2175 0.0 0.0 1.7825 0.0 0.0 0.0 
9 0.2220 0.0282 0.0000 0.0 0.0000 1.4518 0.0 0.0000, 0.3017 -0.0037 
10 0.0000 0.0000 0.0264 0.0 0.6891 0.0000 0.0 1.2845 0.0000 0.0000 
345
346 Chapter 10 The Extended H¨uckel Method 
Reduced Overlap Population Matrix Atom by Atom 
1 2 3 4 
1 0.6876 -0.0702 0.7945 -0.0615 
2 -0.0702 0.6876 0.7945 -0.0615 
3 0.7945 0.7945 1.4866 0.8148 
4 -0.0615 -0.0615 0.8148 6.9277 
Orbital Charges 
1 1.018973 6 0.879574 
2 1.018973 7 1.767672 
3 1.026764 8 1.782529 
4 0.217471 9 1.749778 
5 0.564637 10 1.973630 
Net Charges 
1 -0.018973 3 1.311555 
2 -0.018973 4 -1.273609 
Total charge = 0.000000. 
10-1. Use the output to determine the orientation of the molecule with respect to Cartesian 
coordinates. Sketch the molecule in relation to these axes and number the 
atoms in accord with their numbering in the output. 
10-2. Use the orbital numbering data together with the overlap matrix to figure out 
which of the labels 1s, 2s, 2px, 2py, 2pz goes with each of the ten AOs (i.e., it is 
obvious that AO 1 is a 1s AO, but it is not so obvious what AO 5 is). 
10-3. Use your conclusions from above, together with coefficients in the eigenvector 
matrix, to sketch the MOs having energies of -0.756, -0.611, and -0.597 a.u. 
Which of these are p MOs? Which are s MOs? 
10-4. Label each of the ten MOs “p” or “s” by inspecting the coefficient matrix. 
10-5. What is the Mulliken overlap population between C and O 2pp AOs in this 
molecule? Should removal of an electron from MO 7 cause the C=O bond to 
shorten or to lengthen? 
10-6. Demonstrate that MO 7 satisfies Eq. (10-24). (Note that Hii values in the first 
table are in units of electron volts, while orbital energies are in atomic units.) 
10-7. Use the reduced overlap population matrix to verify that the sum of AO and 
overlap populations is equal to the number of valence electrons. 
10-8. Using MO 7 as your example, verify that the charge matrix table is a tabulation 
of the contributions of each MO to gross atom populations. 
10-9. In the list of “orbital charges,” are the “orbitals” MOs orAOs? Demonstrate how 
these numbers are derived from those in the charge matrix.
Section 10-4 Extended H¨uckel Energies and Experimental Energies 347 
10-10. What is the physical meaning of the “net charges” in the data? Would you characterize 
these results as indicative of low polarity? Which end of the molecule 
should correspond to the negative end of the dipole moment? 
10-11. How many MOs will be produced by an EHMO calculation on butadiene? 
References 
[1] R. Hoffmann, J. Chem. Phys. 39, 1397 (1963). 
[2] C. E. Moore, Atomic Energy Levels, Natl. Bur. Std. (U.S.) Circ. 467. Natl. Bur. 
Std.,Washington D.C., 1949. 
[3] H. A. Skinner and H. O. Pritchard, Trans. Faraday Soc. 49, 1254 (1953). 
[4] J. Hinze and H. H. Jaff´e, J. Amer. Chem. Soc. 84, 540 (1962). 
[5] H. Basch, A. Viste, and H. B. Gray, Theoret. Chim. Acta 3, 458 (1965). 
[6] T. Anno, Theoret. Chim. Acta 18, 223 (1970). 
[7] J. A. Pople and G. A. Segal, J. Chem. Phys. 43, S136 (1965). 
[8] R. S. Mulliken, J. Chem. Phys. 23, 1833, 1841, 2338, 2343 (1955). 
[9] L. Pauling, J. Chem. Phys. 1, 56 (1933). 
[10] B. L. Kalman, J. Chem. Phys. 60, 974 (1974). 
[11] D. B. Boyd, Theoret. Chim. Acta 30, 137 (1973). 
[12] A. B. Anderson and R. Hoffmann, J. Chem. Phys. 60, 4271 (1974). 
[13] A. B. Anderson, J. Chem. Phys. 62, 1187 (1975).
Chapter 11 
The SCF-LCAO-MO Method 
and Extensions 
11-1 Ab Initio Calculations 
A rigorous variational calculation on a system involves the following steps: 
1. Write down the hamiltonian operator Hˆ for the system. 
2. Select some mathematical functional form . as the trial wavefunction. This form 
should have variable parameters. 
3. Minimize 
E¯ = 
 .*Hˆ .dt 

 .*.dt 
(11-1) 
with respect to variations in the parameters. 
The simple and extended H¨uckel methods are not rigorous variational calculations. 
Although they both make use of the secular determinant technique from linear variation 
theory, no hamiltonian operators are ever written out explicitly and the integrations in 
Hij are not performed. These are semiempirical methods because they combine the 
theoretical form with parameters fitted from experimental data. 
The term ab initio (“from the beginning”) is used to describe calculations in which 
no use is made of experimental data. In an ab initio variational method, all three steps 
listed above are explicitly performed. In this chapter we describe a certain kind of 
ab initio calculation called the self-consistent field (SCF) method. This is one of the most 
commonly encountered types of ab initio calculation for atoms or molecules. We also 
describe a few popular methods for proceeding beyond the SCF level of approximation. 
The SCF method and extensions to it are mathematically and physically considerably 
more complicated than the one-electron methods already discussed. Thus, one 
normally does not perform such calculations with pencil and paper, but rather with 
complicated computer programs. Therefore, in this chapter we are not concerned with 
how one does such calculations because, in most cases, they are done by acquiring a 
program written by a group of specialists. Rather we are concerned with a description 
of the mathematical and physical underpinnings of the method. Because the method 
is simultaneously complicated and rigorously defined, a special jargon has developed. 
Terms like “Hartree–Fock,” or “correlation energy” have specific meanings and are 
pervasive in the literature. Hence, a good deal of emphasis in this chapter is put on 
defining some of these important terms. 
348
Section 11-3 The Form of theWavefunction 349 
11-2 The Molecular Hamiltonian 
In practice, one usually does not use the complete hamiltonian for an isolated molecular 
system. The complete hamiltonian includes nuclear and electronic kinetic energy operators, 
electrostatic interactions between all charged particles, and interactions between 
all magnetic moments due to spin and orbital motions of nuclei and electrons. Also 
an accounting for the fact that a moving particle experiences a change in mass due to 
relativistic effects is included in the complete hamiltonian. The resulting hamiltonian 
is much too complicated to work with. Usually, relativistic mass effects are ignored, 
the Born–Oppenheimer approximation is made (to remove nuclear kinetic energy operators), 
and all magnetic interactions are ignored (except in special cases where we are 
interested in spin coupling). The resulting hamiltonian for the electronic energy is, in 
atomic units, 
Hˆ =-
1
2 
n 

i=1 
.2
i - 
N 

µ=1 
n 

i=1 
Zµ/rµi +
n-1 

i=1 
n 
 
j=i+1 
1/rij (11-2) 
where i and j are indices for the n electrons and µ is an index for the N nuclei. The 
nuclear repulsion energy Vnn is 
Vnn =
N-1 

µ=1 
N 
 
.=µ+1
ZµZ./rµ. (11-3) 
In choosing this hamiltonian, we are in effect electing to seek an energy of an 
idealized nonexistent system—a nonrelativistic system with clamped nuclei and no 
magnetic moments. If we wish to make a very accurate comparison of our computed 
results with experimentally measured energies, it is necessary to modify either the 
experimental or the theoretical numbers to compensate for the omissions in Hˆ . 
11-3 The Form of theWavefunction 
The wavefunction for an SCF calculation is one or more antisymmetrized products of 
one-electron spin-orbitals. We have already seen (Chapter 5) that a convenient way to 
produce an antisymmetrized product is to use a Slater determinant. Therefore, we take 
the trial function . to be made up of Slater determinants containing spin-orbitals f. 
If we are dealing with an atom, then the f’s are atomic spin-orbitals. For a molecule, 
they are molecular spin-orbitals. 
In our discussion of many-electron atoms (Chapter 5), we noted that certain atoms 
in their ground states are fairly well described by assigning two electrons, one of each 
spin, to each AO, starting with the lowest-energy AO and working up until all the 
electrons are assigned. If the last electron completes the filling of all the AOs having a 
given principal quantum number, n, we have a closed shell atomic system. Examples 
are He(1s2) and Ne(1s22s22p6). Atoms wherein the last electron completes the filling 
of all AOs having a given l quantum number are said to have a closed subshell. An 
example is Be(1s22s2). Both types of system tend to be well approximated by a single 
determinantal wavefunction if the highest filled level is not too close in energy to the
350 Chapter 11 The SCF-LCAO-MO Method and Extensions 
lowest empty level. (Beryllium is the least successfully treated of these three at this level 
of approximation because the 2s level is fairly close in energy to the 2p level.) A similar 
situation holds for molecules; that is, the wavefunctions of many molecules in their 
ground states are well represented by single determinantalwavefunctions with electrons 
of paired spins occupying identical MOs. Such molecules are said to be closed-shell 
systems. We can represent a trial wavefunction for a 2n-electron closed-shell system as 
.closed shell =
f1 (1) ¯ f1 (2)f2 (3) ¯ f2 (4) ···fn (2n-1) ¯ fn (2n)
(11-4) 
where we have used the shorthand form for a Slater determinant described in Chapter 5. 
For the present, we restrict our discussion to closed-shell single-determinantal wavefunctions. 
11-4 The Nature of the Basis Set 
Some functional form must be chosen for theMOsf. The usual choice is to approximate 
f as a linear combination of “atomic orbitals” (LCAO), theseAOs being located on the 
nuclei. The detailed nature of these AOs, as well as the number to be placed on each 
nucleus, is still open to choice. We consider these choices later. For now, we simply 
recognize that we are working within the familiar LCAO-MO level of approximation. 
If we represent the basis AOs by ., we have, for the ith MO, 
fi =
j 
cji.j (11-5) 
where the constants cji are as yet undetermined. 
11-5 The LCAO-MO-SCF Equation 
Having a hamiltonian and a trialwavefunction, we are nowin a position to use the linear 
variation method. The detailed derivation of the resulting equations is complicated 
and notationally clumsy, and it has been relegated to Appendix 7. Here we discuss the 
results of the derivation. 
For our restricted case of a closed-shell single-determinantal wavefunction, the variation 
method leads to 
Fˆfi =ifi (11-6) 
These equations are sometimes called the Hartree–Fock equations, and Fˆ is often 
called the Fock operator. The detailed formula for Fˆ is (from Appendix 7) 
Fˆ(1)=-
1
2.2
1 -
µ 
Zµ/rµ1 + 
n 

j=1
(2Jˆj -Kˆj ) (11-7) 
The symbols Jˆj and Kˆj stand for operators related to the 1/rij operators in Hˆ . Jˆj is 
called a coulomb operator because it leads to energy terms corresponding to charge 
cloud repulsions. It is possible to write Jˆj explicitly: 
Jˆj = f*j (2) 1/r12)fj (2)dt (2 (11-8)
Section 11-6 Interpretation of the LCAO-MO-SCF Eigenvalues 351 
Kˆj leads ultimately to the production of exchange integrals, and so it is called an 
exchange operator. It is written explicitly in conjunction with a function on which it 
is operating, viz. 
Kˆjfi(1)= f*j (2) (1/r12)fi(2)dt (2)fj (1) (11-9) 
Notice that an index exchange has been performed. It is not difficult to see that the 
expression (see Appendix 9 for bra-ket notation) 
fi |Fˆ|fi =i (11-10) 
will lead to integrals such as 
fi |Jˆj |fi  = fi (1)fj (2) |1/r12|fi (1)fj (2)=Jij (11-11) 
fi |Kˆj |fi  = fi (1)fj (2) |1/r12|fi (2)fj (1)=Kij (11-12) 
which are formally the same as the coulomb and exchange terms encountered in 
Chapter 5 in connection with the helium atom. Notice that, if the spins associated with 
spin-orbitals fi and fj differ, Kij must vanish. This arises because integrations over 
space and spin coordinates of electron 1 (or 2) in Eq. (11-12) lead to integration over 
two different (and orthogonal) spin functions. On the other hand, Jij is not affected by 
such spin agreement or disagreement. 
It would appear from Eq. (11-6) that the MOs f are eigenfunctions of the Fock 
operator and that the Fock operator is, in effect, the hamiltonian operator. There is an 
important qualitative difference between Fˆ andHˆ , however. The Fock operator is itself 
a function of the MOs f. Since the summation index j in Eq. (11-7) includes i, the 
operators Jˆi and Kˆi must be known in order to write down Fˆ, but Jˆi and Kˆi involve fi , 
and fi is an eigenfunction of Fˆ. Hence, we need Fˆ to find fi , and we need fi to know 
Fˆ. To circumvent this problem, an iterative approach is used. One makes an initial 
guess at the MOs f. (One could use a semiempirical method to produce this starting 
set.) Then these MOs are used to construct an operator Fˆ, which is used to solve for 
the new MOs f. These are then used to construct a new Fock operator, which is in 
turn used to find new MOs, which are used for a new Fˆ, etc., until at last no significant 
change is detected in two successive steps of this procedure. At this point, the f’s 
produced by Fˆ are the same as the f’s that produce the coulomb-and-exchange fields 
in Fˆ. The solutions are said to be self-consistent, and the method is referred to as the 
self-consistent-field (SCF) method. 
11-6 Interpretation of the LCAO-MO-SCF Eigenvalues 
The physical meaning of an eigenvalue i is best understood by expanding the integral 
i =fi |Fˆ|fi  (11-13) 
with Fˆ given by Eq. (11-7). We obtain 
i =fi 

-
1
2.2
1 

fi-
µ 
fi
Zu/rµ1
fi+ 
n 

j=1 
2Jij -Kij  (11-14)
352 Chapter 11 The SCF-LCAO-MO Method and Extensions 
It is common practice to combine the first two terms of Eq. (11-14), which depend 
only on the nature of fi , into a single expectation value of the one-electron part of the 
hamiltonian, symbolized Hii . Thus, 
i =Hii + 
n 

j=1
(2Jij -Kij ) (11-15) 
The quantity Hii is the average kinetic plus nuclear-electronic attraction energy for the 
electron in fi . 
The sum of coulomb and exchange integrals in Eq. (11-15) contains all the electronic 
interaction energy. Observe that the index j runs over all the occupied MOs. For a 
particular value of j , say j =k =i, this gives 2Jij -Kij as an interaction energy. This 
means that an electron in fi , experiences an interaction energy with the two electrons 
in fk of 
2 fi(1)fk(2)|1/r12|fi(1)fk(2)-fi(1)fk(2) |1/r12|fk(1)fi(2) (11-16) 
The first part is the classical repulsion between the electron having an orbital charge 
cloud given by |fi |2 and the two electrons having charge cloud |fk|2. The second part is 
the exchange term which, as we saw in Chapter 5, arises from the antisymmetric nature 
of the wavefunction. It enters (11-16) only once because the electron in fi , agrees in 
spin with only one of the two electrons in fk, [Equation (11-15) applies because we 
have restricted our discussion to closed-shell systems.] 
The summation over 2Jij -Kij includes the case j = i. Here we get 2Jii -Kii . 
However, examination of Eqs. (11-11) and (11-12) shows that Jii =Kii , and so we are 
left with Jii . This corresponds to the repulsion between the electron in fi (the energy 
of which we are calculating) and the other electron in fi . Because these electrons must 
occur with opposite spin, there is no exchange energy for this interaction. 
In brief, then, the quantity i , often referred to as an orbital energy or a one-electron 
energy, is to be interpreted as the energy of an electron in fi , resulting from its kinetic 
energy, its energy of attraction for the nuclei, and its repulsion and exchange energies 
due to all the other electrons in their charge clouds |fj |2. 
11-7 The SCF Total Electronic Energy 
It is natural to suppose that the total electronic energy is merely the sum of the oneelectron 
energies, but this is not the case in SCF theory. Consider a two-electron system. 
The energy of electron 1 includes its kinetic and nuclear attraction energies and its 
repulsion and exchange energies for electron 2. The energy of electron 2 includes its 
kinetic and nuclear attraction energies and its repulsion and exchange energies for electron 
1. If we sum these, we have accounted properly for kinetic and nuclear attraction 
energies, but we have included the interelectronic interactions twice as much as they 
actually occur. (The energy of repulsion, say, between two charged particles, 1 and 2, is 
given by the repulsion of 1 for 2 or of 2 for 1, but not by the sum of these.) Therefore, if 
we sum one-electron energies, we get the total electronic energy plus an extra measure
Section 11-8 Basis Sets 353 
of electron repulsion and exchange energy. We can correct this by subtracting this extra 
measure away. Thus, for our closed-shell system 
Eelec = 
n 

i=1 
.
.
2i - 
n 

j=1
(2Jij -Kij )
.
. 
(11-17) 
where the summation is over the occupied orbitals. Comparing Eq. (11-17) with (11-18) 
makes it evident that we can also write 
Eelec = 
n 

i=1 
.
.
2Hii + 
n 

j=1
(2Jij -Kij )
.
. 
(11-18) 
or 
Eelec = 
n 

i=1
(i +Hii) (11-19) 
To obtain the total (electronic plus nuclear) energy, we add the internuclear repulsion 
energy for the N nuclei: 
Etot = Eelec +Vnn (11-20) 
Vnn = 
N-1 

µ=1 
N 
 
.=µ+1 
ZµZ. 
rµ. 
(11-21) 
11-8 Basis Sets 
A great deal of research effort has gone into devising and comparing basis sets for 
ab initio calculations. There are essentially two important criteria: 
1. Wewant a basis set that is capable of describing the actualwavefunction well enough 
to give chemically useful results. 
2. Wewant a basis set that leads to integrals Fij and Sij that we can evaluate reasonably 
accurately and quickly on a computer. 
Many types of basis set have been examined and two of these have come to dominate 
the area of ab initio molecular calculations. These two, which we refer to as the gaussian 
and the Slater-type-orbital (STO) basis sets, are actually very similar in many important 
respects. 
Let us consider the STO basis set first. The essence of this basis choice is to place 
on each nucleus one or more STOs. The number of STOs on a nucleus and the orbital 
exponent of each STO remain to be chosen. Generally, the larger the number of STOs 
and/or the greater the care taken in selecting orbital exponents, the more accurate the 
final wavefunction and energy will be. 
At the least sophisticated end of the spectrum of choices is the minimal basis set of 
STOs, which we encountered in Chapter 7. This includes only those STOs that correspond 
to occupied AOs in the separated atom limit. If we choose a minimal basis set,
354 Chapter 11 The SCF-LCAO-MO Method and Extensions 
then we must still decide how to evaluate the orbital exponents in the STOs. One way 
is to use Slater’s rules, which are actually most appropriate for isolated atoms. Another 
way is to vary the orbital exponents until the energy of the molecular system is minimized. 
This amounts to performing a nonlinear variational calculation along with the 
linear variational calculation. For molecules of more than a few atoms, this procedure 
consumes much computer time, for reasons we will describe shortly, but for small 
molecules (two or three first-row atoms plus a few hydrogens) it is possible to accomplish 
this task. From this, one discovers what orbital exponent best suits an STO in a 
molecular environment. This leads us to the third way of choosing orbital exponents— 
choose the values that were found best for each type of atom in nonlinear variational 
calculations in smaller molecules. 
One may improve the basis by adding additional STOs to various nuclei. Suppose, 
for example, each carbon 2pAO were represented as a linear combination of two p-type 
STOs, each having a different orbital exponent. An example of the basic principle 
involved is indicated in Fig. 11-1. If we treat these functions independently and do a 
linear variational calculation, they will both be mixed into the final wavefunction to 
some degree. If the linear coefficient for the “inner” STO is much larger, it means that 
the p-type charge cloud around this atom in the molecule is calculated to be fairly contracted 
around the nucleus. To describe a more diffuse charge cloud, the wavefunction 
would contain quite a lot of the “outer” STO, and not so much of the “inner” STO. 
Thus, we have a linear variation procedure that, in effect, allows for AO expansion 
and contraction. It is akin to optimizing an orbital exponent, but it does not require 
nonlinear variation. Of course, one still has to choose the values of the “larger . ” and 
“smaller . ” of Fig. 11-1. This is normally done by optimizing the fit to very accurate 
atomic wavefunctions or by a nonlinear variation on atoms. A basis set in which every 
minimal basis AO is represented by an “inner-outer” pair of STOs is often referred to 
as a “double-zeta” basis set. 
Afurther kind of extension is frequently made. In addition to the above types of STO, 
one includes STOs with symmetries different from those present in the minimal basis. 
This has the effect of allowing charge to be shifted in or out of bond regions in new 
ways. For example, one could add p-type STOs on hydrogen nuclei. By mixing this 
with the s-type STOs there, one can describe a skewed charge distribution in the regions 
of the protons. We have already seen (Chapter 7) that a hydrogen atom in a uniform 
Figure 11-1 
 Radial functions R(r)=r exp(-.r) for 2p-type STOs. The larger . value gives an 
STO more contracted around the nucleus. Hence, it is sometimes called the “inner” STO.
Section 11-8 Basis Sets 355 
electric field is polarized in a way that is reasonably well described by an s-p linear 
combination. Since the hydrogen atom in a molecule experiences an electric field due 
to the remainder of the molecule, it is not surprising that such p functions are indeed 
mixed into the wavefunction by the variational procedure if we provide them in the 
basis set. Similarly, d-type STOs may be added to atoms that, in the minimal basis set, 
carried only s- and p-type STOs. Functions of this nature are often called polarization 
functions because they allow charge polarization to occur within the molecule as a 
result of the internally generated electric field. 
It should be evident that one could go on indefinitely, adding more and more STOs to 
the basis, even placing some of them in bonds, rather than on nuclei. This is not normally 
done because the computing task goes up enormously as we add more basis functions. 
In fact, the number of integrals to be calculated eventually increases as N4, where N is 
the number of basis functions. The evaluation of integrals can be a logistic bottleneck in 
ab initio calculations, and for this reason nonlinear variations (of orbital exponents) are 
impractical for any but smaller molecules. Each new orbital exponent value requires 
re-evaluation of all the integrals involving that orbital. In essence, a change of orbital 
exponent is a change of basis set. In linear variations, the basis functions are mixed 
together but they do not change. Once all the integrals between various basis functions 
have been evaluated, they are usable for the remainder of the calculation. 
The STO basis would probably be the standard choice if it were not for the fact 
that the many integrals encountered in calculating Fij elements are extremely time 
consuming to evaluate, even on a computer. This problem has led to the development of 
an alternative basis set class that is based on gaussian-type functions. Gaussian functions 
include an exponential term of the form exp(-ar2). The radial dependence of 
such a function is compared to that for a hydrogenlike 1s function (which is identical to 
a 1s STO) in Fig. 11-2. There are two obvious problems connected with using gaussian 
functions as basis functions: 
1. They do not have cusps at r =0 as s-type hydrogen-like AOs do. 
2. They decay faster at larger r than do hydrogen-like AOs. 
Both of these deficiencies are relevant in molecules because, at r =0 (on a nucleus) 
and at r=8, the molecular potential is like that in an atom, so similar cusp and asymptotic 
behavior are expected for molecular and atomic wavefunctions. Balanced against 
these deficiencies is an advantage: gaussian functions have mathematical properties 
that make it extremely easy to compute the integrals they produce in Fij . This has led 
to a practice of replacing each STO in a basis set by a number of gaussian functions. 
Figure 11-2 
 Radial dependence of hydrogen-like and gaussian functions.
356 Chapter 11 The SCF-LCAO-MO Method and Extensions 
By choosing several values of a in exp(-ar2), one can create a set of “primitive” 
gaussian functions ranging from very compact to very diffuse, and then take a linear 
combination of these to build up an approximation to the radial part of an STO function. 
Multiplication by the standard . and f dependences (spherical harmonics) generates 
p, d, etc. functions. Once this approximation is optimized, the linear combination of 
gaussian functions is “frozen,” being treated thereafter as a single function insofar as 
the subsequent molecular variational calculation is concerned. This linear combination 
of primitive gaussian functions is called a contracted gaussian function. 
Once we have a contracted gaussian function corresponding to each STO, we can go 
through the same hierarchy of approximations as before—minimal basis set, double- 
. basis set, double-. plus polarization functions-only now using contracted gaussian 
functions in place of STOs. Typically, ab initio calculations on systems involving only 
light elements, e.g., H–Ne, involve anywhere from 1 to 15 primitive gaussian functions 
for each contracted gaussian function. Basis sets for heavier elements, however, can 
contain more than 30 primitive gaussians for each contracted gaussian. 
We have already described a certain amount of quantum-chemical jargon. Some of 
the basis set descriptions that one commonly encounters in the modern literature are as 
follows: 
• DZP [double-. gaussian basis with polarization] 
• STO-3G [each STO approximated as a linear combination of three gaussian primitives] 
• 6-31G [each inner shell STOrepresented by a sum of six gaussians and each valence 
shell STO split into inner and outer parts (i.e., double-. ) described by three and 
one gaussian primitives, respectively] 
• 6-31G* [the 6-31G basis set augmented with six d-type gaussian primitives on each 
heavy (Z >2) atom, to permit polarization] 
• 6-31G** [same as 6-31G* but with a set of gaussian p-type functions on H and He 
atoms. Good for systems where hydrogen is a bridging atom, as in diborane or in 
hydrogen bonds] 
• 6-31+G* [the 6-31G* basis set augmented with a set of diffuse s- and p-type gaussian 
functions on each heavy atom, to permit representation of diffuse electronic 
distribution, as in anions] 
• cc-pVnZ, n=D,T, Q, 5 [correlation consistent polarized valence n-. gaussian basis 
sets. The inner shell STOs are described by single contracted gaussian functions 
while the valence STOs are described by n contracted gaussian functions, n=D 
for double-. , n = T for triple-. , etc. Both the number and angular momentum 
symmetry type of the polarization functions are increased with each successive 
correlation consistent basis set in a systematic manner. For example, the cc-pVDZ 
basis set has a set of 5 d-type gaussian primitive functions on each heavy atom, while 
the cc-pVTZ basis set has 2 sets of d-type gaussian functions and one set of 7 f-type 
gaussian primitives. These families of basis sets are designed to converge the total 
energy to the complete basis set limit for the SCF method and its extensions.]
Section 11-10 Correlation Energy 357 
• aug-cc-pVnZ [cc-pVnZ basis sets augmented with one set of diffuse primitive 
gaussian functions for each angular momentum symmetry present in the cc-pVnZ 
basis set, to provide an accurate description of anions and weak interactions, e.g., 
van derWaals forces and hydrogen bonding.] 
11-9 The Hartree–Fock Limit 
It should be apparent that different choices of basis set will produce different SCF 
wavefunctions and energies. Suppose that we do an SCF calculation on some molecule, 
using a minimal basis set and obtain a total electronic energy E1. If we now choose 
a double-. basis and do a new SCF calculation, we will obtain an energy E2 that 
normally will be lower than E1. (If one happens to choose the first basis wisely and 
the second unwisely, it is possible to find E2 higher than E1. We assume here that 
each improvement to the basis extends the mathematical flexibility while including the 
capabilities of all preceding bases.) If we now add polarization functions and repeat 
the SCF procedure, we will find E3 to be lower than E2. We can continue in this way, 
adding new functions in bonds and elsewhere, always increasing the capabilities of our 
basis set, but always requiring that the basis describe MOs in a single determinantal 
wavefunction. The electronic energy will decrease with each basis set improvement, but 
eventually this decrease will become very slight for any improvement; that is, the energy 
will approach a limiting value as the basis set approaches mathematical completeness. 
This limiting energy value is the lowest that can be achieved for a single determinantal 
wavefunction. It is called the Hartree–Fock energy. The MOs that correspond to this 
limit are called Hartree–Fock orbitals (HF orbitals), and the determinant is called the 
HF wavefunction. 
Sometimes the term restricted Hartree–Fock (RHF) is used to emphasize that the 
wavefunction is restricted to be a single determinantal function for a configuration 
wherein electrons of a spin occupy the same space orbitals as do the electrons of 
ß spin. When this restriction is relaxed, and different orbitals are allowed for electrons 
with different spins, we have an unrestricted Hartree–Fock (UHF) calculation. This 
refinement is most likely to be important when the numbers of a- and ß-spin electrons 
differ. We encountered this concept in Section 8-13, where we noted that the unpaired 
electron in a radical causes spin polarization of other electrons, possibly leading to 
negative spin density. 
11-10 Correlation Energy 
The Hartree–Fock energy is not as low as the true energy of the system. The mathematical 
reason for this is that our requirement that . be a single determinant is restrictive 
and we can introduce additional mathematical flexibility by allowing. to contain many 
determinants. Such additional flexibility leads to further energy lowering. 
There is a corresponding physical reason for the HF energy being too high. It is connected 
with the independence of the electrons in a single determinantal wavefunction. 
To understand this, consider the four-electron wavefunction 
. =
f1(1)f1(2)f2(3)f2(4)
(11-22)
358 Chapter 11 The SCF-LCAO-MO Method and Extensions 
Recall from Chapter 5 that the numbers in parentheses stand for the spatial coordinates 
of an electron; that is, f1(1) really means f1(x1, y1, z1)a(1) or f1(r1, .1,f1)a(1).1 In 
other words, if we pick values of r, ., and f for each of the four electrons and insert 
them into Eq. (11-22) we will be able to evaluate each function and we will obtain 
a determinant of numbers which can be evaluated to give a numerical value for . 
and .2. The latter number (times d.) can be taken as the probability for finding one 
electron in the volume element around r1, .1, and f1, another electron simultaneously 
in d.2 at r2, .2, and f2, etc. The important point to notice is that the effect on .2 of 
a particular choice of r1, .1, and f1, is not dependent on choices of r, ., f for other 
electrons because the form of the wavefunction is products of functions of independent 
coordinates. Physically, this corresponds to saying that the probability for finding an 
electron in d.1, at some instant is not influenced by the presence or absence of another 
electron in some nearby element d.2, at the same instant. This is consistent with the 
fact that the Fock operator Fˆ [Eq. (11-7)] treats each electron as though it were moving 
in the time-averaged potential field due to the other electrons. 
Because electrons repel each other, there is a tendency for them to keep out of 
each other’s way. That is, in reality, their motions are correlated. The HF energy 
is higher than the true energy because the HF wavefunction is formally incapable of 
describing correlated motion. The energy difference between the HF and the “exact” 
(for a simplified nonrelativistic hamiltonian) energy for a system is referred to as the 
correlation energy. 
11-11 Koopmans’ Theorem 
Despite the fact that the total electronic energy is not given by the sum of SCF oneelectron 
energies, it is still possible to relate the i ’s to physical measurements. If 
certain assumptions are made, it is possible to equate orbital energies with molecular 
ionization energies or electron affinities. This identification is related to a theorem due 
to Koopmans. 
Koopmans [1] proved2 that the wavefunction obtained by removing one electron 
from fk, or adding one electron to the virtual (i.e., unoccupied) MO fj in a Hartree– 
Fock wavefunction is stable with respect to any subsequent variation in fk, or fj . 
Notice that this ignores the question of subsequent variation of all of the MOs f with 
unchanged occupations. It is not necessarily true that they remain optimized, since 
the potential they experience is changed by addition or removal of an electron. Nevertheless, 
Koopmans’ theorem suggests a model. It suggests that we approximate the 
wavefunction for a positive ion by removing an electron from one of the occupied HF 
MOs for a neutral molecule without reoptimizing any of the MOs. Let us do this and 
compare the electronic energies for the two wavefunctions. 
For the neutral molecule, which we assume is a closed-shell system, 
E =
i 
.
.
2Hii +
j 
(2Jij -Kij )
.
. 
(11-23) 
1Note that f1 in parentheses represents a coordinate of electron 1, whereas f1 outside the parentheses represents 
an MO. 
2See also Smith and Day [2].
Section 11-11 Koopmans’ Theorem 359 
For the cation, produced by removing an electron from fk, 
E+k =
i=k 
.
.
2Hii +
j =k
(2Jij -Kij )
.
.+Hkk +
i=k 
(2Jik -Kik) (11-24) 
The first sum in Eq. (11-24) gives the total electronic energy due to all but the unpaired 
electron in fk. Hkk gives the kinetic and nuclear attraction energies for the unpaired 
electron and the final sum gives the repulsion and exchange energy between this electron 
and all the others. Nowwe note that the last sum is exactly equal to the void produced in 
the first sum due to the restriction j =k. Therefore, we can combine these by removing 
the index restriction and deleting the last sum. This gives 
E+k =
i=k 
.
.
2Hii +
j 
(2Jij -Kij )
.
.+Hkk (11-25) 
To compare this with E of (11-23) we should remove the remaining index restriction. 
We do this by allowing i to equal k in the sum and simultaneously subtracting the new 
terms thus produced: 
E+k =
i 
.
.
2Hii +
j 
(2Jij -Kij )
.
.-Hkk -
j 
(2Jkj -Kkj ) (11-26) 
But, by virtue of Eqs. (11-15) and (11-23), this is 
E+k =E -k (11-27) 
Hence, the ionization energy I 0 
k , for ionization from the fk is 
I 0 
k =E+k -E=-k (11-28) 
This illustrates that, within the context of this simplified model, the negative of the 
orbital energies for occupied HF MOs are to be interpreted as ionization energies. 
Another way to see the relation between I 0 
k and-k, is to recognize that the physical 
interactions lost upon removal of an electron from .k, are precisely those that constitute 
k, [See Eq. (11-15).] 
A similar result holds for orbital energies of unoccupied HF MOs and electron 
affinities. (However, this is less successful in practice; see Problem 11-3.) 
In actuality, the relation (11-28) is only approximately obeyed. One reason for this 
has to do with our assumption that doubly occupied SCF MOs produced by a variational 
procedure on the neutral molecule will be suitable for the doubly occupied MOs of the 
cation as well. These MOs minimize the energy of the neutral molecule but give an 
energy for the cation that is higher than whatwould be produced by an independent variational 
calculation. For this mathematical reason, we expect the Koopmans’ theorem 
prediction for the ionization energy to be higher than the value predicted by taking the 
difference between separate SCF calculations on the molecule and cation (which we 
will symbolizeSCF). The corresponding physical argument is that use of Eq. (11-28) 
views ionization as removal of an electron without any reorganization of the remaining
360 Chapter 11 The SCF-LCAO-MO Method and Extensions 
TABLE 11-1 
 Ionization Energies (in electron volts) of Water as 
Measured Experimentally and as Predicted from SCF Calculations 
SCF (near HF limit)b 
Cation state Observeda Koopmans SCF 
2B2 12.62 13.79 11.08 
2A1 14.74 15.86 13.34 
2B2 18.51 19.47 17.61 
aFrom Potts and Price [3]. 
bFrom Dunning et al. [4]. 
electronic charge. This neglects a process that stabilizes the cation and lowers the ionization 
energy. Whichever argument we choose, we have here a reason for expecting 
- to be an overestimate of the value obtained by independent calculations, SCF. 
Another error results from the neglect of change in correlation energy. We have seen 
that the total SCF energy for the molecule is too high because the single determinantal 
form of thewavefunction cannot allowfor correlated electronic motion. The SCF energy 
for the cation is too high for the same reason, but the error is different for the two 
cases because there are fewer electrons in the cation. We expect the neutral molecule 
to have the greater correlation energy (since it has more electrons)3 so that proper 
inclusion of this feature would lower the energy of the neutral molecule more than the 
cation, making the true I 0 
k larger than that obtained by neglect of correlation. Hence, 
this leads us expect SCF to underestimate I 0 
k . Since - overestimates SCF, and 
SCF underestimates the ionization energy, we can expect some cancellation of errors 
in using Eq. (11-28). 
An illustration of these relations is provided in Table 11-1, where observed vertical 
ionization energies (i.e., no nuclear relaxation), the appropriate values of -, and the 
values of SCF are compared. 
11-12 Configuration Interaction 
There are several techniques for going beyond the SCF method and thereby including 
some effects of electron correlation. Some extremely accurate calculations on small 
atoms and molecules, making explicit use of interparticle coordinates, were described 
in Section 7-8. There is one general technique, however, that has traditionally been used 
for including effects of correlation in many-electron systems. This technique is called 
configuration interaction (CI). 
The mathematical idea of CI is quite obvious. Recall that we restricted our SCF 
wavefunction to be a single determinant for a closed-shell system. To go beyond the 
optimum (restricted Hartree–Fock) level, then, we allow the wavefunction to be a linear 
3This reasoning is rather naive. Significant correlation energy contribution can result from a small energy-level 
separation between filled and empty MOs (rather than from merely the number of electrons), but production of a 
cation should normally increase this gap and lead to reduced correlation.
Section 11-12 Configuration Interaction 361 
combination of determinants. Suppose we choose two determinants D1 and D2, each 
corresponding to a different orbital occupation scheme (i.e., different configurations). 
Then we can let 
. =c1D1 +c2D2 (11-29) 
and minimize E as a function of the linear mixing coefficients c1 and c2. 
If we go through the mathematical formalism and express E¯ as .|Hˆ |./.|., 
expand this as integrals over D1 and D2, and require .E¯/.ci =0, we obtain the same 
sort of 2×2 determinantal equation that we find when minimizing an MO energy as a 
function of mixing of two AOs. That is, we obtain 
 
H11 -E¯S11 H12 -E¯S12 
H21 -E¯S21 H22 -E¯S22

=0 (11-30) 
where now 
Hij = Di 
ˆ H
Dj  (11-31) 
Sij = 
Di |Dj  (11-32) 
We see that, whereas before we might have had two AOs interacting to form two 
MOs, here we have two configurations (i.e., two determinantal functions) interacting to 
form two approximate wavefunctions. Our example involves only two configurations, 
but there is no limit to the number of configurations that can be mixed in this way. 
Since each configuration D contains products of MOs, each of which is typically a 
sum ofAOs, the integrals Hij and Sij can result in very large numbers of integrals over 
basis functions when they are expanded. This is the sort of situation where a computer 
is essential, and CI on atoms and molecules, while still expensive compared to SCF, 
have become routine on modern computers. 
Our purpose in this chapter is not to describe how to carry out a CI calculation, but 
rather to convey what a CI calculation is and what its predictive capabilities are. Therefore, 
we will not concern ourselves with the mathematical complexities of evaluating 
Hij and Sij .4 But we will consider one practical aspect of CI calculations, namely, how 
one goes about choosing which configurations should be mixed together, and which 
ones may be safely ignored. 
We begin by considering the H2 molecule. The LCAO-MO-SCF method expresses 
the ground state wavefunction for H2 as 
.(1, 2)=
 
1sg(1)a(1) 1sg(2)a(2) 
1sg(1)ß(1) 1sg(2)ß(2)
 
(11-33) 
that is, as the configuration 1s2 
g . The SCF procedure mixes the AO basis functions 
together in the optimum way to produce the 1sg MO. 
We have noted at several points in this book that, if one begins with a basis set of n 
linearly independent functions, one ultimately arrives at n independent MOs. Hence, 
the 1sg MO of Eq. (11-33) is but one of several MOs produced by the SCF procedure. 
4In most actual calculations, the D’s are orthonormal, and Sij = dij .
362 Chapter 11 The SCF-LCAO-MO Method and Extensions 
It is called an occupied MO because it is occupied with electrons in this configuration. 
All the other MOs in this case are unoccupied or virtual MOs. The virtual MOs of 
H2 have symmetry properties related to the molecular hamiltonian, just as does the 
occupied MO. Thus, we can refer to 1su, 2sg, 2su, 1pu, 1pg, etc., virtual MOs of H2. 
Which of these virtual MOs are produced by an SCF calculation depends on the number 
and nature of the AO basis set provided at the outset. If no p-type AOs are provided, 
no p-type MOs will be produced. If only a minimal basis (1sa and 1sb) is provided, 
1su will be the only virtual MO produced. 
It is important to distinguish between the physical content of occupied versus virtual 
SCF MOs. The SCF procedure finds the set of occupied MOs for a system leading to the 
lowest SCF electronic energy. The virtual orbitals are the residue of this process. The 
virtual MOs span that part of the basis set function space that the SCF procedure found 
least suitable for describing.. The subspace is sometimes referred to as the orthogonal 
complement of the occupied orbital subspace. (Note that this situation differs from that 
pertaining to H¨uckel-type calculations, where MOs and energy levels are calculated 
without regard for electron occupancy. Only after the variational procedure are electrons 
added.) 
Our concern with virtual MOs is due to the fact that they provide a ready means for 
constructing new configurations to mix with our 1s2 
g configuration for H2. Thus, using 
some of the above-mentioned virtual MOs, we could write determinantal functions corresponding 
to the excited configurations 1sg1su, 1sg2sg, 1sg2su, 1sg1pu, etc.5 These 
are commonly referred to as singly excited configurations because one electron has 
been promoted from a ground-state-occupied MO to a virtual MO. (This is not meant 
to imply that the orbital energy difference is equal to the expected spectroscopic energy 
of the transition.) It is also possible to construct doubly excited configurations, such as 
1s2 
u , 1su2sg, 2s2 
g , 1su2su, 1su1pu, etc. For systems having more electrons, one can 
write determinants corresponding to triple, quadruple, etc., excitations. If one has a 
reasonably large number, say 50, of virtual orbitals and, say, 10 electrons to distribute 
among them, then there is an enormous number of possible configurations. A major 
step in doing a CI calculation is deciding which configurations might be important in 
affecting the results and ought therefore to be included. 
We can gain insight into this problem by considering our minimal basis set H2 
problem in more detail. We have
1sg = Ng(1sA +1sB) (11-34) 
1su = Nu(1sA -1sB) (11-35) 
where Ng and Nu are normalization constants. The spatial part of the ground configuration 
is 
.space =1sg(1)1sg(2) (11-36) 
5As was shown in Chapter 5, the symmetry requirements of the wavefunction require that each of these open 
shell configurations be expressed as a linear combination of two 2×2 determinants; for example, 1sg2su stands 
for the combination 
1/v2 
1sg(1)2 ¯su(2)
±
1 ¯sg(1)2su(2)

Section 11-12 Configuration Interaction 363 
which expands to 
.space =N2 
g [1sA (1) 1sA (2)+1sB (1) 1sB (2)+1sA (1) 1sB (2)+1sB (1) 1sA (2)] 
If both electrons are near nucleus A, the first term is quite large. This may be 
rephrased to say that .2 gives a sizable probability for finding both electrons near 
nucleusA. The second term gives a similar likelihood for finding both electrons near B. 
These two terms are referred to as ionic terms because they become large whenever the 
instantaneous electronic dispositions correspond to H-AH+B and H+AH-B , respectively. 
The last two terms cause .2 to be sizable whenever an electron is near each nucleus. 
Hence, these are called covalent terms, and their presence means that . contains significant 
“covalent character.” In fact, because all four terms have the same coefficient, 
the configuration 1s2 
g is said to have 50% covalent and 50% ionic character. 
Is this bad? It turns out to be no problem at all when the nuclei are close together. 
Indeed, in the united-atom (helium) limit, the ionic-covalent distinction vanishes. But 
at large internuclear separations it is very inaccurate to describe H2 as 50% ionic. In 
reality, H2 dissociates to two neutral ground state H atoms—that is, 100% “covalent,” 
with an electron near each nucleus. In short, the SCF-MO description does not properly 
describe the molecule as it dissociates. This means that the calculation of E¯ versus 
RAB for H2 will deviate from experiment more and more as RAB increases. This defect 
in the SCF treatment of H2 occurs for many other molecular species also. 
Can we correct this defect through use of CI?We ask the question this way: “What 
configuration could we mix with 1s2 
g in order to make the mixture of covalent and ionic 
character variable?” Since 1s2 
g expands to give us covalent and ionic terms of the same 
sign, we need an additional configuration that will give them with opposite sign. Then 
admixture of the two configurations will affect the two kinds of term differently. The 
configuration that will accomplish this is 1s2 
u : 
1su (1) 1su (2) = N2 
u [1sA (1) 1sA (2)+1sB (1) 1sB (2)-1sA (1) 1sB (2) 
-1sB (1) 1sA (2)] (11-37) 
Mixing these two configurations together gives 
. (c1/c2) = c11sg (1) 1sg (2)+c21su (1) 1su (2) 
= c1N2 
g +c2N2 
u  [1sA (1) 1sA (2)+1sB (1) 1sB (2)] 
+ c1N2 
g -c2N2 
u  [1sA (1) 1sB (2)+1sB (1) 1sA (2)] (11-38) 
If c1/c2 is readjusted at each value of RAB to minimize E¯, it is evident that the 
relative weights of covalent and ionic character in Eq. (11-38) will change to suit 
the circumstances. Actual calculations on this system show that, as RAB gets large, 
c1/c2 approaches a value such that c1N2 
g + c2N2 
u approaches zero, so that the ionic 
component of . vanishes. 
This example illustrates that CI of this sort has an associated physical picture. 
It suggests that, in any CI calculation involving the dissociation (or extensive stretching) 
of a covalent bond, important configurations are likely to include double excitations 
into the antibonding virtual “mates” of occupied bonding MOs.
364 Chapter 11 The SCF-LCAO-MO Method and Extensions 
What about other configurations for H2? What will 1sg2sg do for the calculation, 
assuming now an extended basis set has produced a 2sg MO? Suppose we take as our 
trial function 
. =c11s2 
g +c21sg2sg (11-39) 
where the configurations are understood to stand for determinants. If the 1sg MO has 
been produced by an SCF calculation on the ground state, and 2sg is a virtual MO from 
that SCF calculation, then it is possible to show that the CI energy minimum occurs 
when c2 in Eq. (11-39) is zero. In other words, these determinants will not mix when 
they are combined in this way. An equivalent statement is that the mixing element 
H12 = 1s2 
g |Hˆ |1sg2sg vanishes. Hence, the CI determinant (11-33) is already in 
diagonal form, and no variational mixing will occur. This is an example of Brillouin’s 
theorem, which may be stated as follows: 
EXAMPLE 11-1 If D1 is an optimized single determinantal function and Dj is a 
determinant corresponding to any single excitation out of an orbital fj occupied in 
D1 and into the virtual subspace (orthogonal complement) ofD1, then no improvement 
in energy is possible by taking . =c1D1 +c2Dj . 
The proof of Brillouin’s theorem is very simple. We start with a basis set that 
spans a function space. An SCF calculation is performed, which produces the best 
single-determinantal wavefunction we can possibly get within this function space. 
This isD1. Dj differs fromD1 in only one orbital, which means they differ in only 
one row. A general property of determinants is that, if two of them differ in only 
one row or column, any linear combination of the two can be written as a single 
determinant (see Problem 11-4). This means that any combination c1D1 +c2Dj 
is still expressible as a single determinant. Since Dj makes no use of functions 
outside our original basis set, c1D1 + c2Dj is a single determinant within our 
original function space. However,D1 is already known to be the single determinant 
within this function space that gives the lowest energy, and c1D1 +c2Dj cannot 
do better. QED. 
A doubly excited configuration differs from D1 in two rows, and mixing such 
a configuration with D1 produces a result that cannot be expressed as a single 
determinant. 
Because of Brillouin’s theorem, one might decide to omit all single excitations from 
CI calculations. But it is important to recognize that singly excited configurations can 
affect the results of CI calculations in the presence of doubly excited configurations. 
This comes about because nonzero mixing elements can occur between singly and 
doubly excited configurations in the CI determinant. To illustrate, let .0 be an SCF 
single determinant, .1 be a singly excited configuration, and .2 be a “double.” Then 
the CI determinant could be, assuming orthogonal determinants, 
 
H00 -E 0 H02 
0 H11 -E H12 
H02 H12 H22 -E

=0 (11-40)
Section 11-13 Size Consistency 365 
The zeros result from Brillouin’s theorem. However, H12 does not necessarily vanish, 
and solution of this 3×3 determinantal equation leads to a wavefunction of the form 
. =c0.0 +c1.1 +c2.2 (11-41) 
with c1 not zero. .1 comes in on the coattails of .2 and is referred to as a secondorder 
correction. This is not a guarantee that it will be unimportant, however. 
(See Example 7-4 for similar behavior in a different context.) 
Another rule that is useful for recognizing configurations that may be omitted is 
the rather obvious one that each configuration must share the same set of eigenvalues 
for operators commuting with the hamiltonian. That is, if . is to be associated 
with a particular symmetry, angular momentum, spin angular momentum, etc., then 
each configuration in . must have that same symmetry, angular momentum, etc. This 
means that, for the ground state of H2, 1s2 
g will not mix with 1sg1su because the 
latter has overall u symmetry. 1su2su could contribute, but the symmetrized combination 
corresponding to the singlet state (|1su2 ¯su|-|1 ¯su2su|) must be used rather than 
the (positive) triplet state combination. The configuration 1su1pu will not contribute 
because it has the wrong total angular momentum. 
Even with the aid of all these rules, a calculation on a molecule such as N2 or O2 
using a reasonably extended basis set gives rise to an enormous number of possible 
configurations. Additional rules of thumb have been found to help choose the major 
configurations. It has been found, for example, that triply or higher excited configurations 
are usually of lesser importance than doubly excited configurations. [Since the 
hamiltonian contains only one- and two-electron operators, interaction elements must 
vanish between the ground-state configuration and all triply or higher-excited configurations. 
But, like singly excited configurations, these can, in principle, come in on the 
coattails of doubly (or other) excited configurations.] In addition, a study of the energy 
change in some process involving primarily the valence electrons (e.g., stretching N2) 
really does not require calculation of the correlation energy of the 1s electrons since 
they are fairly unaffected by the change. Any correlation energy for these electrons 
tends to cancel itself when initial and final state energies are subtracted. Therefore, in a 
CI calculation of such a process, it is reasonable to omit configurations corresponding 
to excitation of a 1s electron unless high accuracy is desired. 
The acronym CID refers to a CI calculation in which only all doubly excited configurations 
are included. Inclusion of all singly and doubly excited configurations is 
referred to as a CISD calculation. Full CI (FCI) means all excited configurations have 
been included, and this is the limit that gives all of the correlation energy within the 
chosen basis set. The combination of full CI and a complete basis gives the exact energy 
(generally nonrelativistic and within the Born–Oppenheimer approximation). 
11-13 Size Consistency and the Møller–Plesset 
and Coupled Cluster Treatments of Correlation 
Whenever certain parts of a well-defined procedure are omitted, as when full CI is truncated 
to CID or CISD, one must consider whether systematic errors are introduced. This 
is indeed the case in the above example. Suppose CID calculations are made for the 
energy of N2 as a function of internuclear distance. At short distances, we treat the
366 Chapter 11 The SCF-LCAO-MO Method and Extensions 
system as a 14-electron molecule, including configurations in which 12 of the electrons 
are in their HF-occupied MOs. At very large distances we have two nitrogen atoms, 
which we normally treat as having twice the energy of one atom. Now CID on atomA 
includes the HF configuration, DA
0 , as well as doubly excited configurations in which 
five of the seven electrons are in their HF-occupied AOs. Let DA
2 represent this class 
of configuration. Then .A = c0DA
0 + c2DA
2 . Atom B has a similar CID wavefunction: 
.B =c0DB
0 +c2DB
2 . The wavefunction for the overall, noninteracting system is 
the antisymmetrized product of these wavefunctions. It will contain terms like DA
0 DB
0 , 
DA
0 DB
2 , and DA
2 DB
2 . There are no terms present corresponding to a single excitation 
at each atom, DA
1 DB
1 , and such terms would be present in a CID treatment of the 
combined system. Also, there is a class of terms present corresponding to four promoted 
electrons,DA
2 DB
2 , and these termswould not be present in a CID treatment of the 
combined system. The dilemma is that, if we treat the system as a single 14-electron 
“molecule,” which is appropriate at small R, we mix in different terms than if we treat it 
as two separate atoms, which is appropriate at large R. If we choose some fairly large R 
value to redefineN2 as two separate atoms, we change the nature of the CI in a discontinuous 
way at an arbitrary point. This feature of truncated CI is called the problem of size 
consistency; CIDandCISDmethods are not size consistent. Doing separate calculations 
on each of two separated atoms and combining the energies yields a different result from 
doing a calculation on one system made up of two separated atoms. 
A correlation method that is size consistent has been developed by Pople and 
co-workers.6 It is based on perturbation theory that was introduced many years ago 
by Møller and Plesset.7 This approach divides the process of treating correlation into 
a series of corrections to an unperturbed starting point. If one chooses to do such a 
calculation to, say, third order (MP3, standing for Møller–Plesset to third order), then 
the set of configurations to be included is determined by the perturbation formulas.8 It 
does not require further decision by the person doing the calculation and can be wholly 
managed by a computer program. Møller–Plesset perturbation theory is different from 
standard CI in at least two important respects: It is size consistent, and it is not variationally 
bound. One cannot assume, therefore, that going to higher and higher orders 
of perturbation will cause the calculated energy to approach closer and closer to the 
true energy from above. 
Because of the way MP theory defines the unperturbed system, the starting point 
energy (MP0) is the sum of HF one-electron energies. The first-order correction to 
the energy (MP1) brings in the appropriate electronic coulomb and exchange integrals, 
giving the correct HF energy. MP2 brings in contributions wherein doubly excited 
configurations “interact with” (i.e., occur in the same integral with) the ground configuration. 
MP3 adds contributions due to doubly excited configurations interacting 
with each other. MP4 brings in interactions involving also single, double, triple, and 
quadruple excitations. The selection of interaction terms by the perturbation formalism 
is what produces size consistency, but it leaves out certain terms at each level that would 
be included in a variational calculation. 
In coupled cluster (CC) approaches, which are also size consistent and generally 
not variationally bound, instead of including all configurations to a particular order as 
6See Binkley and Pople [5]. 
7See Møller and Plesset [6]. 
8Perturbation theory is presented in Chapter 12. The present discussion avoids mathematical details.
Section 11-14 Multideterminant Methods 367 
in MP theory, each class of excited configurations is included to infinite order. This is 
accomplished via an exponential excitation operator, 
CC =eTˆ.0 =
1+Tˆ + ˆ T 2 
2! + ˆ T 3 
3! +. . .

.0 (11-42) 
where .0 is the HF determinant for an N-electron system, and Tˆ = Tˆ1 + Tˆ2 + 
Tˆ3 +···+TˆN. Tˆ1 produces singly excited determinants, Tˆ2 doubly excited ones, and 
so on. Because of the exponential nature of the excitation operator, each class of 
excitations is included to all orders, e.g., terms in Tˆ2 would include products of double 
excitations (Tˆ2 
2 ) that would be considered a subset of the possible quadruple excitations 
in CI. This is what makes CC theory size consistent. Usually coupled cluster theory 
is truncated to include just Tˆ1 and Tˆ2, i.e., CCSD. One of the most accurate post-HF 
methods has been shown to be the CCSD(T) method,in which a CCSD calculation is 
followed by a contribution due to triple excitations (Tˆ3) via perturbation theory. 
11-14 Multideterminant Methods 
Up to this point, the methods that have been presented for describing electron correlation 
effects have been constructed with the single determinant SCF wavefunction 
as a starting point. For most molecules near their equilibrium geometries, this is a 
very good zeroth-order approximation, but as we saw earlier for the H2 molecule, as 
covalent bonds are stretched towards dissociation multiple determinants are required 
for even a qualitative description. This puts much stronger demands on these so-called 
single reference methods, and their accuracy can be much degraded or even unphysical 
in these regions. In a multiconfigurational SCF (MCSCF) calculation one writes the 
wavefunction as a linear combination of determinants exactly as in a CI calculation, 
and the energy is minimized as a function of the linear CI coefficients. However, in 
an MCSCF calculation one also simultaneously optimizes the MO coefficients of the 
orbitals that are used to construct the determinants, using methods analogous to SCF 
theory. Because this greatly adds to the complexity of the calculation, the number of 
determinants used in MCSCF is generally much smaller than in a standard HF-based CI 
calculation. In the simplest case, only the additional determinants that allow for a qualitative 
treatment of the process under study are included, e.g., one would include only 
the determinants corresponding to excitations of bonding electrons into their respective 
antibonding orbitals when stretching the triple bond of N2. This procedure results in 
a set of MCSCF molecular orbitals (some strongly occupied, some weakly occupied) 
that smoothly changes in character from equilibrium to dissociation. 
In multireference CISD (MRCISD) calculations, the wavefunction is written as 
. =
i 
ci.i +
s 
cs.s +
d 
cd.d (11-43) 
wherei ci.i is the set of MCSCF reference determinants, .s are new determinants 
formed by single excitations into the virtual orbitals relative to all of the reference 
determinants, and .d are doubly excited determinants. An MRCISD calculation of this 
type can yield a very balanced and accurate description of a molecule’s potential energy 
surface, but often at a relatively steep cost in terms of computational requirements.
368 Chapter 11 The SCF-LCAO-MO Method and Extensions 
11-15 Density Functional Theory Methods 
The wavefunction . for an n-electron molecule is a function of 3n spatial coordinates 
and n spin coordinates. From . we can produce the molecule’s spin-free electron 
density function, .(1), by integrating .*. over all of the spin coordinates and all the 
the space coordinates except those for one of the electrons:9 
.(1)= |.(1, 2, . . . ,N)|2d.1dt2 . . . dtn (11-44) 
which is a function of only the three spatial coordinates.10 We have seen that, in the 
early days of quantum chemistry, a major challenge was the evaluation of integrals 
over the interelectronic-repulsion term in the hamiltonian, as well as dealing with the 
related problem of electron correlation. Several methods were devised that attempted 
to approximate these quantities from the density function .(1), with moderate success. 
However, the continuing progress in computer speed and the development of sophisticated 
ab initio methods gradually shifted attention away from approaches using the 
density function. 
In 1964, proof by Hohenberg and Kohn [7] of a connection between the ground 
state energy, E0, for a system and .0, the ground state density function,11 sparked 
new interest in finding a rigorous way to go from knowledge of the attractively simple 
three-dimensional density function to a value for E0. 
Recall that, for a system having n electrons and N nuclei, the hamiltonian operator 
for the electronic energy is 
H =-
1
2 
n 

i=1 
.2
i + 
n 

i=1 
N 

a=1 
-Za 
ria +
n-1 

i=1 
n 
 
j=i+1 
1 
rij 
(11-45) 
The first and last terms can be written down immediately if we know how many 
electrons are present, but the middle term depends onN
a=1 -Za 
ria 
, which is a function of 
nuclear charges and locations. This quantity is called the external potential, symbolized 
vext (	r), because it results from the presence of fields produced by particles not included 
in the group of electrons. 
Hohenberg andKohn were able to prove that there is a uniqueness relation between .0 
and the external potential: Notwo external potentials could give the same .0. This raises 
the possibility that one could work backwards from .0 to find vext (	r) and then E0. The 
following route comes first to mind: Integrate .0 to get the number of electrons n. Figure 
out vext (	r) from .0. This would allow one to write down the hamiltonian operator. 
Then, using ab initio methods, one could get to an accurate E0 and .0, and from .0 
one could calculate T0, Vne0 , Vee0 , and all the other properties of interest for the system. 
Two problems exist with this scenario. First, there is no generally applicable procedure 
known for getting from .0 to vext (	r). We can posit that vext (	r) is a functional of 
.0, which we symbolize vext [.0], but we don’t know what the functional relationship 
is. Second, even if we could get back to the hamiltonian operator, it would simply 
9Because . is antisymmetric for exchange of electrons, the density function is independent of our choice as to 
which electron’s coordinates should be spared from integration. 
10Recall that . is the spin coordinate and t is the coordinate for space and spin. dt = dv d.. 
11We henceforth suppress the electron index in ..
Section 11-15 Density Functional Theory Methods 369 
land us back on square one: We would still have to solve the whole problem in the 
traditional way. Nevertheless, the hopes raised by this uniqueness theorem have led to 
the current goal of density functional theory, which is to find a procedure that takes us 
from . to E in a rigorous way that avoids the complexities of landing on square one 
and proceeding using standard ab initio methods. [8] 
Asubsequent relation proved by Hohenberg andKohn [7] indicated away to proceed. 
They proved that an approximate density function, .0,approx, when subjected to the 
(unknown) procedure that relates the exact .0 to the exact E0, must yield an energy 
higher than the exact E0 :E0,approx =E0, so a variational bound exists. Note that the 
unknown process referred to here is one that assumes vext (	r) to be the same for the 
analysis of .0 and .0,approx, which means that the same nuclear framework applies in 
both cases. 
If a procedure were known for finding E from ., then the existence of a variational 
bound would allow a variational procedure analogous to what we have applied earlier. 
One would start with a trial ., calculate its energy, and vary . to locate the . that gives 
the lowest energy. 
The barrier to proceeding is the lack of a way to get E from .. Hence, the development 
of approximate functionals that relate the energy to the electron density is an 
extremely active area of current research and probably will be for some time to come. 
In analogy to wavefunction methods, the functional that connects E to ., E[.], can 
be separated into an electronic kinetic energy contribution, T [.], a contribution due 
to nuclear-electron attractions, Ene[.], and the electron-electron repulsions, Eee[.]. 
The latter term can be further decomposed into Coulomb and exchange terms, J [.] and K[.]. Both the nuclear-electron attraction and the interelectronic Coulomb terms 
can be easily written in terms of the density using their classical expressions as in 
wavefunction methods. For an accurate treatment of the electronic kinetic energy term, 
however, one must differentiate a wavefunction,12 and this has led to the practice first 
proposed by Kohn and Sham [9] of expressing the density in terms of one-electron 
orbitals f (constructed numerically or from a basis set of Slater or gaussian functions). 
These orbitals serve two purposes. They allow us to calculate a value of the kinetic 
energy within a single Slater determinant framework similar to Hartree–Fock theory, 
TS = 
n 

i=1
fi|- 
1
2.2|fi  (11-46) 
and to obtain the electron density, defined in terms of these Kohn–Sham orbitals as 
.s = 
n 

i 
|fi |2 (11-47) 
The final DFT energy expression is then written as 
EDFT [.]=TS[.]+Ene[.]+J [.]+Exc[.] (11-48) 
where the exchange correlation functional Exc[.] contains the difference between the 
exact kinetic energy and TS, the nonclassical (exchange) part of electron-electron repulsions, 
K[.], and correlation contributions to both K[.] and J [.]. The Kohn–Sham 
12As far as we know, we must differentiate a wavefunction to get kinetic energy. If there is a functional that 
permits us to obtain kinetic energy directly from the density function, we might avoid having to use orbitals.
370 Chapter 11 The SCF-LCAO-MO Method and Extensions 
orbitals are eigenfunctions of an effective one-electron hamiltonian that is nearly identical 
in form to the Fock operator in the SCF equations. In theKohn–Sham case, however, 
the HF exchange operators are replaced by the functional derivative of the exchange 
correlation energy. Assuming the existence of Exc[.] and an initial guess for the electron 
density, one then solves the Kohn–Sham eigenvalue equations for the orbitals, 
which can then be used to define a new electron density and effective hamiltonian. 
These iterations continue until the density is converged to within a specified threshold. 
The exact form of Exc[.] is not currently known, however, and a rapidly growing 
list of approximate exchange correlation functionals have appeared in the literature. 
Because these are all estimates of a part of the overall energy, the total energy 
finally calculated is not an upper bound to the true energy. Also, DFT is not sizeconsistent. 
Generally, most existing exchange correlation functionals are split into a pure 
exchange and correlation contribution, Ex[.] and Ec[.] and the current functional 
nomenclature often reflects this with two-part acronyms, e.g., the BLYP DFT method 
uses an exchange functional from Becke (B) [10] and a correlation functional by Lee, 
Yang, and Parr (LYP) [11]. In principle, the exchange contribution could be calculated 
exactly (for a single determinant) in the same manner as TS, but this is generally not 
done since this disturbs the balance between Ex[.] and Ec[.]. In hybrid DFT, a 
percentage of this exact exchange is included in Exc[.]. 
The great benefit of present day DFT methods is computational cost. With the 
exchange correlation functionals commonly used, the computational effort is similar 
to a SCF calculation, but since Exc[.] implicitly includes some amount of electron 
correlation, the accuracy of DFT (depending on the chosen functional) is often similar 
to that obtained with MP2 or better. The great weakness of DFT at the present time, 
however, is the inability to systematically improve upon Exc[.] and converge towards 
the exact Born–Oppenheimer energy like one might conceptually do in a wavefunctionbased 
CI or CC calculation, e.g., SCF, CCSD, CCSDT, CCSDTQ, etc. with sequences 
of correlation consistent basis sets. 
One of the simplest DFT methods is the local density approximation (LDA), which 
assumes the density behaves locally like a uniform electron gas. Generally this does 
not lead to an accurate description of molecular properties, but if one makes Ex[.] and 
Ec[.] depend also on the gradient of the density, yielding gradient corrected DFT or 
the generalized gradient approximation (GGA), the results are much more accurate. 
Finally, the definition of Exc[.] also lends itself to semiempirical contributions. One 
such parameterization that has been very successful is the B3LYP hybrid DFT method, 
which includes 20% exact exchange and involves three semiempirical parameters that 
were obtained by fits to experimental thermochemical data (heats of formation, etc.) 
of small molecules [12]. 
11-16 Examples of Ab Initio Calculations 
Self-consistent-field and correlated calculations have now been made for a very large 
number of systems. The best way to judge the capabilities of these methods is to survey 
some of the results.13 
13For extensive surveys, see Schaefer [13], Hehre et al. [14], and Raghavachari [15].
Section 11-16 Examples of Ab Initio Calculations 371 
Table 11-2 provides information on energies for a number of atoms in their ground 
states. Self-consistent-field energies are presented for three levels of basis set complexity. 
In the STO single-. level, a minimal basis set of one STO per occupied AO is 
used, and the energy is minimized with respect to independent variation of every orbital 
exponent . . The STO double-. basis set is similar except that there are two STOs for 
each AO, the only restriction being that the STOs have the same spherical harmonics 
as the AOs to which they correspond. 
The Hartree–Fock energies are estimated by extrapolating from more extensive basis 
sets, and represent the limit achievable for the SCF approach using a complete basis 
set. We can make the following observations: 
1. The improvement in energy obtained when one goes from a single-. to a double-. 
STO basis set is substantial, especially for atoms of higher Z. 
2. The agreement between the optimized double-. data and the HF energies is quite 
good. Even for neon, the error is only about 10-2 a.u. (0.27 eV). Thus, for atoms, 
the double-. basis is capable of almost exhausting the energy capabilities of a 
single-configuration wavefunction. 
3. The disagreement between HF and “exact” energies (i.e., the correlation energy) 
grows progressively larger down the list. For neon it is almost 0.4 a.u. (10 eV), 
which is an unacceptable error in chemical measurements. 
One might think that the magnitude of the correlation energy in these examples 
would make SCF calculations on heavy atoms useless for quantitative purposes, but this 
TABLE 11-2 
 Ground-State Energies (in atomic units) for Atoms, as Computed by the SCF 
Method and from Experiment 
STO 
Atom Single . a Double . a Hartree–Focka Exactb 
Correlationc 
energy 
He -2.8476563 -2.8616726 -2.8616799 -2.9037 -0.0420 
Li -7.4184820 -7.4327213 -7.4327256 -7.4774 -0.0447 
Be -14.556740 -14.572369 -14.573021 -14.6663 -0.0933 
B -24.498369 -24.527920 -24.529057 -24.6519 -0.1228 
C -37.622389 -37.686749 -37.688612 -37.8420 -0.1534 
N -54.268900 -54.397951 -54.400924 -54.5849 -0.1840 
O -74.540363 -74.804323 -74.809370 -75.0607 -0.2513 
F -98.942113 -99.401309 -99.409300 -99.7224 -0.3131 
Ne -127.81218 -128.53511 -128.54705 -128.925 -0.378 
Ar -525.76525 -526.81511 -526.81739 -527.542 -0.725 
aFrom Roetti and Clementi [16]. 
b“Exact” equals experimental with relativistic correction but without correction for Lamb shift. See Veillard 
and Clementi [17]. 
cCorrelation energy is “exact” minus HF energy.
372 Chapter 11 The SCF-LCAO-MO Method and Extensions 
is not the case. Most frequently we are not concerned with the value of the total energy 
of a system so much as with energy changes (e.g., in spectroscopy or in reactions) or 
else with other properties such as transition moments (for spectroscopic intensities) or, 
in molecular systems, dipole moments. 
Let us, therefore, see how well SCF calculations can predict atomic ionization energies. 
We have already indicated (Section 11-11) that there are two ways we can get 
ionization energies from SCF calculations. The first, and simplest, is to take the various 
-i , as suggested by Koopmans’ theorem. Table 11-3 shows that this gives only 
rough agreement with experimental values for neon. Another way is to do separate SCF 
calculations for each excited state produced by removal of an electron from an orbital 
(i.e., for each “hole state”) and equate the ionization energies to the energy differences 
between these and the neutral ground state (SCF). This second method requires much 
more effort. As Table 11-3 indicates, however, the extra effort leads to great improvement 
in agreement between theoretical and experimental values. We conclude that SCF 
calculations on atoms and ions give quantitatively useful data on ionization energies, 
even for ionization out of deep-lying levels. The Koopmans’ theorem approach is less 
accurate, although still qualitatively useful. 
A related problem is the calculation of energies of excited states of atoms. Weiss 
[19] has reported calculations on some of the excited states of carbon, and his results are 
summarized in Fig. 11-3. Inspection of this figure reveals that near-HF calculations only 
roughly reproduce the energy spectrum, but CI (with four or five configurations) brings 
about marked improvement. Weiss has omitted configurations involving excitations of 
1s electrons, and so these results ignore correlation energy for the inner-shell electrons. 
The agreement suggests that these electrons experience almost no change in correlation 
for transitions among these states. Weiss has also calculated oscillator strengths14 
associated with atomic transitions and he finds that CI is necessary before reasonable 
agreement with experiment is achieved. 
TABLE 11-3 
 Ionization Energies of Neona 
Ionization potential (a.u.) 
Ion configuration Koopmans SCF Experiment 
1s2s22p6 32.7723 31.9214 31.98 
1s22s2p6 1.9303 1.8123 1.7815 
1s22s22p5 0.8503 0.7293 0.7937 
aFrom Bagus [18]. The basis set includes 5 s-type and 12 p-type STOs 
(4 of each m quantum number). . ’s were varied as well as linear coefficients. 
The neutral ground state gives E=-128.547 a.u. (compare 
Table 11-2). 
14The oscillator strength is a measure of the probability (i.e., intensity) of a transition. For a transition between 
states a and b in a 2n-electron system it is commonly given by the formula 
2
3 
(Eb -Ea)
 
 .*a
.. 
. 
2n 

i=1 
ri
.. 
.
.b dt
 
2
Section 11-16 Examples of Ab Initio Calculations 373 
Figure 11-3 
 Transition energies in the C+ ion as calculated by HF, CI, and as measured. (From 
Weiss [19].) Ionization from the ground state of C+ occurs at 0.8958 a.u. 
In brief, then, the evidence indicates that reasonably accurate atomic ionization 
energies can be obtained by high-quality SCF calculations on the neutral and ionized 
species (SCF, not-), but that transition energies and intensities require CI sufficient 
to account for much of the valence electron correlation. 
Before we leave the subject of atoms, it should be pointed out that, for any atom, the 
expectation value T¯ of the kinetic energy operator is equal to -E¯ if the wavefunction 
has been optimized with respect to a scale factor in the coordinates r1, r2, etc. This 
relation, called the virial relation, is proved in Appendix 8. It is necessarily satisfied 
for any level of calculation that cannot be improved by replacing every ri in . by .ri 
and allowing . to vary. Since the single- and double-. STO solutions have already 
been optimized with respect to such scale parameters, they satisfy the virial relation. 
Thus, for the berylliumatom, the single-. STO value for E¯ is (Table 11-2)-14.556740 
a.u., and so we know that T¯ = +14.556740 a.u. and V¯ = -29.113480 a.u. for this 
wavefunction (since E¯ = T¯ + V¯ ). For the double-. wavefunction T¯ =+14.572369 
a.u., etc. The Hartree–Fock wavefunction is, by definition, the lowest-energy solution 
achievable within a restricted (single determinantal15) wavefunction form. Use of a 
scale factor does not affect the wavefunction form. Hence, no further lowering of E¯ 
below the HF level is possible in this way, and the HF energies E¯, T¯, and V¯ must 
satisfy the virial relation also. Finally, the exact energies are the lowest achievable for 
15For open-shell systems, more than one determinant may be needed to satisfy symmetry requirements. This 
is still considered a HF wavefunction.
374 Chapter 11 The SCF-LCAO-MO Method and Extensions 
any wavefunction. Again, scaling cannot lower the energy further, so these energies 
also satisfy the virial relation. 
We turn next to ab initio calculations on molecules. First, let us compare HF and 
exact energies for molecules as we did for atoms and see how large the errors due 
to correlation are. The results are not too different from those for atoms having the 
same number of electrons, as shown in Table 11-4; that is, the correlation energies 
for molecules having ten electrons (CH4, NH3, H2O, HF) are about the same as that 
for neon, whereas that for the 18-electron molecule H2O2 is more like the correlation 
energy for argon. But this is only a very rough rule of thumb. We have already 
indicated that the correlation energy in a molecule varies with bond length, a factor 
not present in atomic problems. In order to get a more meaningful idea of the capabilities 
of ab initio calculations on molecules, we must look more closely at specific 
examples. 
A calculation on the OH radical, reported by Cade and Huo [21], provides a good 
example of the capabilities of the extended basis set LCAO-MO-SCF technique on a 
small molecule. Their final wavefunction for the ground state at an internuclear separation 
R =1.8342 a.u. is presented in Table 11-5. A minimal basis set of STOs for 
OH would include 1s, 2s, 2px, 2py, and 2pz STOs on oxygen and a single 1s AO on 
hydrogen. Cade and Huo chose a much more extensive basis. Oxygen is the site 
for two 1s, two 2s, one 3s, four 2p, one 4f, eight 2pp, two 3dp , and four 4fp STOs. 
On hydrogen, there are two 1s, one 2s, one 2ps , two 2pp , and two 3dp STOs. (The 
p-type basis functions are indicated in Table 11-5 for only one of the two directions 
perpendicular to the O–H axis.) The orbital exponents for all of these STOs have 
been optimized, and the resulting wavefunction is of “near-Hartree–Fock” quality. The 
optimized . values appear in Table 11-5. The STO labeled s2po is located on oxygen 
and has the formula 
s2po =(2.13528)5/2p-1/2r exp(-2.13528r) cos . (11-49) 
TABLE 11-4 
 Estimated Hartree–Fock and Correlation Energies for Selected Moleculesa and 
Atomsb 
Molecule 
or atom 
E (HF) 
(a.u.) 
E (correlation) 
(a.u.) 
Molecule 
or atom 
E (HF) 
(a.u.) 
E (correlation) 
(a.u.) 
H2 -1.132 -0.043 Ne -128.547 -0.378 
He -2.862 -0.042 CO -112.796 -0.520 
BH3 -26.403 -0.195 N2 -108.994 -0.540 
O(1D) -74.729 -0.262 Si (1D) -288.815 -0.505 
CH4 -40.219 -0.291 B2H6 -52.835 -0.429 
NH3 -56.225 -0.334 S (1D) -397.452 -0.606 
H2O -76.067 -0.364 H2O2 -150.861 -0.688 
HF -100.074 -0.373 Ar -526.817 -0.725 
aFrom Ermler and Kern [20]. 
bSee Table 11-2.
Section 11-16 Examples of Ab Initio Calculations 375 
TABLE 11-5 
 Near Hartree–Fock Wavefunction for the OH Molecule in Its Ground-State 
Configuration (1s22s23s21p3) at R = 1.8342 a.u.a 
.s C1s C2s C3s .p C1p 
s1so (. = 7.01681) 0.94291 -0.25489 0.07625 p2po (. = 1.26589) 0.37429 
s1so (12.38502) 0.09313 0.00358 -0.00153 p2po (2.11537) 0.46339 
s2so (1.71794) -0.00162 0.46526 -0.20040 p2po (3.75295) 0.23526 
s2so (2.86331) 0.00418 0.55854 -0.18328 p2po (8.41140) 0.01023 
s3so (8.64649) -0.03826 -0.02643 0.00550 p3do (1.91317) 0.02871 
s2po (1.28508) -0.00055 0.05179 0.30153 p4fo (2.19941) 0.00506 
s2po (2.13528) -0.00056 -0.07538 0.37791 p2pH (1.76991) 0.02442 
s2po (3.75959) 0.00115 0.01874 0.18390 p3dH (3.32513) 0.00282 
s2po (8.22819) 0.00059 0.00229 0.00952 
s3do (1.63646) -0.00047 0.02437 0.04676 
s3do (2.82405) 0.00016 0.00845 0.01595 
s4fo (2.26641) -0.00013 0.00882 0.01232 
s1sH (1.31368) 0.00150 -0.04651 0.21061 
s1sH (2.43850) -0.00034 0.09413 0.05113 
s2sH (2.30030) 0.00000 0.07654 0.04539 
s2pH (2.8052) 0.00018 0.01182 0.00999 
aFrom Cade and Huo [21]. 
There are three s-type MOs and two p-type MOs to accommodate the nine electrons 
of this radical. One p-type MO is 
f1py = 0.37429p2poy +0.46339p2poy +0.23526p2poy 
+0.01023p2poy + 0.02871p3doy +0.00506p4foy 
+0.02442p2pHy +0.00282p3dHy (11-50) 
and the other occupied p-type MO would be the same except with x instead of y. (The 
z axis is coincident with the internuclear axis.) It is evident that writing out the complete 
wavefunction given in Table 11-5 would result in a very cumbersome expression. It is 
a nontrivial problem to relate an accurate but bulky wavefunction such as this to the 
kinds of simple conceptual schemes chemists like to use. One solution is to have a 
computer produce contour diagrams of the MOs. Such plots for the valence MOs 2s, 
3s, and 1p of Table 11-5 are presented in Fig. 11-4. 
Cade and Huo [21] carried out similar calculations for OH at 13 other internuclear 
distances and also for the united atom (fluorine) and the separated atoms in the states 
with which the Hartree–Fock wavefunction correlates. Some of their data are reproduced 
in Table 11-6. A plot of the electronic-plus-nuclear repulsion energies is given 
in Fig. 11-5 along with the experimentally derived curve. It is evident that the near HF 
curve climbs too steeply on the right, leading to too “tight” a potential well for nuclear 
motion and too small an equilibrium internuclear separation. This comes about because, 
as mentioned earlier, the HF solution dissociates to an incorrect mixture of states, some
376 Chapter 11 The SCF-LCAO-MO Method and Extensions 
Figure 11-4 
 Contour plots of HF valence orbitals for OH as given in Table 11-5. (From 
Stevens et al. [22].) 
TABLE 11-6 
 Spectroscopic Parameters and Dipole Moment for OH from Theoretical Curves 
and from Experimenta 
Wavefunction 
Dipole 
moment 
(debyes) 
Re 
(a.u.) 
De 
(eV) 
.e 
(cm-1) 
.exe 
(cm-1) 
ae 
(cm-1) 
SCF 1.780 1.795 8.831 4062.6 165.09 0.661 
CI 1.655 1.838 4.702 3723.6 83.15 0.628 
Experimental 1.66±.01 1.834 4.63 3735.2 82.81 0.714 
aFrom Stevens et al. [22]. 
of which are ionic. It is possible to use the HF curve of Fig. 11-5 to derive theoretical 
values for molecular constants that can be compared to spectroscopic data. The results 
are displayed in Table 11-6, and they reflect the inaccuracy in the HF energy curve. 
Included there are the SCF and experimental values for the molecular dipole moment. 
We turn now to the behavior of V¯ /T¯ for the HF wavefunctions of Cade and Huo 
at various internuclear separations. The data appear in Table 11-7. Observe that the
Section 11-16 Examples of Ab Initio Calculations 377 
Figure 11-5 
 Theoretical and experimental energy curves for OH (from Stevens et al. [22].) 
value of -2.00000 for V¯ /T¯ occurs at three values of R: 0,8, and the point where E¯ 
is a minimum. At R =0 and 8, we are dealing with one or two atoms, for which we 
have already seen the HF solution should give V¯ /T¯ =-2. At intermediate R we have 
a diatomic molecule, for which the virial relation is (see Appendix 8) 
2T¯ +V¯ +R
.E¯ 
.R =0 (11-51) 
There are three cases to consider. If .E¯/.R = 0, then V¯ /T¯ = -2. This will occur 
at the minimum of the potential energy curve (and also at any subsidiary maxima or 
minima). If .E¯/.R is negative, then, since V¯ /T¯ = -2 - (R/T¯)(.E¯/.R) and T¯ is 
TABLE 11-7 
 HF Total Energies and V¯/T¯ for OH as a Function of Internuclear Distancea 
R (a.u.) E (a.u.) V¯/T¯ R (a.u.) E (a.u.) V¯/T¯ 
0 -99.40933 -2.00000 1.90 -75.41837 -2.00129 
1.40 -75.34382 -1.99076 2.00 -75.41140 -2.00225 
1.50 -75.38378 -1.99398 2.10 -75.40163 -2.00300 
1.60 -75.40696 -1.99651 2.25 -75.38372 -2.00380 
1.70 -75.41829 -1.99850 2.40 -75.36367 -2.00433 
1.75 -75.42065 -1.99933 2.60 -75.33582 -2.00474 
1.795 -75.42127 -2.00000 2.80 -75.30822 -2.00492 
1.8342 -75.42083 -2.00052 8 -75.30939 -2.00000 
aFrom Cade and Huo [21].
378 Chapter 11 The SCF-LCAO-MO Method and Extensions 
positive, V¯ /T¯ will be algebraically higher than -2 (e.g.,-1.98). If .E¯/.R is positive, 
V¯ /T¯ will be lower than -2. Thus, the values of V¯ /T¯ in Table 11-7 reflect the slope of 
a line tangent to the potential energy curve at each R value. 
As mentioned earlier, it is possible to at least partly include the effects of electron 
correlation by allowing determinants corresponding to other configurations to mix into 
the wavefunction. Such calculations have been performed for the OH radical by several 
groups, and the results of Stevens et al. [22] are included in Table 11-6 and Fig. 11-5. 
These data come from intermixing 14 configurations. It is evident that the inclusion of 
correlation through CI has markedly improved the agreement with experiment. 
Many diatomic molecules have been treated at a comparable and higher level, and it 
is clear that ab initio calculations including electron correlation are capable of giving 
quite accurate molecular data. In cases in which the diatomic system is experimentally 
elusive, such calculations may be the best source of data available. Afurther example of 
this is provided in Table 11-8, in which are listed dipole moments for ground and some 
excited states of diatomic molecules. The dipole moments computed from near-HF 
wavefunctions contain substantial errors. It can be seen that CI greatly improves dipole 
moments. It has been observed that inclusion of singly excited configurations is very 
important in obtaining an accurate dipole moment. 
As a general rule, CID correlates electron motion and therefore has a significant 
energy-lowering effect but has little effect on the one-electron distribution or related 
properties, like dipole moment. Inclusion of singly excited configurations (CISD) 
allows the one-electron distribution to shift in response to the change in calculated 
interelectronic repulsion. For example, the value of the ground-state dipole moment 
of CO (entry 4 of Table 11-8) is calculated at the CID level to be -0.20D and at the 
CISD level to be +0.12D. Thus CID may be a suitable level of computational effort if 
the interest is in energy, but CISD is better if the interest is in one-electron properties. 
TABLE 11-8 
 Calculated and Experimental Dipole Moments of Diatomic Molecules 
(in Debyes) 
Molecule and polarity State HF at Re
a CI at Re
a Experiment Reference 
Li+H- X 1+ 6.002 5.86 5.83 [23] 
C+N- X 2+ 2.301 1.465 1.45±0.08 [24] 
C-N+ B 2+ — 0.958 1.15±0.08 [24] 
C-O+ X 1+ -0.274 0.12 0.112±0.005 [23] 
C+O- A 3 2.34 1.43 1.37 [25] 
C-S+ X 1+ 1.56 2.03 1.97 [23] 
C-S+ A 1 -0.09 0.63 0.63±0.04 [26] 
C-H+ X 2 1.570 1.53 1.46±0.06 [27] 
O-H+ X 2 1.780 1.655 1.66±0.01 [27] 
F-H+ X 1 1.942 1.805 1.797 [27] 
N-H+ X 3- 1.627 1.537 Unknown [27] 
aExperimental Re Value used.
Section 11-16 Examples of Ab Initio Calculations 379 
Highly accurate properties can be obtained with more sophisticated electron correlation 
methods, such as CCSD(T) or MRCISD. 
Other examples of the use of ab initio methods on small molecules are shown in 
Table 11-9, which displays some calculated properties for the electronic ground states of 
H2 andN2 as a function of method with the cc-pVTZ basis set. In the case ofH2, the SCF 
bond length and harmonic frequency are both slightly too large compared to experiment, 
but the dissociation energy is underestimated by more than 30 kcal/mole due to the lack 
of electron correlation. For this two-electron system, the CISD and CCSD methods are 
equivalent to a FCI and exhibit marked improvement compared to SCF. The remaining 
deviations from experiment at this level of theory can be attributed to the use of the 
finite cc-pVTZ basis set. The MP methods show systematic improvement with each 
order of perturbation theory, but even a fourth-order treatment of single and double 
excitations results in non-negligible errors compared to FCI for this simple system. The 
B3LYP hybrid density functional method is observed to yield very reliable properties 
in this case. 
As might be expected due to its triple bond, the N2 molecule is considerably more 
challenging for ab initio methods. With the cc-pVTZ basis set, the SCF dissociation 
energy is smaller than experiment by nearly a factor of 2. Appreciable differences are 
now observed between the CISD and CCSD results, with the latter being somewhat 
closer to experiment. In addition, triple excitations, as measured by the difference 
between CCSD and CCSD(T), are relatively important for N2, raising the dissociation 
energy by nearly 9 kcal/mole. In contrast to the H2 case, the results for N2 using 
perturbation theory (MP2, MP3, MP4) display a disturbing oscillatory behavior. This 
type of result with MP methods has been the subject of several previous studies.16 
TABLE 11-9 
 Calculated Equilibrium Bond Lengths, Harmonic Vibrational Frequencies, and 
Equilibrium Dissociation Energies for the Ground States of H2 and N2 with the cc-pVTZ Basis Set 
Compared to Experiment 
H2 N2 
re(Å) .e(cm-1) 
De 
(kcal/mole) re(Å) .e(cm-1) 
De 
(kcal/mole) 
SCF 0.734 4587 83.7 1.067 2732 120.4 
CISD 0.743 4409 108.4 1.089 2509 193.1 
MP2 0.737 4526 103.6 1.114 2195 228.7 
MP3 0.739 4476 107.1 1.090 2532 206.0 
MP4 0.741 4441 108.0 1.113 2192 221.2 
CCSD 0.743 4409 108.4 1.097 2424 207.7 
CCSD(T) 1.104 2346 216.5 
B3LYP 0.743 4419 110.3 1.092 2449 229.6 
Expt. [28] 0.741 4403 109.5 1.098 2359 228.4 
16See Dunning and Peterson [29] and references therein.
380 Chapter 11 The SCF-LCAO-MO Method and Extensions 
Lastly, it is again the case that the B3LYP method yields relatively accurate results for 
this molecule and appears to be comparable in quality to MP2. 
The results shown in Table 11-10 explore the choice of basis set with the CCSD and 
CCSD(T) methods for the H2 and N2 molecules, respectively. A large dependence on 
basis set is observed in each case. The use of a minimal basis set, STO-3G, leads to large 
errors since it provides very fewvirtual orbitals for electron correlation. Just a double-. 
basis set, either 6-31G** or cc-pVDZ, is observed to be a great improvement. The 
systematic convergence of the correlation consistent basis sets is readily observed in 
these results. One should note that increasing the size of the basis set from cc-pVTZ to 
cc-pV5Z in N2 results in an increase in De by nearly 9 kcal/mole. This implies that the 
highly accurate result for De shown in Table 11-9 for the MP2 level of theory with the 
cc-pVTZ basis set was clearly fortuitous. From these results it should be obvious that 
errors due to basis set incompleteness can often rival those due to inadequate electron 
correlation. 
As shown above, the hybrid DFT method B3LYP can be competitive in accuracy to 
more computationally expensive methods, such as CCSD(T). In fact, recent benchmark 
calculations [30] have shown that for the calculation of thermochemical quantities like 
enthalpies of formation, B3LYP exhibits average errors of only 1–5 kcal/mol. While 
these are still more than a factor of two larger than the accuracy obtainable with coupled 
cluster methods, the much lower computational cost of B3LYP makes it a very attractive 
alternative. The accuracy of equilibrium bond lengths and harmonic vibrational 
frequencies calculated by B3LYP have also been shown to be very satisfactory. The 
accurate calculation of some molecular properties, however, is still a great challenge to 
hybrid DFT methods. In particular, reaction activation energies are often too small and 
van derWaals interactions can be qualitatively incorrect. Correcting these deficiencies 
is the goal of many second generation hybrid DFT functionals.17 
We have seen that inclusion of electron correlation often improves the E¯ versus 
R curve because it allows for variable ionic-covalent character in the wavefunction. 
TABLE 11-10 
 Dependence on Basis Set Choice for the CCSD and CCSD(T) Properties of H2 
and N2, respectively 
H2/CCSD N2/CCSD(T) 
re(Å) .e(cm-1) 
De 
(kcal/mole) re(Å) .e(cm-1) 
De 
(kcal/mole) 
STO-3G 0.735 5002 128.1 1.190 2145 147.8 
6-31G** 0.738 4504 105.9 1.120 2342 201.6 
cc-pVDZ 0.761 4383 103.6 1.119 2339 200.6 
cc-pVTZ 0.743 4409 108.4 1.104 2346 216.5 
cc-pVQZ 0.742 4403 109.1 1.100 2356 222.9 
cc-pV5Z 0.742 4405 109.3 1.099 2360 225.1 
Expt. [28] 0.741 4403 109.5 1.098 2359 228.4 
17See, for instance, Zhao et al. [31] and Xu et al. [32].
Section 11-16 Examples of Ab Initio Calculations 381 
However, there are some diatomic molecules that maintain a high degree of ionic 
character even when the nuclei are quite widely separated. NaCl is an example. For 
such systems, the Hartree–Fock energy curve is quite nearly parallel to the exact energy 
curve throughout the minimum energy region (i.e., the correlation energy is almost 
constant) and the theoretical values of spectroscopic constants agree quite well with 
experimental values. (In vacuo, an electron ultimately transfers from Cl- to Na+, and 
the experimental curve leads to neutral dissociation products, whereas the HF curve 
does not. This theoretical error affects the curve only at large R, however, and so has 
little effect on spectroscopic constants.) Schaefer [13] has reviewed this situation. 
Of course, ab initio calculations have been performed on molecular systems much 
larger than the molecules referred to above. However, as one moves to molecules 
having four or more nuclei, one encounters a new difficulty: Integrals now appear that 
have the form 
ab | cd=.a(1).b(2) |1/r12|.c(1).d (2) (11-52) 
where .a is a basis function located on nucleus a, etc. Such integrals have basis functions 
on four different nuclei and are referred to as four-center integrals. If the basis 
functions . are STOs, such integrals are relatively slow to evaluate on a computer. If 
they are gaussian functions, the computation is much faster, and this is the main reason 
for using gaussian basis functions. But the number of such integrals becomes enormous 
for a reasonable basis set and a medium sized molecule. In fact, the number of such integrals 
grows as the fourth power of the number of basis functions. Thus, replacing each 
STO by, say, three gaussian functions, will lead to 34 times as many integrals to evaluate. 
Even though such integrals can be evaluated very rapidly, we eventually come to 
molecules of such a size that the sheer number of integrals makes for a substantial computing 
effort. The efficient calculation of molecular integrals continues to be an active 
research area, however, and newtechniques have nowdiminished the importance of this 
bottleneck with reasonably sized gaussian basis sets on systems up to hundreds of atoms. 
Modern quantum chemical programs have made high-quality calculations on reasonably 
large molecules tractable, but one is always balancing the level of accuracy 
against the computer time needed to achieve it. While a Hartree–Fock calculation on 
benzene with a cc-pVTZ basis set (264 contracted gaussian functions) might require 
just 4 minutes to complete on a given computer, inclusion of electron correlation at 
the MP2, CCSD, and CCSD(T) levelswould require an additional 0.1, 4.3, and 11 times 
4 minutes, respectively. 
Numerous calculations have been reported for barriers to internal rotation in various 
molecules. The theoretical barriers agree best with experiment for molecules having 
threefold symmetry in the rotor. Self-consistent-field values are compared with experimental 
barrier values inTable 11-11. In every case, the theoretical energy curve predicts 
the correct stable conformation and even does reasonably well at predicting barrier 
height. The disagreement between different computed values of the barrier for the 
same molecule reflects differences in basis sets and, sometimes, differences in choices 
for bond length and angle made by different workers. The evidence to date suggests 
that ab initio calculations approaching the HF limit will ordinarily be within 20% of 
the experimental barrier. Even this level of accuracy is useful because experimental 
measurements of barriers in transient molecules or for excited molecules are often very 
rough, ambiguous, or nonexistent. Given the favorable cost and relative accuracy of
382 Chapter 11 The SCF-LCAO-MO Method and Extensions 
TABLE 11-11 
 Internal Rotation Barriers from Experiment and as 
Calculated by the LCAO-MO-SCF Method 
Barrier (kcal/mole) 
Molecule SCF Experiment Reference 
CH3–CH3 2.58 2.88 [33] 
2.88 — [34] 
CH3–NH2 1.12 1.98 [35] 
2.02 — [34] 
CH3–OH 1.59 1.07 [34] 
CH3–CH2F 2.59 3.33 [33] 
CH3–N=O 1.05 1.10 [36] 
CH3–CH=CH2 1.25 1.99 [37] 
cis-CH3–CH=CFH 1.07 1.06 [37] 
trans-CH3–CH=CFH 1.34 2.20 [37] 
CH3–CH=O 1.09 1.16 [38] 
DFT approaches compared to HF, even higher quality results might be expected with 
the use of methods such as B3LYP; hence, DFT is often now the method of choice for 
calculations on medium to large organic molecules. 
A large number of ab initio calculations have been made on clusters of molecules. 
Many of these have sought to delineate the distance and angle dependence of the hydrogen 
bond strength between molecules like water or hydrogen fluoride. Xantheas et al. 
[39] have reported large basis set MP2 calculations on small water clusters, (H2O)n, 
where n ranged from 2–6. These calculations predict that there are four distinct isomers 
of the water hexamer (n = 6) whose relative energies lie within ~1 kcal/mole of each 
other. These kinds of results are of great usefulness in defining new effective interaction 
potentials involving water that can be used in large-scale molecular simulations of 
solvation phenomena. Re et al. [40] have calculated the structures and relative energies 
of sulfuric acid solvated by 1–5 water molecules using the B3LYP method to provide a 
fundamental understanding of acid ionization. In addition to investigating the interaction 
of water with both the cis and trans conformers of H2SO4, they found that just five 
water molecules were sufficient tomake dissociation intoHSO-4
andH3O+ energetically 
favorable. The field of materials science is also benefitting from ab initio calculations, 
and studies of metal clusters and their absorbates are currently areas of high interest. 
A great deal of attention has been given to the calculation by ab initio methods of 
energy surfaces for chemical reactions. For many years, such efforts were limited to 
reactions, such as D+H2.HD+H, which involve only a small number of electrons 
and nuclei. Much more complicated systems are now being explored. 
In setting out to perform such a calculation, one likes to have some idea of whether the 
correlation energy of the system will change significantly with nuclear configuration. 
If it does not, then a Hartree–Fock or MCSCF calculation will parallel the true energy 
surface. If the correlation energy does change, it is necessary to include some treatment 
of electron correlation in the calculation.
Section 11-16 Examples of Ab Initio Calculations 383 
As a rough rule of thumb, one expects the correlation energy to change least when the 
reactants, the intermediate or transition state complex, and the products are all closedshell 
systems, hence all approximately equally well described by a single-determinantal 
wavefunction. Some calculations on SN2 and radical reactions are summarized in 
Table 11-12. It can be seen that the SN2 reactions, which do involve closed-shell 
systems in the three stages mentioned above, are fairly insensitive to the inclusion of CI, 
whereas the radical reactions undergo extensive change of correlation energy. 
The determination of the potential energy surface for the unimolecular rearrangement 
HOCl
HClO by Peterson et al. [43] provides an example of a very accurate 
and exhaustive calculation on a fairly small molecule. Because there are only three 
nuclei, there are only three structural variables to explore, so the number of calculations 
needed to map out the surface is not too large. (Note that, with three geometric variables, 
the energy “surface” is really a four-dimensional hypersurface.) These authors 
were also interested in the reactions occurring on this surface, i.e., Cl+OH.HCl+O 
and Cl +OH.ClO+H, which required a global representation of the surface that 
was constructed from over 1500 individual energies. Since the full energy surface 
involves bond breaking processes, MCSCFand MRCISD methods were utilized. Accurate 
relative energetics between HOCl, HClO, and the various dissociation asymptotes 
were obtained by carrying out calculations with a series of three correlation consistent 
basis sets at each geometry. This produced an approximate complete basis set (CBS) 
MRCISD energy surface. At the MRCISD CBS limit, HOCl was found to be more 
stable than HClO by 53.7 kcal/mole and the barrier for HOCl.HClO was predicted 
to be 73.5 kcal/mole above the HOCl minimum. 
After determining an analytical representation of this surface from the individual 
energies, these authors carried out calculations of the full anharmonic vibrational spectrum 
of HOCl and HClO by solving the Schr¨odinger equation for nuclear motion. 
The HClO molecule has not yet been experimentally observed, but these calculations 
predict that the lowest three vibrational levels of this species lie below its dissociation 
threshold, so it should be detectable. 
TABLE 11-12 
 Reaction Barrier Energies for Reactions as Calculated by ab Initio Methods 
Reaction barrier 
(kcal/mole) 
Reactant Transition Product Reaction type SCF 
CI 
(no.config.) Reference 
H- +CH4 (CH5)- CH4 + H- SN2 59.3 55.2(6271) [41] 
F- +CH3F (FCH3F)- CH3F + F- SN2 5.9 5.9(26910) [41] 
H• +CH4 CH•5 CH•3 + H2 Radical 
abstraction 
(axial) 35.2 18 (692) [42] 
H• +CH4 CH•5 CH4 + H• Radical 
exchange 
(inversion) 63.7 41.7(692) [42]
384 Chapter 11 The SCF-LCAO-MO Method and Extensions 
The decisions regarding basis set and level of correlation can be daunting and in 
the past this sometimes discouraged nonspecialists from taking advantage of ab initio 
methods. However, there are now a wide range of programs that are available, which 
have made ab initio calculations amenable to theoreticians and experimentalists alike. 
The best known of these is undoubtedly the GAUSSIAN series of programs originally 
developed by the group of J.A. Pople. In this and other programs, one can conveniently 
choose from a large variety of available basis sets and methods to carry out energy 
evaluations or geometry optimizations and harmonic frequency calculations. These 
programs have brought about a revolution in the way that chemical research is done. 
For small molecules (~1–5 nonhydrogen atoms) ab initio methods are sometimes 
more precise and reliable than experiment, especially for unstable systems. The saga 
of the energy difference between ground and excited CH2 is one of the best known of 
these experimental–theoretical confrontations.18 
In summary, ab initio calculations provide useful data on bond lengths and angles, 
molecular conformation and internal rotation barriers, for ground and excited states 
of molecules. They are also very useful for calculating accurate thermochemistry, 
ionization energies, oscillator strengths, dipole moments (as well as other one-electron 
properties) and excitation energies. If one has access to large blocks of computer time, 
ab initio calculations can reveal the nature of energy surfaces pertaining to chemical 
reactions or molecular associations, as in fluids. The accuracy of the calculation and 
the magnitude of the system are limited ultimately by computer speed and capacity. 
11-17 Approximate SCF-MO Methods 
At the beginning of this chapter it was stated that ab initio calculations require exact 
calculation of all integrals contributing to the elements of the Fock matrix, but we have 
seen that, as we encounter systems with more and more electrons and nuclei, the number 
of three- and four-center two-electron integrals becomes enormous, driving the cost of 
the calculation out of the reach of most researchers. This has led to efforts to find sensible 
and systematic simplifications to the LCAO-MO-SCF method—simplifications 
that remain within the general theoretical SCF framework but shorten computation of 
the Fock matrix. 
Since many of the multicenter two-electron integrals in a typical molecule have very 
small values, the obvious solution to the difficulty is to ignore such integrals. But we 
wish to ignore them without having to calculate them to see which ones are small since, 
after all, the reason for ignoring them is to avoid having to calculate them. Furthermore, 
we want the selection process to be linked in a simple way to considerations of basis set. 
That is, when we neglect certain integrals, we are in effect omitting certain interactions 
between basis set functions, which is equivalent to omitting some of our basis functions 
part of the time. It is essential that we know exactly what is involved here, or we may 
obtain strange results such as, for example, different energies for the same molecule 
when oriented in different ways with respect to Cartesian coordinates. 
A number of variants of a systematic approach meeting the above criteria have been 
developed by Pople and co-workers, and these are nowwidely used. The approximations 
18See Goddard [44] and Schaefer [45].
Section 11-17 Approximate SCF-MO Methods 385 
are based on the idea of neglect of differential overlap between atomic orbitals in 
molecules. 
Differential overlap dS between twoAOs, .a and .b, is the product of these functions 
in the differential volume element dv: 
dS =.a(1).b(1)dv (11-53) 
The only way for the differential overlap to be zero in dv is for .a or .b, or both, to 
be identically zero in dv. Zero differential overlap (ZDO) between .a and .b in all 
volume elements requires that .a and .b can never be finite in the same region, that is, 
the functions do not “touch.” It is easy to see that, if there is ZDO between .a and .b 
(understood to apply in all dv), then the familiar overlap integral S must vanish too. 
The converse is not true, however. S is zero for any two orthogonal functions even if 
they touch. An example is provided by an s and a p function on the same center. 
It is a much stronger statement to say that .a and .b have ZDO than it is to say 
they are orthogonal. Indeed, it is easy to think of examples of orthogonal AOs but 
impossible to think of any pair ofAOs separated by a finite or zero distance and having 
ZDO. Because AOs decay exponentially, there is always some interpenetration. 
The attractive feature of the ZDO approximation is that it causes all three- and fourcenter 
integrals to vanish. Thus, in a basis set of AOs . having ZDO, the integral 
.a(1).b(2) |1/r12|.c(1).d (2) will vanish unless a =c and b =d. This arises from 
the fact that, if a =c, .*a (1).c(1) is identically zero, and this forces the integrand to 
vanish everywhere, regardless of the value of (1/r12).*b (2).d (2). 
It is not within the scope of this book to give a detailed description or critique of 
the numerous computational methods based on ZDO assumptions. An excellent monograph 
[46] on this subject including program listings is available. Some of the acronyms 
for these methods are listed in Table 11-13. In general, these methods have been popular 
because they are relatively cheap to use and because they predict certain properties 
(bond length, bond angle, energy surfaces, electron spin resonance hyperfine splittings, 
molecular charge distributions, dipole moments, heats of formation) reasonably well. 
However, they generally do make use of some parameters evaluated from experimental 
data, and some methods are biased toward good predictions of some properties, while 
other methods are better for other properties. For a given type of problem, one must 
exercise judgment in choosing a method. 
As an example of the sort of chemical system that becomes accessible to study using 
such methods, we cite the valence-electron CNDO/2 calculations of Maggiora [56] 
on free base, magnesium, and aquomagnesium porphines. Such calculations enable 
us to examine the geometry of the complex (i.e., is the metal ion in or out of the 
molecular plane, and how is the water molecule oriented?), the effects of the metal 
ion on ionization energies, spectra, and orbital energy level spacings, and the detailed 
nature of charge distribution in the system. 
Use of a combination of methods is often convenient. Novoa andWhangbo [57] studied 
theoretically the relative stabilities of di- and triamides in various hydrogen-bonded 
and nonhydrogen-bonded conformations, in both the absence and presence of solvent 
(CH2Cl2) molecules. There are many structural parameters to optimize in each of 
the conformations, and so high-level ab initio calculations for energy minimization of 
each class of structure would be prohibitively expensive. Instead, AM1 was used to 
determine the optimum geometry for each configuration, and then ab initio calculations
386 Chapter 11 The SCF-LCAO-MO Method and Extensions 
TABLE 11-13 
 Acronyms for Common Approximate SCF Methods 
Acronym Description 
CNDO/1 Complete neglect of differential overlap. Parametrization Scheme no. 1 
(Pople and Segal [47]). 
CNDO/2 Parametrization scheme no. 2. Considered superior to CNDO/1 (Pople 
and Segal [48]). 
CNDO/BW Similar to above with parameters selected to give improved molecular 
structures and force constants. (See Pulfer and Whitehead [49] and 
references therein.) 
INDO Intermediate neglect of differential overlap. Differs from CNDO in that 
ZDO is not assumed between AOs on the same center in evaluating 
one-center integrals. This method is superior to CNDO methods for 
properties, such as hyperfine splitting, or singlet-triplet splittings, 
which are sensitive to electron exchange (Pople et al. [50]). 
MINDO/3 Modified INDO, parameter scheme no. 3. Designed to give accurate 
heats of formation (Bingham et al. [51] and also Dewar [52]). 
NDDO Neglect of diatomic differential overlap. Assumes ZDO only between 
AOs on different atoms (Pople et al. [53]). 
MNDO Modified neglect of diatomic overlap. A semiempirically parametrized 
version of NDDO.Yields accurate heats of formation and many 
other molecular properties, but fails to successfully account for 
hydrogen bonding (Dewar and Thiel [54]). 
AM1 “Austin Model 1.” A more recent parametrization of NDDO that 
overcomes the weakness of MNDO in that it successfully treats 
hydrogen bonding. (Dewar et al. [55].) 
(e.g., 6-31G** with MP2) were done for a few near-optimum geometries for each conformation 
to check the AM1 results. 
Additional helpful information on standard programs available at ab initio and 
semiempirical levels—where to get them, how to use them, what they have been used 
for—is available in the very well-written reference handbook by Clark [58]. 
11-17.A Problems 
11-1. Use the data in Table 11-3 to calculate the theoretical transition energies for Ne+ 
when 1s and 2s electrons are excited into the 2p level. The experimental values 
are 2p.2s, 0.989 a.u.; 2p.1s, 31.19 a.u. 
11-2. Use the data in Table 11-1 to estimate separately the errors in ionization energies 
for the three states due to 
a) omission of electron correlation. 
b) failure to allow electronic relaxation. 
11-3. In Section 11-11, it is argued that neglect of electron correlation and electronic 
relaxation in setting I 0 
k =-k causes errors of opposite sign that partly cancel.
Section 11-17 Approximate SCF-MO Methods 387 
Would this also occur whenKoopmans’theorem is used to predict electron affinities? 
Why? 
11-4. Demonstrate that, if 
D1 =
 
a c 
b d
 
and D2 =
 
a e 
b f
 
then D1 +.D2 =
 
a c+.e 
b d+.f
 
11-5. A singly excited configuration .1 differs by one orbital from the ground state 
.0 and also by one orbital from certain doubly excited configurations .2. 
Brillouin’s theorem gives .0|Hˆ |.1=0, but not .1|Hˆ |.2=0. Where does 
the attempted proof to show that .1|Hˆ |.2=0 break down? 
11-6. Show that, if . = c0.0 + c1.1 + c2.2 +···+cn.n, and if . is to be an 
eigenfunction of Aˆ with eigenvalue a1, then it is necessary that all the .i(i = 0, . . . , n) also be eigenfunctions of Aˆ with eigenvalues a1. 
11-7. How many distinct four-center coulomb and exchange integrals result when one 
has four nuclei, each being the site of five basis functions? Make no assumptions 
about symmetry or basis function equivalence or electron spin. 
11-8. For a homonuclear diatomic molecule, which of the following singly excited 
configurations would be prevented for reasons of symmetry from contributing 
to a CI wavefunction for which the main “starting configuration” is 
1s2 
g 1s2 
u 2s2 
g 1p4 
u ? 
a) 1s2 
g 1s2 
u 2s2 
g 1p3 
u 1pg (i.e., 1pu.1pg) 
b) 2sg.3sg 
c) 2sg.1pg 
d) 1sg.3sg 
11-9. Write down the hamiltonian operator for electrons in the water molecule. Use 
summation signs with explicit index ranges. Use atomic units. 
11-10. An SCF calculation on ground state H2 at R =1.40 a.u. using a minimal basis 
set gives a sg and a su MO having energies 
sg =-0.619 a.u. su =+0.401 a.u. 
The nonvanishing two-electron integrals over these MOs are 
  sg(1)sg(2)(1/r12)sg(1)sg(2)dv(1)dv(2) = 0.566 a.u. 
  sg(1)su(2)(1/r12)sg(1)su(2)dv(1)dv(2) = 0.558 a.u. 
  sg(1)su(2)(1/r12)sg(2)su(1)dv(1)dv(2) = 0.140 a.u. 
  su(1)su(2)(1/r12)su(1)su(2)dv(1)dv(2) = 0.582 a.u. 
a) Write down the Slater determinant for the ground state of H2. 
b) Calculate the SCF electronic energy for H2 at R =1.40 a.u.
388 Chapter 11 The SCF-LCAO-MO Method and Extensions 
c) Calculate the total (electronic plus nuclear repulsion) energy for H2. 
d) What is the bond energy for H2 predicted by this calculation, assuming that 
the minimum total energy occurs at R =1.40 a.u.? 
e) Estimate the (vertical) ionization energy for H2. 
f) What is the value of the kinetic-plus-nuclear-attraction energy for one electron 
in ground-state H2 according to this calculation? 
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Section 11-17 Approximate SCF-MO Methods 389 
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390 Chapter 11 The SCF-LCAO-MO Method and Extensions 
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Chapter 12 
Time-Independent 
Rayleigh–Schr ¨ odinger 
Perturbation Theory 
12-1 An Introductory Example 
Imagine a city having one million resident wage earners. The city government plans 
to raise additional revenue by assessing each such resident a wage tax. This tax will 
not apply to wage earners residing in suburbs. The government estimates that a $10 
assessment will bring in new revenues of $10 million. This estimate assumes that the 
city population before imposition of the tax will hold after the tax is imposed as well. 
But the tax will produce a slight change, or perturbation, in the economic climate of 
the city. It is true that the tax is small, and so few people are likely to move to the 
suburbs as a result of it. Therefore, it is probably fairly accurate to use the population 
of the city before the perturbation to calculate the change in revenue brought about 
by the perturbation. But, if the perturbation were large, say $1000 a head, we would 
expect a substantial migration of wage earners to the suburbs, and so the estimate 
produced by using the original unperturbed population would contain substantial error. 
Corrections should be made, therefore, to account for population changes produced by 
the perturbation. 
This use of the unperturbed population to calculate the change in revenue is a crude 
example of a certain level of estimation (called “first order”) in Rayleigh–Schr¨odinger 
perturbation theory.1 We will now proceed to develop the theory more formally in 
the context of wavefunctions and energies. The above example has been presented to 
encourage the reader to anticipate that there is a lot of simple good sense in the results of 
perturbation theory even though the mathematical development is rather cumbersome 
and unintuitive. 
12-2 Formal Development of the Theory 
for Nondegenerate States 
Perturbation theory involves starting with a system with known hamiltonian, eigenvalues, 
and eigenfunctions, and calculating the changes in these eigenvalues and eigenfunctions 
that result from a small change, or perturbation, in the hamiltonian for the 
1Other perturbation methods exist, but the Rayleigh–Schr¨odinger (R-S) theory is the oldest and the most widely 
used in quantum chemistry. 
391
392 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
system. We restrict the discussion to stationary states of systems having hamiltonians 
that are not time dependent. Let the known, unperturbed system have H0 as hamiltonian, 
and let the eigenfunctions .i be orthonormal. Then 
H0.i = Ei.i (12-1) 

 .*i .j dt = dij (12-2) 
and the functions . form a complete set as discussed in Chapter 6. We are interested 
in the system with hamiltonian 
H =H0 +.H (12-3) 
where .H is the perturbation.2 (The parameter . is a scalar quantity that will be 
convenient in the mathematical development of the theory. When the derivation is 
complete, we will set . equal to unity so that it no longer appears explicitly in any 
formula, and then H must account entirely for the perturbation.) The eigenvalues and 
eigenfunctions of the perturbed hamiltonian H are unknown. Let us symbolize them 
as W and f, respectively. Then 
Hfi =Wifi (12-4) 
It is clear that, when .=0, then H =H0, fi =.i and Wi =Ei. As . is increased 
from zero, Wi and fi change in (we expect) a continuous way. In other words, Wi 
and fi are continuous functions of the variable parameter ., and they are known at the 
particular value .=0. Therefore, we can expand them3 as series in powers of . about 
the point .=0. Thus, for a given state i,4 
Wi =.0W(0) 
i +.1W(1) 
i +.2W(2) 
i +.3W(3) 
i +··· (12-5) 
Here we must remember that . is a variable and W(0) 
i , W(1) 
i , etc. are constants. The 
superscript in parentheses is simply a label to tell us for which power of . this constant 
is the coefficient. Since we know that Wi =Ei when .=0, we see at once that W(0) 
i in 
Eq. (12-5) is equal to Ei . Our problem is to evaluate W(1) 
i , W(2) 
i , etc. It is traditional 
to call W(0) 
i (or Ei ) the unperturbed energy or, sometimes, the energy to zeroth order. 
.W(1) 
i (which is just W(1) 
i after . is set equal to unity) is the first-order correction to 
the energy, and W(0) 
i +.W(1) 
i is the energy to first order. .2W(2) 
i , .3W(3) 
i , . . . , etc. 
are the second-, third-, etc., order corrections to the energy. Normally, for expansion 
in a series to be useful, it is necessary for the series to converge at a reasonable rate. 
In most simple applications of perturbation theory, only a few orders of correction are 
made. Thus, one very commonly reads of energies calculated “to first order,” or “to 
second order.” Calculations to much higher orders are also made, but these are not as 
common. 
2Some treatments expand H as H0 +.H +.2H+··· and ultimately achieve working formulas that appear 
different from those we will achieve. In fact, they are equivalent. For an example of this alternative formulation, 
see Pauling andWilson [1, Chapter 6]. 
3In effect, we assume fi and W to be analytic functions of . in the range 0=.=1. 
4It is important to recognize that henceforth the subscript i will refer to the state that we are studying as a 
function of ..
Section 12-2 Formal Development of the Theory for Nondegenerate States 393 
In precisely the same manner, we can expand the unknown eigenfunction fi as a 
power series in .: 
fi =f(0) 
i +.f(1) 
i +.2f(2) 
i +.3f(3) 
i +··· (12-6) 
Since fi is a function of particle coordinates, f(0) 
i , f(1) 
i , etc. are also functions, but 
they are invariant to changes in .. Again, it is clear that f(0) 
i = .i . f(0) 
i (or .i) is 
the unperturbed or zeroth-order wavefunction, .f(1) 
i is the first-order correction to the 
wavefunction, etc. 
Now we substitute Eqs. (12-3), (12-5), and (12-6) into (12-4) and obtain 
(H0 +.H) f(0) 
i +.f(1) 
i +.2f(2) 
i +··· 
=W(0) 
i +.W(1) 
i +.2W(2) 
i +···f(0) 
i +.f(1) 
i +.2f(2) 
i +··· (12-7) 
The variable in Eq. (12-7) is ., and each power of . is linearly independent of all other 
powers of .. As indicated in Section 3-4D, this means that Eq. (12-7) can be satisfied 
for all values of . only if it is satisfied for each power of . separately. Collecting terms 
having the zeroth power of . gives
H0f(0) 
i =W(0) 
i f(0) 
i (12-8) 
However, we have already recognized that 
f(0) 
i =.i, W(0) 
i =Ei (12-9) 
and Eq. (12-8) is simply a restatement of Eq. (12-1). Collecting terms from Eq. (12-7) 
containing . to the first power we obtain 
. Hf(0) 
i +H0f(1) 
i -W(0) 
i f(1) 
i -W(1) 
i f(0) 
i =0 (12-10) 
This equality must hold for any value of ., so the term in parentheses is zero. Hence, 
rearranging and making use of Eqs. (12-9), we have the first-order equation 
H -W(1) 
i .i +(H0 -Ei)f(1) 
i =0 (12-11) 
Let us multiply this from the left by .*i and integrate: 

 .*i H.i dt -W(1) 
i 

 .*i .i dt +
 .*i H0f(1) 
i dt -Ei 
 .*i f(1) 
i dt =0 
(12-12) 
Using the hermitian property of H0, it is easy to show that the third and fourth terms 
cancel, leaving 
W(1) 
i =
 .*i H.i dt (12-13a)
394 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
Thus we have arrived at an expression for the first-order correction to the energy in 
terms of known quantities. It is the expectation value for the perturbation operator 
calculated using the wavefunction of the unperturbed system. The analogy between 
this formula and the use of the population of the unperturbed city to calculate additional 
revenues from a new tax should be apparent. H corresponds to the tax per wage earner, 
and .*i .i dt corresponds to the sum of wage earners in the city before the tax was 
imposed. 
EXAMPLE 12-1 A one-dimensional box potential is perturbed so that it is raised by 
a constant amount, d, in the left half of the box, and lowered by d in the right half. 
What is the first-order change in energy for the lowest-energy state .1? for .2? 
SOLUTION 
 H is antisymmetric for reflection through the box center, and .2
1 is symmetric, 
so the integral givingW(1) 
1 equals zero by symmetry. The same argument applies to .2. Another 
way to argue is to recognize that .2
n is symmetric in the box (for all n), which means that half of 
the distribution responds to the energy increase and half to the equal energy decrease, so the net 
energy change to first order is zero for all states.  
Equation (12-13a) can be written in bra-ket notation (see Appendix 9): 
W(1) 
i =.i
H
.i  (12-13b) 
or in the notation wherein an integral is indicated by affixing subscripts to the operator 
(i.e., as a matrix element) 
W(1) 
i =H ii (12-13c) 
In most discussions of perturbation theory one of these alternative notations is used. 
We will continue our formal development from this point using bra-ket notation for 
integrals. 
To find an expression for f(1) 
i , the first-order correction to the wavefunction for the 
ith state, we first recognize that we can expand f(1) 
i in terms of the complete set of 
eigenfunctions .: 
f(1) 
i =
j 
c(1) 
ji .j (12-14) 
The summation symbol suggests that . is a discrete set of functions. This need not be 
true. Contributions from functions whose eigenvalues are in a continuumwould require 
integration rather than summation. However, we will use the sum symbol since most 
actual applications of perturbation theory in quantum chemistry invoke only discrete 
functions. 
We now insert Eq. (12-14) into (12-11) to obtain 
H -W(1) 
i .i +(H0 -Ei)
j 
c(1) 
ji .j =0 (12-15)
Section 12-2 Formal Development of the Theory for Nondegenerate States 395 
Multiplying from the left by .*k and integrating yields 
.k|H|.i-W(1) 
i .k|.i+
j 
c(1) 
ji 	.k|H0|.j-Ei .k|.j 
=0 (12-16) 
If k=i, this reduces to Eq. (12-13), as was shown above. If k =i, the terms in the sum 
all vanish except when j =k. Thus, 
.k|H|.i+c(1) 
ki 	.k|H0|.k-Ei .k|.k
=0, (k=i) (12-17) 
or 
c(1) 
ki = .k|H|.i  
Ei -Ek 
, k=i (12-18) 
Inserting this into Eq. (12-14) gives an expression for f(1) 
i : 
f(1) 
i =
j =i 
.j |H|.i  
Ei -Ej 
.j (12-19) 
This formula prescribes the way the first-order correction to the wavefunction is to be 
built up from eigenfunctions of the unperturbed system. We discuss this formula in 
detail later when considering an example. 
Note that, if the perturbed state of interest (the ith) is degenerate with another state 
(the lth), Eqs. (12-18) and (12-19) blow up for k or j equal to l unless .l |H|.i  vanishes. Therefore, we restrict the theoretical discussion of this section to states of 
interest that are nondegenerate and discrete. (States of the system other than the ith 
may be degenerate or continuum states, however.) 
If we extract the terms containing .2 from Eq. (12-7) and proceed, in the same way 
as above, to expand f(2) 
i as a linear combination of unperturbed eigenfunctions, 
f(2) 
i =
j 
c(2) 
ji .j (12-20) 
we arrive, after some manipulation, at the following formula for W(2) 
i : 
W(2) 
i = 
j (=i) 
.i |H|.j .j |H|.i  
Ei -Ej 
(12-21) 
Comparing this with Eq. (12-18) allows us to write 
W(2) 
i = 
j (=i)

c(1) 
ji  
2 
(Ei -Ej ) (12-22) 
or comparing Eqs. (12-19) and (12-21) gives 
W(2) 
i =.i |H|f(1) 
i  (12-23)
396 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
The formula that emerges for c(2) 
ji of Eq. (12-20) is 
c(2) 
ji = 
k(=i) 
.j |H|.k.k|H|.i  
	Ei -Ek
 	Ei -Ej 
 - .i |H|.i .j |H|.i  
(Ei -Ej )2 , i=j (12-24) 
The boxed equations are “working equations” since they enable us to calculate the 
correction terms from the known eigenvalues and eigenfunctions of the unperturbed 
system. Equation (12-23) is interesting because it indicates that the second-order correction 
to the energy is calculable if we know the zeroth and first-order functions.5 
Comparing Eqs. (12-13) and (12-23) shows that, whereas the first-order correction for 
the energy is the average value for the perturbation operator with the unperturbed wavefunction, 
the second-order correction is an interaction element between two functions, 
and not an average value in the usual sense. 
Higher-order correction terms may be found by proceeding in a similar way with 
.3,.4, etc. terms from Eq. (12-7). The equations become progressively more cumbersome 
and will be of no interest to us for applications to be considered in this book. 
Having made use of the parameter . to keep terms properly sorted, we can now 
dispense with it by setting it equal to unity. Then 
H = H0 +H (12-25) 
Wi = Ei +W(1) 
i +W(2) 
i +··· (12-26) 
fi = .i +f(1) 
i +f(2) 
i +··· (12-27) 
Notice that fi is not normalized. The normalization coefficient needed will depend on 
the order to which fi has been calculated.6 
12-3 A Uniform Electrostatic Perturbation of 
an Electron in a “Wire” 
12-3.A Description of the System 
Suppose that a small uniform electric field is applied to an electron somehowconstrained 
to move on a line segment of length L. In the absence of this field, we assume the 
electron states to be described by the one-dimensional “box” wavefunctions discussed 
5L¨owdin [2] has shown that, if we know all the fi ’s up to f
(n) 
i , we can calculate all theWi ’s up to and including 
W
(2n+1) 
i . 
6One can guarantee normality up to second order in f2
i by setting c
(1) 
ii =0, c
(2) 
ii =-1
2k
c
(1) 
ki  
2 
(see Schiff [3]).
Section 12-3 A Uniform Electrostatic Perturbation of an Electron in a “Wire” 397 
Figure 12-1 
 Potential of an electron in a line segment of length L in the presence of a uniform 
electric field. The perturbation is “small” if the potential change u across L is small compared to 
p2/2L2, the energy in atomic units of the lowest unperturbed state. 
in Chapter 2. The electric field will be treated as a perturbation. The potential energy 
of the electron is sketched in Fig. 12-1. A uniform field produces a constant gradient 
(in potential energy) along the line segment. For purposes of discussion, we let the 
perturbation rise from zero at x =0 to u at x =L, but we shall see later that there is a 
degree of arbitrariness here. The perturbation, then, is given by 
H =ux/L, 0=x =L (12-28) 
where u, x, and L are measured in atomic units. 
12-3.B The Energy to First Order 
We now ask, what is the effect, to first order, of H on the energy of the lowestenergy 
state of the electron? As described in the preceding section, this is obtained by 
calculating the average value of H using the unperturbed wavefunction: 
W(1) 
1 =
 L 
o 
2/Lsin(px/L)
 (ux/L) 2/Lsin(px/L)
dx (12-29) 
We now describe three ways to evaluate this integral. One way is to integrate explicitly, 
the other two ways involve simple inspection. 
1. Explicit Integration. Factoring constants from Eq. (12-29) gives 
W(1) 
1 =2u/L2 
 L 
0 
x sin2(px/L) dx (12-30) 
To achieve a common variable, we multiply x and dx each by p/L and outside by 
L2/p2, thereby keeping the value unchanged: 
W(1) 
1 =2u/p2 
 x=L 
x=0 
(px/L) sin2(px/L)d(px/L) (12-31) 
Letting px/L=y and noting that y =0, p when x =0, L, we have 
W(1) 
1 =2u/p2 
 p 
0 
y sin2 y dy (12-32)
398 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
Standard tables (see also Appendix 1) lead to a value of p2/4 for the integral, and 
W(1) 
1 =(2u/p2)(p2/4)=u/2 (12-33) 
To first order, then, the energy of the lowest-energy perturbed state is, in atomic 
units, 
W =(p2/2L2)+u/2 (12-34) 
2. Evaluation by Inspection: First Method. Equation (12-34) is certainly a reasonable 
result since, as the potential is increased everywhere in the box (except at x =0), 
we expect that the energy of the electron should also increase. The fact that the 
increase is such a simple quantity (u/2) suggests that there might be a simple way 
to understand this result, and this is indeed the case. Consider the distribution of 
the electron in the lowest unperturbed state, shown in Fig. 12-2. This distribution is 
symmetric about the midpoint of the line segment. Consequently, for each instant of 
time the unperturbed electron spends in element dx1 of Fig. 12-2, it spends an equal 
instant in the symmetrically related element dx2. In other words, for each instant the 
electron experiences a perturbation potential less than u/2, it experiences an instant 
of potential greater (by an equal amount) than u/2. Hence, the average potential 
must be precisely u/2. Since we know that .2 is symmetric for every state in the 
unperturbed box, we can immediately extend our result and say that the first-order 
correction to the energy of every state is u/2. 
The ability to evaluate first-order energies by inspection is very useful. Even in 
cases where exact evaluation by this technique is not possible, it may still be useful 
in making an estimate or in checking the reasonableness of a computed result. 
3. Evaluation by Inspection: Second Method. A variation of the above approach is 
sometimes useful. We begin by recognizing that, whereas H is neither symmetric 
nor antisymmetric about the midpoint of the wire, we can make it antisymmetric by 
subtracting the constant u/2, as indicated in Fig. 12-3. 
Figure 12-2 
 .2(n=1) for a particle in the unperturbed box.
Section 12-3 A Uniform Electrostatic Perturbation of an Electron in a “Wire” 399 
Figure 12-3 
 The function ux/L is unsymmetric about L/2, but ux/L-u/2 is antisymmetric 
about L/2. 
By writing 
H =ux/L-u/2+u/2=Hanti +u/2 (12-35) 
we express H as an antisymmetric function plus a constant. Our integral for W(1) 
1 
becomes
W(1) 
1 =
 L 
0 
.n=1Hanti.n=1 dx +
 L 
0 
.n=1(u/2).n=1 dx (12-36) 
The first integral vanishes because .2 is symmetric. The second integral is just u/2 
times unity since . is normalized. Hence, W(1) 
1 =u/2. 
EXAMPLE 12-2 A one-dimensional box potential is perturbed so that it is raised 
by a constant amount, d, on the left side of the box but is unchanged on the right 
side. What is the first-order change in energy for .1? for .2? 
SOLUTION 
 One can argue, similarly to the reasoning in Example 12-1, that half of the 
distribution responds to the increase in potential, half to zero change, giving a net first-order 
increase of d/2. Or, one can turn the perturbation into an antisymmetric function plus a constant: 
H =H -d/2+d/2=Hanti +d/2, leading toW(1) 
1 =d/2. The same result applies to .2 and all 
higher states because .2
n is symmetric in the box for all n.  
12-3.C The First-Order Correction to .1 
How should we expect the lowest-energy wavefunction to change in response to the 
perturbation? Since we are dealing with the lowest-energy state, we might expect 
the electron to spend more time in the low-potential end of the box (the nonclassical 
result), and so the wavefunction should tend to become skewed, as shown in 
Fig. 12-4. In this figure it is demonstrated how the perturbed wavefunction can be 
resolved into an unperturbed wavefunction and a correction, or difference, function. 
Since this correction function must increase . on the low potential side and decrease 
. on the high potential side, it is clear that it must be close to antisymmetric in nature. 
According to Eq. (12-19), the first-order approximation f(1) 
1 to this correction function 
is formed by adding together small amounts of higher-energy wavefunctions. Comparing 
the correction function in Fig. 12-4 and these higher-energy wavefunctions 
(Fig. 12-5) leads us to expect that .2 will be a heavy contributor, whereas .3, being
400 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
Figure 12-4 
 Sketches of the lowest-energy wavefunction (a) before and after perturbation and 
(b) the difference between them. 
Figure 12-5 
 Sketches of the (a) second, (b) third, and (c) fourth (n=2, 3, 4, respectively) wavefunctions 
. for the particle in the unperturbed box. 
symmetric, will not contribute strongly. The mixing coefficient for .2 is, according to 
Eq. (12-18), 
c(1) 
21 =  L 
0 	v2/Lsin 2px/L
 	ux/L
 	v2/Lsinpx/L
dx 
-3p2/2L2 = 
32L2u 
27p4 (12-37) 
Thus, c(1) 
21 is positive, and .2 contributes to f(1) 
1 in the manner expected. The mixing 
coefficient for .3 is given by 
c(1) 
31 = .1|Hanti|.3 
E1 -E3 + .1|u/2|.3 
E1 -E3 =0+0 (12-38) 
where we have used expression (12-35) for H. The first integral vanishes because 
.1 and .3 are symmetric. The second integral vanishes because they are orthogonal. 
Clearly, no symmetric state will contribute to f(1) 
1 . 
Since .4 is antisymmetric, it can contribute to f(1) 
1 . On evaluation, we find that c(1) 
41 
is about 2% of c(1) 
21 (Problem 12-5). c(1) 
41 is so much smaller than c(1) 
21 for two reasons. 
First, the integral .1|H|.4 in the numerator of c(1) 
41 is much smaller than .1|H|.2 
in c(1) 
21 . This means that the shifting of charge produced by adding .4 to .1 is much less 
helpful for lowering the energy than is that produced by adding .2 to .1. Examination 
of .2 and .4 (Fig. 12-5) reveals why this is so. .2 causes removal of charge from 
the right-hand half of the box and accumulation of charge in the left-hand half. .4 
causes removal of charge from the second and fourth quarters (numbering from the left) 
and buildup of charge in the first and third quarters. On balance .4 helps, but charge 
buildup in the third quarter is not desirable nor is removal of charge from the second 
quarter, and so .4 is much less helpful than .2. The second reason for c(1) 
41 being so 
small is that E1 -E4 in the denominator of c(1) 
41 is five times as big as E1 -E2 in c(1) 
21 .
Section 12-3 A Uniform Electrostatic Perturbation of an Electron in a “Wire” 401 
In general, mixing between states of widely different energies is discouraged by the 
formula for f(1) 
i . 
The fact that a large contribution by .j to f(1) 
i is favored by large .i |H|.j  and 
small |Ei -Ej | is strikingly similar to the situation found for variational calculations 
(Chapter 7). There, the mixing between two basis functions .i and .j. is favored if 
.i |H|.j=Hij is large and if |Hii -Hjj | is small. 
EXAMPLE 12-3 What sign should be expected for c(1) 
21 for the perturbation of Example 
12-1? of 12-2? 
SOLUTION 
 Assuming that .1 will behave anticlassically, and be skewed so as to decrease the 
probability distribution in the higher-energy region (left side), we expect c
(1) 
21 to have the opposite 
sign of that in Eq. 12-37. Hence, it should be negative. This applies for both Examples 12-1 and 
12-2.  
12-3.D The Role of an Additive Constant in H
 
Review of the results of Sections 12-3B and C will show that addition of a constant 
to H will change W(1) 
i by the same constant for all states and have no effect on f(1) 
i . 
The change in energy of all states by a constant is equivalent to a relocation of the zero 
of energy and is normally of no interest. Our initial statement that the perturbation is 
produced by a uniform electric field left the additive constant in H unspecified. We 
chose to set H = 0 at x = 0, but we could have made any of an infinite number of 
choices for H at x =0. A more sensible choice would have been the antisymmetric 
function H =ux/L-u/2 because this still leads to changes in wavefunctions due to 
the perturbation, but introduces no energy change to first order (W(1) 
i =0 for all states 
using thisH). In other words, the antisymmetric function includes the relevant physics 
of the problem and excludes the trivial effects of a constant (u/2) potential change. 
Our choice of H =ux/L was made for pedagogical reasons. 
12-3.E The Calculation ofW(2) 
1 
Since the constant first-order contribution to the energies of all the states is not physically 
interesting, let us examine the second-order contribution to the ground-state energy, 
W(2) 
1 . Equation (12-23) shows that this is related to the first-order correction to the 
wavefunction, f(1) 
1 , which, as we have already seen, causes thewavefunction to become 
skewed toward the low-potential end of the box. It is clear from Eq. (12-22) that, in 
calculating the coefficients for f(1) 
1 , we have already done most of the work needed to 
find W(2) 
1 . We saw earlier that f(1) 
1 is made primarily from .2. For simplicity, we will 
neglect all higher-energy contributions, and so 
f(1) 
1 ~=
c(1) 
2 .2 =(32L2u/27p4).2 (12-39)
402 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
and, using Eq. (12-22), 
W(2) 
1 ~=
(32L2u/27p4)2(-3p2/2L2) (12-40) 
Thus, the effect of W(2) 
1 is to lower the energy. 
We pause at this point to summarize our results. The perturbation raises the potential 
everywhere in the box. (This depends on our choice of an arbitrary constant.) The 
energy to first order is increased by the same constant amount for every state. This is 
easily seen by inspection, utilizing simple symmetry features ofH and the unperturbed 
wavefunctions. The wavefunction for the lowest-energy state is skewed, to first order, 
in a way that is energetically favorable as far as interaction with H is concerned. 
The second-order contribution to the energy for this state is negative, reflecting this 
energetically favorable shift of charge. Thus far, everything behaves sensibly. We next 
examine the behavior of the second-lowest energy state, where some important new 
features occur. 
EXAMPLE 12-4 For the .1 case of Example 12-1, should the energy to second 
order be higher or lower than the unperturbed energy? 
SOLUTION 
 The energy to second order is E1 +W
(1) 
1 +W
(2) 
1 . We have already seen that 
W
(1) 
1 =0. Thus the question comes down to asking whetherW
(2) 
1 is positive or negative. The firstorder 
correction to the wavefunction shifts charge from the higher-energy regions, hence produces 
a negative W
(2) 
1 .  
12-3.F The Effects of the Perturbation on .2 
We begin by examining f(1) 
2 , the first-order correction to .2. Inspection of Eq. (12-18) 
leads at once to the observation that c(1) 
21 =-c(1) 
12 . This means that, since .2 contributes 
to f(1) 
1 with a positive coefficient, .1 contributes to f(1) 
2 with a negative coefficient. 
A sketch of .2 minus a small amount of .1 will show that this has the effect of shifting 
charge from the left half to the right half of the box. This is just the reverse of what we 
found for the lowest-energy state. 
The antisymmetry of Eq. (12-18) for interchange of i and k allows us to make the 
following general statement. Let a perturbation occur that raises or lowers the potential 
more in one region of space than in another. The first-order correction to a given 
wavefunction will contain higher-energy wavefunctions in a manner to cause charge 
to shift into regions of lowered (or less raised) potential and it will contain lowerenergy 
wavefunctions in a manner to cause charge to shift into regions of raised (or less 
lowered) potential. 
We have not yet completed our construction of f(1) 
2 . We must calculate coefficients 
for contributions from the higher-energy functions .3, .4, etc. The state .3 should 
contribute fairly strongly since it has the proper symmetry and is not too distant in 
energy from E2: 
c(1) 
32 = .2|H|.3 
E2 -E3 = 
3
5 
32L2u 
25p4 (12-41)
Section 12-4 The Ground-State Energy to First-Order of Heliumlike Systems 403 
Therefore, .3 contributes to f(1) 
2 with a coefficient about 35
the magnitude of the contribution 
from .1. The sign of c(1) 
32 is positive and a sketch of .2 plus a small amount 
of .3 will demonstrate that this causes charge shifting to the left in accord with our 
general statement above. 
Since.4 contributes nothing (by symmetry) and.5 is fairly distant in energy, we will 
neglect all contributions to f(2) 
2 above .3. 
We have, then, two sizable contributions to f(1) 
2 , each favoring charge shifts in 
opposite directions. Let us see how W(2) 
2 reflects this: 
W(2) 
2 ~=
c(1)2 
12 (E2 -E1)+c(1)2 
32 (E2 -E3) (12-42a) 
~=
c(1)2 
12 (3)+(
3
5
c(1) 
12 )2(-5)p2/2L2 (12-42b) 
~=
1.2c(1)2 
12 p2/2L2 (12-42c) 
W(2) 
2 is the difference between energy contributions of opposite sign. The net result 
(energy increases) comes about in this case because .1 contributes more heavily 
than .3. 
The fact that contributing wavefunctions from lower and higher energies affectW(2) 
i 
oppositely is made evident by Eq. (12-22). Hence, we can extend our general statement 
above by adding that the second-order contribution to the energy from states below .i 
in energy cause the energy of the j th state to go up, contributions from above cause it 
to go down. 
Because the unperturbed energies of the particle in the box increase as n2, any state 
(except the lowest) is closer in energy to states below than to states above. Hence, for 
at least some kinds of perturbation, we might expect these states to “feel” the effects of 
states at lower energies more strongly and to rise in energy (as far asW(2) is concerned). 
This is what happens in this example. There are no states below the lowest, and so 
W(2) 
1 cannot be positive, but W(2) 
2 is positive and it turns out that W(2) 
i is positive for 
all higher i as well. 
It is interesting to compare these results with those from classical physics. Classically, 
the particle moves most slowly at the top of the potential gradient and therefore 
spends most of its time there. The lowest-energy state has responded in the opposite 
manner, in a way we might call anticlassical. The second and all higher states have 
responded classically. 
12-4 The Ground-State Energy to First-Order 
of Heliumlike Systems 
The hamiltonian for a two-electron atom or ion with nuclear chargeZ a.u. is (neglecting 
relativistic effects and assuming infinite nuclear mass) 
H(1, 2)=-
1
2 	.2
1 +.2
2 
-Z/r1 -Z/r2 +1/r12 (12-43) 
This may be written as a sum of one-electron operators and a two-electron operator: 
H(1, 2)=H(1)+H(2)+1/r12 (12-44)
404 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
where H(i) is simply the hamiltonian for the hydrogenlike system with nuclear 
charge Z: 
H(i)=-
1
2.2
i -Z/ri (12-45) 
Since we know the eigenvalues and eigenfunctions for H(i), we can let H(1)+H(2) 
be the unperturbed hamiltonian with 1/r12 the perturbation. Such a perturbation is not 
very small, but it is of interest to see how well the method works in such a case. We 
have, therefore, for the lowest-energy state of the system: 
H0 = H(1)+H(2) (12-46) 
H = 1/r12 (12-47) 
.1 = (Z3/p) exp [-Z(r1 +r2)] (12-48) 
E1 = -Z2/2-Z2/2=-Z2 (12-49) 
W(1) 
1 = .1|1/r12|.1 (12-50) 
All quantities are in atomic units. The unperturbedwavefunction [Eq. (12-48)] is simply 
the product of two one-electron 1sAOs. Because this is an eigenfunction of the system 
in the absence of interelectronic repulsion, it is too contracted about the nucleus. 
The first-order correction to the energy is the repulsion between the two electrons 
in this overly contracted eigenfunction [Eq. (12-50)]. We have encountered this same 
repulsion integral in our earlier variational calculation on helium-like systems. The 
evaluation of this integral is described in Appendix 3. Its value is 5Z/8. Hence, to first 
order, 
W1 =E1 +W(1) 
1 =-Z2 +5Z/8 (12-51) 
In Table 12-1, this result is compared with exact energies for the first ten members of 
this series. Several points should be noted: 
1. The effect of the perturbation to first order is to increase the ground-state energy. 
This is expected since 1/r12 is always positive (i.e., repulsive). 
2. The energy to first order is never below the exact ground-state energy. This is a 
general property of perturbation calculations as is easily proved (Problem 12-1). 
3. The energy to first order is in error by a fairly constant amount throughout the series. 
For H-, this gives a substantial percentage of error and fails to show H- stable 
compared with anHatom and an unbound electron. For higherZ, this error becomes 
relatively smaller since 1/r12 becomes relatively less important compared withZ/r... 
The assumption that H is a small perturbation is thus better fulfilled at large Z 
and results in W(1) 
1 being a much smaller correction relative to the total energy 
W1. Because “large” and “small” are relative terms, they can be misleading. Since 
energies of chemical interest are often small differences between large numbers (see 
the discussion at the end of Chapter 7), errors that were originally relatively small can 
become relatively large after the subtraction. Therefore, even though perturbation 
terminology would suggest that the results at Z =10 are better than those at Z =1, 
this may not be the case for some practical applications.
TABLE 12-1 
 Comparison of Exact Energy (in atomic units) with Energy to First Order when H =r-1 
12 
Z System E1 W(1) 
1 E1 +W(1) 
1 Ea
exact Eexact -E1 -W(1) 
1 %Error 
1 H- -1.0000 5/8=0.6250 -0.3750 -0.52759 -0.15259 28.92 
2 He -4.0000 5/4=1.2500 -2.7500 -2.90372 -0.15372 5.29 
3 Li+ -9.0000 15/8=1.8750 -7.1250 -7.27991 -0.15491 2.13 
4 Be2 
+ -16.0000 5/2=2.50000 -13.5000 -13.65556 -0.15556 1.14 
5 B3+ -25.0000 25/8=3.1250 -21.8750 -22.03097 -0.15597 0.71 
6 C4+ -36.0000 15/4=3.7500 -32.2500 -32.40624 -0.15624 0.48 
7 N5+ -49.0000 35/8=4.3750 -44.6250 -44.78144 -0.15644 0.35 
8 O6+ -64.0000 5.0000 -59.0000 -59.15659 -0.15659 0.26 
9 F7+ -81.0000 45/8=5.6250 -75.3750 -75.53171 -0.15671 0.21 
10 Ne8+ -100.0000 25/4=6.2500 -93.7500 -93.90680 -0.15680 0.17 
aEexact is the nonrelativistic energy to thirteenth order in Z-1, truncated at the fifth decimal place. See Scherr and Knight [4]. 
405
406 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
12-5 Perturbation at an Atom in the Simple H¨uckel 
MO Method 
Perturbation theory can be used to estimate the effect of a change in the value of the 
coulomb integral Hkk at carbon atom k. This is normally given the value of a, but it 
might be desirable to consider a different value due, for example, to replacement of an 
attached hydrogen by some other atom or group. We take the unperturbed H¨uckel MOs 
and orbital energies as our starting point and let the perturbed value of Hkk be a +da. 
Therefore, 
H =da (at center k only) (12-52) 
The first-order correction for the energy of fi , the ith MO, is 
W(1) 
i =fi |H|fi  (12-53) 
Now fi is a linear combination of the AOs .: 
fi =
j 
cji.j (12-54) 
and so 
W(1) 
1 =
j 

l 
c*jicli .j |H|.l (12-55) 
but H is zero except at atom k, where it is da, and so the sum reduces to one term: 
W(1) 
i =c*kicki da (12-56) 
Summing over all the MOs times the number of electrons in each MO gives the firstorder 
correction to the total energy: 
W(1) =da
i 
nic*kicki =daqk (12-57) 
where qk is the p-electron density at atom k. Thus the energy change to first order is 
equal to the change in Hkk times the unperturbed electron density at that atom. This 
can also be seen to be an immediate consequence of the alternative energy expression 
E =
l 
qlal +2
l<m
plmßlm (12-58) 
which was derived in Chapter 8. 
EXAMPLE 12-5 Consider the methylene cyclopropene molecule, C4H4, in the 
HMO approximation. (See Appendix 6 for data.) At which carbon will a perturbation 
involving a affect the total p energy the most? Calculate the energy to first 
order if the value of a at that atom increases to a +0.1000ß. Calculate the energy 
change, to first order, of each of the four MOs.
Section 12-5 Perturbation at an Atom in the Simple H¨uckel MO Method 407 
SOLUTION 
 The total energy will be most affected if the perturbation occurs at the atom 
having the greatest p-electron density, that is, at atom 4. To first order, the total p energy 
changes by (0.1000ß)(1.4881) = 0.1488ß, giving an energy to first order of 4a + 4.9264ß + 0.1488ß =4a +5.1112ß. Each MO energy drops by (using Eq. (12-56), from lowest to highest, 
0.00794ß, 0.06646ß, 0.0000ß, 0.02559ß. The sum of the first two of these, times 2 (because of 
having two electrons each), gives 0.1488ß, which equals the first-order change in total p energy 
found above.  
Calculation of first-order corrections to the MOs proceeds in a straightforward manner 
using Eq. (12-19) (see Problem 12-17). One of the results of interest from such a 
calculation is the change in p-electron density at atom l due to a change in the coulomb 
integral at atom k. The differential expression is 
dq1 =(.ql/.ak)dak =pl,kdak (12-59) 
The quantity of interest pl,k is called the atom-atom polarizability. We will now derive 
an expression for pl,k in terms of MO coefficients and energies. 
We assume that our MOs fall into a completely occupied set f1, . . . ,fm and a 
completely empty set fm+1, . . . ,fn. We take da to be positive, which means that center 
k is made less attractive for electrons. According to Eq. (12-19) the j th MO is, to first 
order, 
fj +f(1) 
j =fj + 
n 
 
i=1,i=j 
fj |H|fi  
Ej -Ei 
fi (12-60) 
where Ej is the orbital energy of fj . As indicated above, the integral vanishes except 
over AO k. Therefore, 
f(1) 
j = 
n 
 
i=1,i=j 
c*kj cki dak 
Ej -Ei 
fi (12-61) 
Expanding fi in terms of AOs, we have 
f(1) 
j =dak 
n 
 
i=1,i=j 
ckj cki 
Ej -Ei 
n 

l=1 
cli.l (12-62) 
We can write this as 
= 
n 

l=1 
c(1) 
lj .l (12-63) 
where 
c(1) 
lj =dak 
n 
 
i=1,i=j 
ckj ckicli 
Ej -Ei 
(12-64) 
The change in density at atom l equals the square of the perturbed wavefunction minus 
that of the unperturbedwavefunction and involves only the doubly occupiedMOs1-m: 
dql = 2 
m 

j=1 
(clj +c(1) 
lj )2 -c2
lj 
 (12-65) 
= 4 
m 

j=1 
clj c(1) 
lj (12-66)
408 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
where we have neglected (c(1) 
lj )2 because the perturbation is assumed small. Substituting 
Eq. (12-64) into (12-66) gives 
dql =4dak 
m 

j=1 
n 
 
i=l, i=j 
clj ckj ck..cli 
Ej -Ei 
(12-67) 
This equation can be simplified further by recognizing that, for j and i <m+1, each 
term with denominator Ej. - Ei has a mate with denominator E1. - Ej . Hence, the 
terms differ only in sign and cancel. Therefore, 
m 
 
j=1,i=j 
clj ckj ckicli 
Ej -Ei =0 (12-68) 
and so 
dql =4dak 
m 

j=1 
n 
 
i=m+1 
clj ckj ckicli 
Ej -Ei 
(12-69) 
Comparing Eqs. (12-69) and (12-59) gives 
plk =4 
m 

j=1 
n 
 
i=m+1 
clj ckj ckicli 
Ej -Ei 
(12-70) 
Related quantities, derivable in a similar manner, are the bond-atom polarizability, 
pst,r , and the bond-bond polarizability, ptu,rs : 
pst,r =2 
m 

j=1 
n 
 
k=m+1 
crj crk(csj ctk +ctj csk) 
Ej -Ek 
(12-71) 
ptu,rs =2 
m 

j=1 
n 
 
k=m+1 
(crj csk +csj crk)(ctj cuk +cuj ctk) 
Ej -Ek 
(12-72) 
These refer, respectively, to the change in bond order pst induced by a perturbation 
dar , and to the change in bond order ptu induced by a change in ßrs . 
The atom–atom polarizability formula (12-70) makes it evident that a large contribution 
to the change in electron density at atom l due to a change in the coulomb 
integral at atom k is favored if a filled and an empty MO exist that have large absolute 
coefficients on both atoms and are close in energy. However, another factor comes into 
play that usually confounds this expectation, as the following example shows: 
EXAMPLE 12-6 Consider the naphthalene molecule, C10H8, in the HMO approximation. 
(See Appendix 6 for data.) If atom 1 were perturbed in a way to make it 
more attractive to p electrons, which of the atoms 2, 3, or 4, would you expect to 
experience the greatest population change?
Section 12-6 Perturbation Theory for a Degenerate State 409 
SOLUTION 
 The HOMO-LUMO contribution to the charge shift favors atom 4, since its 
coefficients in these MOs are larger than those for atoms 2 or 3. However, the actual data (not given 
in Appendix 6) are: p1,2 = 0.21342, p1,3 =-0.01771, p1,4 = 0.13937. So atom 2 experiences 
the greatest population change, atom 3 the least. Why is this? It results from the fact that we 
are summing products that can be positive or negative. Two atoms near each other are less likely 
to have nodal planes separating them, so their coefficients are more likely to be of the same sign 
giving fewer sign changes in their products. So the sum has less tendency to cancel itself. This 
is the way in which MO theory predicts a tendency for polarization effects to die off as distance 
increases—the well-known “inductive effect.” This dying-off is not monotonic, as can be seen 
in the remaining atom-atom polarizabilities: p1,5 =0.02325, p1,6 =-0.00643, p1,7 =0.03228, 
p1,8=-0.02674, p1,9 =0.08889, p1,10=-0.00356.  
In general,plk can have either sign. This means that making atom k more attractive for 
electrons will, to first order, cause the p electron density to decrease at some carbons and 
increase at others. (This is necessary since the totalp-charge density must be conserved.) 
12-6 Perturbation Theory for a Degenerate State 
Equations (12-19), (12-21), and (12-24) have denominators containing Ei -Ej. If .i 
and .j are degenerate, this leads to difficulty, alerting us to the fact that our earlier 
derivations do not take proper account of states of interest that are degenerate. The 
problem results from our initial expansion of f.l as a power series in .. The leading 
term in the expansion, f(0) 
i , is the wavefunction that fi becomes in the limit when 
.=0. In nondegenerate systems there is no is ambiguity; f(0) 
i has to be .i . But if .i is 
degenerate with .j , any linear combination of them is also an eigenfunction. We need 
a method to determine how .i and .j should be mixed together to form the correct 
zeroth-order functions f(0) 
i and f(0) 
j . 
To find the conditions that f(0) 
i and f(0) 
j must satisfy, we return to the first-order 
perturbation equation (12-11) but with f(0) 
i in place of .i : 
(H -W(1) 
i )f(0) 
i +(H0 -Ei)f(1) 
i =0 (12-73) 
Multiplying from the left by f(0)* j and integrating gives (taking f(0) 
i and f(0) 
j to be 
orthogonal and H0 hermitian) 
f(0) 
j |H|f(0) 
i +(Ej -Ei)f(0) 
j |f(1) 
i =0 (12-74) 
If Ei =Ej , this gives 
f(0) 
j |H|f(0) 
i =0 (12-75) 
This means that the first-order perturbation equation is satisfied for degenerate states 
only when the wavefunctions “diagonalize” the perturbation operator H. Therefore, 
our problem reduces to finding those linear combinations of .i and .j that will diagonalize 
H. We have already seen two ways to accomplish this in earlier chapters. One 
way is to construct the matrix H, where (H)ij =.i |H|.j , and then find the unitary 
matrix C that diagonalizes H in the similarity transformation C†HC. The elements in 
C are then the coefficients for the linear combinations of .i and .j , and the diagonal
410 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
elements of the diagonalized matrix are equal to f(0) 
i |H|f(0) 
i  and f(0) 
j |H|f(0) 
j ; 
they are the first-order corrections to the energy of f(0) 
i and f(0) 
j . The other method, 
equivalent to the above, is to set up the determinantal equation 
 
H11 -E H12 
H21 H22 -E 

=0 (12-76) 
and solve for the roots E. Substituting these back into the simultaneous equations 
corresponding to the determinant gives the coefficients with which .i and .j should 
be combined. The roots E are the first-order corrections to the energy and are identical 
to the diagonal matrix elements mentioned above. 
In the event that the roots are equal, the perturbation has failed to split the degeneracy 
to first order and it is then necessary to proceed to higher order. We will not pursue this 
problem to higher orders, however. 
If there are n degenerate states, the above procedure is simply carried out with an 
n×n matrix or determinant. 
We will now give an example of perturbation theory for a degenerate state. 
12-7 Polarizability of the Hydrogen 
Atom in the n=2 States 
In Chapter 7, the polarizability of the hydrogen atom in the 1s statewas calculated by the 
variation method (see Section 7-4 and Problem 7-13). We now use perturbation theory 
for degenerate states to calculate to first order the polarizabilities of the n=2 states. 
The perturbation due to a z-directed uniform electric field is, in atomic units, 
H=-Fr cos . (12-77) 
There are four degenerate states at the n=2 level, giving us a 4×4 secular determinant. 
If we choose the real AOs as our basis set and let s, x, y, z represent these functions, 
the determinantal equation is 
 
H ss -E H sz H sy H sx 
H zs H zz -E H zy H zx 
H ys H yz H yy -E H yx 
H xs H xz H xy H xx -E 
 
=0 (12-78) 
where 
H yz =y|H|z, etc. (12-79) 
The operator H is antisymmetric for reflection in the xy plane, but symmetric for 
reflection in the xz or yz planes. It follows from this that all integrals in Eq. (12-78) 
vanish except for H sz and H zs , which are equal. Assigning a value of x (not to be 
confused with the x coordinate) to these integrals, our equation becomes 
 
-E x 0 0 
x -E 0 0 
0 0 -E 0 
0 0 0 -E 
 
=0 (12-80)
Section 12-7 Polarizability of the Hydrogen Atom in the n=2 States 411 
The block diagonal form indicates that this is a product of one 2×2 determinant and 
two 1×1’s. Evidently, the proper zeroth-order wavefunctions must be a mixture of 2s 
and 2pz AOs, with 2px and 2py being acceptable as they stand. The roots of the 1×1 
determinants are zero, the roots of the 2 ×2 are E+=+x, E-=-x. These lead to 
coefficients that produce the zeroth-order wave-functions 
f(0) 
+ = 1/v2 (2s+2pz) (12-81) 
f(0) 
- = 1/v2 (2s-2pz) (12-82) 
The roots +x and -x are the first-order corrections to the energy. We now proceed 
to calculate these quantities: 
x =2s|H|2pz=···=3F (12-83) 
Our perturbation calculation indicates that the n = 2 level splits into three levels 
under the influence of an electric field. Since this splitting occurs to first order, it is 
sometimes called a first-order Stark effect. Since x is proportional to F, the splitting 
is linear in F, as indicated in Fig. 12-6. 
It is instructive to compare these results with the behavior of the 1s state. In the first 
place, the effect of a uniform electric field on the 1s level is zero, to first order, because 
1s|H|1s vanishes for reasons of symmetry. Only when first-order corrections are 
made to the 1swavefunction are energy effects seen, and these occur in the second-order 
energy terms. Therefore, the 1s state gives a second-order Stark effect, but no firstorder 
effect. The 1s state gives no first-order effect because the spherically symmetric 
zeroth-order wavefunction has no electric dipole to interact with the field. But the 
proper zeroth-order wave functions for some of the n=2 states, given by Eqs. (12-81) 
and (12-82), do provide electric dipoles in opposite directions that interact with the 
field to produce first-order energies of opposite signs. 
Notice that the energy change for the second-order effect goes as F2 [(Eq. (7-62)], 
whereas that for the first-order effect goes as F. These dependences are indicative of 
an induced dipole and a permanent dipole, respectively. The induced dipole for the 
1s state depends on the field strength F, since, as F increases, the dipole moment 
increases (due to mixing in higher states). This induced dipole, which depends on F, 
Figure 12-6 
 Energy to first order of n = 2 level of hydrogen as a function of uniform electric 
field strength (z-directed field).
412 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
then interacts with the field of strength F. Since both the size of the induced dipole and 
the energy of interaction with the field depend on F to the first power, the energy goes 
as F2. For the n=2 wavefunctions f(0) 
+ and f(0) 
- , however, the dipole is permanent.7 
(The mixing to produce these states occurs even in the limit as F goes to zero.) This 
permanent dipole interacts with the field to give an energy depending on F instead 
of F2. 
The 2px and 2py AOs are not affected to first-order in the perturbation. At higher 
orders, these AOs (as well as f(0) 
+ and f(0) 
- ) mix in higher-energy AOs of symmetries 
appropriate to produce induced dipoles. However, due to the equivalence of 2px and 2py 
with respect to H, the induced dipole is the same in both cases so that the degeneracy 
is still not lifted. 
The qualitatively different behaviors of the n = 1 and n = 2 levels of hydrogen 
result from the fact that degenerate eigenfunctions can be mixed together at no energy 
expense, whereas the n=1 state can produce a dipole only by mixing in higher-energy 
states. This suggests that the polarizability of an atom or molecule should be strongly 
dependent on the availability of fairly low-energy unoccupied orbitals orwavefunctions, 
and this is indeed the case. As we proceed down the series H-, He, Li+, Be+2, . . . , etc., 
we find that the distance in energy between the occupied state and the higher-energy 
states increases. Also, we find that the systems become less and less polarizable. Again, 
if we compare helium and beryllium, both of which are ordinarily considered to have 
s2 valence-state configurations, we find beryllium to be much more polarizable. This 
is due to the presence of empty 2p orbitals lying fairly close in energy to the 2s AO 
in beryllium. In fact, these two atoms are strongly reminiscent of the n = 1, n = 2 
polarizabilities in hydrogen, the chief difference being that the 2s and 2p levels of 
beryllium are only close in energy—not actually degenerate. 
12-8 Degenerate-Level Perturbation Theory by Inspection 
Degenerate and nondegenerate perturbation theories share the feature that the first-order 
correction to the energy is the expectation value of the perturbation operator using the 
unperturbed wavefunction. Often we can evaluate or estimate this expectation value by 
inspection. The special problem for degenerate wavefunctions or orbitals is that there 
is an infinite number of ways to express the unperturbed wavefunctions, and only one 
of these ways is proper for finding W(1). The mathematical procedure for finding the 
proper set of functions tends to complicate and obscure the situation. However, it is 
often easy to deduce the proper unperturbed wavefunctions by inspection. 
The clue for doing this comes from Eq. (12-76). It is apparent that this equation, 
which leads to the first-order corrections to the energy and to the proper unperturbed 
(or zeroth-order) wavefunctions, is exactly like the secular determinantal equation we 
would use for a variational calculation. Since the secular equation comes from a 
calculus-based determination of energy extrema, we know that the energies that come 
out of Eq. (12-76) must be the minimum and maximum possible values and that the 
proper zeroth-order functions must be those that give the maximum and minimum 
expectation values for H. This leads immediately to a powerful insight: For a given 
7In the language of hybridization (Chapter 13), f
(0) 
+ and f
(0) 
- are sp hybrids.
Section 12-8 Degenerate-Level Perturbation Theory by Inspection 413 
perturbation, the proper zeroth-order degenerate wavefunctions to use for predicting 
the effect on energies are that set giving the greatest difference in response. 
This rule can be seen operating in the hydrogen atom polarization example of the 
preceding section. The proper combination of 2s and 2pz AOs is the pair of sp hybrid 
combinations giving the greatest energy increase and decrease. (The 2px and 2py AOs 
do not mix any 2s or 2pz into themselves because to do so will have no energy-raising 
or -lowering effect.) 
Another example is seen when changing the coulomb integral at an atom in theHMOlevel 
treatment of a cyclic molecule. Such molecules always have some degenerate 
MOs, and so one must decide which linear combination is “proper” for evaluating 
the effect of this perturbation. The situation is analyzed most easily by recognizing 
that mixing such degenerate MOs together corresponds to rotating the positions of 
the nodes and antinodes in the molecule. So the question of finding the proper MOs 
becomes one of asking where the nodes and antinodes should be located. The greatest 
difference in response to changing the coulomb integral at an atom comes when one 
of the degenerate pair of MOs has a node (i.e., is zero) at that atom and the other 
MO has an antinode (i.e., has maximum absolute value) there. Suppose, for example, 
you are asked to evaluate the effect on a particular degenerate pair of MOs of such a 
perturbation at carbon number 3 in a cyclic molecule. When you look at those MOs 
in a tabulation, you may find that atom 3 is not at the node or antinode position for 
either MO. Instead, you might observe that the node of one MO and antinode of the 
other occur at atom number 4. You could set about mixing these MOs to move the 
node-antinode positions to atom 3, but it is much easier and just as correct to pretend 
that the perturbation is occurring at atom 4, since, in a cycle, all atoms are equivalent. 
This allows you to use the tabulated MOs to evaluate the perturbation (now at atom 4) 
without any modification. Once this is recognized, the problem can be completed by 
inspection, just as in nondegenerate cases: TheMOenergy for theMOthat has its node 
at atom 4 is unaffected to first order by the perturbation, and the other is affected by an 
amount proportional to the square of the coefficient at atom 4. Thus, the MOs become 
nondegenerate and are described to zeroth order by these “proper” functions. 
It is important to realize that the identity of the proper unperturbed wavefunctions 
depends on the nature of the perturbation. Thus, for the cyclic system discussed above 
(which we now assume has an even number of atoms), stretching the molecule so that 
two bonds on opposite sides of the cycle get longer is treated perturbatively with a 
different pair of proper MOs. We can quickly guess which MOs those are by seeking 
the pair that respond most differently to the bond stretching. Since any degenerate MO 
in this even–alternant cycle has at least one nodal plane perpendicular to the molecular 
plane, there is always a rotational position that places a nodal plane across the center 
of the two bonds being stretched. The mate to this MO then has an antinode in these 
bonds. That means that one MO is antibonding in these bonds and the other MO is 
bonding. This in turn means that one MO’s energy will drop as the bond lengthens and 
the other MO’s energy will rise. This, then, is the proper MO set to use for evaluating 
this perturbation to first order. Furthermore, these MOs are nondegenerate after the 
perturbation and are the zeroth-order approximation to theMOsof the perturbed system. 
The two perturbations described above bring up a final important point. Suppose 
we have two different perturbations occurring simultaneously in a system involving 
degenerate MOs or state functions. The question then arises as to whether the proper
414 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
zeroth-order functions are the same or different for the two perturbations. We have just 
seen that, if the perturbations involve substituting a new atom for a carbon and stretching 
a pair of bonds in a cyclic molecule, the proper zeroth-order MOs are different. This 
means that one of these perturbations favors a postperturbative pair of nondegenerate 
MOs that is different from the pair being favored by the other perturbation. In other 
words, these perturbations compete to produce different final MOs. The other possibility 
is that two perturbations have the same set of proper zeroth-order wavefunctions, 
which means that they cooperate to produce the same final MOs. The cooperative 
situation is the one that gives the largest energy effects, since a single set of proper 
zeroth-order functions is selected that produces the greatest difference in response to 
both perturbations simultaneously. Competing perturbations require a compromise 
among zeroth-order functions which involves diminution in the differences in energy 
response. 
The reader may have realized that there is another way to select the proper zerothorder 
wavefunctions, and that is to use symmetry. Recall that the proper wavefunctions 
.(0) 
i and .(0) 
j are those that “diagonalize the determinant,” i.e., that cause 
.(0) 
i |H|.(0) 
j  to equal zero. In each of the examples cited above, this can be seen to 
be true. The H operator corresponding to changing the coulomb integral at an atom, 
and nowhere else, is unaffected by reflection through a plane containing that atom. The 
MO placing a maximum at that atom is symmetric for such a reflection, and the MO 
placing a node there is antisymmetric. Hence the integrand is overall antisymmetric and 
the integral vanishes, in accord with the requirement that must be satisfied by proper 
zeroth-order wavefunctions. A similar argument applies for stretching or shrinking 
bonds, with the reflection plane nowoccurring at the bond midpoints. As a useful guide, 
then, proper zeroth-order wavefunctions may be recognized as those having opposite 
symmetry for an operation by which the perturbation operator is unaffected. 
EXAMPLE 12-7 Consider the cyclopentadienyl radical in theHMOapproximation. 
Assume it to be planar and pentagonal. (SeeAppendix 6 for data.) If a3 is perturbed 
to become a +0.1000ß, what are the MO energy changes, to first order? 
SOLUTION 
 The lowest-energyMOis nondegenerate, and all atoms have the same coefficient, 
so E1 = (0.4472)2(0.1000ß) = 0.0200ß. The second two MOs are degenerate, so we need to 
identify the proper zeroth-order combinations. One should be symmetric with respect to reflection 
through atom 3, the other antisymmetric. None of the tabulated MOs have these properties about 
atom 3, but they do have these properties for atom 1. Therefore, we proceed by pretending that atom 
1 is the one being perturbed (it makes no difference). Then,E2=(0.6325)2(0.1000ß)=0.0400ß, 
E3 =0.0000ß, E4 =E2, E5 =E3.  
12-9 Interaction Between Two Orbitals: An Important 
Chemical Model 
Qualitative quantum-chemical discussion often relies on a simplified model wherein all 
interactions are neglected except for the primary one. For example, in considering how 
a Lewis base and a Lewis acid interact, one might consider only the highest occupied
Section 12-9 Interaction Between Two Orbitals: An Important Chemical Model 415 
MO (HOMO) of the base (electron donor) and the lowest unfilled MO (LUMO) of the 
acid (electron acceptor). Consideration of the ways two levels interact will provide 
some useful rules of thumb and also consolidate some of our earlier findings.8 
We label the unperturbed orbitals and energies .a, .b, and Ea, Eb. We will work 
within the H¨uckel type of framework so that state energy differences may be written as 
orbital energy differences. The orbitals are allowed to interact with each other by virtue 
of close approach. We will assume that the perturbation felt by the orbitals is proportional 
to the overlap between them. (If the systems involved are ions, strong electrostatic 
interactions will exist as well. Therefore, we assume all systems to be neutral.) 
Since the overlap of .a on .b is the same as that of .b on .a, the energy change 
to first order is the same for both levels and hence we ignore it (unless the levels are 
degenerate—an eventuality we discuss shortly). 
The second-order contributions to the energies are: 
W(2) 
a = |.a|H|.b|2 
Ea -Eb 
(12-84) 
W(2) 
b = |.a|H|.b|2 
Eb -Ea 
(12-85) 
Thus, the second-order energy W(2) 
a is positive if Ea >Eb, negative if Ea <Eb. In 
other words, the higher-energy level is pushed up, or destabilized, through interaction 
with the level below, and the lower-energy level is stabilized by interaction with the 
level above. This behavior, which we noted also in connection with the perturbed 
particle in a wire, may be summarized by the statement: interacting levels repel each 
other. Equations (12-84) and (12-85) also make clear that the levels repel each other 
more strongly the greater their interaction (numerator) and the smaller their energy 
separation (denominator). 
If Ea = Eb, we must set up a 2 ×2 first-order perturbation determinant using .a 
and .b as basis, and find the two roots. Just as was true in the polarizability calculation 
of Section 12-7, we will find that one root lies above Ea, and the other lies an equal 
distance below. Thus, for degenerate levels, the repulsion between levels becomes a 
first-order effect. 
The first-order mixing of .a and .b parallels the energy results in the expected 
way. If the second-order energy stabilizes the level, the orbitals are mixed in a bonding 
fashion. Since we have seen that it is the lower-energy orbital that is stabilized, we can 
state that, if two orbitals interact, the lower-energy one of the two mixes into itself the 
higher-energy one in a bonding way, while the higher-energy orbital mixes into itself 
the lower one in an antibonding way [5]. In short, “the upper combination takes the 
node” [5]. 
As an example of the utility of this model, consider the norbornadiene molecule 
(see Fig. 12-7a). According to our understanding of unsaturated systems, the double 
bonds in this molecule should behave like isolated ethylene double bonds and not 
like the conjugated bonds of butadiene. But these bonds are at an orientation and 
proximity allowing significant overlap between 2pp AO lobes beneath the molecule. If 
we treat this system with our two-level model, the unperturbed MOs are the bonding 
8The discussion in this section follows closely that of Hoffmann [5]. That article reviews a variety of illustrative 
applications of the two-orbital model.
416 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
Figure 12-7 
 p MOsin norbornadiene: (a) norbornadiene; (b) basis of unperturbed MOs; (c) basis 
of perturbed MOs; (d) quadricyclene (after R. Hoffmann [5].) 
and antibonding p MOs of two ethylene molecules (Fig. 12-7b). We expect the MO 
overlap across the molecule to split each of these levels as indicated in Fig. 12-7c, the 
bonding interaction producing stabilization in each case. (Combinations involving a 
bonding MO on one side and an antibonding MO on the other are ruled out because 
symmetry forces the interaction term to vanish. Even if this were not the case, the 
energy gap between these unperturbed MOs would make such contributions negligible 
compared to those from the degenerate MOs.) 
A careful look at Fig. 12-7c reveals certain implications about norbornadiene. For 
instance, the ionization energy for norbornadiene should be smaller than that for
Section 12-10 Spectroscopic Selection Rules 417 
norbornene, which has but one double bond. Experimental measurements support this 
contention. Also, since the highest occupied MO of norbornadiene is antibonding 
between carbons 3 and 5 and also 6 and 2, whereas the lowest empty MO is bonding 
between these carbons, excitation of an electron from the former MO to the latter 
should promote formation of quadricyclene (Fig. 12-7d). This compound is a common 
product in the photochemistry of norbornadienes. 
12-10 Connection Between Time-Independent 
Perturbation Theory and Spectroscopic 
Selection Rules 
Molecules may change to higher-or lower-energy states under the influence of incident 
light. Such processes are called, respectively, absorption and induced emission. Perturbation 
theory can be used to study such transitions induced by an external oscillating 
electromagnetic field. Here we briefly describe a rather simple connection between 
selection rules and the perturbation theory we have discussed in this chapter. 
Let a molecule initially be in a state with wavefunction .i . We are interested in the 
probability of a transition occurring to some final state with wavefunction .f . A timedependent 
perturbation treatment (not given here, but see Section 6-16) indicates that 
such a transition is probable only when the external field frequency . satisfies the 
conservation of energy relation 
. =|Ef -Ei |/h (12-86) 
Even when this condition is satisfied, however, we may find experimentally that the 
transition is so improbable as to be undetectable. Evidently some factor other than 
satisfaction of Eq. (12-86) is also involved. 
Since the molecule is being subjected to light, it experiences fluctuating electric and 
magnetic fields. For ordinary (as opposed to magnetic) spectroscopy, the effects of 
the magnetic field are negligible compared to those of the electric field. Therefore, 
we ignore the former and imagine the molecule at a particular instantaneous value 
of an external electric field. As we have seen from earlier sections, this field causes 
polarization through the admixture of unperturbed wavefunctions .1, .2, etc., with .i 
to form a perturbed wavefunction fi . For a normalized fi , we have, therefore, 
fi =ci.i +c1.1 +c2.2+···+cf .f +··· (12-87) 
At a later time, when the perturbation is over, the system returns to an unperturbed 
state. The probability of returning to the initial state is given by c*i ci . The probability 
of going instead to the final state described by .f is given by c*f cf . If cf is zero, there 
is no tendency for a transition to that final state to occur and the transition is said to be 
forbidden. If cf is nonzero, the transition is allowed. 
Since the coefficient cf is given to first order by [see Eq. (12-18)], 
cf = .f |H|.i  
Ei -Ef 
(12-88) 
where H is the perturbation operator for the electric field component of the light, 
we can say at once that a transition between two states is forbidden (to first order) if 
.f |H|.i  vanishes.
418 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
One example of a forbidden transition is that between a singlet state and a triplet 
state of a system. We can see this at once since H is an electric field and does not 
interact with spin magnetic moment. Therefore the orthogonality between .i and .f 
due to spin functions is uninfluenced by H and the integral .f |H|..1 must vanish. 
In effect, the perturbation H causes the initial singlet state to become polarized by 
mixing in other singlet state functions, but gives no impetus for mixing in states of 
different multiplicity. Transitions between states of different multiplicity are said to be 
spin forbidden. 
Another example of a forbidden transition is that between two different s-type states 
of a hydrogen atom. Such states have spherically symmetric wavefunctions, but H 
(the electric field) is antisymmetric for reflection through a plane (to within an additive 
constant), and so .f |H|.i  must vanish for reasons of symmetry. It is easy 
to generalize this argument to other states of the hydrogen atom (p to p, d to d, 
etc.) and also to certain other atoms (e.g., the alkali metals) electronically similar to 
hydrogen. 
Molecular electronic transitions can be understood from the same standpoint. The 
intense p*.p transition in ethylene is a simple example.9,10 Let us imagine that the 
molecule is oriented as shown in Fig. 12-8. Suppose we could somehow orient all our 
molecules thisway (in a host matrix or in a crystal) and that we then subjected the sample 
to plane-polarized light. We will consider what should happen for light polarized in 
each of the directions x, y, z. First, let us consider the integral p|z|p*. The ethylene 
molecule has three reflection planes of symmetry, so we can examine the symmetry 
of the integrand with respect to each of these three reflections. Remember that, if the 
integrand is antisymmetric for any one of these reflections, the integral vanishes. The 
function p can be seen, from inspection of Fig. 12-8, to be symmetric for reflection in 
the xz and yz planes, antisymmetric for reflection in the xy plane. These observations 
are shown in Table 12-2, along with similar conclusions regarding the symmetries of 
the functions p*, x, y, and z. The symbols “s” and “a” stand for “symmetric” and 
“antisymmetric.” Our integral p|z|p* can now be seen to have an integrand that 
is symmetric for reflection in the xz plane but antisymmetric for reflection in the xy 
Figure 12-8 
 p and p* MOs for ethylene. The C–C bond is coincident with the x axis and all 
nuclei lie in the xy plane. 
9Here, p* refers to the antibonding pg MO, not to complex conjugation. 
10In this discussion, we consider molecules of high symmetry, so that reflection planes of symmetry exist. For 
molecules of lower symmetry, one must resort to actual calculation of the cf in Eq. (12-88).
Section 12-10 Spectroscopic Selection Rules 419 
TABLE 12-2 
 Symmetries of Functions under Reflection through 
Cartesian Coordinate Planes 
Symmetry operation 
Function xy reflection xz reflection yz reflection 
p a s s 
p* a s a 
z a s s 
x s s a 
y s a s 
p z p* a a a = a s s s = s s s a = a 
p x p* a s a = s s s s = s s a a = s 
p y p* a s a = s s a s = a s s a = a 
and yz planes. Hence, this integral vanishes, and this transition is forbidden insofar 
as light polarized perpendicular to the molecular plane is concerned. According to 
Table 12-2, symmetry also causes p|y|p* to vanish. However, symmetry does not 
force p|x|p* to vanish. Therefore, the p*.p transition is allowed and is polarized 
along the C–C axis. Indeed, because of the significant spatial extension of the p and 
p* MOs (compared toAOs), the integral p|x|p* is relatively large, which means that 
the transition is not only allowed, but is intense. 
It is physically reasonable that the p*.p transition should be polarized along the 
C–C axis. If some p* character is mixed with p, it is easy to see that this results in a 
shift of p charge from one carbon to the other—a shift along the x axis. Conversely, an 
electric field in the x direction will cause polarization along the x axis, hence mix p* 
character into the p MO. When the perturbation is removed, a finite probability exists 
that the molecule will go to the state wherein p* is occupied. 
In the above example, we imagined all ethylene molecules to be identically oriented. 
Under those conditions, we would observe a maximum in p*.p absorption when our 
incident light was polarized parallel to the molecular axis and zero absorption (ideally) 
when the polarization axis was perpendicular to the molecular axis. If the light is 
unpolarized, or if the ethylene is randomly oriented, as in liquid or gaseous states, the 
absorption is isotropic and allowed because there is always a certain degree of “overlap” 
between the molecular axes of most of the molecules and the direction of the electric 
field due to the light. 
The example discussed above illustrates a simple and powerful rule for recognizing 
whether or not a given atomic or molecular orbital dipolar transition is allowed (assuming 
electron spin agreement): A dipole transition is allowed between orbitals that have 
opposite symmetry for one and only one reflection plane.11 The polarization of the 
transition will be perpendicular to that reflection plane. 
11This means that, for an allowed transition, one orbital has one and only one nodal plane where the other 
does not.
420 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
EXAMPLE 12-8 Selection rules for allowed dipolar transitions in a hydrogen atom 
include l=±1, ml =0,±1. Show that these rules are equivalent to the symmetry 
rule stated above.a Consider the following cases: 2s .1s, 2px .1s, 
3dxy.2s, 3dxz.2pz, 3dxy.2pz . 
aThere is yet another way to rationalize these selection rules. Photons are bosons with spin quantum 
number of 1, hence they carry spin angular momentum. The total angular momentum of the atom or molecule 
that absorbs a photon must change to obey the conservation requirement consistent with the vector addition 
of orbital and photon angular momenta. The l rule handles that for one-electron atoms. Furthermore, 
for a unit change of l, the z-component of the absorber’s orbital angular momentum can change by +1, 0, 
or -1 a.u. The ml rule takes care of that. This rationalizes the atomic selection rules, but doesn’t yield 
information on polarization or intensity as readily as the more physical approach described in the text, and 
it is less easily applied to molecules. Generally, however, the more ways one understands a thing, the better. 
SOLUTION 
 The 2s.1s transition is forbidden by the l rule (because l =0) and also by 
the symmetry rule (because neither wavefunction has a nodal plane). This makes it easy to see 
that the l rule applies because it forces one of the orbitals to have one and only one more nodal 
plane than the other. The 2px .1s transition is allowed by the l rule. Since 2px is a linear 
combination of 2p+1 and 2p-1, the ml rule is also satisfied, so the transition is allowed. When 
we examine symmetry, we note that there is only one plane of disagreement–the yz plane, so this 
approach also indicates that the transition is allowed, and that it is polarized in the x direction 
(perpendicular to the yz plane). The 3dxy.2s transition is forbidden by the l rule since l =2. 
The symmetry approach indicates that a transition is forbidden because the 3dxy AO has two nodal 
planes, whereas 2s has none. The 3dxz.2pz transition has l=1. 3dxz is a combination of 3d+1 
and 3d-1, and 2pz is 2p0, so ml =±1. Therefore, this is an allowed transition. The symmetry 
approach shows 3dxz with two nodal planes (xy and yz) and 2pz with one (xy). Therefore, there 
is only one plane of symmetry disagreement (yz), so the transition is allowed and is x polarized. 
The 3dxy .2pz transition has l =+1. 3dxy is a linear combination of 3d+2 and 3d-2, and 
2pz is 2p0, so ml =±2. Therefore, the transition is forbidden. The symmetry analysis shows 
3dxy having two nodal planes (xz and yz), and 2pz having one (xy). Therefore, they disagree in 
symmetry in three planes, and the transition is forbidden. This last case makes it clear why theml 
rule is needed: When satisfied, it guarantees that all but one of the nodal planes in the two AOs 
coincide. (If ml =±2, then, taking the z axis to be vertical, one AO has two more vertical nodal 
planes than the other, so there must be at least two reflection planes of symmetry disagreement.)  
12-10.A Problems 
12-1. Prove that the energy to first order for the lowest-energy state of a perturbed 
system is an upper bound for the exact energy of the lowest-energy state of the 
perturbed system, that is, that E0 +W(1) 
0 =W0. 
12-2. A one-dimensional box potential is perturbed as sketched in Fig. P12-2. From a 
consideration of the first three wavefunctions of the unperturbed system, .1, .2, 
Figure P12-2 

Section 12-10 Spectroscopic Selection Rules 421 
.3, which will have its energy increased most, to first order, and which least? 
No explicit calculation is necessary to answer this question. 
12-3. Evaluate by inspection the first-order contributions to the energies for the states 
shown in Fig. P12-3. In every case the unperturbed state is a particle in the 
indicated state in the one-dimensional, infinitely deep, square well. 
Figure P12-3 
 a)H =sine function as shown, .i =.1. b)H as shown, .i =.1. c)H as shown; 
.i =.1. d) H as shown; .i =.2. e) H = cosine as shown, ... =.2. f) H as shown; .i =.2. 
12-4. a) Evaluate W(1) 
2 for the particle in a box with the perturbing potential shown in 
Fig. P12-4. 
b) For the perturbation above: 
1) What sign would you expect for c(1) 
21 ? Describe the effect of this on f1; 
on W(2) 
1 . 
2) What sign would you expect for c(1) 
12 ? Describe the effect of this on f2; 
on W(2) 
2 .
422 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
Figure P12-4 
 
12-5. Calculate c(1) 
41 for the perturbed particle in a wire example discussed in Section 
12-3. Show that c(1) 
41 is only about 2% as large as c(1) 
21 . Use 

 p 
0 
y sin my sin ny dy =(-1)m+n -1 2mn/(m2 -n2)2, m,n=1, 2, . . . ,m=n 
12-6. An electron moves in a harmonic potential, V = 1
2kx2. What is the effect, to first 
order, on the energies of superimposing an electric field, V =Ex? Explain your 
reasoning. 
12-7. Calculate the energy to first order of He+ in its lowest-energy state. Use the 
hydrogen atom in its ground state as your zeroth-order approximation. Use 
atomic units. Predict the signs (plus, zero, minus) of c(1) 
2s,1s and c(1) 
2p0,1s. Explain 
your reasoning. 
12-8. The previous problem gives an energy to first order of -3/2 a.u. for He+, using 
H as a starting point. Now reverse the process and try to get the ground-state 
energy ofHto first order, using He+ as starting point. Discuss the reasonableness 
of your answer in terms of a lower bound. 
12-9. A hydrogen atom in its ground state is perturbed by imposition of a uniform 
z-directed electric field of strength F atomic units: Field =-Fz=-Fr cos .. 
a) What is the first-order change in energy experienced by the atom? Show 
your logic. 
b) For the first-order correction to the ground-state wavefunction, we can consider 
c(1) 
2s,1s and c(1) 
2pz ,1s to be the coefficients for mixing in 2s and 2pz character. 
Predict the signs of these coefficients and sketch their effects on the 
ground-state wavefunction. 
c) Predict the sign of W(2) 
1s and explain your reasoning.
Section 12-10 Spectroscopic Selection Rules 423 
12-10. What is the H¨uckel MO p-electron energy to first order of the molecule (I) if 
C4 is perturbed by H =0.1ß? (See Appendix 6 for H¨uckel data.) 
12-11. Consider the fulvene molecule (II) in theHMOapproximation (seeAppendix 6). 
At which carbon will a perturbation involvinga affect the totalp-electron energy 
the least? Calculate to first order the energy of fulvene with a6 =a +0.5ß. If 
a computer and H¨uckel program are available to you, calculate the energy for 
this perturbed molecule directly and compare with your first-order result. 
12-12. Produce an expression for pkk, the atom–atom self-polarizability, from 
Eq. (12-70). Can this, like plk, have either sign? What can you infer about the 
sign of dqk when dak is positive? Discuss the physical sense of this. 
12-13. Using data from Example 12-6, calculate the self-atom polarizability, p1,1 for 
carbon 1 in naphthalene. 
12-14. Consider the hexatriene system. (See Appendix 6 for data.) One of the two 
central atoms is perturbed so that a becomes replaced by a +hß on that one 
atom only. 
a) What is the energy change, to first order, of the total p energy of hexatriene? 
b) Which of the occupied MOs is perturbed the most (i.e., has the largest firstorder 
contribution to the energy) by this perturbation? 
12-15. Answer the following questions from examination of HMO data for benzcyclopentadienyl 
radical, C9H7, given in Appendix 6. 
a) What is the total p energy to first order if atom 5 is replaced by a new atom 
X which contributes the same number of p electrons a carbon did and has 
ax =a +0.5ß and also has ßCX =0.9ß? 
b) For the positive ion of this system, how do you think the self-atom polarizability 
should compare qualitativeIy at atoms 1 and 2? Make your prediction 
by inspection, rather than computation, and explain your thinking. 
12-16. Use simple H¨uckel MOs for butadiene to calculate to first order the change in 
p energy that would result from closing cis-butadiene to cyclobutadiene (III). 
Repeat the approach for closing hexatriene to benzene. (IV). Which of these 
two systems benefits most from cyclic as opposed to linear topology? For each
424 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
system, compare your energy to first order with the actual H¨uckel energy (see 
Appendix 6 for data). 
12-17. The unperturbed H¨uckel MO energy levels of the allyl system are sketched in 
Fig. P12-17. The system is perturbed so that H22 =a +cß where c is positive. 
a) Sketch the effects of this perturbation, to first order, on the energy levels. 
b) Calculate to first order the perturbed MOs f(1) 
1 , f(1) 
2 , f(1) 
3 in terms of the 
unperturbed MOs .1, .2, .3. Sketch the results in a manner that makes 
clear the nature of the change in each MO. 
Figure P12-17 
 
12-18. Calculate the atom–atom polarizabilities p1,2 and p1,3 for the allyl cation. What 
do your results indicate will happen to the p-electron densities at atoms 2 and 
3 if atom 1 becomes more attractive? 
12-19. The cyclopropenyl system is perturbed so that the H¨uckel matrix elementH22= 
a +cß, where c is positive. 
a) Ascertain, by calculation or inspection, the appropriate zeroth-order degenerate 
wavefunctions for this situation. 
b) Sketch the effects of the perturbation on the orbital energies to first order. 
12-20. Given the HMO data in Appendix 6 for the cyclopentadienyl radical, calculate 
to first order all the orbital energies you would obtain if atom 2 were changed 
so that a2 became a +0.5ß. 
12-21. Imagine two allyl radicals coming together to form benzene. Each allyl has 
three p MOs and benzene has six. 
a) Sketch an energy level diagram showing how interaction between allyl 
orbitalscanberelatedtotheMOenergiesofbenzene,atthesimpleH¨uckellevel. 
b) Show with sketches how the three lowest-energy zeroth-order allyl combination 
orbitals are related to the three lowest-energy benzene MOs. 
c) Estimate to first order the energies of the benzene-like MOs coming from 
the nonbonding MOs of the allyl pair. Compare these to the relevant MO 
energies of benzene. (Remember that, when combining fragments, you must 
renormalize the functions) 
d) What is the energy to first order of the lowest MO of the allyl combination, 
and how does it compare to the lowest benzene orbital energy? 
12-22. Discuss the possible transitions p3.p2 and p4.p2 in butadiene (Here p1 is 
the lowest-energy MO, etc.) 
a) Are they dipole-allowed transitions? 
b) If yes, what is the polarization?
Section 12-10 Spectroscopic Selection Rules 425 
TABLE P12-24a 
 
Level Energy c1 c2 c3 c4 c5 c6 c7 
-2.01ß -0.238 0.500 -0.406 0.354 0.336 0.354 -0.406 
-1.26ß -0.397 0.500 -0.116 -0.354 0.562 -0.354 -0.116 
-1.00ß 0 0 -0.500 0.500 0 -0.500 0.500 
0.00 -0.756 0 0.378 0 -0.378 0 0.378 
1.00ß 
1.26ß 
2.10ß 
aCoefficients for 2pz basis function forming the four highest MOs are listed. Overlap has been 
assumed to be negligible. 
12-23. Calculate the dipole moment in the z direction for the states f(0) 
± of Eq. (12-81). 
Now construct a variational function of the form 
f =cos(a)2s+sin(a)2pz 
and maximize the z component of the dipole moment as a function of a. Are the 
states f(0) 
± those of maximum dipole magnitude? Discuss why this is reasonable. 
12-24. The benzyl radical, C7H7(V), has the H¨uckel energies and ground-state configuration 
given in Table P12-24. 
The radical is trapped and oriented in an external reference system as shown. 
Light polarized in the x direction is beamed on the system, and the frequency 
varied until an absorption is observed. Assuming this to result from excitation 
of the unpaired electron, to which level has the electron been promoted? 
12-25. The simple H¨uckel energies, occupation numbers, and coefficients for MOs 
in naphthalene (VI) are listed in Table P12-25. Assume that a single crystal 
of naphthalene is oriented so that each molecule is aligned with respect to an 
external coordinate system (VI). 
a) Light polarized in the x direction is beamed on the crystal. Assuming that 
the electron is excited from the highest occupied MO, to which empty MOs 
could it be promoted by the x-polarized light? 
b) Which transitions from the highest occupied MO would be allowed for 
y-polarized light? 
c) Which ones would be allowed for z-polarized light? 
d) Are any transitions from the highest occupied MO forbidden for nonpolarized 
light?
TABLE P12-25 
 
Energies 
-ß 
Occupation 
no. 
Coefficients 
1 2 3 4 5 6 7 8 9 10 
-2.303 2 
-1.618 2 
-1.303 2 
-1.000 2 
-0.618 2 -0.42 -0.26 0.26 0.42 0 -0.42 -0.26 0.26 0.42 0 
+0.618 2 -0.42 0.26 0.26 0.42 0 -0.42 -0.26 0.26 0.42 0 
+1.000 0 0 0.41 -0.41 0 0.41 0 -0.41 0.41 0 -0.41 
+1.303 0 0.40 -0.17 -0.17 0.40 -0.35 0.40 -0.17 -0.17 0.40 -0.35 
+1.618 0 -0.26 0.42 -0.42 0.26 0 -0.26 0.42 -0.42 0.26 0 
+2.303 0 -0.30 0.23 -0.23 0.30 -0.46 0.30 -0.23 0.23 -0.30 0.46 
426
Section 12-10 Spectroscopic Selection Rules 427 
12-26. A hydrogen atom in the 2s state is metastable, but, when passed through an 
electric field, its tendency to relax to the 1s state is greatly enhanced. Explain. 
12-27. Using EHMO data for formaldehyde given in the Problems section of Chapter 
10, decide which empty MOs can be reached by allowed dipolar transitions 
of an electron out of MO 7. For each allowed transition, state the polarization 
with respect to the molecular axis. 
12-28. Square cyclobutadiene has a degenerate pair of nonbonding MOs. Calculate 
the energies of these MOs, to first order, when two carbons are replaced by 
atoms that contribute the same number of pi electrons as carbon did and are 
treated at the HMO level by replacing a with a +0.5ß, and the substitutions 
occur at a) opposite and b) adjacent positions. Discuss your results in terms 
of cooperative vs. competing perturbations. Which compound should be more 
stable? 
12-29 Calculate to first order the HMO energy changes for the occupied MOs of 
benzene that occur when the molecule is stretched so that an opposing pair of 
bonds have resonance integrals of 0.9ß instead of 1.0ß. 
12-30 The intensity of a transition between initial (.i) and final (.f ) states is dependent 
upon the transition dipole moment: 
µ = e  .*f 
r.i dt. Calculate 
µ in 
atomic units for 2p0.1s in a hydrogen-like ion having atomic number Z. 
Assume that the electron has the same spin in both states. (See Appendix ?? 
for the definition of 1 a.u. of electric dipole moment.) 
Multiple Choice Questions 
(Try to answer these without referring to the text.) 
1. For the n=2 states of a hydrogen atom in a uniform electric field of strength F, 
a) the first-order correction to the energy equals zero for all four states. 
b) there are four different values of the energy to first order. 
c) the first-order correction to the energy is proportional to F for two states. 
d) the first-order correction to the energy is proportional to F2 for two states. 
e) None of the above is a true statement. 
2. Which one of the following statements is true for a hydrogen atom in its ground 
state that is placed in a uniform z-directed electric field? 
a) The first-order correction to the energy equals zero. 
b) The proper zeroth-order wavefunctions are (1/v2)(1s ±2pz). 
c) The second-order correction to the energy is positive. 
d) The first-order correction to the wavefunction contains some 2s AO. 
e) None of the above statements is true.
428 Chapter 12 Time-Independent Rayleigh–Schr ¨ odinger Perturbation Theory 
3. The hydrogen atom transition 3dxy.2px is 
a) dipole allowed and z-polarized. 
b) dipole allowed and x-polarized. 
c) dipole allowed and y-polarized. 
d) dipole allowed and xy-polarized. 
e) forbidden. 
4. Which of the following integrals over all space vanish by symmetry for a hydrogen 
atom with nucleus at the coordinate origin? 1)  1s y 2pzdv 2)  2pz z 3dxydv 
3)  2pz x 3dxzdv 
a) 1 only. 
b) 1 and 2 only. 
c) 1 and 3 only. 
d) 2 and 3 only. 
e) 1, 2, and 3. 
References 
[1] L. Pauling and E. B. Wilson, Introduction to Quantum Mechanics. McGraw-Hill, 
NewYork, 1935. 
[2] P. O. L¨owdin, J. Mol. Spectry. 13, 326 (1964). 
[3] L. I. Schiff, Quantum Mechanics, 3rd ed. McGraw-Hill, NewYork, 1968. 
[4] C.W. Scherr and R. E. Knight, Rev. Mod. Phys. 35, 436 (1963). 
[5] R. Hoffmann, Accounts Chem. Res. 4, 1 (1971).
Chapter 13 
Group Theory 
13-1 Introduction 
It should be evident by this point that symmetry requirements impose important constraints 
on MOs, and that one can often use symmetry arguments to tell what the MOs 
of a molecule must look like, in advance of calculation. We have also seen that one 
can often use symmetry arguments to tell whether or not an integral vanishes. Thus, 
we have made much use of symmetry already, but in an informal way. The formal 
use of symmetry, through group theory, will be described in this chapter. Knowledge 
of group theory augments one’s power to use symmetry as a shortcut, and familiarity 
with formal symmetry notation is necessary to follow the literature of many areas of 
chemistry, particularly inorganic and quantum chemistry, and molecular spectroscopy. 
13-2 An Elementary Example 
Consider the following four operations, or commands: 
1. Left face (L). 
2. Right face (R). 
3. About face (A). 
4. Remain as you are (E). 
These four operations constitute a group in the mathematical sense, and provide a 
convenient example with which to illustrate some definitions and terminology of group 
theory. The operations are called the elements of the group. Because there are four 
elements, this group is said to be of order four. 
Howdo we knowthat these four operations constitute a group? Before answering this 
question, it is necessary to describe what is meant by a product of elements. Products 
of group elements are written in the usual algebraic manner. That is, the symbols are 
written side by side without an algebraic symbol. Just as xy is understood to be the 
product of x and y,LR is understood to be the product of “left face” and “right face.” 
But elements of a group need not commute, and so the ordering of symbols in a product 
is important in group theory. The sequence of operations is understood to read from 
right to left in the product. Thus, LR means “right face” followed by “left face.” 
If we imagine a drill soldier carrying out the sequence of operations implied by 
LR, the result is a return of the soldier to his original position. That is, the product of 
429
430 Chapter 13 Group Theory 
TABLE 13-1 
 Products of 
Elements in the Group of Four 
Commands
First operation 
Second operation 
E L R A 
E E L R A 
L L A E R 
R R E A L 
A A R L E 
operations LR gives the same result as the single operation E. This is written LR=E. 
It is possible to write similar equations for all the product combinations in our group, 
and to arrange the results as in Table 13-1. An important fact to notice from this group 
multiplication table is that there is no sequence of operations in the group that cannot 
be accomplished by a single operation. All the possible positions open to the drill 
soldier through a sequence of two moves are also achievable through a single move. 
In other words, no product of elements takes us out of the group—the group exhibits 
closure. 
The reciprocal, or inverse of an operation is that subsequent operation which returns 
the soldier to his original position. Hence, L is the reciprocal of R, which is expressed 
as LR =E, or L=R-1. Examination of Table 13-1 shows that every column has E 
appearing once, which means that every element of our group has an inverse in the 
group. 
The associative law is obeyed if, in general, the sequence of operations A(BC)= 
(AB)C, where the parentheses enclose the pair of elements the product of which is to 
be evaluated first. We can test whether our elements satisfy this law by trying out all the 
combinations. For example L(RA)=LL=A and (LR)A=EA=A. One can quickly 
show that this group satisfies the associative law for all combinations of elements. 
We are now in a position to define a group. A group is a set of elements that meets 
the following requirements: 
1. The group contains the identity element (traditionally symbolized E) which corresponds 
to “make no change.” 
2. The group exhibits closure with respect to “multiplication.” The product of any two 
elements in the group is a single element in the group. 
3. There is a reciprocal in the group for every element of the group. 
4. The elements of the group obey the associativity law (AB)C =A(BC). 
We mentioned that, in general, group elements need not commute. That is, it is 
possible that AB 
=BA. However, in this example, the elements do all commute. This 
results in Table 13-1 being symmetric about the main diagonal. Groups in which all 
the elements commute are called abelian groups.
Section 13-3 Symmetry Point Groups 431 
13-3 Symmetry Point Groups 
Our example of four operations of a drill soldier differs from the kinds of groups most 
commonly used in quantum chemistry because it is not a symmetry group. Assuming 
that the soldier begins by facing north, we are able to distinguish four distinct possible 
subsequent orientations for him (facing the four compass directions). After a symmetry 
operation, an object is understood to be indistinguishable from the object before the 
operation. 
Whether or not an operation is a symmetry operation must be decided within the 
context of the object being operated on. For example, a square (devoid of identifying 
marks enabling us to distinguish one corner from another; see Fig. 13-1a) may be 
rotated by 90. about an axis perpendicular to the center (Fig. 13-1b), and be indistinguishable 
from its starting configuration. However, rotation by 120. does not lead to 
an indistinguishable configuration (Fig. 13-1c). 
Therefore, rotation by 90. is a symmetry operation for a square, but rotation by 120. 
is not. However, for an equilateral triangle, rotation by 90. is not a symmetry operation, 
while rotation by 120. is. For a circle, both rotations are symmetry operations. The 
only operation that is a symmetry operation for every object is the identity operation E. 
Another sort of symmetry operation pertains to infinite networks such as occur in 
idealized models of crystals. Here we can define operations that move an infinite 
line of cells or atoms to the left or right by a unit number of “steps,” producing a 
configuration indistinguishable from the initial one (Fig. 13-2). Symmetry operations 
can be usefully categorized into those that leave at least one point in space unmoved 
(rotations, reflections, inversion) and those that do not (translations). The latter category 
is of importance in the fields of crystallography and solid-state physics. The former 
category is useful when we deal with systems that do not undergo unending periodic 
repetition in space, and it is with this category that we will be concerned in this chapter. 
If we pick some object or shape of finite size and construct a group from symmetry 
operations for that shape, we have a symmetry point group for that object. We now 
illustrate how this is done for a ball-and-stick model of the ammonia molecule in its 
equilibrium nuclear configuration. It is not difficult to find operations that do no more 
than interchange identical hydrogen nuclei. There are a number of possible rotations 
about the z axis (Fig. 13-3). One could rotate by 120., 240., 360., 480., etc., either 
Figure 13-1 
 Effects of rotations on a square. 
Figure 13-2 
 Segment of an infinite repeating sequence of identical “cells”.
432 Chapter 13 Group Theory 
z y 
x 
1 
10 2 
1112 
3 
4 
7 6 5 
8 
9
Top view (showing clock) 
Figure 13-3 
 Orientation of a model of ammonia in Cartesian space. 
A 
B 
C 
y 
Figure 13-4 
 Three symmetry reflection planes for ammonia. 
clockwise or counterclockwise. Then there are three reflection planes. One of these 
is the xz plane (labeled “A” in Fig. 13-4) and the other two are planes containing the 
z axis and one of the other N–H bonds. Also there is the identity operation E, which 
we know that we need to satisfy the general requirements for a group. 
There is a problem with the rotations, and this is that we could produce an infinite 
number of them. How do we know when to stop? The answer is that we select only 
those that avoid “redundancies” among our indistinguishable configurations. Redundancies 
occur when two operations lead to “identical” indistinguishable configurations. 
For example, rotation by 360. is redundant with the identity operation because both 
operations leave the protons (which we imagine for the moment to be distinguishable) 
in the same locations. Similarly, rotation clockwise by 240. is redundant with rotation 
counterclockwise by 120. (Fig. 13-5). If we remove all such redundancies, we are left 
with only two rotations. 
There is arbitrariness as to how we wish to describe these. If we let the first be 
clockwise rotation by 120., the second could be described as either clockwise rotation 
Figure 13-5 
 Identical configurations result from the two rotations shown.
Section 13-3 Symmetry Point Groups 433 
by 240. or counterclockwise rotation by 120.. We will use the latter description. (Note 
that the clock is assumed to face +8 on the axis about which rotation occurs; see 
Fig. 13-3.) 
Our list of symmetry operations is now as follows: 
1. Identity (E). 
2. Reflection through xz plane (A). 
3. Reflection through plane B (B). 
4. Reflection through plane C (C). 
5. 120. clockwise rotation about z axis (D). 
6. 120. counterclockwise rotation about z axis (F ). 
We now proceed to set up the multiplication table for these operations. It is suggested 
that the reader do this as an exercise, checking the result against Table 13-2. 
Construction of the multiplication table is more challenging in this case than in our 
earlier example. 
Let us now use this table to see if the requirements for a group are satisfied by these 
symmetry elements. Since every product of two operations is equivalent to one of the 
operations of the set (i.e., since every spot in the table is occupied by one of our six 
symbols) the set exhibits closure. Also, the identity element is present in the set and 
occurs once in each column and row (Problem 13-2), and so every element has an 
inverse in the set. The associative law is satisfied, as one can establish by trying various 
examples, e.g., D(CB)=DF =E,(DC)B =BB =E, so that D(CB)=(DC)B. (In 
general, symmetry operations satisfy the associativity law.) Therefore, our set of six 
symmetry operations constitutes a symmetry point group of order 6. 
Inspection of Table 13-2 reveals that it is not symmetric across its principal diagonal. 
For example, CF =B, and FC =A; this is not an abelian group. One might inquire 
whether this group is the smallest that we can set up for the ammonia model. Inspection 
of the multiplication table should convince the reader that the following subsets meet 
the requirements for a group: E; E, D,F;E,A;E,B;E,C. These subgroups can be 
TABLE 13-2 
 Multiplication Table for 
Symmetry Operations of an Ammonia Molecule 
First operation 
Second operation
E A B C D F 
E E A B C D F 
A A E D F B C 
B B F E D C A 
C C D F E A B 
D D C A B F E 
F F B C A E D
434 Chapter 13 Group Theory 
distinguished from the full group by virtue of the fact that none of them exhausts all 
the possible physically achievable, nonredundant, indistinguishable configurations for 
the molecule. 
EXAMPLE 13-1 BH3, like ammonia, has a three-fold symmetry axis. However, 
BH3 is planar. As a result, there is an extra symmetry operation—reflection through 
the molecular plane—that does not move any nuclei. Thus, both the identity 
operation and this reflection leave all nuclei unmoved. Is this an example of 
redundant operations? 
SOLUTION 
 These operations are not redundant. Even though all the nuclei stay fixed for 
both operations, reflection interchanges the spaces above and below the plane, whereas the identity 
operation changes nothing. Two operations are redundant only if they move all points in space in 
the same way.  
13-4 The Concept of Class 
Imagine that we are subjecting our ammonia model to the various symmetry operations 
and that someone in a parallel universe is subjecting his ammonia model to symmetry 
operations too. Suppose that the parallel universe differs from ours in that everything is 
reflected through the xz plane. We shall refer to this as “the mirrorAuniverse.” We now 
pose the following problem. Suppose we initially have a particular configuration and 
our “mirrorA” man has the corresponding configuration (achieved by reflecting through 
plane A) as shown in Fig. 13-6. If the mirror A man now performs some symmetry 
Figure 13-6 
 Operation X in this universe parallels operation F in the mirror A universe.
Section 13-4 The Concept of Class 435 
operation on his model, say, counterclockwise rotation by 120.,F, he ends up with 
the final configuration shown at the lower right of the figure. (We continue to define 
“clockwise” from clocks in our universe.) For us to arrive at the corresponding final 
orientation, what operation must we perform? We symbolize this unknown operation 
X. We can find out what our final configuration must be by taking the reciprocal of 
reflection A on the mirror man’s final configuration. (A takes us from here to there; 
A-1 takes us from there to here.) But A-1 =A, so we obtain the result at the lower 
left of the figure. It is evident that X must be D, clockwise rotation by 120.. This is 
not surprising. It is related to the fact that, when you turn counterclockwise before a 
mirror, your image appears to turn clockwise. 
We can repeat the solution to this problem in a more formal way by requiring that the 
operation X followed by A (into the mirror A universe) bring us to the same configuration 
as operation A followed by F. That is, we seekX such that AX=FA. Multiplying 
both sides from the left by A-1 gives X = A-1FA. This is merely a mathematical 
statement of our above discussion. It says, “What operation is equivalent to the process 
of reflection through plane A followed by counterclockwise rotation by 120. followed 
by the inverse of reflection through plane A?” Using our multiplication table, we see 
that A-1 =A, and so X =AFA=AB =D. Therefore, D =A-1FA. 
We can find the mirror A universe equivalents to all the symmetry operations in the 
group in a similar way. Thus, 
A-1EA = E, A-1AA=A, A-1BA=C, 
A-1CA = B, A-1DA=F, A-1FA=D 
There is no need to stop here. We can imagine other parallel universes corresponding 
to reflections B and C and rotations D and F. (The E universe is the one we inhabit.) 
We can use the same sort of technique to accumulate corresponding operations for all 
these universes. The results are given in Table 13-3. 
If we examine this table, we note certain patterns. The operation E in our universe 
corresponds to E in all the universes. The reflections (A,B, and C) always correspond 
to reflections, and the rotations (D and F) always correspond to rotations. This makes 
physical sense. If we “do nothing” (E) in our universe, we expect the people in the other 
universes to do nothing also. If we rotate by 120., we expect the people to perform 
TABLE 13-3 
 X = Y-1ZY as a 
Function ofY and Z 
Z 
Y 
E A B C D F 
E E A B C D F 
A E A C B F D 
B E C B A F D 
C E B A C F D 
D E C A B D F 
F E B C A D F
436 Chapter 13 Group Theory 
rotations by 120. too, although not necessarily always in the same direction. The 
argument for reflection is the same. These are examples of three classes of operation. 
The formal definition of class is as follows: If P and Q in a group have the property 
that X-1PX =P or Q and X-1QX =P or Q for all members X in the group, then 
P and Q belong to the same class. Physically, operations in the same class are of the 
“same kind”—all reflections, all rotations, etc. 
EXAMPLE 13-2 BH3, like ammonia, has three planes of reflection symmetry that 
contain the three-fold rotational axis. However, BH3, being planar, also has a 
reflection operation through the molecular plane. Is this reflection operation in the 
same class as the other three? 
SOLUTION 
 No. In both molecules, the three planes each interchange two hydrogens and 
keep one fixed in space. The extra reflection plane in BH3 keeps all three hydrogens fixed. If we 
reflect so as to keep all hydrogens fixed, there is no way a person in a symmetry-related universe 
could make a corresponding reflection by using a reflection that interchanges two hydrogens and 
keeps one fixed.  
We shall see that classes are important subdivisions of groups. Note, however, that 
a class need not be a subgroup. For instance, D and F constitute a class, but, as the 
class does not include E, it is not a subgroup. 
In discussing the concept of class, it is unnecessary to postulate parallel universes, 
and the reader should not be disturbed by this pedagogical device. The people in the 
“other universes” are merely working with ammonia models that have been reflected or 
rotated with respect to the model orientation that we chose in Fig. 13-3. Operations in 
the same class are simply operations that become interchanged if our coordinate system 
is subjected to one of the symmetry operations of the group. 
13-5 Symmetry Elements and Their Notation 
There are five kinds of symmetry operations that one can utilize to move an object 
through a maximum number of indistinguishable configurations. One is the trivial 
identity operation E. Each of the other kinds of symmetry operation has an associated 
symmetry element1 in the object. For example, our ammonia model has three reflection 
operations, each of which has an associated reflection plane as its symmetry element. 
It also has two rotation operations and these are associated with a common rotation 
axis as symmetry element. The axis is said to be three-fold in this case because the 
associated rotations are each one-third of a complete cycle. In general, rotation by 
2p/n radians is said to occur about an n-fold axis. Another kind of operation—one we 
have encountered before is inversion, and it has a point of inversion as its symmetry 
element. Finally, there is an operation known as improper rotation. In this operation, 
we first rotate the object by some fraction of a cycle about an axis, and then reflect it 
through a plane perpendicular to the rotation axis. The axis is the symmetry element 
and is called an improper axis. 
The following two examples should help clarify the nature of these symmetry elements. 
Consider first the ethane molecule in its eclipsed conformation (I). 
1This term is not to be confused with a group element.
Section 13-5 Symmetry Elements and Their Notation 437 
The following symmetry elements can be identified. (It is important that you satisfy 
yourself that you see these elements in the sketch.) 
1. One three-fold axis coincident with the C–C bond. 
2. Three two-fold axes perpendicular to the C–C bond and intersecting its midpoint. 
3. Three reflection planes, each containing the C–C bond and a pair of C–H bonds. 
4. One reflection plane perpendicular to the C–C bond and bisecting it. 
5. No point of inversion. 
6. One three-fold improper axis coincident with the C–C bond. 
[Note that these operations are to be applied to the rigid molecule. Movement of 
only one methyl group and not the other (i.e., internal rotation) is not allowed.] 
Now let us consider staggered ethane (II) (i.e., one methyl group rotated 60. from 
its eclipsed position). Its symmetry elements are 
1. One three-fold axis coincident with the C–C bond. 
2. Three two-fold axes perpendicular to the C–C bond and intersecting its midpoint. 
3. Three reflection planes, each containing the C–C bond and a pair of C–H bonds. 
4. No reflection plane perpendicular to the C–C bond. 
5. One point of inversion at the midpoint of the C–C bond. 
6. One six-fold improper axis coincident with the C–C bond. 
In going from the eclipsed to the staggered conformation, we have lost a reflection 
plane perpendicular to the C–C bond, gained a point of inversion, and changed the 
order of the improper axis. 
Usually it is not difficult to “see” most elements of symmetry in molecules, the 
exception being improper axes, which tend to be a little tricky. The six-fold improper 
rotation in staggered ethane is not too hard to envision when it is applied once. If we 
rotate clockwise by 60. and reflect, H1 replaces H5,H5 replaces H2, etc. It is a little
438 Chapter 13 Group Theory 
more difficult when we try to imagine applying this operation twice in succession. The 
reader should recognize that two 60. rotations and two reflections result in H1 replacing 
H2,H2 replacing H3,H4 replacing H5, etc. 
EXAMPLE 13-3 BH3 has a three-fold axis and a reflection plane in the molecular 
plane. Does BH3 have a three-fold improper rotation axis as a necessary member 
of its group? 
SOLUTION 
 Athree-fold improper axis coincident with the “regular” three-fold rotational axis 
appears to have the correct symmetry characteristics: It rotates each hydrogen to the next position 
and then interchanges the spaces above and below the plane. Do we need it in the group? Only if 
there is no other single operation that accomplishes the same thing. There is no such operation, so 
this improper axis is indeed a nonredundant symmetry element.  
A notation for these symmetry elements (and for the operations related to them) has 
come to be generally accepted in chemistry (exclusive of crystallography, which uses a 
different notation). This notation is summarized in Table 13-4. The symmetry elements 
of eclipsed ethane would be, by these conventions, C3, 3C2,s (perpendicular to C–C), 
3s (containing C–C), S3. It often occurs that several classes of reflection planes and 
axes for proper or improper rotations are present in a system. In eclipsed ethane, we 
can imagine that the three reflection planes containing the C–C bond might all be in 
the same class, that is, might be interchanged by a symmetry operation. In fact, it is 
easy to see that they are interchanged by rotations about the threefold axis. However, 
none of these could ever be equivalent to the reflection perpendicular to the C–C bond 
because this reflection interchanges the carbons, whereas the others do not. For this 
reason, we keep separate tally of the different classes of reflection in eclipsed ethane, 
rather than simply writing “4s.” 
There is a geometric convention that aids discussion of molecules having axes of 
rotation. One looks for a unique axisCn (usually the axis of highest order) and imagines 
this axis to be vertical (coincident with the z axis). Then a reflection plane perpendicular 
to this axis is horizontal and is labeled sh. Planes containing this axis are necessarily 
vertical and are subdivided into dihedral and vertical planes. Dihedral planes must 
contain the unique reference axis, Cn, and must also bisect the angles between two-fold 
TABLE 13-4 
 Symbols for Symmetry Elements and Operations 
Symmetry operation Symmetry element Symbol 
“Do nothing” (identity) None E 
Rotation by 2p/n radians An n-fold (proper) axis Cn 
Reflection through plane A plane s 
Inversion through a point A point i 
Rotation through 2p/n radians 
followed by reflection 
through a plane 
perpendicular to 
the rotation axis 
An n-fold (improper) axis Sn
Section 13-5 Symmetry Elements and Their Notation 439 
axes perpendicular to Cn. Such planes are labeled sd. Vertical planes that do not 
bisect two-fold axes2 are labeled sv. In eclipsed ethane, the C3 axis is the principal 
axis, so the C–C bond is oriented vertically. Therefore, our symmetry elements are 
labeled C3, 3C2,sh, 3sv, S3. (The vertical planes are labeled sv because they contain, 
rather than bisect, the two-fold axes that are perpendicular to the principal axis.) For 
staggered ethane, we have C3, 3C2, i, 3sd, S6. (Here the vertical planes do not contain 
the two-fold axes, but do bisect them. Hence, we label them sd. Making a simple 
sketch will aid in clarifying this distinction.) 
Once one has recognized the set of symmetry elements associated with a given 
object, it is a straightforward matter to list the symmetry operations associated with the 
set. Simplest are the operations associated with elements s and i, because each such 
element gives rise to only one operation. Proper and improper axes are somewhat more 
complicated. Let us return to our ammonia molecule for illustration of this. There we 
had a threefold axis C3 and we noted that we could rotate by 2p/3 (C+3 ) to get one configuration, 
and 4p/3 [C+3 C+3 = (C+3 )2] to get another. Alternatively we could choose 
to rotate by 2p/3 and -2p/3 (C-3 ), the latter easily being shown to be equivalent to 
(C+3 )2. But if we rotate by (C+3 )3, we return to our original configuration. That is, 
(C+3 )3=E. Therefore, aC3 axis produces two unique operations. The reader can easily 
generalize this to the statement that a Cn axis yields n-1 unique symmetry operations 
Cn,Cn2,Cn3, ... , Cn-1 
n , where we assume rotation to be in the clockwise direction in 
all cases. Benzene has aC6 axis with five associated nonredundant symmetry operations 
(C6,C2
6,C3
6,C4
6,C5
6 ). NowC3
6 is equivalent to a rotation by p radians, an operation 
we normally write as C2. Similarly, C2
6 =C3. The result of such reductions is the set of 
operations C6,C3,C2,C2
3,C5
6 for the C6 axis of benzene. There are several classes of 
rotation here. If we use a compressed notation, we can write this set as 2C6, 2C3,C2. 
This indicates that the presence of a sixfold axis implies the existence of coincident 
twofold and threefold axes. However, these implied elements are not listed for such a 
system since their operations are all contained in the set of operations of the C6 axis. 
Improper axes can also be associated with several symmetry operations. We noted 
earlier that S6, applied twice in succession, results in a simple 2p/3 rotation about 
the S6 axis. In other words, we can write the set S6, S2
6 , S3
6 , S4
6 , S5
6 as S6,C3,S2,C2
3 . 
We stop at S5
6 because S6
6 = E due to the combination of C6
6 and an even number of 
reflections. S3
6 is equivalent to S2 because it contains three rotations by 2p/6 and an 
odd number of reflections, and S2 means one rotation by p and one reflection. The 
operation S2, however, is easily shown to be equivalent to an inversion, and so we 
have, using a compressed notation, 2S6, 2C3, i associated with the S6 axis. Since we 
have already explicitly listed the elements C3 and i in our set of elements for staggered 
ethane (or any other system containing an S6 axis) only the 2S6 operations are unique 
to the S6 axis. The generalization of this case is that an S2n axis with odd n implies that 
elements Cn and i are also present. Of the 2n-1 operations associated with S2n, n-1 
are preempted by the Cn axis and 1 by the element i leaving n - 1 operations to be 
attributed to the S2n axis. [If n=1, we have S2, C1, and i as elements. But C1 =E, so 
we ignore it. There is only one operation here (2 · 1-1=1) and it is preempted by i. 
Therefore, S2 has no unique operations and it is not listed as a symmetry element.] For 
S2n with n even, i is not implied and there are n unique operations. 
2In certain cases there will be two geometrically nonequivalent sets of vertical planes that cannot be distinguished 
on the basis of bisecting two-fold axes (e.g., in square-planar XeF4). In such cases the convention is to take the 
sv planes to be those containing the larger number of atoms.
440 Chapter 13 Group Theory 
EXAMPLE 13-4 How many operations must be associated with S4 in the squareplanar 
molecule XeCl4? 
SOLUTION 
 This molecule possesses a C2 and a C4 axis coincident with S4. C2 usurps S24 
since the latter has two reflections that cancel, making it just like C2. (C2 also usurps C24
.) S44 
=E. 
Only two operations, S14
and S34
, survive as unique symmetry operations  
In eclipsed ethane, we have the element S3, which generates operations S3, 
S2
3,S3
3,S4
3,S5
3,S6
3 , . . . In deciding where to stop here, we note that S3
3 
=E because we 
have here an odd number of reflections. Therefore, S3
3 =s, where we understand this 
to be reflection through the plane perpendicular to the S3 axis. For eclipsed ethane, 
this is sh. The element S6
3 has an even number of reflections and is equal to E. How 
about S4
3? This has an even number of reflections, so it is identical to C4
3 which is the 
same as C3
3C3 = EC3 = C3. Thus, our S3 element has operations that are consistent 
with the presence of C3 and sh elements. We note that eclipsed ethane was indeed 
found to have these elements. Once again, we allow these elements to preempt their 
operations from S3. This leaves only two unique operations, S3 and S5
3 , corresponding 
to clockwise and counterclockwise improper rotations by 2p/3 radians. In general, 
S2n+1 will occur only when C2n+1 and s (perpendicular to C2n+1) are also present. 
These preempt a total of 2n+1 operations from the total of 2(2n+1)-1 = 4n+1 
we can achieve with S2n+1 alone; and we are left with 2n operations for S2n+1. The 
above conclusions are summarized in Table 13-5. 
The set of symmetry operations is contrasted to the set of symmetry elements for 
eclipsed and staggered ethane inTable 13-6. Note that there are 12 symmetry operations 
in each case. By setting up the multiplication table for either of these sets of 12 
operations, we can show that the mathematical requirements for a group are satisfied. 
Thus, each of these sets of symmetry operations constitutes a separate group of order 12. 
The multiplication table for the ammonia model is given in Table 13-7 in terms of 
the symmetry symbols just described. (A,B,C are taken to be s1,s2,s3, respectively.) 
It would be possible, using what has been described up to this point, to construct 
a list of symmetry operations for any object we please. Then we could test the set to 
see if it satisfied the various requirements for a group. In practice, this is not done. 
It turns out that (1) only a rather limited number of distinct kinds of symmetry are 
possible, and (2) in each case, the symmetry operations for the object do form a group. 
The practical question, then, is, given an object, to which of the known groups does it 
TABLE 13-5 
 Number of Operations Associated with Symmetry Elements 
Element (Other elements present) No. operations 
i — 1 
s — 1 
Cn — n-1 
S2n 
n=odd :Cn and i present n-1 
n=even :Cn present n 
S2n+1 C2n+1,s 2n
Section 13-6 Identifying the Point Group of a Molecule 441 
TABLE 13-6 
 Symmetry Operations and Elements in Ethane 
Molecule Elements Operations 
E 
C3 2C3 
3C2 3C2 
sh sh 
3sv 3sv 
S3 2S3 
Group order: 12 
E 
C3 2C3 
3C2 3C2 
i i 
3sd 3sd 
S6 2S6 
Group order: 12 
TABLE 13-7 
 Multiplication Table for the Ammonia Molecule 
First operation 
Second operation
E s1 s2 s3 C + 3 C - 3 
E E s1 s2 s3 C+ 3 C - 3 
s1 s1 E C+ 3 C - 3 s2 s3 
s2 s2 C - 3 E C+ 3 s3 s1 
s3 s3 C + 3 C - 3 E s1 s2 
C + 3 C + 3 s3 s1 s2 C - 3 E 
C - 3 C - 3 s2 s3 s1 E C+ 3 
belong? Once this is settled, we merely look up the tabulated properties of that group 
and save ourselves the effort of working through all the details. 
13-6 Identifying the Point Group of a Molecule 
One can decide to which point group a molecule belongs by systematically looking for 
certain symmetry elements. Each symmetry point group has a unique group symbol, so 
basically one tries to figure out which group symbol is appropriate for a given molecule. 
A flowchart that serves this purpose is displayed in Fig. 13-7. As an illustration, we will 
work out the group symbol for staggered ethane. The first fewquestions in the flowchart
442 Chapter 13 Group Theory 
Is the molecule linear? 
No 
No 
Is there a reflection plane 
perpendicular to the axis? 
T, T h, T d, O, or O h 
(Examine character tables 
to choose among these) 
Is there a 
reflection plane? 
Is there a point 
of conversion? 
C 1 
C I 
C s 
I 
or 
I h 
C v D h 
Are there one or more 
proper rotation axes? 
Does the molecule 
possess six C5 axes? 
Does the molecule 
possess four C3 axes? 
Yes 
No 
No 
No 
No 
No 
Yes Yes 
Yes 
Yes 
Yes 
Yes 
Selection of reference axis 
Cn. (Defines vertical direction.) 
Is there an axis is having 
higher order than any 
other proper axis? 
No 
No 
No 
Yes 
Yes 
Yes 
Yes 
Yes
Yes 
Yes 
Yes 
No 
No 
No 
No 
No 
Of the several axes having 
highest order, is there one 
which is geometrically unique? 
Choose any of the 
equivalent axes as Cn 
(Cn and n now defined) 
Select this as 
reference axis Cn 
Is there an S2n axis 
collinear with Cn? 
S2n 
Cn h
Cn v 
Cn 
Dnh
Dnd 
Dn 
Are there n twofold axes 
perpendicular to Cn? 
Are there any symmetry 
elements present other 
than Cn, S2n, or i ? 
Is there a reflection 
plane (s h) perpendicular 
to Cn ? 
Is there a set of 
n reflection planes 
(sv) containing Cn ? 
Is there a set of 
n reflection planes 
containing Cn and 
bisecting the n 
twofold axes? 
Is there a reflection 
plane (sh) perpendicular 
to Cn? 
Yes (D branch) 
No (C branch) 
Figure 13-7 
 Flow scheme for group symbols.
Section 13-7 Representations for Groups 443 
check to see whether the molecule belongs to one of several special groups of very 
high symmetry. Since staggered ethane is not linear, and since it does not possess the 
symmetry of any of the regular polyhedra, we move to the question concerning proper 
rotation axes. We have noted earlier that there are several proper rotation axes in this 
molecule. We next consider which one should be selected as the reference axis. The 
axis of highest order is threefold and there is only one of these; therefore, we choose 
it as our reference axis. This defines n as 3 for the remainder of the flowchart, and 
determines the vertical direction. Next we look to see if there is an S6 axis coincident 
with the C3 axis, and we find that there is. Checking to see if symmetry elements other 
than C3, S6, and i exist, we find that they do indeed, and that among them are three 
twofold axes perpendicular to the reference axis C3. This leads us along the “D branch” 
of the diagram, which means that, whatever else we find, the major symbol for our 
group will be D. Next we look for a sh reflection plane. Finding none, we next seek 
a set of reflection planes containing C3 and also bisecting the various C2 axes. Such 
dihedral planes are present, so our group symbol is D3d for staggered ethane. The 
reader may verify that eclipsed ethane follows the same route out to the D branch, but 
there we do find a sh (horizontal) plane, and so eclipsed ethane has D3h symmetry. 
A simple exercise, suggested at this point, is to ascertain the symmetry symbol for 
ammonia. (The answer is given in the next section.) 
EXAMPLE 13-5 What is the group symbol for (planar) BH3? 
SOLUTION 
 Starting at the top of Fig. 13-7, the answers “no, no, no, yes” bring us to the box 
for selection of the reference axis. There is one three-fold (proper) rotational axis, and no other of 
higher order, so this defines n (= 3) and “vertical.” Coming out of the box we get answers “no, 
yes, yes,” giving D3h as the group symbol. Notice that we arrive at this conclusion without having 
had to use all the symmetry elements of the molecule (S3 or vertical reflection planes).  
13-7 Representations for Groups 
We have seen that the group of symmetry operations for the ammonia molecule leads 
to a particular group table of product operations (Table 13-7). Let us now see if we 
can assign a number or matrix to each symmetry operation such that the products of 
numbers satisfy the same group multiplication table relationships as do the products 
of symmetry operations. If we can find such a set of numbers or matrices, we say we 
have a representation for the group. 
It is always possible to produce a trivial representation by simply assigning the 
number +1 to each operation. Then any operator relation, such as s1s2 = C+3 is 
necessarily satisfied by the numbers since 1 ·1=1. Nontrivial representations exist too 
(except for the C1 group). For example, the group for molecules (like ammonia) having 
C3v symmetry can be represented by the set of numbers +1, -1, -1, -1, +1, +1, 
respectively, for the operations E, s1, s2, s3, C+3 , C-3 in Table 13-7. Then the relation 
s2C+3 = s3 is paralleled by the representation since (-1)(+1)=-1, etc. It is also 
possible to find a representation for the C3v group wherein each operation corresponds 
to a 2×2 matrix. This is shown in Table 13-8, along with the two other representations 
already mentioned. The reader is encouraged to test that, for instance, s2C+3 =s3 in 
this matrix representation.
444 Chapter 13 Group Theory 
TABLE 13-8 
 Representations for the C3v Group 
E s1 s2 s3 C+3 C-3 
1 1 1 1 1 1 1 
2 1 -1 -1 -1 1 1 
3 
1 0 
0 1
 1 0 
0 -1
 -1/2 v3/2 
v3/2 +1/2
  -1/2 -v3/2 
-v3/2 1/2
  -1/2 v3/2 
-v3/2 -1/2
 -1/2 -v3/2 
v3/2 -1/2 
 
TABLE 13-9 
 A Two-Dimensional Unit Matrix Representation for the C3v Group 
E s1 s2 s3 C+3 C-3 
 1 0 
0 1
 1 0 
0 1
 1 0 
0 1
 1 0 
0 1
 1 0 
0 1
 1 0 
0 1
 
The symbol i is often used to stand for the ith representation of a group. i(R) 
refers to the representation of the particular operation R in the ith representation. Thus, 
2(s3)=-1. In our ammonia example, 1 and 2 are one-dimensional representations 
and 3 is two-dimensional. The representation 1 is really the simplest member of a set 
of equally trivial representations, namely all representations made from unit matrices. 
Thus, the representation  (Table 13-9) is just as successful a representation as is 
1. We could obviously use unit matrices of arbitrary dimension in constructing a 
representation. We can think of such representations as being “built up” from 1: 
(R)=
1(R) 0 
0 1(R)
 etc. 
It is possible to extend this approach by building up representations from mixtures 
of 1,2,3. Thus, 
would produce the representation given in Table 13-10. It is easy to show that these 
matrices multiply together in a way that parallels the group table for symmetry oper- 
TABLE 13-10 
 A C3v Group Representation Built from Other Representations 
E s1 s2 
 (R) 
.
.... 
1 0 0 0 
0 1 0 0 
0 0 1 0 
0 0 0 1
.
.... 
.
.... 
-1 0 0 0 
0 1 0 0 
0 0 1 0 
0 0 0 -1 
.
.... 
.
.... 
-1 0 0 0 
0 1 0 0 
0 0 -1
2 v3/2 
0 0 v3/2 1
2 
.
....
Section 13-7 Representations for Groups 445 
ations, since block diagonal matrices always multiply in a block-for-block manner. 
Hence, the upper left 1×1 blocks (2) multiply among themselves, the 1×1 blocks 
in the 2,2 positions (1) multiply among themselves, and the 2×2 blocks at the lower 
right (3) multiply among themselves. Since each of these sets conforms to the group 
table, it follows that their composite  will also. It is evident that there is no limit to 
the number of representations we could build up in this manner. 
Representations like  and  are called reducible representations because they 
are composites of one or more smaller-dimensional representations and hence can be 
decomposed into those smaller representations. Any representation that cannot be 
reduced, or decomposed, into smaller representations is said to be irreducible. 
How can we tell whether a given representation is reducible or not? If it is already 
one dimensional, it is obviously irreducible. If it is multidimensional and, like  or 
, uniformly block diagonalized throughout the whole set of symmetry operations, 
it is plainly reducible. The trouble comes when it is multidimensional but not block 
diagonalized, for example, 3. It is sometimes the case that such representations are 
reducible. To illustrate this point, let us go back to , which we know is reducible, and 
imagine transforming this representation by multiplying every matrix from the right by 
ß and from the left ß-1, where ß is some 4×4 unitary matrix (see Chapter 9). This 
will give us a new set of six 4 × 4 matrices that no longer will necessarily be block 
diagonalized. Yet it is easy to showthat these newmatrices, ß, are still a representation 
for our group. We prove this by showing that, if three matrices from  (call them A, 
B, C) are related by AB =C, then the corresponding transformed matrices Aß, 
Bß, Cß, satisfy AßBß =Cß. If this is true, then the ability of  to correspond to the 
group table relationships is retained by ß. The proof is as follows: 
Let 
AB =C 
Then 
AßBß =ß-1Aßß-1Bß =ß-1ABß =ß-1Cß =Cß 
The point we are trying to make here is that a reducible representation like  can 
be put forth in many guises, each corresponding to a different choice of ß, and many 
of these will not be block diagonal in form. (Representations that differ only by a 
unitary transformation are said to be equivalent. Thus, a and ß are equivalent 
representations.) To show that such a representation as ß is reducible and also to 
reveal its component representations, we could back transform it by finding the matrix 
ß and calculating 
ßßß-1 =ßß-1ßß-1 = 
s3 C+3 C-3 
.
.... 
-1 0 0 0 
0 1 0 0 
0 0 -1
2 -v3/2 
0 0 -v3/2 1
2 
.
.... 
.
.... 
1 0 0 0 
0 1 0 0 
0 0 -1
2 v3/2 
0 0 -v3/2 -1
2 
.
.... 
.
.... 
1 0 0 0 
0 1 0 0 
0 0 -1
2 -v3/2 
0 0 v3/2 -1
2 
.
....
446 Chapter 13 Group Theory 
Therefore, deciding whether or not a multidimensional representation is reducible is the 
same as deciding whether it can be uniformly block-diagonalized through a common 
unitary transformation. It turns out that there is no single unitary transformation capable 
of diagonalizing all the 2×2 matrices of3, and so3 is an irreducible two-dimensional 
representation. (We shall show later that this can be established without “trying out” an 
infinite number of transformations.) For any group, a goal is to find all the inequivalent, 
irreducible representations possible. Anything beyond this is superfluous information. 
For our C3v (ammonia) group, 1, 2, and 3 exhaust the possibilities and constitute 
a complete set of inequivalent irreducible representations. 
13-8 Generating Representations from Basis Functions 
The reader may wonder how one goes about discovering nontrivial representations like 
3.3 A convenient way to do this will now be described, and we will show at this 
stage the connection between quantum mechanics and the group theory of symmetry 
operations. 
Consider the molecule shown in (III). This molecule has but one nontrivial symmetry 
element—a point of inversion. According to our flowchart, this places it in the Ci point 
group. The only symmetry operations here are E and i. Now consider two functions, 
f1 and f2. Let f1 be located on one end of the molecule. For instance, let f1 be 1sFa , 
a 1s AO centered on the fluorine atom on the left side of the molecule. Let f2 be a 
similar function on the other side of the molecule, 1sFb . Now let us see what happens 
to these functions when they are acted upon by our symmetry operations E and i: 
Ef1 =f1, Ef2 
=f2, if1 
=f2, if2 
=f1 
We see that f1 and f2 are interchanged by inversion. Let us try to find numbers to 
represent these results. Clearly, replacingE by+1 will give the correct result. However, 
to obtain the effect of operation by i we need to interchange f1 and f2. We cannot 
achieve this by multiplying by a number, since f1 and f2 are linearly independent. If, 
however, we rewrite the effect of i as 
i 
f1 
f2

=
f2 
f1
 
it becomes clear that i can be represented by the matrix (01 
10). Thus, use of the functions 
1sFa and 1sFb has generated a two-dimensional representation shown in Table 13-11. 
(The representation for E has been put into a 2×2 matrix form to be in dimensional 
agreement with the representation for i.) 1sFa and 1sFb are called a basis for . Clearly, 
3Indeed, the reader may wonder why we even worry about all this. Patience.
Section 13-8 Generating Representations from Basis Functions 447 
TABLE 13-11 
 A Representation for the Ci Group 
Ci E i Basis 
 
 1 0 
0 1
 0 1 
1 0 
 
(1sFa , 1sFb ) 
any pair of identical functions symmetrically placed with respect to the point of inversion 
would generate the same two-dimensional representation, and so there is nothing 
very special about 1sFa and 1sFb as a basis. Is  reducible? It is easily argued that it 
must be. Any unitary 2×2 transformation will have no effect on the matrix for E since 
ß-11ß =ß-1ß =1. We are thus at liberty to look for any unitary transformation ß that 
diagonalizes the second matrix. Since this is a nonsingular matrix (its determinant is 
unequal to zero), it should be diagonalizable, and it is if we take 
 1/v2 1/v2 
-1/v2 1/v2
0 1 
1 0
1/v2 -1/v2 
1/v2 1/v2 

=
1 0 
0 -1
 
Now that our 2×2 representation has been diagonalized to two 1×1 “blocks,” we can 
rewrite our representations as shown in Table 13-12. 
Our first choice of basis generated a reducible representation. What bases would we 
need to generate the 1×1 representations 1 and 2? To generate 1, we need a basis 
function that turns into itself when it is inverted. One possibility is f1 =1sFa +1sFb . 
If we sketch this function (Fig. 13-8a), it is evident that it is regenerated unchanged by 
inversion. The mathematical demonstration is 
if1 =i(1sFa +1sFb )=i1sFa +i1sFb =1sFb +1sFa =f1 
TABLE 13-12 
 Reduced Representations for 
the Ci Group 
Ci E i Basis 
1 1 1 ? 
2 1 -1 ? 
Figure 13-8 
 Basis functions for irreducible representations of the Ci group.
448 Chapter 13 Group Theory 
f1 is symmetric for operations E and i. For 2, we need some function f2 that turns 
into minus itself upon inversion. f2 =1sFa -1sFb would serve, as Fig. 13-8b shows, 
and as is demonstrated mathematically by 
if2 =i(1sFa -1sFb )=i1sFa -i1sFb =1sFb -1sFa =-f2 
f2 is symmetric for E and antisymmetric for i. (Notice that the way in which 1sFa 
and 1sFb needed to be mixed to produce a diagonal representation is indicated by the 
coefficients in the columns of the matrix we used to diagonalize the original 2 × 2 
representation. The matrix ß not only diagonalizes our , it also tells us how to mix 
the original bases in order to arrive at bases for the irreducible representations.) 
This example demonstrates an important fact: In order to generate a onedimensional 
representation for a group, we need a basis function that is either 
symmetric or antisymmetric for every symmetry operation in the group. This provides 
the point of connection with quantum mechanics. We showed much earlier (Chapter 2) 
that any nondegenerate wavefunction (or MO) must be symmetric or antisymmetric 
for every operation that leaves the hamiltonian unchanged. Since symmetry operations 
leave the hamiltonian unchanged (they merely interchange identical nuclei), it follows 
that nondegenerate wavefunctions or orbitals are symmetric or antisymmetric for 
every operation in the symmetry point group of the molecule. This means that every 
nondegenerate wavefunction or orbital is a basis for a one-dimensional representation 
for its molecular point group. Indeed, it can be proved that every wavefunction or 
orbital, even if degenerate, is a member of a basis for an irreducible representation for 
the point group of the molecule. The irreducible representation produced by an n-fold 
degenerate set of orbitals will be n-dimensional. (We indicate how this comes about in 
a later section.) Therefore, knowing wavefunctions enables one to generate representations. 
More importantly, knowing representations allows us to say something about 
wavefunctions. For example, the representation table (Table 13-12) for the Ci group 
(which is now complete—this group has only two irreducible inequivalent representations) 
tells us the following: (1) The molecule FClBrC–CBrClF in the conformation 
pictured has no degeneracies due to symmetry in its electronic states or in MO energies 
(if we do a calculation at the MO level). We can tell this because only one-dimensional 
representations exist for this group. (2) Every wavefunction for an electronic state 
or orbital must be either symmetric or antisymmetric for inversion. Therefore, the 
following AO combinations are feasible for MOs insofar as symmetry is concerned: 
c(1sFa -1sFb ), 
c1(1sFa +1sFb )+c2(1sCla +1sClb )+c3(1sBra +1sBrb ) 
+c4(1sCa +1sCb )+c5(2sFa +2sFb )+··· 
The following are disallowed: 
c1sFa , 
c1(2sFa +2sFb )+c2(2sCla -2sClb ) 
Our halogenated ethane molecule is a good starting example because it belongs to 
such a simple group. But let us now return to the C3v group of the ammonia molecule 
and continue developing the relations between group theory and MO theory. For 
convenience, the molecule is again sketched (Fig. 13-9) to show its orientation with 
respect to Cartesian axes. (The z coordinate is the principal axis.)
Section 13-8 Generating Representations from Basis Functions 449 
y 
z 
Rz 
x 
1 
10 2 
1112 
3 
4 
7 6 5 
8 
9 
f 
Figure 13-9 
 Orientation of a model of ammonia in Cartesian space. 
The normal practice in group theory is to use Cartesian coordinates or linear combinations 
of such coordinates as basis functions4 for generating many of the representations 
of a group. Therefore, we begin by examining the z coordinate to see what becomes of it 
under the various symmetry operations in theC3v group. The results are easily seen to be 
Ez =+1z, s1z =+1z, s2z =+1z, 
s3z =+1z, C+3 z =+1z, C-3 z =+1z 
Thus, z is a basis for the totally symmetric representation 1 of Table 13-8. The 
coordinates x and y can also be used as bases. Here things get more complicated. 
Clockwise rotation of x by 2p/3 radians (to give x) causes it to end up in a position 
where it must be expressed as a resultant of both x and y (Fig. 13-10). Simple 
trigonometry requires that, for a general rotation through the angle f, the unit vector 
x has an x coordinate of cosf and a y coordinate of sin f. Similarly, a rotation of 
y to a new position designated y must yield a new x coordinate of -sin f and a new 
y coordinate of cosf. Thus, for a rotation through the angle f, we have 
Cf 
x
y

=
x 
y

=
 x cosf +y sin f 
-x sin f +y cosf 
 
Figure 13-10 
 Result of clockwise rotation of x by 2p/3 to produce the new vector x. 
4The function corresponding to a coordinate is not exactly the same thing as the coordinate itself. The z 
coordinate is a ray running perpendicular to the xy plane through the coordinate origin. The function z is the 
altitude above (or below) the xy plane at every point, regardless of whether it is on the z axis; z2 is the square of 
the altitude at every point, etc. The behavior of these functions upon rotation, reflection, etc. is the same as that 
of the coordinate or product of coordinates.
450 Chapter 13 Group Theory 
The rotation operator Cf is thus represented as 
Cf =
 cosf sin f 
-sin f cosf 
 (13-1) 
In the case at hand, clockwise rotation by 2p/3 radians is a decrease of 2p/3 in f. 
Substituting f=-2p/3 for C3+ and f=+2p/3 for C3- gives us the following 2×2 
representations: 
C+3 :
.
. 
-1
2 +
v3 
2 
-
v3 
2 -1
2 
.
. 
C-3 :
.
. 
-1
2 -
v3 
2 
+
v3 
2 -1
2 
.
. 
For reflection s1, it is easy to see that x is unmoved and y goes into minus itself. 
Maintaining our dimensionality of two, this gives 
s1 :
1 0 
0 -1
 
For s2, x, and y again move into positions x and y, expressible as resultants of the 
original x and y vectors (Fig. 13-11). Thus, x=-1
2x + 
v3 
2 y,y = 
v3 
2 x + 1
2y, and s2 
has the representation 
s2 :
.
.
-1
2 
v3 
2 
v3 
2 
1
2 
.
. 
Similarly, s3 is easily shown to have the representation 
s3 :
.
. 
-1
2 -
v3 
2 
-
v3 
2 
1
2 
.
. 
The operation E does not move either x or y and is represented by the two-dimensional 
unit matrix. This completes our use of x and y, and we see that together they generate 
the two-dimensional representation 3. What should we use next? We have not yet 
generated 2, and so we know that we need to look for another basis. A basis that is 
Figure 13-11 
 Result of reflection through the s2 plane of x and y to produce new vectors x 
and y.
Section 13-9 Labels for Representations 451 
Figure 13-12 
 Effect of a reflection on the direction of f. 
frequently employed in group theory is a direction of rotation about some symmetry 
axis. In our case, the symmetry axis is the z axis, and we let Rz stand for a direction of 
rotation about this axis. (Which direction we choose, clockwise or counterclockwise, 
is arbitrary.) Now we consider how Rz is affected by the symmetry operations. C+3 
and C-3 have no effect since they merely shift the origin of the f coordinate but do not 
affect the direction in which f increases or decreases. Put another way, the direction 
of motion of the hands of a clock is not affected by rotating the clock about an axis 
perpendicular to its face. However, reflections s1,s2, and s3 will cause the direction 
to be reversed (Fig. 13-12). Therefore, 
ERz = +1Rz, s1Rz =-1Rz, s2Rz =-1Rz 
s3Rz = -1Rz C+3 Rz =+1Rz C-3 Rz =+1Rz 
and we see that Rz is a basis for 2. 
13-9 Labels for Representations 
Thus far, we have labeled our representations 1,2, etc. We will now describe rules 
for a more meaningful symbolism—one that has become standard. The rules are as 
follows: 
1. A one-dimensional representation is given the main symbol A or B. A is used if the 
representation is symmetric for rotation by 2p/n about the n-fold principal axis, B 
if it’s antisymmetric. (For C1,Cs , and Ci groups, which have no principal axis, the 
symbol is always A.) A two-dimensional representation has the main symbol E. 
Three-dimensional representations are symbolized T (or sometimes F), and fourdimensional 
representations are symbolized G. 
2. Subscripts may be applied as follows. If there are C2 axes perpendicular to the 
principal axis, then a subscript 1 (2) means that the basis for the representation is 
symmetric (antisymmetric) for such a rotation. In the absence of such C2 axes, 
vertical reflection planes are used instead, if present. If there is an inversion center, 
subscripts g (gerade) and u (ungerade) refer to the basis for the representation being 
respectively symmetric or antisymmetric for inversion. 
3. Superscripts may be applied as follows: If there is a sh plane, a single prime means 
the basis for the representation is symmetric for that reflection; a double prime 
means it is antisymmetric.
452 Chapter 13 Group Theory 
TABLE 13-13 
 Representation Table for the C3v Group 
C3v E s1 s2 s3 
A1 1 1 1 1 
A2 1 -1 -1 -1 
E 
1 0 
0 1
 1 0 
0 -1
 -1
2 v3/2 v3/2 1
2 
  -1
2 -v3/2 
-v3/2 1
2 
 
Use of these conventions enables us to write our C3v representation as shown in 
Table 13-13. (The symbolE in the left-most column of the table should not be confused 
with theE in the top row. The former labels a two-dimensional representation, the latter 
refers to the identity operation.) At the right-hand side of the table are listed, for each 
representation, the bases described above plus a fewothers. It is convenient for chemical 
applications to list all Cartesian combinations up to the second power, as has been done 
here. One can go to powers as high as one pleases (z99 is a basis for A1), but first and 
second powers are of most common practical use in chemistry. The use of parentheses 
and commas for bases for the E representation [e.g., (xz, yz)] indicate which pairs of 
functions may be selected as bases for this two-dimensional representation. 
13-10 Some Connections Between the Representation 
Table and Molecular Orbitals 
It is possible, by inspecting Table 13-13, to predict certain properties of MOs or wavefunctions 
for a molecule having C3v symmetry. Suppose that we did anMOcalculation 
on ammonia. What does this table tell us to expect? In the first place, it tells us that 
there are only three MO symmetry types possible. We might find nondegenerate MOs 
having A1 symmetry (totally symmetric), or A2 symmetry (antisymmetric for reflections). 
It is also possible for MOs to exist that form bases for E representations. Since 
such bases are intermixed by some operations, however, these MOs cannot be either 
symmetric or antisymmetric for every operation. Therefore, they must be degenerate. 
And, since the intermixing occurs only within pairs of such bases, the MOs must be 
doubly degenerate. Thus, inspection of the representation table tells us at once that nondegenerate 
and doubly degenerate MOs are possible for ammonia. Since the symmetry 
requirements apply to wavefunctions as well as to one-electron MOs, the table tells 
us also that ammonia can have states with wavefunctions whose spatial symmetries 
are A1,A2, or E. It is conventional to label MOs with lowercase symmetry symbols, 
a1, a2, e, and state functions with uppercase symbols. 
The representation table also tells us which basis AOs can appear in the various 
MOs. For instance, if we used a set of valence AOs for NH3, we would have a 2pz 
AO on nitrogen, oriented along the C3 axis of the molecule, and also 2px and 2py AOs 
perpendicular to C3. Now2px
, 2py, and 2pz transform like x,y, and z, respectively 
(if they are centered at a common point on the C3 axis); just as z is symmetric for 
all symmetry operations of the group, so is 2pz . This means that the 2pz AO can be 
expected to appear only in MOs with a1 symmetry. 2px and 2py must appear together
Section 13-11 Representations for Cyclic and Related Groups 453 
C3+ C3- 
1 1 z, x2 +y2, z2 
1 1 R2 
 -1
2 v3/2 
-v3/2 -1
2 
 -1
2 -v3/2 v3/2 -1
2 
 
(x, y)(Rx,Ry)(x2 -y2, xy)(xz, yz) 
in e-type MOs. What about 2s on nitrogen? Intuitively, we know that it, like 2pz, is 
unaffected by all the operations, and so it should appear only in a1 MOs. The table 
indicates this by listing z2 and x2 +y2 as bases for the A1 representation. Since these 
have the same symmetry, their sum retains the symmetry, and so x2 +y2 +z2 =r2 is 
also a basis forA1. Since r2 is spherically symmetric, this indicates that any spherically 
symmetric function on nitrogen is a basis for the A1 representation. 
If we were to use an expanded basis set of AOs, including 3d AOs on nitrogen, 
the group table tells us that 3dz2 would appear in the nondegenerate a1 MOs and that 
3dx2-y2 , 3dxy, 3dxz, and 3dyz would appear in degenerate e-type MOs. (Recall that 
3dz2 is really 3dz2-r2 or 3d2z2-x2-y2 .) 
What about the 1sAOs on the H’s? Can the representation table tell us in which MOs 
these will appear? It turns out that it can, and that they go into both a1- and e-type MOs, 
but we will defer showing how this can be told from the table until later in the chapter. 
Readers may begin to appreciate the usefulness of group theory in quantum chemistry 
when they consider that, simply by assigning ammonia to the C3v group and 
looking up the representation table, we are able to say that a minimum-valence basis 
set MO calculation will produce nondegenerate, totally symmetric (a1) MOs containing 
N2s,N2pz , and H1s AOs, and doubly degenerate (e) MOs containing N2px ,N2py , 
and H1s AOs. (No a2 MOs will appear because none of our AOs are a basis for that 
representation.) 
13-11 Representations for Cyclic and Related Groups 
Cyclic groups are the groups C2,C3,C4, . . . ,Cn containing only the n - 1 rotation 
operations and the identity operation E. We devote a separate section to these because 
there are some special problems connected with finding and labeling representations 
for these groups. (This section is off the mainstream of development of this chapter 
and may be skipped if desired.) 
Let us consider the operations associated with the n-fold proper axis oriented along 
the z axis and ask what will become of the function f =exp(if) as it is rotated clockwise 
about this axis by 2p/n radians. (We entertain the idea that exp(if) might be a convenient 
basis for a representation since such functions were found to be eigenfunctions for 
the particle-in-a-ring problem in Chapter 2.) Since the clockwise direction is opposite 
to the normal direction of the f coordinate, the effect of the rotation is to put f (f) where 
f (f -2p/n) used to be. To see how the function after rotation compares to that before 
rotation, we must compare exp(if) with exp[i(f -2p/n)]. That is, the representation 
Rf , such that C+n exp(if) = Rf exp(..f), is given by exp(if)/ exp[i(f - 2p/n)], or
454 Chapter 13 Group Theory 
TABLE 13-14a 
 Partial Representation for the C3 Group 
C3 E C3 C2
3 
A 1 1 1 z,Rz 
1 exp(2pi/3) exp(4pi/3) exp(if) 
TABLE 13-14b 
 Partial Representation for the C3 Group 
C3 E C3 C2
3  =exp(2pi/3) 
A 1 1 1 z,Rz 
1  * exp(if) 
exp(2pi/n), which equals cos(2p/n)+i sin(2p/n). For various fractions of a cycle 
(i.e., various n), Rf takes on different values: 
n :1 2 3 4 5 6 7 8 
Rf : 1 -1 exp(2pi/3) i exp(2pi/5) exp(pi/3) exp(2pi/7) exp(pi/4) 
The important point here is that f =exp(if) is a basis for a one-dimensional representation 
for Cn since a rotation turns f into a constant times f and not into some other 
function. For the C3 group, then, we could write a partial representation table as shown 
in Table 13-14a. Now exp(4pi/3) is equal to exp(-2pi/3), and exp(-2pi/3) is the 
complex conjugate of exp(2pi/3). If we let  =exp(2pi/3), we can write the table as 
shown in Table 13-14b. If f =exp(if) is a satisfactory basis, f * =exp(-if) is also 
acceptable, since it is linearly independent of f . If one calculates the representations 
for n=1, 2, 3, . . . as before, one finds that the numbers Rf * are the complex conjugates 
of Rf . Therefore, we can immediately expand our table as shown in Table 13-14c. 
Since the existence of exp(if) as a basis for a one-dimensional representation always 
implies that exp(-if) exists as a basis, these sorts of one-dimensional representations 
always occur in pairs. It is conventional to combine these with braces and refer to 
them with the symbol E, which we claimed earlier is reserved for two-dimensional 
representations. Note, however, that a pair of one-dimensional representations is not 
the same as a two-dimensional representation, and we must broaden our definition of 
the symbol E to include both types of situation. 
Our representation table for the C3 group now looks almost the way one would find 
it in a standard tabulation. However, instead of listing exp(if) and exp(-if) as bases, 
TABLE 13-14c 
 Representation for the C3 Group 
C3 E C3 C3
2  =exp(2pi/3) 
A 1 1 1 z,Rz 
E 
1  * exp(if) 
1 *  exp(-if)
Section 13-11 Representations for Cyclic and Related Groups 455 
TABLE 13-15 
 Representation for the C3 Group 
C3 E C3 C2
3 
A 1 1 1 z,Rz 
E 
1 0 
0 1
 -1
2 
v3 
2 
-
v3 
2 
1
2
 -1
2 -
v3 
v 2 3 
2 -1
2 
 
(x, y) 
the convention is to list x and y. In relating Cartesian to spherical polar coordinates, 
x = r sin . cosf and y = r sin . sin f. When we are concerned only with changes in 
f,x goes as cos f,y as sin f, and, since sin f and cosf are expressible as linear combinations 
of exp(if) and exp(-if), it follows that x and y are equivalent to exp(±if) 
as bases. If we had started out with x and y as bases, we would have found that, for 
rotations, these are intermixed, leading to a truly two-dimensional representation. In 
fact, we worked out the effects of C3 and C2
3 on x and y earlier for the C3v group. There 
we found that C3 and C2
3 (= C-3 ) were represented by the two-dimensional matrices 
shown inTable 13-15. But thisE representation is reducible to the two one-dimensional 
representations through the unitary matrix 
.= 
1 
v2 
1 i 
i 1
 
The resulting block-diagonalized representation is 
E C3 C2
3 
1 0 
0 1
 -1
2 - iv3 
2 0 
0 -1
2 + iv3 
2 
 -1
2 + iv3 
2 0 
0 -1
2 - iv3 
2 
 
which can be separated into two one-dimensional representations: 
E C3 C2
3 
1 -1
2 - iv3 
2 -1
2 + iv3 
2 
1 -1
2 + iv3 
2 -1
2 - iv3 
2 
Since  =exp(2pi/3)=cos(2p/3)+i sin(2p/3)=-1
2 + iv3 
2 , we recognize that this 
pair of one-dimensional representations is the same as the pair we found earlier. Furthermore, 
we note that . is precisely the transformation that turns x (i.e., cosf) and y 
(i.e., sin f) back into exp(±if) (to within a constant multiplier): 
.
x
y

= 
1 
v2 
1 i 
i 1
cosf 
sin f

= 
1 
v2 
cosf +i sin f 
i cosf +sin f

=
 (1/v2) exp(if) 
(-i/v2) exp(-if)
 
Thus, (x, y) produce a reducible, two-dimensional representation equivalent to the 
irreducible representation given by exp(±if). The reason for listing (x, y) as bases
456 Chapter 13 Group Theory 
TABLE 13-16 
 Irreducible Representations for the C3 Group 
C3 E C3 C2
3  = exp(2pi/3) 
A 1 1 1 z,Rz z2, x2 +y2, x2 -y2, xy 
E 
1  * (x, y) 
1 *  (Rx,Ry) (yz, xz) 
is simply that most applications of the table are made to real functions (e.g., chemists 
usually prefer to work with 2px and 2py AOs rather than with 2p+1 and 2p-1). Our 
final form for the table, then, is that shown in Table 13-16, where additional bases have 
been listed. 
The preceding discussion has been within the context of the Cn groups. One might 
ask what sort of symmetry operation is needed to make it impossible for exp(if) to 
be a basis for a one-dimensional representation. The answer is, any operation that 
causes a reversal in the direction of the coordinate f. For then, exp(if).exp(-if), 
and our basis function has turned into another independent function rather than into a 
constant times itself. Therefore, the presence of any symmetry operation that reverses 
the direction of motion of the hands of a clock will suffice to prevent representations of 
the , * sort. Operations that reverse clock direction are sd,sv, and C2
(perpendicular 
to the principal axis). Clock direction is unaffected by sh, i, and Sn. Therefore, we 
can expect , * types of representations to occur in groups of types Cn,Cnh,Sn, all of 
which have a Cn axis but no sd,sv, or C2
elements. 
13-12 Orthogonality in Irreducible Inequivalent 
Representations 
We come now to a very important point regarding representations. We will illustrate 
our arguments with the representation table (Table 13-13) for the C3v group. Notice 
the following features of that table: 
1. If we choose the A1 representation, square all the numbers, and sum over all six 
symmetry operations, we get 6 as a result. 
2. If we do the same thing with the A2 representation, we get the same result. 
3. If we do the same thing for the upper left-hand elements (the 1, 1 elements) of the 
E representation, we get 3 as a result. 
4. If we do the same thing for each of the other positions in the E representation, the 
result is 3 each time. 
In general, the result of this procedure for any irreducible representation in any 
group will be the order of the group divided by the dimension of the representation. That 
is, if h is the order of the group, li is the dimension of representation i , and 
(j,k) 
i (R)
Section 13-12 Orthogonality in Irreducible Inequivalent Representations 457 
is the number in the (j, k) position of the matrix representing symmetry operation R, 
then the mathematical formula that corresponds to our general statement is 

R 


(j,k) 
i (R) 
2 
= 
h
li 
(13-2) 
Note that the absolute square is used to accommodate the , * type representations of 
cyclic groups. 
We can conceive of the set of six numbers for A1 as being a vector of six elements. 
The six numbers of A2 constitute a second vector, and E provides four more sixdimensional 
vectors. If each such vector is multiplied by vli/h, then each vector is 
normalized. 
Now let us examine these vectors regarding their orthogonality. If we take the scalar 
product of the unnormalized vectors A1 and A2, the result is zero: 

1 1 1 1 1 1 
 A1 
.......... 
......... 
1 
-1 
-1 
-1
1
1
.......... 
.........
A2 
=0 
The reader may quickly verify that the scalar product of any two different vectors from 
among the set of six in the C3v representation table is zero. Thus these six vectors are 
orthogonal. Once again, this result always holds between “representation vectors” in 
irreducible inequivalent representations for any group. Combining this orthogonality 
property with the normality property mentioned earlier, we have 

R 
li/h(k,l) 
i (R)* li/h(m,n) 
j (R)=di,j dk,mdl,n (13-3) 
where i and j are understood to be irreducible and, if i 
= j , inequivalent. This 
relation, sometimes called “the great orthogonality theorem,” is of central importance 
in group theory. Its essence is captured by the statement that “irreducible inequivalent 
representations are comprised of orthogonal vectors.” We do not prove the theorem in 
this book,5 but we do make considerable use of Eq. (13-3). 
One immediate result of the relation is that it enables us to tell when we have 
completed the task of finding all the inequivalent irreducible representations of a group. 
If we consider the C3v group, for example, we note that it is of order six, since there 
are six symmetry operations. This means that each “representation vector” will have 
six elements, i.e., is a vector in six-dimensional space. The maximum number of 
orthogonal vectors we can have in six-dimensional space is six. Therefore, the number 
of representation vectors cannot exceed the order of the group. Furthermore, since 
the number of vectors provided by an n-dimensional representation is n2 (e.g., E is 
two-dimensional and gives four vectors), we can state that the sum of the squares of the 
5See Bishop [1] or Eyring et al. [2, Appendix VI].
458 Chapter 13 Group Theory 
dimensions of the inequivalent irreducible representations of a group cannot exceed 
the order of the group. 
In fact, it can be proved5 that this sum of squares of dimensions must equal the order 
of the group when all such representations are included. That is, 
all inequivalent 
irreproducible representations 

i 
l2
i =h (13-4) 
Thus, the fact that the squares of the dimensions of the A1,A2, and E representations 
for the C3v group add up to six, which is the order of the group, indicates that no more 
irreducible representations exist (except those that are equivalent to those we already 
have). 
EXAMPLE 13-6 Count up all the symmetry operations you can think of for (planar) 
BH3. Can you guess from this how many irreducible representations exist for this 
molecule and what their dimensions are? 
SOLUTION 
 The operations are: E, C+3
,C-3
,s1,s2,s3,sh,S+3
,S-3
,C2,C2,C2 or 12 operations, 
so the group order is 12. There must be at least one irreducible representation of order one, so 
we cannot have three two-dimensional representations. We could have: 12 one-dimensional representations; 
eight one-dimensional and one two-dimensional representations; four one-dimensional 
and two two-dimensional representations. From what has been presented so far, all three of these 
are possible.  
13-13 Characters and Character Tables 
Thus far, we have defined representations and shown how they may be generated from 
basis functions. We have distinguished between reducible and irreducible representations 
and have indicated that there is an unlimited number of equivalent representations 
corresponding to any given two- or higher-dimensional representation. An example 
of a pair of equivalent, reducible, two-dimensional representations, derived in Section 
13-11, is given in Table 13-17. Equivalent representations are related through 
unitary transformations, which are a special kind of similarity transformation (see Chapter 
9), and two matrices that differ only by a similarity transformation have the same 
TABLE 13-17 
 Equivalent Representations for the C3 Group 
C3 E C3 C2
3 
x,y 
1 0 
0 1
 -1
2 -v3/2 v3/2 -1
2
 -1
2 v3/2 
-v3/2 -1
2 
 
exp(±f) 
1 0 
0 1
 -1
2 -iv3/2 0 
0 -1
2 +iv3/2
 -1
2 +iv3/2 0 
0 -1
2 -iv3/2

Section 13-13 Characters and Character Tables 459 
TABLE 13-18 
 Characters for the C3v Group 
c3v E s1 s2 s3 C3+ C3- 
A1 1 1 1 1 1 1 z, x2 +y2, z2 
A2 1 -1 -1 -1 1 1 Rz 
E 2 0 0 0 -1 -1 (x, y)(Rx,Ry) 
(x2 -y2, xy)(xz, yz) 
trace, or character (Problem 9-7), which is defined as the sum of the diagonal elements 
of a matrix. The matrices in Table 13-17 exemplify this fact, their characters 
being respectively 2,-1,-1 for E,C3, and C2
3 in both representations. The generally 
accepted symbol for the character of operation R in representation i is .i(R), and the 
mathematical definition is 
.i(R)=
j 

j,j 
i (R) (13-5) 
In practice, it is the characters of irreducible representations that are used in most 
chemical applications of group theory. This means that one needs only the character 
table for a group, rather than the whole representation table. For the C3v group, the 
character table is displayed in Table 13-18. Comparison with Table 13-13 will make 
clear that the character is merely the sum of diagonal elements. (For one-dimensional 
representations, the character and the representation are identical.) Since the representation 
for the identity operation is always a unit matrix, the character for this operation 
is always the same as the dimension of the representation. Hence, the first character in 
a row tells us the dimension of the corresponding representation. 
An immediate consequence of the orthogonality theorem for representations is that 
the vectors resulting from characters are orthogonal too. This is trivially obvious for 
characters of one-dimensional representations. For multidimensional representations, 
the character vector is simply the sum of the representation vectors in diagonal positions. 
If a given vector (say, the A1 vector) is orthogonal to each of these (say, E11 and E22), 
then it is orthogonal to their sum; that is, in terms of the vector notation of Chapter 9, 
if ˜a1e11 =0 and ˜a1e22 =0, then ˜a1(e11+e22)=0. 
Inspection of Table 13-18 reveals a curious thing. For any given row of characters, 
all operations in the same class have the same character. There is a fairly simple reason 
for this. We have indicated already that operations in the same class are operations 
that can be interchanged merely by group reflections, rotations, etc., of the symmetry 
elements in the group, but we have seen that such changes are mathematically effected 
through similarity transformations. This means that representations for operations in 
the same class are interchangeable via similarity transformations. That is, the matrix 
representing, say, s1 (in the E representation of C3v) can be made equal to the matrix 
representing s2 through a similarity transformation: 
T-1s1T=s2 (13-6) 
(From our group Table 13-7, we can ascertain that T must be the matrix representing 
s3.) Now, the two sides of Eq. (13-6) must have the same character since they 
are identical 2×2 matrices, but the left-hand side must have the same character as s1
460 Chapter 13 Group Theory 
TABLE 13-19 
 The Standard Short-Form Character Table for the C3v Group 
C3v E 3s 2C3 
A1 1 1 1 z, x2 +y2, z2 
A2 1 -1 1 Rz 
E 2 0 -1 (x, y)(Rx,Ry)(x2 -y2, xy)(xz, yz) 
since character is unchanged by a similarity transformation. Therefore, s1 and s2 have 
the same character. 
We can take advantage of the above rule to write our character table in abbreviated 
form, illustrated for theC3v group inTable 13-19. This is the standard form for character 
tables. A collection of such tables appears in Appendix 11. 
That characters must be equal in the same class is a restriction on our character 
vectors. In Table 13-19 it is made evident that, in the C3v group, our vectors really only 
have three independent variables each, one for each class. These are properly thought 
of, then, as vectors in three-dimensional space (with weighting factors 1, 3, and 2 for 
E,s, and C3, respectively). There can be no more than three such vectors that are 
orthogonal, and so we are left with the result that the number of inequivalent irreducible 
representations in a group cannot exceed (and is in fact equal to6) the number of classes 
in the group. This result, together with the fact that the sum of squares of dimensions of 
inequivalent irreducible representations must equal the order of the group, often suffices 
to tell us in advance how many representations there are and what their dimensions 
are. For our C3v group, the order is six and there are three classes. Hence, we know 
that there are three representations and that the squares of their dimensions sum to six. 
The problem reduces to: “What three positive integers squared, sum to six?” There is 
only one answer: 1, 1, and 2. The fact that there will always be a totally symmetric 
one-dimensional representation also helps pin down the possibilities. For example, can 
one have a group of order eight and only two classes? There is no way this can happen. 
Two classes would mean two representations. If both were E type, their dimensions 
squared would indeed sum to eight. But one of them must be one-dimensional, and 
there is no way the other can square to seven. 
EXAMPLE 13-7 How many classes of symmetry operation can you count for (planar) 
BH3? Can you use this to select one of the possibilities from Example 13-12? 
SOLUTION 
 The classes are E, sv, C3, C2, sh, S3. That’s six classes, so there are only six irreducible 
representations. There are four one-dimensional and two two-dimensional representations. 
 
Our collected list of conditions that the characters in a completed table must satisfy 
is as follows: 
1. There must be a one-dimensional representation having all characters equal to +1. 
2. The leading character in each row (i.e., the character for operation E) must equal 
the dimension of the representation. 
6See Bishop [1].
Section 13-13 Characters and Character Tables 461 
3. The sum of the squares of the leading characters must equal the order of the group. 
4. The number of rows in the character table must equal the number of classes in 
the group. 
5. The absolute squares of the characters in a given row (times the weighting factor for 
each class if the abbreviated form is used) equals the order of the group. (Character 
vectors are normalized) This results directly from Eq. (13-2). 
6. The character vectors are orthogonal (again, using weighting factors, if appropriate). 
This is a fairly large number of restrictions, and may suffice to allow one to produce 
the character table for a group without ever actually producing representations. For 
example, consider the C4v group, which has the operations E, 2C4,C2, 2sv, and 2sd. 
The group thus has order eight and five classes. There must be five representations, and 
the only way their dimensions can square to eight is if four of them are one-dimensional 
and one is two-dimensional. This already enables us to write Table 13-20. It is not 
TABLE 13-20 
 Partial C4v Character Table 
C4v E 2C4 C2 2sv 2sd 
A1 1 1 1 1 1 
1 1 
2 1 
3 1 
E 2 
difficult to find a way to make 1 orthogonal to A1. We simply place -1 in some 
places to produce four products of -1 and four of +1. Three possibilities are shown in 
Table 13-21. These are orthogonal not only to A1, but to each other as well, and so we 
TABLE 13-21 
 Partial C4v Character Table 
E 2C4 C2 2sv 2sd 
1 1 1 -1 -1 
1 -1 1 1 -1 
1 -1 1 -1 1 
have found the characters for 1,2, and 3. The characters for the E representation 
must have squares that sum to eight and also be orthogonal to all four one-dimensional 
representation vectors. One possibility is fairly obvious. Since the characters for 
operations E and C2 are +1 in all the one-dimensional representations, we could take 
the characters for the E representation to be 
E 2C4 C2 2sv 2sd 
2 0 -2 0 0
462 Chapter 13 Group Theory 
Other cases that meet the normality condition are 
2 ±1 0 ±1 0 
2 0 0 ±1 ±1 
2 ±1 0 0 ±1 
But none of these is orthogonal to all the one-dimensional sets. Therefore, the complete 
character table (except for the basis functions) for theC4v group is showninTable 13-22, 
where the symbols A2,B1,B2 are consistent with symmetry or antisymmetry for C4 
and sv as described in Section 13-9. 
TABLE 13-22 
 Completed C4v Character Table 
C4v E 2C4 C2 2sv 2sd 
A1 1 1 1 1 1 
A2 1 1 1 -1 -1 
B1 1 -1 1 1 -1 
B2 1 -1 1 -1 1 
E 2 0 -2 0 0 
13-14 Using Characters to Resolve Reducible 
Representations 
It was pointed out earlier that several irreducible representations can be combined into 
a larger-dimensional reducible representation. Our example was 
.
.. 
2 0 0 
0 1 0 
0 0 3
.
..
=
.
.. 
A2 0 0 
0 A1 0 
0 0 E
.
..
= 
which is a four-dimensional representation (since 3 is two-dimensional). A matrix 
built up in this way is symbolized A2 .A1 .E. It is easy to see that the characters 
of the reducible representation  are simply the sums of characters for the individual 
irreducible component representations (since the diagonal elements of A2,A1, and E 
all lie on the diagonal of ). Thus, the characters of  are 
E 3s 2C3 
 : 4 0 1 
Furthermore, no matter how  is disguised by a similarity transformation, its character 
vector is unchanged. Now, suppose you were given the representation , disguised 
through some similarity transformation so as to be nonblock diagonal and asked to tell 
which irreducible representations were present. How could you do it? One way would 
be to find the similarity transformation that would return the representation to block
Section 13-15 Identifying Molecular Orbital Symmetries 463 
diagonal form. But there is a much simplerway. We can test to see if the character vector 
of  is orthogonal to the character vectors of each of our irreducible representations. 
For instance, if  contained only A2 and E, its character vector would be orthogonal 
to that of A1 because the character vectors of A2 and E are orthogonal to that of A1. 
If A1 is present in , then  and A1 character vectors are not orthogonal. Indeed, the 
amount of A1,A2, or E present can be found by making use of the character vector 
normality relation. This is illustrated for  by the following equations: 
A1 : 
1
6
(1 · 4+3 · 1 · 0+2 · 1 · 1)=1 
A2 : 
1
6
(1 · 4+3·-1 · 0+2 · 1 · 1)=1 
E : 
1
6
(2 · 4+3 · 0 · 0+2·-1 · 1)=1 
In general, for  =c11 .c22.···.cii .+··· 
ci =(1/h)
R 
.i(R).(R) (13-7) 
where h is the order of the group. This technique for resolving a reducible representation 
into its component irreducible representations is very useful in quantum chemistry, as 
we shall see shortly. 
13-15 Identifying Molecular Orbital Symmetries 
We stated earlier that any MO must be a basis for an irreducible representation. Given 
a set of computed MOs, how does one decide which representation each MO is a 
basis for? One does this by comparing the MOs with the characters in the character 
table. Some examples will make this clear. In Table 13-23 are extended H¨uckel data 
for NH3, oriented as shown in Fig. 13-13. NH3 belongs to the C3v group, having 
the character table shown in Tables 13-18 and 13-19. The minimal valence basis set 
of seven AOs leads to seven MOs. Notice that MOs 5 and 6 and also 2 and 3 are 
degenerate. Therefore, these MOs must be bases for two-dimensional representations, 
and are assigned the symbol e (lower case for MOs). The other MOs are all given the 
main symbol a. There are two possibilities for these MOs—a1 or a2. These differ in 
their characters for reflection, a1 being symmetric, and a2 antisymmetric. If we look 
at the eigenvectors for MOs 1, 4, and 7, we see that they contain the 1s AOs on each 
hydrogen with equal sign and magnitudes. Since reflection always interchanges two 
hydrogens, these MOs are clearly all symmetric for reflection, and so we label them 
all a1. Our result, then, is 
MO: 1 2 3 4 5 6 7 
Symmetry : a1 e e a1
e e a1 
Next, consider staggered ethane, which we have earlier assigned to the D3d point 
group. Orbital energy levels and sketches of the MOs appear in Fig. 13-14. The 
character table for the D3d group is given in Table 13-24. 
As before, we observe that certain of the orbitals have the same energies, so we assign 
such doubly-degenerate MOs the main symbol e. These MOs are either symmetric or
464 Chapter 13 Group Theory 
TABLE 13-23 
 Extended H¨uckel Data for NH3 
MO no. MO energy (a.u.) MO occupancy 
1 0.7494 0 
2 0.1279 0 
3 0.1279 0 
4 -0.4964 2 
5 -0.5955 2 
6 -0.5955 2 
7 -1.0178 2 
Eigenvectors 
AO\MO 1 2 3 4 5 6 7 
N(2s) 1.2946 0.0000 0.0000 -0.1715 0.0000 0.0000 0.7387 
N(2pz ) -0.4369 0.0000 0.0000 -0.9628 0.0000 0.0000 0.0214 
N(2px) 0.0000 1.0275 0.0000 0.0000 0.6498 0.0000 0.0000 
N(2py) 0.0000 0.0000 -1.0275 0.0000 0.0000 0.6498 0.0000 
H1(1s) -0.7166 -1.0399 0.0000 0.0656 0.4422 0.0000 0.1561 
H2(1s) -0.7166 0.5200 0.9006 0.0656 -0.2211 0.3829 0.1561 
H3(1s) -0.7166 0.5200 -0.9006 0.0656 -0.2211 -0.3829 0.1561 
antisymmetric for inversion and are accordingly subscripted g or u, respectively. All of 
the nondegenerate MOs must have the main symbol a, since b does not appear in the 
D3d table. The character table indicates that a1 and a2 differ in that they are respectively 
symmetric and antisymmetric for two-fold rotations that switch the molecule end for 
end. The u, g subscripts again refer to inversion. Inspection of the figures enables 
us to decide which symbols are appropriate in each case. The resulting symmetry 
designations are included in Fig. 13-14. 
Figure 13-13 
 Orientation of ammonia with respect to Cartesian axes.
Section 13-16 Determining in Which Molecular Orbital an Atomic Orbital Will Appear 465 
Figure 13-14 
 Valence MOs for staggered ethane. The energy level spacings have been altered 
for convenience. The hatched areas have a negative sign. 
TABLE 13-24 
 Characters for the D3d Point Group 
D3d E 2C3 3C2 i 2S6 3sd 
A1g 1 1 1 1 1 1 
A2g 1 1 -1 1 1 -1 
Eg 2 -1 0 2 -1 0 
A1u 1 1 1 -1 -1 -1 
A2u 1 1 -1 -1 -1 1 
Eu 2 -1 0 -2 1 0 
13-16 Determining in Which Molecular Orbital an Atomic 
Orbital Will Appear 
Earlier, we noted that the 2pz AO in ammonia will contribute to a1 MOs because 
2pz transforms like z, and z is listed as a basis for the a1 representation. If the 
basis functions were not listed, we could have reached the same conclusion simply by
466 Chapter 13 Group Theory 
Figure 13-15 
 
taking 2pz and putting it through all the symmetry operations to produce a representation: 
E2pz = +1 · 2pz 
si2pz = +1 · 2pz, i=1, 2, 3 
C±3 2pz = +1 · 2pz 
The representation contains only +1, so 2pz obviously “has” a1 symmetry. 
When we come to the 1s AOs on hydrogens in NH3, we cannot use the list of 
basis functions. It includes only coordinates originating on the principal axis, and the 
hydrogens are not on that axis. Here we must generate a representation. Let us try 
to do this by putting one of the hydrogen atoms, H1, through the various symmetry 
operations (also see Fig. 13-15): 
EH1=+1H1 
s1H1=+1H1 
s2H1 = H3 
s3H1 = H2 
C+3 H1 = H2 
C-3 H1 = H3 
Since some of these operations interchange H1 with H2 or H3, we are not achieving 
a one-dimensional representation. We must take all three functions together and work 
out a three-dimensional representation. Thus, 
E
.
.. 
H1 
H2 
H3
.
..=
.
.. 
H1 
H2 
H3
.
..
, E:
.
.. 
1 0 0 
0 1 0 
0 0 1
.
..
, .(E)=3 
s1
.
.. 
H1 
H2 
H3
.
..
=
.
.. 
H1 
H3 
H2
.
..
, s1 :
.
.. 
1 0 0 
0 0 1 
0 1 0
.
..
, .(s1
)=1 
C+3 
.
.. 
H1 
H2 
H3
.
..
=
.
.. 
H2 
H3 
H1
.
..
, C+3 :
.
.. 
0 1 0 
0 0 1 
1 0 0
.
..
, .(C+3 )=0
Section 13-17 Generating Symmetry Orbitals 467 
and similarly for s2,s3, and C-3 . But we are only going to use the characters ., and 
we know that s2 and s3 must have the same character as s1 (i.e., 1) and C-3 must have 
the same character as C+3 (i.e., 0). Our character table for the representation resulting 
from the basis of three hydrogen 1s AOs then, is 
C3v E 3s 2C3 
3H 3 1 0 (H1,H2,H3) 
Notice that a “one” on the diagonal of a 3 × 3 representation matrix has the effect 
of keeping a hydrogen 1s AO in place. Thus, the E operation keeps all three H’s 
unmoved, has three “ones” on the diagonal, and has a character of 3. The s operations 
each leave but one hydrogen unmoved, have a single “one” on the diagonal, and have a 
character of 1. C3 leaves no hydrogen unmoved and has a character of zero. In general, 
the characters of such representations are the numbers of functions not moved by the 
various operations. Thus, we could have written down the above characters for 3H 
without figuring out the representation matrices. 
It is evident that 3H must be reducible, since it is of higher dimension than anything 
in the C3v character table. To resolve 3H into its irreducible components, we use the 
relation (13-7): 
A1 : 
1
6
(1 · 3 · 1+3 · 1 · 1+2 · 0 · 1)=1 
A2 : 
1
6
(1 · 3 · 1+3 · 1·-1+2 · 0 · 1)=0 
E : 
1
6
(1 · 3 · 2+3 · 1 · 0+2 · 0·-1)=1 
Therefore, 3H =A1 .E. We conclude from this that hydrogen 1s AOs can appear in 
MOs having a1 or e symmetry. Table 13-23 indicates that this is correct. 
EXAMPLE 13-8 Given only the MO sketches of Fig. 13-14 (and no energies), is 
there an easy way to select the MOs that are degenerate? 
SOLUTION 
 Nondegenerate MOs must be either symmetric or antisymmetric for every symmetry 
operation of the molecule. Failure to obey this criterion indicates a degenerate MO. For 
instance, the two highest energy MOs in the figure are symmetric or antisymmetric for every 
reflection, rotation, improper rotation, or inversion, hence are nondegenerate. The next two MOs, 
however, are unsymmetrical for some of the operations. Most easily seen is their unsymmetrical 
behavior for rotation by 120. about the C–C axis. These MOs must be degenerate.  
13-17 Generating Symmetry Orbitals 
Consider the lowest-energy extended H¨uckel MO for NH3. It is 
f7 =0.7387N2s +0.0214N2pz +0.1561 1s1 +0.1561 1s2 +0.1561 1s3 
We see that the hydrogen AOs all have the same sign and magnitude in this MO. 
In fact, we noted above that this must happen in all a1 MOs of NH3 for reasons of
468 Chapter 13 Group Theory 
symmetry. If an MO is to be symmetric for all the reflection and rotation operations 
of the C3v group, there is no other combination that will be adequate. When we 
have several equivalent atoms, and symmetry forces their AOs to appear in MOs in 
certain combinations, we refer to those combinations as symmetry orbitals. Thus, 
fH
a1 = N(1s1 + 1s2 + 1s3) is the a1 symmetry orbital for the hydrogens in ammonia 
(N is a normalizing constant). 
One can use the character table to generate symmetry orbitals. This is done by 
picking any one of theAOs, say 1s1, and operating on it with each symmetry operation 
times the character for each operation in the representation of interest. The sum of 
all these operations is an unnormalized symmetry orbital. Thus, for the a1 representation 
in NH3, 
fH
a1 = E · 1 · 1s1 +s1 · 1 · 1s1 +s2 · 1 · 1s1 +s3 · 1 · 1s1 +C+3 · 1 · 1s1 +C-3 · 1 · 1s1 
= 1s1 + 1s1 + 1s3 + 1s2 + 1s2 + 1s3 
= 2(1s1 +1s2 +1s3) 
If we normalize (ignoring overlap between 1s AOs on different centers), we 
obtain 
fa
H1 
= 
1 
v3
(1s1 +1s2 +1s3) 
To generate symmetry orbitals of e symmetry is a little more involved. First we pick 
a 1s AO and do just as before, using now the characters for e rather than a1: 
fe
H =E · 2 · 1s1 +0 · all reflections +(-1) ·C+3 · 1s1 +(-1) ·C-3 · 1s1 
=21s1 -1s2 -1s3 
Normalization yields 
fe
H = 
1 
v6
(2 1s1 -1s2 -1s3) 
Because e symmetry is manifested by doubly-degenerate MOs, we need to find a mate 
for fH
e . We can try to do this by repeating the above procedure except operating on 
1s2 instead of 1s1. This yields 
.H
e = 
1 
v6
(2 1s2 -1s1 -1s3) 
But this function is not orthogonal to fH
e . (The overlap is-1
2 .)To achieve orthogonality, 
we resort to Schmidt orthogonalization (Chapter 6): 
.H 
e =.H
e -SfH
e 
Upon expansion and renormalization, this gives 
.H 
e = 
1 
v2
(1s2 -1s3)
Section 13-17 Generating Symmetry Orbitals 469 
The data in Table 13-23 show that the e-type MOs do contain the hydrogen 1s AOs in 
just the manner prescribed by symmetry. For example, MO 5 has hydrogen 1s coefficients 
0.4422, -0.2211, -0.2211, a combination similar to that in fH
e . The degenerate 
mate, MO 6, has hydrogen coefficients of 0.0000, 0.3829, -0.3829, similar to .H 
e . 
Chemists who have had some experience in these matters are likely to prefer thinking 
in terms of symmetry orbitals. Thus, in thinking of ammonia, they are likely to take 
as a minimal valence basis set the functions: N2s,N2px,N2py,N2pz , 1 v3 
(1s1 + 1s2 + 
1s3), 1 v6 
(21s1 -1s2 -1s3), 1 v2 
(1s2 -1s3). (We continue to ignore overlap between 
1s AOs.) Since they know that it is not possible for bases of different symmetries 
to mix (it would produce an MO of mixed symmetry), they know at once that they 
can have a1 MOs from mixtures of N2s,N2pz and (1/v3)(1s1 +1s2 +1s3) and e-type 
MOs from the remaining functions. In effect, they have used symmetry to partition 
their functions into two subsets that do not interact with each other. This means that, 
in the MO calculation, the hamiltonian matrix will have no mixing elements between 
members of different subsets. This is indicated schematically in Fig. 13-16. They also 
know in advance that there will be three MOs of a1 symmetry (since only three basis 
functions have that symmetry) and four of e symmetry (two degenerate pairs). It is 
interesting to see how strongly symmetry controls the nature of NH3 MOs. 
Because symmetry orbitals depend on symmetry and not on finer details of molecular 
structure, they recur again and again in molecules of similar symmetry. For instance, 
the a1 and e combinations of hydrogen coefficients discussed above for ammonia will be 
found for hydrogenAOs in staggered or eclipsed ethane (see the drawings in Fig. 13-14, 
for instance) and for carbonAOs in MOs for cyclopropenyl (Chapter 8). Because symmetry 
requirements transcend the differences between various approximate methods 
Figure 13-16 
 Partitioned hamiltonian matrix.
470 Chapter 13 Group Theory 
for solving the Schr¨odinger equation, we expect these symmetry patterns to appear 
in MOs for, say, ammonia, at extended H¨uckel, CNDO, INDO, MINDO, or ab initio 
levels of computation. 
13-18 Hybrid Orbitals and Localized Orbitals 
The concept of a hybridized orbital is often encountered in the literature, especially 
in introductory discussions of bonding theory. While this concept is not essential to 
MO theory, it is used enough to justify a brief discussion. 
In the previous section, we showed how one could transform from a basis set of 
STOs to a basis set of symmetry orbitals. Since these two sets are related through 
a unitary transformation, they are equivalent and must lead to the same MOs when 
we do a linear variation calculation. However, there are an infinite number of unitary 
transformations available, and so the set of symmetry orbitals is only one of an infinite 
number of possible equivalent bases. Of course, this set has the unique advantage of 
being a set of bases for representations of the symmetry group, which makes it easy 
to work with. Another set of equivalent basis functions are the hybrid orbitals. These 
have the distinction of being the functions that are concentrated along the directions of 
bonds in the system. Consider, for example, methane, which was discussed in detail in 
Chapter 10. The minimal basis set of valence STOs on carbon can be transformed to 
form four tetrahedrally directed hybrids: 
f1 = 
1
2
(s+px -py +pz), f2 = 
1
2
(s-px +py +pz) 
f3 = 
1
2
(s-px -py -pz), f4 = 
1
2
(s+px +py -pz) 
One of these hybrids, f4, is shown in Fig. 13-17 and can be seen to point toward one 
of the hydrogen atoms. Because the square of each hybrid consists of one part s AO 
to three parts p AO, these are called sp3 hybrids (pronounced s-p-three). The reason 
for focusing on sp3 hybrids in this case is that they have physical appeal since they 
point along the C–H bonds and therefore seem to have a more natural relation to the 
electron-pair bond approach of G. N. Lewis. This is deceptive, however, because the 
set of four carbon sp3 orbitals is completely equivalent to the set of four carbon STOs. 
The sum of squares of the hybrids is spherically symmetric just as is the sum of squares 
of STOs. Thus, even though each hybrid is directed toward a hydrogen, the electron 
density due to all four occupied hybrids is spherically symmetric. Furthermore, after 
the linear variation is performed, the MOs that are produced contain mixtures of hybrids 
to give us the exact same delocalized MOs produced from STOs. No single MO 
consists of just one hybrid and one hydrogen 1s AO, and therefore no single MO can 
be identified with one C–H bond. We conclude then that hybrid orbitals are one of an 
infinite number of choices of basis, that they have an appealing appearance because of 
their concentration in bond regions, but that no concentration of charge in the molecule 
results as a consequence of using hybrids rather than STOs. (Some concentration of 
charge in the bonds does result from overlap between basis functions on carbon and 
those on hydrogens, but this occurs to exactly the same degree for the various equivalent 
basis sets.)
Section 13-18 Hybrid Orbitals and Localized Orbitals 471 
Figure 13-17 
 A sketch of the hybrid f4 = 1
2 (s+px +py -pz ). The direction of the hybrid is 
coincident with that of a vector having the same x, y, z dependence as f4. The other three hybrids 
are identical except that they point toward the other three H atoms. (The coordinate system here is 
rotated with respect to that used in Chapter 10.) 
Another example is planar CH+3
(D3h). As before, we can mix our minimal valence 
STOs on carbon to produce hybrids pointing toward the hydrogens. If one hydrogen is 
on the +y axis, the hybrids are 
.1 = 
1 
v3
s+ 
v2 
v3
py, 
.2 = 
1 
v3
s+ 
1 
v2
px - 
1 
v6
py, 
.3 = 
1 
v3
s- 
1 
v2
px - 
1 
v6
py 
Each of these hybrids, when squared, is one part s to two parts p and is called an 
sp2 hybrid. The coefficients for the p AOs are determined from simple vector considerations: 
One merely calculates how x and y vectors must be combined to produce 
resultant vectors pointing toward the corners of an equilateral triangle. The resulting 
hybridized basis set forCH+3
is.1,.2,.3, plus the 2pz STOon carbon and a 1s STOon 
each hydrogen. As before, the sum of the squares of .1,.2,.3, and 2pz is spherically 
symmetric.
472 Chapter 13 Group Theory 
We turn now to localized orbitals. We have been emphasizing that one can subject 
basis sets to unitary transformations without making any physical difference. A similar 
rule applies for filled molecular orbitals in a determinantal wavefunction. These too 
can be subjected to unitary transformations without affecting the total energy or total 
electronic distribution for the system. (We have encountered this fact before. See, 
for instance, Appendix 7.) Thus, we have the capability of altering the appearance of 
the individual orbitals in the wavefunction without affecting the wavefunction itself. 
Chemists tend to think of the electrons in molecules as being paired in bond regions, 
lone pairs, and inner shells, but MOs are delocalized and do not reflect this viewpoint. 
But by carrying out unitary transformations, we can attempt to produce orbitals that are 
more localized without sacrificing any of the properties of the overallwavefunction. For 
methane, we could mix our four delocalized occupied MOs together to try to produce 
four new orbitals, each one concentrated in a different C–H bond region. One can do 
this, but it is important to realize that these localized orbitals are not eigenfunctions 
of an energy operator, for they have been produced by mixing eigenfunctions having 
different energies. Furthermore, the localization is never complete in any system of 
physical interest. Each localized orbital always contributes at least slightly to charge 
buildup in regions outside that of its primary localization. For instance, a localized 
C–H1 orbital in methane will have small “residual” components at H2,H3, and H4. 
What we have been discussing in this section are various kinds of equivalent orbitals. 
At the level of basis sets, we have indicated that a minimal valence basis set of STOs 
is equivalent to a set of symmetry orbitals and also to a set of hybrid orbitals (as well 
as an infinite number of other possibilities). At the level of molecular orbitals we 
have indicated that the set of occupied delocalized MOs is equivalent to an infinity of 
transformed sets, some of which will tend to be localized in regions chemists associate 
with bonds, lone pairs, or inner shells. One’s choice among the possibilities for basis 
is a matter of taste. However, at the MO level, the delocalized MOs have two features 
that are sometimes advantageous. The first is that their energies are eigenvalues for 
some energy operator (the Fock operator in SCF theory). These are related in a simple 
way to ionization energies and electron affinities via Koopmans’theorem. Hence, delocalized 
MOs are more appropriate when considering photoelectron spectra, etc. The 
second advantage of delocalized MOs is that they display in a clear way the symmetry 
requirements on the system because they are bases for representations. Hence, these 
MOs are the most appropriate to use when one is using MO phase relations to infer 
the nature of certain intra- or intermolecular interactions. (See Chapter 14 for examples.) 
When delocalized MOs are mixed to form localized orbitals, these energy and 
symmetry features become partially disguised. 
13-19 Symmetry and Integration 
Throughout this book, the usefulness of symmetry to determine whether an integral 
vanishes has been emphasized. It should come as no surprise, therefore, that the formal 
mathematics of symmetry—group theory—is also useful for this purpose. 
The basic idea we have used all along is that, if an integrand is antisymmetric for any 
symmetry operation, it must have equal positive and negative regions, which cancel on 
integration. If there is no symmetry operation for which the integrand is antisymmetric, 
then the integral need not vanish. (It still might vanish, but not because of symmetry.)
Section 13-19 Symmetry and Integration 473 
The group theoretical equivalent of this is as follows. Suppose that we have an integral 
over the integrand f : 
 f dv =? 
The function f is identified as being related to some symmetry point group. (Examples 
are given shortly.) We want to know what representation f is a basis for. If f produces 
a representation containing A1, then f has some totally symmetric character and the 
integral need not vanish. But if f is devoid of A1 character, the integral vanishes by 
symmetry since all other representations are antisymmetric for at least one operation. 
Our problem, therefore, is to decide which irreducible representations are present in 
the representation that is produced by the integrand f . 
In quantum chemistry, the integrand of interest is often a product of wavefunctions 
(or orbitals) and operators. For example, the hamiltonian matrix Hˆ contains integrals 
of the form 
Hij = .*i Hˆ .j dt 
We know that Hˆ is invariant for any symmetry operation of the group, and so Hˆ has 
A1 symmetry. .i and .j are assigned symmetries by comparing their behaviors under 
various operations with the group character table, as illustrated earlier. Thus, it is fairly 
easy to ascertain the symmetries of the various parts of the integrand. The problem is 
to determine the symmetry of the product .*i Hˆ .j . 
To develop a rule for products, let us consider the simplest case—the one-dimensional 
representations. Suppose that .1 and .2 are bases for one-dimensional representations 
1 and 2. Then, for some symmetry operation R 
R.1 =.1(R).1, R.2 
=.2(R).2 
where . is a character (1, -1, , or *). If we operate on the product .1.2 with R, we 
obtain7 
R.1.2 =(R.1)(R.2)=.1(R).1.2(R).2 =.1(R).2(R).1.2 
That is, the characters for the product .1.2 are equal to the products of the characters 
for .1 and .2. We have demonstrated the rule for one-dimensional representations, 
but it can be proved for higher-dimensional cases as well. In group theory, the product 
of two functions, like .1.2, is referred to as a direct product to distinguish it from a 
product of symmetry operations, like s3C+3 . The symbol for a direct product is .. 
For the C3v group, the characters for some direct products of bases for irreducible 
representations are shown in Table 13-25. The direct product x2 has as characters 
the product of characters of E times itself. These characters (4, 0, 1) do not agree 
with any of the irreducible representation character sets, and so E .E is reducible. 
We can tell, in fact, that E . E is four-dimensional from the leading character. To 
resolve E.E, we employ the formula (13-7), which gives E.E=A1.A2.E, and 
fits the observation that E .E is four-dimensional. The other direct products listed 
7ThatR.1.2 =(R.1)(R.2) is not always obvious to the student, but it should be evident that operating on (say 
reflecting) the function .1.2 gives the same result as reflecting .1 and .2 separately and then taking the product.
474 Chapter 13 Group Theory 
TABLE 13-25 
 Characters for Direct Products of Bases for Irreducible Representations of C3v 
C3v E 3s 2C3 
A1 1 1 1 z x2 +y2, z2 
A2 1 -1 1 Rz 
E 2 0 -1 (x,y), (Rx,Ry) (x2 -y2, xy)(xz, yz) 
E .E 4 0 1 x2 y2 x2z2 xy 
A2 .E 2 0 -1 Rzx Rzy 
A2 .A2 1 1 1 R2
z 
A1 .A2 .E 2 0 -1 zRzx 
in Table 13-26 ( Rzx,R2
z , zRzx, etc.) all give character sets indicative of irreducible 
representations. We see that A2 .E =E,A2 .A2 =A1,A2 .A1 .E =E. 
There is a general rule that is illustrated by these examples: A direct product of 
bases for two irreducible representations contains A1 character if and only if the two 
irreducible representations are the same. That is, if fi is a basis for i and fj is a basis 
for j and fifj is a basis for i,j , where 
i,j. =i .j =c1A1 .c22 .···.cii .cjj .··· 
then c1 
=0 if and only if i =j . 
Another rule is that, if i is A1, then i,j =j ; that is, multiplying a function f2, 
by a totally symmetric function f1 gives a product with the symmetry of f2. 
Now we are in a position to decide whether the integral of a product of functions and 
operators will vanish. For our examples, we will continue to use orbitals, operators, 
and coordinates from the ammonia molecule. Some of these quantities, segregated 
according to symmetry, are given in Table 13-26. 
TABLE 13-26 
 Operators and Orbitals for Ammonia Classified by Symmetry 
a1 e 
N2s N2px 
N2pz N2py 
(1/v3)(1s1 +1s2 +1s3) (1/v6)(2 · 1s1 -1s2 -1s3) 
f1 
f4 
f7
... 
.. 
MOs(see Table 13-23) (1/v2)(1s2 -1s3) 
Hˆ 
f2 
f3 
f5 
f6
..... 
.... 
MOs 
z 
x
y
Section 13-19 Symmetry and Integration 475 
EXAMPLE 13-9 Indicate whether each of the following integrals must vanish due 
to symmetry. 
1.  N2pzHˆN2pxdv 
2.  N2px Hˆ 1 v2 
(1s2 -1s3)dv 
SOLUTION 
 
1.  N2pz HˆN2px dv: The symmetries of the three functions in the integrand are respectively 
A1,A1,E. The direct product has symmetry E. There is no A1. The integral vanishes. 
2.  N2px Hˆ 1 v2 
(1s2 - 1s3)dv: The symmetries are E,A1, E. The direct product is therefore 
E .E .A1 =A1 .A2 .E. Since A1 is present, the integral need not vanish. 
 
These two examples are related to the block diagonalization of the matrix H, 
discussed in a previous section. The zero blocks in that matrix correspond to integrals 
between functions of different symmetry. Since Hˆ is of A1 symmetry, it has no influence 
on the symmetry of the integrand. If .i and .j have different symmetries, their 
direct product cannot contain A1 symmetry and the integral over .iHˆ .j must vanish. 
The reader can now understand how the computational procedure guarantees MOs of 
“pure” symmetry (i.e., bases of irreducible representations). If two basis functions .i 
and .j differ in symmetry, there will be a zero value for Hij . A zero Hij means that 
mixing .i and .j will produce no energy lowering. Hence, the variation procedure 
will not mix these functions together in the same MO, so the MO will not be of mixed 
symmetry. 
EXAMPLE 13-10 Indicate whether each of the following integrals must vanish due 
to symmetry. 
1.  f1xf4 dv 
2.  f3yf5 dv 
SOLUTION 
 
1.  f1xf4 dv: Integrals of this sort are involved in calculating spectral intensities. The symmetries 
are A1 .E .A1 =E and the integral vanishes. This means a transition between states 
corresponding to an electron going from f1 to f4 (or f4 to f1) is forbidden for x-polarized light 
and an oriented molecule (see Section 12-9). 
2.  f3yf5 dv: The symmetry here is E .E .E, which gives characters 8, 0,-1. This resolves 
into A1.A2 .3E and the integral need not vanish. Corresponding transitions are “y allowed.” 
 
EXAMPLE 13-11 In Chapter 12 , we indicated that the p*.p transition in ethene 
is dipole-allowed and polarized parallel to the molecular axis. Verify this from the 
character table for ethene, after determining the group symbol.
476 Chapter 13 Group Theory 
SOLUTION 
 Referring to Fig. 13-7, ethene yields the answers “no, no, no, yes, {no,” so choose 
any one. We arbitrarily take C2 along the C–C axis} “no, yes,” (go to D branch) “yes”, so D2h. Our 
choice of principal axis puts the C–C bond vertical. Using that we can identify the p MO as B2u 
and the p* MO as B3g. The product of characters for B2uB3g =1 1-1-1-1-1 1 1, which 
can be seen to have B1u symmetry. So  p*(x, y, or z)p dv 
= 0 only if x, y, or z also has B1u 
symmetry. z does, so the transition is allowed and is z-polarized (i.e., it is a parallel transition.)  
In this chapter we have seen how formal group theory can be used to characterize 
MO symmetries, construct symmetry orbitals, and indicate whether integrals vanish by 
symmetry. It is true that one can performMOcalculations and get correct results without 
explicitly considering symmetry or group theory, since the computational procedures 
satisfy symmetry considerations automatically. But group theory allows a much deeper 
understanding of the constraints that symmetry places on a problem and often leads to 
significant shortcuts in computation. 
A notable feature of group theory is its hierarchy of concepts. At the lowest level are 
the symmetry operations and the basis functions they operate on. At the intermediate 
level are the representations for the group, produced from the basis functions. At the 
highest level are the characters, produced from the representations. The characters 
provide the “handles” that we actually work with, but our interest is often focused on 
the basis functions to which they are related. This tends to lend an air of unreality to 
the use of group theory. An aim of this chapter has been to avoid this feeling of unreality 
by dispensing with formal proofs, and instead illustrating relationships through 
investigation of examples. Further insight should come from solving the problems at 
the end of this chapter. 
13-19.A Problems 
13-1. Do the following operations constitute a group? “come 90. to port” (P ) “come 
90. to starboard” (S) “steady as she goes” (E). 
13-2. The text indicates that every element in the group has an inverse if E appears in 
each column of the multiplication table. But E also appears once in each row. 
What does this mean? 
13-3. Consider the group of four operations of the drill soldier (Section 13-2). 
a) To which symmetry point group is this set of four operations isomorphic (i.e., 
which group has the same product relationship)? 
b) Based on the mathematical definition of class and Table 13-1, how many 
classes are there in this group? 
c) Based on your physical intuition about kinds of operation, how many classes 
would you have anticipated for this group? If there is a discrepancy between 
(b) and (c), try to explain it. 
13-4. The C3v (ammonia) group is of order six. This is the same as the number of 
ways one can place three hydrogens at the three corners of an equilateral triangle 
(3 · 2 · 1). When we consider the C4v group, we have 24 ways we can place four 
hydrogens at the corners of a square (4 · 3 · 2 · 1). But the C4v group only has 
order eight. Explain.
Section 13-19 Symmetry and Integration 477 
13-5. For each of the molecules (IV)-(VII), 
a) list the symmetry elements, 
b) calculate the number of symmetry operations for each element, 
c) obtain the order of the group, 
d) determine the group symmetry symbol, 
e) check your results for (a-c) against the appropriate character table in 
Appendix 11. 
13-6. a) Demonstrate that U is a unitary matrix. 
b) Demonstrate that U†AU is diagonal. 
U=
 1 v2 - 1 v2 
1 v2 
1 v2 

, A=
0 1 
1 0
 
13-7. Assign the following molecules to point groups, look up their character tables, 
and indicate in each case whether one could expect doubly degenerate MOs. 
a) C6H6 
b) CH2Cl2 
c) B2H6 (see Problem 13-5) 
d) Staggered C2H6 
e) Staggered CH3CCl3 
13-8. Consider the planar molecule CO2- 3 (VIII). The oxygen atoms are at the corners 
of an equilateral triangle. 
a) What is the point group of this molecule? 
b) Using the appropriate character table, assign a symmetry symbol to each of 
MOs (IX)–(XII).
478 Chapter 13 Group Theory 
13-9. Table P13-9 gives the eigenvalues and eigenvectors resulting from an extended 
H¨uckel calculation of the allene molecule (XIII). The molecule is aligned 
as shown with respect to Cartesian axes. Ascertain the point group for this 
molecule. Using the character table for this group, assign a symmetry symbol 
to each MO. Is a transition from the highest occupied MO level to the lowest 
unoccupied level allowed by symmetry for any direction of polarization? 
13-10. Consider the water molecule, oriented as shown in Fig. P13-10 with the y axis 
perpendicular to the molecular plane and the z axis bisecting the H–O–H angle. 
a) Figure out as many nonredundant symmetry operations for this molecule 
as you can, and set up their multiplication table. Ascertain that you have a 
group of operations by checking closure, etc. 
b) Use the functions z,Rz, x, and y as bases to set up a character table for 
this group. 
TABLE P13-9 
 Extended H¨uckel Molecular Orbitals for Allene 
C1 C2 
MO 
no. 
Energy 
(a.u.) 2s 2pz 2px 2py 2s 2pz 2px 
1 1.7868 0 0 1.58 0 -1.23 0 0.58 
2 1.4916 1.52 0 0 0 -1.01 0 0.57 
3 0.4927 -0.54 0 0 0 -0.51 0 -0.75 
4 0.3681 0 0 0.67 0 0.36 0 0.83 
5 0.3231 0 0 0 -0.34 0 0 0 
6 0.3231 0 0.34 0 0 0 -0.11 0 
7 -0.2619 0 0 0 0.85 0 0 0 
8 -0.2619 0 0.85 0 0 0 -0.78 0 
9 -0.4326 0 0.56 0 0 0 0.67 0 
10 -0.4326 0 0 0 -0.56 0 0 0 
11 -0.4881 0 0 -0.49 0 -0.08 0 0.41 
12 -0.5558 0 -0.16 0 0 0 -0.04 0 
13 -0.5558 0 0 0 0.16 0 0 0 
14 -0.6221 0.43 0 0 0 -0.14 0 -0.32 
15 -0.8102 0 0 -0.13 0 -0.44 0 -0.02 
16 -0.9363 0.47 0 0 0 0.36 0 -0.02
Section 13-19 Symmetry and Integration 479 
Figure P13-10 
 Relation of the water molecule to cartesian axes. 
c) Use the resulting table to find out the symmetries of all MOs that can contain 
1s AOs on the hydrogens. 
d) Produce the symmetry combinations of these AOs that can appear in the 
MOs of water. 
13-11. Consider the trans-chlorobromotetramine cobalt (III) ion (XIV). 
a) Find the appropriate symmetry elements for this molecule and set up their 
group multiplication table. (Ignore the hydrogens on the ammonias.) Check 
the multiplication table to be sure all requirements for a mathematical group 
are satisfied. 
Atomic Orbital Coefficients C5 
2py h3 h4 2s 2pz 2px 2py h6 h7 
0 0.30 0.30 1.23 0 0.58 0 -0.30 -0.30 
0 0.24 0.24 -1.01 0 -0.57 0 0.24 0.24 
0 0.61 0.61 -0.51 0 0.75 0 0.61 0.61 
0 -0.58 -0.58 -0.36 0 0.83 0 0.58 0.58 
1.27 -0.88 0.88 0 0 0 0.11 0 0 
0 0 0 0 -1.27 0 0 0.88 -0.88 
0.01 -0.21 0.21 0 0 0 -0.78 0 0 
0 0 0 0 0.01 0 0 -0.21 0.21 
0 0 0 0 -0.10 0 0 -0.16 0.16 
0.10 0.16 -0.16 0 0 0 -0.68 0 0 
0 0.17 0.17 0.08 0 0.41 0 -0.17 -0.17 
0 0 0 0 -0.48 0 0 -0.44 0.44 
-0.48 -0.44 0.44 0 0 0 -0.04 0 0 
0 -0.24 -0.24 -0.14 0 0.32 0 -0.24 -0.24 
0 -0.18 -0.18 0.44 0 -0.02 0 0.18 0.18 
0 0.07 0.07 0.36 0 0.02 0 0.07 0.07
480 Chapter 13 Group Theory 
b) Use the following as bases for representations: z,Rz, x, y, x2 - y2, xy. 
Make sure you get all the inequivalent irreducible representations allowed 
in the group by checking i l2
i =h. 
c) Set up the character table. Now ascertain which symmetry orbitals contain 
2s orbitals of nitrogen. Give the symmetry combinations of those AOs that 
appear in these symmetry orbitals. 
13-12. Use the relationships that must exist among characters to complete the following 
tables. Include proper symbols for the representations, but do not include bases. 
(a) 
E C2 sv s  v (b) 
E 2C3 3C2 sh 2S3 3sv 
13-13. The D5 group has four classes of operation and has order 10. How many 
inequivalent irreducible representations are there and what are their dimensions? 
13-14. Consider the structure shown in Fig. P13-14. 
Figure P13-14 
 Square pyramid. 
a) Figure out the symmetry elements and operations for this molecule. 
b) What is the group order and number of classes? 
c) How many inequivalent irreducible representations are there and what are 
their dimensions? 
d) Ascertain the group symbol and compare your answers with the character 
table in Appendix 11. 
13-15. A group has the following representations: A1,A2,B1,B2,E1,E2. What is the 
group order and how many classes are there? 
13-16. Find the matrices that transform s and p STOs into sp3 and sp2 hybrids (Section 
13-18). Demonstrate that these are unitary matrices. 
13-17. It has been argued (Section 13-18) that sp2 hybrid orbitals are appropriate basis 
functions for CH+3
(D3d). Could one use a basis set of sp3 hybrid orbitals for 
this system?
Section 13-19 Symmetry and Integration 481 
13-18. In each of the following cases, resolve the given character set. If these were 
characters of integrands, would the integral vanish by symmetry? For example, 
C3v :4 0 1 A1 .A2 .E No. 
a) C4v: 5 -1 1 1 -3 
b) D2h: 3 -1 -1 3 -1 3 3 -1 
c) D3d: 8 2 0 0 0 0 
13-19. Referring to the data in Problem 13-9, which integrals below must vanish by 
symmetry? 
a)  f6xf10 dv 
b)  f6yf10 dv 
c)  f1xf2 dv 
d)  f3xf14 dv 
e)  f2zf4 dv 
13-20. Consider the possible electronic excitations of staggered ethane from its occupied 
1eg MOto the various empty MOs of symmetry a1g, a2u, eg, and eu. Which 
of these are symmetry allowed and how are they polarized? 
13-21. Referring to the data in Problem 12-24, which transitions from MO 4 can be 
induced by y-polarized light? (Note: the molecular y axis is coincident with 
the symmetry z axis.) 
13-22. The order of a group for a given object is equal to the number of equivalent 
locations for any nonspecial point in space about the object. Without referring 
to tables, predict the order of the group for 
a) a square (both faces identical). 
b) a cube (all faces identical). 
Multiple Choice Questions 
(Answer the following questions without referring to text or tables.) 
1. Generate the symmetry operations for water, and choose the true statement from the 
set below. 
a) H2O has two C2 axes. 
b) H2O belongs to a group of order three. 
c) H2O belongs to a group having exactly three classes. 
d) H2O can have only nondegenerate MOs. 
e) None of the above statements is true. 
2. Which one of the following statements is true? 
a) All reflections in a group must belong to the same class. 
b) All linear molecules must have an inversion center. 
c) An S6 axis has exactly two nonredundant operations.
482 Chapter 13 Group Theory 
d) Use of a basis set of hybrid orbitals built from a minimal basis set of STOs leads 
to a different variationally minimized energy than does use of the minimal basis 
set of STOs without hybridization. 
e) None of the above is a true statement. 
3. Which one of the following is not a symmetry element for eclipsed ethane? 
a) i 
b) S3 
c) C2 
d) sv 
e) sh 
4. Which of the following nondegenerate MOs is (are) possible for the CO2 molecule? 
+ – + 
1 2 3 4 
+ + 
–
+ 
+ – + – 
+
– 
–
+ 
– + 
a) 1 only. 
b) All four are possible. 
c) 1, 2, and 3 only. 
d) 2 and 3 only. 
e) 1 and 3 only. 
5. The presence of an S4 axis guarantees the presence of 
a) a point of inversion. 
b) a sh reflection plane. 
c) a C4 axis. 
d) a C2 axis. 
e) None of the above is correct. 
6. A molecule having only the E symmetry operation can have 
a) only nondegenerate MOs. 
b) MOs of any degeneracy. 
c) an infinitely degenerate level. 
d) only doubly-degenerate MOs. 
e) None of the above is correct. 
7. An MO that is symmetric or antisymmetric for every symmetry operation of a 
molecule 
a) must be degenerate. 
b) cannot be variationally altered. 
c) must be the lowest- or highest-energy MO. 
d) must be nondegenerate. 
e) None of the above is correct.
Section 13-19 Symmetry and Integration 483 
8. The character for a reducible representation of C4 on a square planar molecule 
equals 
a) 0 
b) 1 
c) 2 
d) -2 
e) 4 
9. Consider the planar molecule-ion NOCl+2
. 
Cl 
Cl 
N O 
+ 
The number of reflection planes, s, three-fold rotation axes, C3, two-fold rotation 
axes, C2, and inversion centers, i, that are symmetry elements for this moleculeion 
axis 
s C3 C2 i 
a. 1 0 1 0 
a) 2 0 1 0 
b) 2 0 1 1 
c) 2 1 2 0 
d) None of the above is correct. 
References 
[1] D. M. Bishop, Group Theory in Chemistry. Oxford University Press, London and 
NewYork, 1973. 
[2] H. Eyring, J. Walter, and G. E. Kimball, Quantum Chemistry. Wiley, New 
York, 1944.
Chapter 14 
Qualitative Molecular Orbital Theory 
14-1 The Need for a Qualitative Theory 
Ab initio and semiempirical computational methods have proved extremely useful. But 
also needed is a simple conceptual scheme that enables one to predict the broad outlines 
of a calculation in advance, or else to rationalize a computed result in a fairly 
simple way. Chemistry requires conceptual schemes, simple enough to carry around 
in one’s head, with which new information can be evaluated and related to other information. 
Such a theory has developed alongside the mathematical methods described 
in earlier chapters. We shall refer to it as qualitative molecular orbital theory (QMOT). 
In this chapter we describe selected aspects of this many-faceted subject and illustrate 
QMOT applications to questions of molecular shape and conformation, and reaction 
stereochemistry. 
14-2 Hierarchy in Molecular Structure 
and in Molecular Orbitals 
We seek a simple qualitative approach to the question, “How does the total energy 
of a system change as the nuclei move with respect to each other?” This question 
is very broad, encompassing the phenomena of molecular structure and chemical 
reactivities. 
It is useful to distinguish three kinds of process that can occur as nuclei move. One 
of these is the process in which two nuclei move closer together or farther apart, with 
their separation being somewhere around a bond length, either at the outset or the 
conclusion of the motion (or both). This process includes the breaking or forming of 
bonds and also the stretching or compressing of bonds. It also includes the forcing 
together of two species that will not bond (e.g., He with He). Let us refer to this as 
a nearest-neighbor interaction, even though the two nuclei need not be bonded in the 
usual chemical sense. The second process is the changing of the bond angle between 
two nuclei bonded to a third. The changing of the H–O–H angle in water is an example. 
A necessary consequence of such a change in angle is a change in distance between 
the two moving nuclei (here H–H). In geometries normally of interest, however, this 
distance is somewhat greater than a typical bond length throughout the entire process. 
We refer to this process as bond-angle change. The third process is the rotation of one 
part of a system with respect to the other about some axis (usually a single bond in the 
system). An example is rotation of one methyl group in ethane with respect to the other. 
484
Section 14-3 H+2
Revisited 485 
We refer to this process as a torsional angle change, or an internal rotation. Such a 
change will produce changes in distances between nuclei located on opposite ends of 
the torsional axis, but these distances typically remain several times as great as a bond 
length throughout the entire process. 
Chemists have long recognized that the energies associated with these three kinds of 
change fall into a loose hierarchy, with nearest-neighbor interactions having the greatest 
effect on energy, bond-angle changes having a smaller effect, and torsional angle 
changes the least. Indeed, for this reason spectral transitions corresponding to stretching, 
bending, and torsional modes are found in different regions of the electromagnetic 
spectrum. 
Most nearest-neighbor interactions are a consequence of the connectedness, or 
topology, of a molecule. This aspect of molecular structure, sometimes called the firstorder, 
or primary structure, is the first aspect one considers when establishing structure, 
and it is the first aspect that chemists became aware of, historically. The second-order 
aspect of structure concerns the bond angles. Once those are at least roughly known, 
one can go on to consider the third-order, or tertiary structure resulting from torsional 
energetics. The last aspect is usually referred to as the conformation of the system. 
It is advantageous to discuss MOs from a similar viewpoint. If we wish to guess 
the nature of the MOs of a system (i.e., where they have their nodes) and the MO 
energy order, we first consider the topology of the system. (Recall that this is the 
only thing that the simple H¨uckel method considers.) Elementary arguments lead to a 
fair approximation of the appearance and relative energies of the MOs. Next, we can 
consider bending the system, bending the MOs along with it. By judging whether MO 
energies will rise or fall in this process, we shall showthat one can often predict whether 
the molecule will be more stable in the linear or bent form. Finally, when we know 
the second-order structure, we can imagine the various conformational possibilities, 
allowing the MOs to be carried along with the nuclei. Again, by judging how the MO 
energies respond, it is possible to make predictions as to which conformation is most 
stable. 
In order to formulate rules for QMOT, we will return to the H+2
molecule ion and the 
H2 molecule. Then, using insights gained there, we shall consider more complicated 
systems. 
14-3 H+2
Revisited 
In Chapter 7 we used the linear variation method to solve the minimal basisH+2 problem. 
However, symmetry conditions alone suffice to force the solutions to be 
.sg = 1/
2(1+S)(1sa +1sb), Eg(el)=(Haa +Hab)/(1+S) (14-1) 
.su = 1/
2(1-S)(1sa -1sb), Eu(el)=(Haa -Hab)/(1-S) (14-2) 
where Haa and Hab are negative energies, and S is the (positive) overlap integral 
between 1s AOs on nuclei a and b. The energies Eg(el) and Eu(el) are upper bounds 
for the electronic energy of the sg and su states. The total energies are obtained by 
adding the internuclear repulsion energy 1/R a.u., where R is the internuclear distance 
in atomic units.
486 Chapter 14 Qualitative Molecular Orbital Theory 
Figure 14-1 
 Steps in compiling total energies for minimal basis LCAO-MO calculation on H+2 
at R =2 a.u. 
It is instructive to consider the total energies, Eg(tot) and Eu(tot) as the result of an 
artificial step-by-step procedure as indicated in Fig. 14-1. We begin with a 1s AO on 
nucleus a at E=-1
2 a.u. If we now bring proton b up to a distance of 2 a.u., but do 
not allow the 1sAO to respond, we get an energy lowering due to the increased nuclear 
attraction now felt by the electron in 1sa. At R=2 a.u., this additional attraction lowers 
the energy by 0.472 a.u., givingHaa=-0.972 a.u. Let us refer to this as the frozenAOin 
molecule energy. (It may be thought of as the energy to first order due to a “perturbation” 
caused by the approach of proton b, although this is hardly a small perturbation.) As 
our next step, we include Hab. This term equals -0.699 a.u., and it splits the energy 
evenly around Haa, as shown in the figure. The resulting energies are the expectation 
values for the electronic energies of the unnormalized (due to omission of overlap) 
functions 2-1/2(1sa ± 1sb). Two things have happened in this step: The electronic 
charge has become delocalized over both centers, and it has changed in amount so that 
we no longer have one electron in eachMO. Because the overlap term S (equal to 0.586 
at R =2 a.u.) has not yet come into the calculation, we have here the energies due to 
1+S, or 1.586 electron in the sg MO, and 1-S, or 0.414 electron in the su MO. 
Before proceeding to the next step, let us examine the two energies at this point to 
see how they compare for equal amounts of charge. The 0.414 electronic charge in su is
Section 14-3 H+2
Revisited 487 
associated with an energy of -0.273 a.u., whereas the 1.586 charge in sg corresponds 
to-1.671 a.u. It is clear that the sg MOis inherently of lower energy (relative to E=0) 
per unit charge, due to the detailed nature of kinetic and nuclear electronic energies. 
(We will not concern ourselves with these detailed aspects, however.) 
Our next step is to normalize the charges by dividing by 1±S. This essentially adds 
0.586 electron to su and subtracts 0.586 from sg. From the figure, we see that this lowers 
the energy of su and raises that of sg. This is reasonable: Adding more charge to an 
MO of negative energy should make its energy contribution more negative, removal of 
charge should make it less negative. A very important fact, though, is that sg rises more 
than su goes down. This simply reflects our observation in the preceding paragraph 
that sg charge “has lower energy per unit charge,” so removal of it “costs more” than 
is gained by putting it into su. A useful summary of the effect of renormalization is: 
Renormalization tends to cancel the energy level changes due to nuclear motion. The 
higher an energy level is, the less effective is this cancellation. Therefore, net energy 
changes (due to nuclear motion) tend to be greatest for the antibonding member of a 
bonding-antibonding pair of levels, and, in general, greater for a higher-energy MO 
(bonding or antibonding) than for a lower-energy MO. 
The final step is to add the internuclear repulsion energy of 1/2 a.u. to each level. It 
is important to notice that the final total energy levels are related to the initial H atom 
energies in almost the identical way that Eg(el) and Eu(el) are related to the “atom-inmolecule” 
energy Haa (see Fig. 14-2) because the energy lowering due to the original 
1sa electron being attracted by proton b (-0.472 a.u. atR=2 a.u.) is fairly close in magnitude 
to the repulsion between the protons (+0.5 a.u.). We expect this near cancellation 
to hold as long as R is large enough that proton b is “outside” the charge cloud due to the 
1sa electron. In other words, the total energies should lie above or below the separated- 
Figure 14-2 
 Total and electronic energies for bonding and antibonding states of H+2 compared 
to appropriate reference energies.
488 Chapter 14 Qualitative Molecular Orbital Theory 
atom energy in much the same way that the electronic energies lie above or below the 
“atom-in-molecule” energy, if the internuclear separation is not too small. [We know 
that, as R approaches zero, the internuclear repulsion approaches infinity, whereas the 
exact values of Eg(el) and Eu(el) respectively, approach the united atom He+ energies 
of -2 a.u. (1s) and -0.5 a.u. (2ps ), and the cancellation ultimately breaks down.] 
From this discussion of H+2 , the following ideas emerge: 
1. The energy of an MO is lowered by bonding interactions, raised by antibonding 
interactions, relative to the appropriate reference energy. 
2. Antibonding interactions are inherently more destabilizing than bonding interactions 
are stabilizing. 
3. A parallel exists between electronic energy change and total energy change if we 
are careful about reference energies and internuclear distances. 
But H+2 is atypical of the systems we will be interested in treating by QMOT. We 
will usually be considering neutral, many-electron systems, and H+2 is a charged, oneelectron 
system. Therefore, we turn next to the molecule H2 to consider how well the 
above ideas carry over to a more typical system. 
EXAMPLE 14-1 The simplest MO treatment of H2 would follow an HMO-like 
procedure, and assume that two electrons in the 1sg MO should give a total energy 
for H2 of 2Eg(tot,H+2 ), or 2(-0.554 a.u.)=-1.108 a.u. The correct total energy 
forH2 is-1.1745 a.u. (Table 7-4), so the agreement is better than one might expect. 
Is this because our procedure is physically correct? 
SOLUTION 
 There are several errors in the procedure. Most obvious is that E(tot) for H+2 
includes the repulsion between two protons, so our value for H2 contains double that nuclear 
repulsion. But both molecules have a pair of protons. So how come the result agrees so well with 
the exact value? For one thing, doubling E(tot) for H+2 fails to include interelectronic repulsion, 
so we fail to include any interelectronic repulsion but we include too much internuclear repulsion. 
In addition, Re for H2 is about 3/4 that for H+2 , so Vnn in H2 is really about 4/3 Vnn in H+2 . Also, 
Vne in these two species differs, both because the protons are not separated by the same distance 
and because the electron cloud is swollen by the presence of interelectronic repulsion. Because 
the electronic potential energy changes, the kinetic energy changes too. We are observing here 
the result of partial cancellation of several errors.  
14-4 H2: Comparisons with H+2 
We first consider the ramifications of the neutrality and nonpolarity of H2. Imagine that 
we have a ground-state hydrogen atom Ha and we allow another similar atom, Hb, to 
approach it. At values of R in excess of, say, 2 a.u., we expect the perturbation felt by 
Ha to be much smaller than was the case in our H+2 discussion. This is because, where 
before we had approach by a charged particle, here we have approach by a neutral atom. 
The attraction between the electron on Ha and the proton on Hb is counterbalanced 
by repulsion between the electron on Ha and that on Hb. This means that the frozen 
atom-in-molecule energy for H2 is fairly close to the energy of the isolated atom. 
This simplifies our qualitative treatment for neutral molecules since it means we can
Section 14-4 H2: Comparisons with H+2
489 
meaningfully compare molecular electronic energies directly with unperturbed atomic 
electronic energies. 
We now come to a consideration of the orbital energies of H2. Here we will find 
that there are important differences between a one-electron system like H+2 and multielectronic 
systems. Consider the lowest (1sg) MO of H2. According to the description 
of ab initio theory in Chapter 11, the “energy of” this MO at some internuclear separation 
R [call it E1sg(R)] is equal to the energy of an electron in this MO (i.e., the 
orbital energy is equal to the one-electron energy). This energy results from kinetic 
and nuclear attraction components and from repulsion for the other electron. But where 
is this other electron? If we are considering the ground state of H2, it is in the 1sg 
MO also. That gives us an energy we might call E1sg(R, 1s2 
g ). However, we might 
instead be considering an excited state of H2, perhaps with the configuration 1sg1su. 
Depending on whether we choose the symmetric or antisymmetric combination for the 
spatial part of the wavefunction, we shall be considering an excited singlet or triplet 
state. In each case, repulsion for the 1sg electron will be different. Thus, we already 
have three orbital energies for this lowest MO, and we could continue getting more by 
considering other excited states of H2. Clearly, the orbital energies in a multielectronic 
molecule are dependent on the state being considered. 
Another important feature of one-electron energies in multielectronic systems is that 
they do not add up to the total electronic energy. That is, 2E1sg(R, 1s2 
g ) is not equal 
to the electronic energy of H2 at R. As pointed out in Chapter 11, this is because 
E1sg(E, 1s2 
g ) includes the interelectronic repulsion, so 2E1sg(R, 1s2 
g ) counts the interelectronic 
repulsion twice. 
One might imagine that, since MO energies in multielectronic systems are statedependent 
and cannot be simply summed to give the electronic energy, it is hopeless 
to use them as a basis for explanation or prediction. However, this would be too 
pessimistic. It turns out that the qualitative features of MO energies are not all that 
sensitive to change of state (see Fig. 14-3) and, besides, we are usually concerned with 
the ground state or with excited states wherein most of the electrons remain in orbitals 
occupied in the ground state. Also, the extra measure of interelectronic repulsion that is 
Figure 14-3 
 Orbital energies as function of internuclear separation for He2. The two curves for 
1sg are qualitatively similar, although far from identical. (FromYarkony and Schaefer [1].)
490 Chapter 14 Qualitative Molecular Orbital Theory 
included in the sum of one-electron energies can sometimes be expected to compensate 
roughly for the as yet unincluded internuclear repulsion energy. This rough equality 
requires that our system be neutral (so that the total number of repelling negative charges 
is equal to the total amount of positive charge) and nonpolar (so that the loci of charge 
are similar). We expect the equality to be quite good at large internuclear separation, 
where the electrons of each atom repel those of others as though they were centered 
on the nuclei, and to become progressively poorer as the atoms get closer together and 
undergo interpenetration of charge clouds. If we accept this rough equality, then we 
are permitted to approximate the change in total energy of the system (when the nuclei 
move), as equal to the change in the sum of one-electron energies.1 
14-5 Rules for Qualitative Molecular Orbital Theory 
The considerations discussed in the two preceding sections provide a rather loose justification 
for the following QMOT rules related to total energy changes when nuclei 
are moved. 
1. An increase in overlap population between twoAOs tends to lower the energy of an 
MO; a decrease tends to raise it. 
2. The effects of a given amount of orbital overlap population change on orbital energy 
are much more pronounced for higher-energy MOs. A corollary is: The destabilizing 
effect of an antibonding relation between two AOs tends to be greater than 
the stabilizing effect of a bonding relation, other factors (AO identities, distances) 
being equal. 
3. The sum of changes of one-electron energies should approximately equal the total 
energy change if we are treating a neutral, nonpolar system and if the nuclei that 
are moving with respect to each other are separated by a distance of a normal bond 
length or more. 
Rule (2) and its corollary are due to the different effect of renormalization on orbitals 
of different energy. Note that rule three refers only to energy changes. Wedonot expect 
the sum of one-electron energies to equal the total energy. This becomes obvious in 
the limit of infinite separation of atoms (e.g., in CO2) in which there is no internuclear 
repulsion, but interelectronic repulsion within each atom is still being counted twice. 
These rules are normally applied only to MOs made from valence-shellAOs. Innershell 
electrons are only very weakly perturbed in normal nuclear motions. The resulting 
small changes, while detectable and useful for analytical purposes, are inconsequential 
compared with valence electron effects. 
14-6 Application of QMOT Rules to Homonuclear 
Diatomic Molecules 
If we apply our rules from the preceding section to the motion of a pair of identical 
nuclei, each carrying a 1s AO, we obtain the orbital energy versus R curve 
shown in Fig. 14-4. The sg MO energy is predicted to drop as overlap population 
1For comments on this point, see Ruedenberg [2].
Section 14-6 Application of QMOT Rules to Homonuclear Diatomic Molecules 491 
Figure 14-4 
 Energy versus internuclear separation R for two nuclei with 1s AOs, derived from 
QMOT rules. 
increases, the su to rise as out-of-phase overlap increases, and su rises faster than 
sg drops. In QMOT, we argue from this figure that one-, two-, or three-electron 
homonuclear diatomic molecules should be stable (H+2 ,H2,H-2 ,He+2
) and that 
four-electron molecules (He2) should be unstable. Since a similar figure holds for 
molecules in which the valence orbital is 2s, 3s, etc., we might also expect systems 
like Li+2
, Li2, Li-2
, Be+2
,Na+2
,Na2,Na-2
,Mg+2
to exist, but not Be2 or Mg2. This set 
of predictions is a reasonable starting point, and in many cases agrees with observation 
(see e.g., Table 7-2). By the time we get to the Na–Mg series, however, the energy 
difference between 3s and 3p AOs is so small that it is unrealistic to treat these as 
“pure” 3s cases. Consequently, we must put less faith in simple predictions for those 
molecules. The comparison between our QMOT guess about these molecules and 
experimental or ab initio observation is shown in Table 14-1. 
The reader may have noticed that we have violated one of our QMOT conditions by 
considering nonneutral systems. One is presumably less safe in including such systems, 
although several of them have been studied and found to fit the QMOT prediction. 
TABLE 14-1 
 Stability of Some Homonuclear Diatomics as Predicted by QMOT and as 
Observed from Experiment or ab initio Calculation 
Molecule QMOT 
Experimental 
or ab initio Molecule QMOT 
Experimental 
or ab initio 
H+2
Stable Stable Be+2
Stable Stable 
H2 Stable Stable Be2 Not Stable Not Stable 
H-2
Stable Stablea Na+2
Stable Stable 
He+2
Stable Stable Na2 Stable Stable 
He2 Not Stable Not Stable Na-2
Stable ? 
Li+2
Stable Stable Mg+2
Stable ? 
Li2 Stable Stable Mg2 Not Stable Stable (?) 
Li-2
Stable Stable 
aThis molecule–ion is unstable with respect to losing an electron and forming this neutral molecule at its lowest 
energy, but it is stable with respect to dissociation into a neutral atom and a negative ion in their ground states.
492 Chapter 14 Qualitative Molecular Orbital Theory 
Apparently, if one is concerned only with the question of the presence or absence of 
a valley in the total energy curve (and not with relative depths of valleys or relative 
Re values), deviation from neutrality by one electron is not too damaging (as we saw 
for H+2 ). However, for doubly positive diatomics, like He2+ 2 , the internuclear repulsion 
dominates so that there is no stable species. 
One can add p-type AOs to the two nuclei and expand the orbital energy plot, as 
illustrated in Fig. 14-5. Filling in the orbital levels with electrons leads to the prediction 
that B2 is singly bonded, C2 is doubly bonded, N2 triply bonded, O2 doubly bonded, 
F2 singly bonded, and Ne2 not bonded. These systems, with their singly charged 
relatives, have been described in Chapter 7, and we will say no more about them here 
except that the agreement between QMOT rules and observations is quite respectable 
(see Fig. 14-6). 
An important distinction exists between the energy versus R curves drawn in 
Fig. 14-5 and those in Fig. 7-17. The latter set was constructed by sketching orbital 
energies for the united-atom and separated-atom limits, and linking these energies 
together using symmetry agreement and the noncrossing rule. The former set was constructed 
by sketching orbital energies for the separated atoms, and then using QMOT 
rules to decide which curves go up, and which go down in energy as R decreases. 
Figure 14-5 
 Qualitative sketches of homonuclear diatomicMOenergies as a function of R based 
on QMOT rules.
Section 14-6 Application of QMOT Rules to Homonuclear Diatomic Molecules 493 
Figure 14-6 
 De versus aufbau sequence for homonuclear diatomic molecules and ions. The 
tendency for systems having more net bonding electrons to have a greater De is adhered to fairly well 
(see Table 7-2 for data). 
When we wish to emphasize this distinction, we will refer to these as two-sided and 
one-sided correlation diagrams, respectively. 
According to the simple QMOT viewpoint, we should expect a bonding MO to be 
lower in energy than the AOs that comprise it. Likewise, an anti-bonding MO should 
be higher in energy. We can test this by comparing first ionization energies (IEs) for 
molecules against those for the constituent atoms. (Recall that Koopmans’ theorem 
equates orbital energy to ionization energy.) If the highest occupied MO (HOMO) of 
the molecule is bonding, the IE for the molecule should be greater than that for the 
atom. If it is antibonding, the IE of the molecule should be smaller. Strictly speaking, 
the HOMO of the molecule may consist of mixtures of AOs on each atom. In that 
case, we should compute a “valence state ionization energy” for the atom. Also, for 
heteronuclear moleculesAB, we need to knowwhat percentage of theHOMOto identify 
with each atom in order to obtain a suitable average atomic IE to compare with the IE 
of the molecule. Even if we ignore these corrections, however, and simply compare 
experimental IEs, the anticipated behavior is shown nicely by diatomic molecules, as 
indicated in Table 14-2. 
There is much similarity between the QMOT rules and the assumptions inherent in 
the extendedH¨uckel method described in Chapter 10. There, also, a bonding interaction 
is equated with energy lowering and an antibonding interaction with an energy rise. 
Furthermore, the sum of orbital energies is assumed to change in parallel with the 
total energy change, even though the internuclear repulsion is not included. We noted 
that this assumption limits the range of validity for EHMO calculations to relative 
nuclear motions at distances that are on the order of a bond length or greater. The 
EHMO method is, in essence, the numerical equivalent to the qualitativeMOapproach, 
and such calculations can serve as a guide for developing qualitative explanations in 
complicated situations or for producing numbers that enable one to compare QMOT 
with experimental or ab initio results.
TABLE 14-2 
 Comparison of Molecular and Atomic First Ionization Energiesa 
Molecule 
HOMO bonding 
(b) or 
anitbonding (a) IE (eV) 
IE of molecule 
expected > or < 
than atomic average IE atoms 
Does experiment 
agree with theory? 
H2 b 15.427 > 13.598 Yes 
He2 a ~22.0 < 24.46 Yes 
Li2 b 5.12 > 5.363 No 
B2 b ~9.5 > 8.257 Yes 
C2 b 12.0±0.6 > 11.267 Yes 
N2 b 15.576 > 14.549 Yes 
O2 a 12.06 < 13.618 Yes 
F2 a 15.7 < 17.426 Yes 
Ne2 a 20.1 < 21.47 Yes 
Si2 b 7.4±0.3 > 8.15 No 
Cl2 a 11.48±0.1 < 13.02 Yes 
Br2 a 10.53 < 11.85 Yes 
I2 a 9.3 < 10.457 Yes 
CN b 14.5±0.5 > 11.27 (C), 14.55 (N) Yes 
NO a 9.25 < 13.62 (O), 14.55 (N) Yes 
CO b 14.013 > 13.62 (O), 11.27 (C) Yes 
CS b 11.8, 119 > 11.27 (C), 10.36 (S) Yes 
ICl a 10.3 < 10.46 (I), 13.02 (Cl) Yes 
IBr a 9.98 < 10.46 (I), 11.85 (Br) Yes 
aData are from [3] or from Table 7-2. 
494
Section 14-7 Shapes of Polyatomic Molecules: Walsh Diagrams 495 
14-7 Shapes of Polyatomic Molecules: Walsh Diagrams 
In this section we will describe howthe rules and concepts ofQMOT enable one to rationalizeandpredictmolecularshapes. 
Theearliestsystematictreatmentofthisproblemwas 
givenbyWalsh,2 whoseapproachhasbeenextendedbyothers, particularlyGimarc[6].3 
We begin by considering the symmetric triatomic class of molecules HAH, where 
A is any atom. Such molecules can be linear or bent. Walsh’s approach predicts which 
are linear, which are bent, and sometimes which of two bent molecules is more bent. 
We approach the problem in the following way. First we sketch the valence MOs 
for the generalized linear molecule HAH, deciding which is lowest, second lowest, etc. 
in energy. Then we imagine bending the molecule and argue whether each MO should 
go up or down in energy on the basis of our QMOT rules. This produces a chart of 
orbital energies versus bond angle—a one-sided correlation diagram. Finally, we use 
this diagram to argue that HAH will be linear or bent, depending on how many valence 
electrons HAH has and on how they are distributed among the MOs. 
The first problem is to sketch the MOs for linear HAH and decide their energy order. 
Probably the simplest way to do this is through use of symmetry and perturbation arguments. 
We know that the linear molecule belongs to the D8h point group, possessing 
a center of inversion and a reflection plane through the central atom and perpendicular 
to the HAH axis. This means that the two hydrogen 1sAOs (1s1 and 1s2) will appear in 
MOsin the symmetry combinations fg=1s1+1s2,fu =1s1-1s2, where g and u stand 
for gerade and ungerade, respectively (see Chapters 7 and 13 for a background discussion). 
Recognizing this, we can next consider which MOs will result from interactions 
between the valenceAOson atomAand the symmetry orbitalsfg andfu. Aperturbationtype 
diagram for this appears in Fig. 14-7. We have assumed that only the valence 
s and pAOs onA are involved in bonding. Extension to include dAOs is possible. 
On the right side of Fig. 14-7, the symmetry orbitals fg and fu are shown to be 
slightly split. This reflects the greater stability of the in-phase, or bonding, combination. 
However, the splitting is slight because the hydrogen atoms are quite far apart (so that 
atom A can fit between them). These two levels sandwich the separated-atom limit of 
-1
2 a.u. or -13.6eV. 
The AOs of atom A are sketched on the left. Their energies are arranged so 
that the 2p energies are about the same as the fg,fu symmetry orbital energies on 
the right. (For example, AO energies used for nitrogen in EHMO calculations are: 
2s~-25eV, 2p~-13 eV.) For the linear molecule, we can label the s and p AOs on 
atom A as s or p and g or u. 
To generate the MO energy-level pattern from the interactions between these AOs 
onA and fg,fu, we use the following rules from perturbation theory (see Chapter 12): 
1. Interactions occur only between orbitals of identical symmetry. 
2. Interactions lead to larger splittings if the interacting orbitals are closer in energy 
(overlap considerations being equal). 
The resulting energy levels appear in the central column of Fig. 14-7. 
We need sketches of the MOs whose energy level pattern we have just approximated. 
We can guess the qualitative appearance of these by recalling that, when two orbitals 
2SeeWalsh [4] and the papers immediately following. See also Mulliken [5]. 
3For a critical review of the theoretical validity ofWalsh’s method, see Buenker and Peyerimhoff [7].
496 Chapter 14 Qualitative Molecular Orbital Theory 
Figure 14-7 
 Orbital energies of MOs formed by interaction of su and sg symmetry orbitals on 
H1 . . .H2 with valence s and p AOs on central atom A. The p AOs and MOs are degenerate but are 
sketched as slightly split. 
interact to give splitting, the lower energy corresponds to a bonding interaction, the 
higher energy to antibonding. Thus, for example, Fig. 14-7 indicates that the 1sg MO 
is a bonding combination of the 2sAO onA and the sg symmetry orbital, whereas 2sg 
is the antibonding combination. The pu MOs are simply the pp AOs on A, since there 
is nothing of the same symmetry for them to interact with. The six MOs for the HAH 
molecule are sketched in Fig. 14-8a. 
Before proceeding, notice that the lowest two MOs are A–H bonding, the next two 
are nonbonding, the highest two are antibonding. This is an example of the way in 
which topological, or nearest-neighbor, interactions govern the gross features of energy 
ordering. In fact, we could have generated this same set of MOs and energy order by 
simply sketching all of theMOswe could think of that were symmetric or antisymmetric 
for relevant symmetry operations and then putting the bonding ones lowest (with s lower 
than p), nonbonding next, and antibonding highest (again recognizing that greater s 
character should yield lower energy). (Note that the nonbondingMOsare not symmetric 
or antisymmetric for arbitrary rotations about the C8 axis. Hence, they must form a 
basis for a representation of dimension greater than one. Hence, they are degenerate. 
See Chapter 13 for detailed discussion.) Electrons in the two lowest MOs produceA–H 
bonding. Electrons in the next two MOs have little effect on A–H bonding and, in fact, 
constitute what a chemist normally thinks of as lone pairs. Electrons in the two highest 
MOs tend to weaken the A–H bonds. We normally do not worry about questions of
Section 14-7 Shapes of Polyatomic Molecules: Walsh Diagrams 497 
Figure 14-8 
 The Walsh-type correlation diagram for HAH: (a) linear D8h; (b) bent C2v. The 
cross-hatched parts of MOs have opposite sign from open parts. 
shape for systems where these highest two MOs are filled because such an HAH system 
is not even bonded (i.e., we do not worry about second-order structure if there is no 
stable first-order structure). Consequently, these two highest MOs will be omitted in 
many energy-versus-angle diagrams, although there are certain cases where they can 
be useful (e.g., in singly excited configurations) (see Problem 14-5). 
EXAMPLE 14-2 Referring to Fig. 14.8, give the QMOT rationale for the MOs in 
the lower-half of the figure having energies below those in the upper half. What is 
the rationale for the lowest (a1g) MO being at lower energy that the second-lowest 
(a2u) MO, and why are these both below the next (eu) MOs?
498 Chapter 14 Qualitative Molecular Orbital Theory 
SOLUTION 
 The lowest-energy half of the set are all C–H bonding MOs, the other half all 
being C–H antibonding. Because there are so many C–H bonds, their nature determines this 
separation. Within the C–H bonding set, the lowest-energy pair differ from the others in having 2s 
AOs on carbon, which are sufficiently lower in energy than 2pAOs to give them lower energy. The 
a1g MO is C–C bonding, hence has lower energy than the a2u MO, which is C–C antibonding.  
We now consider how the MO energies change upon bending the molecule. As 
1sg is bent, the two hydrogen AOs move closer together. Because they have the same 
phase, this leads to an overlap increase, but it occurs over a fairly long distance, being 
a second-nearest neighbor interaction. Therefore, there is an energy lowering, but it 
is not very large. Bending 1su leads to two changes. First, the 1s AOs move away 
from the axis of maximum concentration of the 2p AO. This causes a substantial loss 
of overlap and a substantial increase in energy. Second, the 1s AOs move closer to 
each other. They disagree in phase, and this also tends to increase the energy, although 
it is a relatively small effect because it occurs between second-nearest neighbors. In 
the linear molecule, 1pux and 1puy MOs contain no contribution from hydrogen 1s 
because the hydrogens are in a nodal plane in each case. If we imagine that, in bending 
the molecule, we keep the hydrogens in the xz plane, then we see that the hydrogens 
are remaining in the nodal plane for 1puy, but have moved away from the nodal plane 
of 1pux. Once this happens, the 1s AOs are no longer forbidden by symmetry from 
contributing to the px MO, and a “growing in” of 1sAOs occurs, leading to an MO like 
the 2a1 MO drawn in Fig. 14-8. This behavior is more complicated than we observed 
for the lower-energy MOs because it involves more than a mere distortion of an existing 
MO. With a little experience, this additional complication is easily predicted (or, one 
can do an EHMO calculation and “peek” at the answer by sketching out the MOs 
contained in the output; see Problem 14-6). The effect on the energy of this 2a1 orbital 
is quite large because several things happen, all of which are energy lowering: 
1. For a given amount of 1s AO, the bonding overlap with 2p increases with bending. 
2. For a given amount of 1s AO, the bonding overlap between the two hydrogens 
increases with bending. 
3. Since the amount of 1sAO present is not constant, but increases with bending (from 
zero in the linear configuration) the rate of energy lowering due to (1) and (2) is 
further augmented. 
Also, this is a fairly high-energy MO, and QMOT rule (3) tells us to expect such MOs 
to respond more dramatically to overlap changes. Finally, the overlap of a 1s AO with 
a 2p AO on another nucleus varies as cos ., where . is zero when the 2p AO points 
directly at the 1sAO. This means that the rate of change of overlap with angle is much 
greater around . =90. than at . =0. (see Problem 14-8). This is another reason for 
thinking that 1pux.2a1 will drop in energy much faster than 1su.1b2 will rise. 
The other p MO, 1puy, undergoes no changes in overlap since the hydrogen atoms 
remain in the nodal plane throughout the bending process. Therefore,QMOTarguments 
predict no energy change for this MO. 
The highest two MOs change in energy in ways that should be obvious to the reader, 
based on the above examples. Since these are the highest-energy MOs, they should 
show further enhanced sensitivity to overlap changes. 
The MOs for the bent form are labeled in accordance with the symmetry notation for 
the C2v point group, with a and b meaning symmetric and antisymmetric, respectively,
Section 14-7 Shapes of Polyatomic Molecules: Walsh Diagrams 499 
for rotation about the two-fold axis and 1 and 2 being analogous symbols for reflection 
in the plane containing theC2 axis and perpendicular to the molecular plane. The lowest 
valence MO of each symmetry type is numbered “1” despite the fact that lower-energy 
inner-shell orbitals exist. 
Because of the very qualitative nature of the arguments leading to Fig. 14-8, no effort 
is made to attach a numerical scale, either for energy or angle. 
We are now in a position to see how predictions based on ourWalsh-type correlation 
diagram compare with experimental data. For molecules having only one or two valence 
electrons, we expect the preferred shape in the ground state to be bent. Examples areH+3 
and LiH+2
, both of which have been shown by experiment and/or accurate calculation 
to be bent. Molecules with three or four valence electrons should be linear, since 
the 1su -1b2 energy rise is much greater than the energy change for the lower MO. 
Examples are BeH+2
,BeH2, and BH+2
, which are indeed linear. Addition of one or two 
more electrons nowbrings the 1pu-2a1 MOinto play, and we have already argued that 
the energy change of this MO should be considerably greater than that of 1su -1b2. 
Basically we have here a competition between a filled MO that favors the linear form 
and a higher, partially or completely filled MO favoring the bent form. We therefore 
might reasonably expect molecules in which 2a1 is singly occupied to be bent, and 
molecules in which it is doubly occupied to be more bent. Occupancy of the 1pu -1b1 
MO should have no effect on angle. Thus, that a molecule like BH2(1a1)2(1b2)2(2a1) 
has an equilibrium bond angle of 131., whereas SiH2(1a1)2(1b2)2(2a1)2 has an angle 
of 97.,NH2(1a1)2(1b2)2(2a1)2(1b1) has 103. and H2O(1a1)2(1b2)2(2a1)2(1b1)2 has 
105. is in pleasing accord with these simple ideas. 
Changes of angle upon electronic excitation also agree well with the correlation diagram. 
The triplet state of SiH2 resulting from the 2a1.1b1 excitation has a wider angle 
(124.) than does the ground state (97.). The excited singlet corresponding to the same 
excitation has a comparable angle (126.). NH2, when excited from . . . (2a1)2(1b1) to 
(2a1)(1b1)2 opens from 103. to 144.. The isoelectronic PH2, under similar excitation, 
opens from 92. to 123.. There are other examples to support the validity of the 
HAH diagram, but these suffice to illustrate that this qualitative approach has considerable 
generality and utility. Note again that singly charged cations appear to fit QMOT 
predictions despite the fact that there is less theoretical basis for success here. 
EXAMPLE 14-3 SiH2 and H2O have rather similar H-A-H angles of 97. and 105., 
respectively. How would you expect these angles to compare for SiH+2
and H2O+? 
SOLUTION 
 SiH2 loses an electron from the 2a1 MO, and should open up to an angle in the 
130. range. H2O loses an electron from the 1b1 MO, so its angle should not be greatly affected. 
(The value for H2O+ is 110.5..)  
Walsh-type correlation diagrams have been constructed and discussed for many 
systems, among them AH3, HAB, HAAH, BAAB, H2AAH2, B2AAB2, H3AAH3.4 It 
is not appropriate that these all be described here. We will briefly discuss two more 
cases that bring in some additional features. 
Molecules with HAB configuration lack the high symmetry of HAH, and this means 
that the MOs are not as highly symmetry-determined as in HAH. Probably the simplest 
4See Gimarc [6].
500 Chapter 14 Qualitative Molecular Orbital Theory 
way to arrive at sketches for linear HAB MOs is to start with AB MOs (similar to 
A2 MOs) and add the 1s AO of H in bonding and antibonding modes to form linear 
MOs. The results for the seven lowest-energy MOs (all A–H bonding or nonbonding) 
are seen at the left side of Fig. 14-9. (Diatomic MOs were discussed in Chapter 7.) 
We now imagine the hydrogen atom to move away from theAB axis, as shown, and 
use our QMOT rules to decide whether the energies should rise or fall. As before, we 
expect overlap changes between H and A (nearest neighbors) to have a greater effect 
on energy than those between H and B. 
The bent molecule has only one symmetry element, namely a reflection plane containing 
the nuclei. An a MO is symmetric under this reflection, a is antisymmetric. 
Figure 14-9 
 TheWalsh-type diagram for the HAB system: (a) linear C8v; (b) bent Cs. (After 
Gimarc [6].) Reprinted with permission from Accounts Chem. Res. 7, 384 (1974). Copyright by the 
American Chemical Society.
Section 14-7 Shapes of Polyatomic Molecules: Walsh Diagrams 501 
This paucity of symmetry types means that many of the correlation lines in the diagram 
refer to the same symmetry. Hence, it is not too surprising that some of our correlations 
run into conflict with the noncrossing rule. In Fig. 14-9, dashed lines are drawn from 3s 
to 4a and from 1px to 3a. These dashed lines connect theMOdrawings in the manner 
expected if we ignore the noncrossing rule and simply bend the MOs along with the 
molecule. They are, as it were, intended correlations. But these lines are associated with 
the same symmetry a and hence cannot cross. Instead we have an avoided crossing, as 
3s switches course and connects with 3a and 1px goes to 4a (solid lines in Fig. 14-9). 
Avoided crossings are not uncommon in quantum chemistry, and they occur in curves 
referring to state energy as well as orbital energy. A generalized sketch exemplifying 
the idea is shown in Fig. 14-10. The actual energy change (as a function of bond length, 
angle change, or whatever process is occurring) may show an intermediate maximum 
or minimum as a result of the avoided crossing. (The dashed lines are energies we 
predict by “forgetting” to allow the two functions of the same symmetry to be mixed 
in the variational procedure. The error involved in this is small if the two functions 
are of dissimilar energy. As they grow closer in energy, the error grows worse, and the 
deviation between solid and dashed lines gets bigger as the dashed lines converge.) 
A classic example of such an intermediate maximum is seen in an excited state 
of H2, illustrated in Fig. 14-11. Such maxima are important in understanding highenergy 
processes because they provide a means for some molecules to exist in bound 
vibrational states even while unstable with respect to dissociation products. Such states 
are called metastable states. 
The HAB Walsh diagram of Fig. 14-9 rationalizes the fact that the ten-valence 
electron HCN is linear in the ground state (due to 1px -4a) but bent in the first excited 
state. Such molecules as HNO, HNF, and HOCl, with 12–14 electrons, are bent in both 
ground and excited states because there is always at least one electron in the 5a level. 
The final system we shall consider is the H3AAH3 system. We will show how 
QMOT can be used to understand why diborane (B2H6) has a bridged structure whereas 
ethane (C2H6) does not, why ethane prefers to be staggered (D3d) rather than eclipsed 
(D3h), and what geometry changes we might expect if ethane is forced into the eclipsed 
conformation. 
Figure 14-10 
 Intended (dashed lines) and actual correlations between orbitals or states having 
the same symmetry.
502 Chapter 14 Qualitative Molecular Orbital Theory 
Figure 14-11 
 The four lowest states of H2. Note the curve for the first excited 1+g state. This 
shape results from an avoided crossing. (From Sharp [8].) 
Gimarc’s diagram relating MOs for A2H6 in D3d staggered and D2h bridged shapes 
is shown in Fig. 14-12. The MOs on the left are all H–A bonding and are arranged 
pretty much in the order one would expect on the basis of theA–A bonding. The lowest 
two are composed mainly of valence s AOs on atoms A. Next come the p bonding 
combinations, then the ps bond, followed finally by the p antibonds. The 1eup bonding 
levels lie below the 2a1gps bonding level because the former have greater ability to 
overlap with the hydrogens. 
There are only two MOs that show much energy change the molecule goes from 
D3d to D2h geometry. These are the 2a1g.2ag and the 1eg.1b2g. In the former 
case, the energy drops because two hydrogens have moved from positions off axis of 
one p lobe into positions off axis of two p lobes, thereby increasing the total amount
Section 14-7 Shapes of Polyatomic Molecules: Walsh Diagrams 503 
Figure 14-12 
 Walsh-type correlation diagram forD3d -D2h shapes ofH3AAH3. (After Gimarc 
[6].) Reprinted with permission from Accounts Chem. Res. 7, 384 (1974). Copyright by theAmerican 
Chemical Society. 
of overlap. The energy in the latter case rises very markedly because the change in 
geometry places all six hydrogens into nodal planes, greatly reducing the overlap. 
The prediction is that a 10- or 12-valence electronA2H6 system should favor a bridged 
D2h geometry over D3d but a 14-valence electron system should prefer D3d over D2h. 
Diborane (12 valence electrons) and ethane (14) have structures consistent with this. 
EXAMPLE 14-4 What structure should be expected for B2H-6 ?
504 Chapter 14 Qualitative Molecular Orbital Theory 
SOLUTION 
 Assuming that the extra electron occupies the 1b2g MO, one might expect that 
B2H-6 would have the D3d structure of ethane. It depends on whether or not one electron in the 
1b2g MO suffices to overcome the effects of the doubly occupied MOs. (Experiments and ab initio 
calculations indicate that B2H-6 has the same shape as ethane.)  
In structural problems such as this, one must be careful that, when comparing two 
possible molecular shapes, one is not overlooking other possibilities that might be even 
more stable. For instance, even though the diagram in Fig. 14-12 indicates that ethane 
should prefer D3d geometry to D2h, it says nothing about D3d relative to D3h, the 
eclipsed form. For this we must construct another diagram, shown in Fig. 14-13. Here 
the principal energy changes occur in the doubly degenerate e- type MOs. Recall that 
Figure 14-13 
 Walsh-type diagram for D3d–D3h A2H6: (a) D3d (staggered); (b) D3h (eclipsed). 
Note that the e-type MOs for the two forms do not turn into each other by rotating about the C–C 
bond. This is really a two-sided correlation diagram, the high symmetry of each form determining 
the MOs. Then QMOT rules are used to decide how similar MOs on the two sides should relate in 
energy. (See Lowe [9].)
Section 14-8 Frontier Orbitals 505 
degenerate MOs need not be symmetric or antisymmetric for all symmetry operations 
of the molecular point group. This results, in this case, in the e-type MOs being much 
more unbalanced, or lopsided, than the nondegenerate a-type MOs. As a result, the 
changes in overlap population upon rotation are much bigger for e-than for a-type 
MOs. The higher-energy eg -e set of MOs dominates the lower energy eu -e set 
for two reasons; it is at higher energy and hence more sensitive to overlap change, 
and the coefficients on the hydrogens are bigger due to the central nodal plane, which 
reduces the size of the MO in the A–A bond, forcing it to be larger elsewhere. In sum, 
a 14-valence electron A2H6 molecule prefers the staggered (D3d) form because the 
long-range H···H antibonding in 1e dominates the long range bonding in 1e. We 
can describe this as “nonbonded repulsion” between hydrogens at opposite ends of the 
molecule. 
The energy changes in Fig. 14-12 are much larger than those in Fig. 14-13. In the 
former, we are charting energy changes associated with changes in bond angle and even 
molecular topology. Overlap changes are large and occur between nearest neighbors. 
In Fig. 14-13 we are charting energy changes associated with internal rotation. Here 
the overlap changes occur between third-nearest neighbors and are very small. The 
order in which the possibilities have been examined—D3d versusD2h followed byD3d 
versusD3h—is thus sensible in that we are considering the grosser energy changes first. 
One can go even further and guess the qualitative changes in C–C, C–H distances, 
and C–C–H angle if ethane is forced into the eclipsed conformation. We argue that the 
1eg.1e pair of MOs suffer the greatest overlap change, losing population between 
the vicinal hydrogens. We must renormalize the MO to compensate for this loss, just 
as we had to in H+2 , discussed earlier. To renormalize, the MO 1e is multiplied by a 
factor slightly greater than unity. This magnifies the p antibond between the carbons 
and the bonding between carbon and hydrogens. Thus, we expect eclipsed ethane to 
have a slightly lengthened C–C bond, slightly shorter C–H bonds, and a larger C–C–H 
angle (the latter presumably mainly due to the increased vicinal repulsion brought 
about by overlap changes in the original rotation). Ab initio calculations5 support these 
predictions. 
14-8 Frontier Orbitals 
We have indicated that higher-energy MOs tend to undergo more pronounced energy 
changes upon overlap change due to distortion of the nuclear frame. This fact has led 
to a shortcut method for guessing the results of full orbital correlation diagrams of the 
sort we have already discussed. One merely considers what the energy behavior will 
be for the highest occupiedMO(HOMO) and bases the prediction entirely on that MO, 
ignoring all the others. Fukui6 was the first to draw attention to the special importance 
of the HOMO. He also noted that, in certain reactions in which the molecule in question 
acted as an electron acceptor, the lowest unfilledMO(LUMO) of the molecule (before it 
has accepted the electrons) is the important one. These two MOs are called the frontier 
MOs. It sometimes happens that the second-highest-energy occupied MO undergoes a 
5Stevens [10] finds that the C–C distance increases by 0.01 Å, the C–H distance decreases by 0.001 Å, and the 
C–C–H angle opens by 0.3.. 
6See Fujimoto and Fukui [11].
506 Chapter 14 Qualitative Molecular Orbital Theory 
Figure 14-14 
 HOMO and LUMO frontier MOs and subjacent and superjacent MOs. 
much bigger overlap change than theHOMOdoes, and therefore dominates the process. 
Recognition of this fact has led to a special name for the MOs just belowtheHOMOand 
just above the LUMO. They are called subjacent and superjacent MOs (see Fig. 14-14). 
An example of the frontier MO approach is provided by reconsidering the staggered 
versus eclipsed conformation for ethane. We expect the highest-energy, lopsided MO 
to be the p antibonding eg -e degenerate pair. Qualitative molecular orbital theory 
rules lead us to expect this pair to have higher energy in the eclipsed form. Therefore, 
we expect ethane to be staggered. 
The same approach can be used to predict the conformation of dimethylacetylene, 
H3C–C=C–CH3. The HOMO is again a degenerate pair of p-type MOs. (Generally 
speaking, one assumes that occupied p MOs are higher in energy than occupied s MOs, 
and this is often true. Even when it is not, however, the p-type MOs often tend to be 
more lopsided and hence to dominate because their overlap changes are greater.) The 
two degenerate HOMOs are delocalized over the entire molecule and can be expected 
to have the following characteristics: 
1. They will be orthogonal to each other. 
2. They will be bonding in the central C=C region, helping to establish multiple bond 
character there. 
3. They will be antibonding in the C–C single bond regions, thereby cancelling out 
double-bond character from a lower set of p-type MOs. 
4. They will be C–H bonding. 
[Use of rules (2)–(4) often suffices to establish the qualitative nature of HOMOs of 
hydrocarbons.] The results of all these conditions are the MOs sketched in Fig. 14-15. 
Observe that the end-to-end hydrogen overlap is most positive in the eclipsed conformation. 
This leads to the prediction that this molecule is more stable in the eclipsed 
conformation. Since the hydrogens are so far apart, the overlap change is expected to 
be very small. Ab initio calculations indicate that dimethylacetylene is more stable in 
the eclipsed form and that it has a barrier of less than 0.02 kcal/mole. 
As another example of frontier orbital usage, consider the methyl rotation barrier 
in propene, H3C–CH=CH2. Here the HOMO should be p-bonding in the double 
bond, antibonding in the single bond, and C–H bonding in the methyl group. This 
MO is sketched in Fig. 14-16 for the two possible conformations. The end-to-end 
antibonding in this MO is greatest for case (b), and so conformation (a) is favored. 
Indeed, it has been observed that, in general, a threefold rotor attached to a double bond
Section 14-8 Frontier Orbitals 507 
Figure 14-15 
 Degenerate HOMOs for dimethylacetylene, (a) eclipsed, (b) staggered, lead to the 
prediction that this molecule should prefer the eclipsed conformation. 
Figure 14-16 
 HOMO of propene in two conformations. Methyl group (a) eclipses and (b) staggers 
the double bond. The energy is lower in (a). 
prefers to eclipse the double bond. A few examples are acetaldehyde (H3C–CH=O), 
N-methylformaldimine (H3C–N=CH2), nitrosomethane (H3C–N=O), and vinyl 
silane (H3Si–CH=CH2). 
Notice that the HOMO of Fig. 14-16 is qualitatively similar to one of the 1eg -1e 
HOMOs of ethane. The QMOT frontier orbital argument for the stability of staggered 
ethane is basically the same as that for the stability of form (a) in propene. Observe that 
this MO also resembles the HOMO of 1,3-butadiene, and would lead to the prediction 
that the trans form of this molecule is more stable than the cis. This is, in fact, observed 
to be the case. 
It is important to bear in mind that the frontier orbital approach is an approximation 
to an approximation. It is not always easy to know when one is on safe ground. Of 
the examples mentioned here, ethane and dimethylacetylene are safest because the 
overlap changes are small. Hence, the perturbation is slight, and our assumption that 
the MOs of the two forms are essentially identical, except for AO overlap changes, is 
quite accurate. Also, symmetry is high, and so we know that s MO overlap changes 
will be smaller than lopsided p MO overlap changes. Propene and butadiene are risky. 
Here the whole molecular framework is lopsided, so overlap changes are large in s as
508 Chapter 14 Qualitative Molecular Orbital Theory 
well as in p MOs. Indeed, if one performs EHMO calculations on these molecules, 
one finds that s-type MO energies change much more than does the p HOMO energy. 
This comes about because of the fairly close approach by some of the hydrogens in 
these molecules. Much of this s energy change cancels out among the several s MOs. 
In propene, the cancellation is so complete that the p HOMO energy change is almost 
the same as the total EHMO energy change. In butadiene, however, the p HOMO 
accounts for only about one-third of the total energy change. Even though the frontier 
orbital method has an astonishing range of qualitative usefulness (we shall see more 
applications shortly), it is clear that caution is needed. 
EXAMPLE 14-5 How should the bond lengths and angles of propene change if the 
methyl group rotates from the stable conformation (Fig. 14-16a) to the unstable 
one (Fig. 14-16b)? 
SOLUTION 
 The HOMO for form b has a bit more negative overlap, which requires a slightly 
larger normalizing coefficient. This increases the influence of the HOMO, so we expect a slight 
shortening of the two out-of-plane C–H bonds and the C=C bond, a lengthening of the C–C single 
bond, and a slight opening of the H–C–H angle for the out-of-plane C–H bonds. No changes 
are predicted for the in-plane C–H bond lengths or angles as a result of HOMO renormalization. 
(However, other MOs are also being renormalized, so other changes will result. However, the 
HOMO-induced changes should dominate.)  
14-9 Qualitative Molecular Orbital Theory of Reactions 
It has been found possible to extend and amplify QMOT procedures so that they apply 
to chemical reactions. One of the most striking examples of this was application 
to unimolecular cyclization of an open conjugated molecule (e.g., cis-1,3-butadiene, 
closing to cyclobutene). This type of reaction is called an electrocyclic reaction. The 
details of the electrocyclic closure of cis-1,3-butadiene are indicated in Fig. 14-17. 
If we imagine that we can keep track of the terminal hydrogens in butadiene (perhaps 
by deuterium substitution as indicated in the figure) then we can distinguish between 
two products. One of them is produced if the two terminal methylene groups have 
rotated in the same sense, either both clockwise or both counterclockwise, to put the 
two inside atoms of the reactant (here D atoms) on opposite sides of the plane of the 
four carbon atoms in the product. This is called a conrotatory (c.on · r ¯o · t ¯a · tory) 
closure. The other mode rotates the methylenes in opposite directions (disrotatory) to 
give a product wherein the inside atoms appear on the same side of the C4 plane. 
A priori, we do not know whether the reaction follows either of these two paths. 
Figure 14-17 depicts processes where both methylene groups rotate by equal amounts 
Figure 14-17 
 Two idealized modes of electrocyclic closure of cis-1,3-butadiene.
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 509 
as the reaction proceeds. This is an extreme case of what is known as a concerted 
process. The two processes occur together, or in concert. The opposite extreme is a 
nonconcerted, or stepwise process, wherein one methylene group would rotate all the 
way (90.) and only after this was completed would the other group begin to rotate. 
This process would lead to an intermediate having a plane of symmetry (ignoring the 
difference between D and H), which means that the second methylene group would be 
equally likely to rotate either way, giving a 50–50 mixture of the two products pictured 
in Fig. 14-17. 
One can make a case for the reaction having some substantial degree of concertedness 
(by which we mean that the second methylene should be partly rotated before the first 
one is finished rotating). The reaction involves destruction of a four-center conjugated 
p system and formation of an isolated p bond and a new C–C s bond. Energy is lost in 
the dissolution of the old bonds, and gained in formation of the newones. Therefore, we 
expect the lowest-energy path between reactants and products to correspond to a reaction 
coordinate wherein the new bonds start to form before the old ones are completely 
broken. But the new s bond cannot form to any significant extent until both methylene 
groups have undergone some rotation. Thus, concertedness in breaking old bonds 
and forming new ones is aided by some concertedness in methylene group rotations. 
(Note that concertedness does not necessarily imply absence of an intermediate. If the 
reaction surface had a local minimum at a point at which both methylenes were rotated 
by 45., it would not affect the argument at all.) 
Because they knewthat many electrocyclic reactions are observed to be stereospecific 
(i.e., give ~100% of one product or the other in a reaction like that in Fig. 14-17), 
Woodward and Hoffmann [12] sought an explanation of a qualitative MO nature. They 
used frontier orbitals and argued how their energies would change with a con- or 
disrotatory motion, due to changes in overlap. For butadiene in its ground state, the 
HOMOis the familiar p MOshown in the center of Fig. 14-18. The figure indicates that 
the interaction between p–p AOs on terminal carbons is favorable for bonding in the 
region of the incipient s bond only in the conrotatory case. Therefore, the prediction 
is that, for concerted electrocyclic closure, butadiene in the ground state should prefer 
to go by a conrotatory path. When the reaction is carried out by heating butadiene 
(thermal reaction), which means that the reactant is virtually all in the ground electronic 
state, the product is indeed purely that expected from conrotatory closure. 
One can also carry out electrocyclic reactions photochemically. The excited butadiene 
now has an electron in a p MO that was empty in the ground state. This MO was 
the lowest unoccupied MO (LUMO) of ground-state butadiene, pictured in Fig. 14-19. 
One can see that the step to the next-higher MO of butadiene has just introduced one 
Figure 14-18 
 The HOMO of ground state cis-1,3-butadiene as it undergoes concerted closure 
by either mode.
510 Chapter 14 Qualitative Molecular Orbital Theory 
Figure 14-19 
 The HOMO of the first excited state of cis-1,3-butadiene as it undergoes closure 
by either mode. 
more node, reversing the phase relation between terminal p AOs, and reversing the 
predicted path from con-to disrotatory. Experimentally., the photochemical reaction 
is observed to give purely the product corresponding to disrotatory closure. (It is 
not always obvious which empty MO becomes occupied in a given photochemical 
experiment. One assumes that the LUMO of the ground state is the one to use, but 
there is some risk here.) 
EXAMPLE 14-6 Is the LUMO . HOMO transition dipole-allowed for cis-1,3- 
butadiene? 
SOLUTION 
 The HOMO is antisymmetric for reflection through the symmetry plane that 
bisects the molecule, and the LUMO is symmetric for this reflection. This is the only symmetry 
reflection plane where the MOs have opposite symmetry, so the transition is dipole-allowed (and is 
polarized from one side of the molecule towards the other). The group theory approach for this C2. 
molecule is that the HOMO has a2 symmetry, the LUMO has b1 symmetry, their product has b1 
symmetry, and, since x also has b1 symmetry, the transition is allowed and is x-polarized (where 
x is colinear with the central C–C bond).  
One might worry about the fact that we are looking at only a part of one MO, thereby 
ignoring a great deal of change in other MOs and other parts of the molecule. However, 
much of this other change, while large, is expected to be about the same for either of the 
two paths being compared. The large overlap changes between p–p AOs on terminal 
and inner carbon atoms, for instance, are about the same for either mode of rotation. 
This approach, then, is focused first on the frontier orbitals, which are guessed as being 
most likely to dominate the energy change, and second on those changes in the frontier 
orbitals that will differ in the two paths. 
This method is trivially extendable to longer systems. Hexatriene closes to cyclohexadiene 
in just the manner predicted by the frontier orbitals. The only significant 
change in going from butadiene to hexatriene is that we go from four to six p electrons. 
This means that the HOMO for hexatriene has one more node than that for butadiene 
(or, the HOMO for a 2n p-electron system is like the LUMO for a 2n-2 p-electron 
system insofar as end-to-end phase relations are concerned). The net effect is that the 
predictions for hexatriene are just the reverse of those for butadiene. That is, hexatriene 
closes thermally by the disrotatory mode and photochemically by the conrotatory mode. 
The general rule, called a Woodward–Hoffmann rule, is this: the thermal electrocyclic
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 511 
reactions of a k p-electron system will be disrotatory for k =4q +2, conrotatory for 
k = 4q (q =0, 1, 2, . . . ); in the first excited state these relationships are reversed.7 
It is possible to treat electrocyclic reactions in another way, namely, via a two-sided 
correlation diagram approach. This was first worked out by Longuet-Higgins and 
Abrahamson [14]. Only orbitals (occupied and unoccupied) that are involved in bonds 
being made or broken during the course of the reaction are included in the diagram. For 
butadiene, these are the four p MOs already familiar from simple H¨uckel theory. For 
cyclobutene, they are the two p MOs associated with the isolated 2-center p bond and 
the two s MOs associated with the new C–C s bond. These orbitals and their energies 
are shown in Fig. 14-20. 
In cyclobutene, the s and s* MOs are assumed to be more widely split than the p 
and p* because the ps AOs overlap more strongly. Also, the s MO is assumed lower 
than p1 of butadiene. However, these details are not essential. All we have to be certain 
of is that we have correctly divided the occupied from the unoccupied MOs on the two 
sides. The dashed line in Fig. 14-20 separates these sets. 
Next we must decide which symmetry elements are preserved throughout the idealized 
reactions we wish to treat. Let us consider first the reactants and products. These 
have C2v symmetry, that is, a two-fold rotational axis, C2, and two reflection planes s1 
Figure 14-20 
 MOs associated with bonds being broken or formed in the electrocyclic closure of 
(a) cis-1,3-butadiene to (b) cyclobutene. 
7SeeWoodward and Hoffmann [13, p. 45].
512 Chapter 14 Qualitative Molecular Orbital Theory 
Figure 14-21 
 Sketches illustrating that the conrotatory mode (b) preserves the C2 axis while the 
disrotatory mode (a) preserves the reflection plane s1. 
and s2 containing the C2 axis (see Fig. 14-21). A conrotatory twist preserves C2, but, 
during the intermediate stages between reactant and product, s1 and s2 are lost as symmetry 
operations. A disrotatory twist preserves s1 but destroys C2 and s2. Therefore, 
when we connect energy levels together for the disrotatory mode, we must connect 
levels of the same symmetry for s1, but for the conrotatory mode, they must agree 
in symmetry for C2. The s2 plane applies to neither mode and is therefore ignored. 
The symmetries for each MO are easily determined from examination of the sketches 
in Fig. 14-20, and are given in Table 14-3. These assignments lead to two different 
correlation diagrams, one for each mode. It is conventional to arrange these as shown 
in Fig. 14-22. 
TABLE 14-3 
 Symmetries for C2v MOs 
MO s1 C2 MO s1 C2 
Butadiene Cyclobutene 
p1 Sa A s S S 
p2 A S p S A 
p3 S A p* A S 
p4 A S s* A A 
aS is symmetric; A antisymmetric.
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 513 
Figure 14-22 
 A pair of two-sided correlation diagrams (one for each mode) for the electrocyclic 
reactions of cis-1,3-butadiene: (a) cyclobutene; (b) butadiene; (c) cyclobutene. 
There is curve crossing in these diagrams, but it is always lines of different symmetry 
that cross, and so no violation of the noncrossing rule occurs. 
If we are considering a thermal reaction, the lowest two p MOs of butadiene are 
occupied. These correlate with the lowest two MOs of cyclobutene if the conrotatory 
mode is followed, and the thermal conversion of cis-butadiene to cyclobutene by a 
conrotatory closure is said to be symmetry allowed. The other mode correlates p2 with 
an empty cyclobutene MO (p*). Taking this route moves the reactant toward doubly 
excited cyclobutene. (Even though we might anticipate de-excitation somewhere along 
the way, the energy required in early stages would still be much higher than would 
be needed for the symmetry-allowed mode.) This is said to be a symmetry-forbidden 
reaction. 
If we now imagine photo-excitation of cis-butadiene to have generated a state associated 
with the configuration p2 
1 p2p3, and trace the fate of this species for the two
514 Chapter 14 Qualitative Molecular Orbital Theory 
modes of reaction, we note that the disrotatory route leads to cyclobutene in the configuration 
s2pp* while the conrotatory mode gives sp2s*. Both of these are excited, 
but the former corresponds to the lowest excited configuration (p .p*) while the 
second corresponds to a very high-energy excitation (s .s*). Therefore, the former 
is “allowed” (since it goes from lowest excited reactant to lowest excited product) and 
the latter is “forbidden.” 
The two-sided correlation diagrams of Fig. 14-22 thus lead to the same predictions 
as the frontier orbital maximization of overlap approach. The difference between 
these approaches is as follows: The frontier-orbital approach requires sketching the 
HOMOand then judging overlap changes upon nuclear motion usingQMOT reasoning. 
The two-sided correlation diagram approach requires sketching all the MOs (occupied 
and unoccupied) of both reactant and product involved in bonds breaking or forming, 
ordering the corresponding energy levels, and finding symmetry elements preserved 
throughout the reaction. Once all this is done, the levels are connected by correlation 
lines without reliance on QMOT reasoning. Some qualitative reasoning enters in the 
ordering of energy levels (levels with more nodes have higher energy), but the two-sided 
correlation diagram technique is the more rigorous method of the two and tends to be 
preferred whenever the problem has enough symmetry to make it feasible. Reliance 
on frontier orbitals is more common for processes of lower symmetry. 
Concern is sometimes expressed about the apparent restrictions resulting from use 
of symmetry in correlation diagram arguments. One can imagine the butadiene cyclization 
occurring with less-than-perfect concertedness, the two methylene groups rotating 
by different amounts as the reaction proceeds. But that would destroy all symmetry 
elements. Will our symmetry-based arguments still pertain to such an imperfectly concerted 
reaction coordinate? Again, if we label certain sites by substituting deuteriums 
for hydrogens as shown in Fig. 14-17, the symmetry will be destroyed. Do our predictions 
still apply? One can answer these questions affirmatively by reasoning in the 
following way. If we had a collection of nuclei and electrons, and we could move 
the nuclei about in arbitrary ways and study the ground-state energy changes, experience 
tells us that the energy would be found to change in a smooth and continuous 
way. We can think of the energy as a hypersurface, with hyperdimensional “hills,” 
“valleys,” and “passes.” Now, in a few very special nuclear configurations, identical 
nuclei would be interrelated by symmetry operations, and we would be able to make 
deductions on group-theoretical grounds. Such deductions would only strictly apply to 
those symmetric configurations, but they would serve as indicators of what the energy 
is like in nearby regions of configuration space. Thus, the correlation diagram indicates 
that a perfectly concerted thermal electrocyclic reaction of butadiene will require 
much less energy to go conrotatory as opposed to disrotatory. The inference that a 
less-perfectly concerted reaction will have a similar preference is merely an assumption 
that it is easier to pass through the mountains in the vicinity of a low pass than 
a high one. Experience also leads us to expect that substituting for H a D (or even a 
CH3) will have little effect on the MOs, even though, strictly speaking, symmetry is 
lost. In essence, we work with an ideal model and use chemical sense to extend the 
results to less ideal situations, just as we do when, in applying the ideal gas equation 
of state to real gases, we avoid the high-pressure, low-temperature conditions under 
which we know the oversimplifications in the ideal gas model will lead to significant 
error.
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 515 
It is possible to combine information on orbital symmetries and energies to arrive 
at state symmetries and energies. Then one can construct a correlation diagram for 
states.8 We now demonstrate this for the dis- and conrotatory reactions just considered. 
Each orbital occupation scheme is associated with a net symmetry for any given symmetry 
operation. Character tables could be used to assign these symmetries, but this is 
not necessary. All we need to use is the fact that, in multiplying functions together, symmetries 
follow the rules: S×S=S,A×A=S,S×A=A. Thus, any doubly occupied 
MO in a configuration will contribute symmetrically to the final result. To ascertain 
the net symmetry, then, we focus on the partly filled MOs. The symmetries for C2 and 
s1 of ground and some excited configurations of butadiene and cyclobutene are listed 
in Table 14-4. Included are the cyclobutene configurations that result from intended 
correlations of various butadiene configurations. (For instance, p2 
1 p2 
2 butadiene has an 
TABLE 14-4 
 Symmetries and Intended Correlations of Some Configurations 
of Butadiene and Cyclobutene 
Cyclobutene 
Symmetry for “intended” configuration 
Configuration s1 C2 Con Dis 
Butadiene 
p1
2p2
2 Sa S p2s2 s2p*2 
p1
2p2p3 A A p2ss* s2p*p 
p1
2p2p4 S S p2sp* s2p*s* 
p1p2
2p3 S S ps2s* sp*2p 
p1p2
2p4 A A ps2p* sp*s*2 
p1
2p2 
3 S S s2s*2 s2p2 
... 
Cyclobutene 
s2p2 S S 
s2pp* A A 
sp2p* A S 
sp2p* A S 
sp2s* A A 
s2p*2 S S 
s2p*s* S A 
... 
aS is symmetric; A antisymmetric. 
8Actually, we shall be looking at simple products of MOs, or configurations. Each configuration is associated 
with one or more states and gives the proper symmetry for these states as well as an approximate average energy 
of all the associated states. Hence, the treatment described here gives a sort of average state correlation diagram. 
It might be more accurately called a configuration correlation diagram.
516 Chapter 14 Qualitative Molecular Orbital Theory 
intended correlation with s2p*2 cyclobutene if the disrotatory mode is followed. This 
is inferred from the orbital correlation diagram, Fig. 14-22.) 
Assuming that the energies of states associated with these configurations fall into 
groups roughly given by sums of orbital energies, we obtain the two-sided diagram 
shown in Fig. 14-23. Only a few of the configurations are interconnected, to keep 
the diagram simple. Note that the ground-state configuration of butadiene correlates 
directly with the ground state of cyclobutene for conrotatory closure, but has an intended 
correlation with a doubly excited configuration in the disrotatory mode. This intended 
correlation would violate the noncrossing rule by crossing another line of S symmetry, 
so that the actual curve turns around and joins onto the ground state level for cyclobutene. 
The effect of the intended correlation with a high-energy state is to produce a significant 
barrier to reaction. The figure shows that, for the first excited configuration, the highenergy 
barrier occurs for the opposite mode of reaction. State correlation diagrams 
Figure 14-23 
 A state or configuration correlation diagram for the electrocyclic closure of cis- 
1,3-butadiene.
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 517 
thus convert a “symmetry-forbidden” orbital correlation diagram into a high-activationenergy 
barrier: the conclusions are the same using either diagram. 
Another kind of reaction that is formally closely related to the electrocyclic reaction 
is the cycloaddition reaction, exemplified by the Diels–Alder reaction between ethylene 
and butadiene to give cyclohexene (I). Such reactions are classified in terms of the 
number of centers between the points of connection. Thus, the Diels–Alder reaction 
is a [4+2] cycloaddition reaction. One can conceive of several distinct geometrical 
possibilities for a concerted mechanism for such a reaction. The two new s bonds can 
be envisioned as being formed on the same face (suprafacial) (II) or opposite faces 
(antarafacial) (III) of each of the two reactants. The various possibilities are illustrated 
in Fig. 14-24. Qualitative MO theory is used to judge which process is energetically 
most favorable. One has a choice between the two-sided correlation diagram and the 
frontier orbital approach. We demonstrate the latter9 since it is simpler. Both methods 
lead to the same conclusion. In the course of this reaction, electrons become shared 
between the p systems of butadiene and ethylene. This is accomplished, to a rough 
approximation, by interaction between the HOMO of butadiene and the LUMO of 
ethylene and also between the LUMO of butadiene and the HOMO of ethylene. Let us 
consider the former interaction. The MOs are shown in Fig. 14-25 and the overlapping 
regions are indicated for the four geometric possibilities. Inspection of the sketches 
indicates that the two MOs have positive overlap in the regions of both incipient s 
bonds only for the [4s+2s] and [4a+2a] modes. Therefore, the prediction is that these 
modes proceed with less activation energy and are favored. Now let us turn to the other 
pair of MOs, namely the LUMO of butadiene (IV) and the HOMO of ethylene (V). 
9See Hoffmann andWoodward [15].
518 Chapter 14 Qualitative Molecular Orbital Theory 
Figure 14-24 
 Four suprafacial–antarafacial combinations possible for the Diels–Alder 2 + 4 
cycloaddition reaction. 
Figure 14-25 
 Overlaps betweenHOMOof butadiene and LUMO of ethylene resulting from four 
interactive modes pictured in Fig. 14-24. The geometries for the four modes would all differ. These 
drawings are highly stylized. 
Note that, for each MO, the end-to-end phase relationship is reversed from what it was 
before. Two symmetry reversals leave us with no net change in the intermolecular phase 
relations. It is easy to see, therefore, that theseMOsalso favor the [s, s] and [a, a] modes. 
In cycloaddition reactions of this sort, one need analyze only one HOMO-LUMO pair
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 519 
in order to arrive at a prediction. Extension to longer molecules or to photochemical 
cycloadditions proceeds by the same kinds of arguments presented for the electrocyclic 
reactions. 
Another type of reaction to which qualitative MO theory has been applied is the 
sigmatropic shift reaction, where a hydrogen migrates from one carbon to another and 
simultaneously a shift in the double bond system occurs. An example is given in 
Fig. 14-26. 
The usual treatment of this reaction10 involves examining the HOMO for the system 
at some intermediate stage in the reaction where the hydrogen has lost much of its 
bonding to its original site and is trying to bond onto its new site. At this stage, the 
HOMO of the molecule becomes like that of the nonbonding MO of an odd alternant 
hydrocarbon (Fig. 14-27) with a slightly bound hydrogen on one end. The suprafacial 
mode is favored in this particular case because the hydrogen can maintain positive 
overlap simultaneously with its old and new sites—the new bond can form as the old 
Figure 14-26 
 The two possible distinct products resulting from a shift of a hydrogen from position 
1 to position 5 in a substituted 1,3-pentadiene. Groups A, B, C,Dare deuterium atoms, methyl groups, 
etc., enabling us to distinguish the products. 
Figure 14-27 
 Phase relations in the HOMO for (a) suprafacial and (b) antarafacial [1, 5] 
sigmatropic shifts. 
10SeeWoodward and Hoffmann [16].
520 Chapter 14 Qualitative Molecular Orbital Theory 
bond breaks. This is not possible for the antarafacial [1, 5] shift. However, the [1, 7] 
shift prefers the antarafacial mode. 
Many other types of chemical reaction have been rationalized using qualitative MO 
theory. The association of SN2 reactions withWalden inversion (i.e., the adding group 
attacks the opposite side of an atom from the leaving group) is rationalized by arguing 
that an approaching nucleophile will donate electrons into the LUMO of the substrate. 
The LUMO for CH3Cl is shown in Fig. 14-28. A successful encounter between CH3Cl 
and a base results in a bond between the base and the carbon atom, so the HOMO of 
the base needs to overlap the p AO of carbon in the LUMO of Fig. 14-28a. Attack at 
the position marked “1” in the figure is unfavorable because any base MO would be 
near a nodal surface, yielding poor overlap with the LUMO. Therefore, attack at site 
2 is favored. As the previously empty LUMO of CH3Cl becomes partially occupied, 
we expect a loss of bonding between C and Cl. Also, negative overlap between the 
forming C-base bond and the three “backside” hydrogens should encourage the latter 
to migrate away from the attacked side, as indicated in Fig. 14-28b. 
The tendency of a high-energy, occupied MO of the base to couple strongly with 
the LUMO of a molecule like CH3Cl depends partly on the energy agreement between 
these MOs. If they are nearly isoenergetic, they mix much more easily and give a 
bonded combination of much lower energy. Molecules where theHOMOis high tend to 
be polarizable bases. A high-energy HOMO means that the electrons are not very well 
bound and will easily shift about to take advantage of perturbations. Such bases react 
readily with molecules having a low-energyLUMO(Fig. 14-29a). This corresponds to a 
“soft-base-soft-acid” interaction in the approach of Pearson [17]. When theHOMOand 
LUMO are in substantial energy disagreement, orbital overlap becomes less important 
as a controlling mechanism, and simple electrostatic interactions may dominate. This 
Figure 14-28 
 (a) The LUMO of CH3Cl. (b) Positive overlap between HOMO of base B and 
LUMO of CH3Cl increases antibonding between C and Cl and also repels H atoms from their original 
positions.
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 521 
Figure 14-29 
 LUMO–HOMOinteractions and splitting for (a) nearly degenerate levels, (b) wellseparated 
levels. 
is a “hard-acid-hard-base” situation. Thus, we expectQMOT rules to apply to soft-soft, 
rather than hard-hard interactions. 
The reader may, by this time, begin to appreciate the very wide scope of QMOT 
and the large number of variations of a common theme that have been used. In this 
chapter we have given only a few representative examples. We have not described 
all the variations or all types of application. For fuller treatment, the reader should 
consult specialized books on this subject, some of which have been referred to in this 
chapter [18]. 
14-9.A Problems 
14-1. Calculate to first order the electronic energy of a hydrogen atom in its 1s state 
and in the presence of an additional proton at a distance of 2 a.u. What is the 
total energy to first order? Repeat for distances of 1 and 3 a.u. (See Appendix 3.) 
14-2. Evaluate and graph the effects of dividing Haa ±Hab by 1±S for each of the 
following cases: Haa = 0,-5,-10,+10,-20. In each case, let Hab =-5, 
S = 0.5. Does the QMOT rule that antibonding interactions are more
522 Chapter 14 Qualitative Molecular Orbital Theory 
destabilizing than bonding interactions are stabilizing apply in all cases? Is the 
bonding level always the lower of the two? Does the QMOT expectation appear 
to be better followed by very low-energy levels, or by higher-energy levels? 
14-3. Table P14-3 is a list of electron affinities (in electron volts) of certain molecules 
and atoms. Can you rationalize the molecular values relative to the atomic values 
using QMOT ideas? 
TABLE P14-3 
 Atomic and Molecular Electron Affinities (in electron volts) 
H (0.75) Cl (3.61) C2 (3.4) O2 (0.45) Cl2 (2.38) 
C (1.26) Br (3.36) CN (3.82) F2 (3.08) Br2 (2.6) 
N (0.0 ± 0.2) I (3.06) N2 (-16) S2 (1.67) I2 (2.55) 
O (1.46) S (2.08) CO (<-1.8) SO (1.13) ICl (1.43) 
F (3.40) H2 (~-2) CS (0.21) FCl (1.5) IBr (2.6) 
14-4. For some time, it was uncertain whether the ground states of CH2 and NH+2 
are singlets (2a1)2 or triplets (2a1)(1b1). The ground-state geometries of these 
systems have HAH angles of 134.(CH2) and 140–150.(NH+2
). Based on other 
data described in the text for HAH systems, would you say these angles are more 
consistent with a singlet or a triplet ground state? Assuming that the first excited 
state is the other multiplicity, should the first excited state be more or less bent 
than the ground state? 
14-5. Based on Fig. 14-8, what should happen to the geometry ofH2Oupon 3a1.1b1 
excitation? 
14-6. Carry out EHMO calculations for CH2 at HCH angles of 180., 150., 120., 
and 90.. (Use a constant C–H bond distance of about 1Å in all cases.) From 
an examination of the MO coefficients, sketch and assign symmetry symbols 
to each MO. Plot the energies versus angle. Critically discuss your computed 
results compared to Fig. 14-8. If there are differences, try to rationalize them. 
(Remember that any EHMO orbital energy change can be analyzed in terms of 
Mulliken population changes, as discussed in Chapter 10.) 
14-7. Using the EHMO energy formula [Eq. (10-25)], analyze the contributions to 
the orbital energy change between 180. and 120. that you calculated for the 
1b2 MO in Problem 14-6. What percentage of the energy change comes from 
loss of overlap between 1s and 2pAOs? From antibonding between hydrogens? 
Compare this with the discussion in the text. 
14-8. If overlap between an s AO and a p AO goes as cos . (for constant and finite 
R) (Fig. P14-8), what is the mathematical expression for the rate of change of 
overlap with angle? Calculate the effect on overlap of a 30. shift, starting from 
. =0. Calculate the effect of a 30. shift from . =90..
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 523 
Figure P14-8 
 
14-9. Use symmetry to help establish an energy level pattern and MO sketches for 
planarAH3 (equilateral triangular). UseQMOT rules to produce the correlation 
diagram for planar versus pyramidal AH3. Based on your reasoning, which of 
the following should be planar? BH3,CH+3
,BeH-3
,NH3,PH3,H3O+,CH-3
. 
Can you think of any other shapes that might be examined as possibilities for 
AH3 systems? 
14-10. Use an EHMO program to generate MOs for CO2 at 180., 150., 120., and 90.. 
Sketch the MOs, characterize their symmetries, and construct an orbital energy 
correlation diagram for this molecule. Indicate what causes each MO energy 
to rise or fall. Based on your figure, would you expect the following molecules 
to be linear or bent? BeCl2,C3,CO2,N-3 ,NO2,O3,F2O. 
14-11. Should a pu.pg electronic transition for ozone cause the O–O–O angle to 
increase or decrease according toWalsh-type arguments? Sketch the MOs and 
indicate your reasoning. [The notation pg and pu refers to the MOs in the 
linear molecule. For the equilibrium bent structure, these MOs are: pg . 
a2, b2;pu.a1, b1.] 
14-12. Consider the electrocyclic reaction wherein the allyl anion closes to form a 
cyclopropenyl p anion (VIII). (The negative charge in the cyclopropyl anion 
may be thought of as resulting from double occupancy of a p–p AO on the 
singly protonated carbon.) Sketch the orbitals being formed or destroyed in 
this process. Determine the orbital symmetries for the symmetry operations 
conserved in conrotatory and disrotatory modes of closure. Set up an orbital 
correlation diagram and decide which mode is more likely for thermal and 
photochemical reactions. 
14-13. Construct an orbital correlation diagram for the “broadside” 2+2 cycloaddition 
of two acetylenes to form cyclobutadiene (IX). Is the reaction likely to proceed
524 Chapter 14 Qualitative Molecular Orbital Theory 
through an intermediate of square planar geometry? Assuming this geometry, 
would reaction be easier thermally or photochemically? 
14-14. For a system having no symmetry elements (except E), what is the result of the 
noncrossing rule? 
14-15. a) Construct an orbital correlation diagram for the 2 + 4 cycloaddition 
(Diels–Alder) reaction discussed in the text. 
b) Construct a state correlation diagram for this reaction. 
14-16. Based on the orbital relations discussed in the text and extensions of these relations 
to other cases, formulate generalized verbal rules (Woodward–Hoffmann 
rules) for cycloadditions and sigmatropic shift reactions. 
References 
[1] D. R.Yarkony and H. F. Schaefer, III, J. Chem. Phys. 61, 4921 (1974). 
[2] K. Ruedenberg, J. Chem. Phys. 66, 375 (1977). 
[3] Ionization potentials, appearance potentials, and heats of formation of gaseous 
positive ions, NSRDS-NBS 26, Natl. Bur. Stand. (1969). 
[4] A. D.Walsh, J. Chem. Soc. p. 2260 (1953). 
[5] R. S. Mulliken, Rev. Mod. Phys. 14, 204 (1942). 
[6] B. M. Gimarc, Accounts Chem. Res. 7, 384 (1974). 
[7] R. J. Buenker and S. D. Peyerimhoff, Chem. Rev. 74, 127 (1974). 
[8] T. E. Sharp, Atomic Data 2, 119 (1971). 
[9] J. P. Lowe, J. Amer. Chem. Soc. 92, 3799 (1970). 
[10] R. M. Stevens, J. Chem. Phys. 52, 1397 (1970). 
[11] H. Fujimoto and K. Fukui, in Chemical Reactivity and Reaction Paths 
(G. Klopman, ed.).Wiley (Interscience), NewYork, 1974. 
[12] R. B.Woodward and R. Hoffmann, J. Amer. Chem. Soc. 87, 395 (1965). See also 
R. Hoffmann, Angew. Chem. Int. Ed. 43, 2, 2004, for historical insights. 
[13] R. B. Woodward and R. Hoffmann, The Conservation of Orbital Symmetry. 
Academic Press, NewYork, 1970. 
[14] H. C. Longuet-Higgins and E. W. Abrahamson, J. Amer. Chem. Soc. 87, 2045 
(1965).
Section 14-9 Qualitative Molecular Orbital Theory of Reactions 525 
[15] R. Hoffmann and R. B.Woodward, J. Amer. Chem. Soc. 87, 2046 (1965). 
[16] R. B.Woodward and R. Hoffmann, J. Amer. Chem. Soc. 87, 2511 (1965). 
[17] R. G. Pearson, ed., Hard and Soft Acids and Bases. Dowden, Hutchison, and Ross, 
Stroudsburg, PA, 1973. 
[18] B. M. Gimarc, Molecular Structure and Bonding. Academic Press, New York 
1979.
Chapter 15 
Molecular Orbital Theory 
of Periodic Systems 
15-1 Introduction 
A structure is periodic in space, and hence has a periodic potential, if a subunit of 
the structure can be found that generates the entire structure when it is repeated over 
and over while traversing one or more spatial coordinates. Thus, a regular polymer 
can be “generated” mentally by translating a unit cell along an axis, sometimes with 
accompanying rotations. A crystalline solid results from such translations along three 
coordinates. In an analogous sense, some molecules are periodic. Benzene is an 
example since it can be generated by rotating a C–H unit in 60. increments about a 
point which ultimately becomes the molecular center. 
We have already seen (Chapters 8 and 13) that the orbital energies, degeneracies, 
and coefficients in benzene are to a great extent determined by symmetry. But these 
symmetry constraints can also be viewed as resulting from the cyclic “periodicity” of 
benzene, and this finite cyclic periodicity is, in essential ways, like the extended infinite 
periodicity of, say, regular polyacetylene or of graphite. It is reasonable, therefore, that 
some of the aspects of the energies and orbitals in these extended periodic structures 
are closely related to those in cyclic periodic systems. 
The concepts commonly used to describe periodic polymers, surfaces, or solids are 
closely related to those we have already developed for molecules, but their terminology 
and depiction are not familiar to most chemists. Our goal in this chapter is to develop 
an understanding of these concepts by progressing from more familiar cyclic systems, 
like benzene, to polymers, and to show how qualitative MO concepts apply to periodic 
systems in general. 
15-2 The Free Particle in One Dimension 
We will first examine the particle moving parallel to the x coordinate with no variation 
in potential energy. (Let V =0.) This system is periodic because every segment of x 
is the same as every other, and it provides a convenient starting point for discussion of 
periodic systems in general. 
526
Section 15-2 The Free Particle in One Dimension 527 
The quantum-mechanical solutions for this system are discussed in Chapter 2. The 
relevant points are: 
1. All non-negative energies are possible because there are no boundary conditions 
(beyond the conditions applying to well-behaved functions). 
2. Except for E =0, all energy levels are doubly degenerate. 
3. The pair of linearly independent wavefunctions associated with each doubly degenerate 
energy E can be combined in an infinite number of ways to produce different 
resultant wavefunction pairs. Two of these pairs are especially convenient. These 
are 
.+ = exp
v2mE 
¯h 
ix
 (15-1a) 
.- = exp
-v2mE 
¯h 
ix
 (15-1b) 
and 
.sin = sin
v2mE 
¯h 
x
 (15-2a) 
.cos = cos
v2mE 
¯h 
x
 (15-2b) 
4. The exponential forms [Eqs. (15-1ba, b)] are also eigenfunctions for the momentum 
operator, pˆx = (h¯/i)d/dx. Thus .+ corresponds to a particle moving parallel to 
the x axis with momentum +v2mE (i.e., toward x =+8), and .- corresponds 
to motion toward x=-8. We can associate the double degeneracy of the energies 
with the fact that there is no difference in the potential felt by the particle, regardless 
of whether it moves from left to right or right to left: The system has two equivalent 
directions. 
5. The real forms [Eqs. (15-2ba, b)] give oscillating particle distributions (except when 
E =0). In a given pair, the sine puts maximum particle density where the cosine 
puts its minimum, and vice versa. As E increases, the frequency of nodes increases. 
These wavefunctions are not “pure” momentum states because they are not eigenfunctions 
of pˆx (except when E=0). They are formed by mixing of the exponential 
forms. However, these functions are sometimes more convenient to work with, 
especially pictorially, and they are just as good as the exponential forms as long as 
we are concerned with energy rather than momentum. 
Equations (15-1) and (15-2) are sometimes written with v2mE/¯h replaced by the 
symbol k, with k having any value from zero to infinity. Evidently, k in this system is 
proportional to particle momentum, hence to the square root of the kinetic energy, T. 
(Since V =0, the kinetic energy is identical to the total energy, E.) For a free particle, 
de Broglie’s relation holds, so p is proportional to 1/., where . is the de Broglie 
wavelength. This means that k is proportional to the number of de Broglie wavelengths
528 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-1 
 Three ways of picturing the energies of the free particle. (b) and (c) show the 
dependence of E on k. In (c), the energies are understood to be degenerate (except at E =0). 
per unit of distance. Consequently, k is often referred to as the wavenumber. It is 
the de Broglie-wave-equivalent to the wavenumber of light or of any other classical 
harmonic wave. 
Summarizing, for the free particle, 
k .px .vE =vT .1/. (15-3) 
There are several ways in which the solutions of the Schr¨odinger equation for periodic 
structures are displayed graphically. One way is the familiar one of drawing all 
the energy levels against a vertical energy scale, as in Fig. 15-1a. (Since all energies 
are allowed for the free particle, this gives a single energy at E = 0 and a degenerate 
continuum, or “band”, from E >0 to 8, rather than a series of discrete lines.) 
Alternatively, we can plot E versus k to obtain the parabolic plot of Fig. 15-1b. This 
can be simplified, though, by plotting only one arm of the parabola (Fig. 15-1c), making 
implicit the fact that, for each solution with wavenumber k, there is a degenerate 
solution with wavenumber -k. Graphs that relate energy to k for periodic polymers, 
surfaces, or solids are called band diagrams. Figure 15-1c is the band diagram for the 
one-dimensional free particle. 
Another quantity of great physical importance is the number of states near a particular 
energy value. This is called the “density of states” (DOS). If we consider Fig. 15-1c, 
we can see that the states associated with 0<|k|<1 all lie within a certain range which 
we call E0 -E1 in Fig. 15-2a. If we assume that all values of k are equally likely (we 
show this to be true in Section 15-5), then there is an equal (infinite) number of states 
associated with 1<|k|<2, and these lie in the larger range E1 -E2. Hence, the states 
in the range E1 -E2 are less densely packed than those in the range E0 -E1. It is 
not hard to show that the density of states for the free particle drops off as 1/vE, as 
plotted in Fig. 15-2b. 
Ultimately we will consider one-dimensional periodic structures (polymers) with 
varying potentials caused by the presence of nuclei and other electrons, and we will find 
that these systems retain some of the above features. In particular, degeneracy due to 
directional equivalence and choice of real or complex forms for orbitals persist. Also, 
the quantity k retains its meaning as a wavenumber (in a restricted sense). However, 
k loses its simple relation to momentum and energy (because the electrons possess 
potential energy as well as kinetic energy), and, typically, certain energies become 
disallowed. This changes the infinite band to band segments separated by gaps in energy.
Section 15-3 The Particle in a Ring 529 
Figure 15-2 
 The states between |k|=0 and 1 are equal in number to those between |k|=1 and 
2, but the former set is packed into a smaller energy range, producing a greater density of states. 
15-3 The Particle in a Ring 
A particle of mass m constrained to move in the angular coordinate f about a ring 
of radius r with V (f)=0 is the cyclic analog of the free particle. This system, also 
described in Chapter 2, is similar in many ways to the one just discussed. Except for the 
case of E =0, all solutions are doubly degenerate and describable with real (trigonometric) 
or complex (exponential) functions. The quantum number j (we will use j in 
cyclic systems, k in linear systems) is proportional to the angular momentum, to the root 
of the energy, to the reciprocal wavelength, and to the number of de Broglie waves in 
one circuit of the ring. However, since the number of waves in the ring must be integral, 
we have a periodic boundary condition that restricts j to integer values and leads to a 
discrete energy spectrum as opposed to the continuous spectrum of the free particle. 
The diagrams summarizing relations between energies, j , and number of states are 
collected in Fig. 15-3. 
–2 –1 0 +1 +2 
j 
E 
(a) (b) 
0
0 1 2 
NOS 
Figure 15-3 
 (a) The energy for a particle in a ring has parabolic dependence on j , but exists only 
when j is an integer. (b) The number of states versus energy. This is the discrete-state analog of the 
density-of-states plot for a continuum of energies.
530 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Cyclic structures with varying potentials exist (e.g., benzene), and such systems 
retain the degeneracy and real or complex orbital features. Also, j continues to have 
meaning (restricted) with respect towavelength and number of nodes. However, angular 
momentum and energy are no longer simply related to j . 
15-4 Benzene 
We next examine benzene—a cyclic system having a nonuniform potential. We consider 
only the p electrons as described by the simple H¨uckel method, since the features we 
wish to point out are already present at that elementary level. A formula is presented 
(without derivation) in Chapter 8 [Eq. (8-52)] for the MOs of molecules like benzene. 
This gives, for the j th MO of benzene, 
fj = 
6 

n=1
{6-1/2 exp[2pij (n-1)/6]}2pp (n), j =0, 1, . . . , 5 (15-4) 
where i =v-1 and 2pp (n) is a 2pp atomic orbital centered on the nth carbon atom. 
The energy for MO fj is equal to [Eq. (8-51)] 
j =a +2ß cos(2pj/6). (15-5) 
If j =0, all the exponentials equal one and Eq. (15-5) gives 0 =a +2ß, which can 
be recognized as the lowest-energy p MO, and Eq. (15-4) gives f0=(1/v6)[2pp (1)+ 2pp (2)+2pp (3)+2pp (4)+2pp (5)+2pp (6)]. This gives all the 2pp AOs the same 
phase, which is what we expect for the totally bonding, lowest-energy p MO. 
If j = 1, 1 = a + 2ß cos(p/3) = a + 2ß(1/2) = a + ß. This is an energy in the 
second-lowest level, which is a degenerate level. The corresponding MO is f1 = 
(1/v6){[exp(0)]2pp (1) + [exp(pi/3)]2pp (2) + [exp(2pi/3)]2pp (3) + [exp(pi)] × 2pp (4) + [exp(4pi/3)]2pp (5) + [exp(5pi/3)]2pp (6)}. This somewhat formidablelooking 
MO is complex. Because f1 is one of a degenerate pair of MOs (the other one 
is f5), we can mix the complex MOs to obtain real ones that are still eigenfunctions. 
The real forms of the benzene MOs are described in Section 8-10D. The MO formulas 
produced by Eq. (15-4) are real for nondegenerate cases but are usually complex for 
degenerate MOs. However, it is always possible to mix any pair of degenerate complex 
MOs to produce a pair of degenerate real MOs. Notice that the complex MOs place 
the same electron density at each carbon whereas the real forms do not. The real MOs 
show nodes at various points in the ring (see Fig. 8-13) just as the real wavefunctions 
for the free particle show nodes at various points in x. 
In these formulas, j is restricted to the integer values ranging from 0 to 5, giving six 
MOs. If one tries other integer values of j in formulas (15-4) and (15-5), one simply 
reproduces members of the above set. Indeed, any six sequential integer values for 
j produces the same set of solutions as does the sequence 0–5. In particular, the set 
-2,-1, 0, 1, 2, 3 is perfectly acceptable and provides a match with the conventions of 
solid-state physics and chemistry. 
The energies for the six unique MOs of benzene are reproduced over and over again 
if we allow the index j to run beyond the specified range. This is depicted in Fig. 15-4a, 
and it is easy to see that the same six energies result for any six contiguous j values.
Section 15-4 Benzene 531 
Figure 15-4 
 (a) Energies for benzene as a function of j . (b) Unique energies of benzene. 
(c) Number of states at each energy. All data refer to p energies at the simple H¨uckel level. 
The set from -2 to 3, alluded to above, can be replotted in condensed form over the 
range 0–3 (Fig. 15-4b) if we keep in mind the fact that the energies are degenerate 
except for the first and last. The “number of states” diagram for benzene is sketched in 
Fig. 15-4c. 
If we consider a monocycle with thousands of carbon atoms, Eq. (15-5) gives (condensed) 
results as sketched in Fig. 15-5a. The very large number of energies still lie 
in the range a +2ß to a -2ß with the cosine wave now stretched over a much larger 
range of integers. The well-separated points of Fig. 15-4b coalesce into the line of 
Figure 15-5 
 (a) H¨uckel energies and (b) density of states for a very large number of carbons in 
a cycle.
532 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Fig. 15-5a, which now appears almost continuous. The shape of the curve in Fig. 15-5a 
makes it evident that the states around j =0 and j =N/2 are closer together in energy 
than are those around j =N/4, giving the density of states curve shown in Fig. 15-5b. 
There is a qualitative difference between what we found for the particle in a ring and 
benzene. In the former case, k can increase without limit, and energy keeps increasing 
as the square of k. In benzene, increasing j beyond the prescribed range simply causes 
the energy to cycle back and forth between a +2ß and a -2ß. Why do these systems 
behave so differently in this regard? The energy of the free particle increases as we 
fit more and more waves into a circle of fixed radius, obtaining, therefore, shorter de 
Broglie wavelengths. The H¨uckel energy of a benzene p MO depends on the extent of 
bonding and antibonding character between adjacent p AOs. When j is zero, all the 
interactions are bonding and =a+2ß. Asj increases, bonding interactions disappear, 
ultimately to be replaced by antibonding interactions. At j = 3, all interactions are 
antibonding and  = a -2ß. It certainly makes sense that this is the highest energy 
an MO can have because this is the most antibonding arrangement imaginable. But 
what happens mathematically to make this work out? Let us examine the exponential 
functions in Eq. (15-4) for two values of j , say j =3 and j =9, to see if we can resolve 
this question. An immediate problem confronts us: The function is complex, and 
sketching a complex function is not convenient. However, because the complex MOs 
can always be mixed to form real MOs, we can let the mixing occur within Eq. (15-4) 
itself to give sine and cosine equivalents. For example, 
fj cos = 
6 

n=1
{6-1/2 cos[2pj (n-1)/6]}2pp (n). (15-6) 
(For some values of j , this expression will yield MOs that are not normalized.) Equation 
(15-6) tells us that MO fj has six terms, each term being a 2pp AO times a 
coefficient. The values of the coefficients are given by the term in curly brackets. This 
term produces a discrete set of values, since n is a discrete set of integers, but we can 
sketch a continuous function [by replacing (n-1)/6 with the continuous variable f] 
and then locate the places on this “coefficient wave” that correspond to the discrete 
points of interest. Sketches for this “coefficient wave” when j =3 and j =9 are shown 
in Fig. 15-6. The special points of interest, where actual coefficient values are given 
Figure 15-6 
 Coefficient waves for j =3 (dashed curve) and j =9 (dotted curve). Both curves 
intercept the same coefficient values (±1/v6) at the positions related to carbon atoms 1–6, so the 
curves produce the same MO.
Section 15-5 General Form of One-Electron Orbitals in Periodic Potentials 533 
for benzene MOs, are also shown (at the positions labeled 1–6), and it is obvious that 
the same set of coefficients is produced by each function. Increasing j has caused the 
coefficient wave to oscillate with a shorter wavelength, but the extra oscillations do not 
register because they occur between the special points of interest where we are sampling 
the function (i.e., at the points where the carbon nuclei reside). In other words, the extra 
oscillations in our benzene coefficient functions are just “empty wiggling” and have 
no effect on the MO, so we cut off the j range where the meaningful wiggling starts 
to become empty wiggling. The reason that the particle in a ring does not show this 
behavior is that the exponential (or sine, cosine) functions in that system actually are 
the wavefunctions, while in benzene these functions are sampled only at discrete points 
where AOs are located, and the resulting coefficients are then used to produce MOs. 
Notice that the MOs for benzene as given by Eq. (15-4) are produced from an 
equation having the following form: There is an exponential term that, for each MO, 
supplies the coefficients for various carbons in the molecule, and there is a basis set of 
functions located on the various carbons. This basis set has the same “periodicity” as 
does the molecule. (In the case of benzene, we have so far taken this to be six identical 
2pp AOs.) Wavefunctions for all periodic systems have this same form—an exponential 
(or equivalent trigonometric) expression times a periodic basis. This is the content of 
Bloch’s theorem, which we prove in the next section. 
EXAMPLE 15-1 What coefficients are generated for a benzene MO by a coefficient 
wave of the type shown in Fig. 15-6 with j =-3? j =0? 
SOLUTION 
 The arguments of the cosine function of Eq. (15-6) will be the negative of those 
of the j =3 case. Because the cosine is symmetric about an argument of zero, we obtain the same 
curve for j =-3 as for j =+3, hence the same MO coefficients. This demonstrates that the MO 
generated by j -6 is the same as that generated by j. For j =0, the cosine arguments all vanish, 
the cosines all equal +1, and the curve of Fig 15-6 becomes a straight line at a coefficient value of 
+1/v6. This produces the lowest-energy MO of benzene.  
15-5 General Form of One-Electron Orbitals in Periodic 
Potentials—Bloch’s Theorem 
We will establish in this section the general mathematical nature of eigenfunctions for 
periodic hamiltonians. This is the content of Bloch’s theorem. The actual proof will be 
carried through for cyclic “periodic” systems, and those results will then be extended to 
noncyclic periodic systems. The rationale for this is that a nearly infinite linear periodic 
structure can be treated as cyclic without introducing error. That is, a sufficiently long 
extended chain of atoms can be assumed to have the same wavefunctions and energies 
as the same chain joined end to end to form a cycle. For short chains and rings, this 
is not the case; the lowest-energy MO for hexatriene is not as low in energy as that 
for benzene, nor is it uniform over the whole chain (as it is in benzene). But for long 
enough chains, the difference becomes negligible. 
Bloch’s theorem, as it applies to periodic cycles, states that eigenfunctions have the 
form 
j (f)=exp(ijf)Uj (f) (15-7)
534 Chapter 15 Molecular Orbital Theory of Periodic Systems 
where i = v-1, j is an integer and Uj (f) has the periodicity of the cycle. If the 
cycle belongs to the Cn point group, then U(f) = U(f + 2p/n). Functions of the 
form of Eq. (15-7) are called Bloch functions. It was stated in Section 13-8 that every 
wavefunction or orbital is a member of a basis for an irreducible representation for the 
point group of the system, so it follows that the wavefunctions we seek to characterize 
have to be bases for irreducible representations of the Cn point group. These groups 
are discussed in Section 13-11, where it is shown that the functions  =exp(if) and 
* =exp(-if) are bases for irreducible representations for these groups, which is to 
say that they are eigenfunctions for the rotation operator, no matter what the value of the 
integer n in Cn. We consider the more general functions exp(±ijf) with j an integer 
(so that the function obeys the cyclic continuity condition) and ask what happens when 
such functions are subjected to a rotation. Let the rotation operator be symbolized Cq
n 
for a rotation of q times 2p/n. The n-fold cyclic system is invariant to this rotation if 
q is an integer. It is not difficult to show that (see Section 13-11) 
Cq
nf (f)=f (f -2pq/n) (15-8) 
(For example, clockwise rotation by 60. brings to each point in the circle the value that 
used to be 60. in the counterclockwise direction.) Then 
Cq
n exp(ijf) = exp[ij (f -2pq/n)] 
= exp(-2piqj/n) exp(ijf) (15-9) 
Equation (15-9) shows that exp(ijf) is an eigenfunction for the rotation operator, so 
exp(ijf) is a basis for an irreducible representation, with j any integer. We can build 
additonal flexibility into this eigenfunction if we write an exponential function in a 
more complicated way: exp[i(j +nN)f] with N also an integer. We know that this 
must still be a basis function because (j +nN) must still be an integer. If we operate 
with Cq
n, we find that (Problem 15-6) 
Cq
n exp [i(j +nN)f]=exp(-2piqj/n) exp [i(j +nN)f] (15-10) 
This shows that our newest, most general, exponential has the same eigenvalue for 
Cq
n no matter what integer value we choose for N. [Observe that the eigenvalue in 
Eq. (15-10) depends on q, j, and n, but not N.] Therefore, since we have an unlimited 
number of choices for the integer N, we have an unlimited number of degenerate 
eigenfunctions for Cq
n for each choice of j . We can mix together such degenerate 
eigenfunctions in any way we please and still have an eigenfunction. Let us take the 
linear combination 
8 
N=-8
AN exp [i(j +nN)f] (15-11) 
This can be factored into the form 
exp(ijf) 
8 
N=-8
AN exp(inNf) (15-12) 
The sum is a function of f which we can call U(f). It has the same periodicity in f 
as the cycle because n is equal to the number of identical cells in the cycle and N is
Section 15-5 General Form of One-Electron Orbitals in Periodic Potentials 535 
an integer. It is actually the general Fourier series expansion of a periodic function. 
For N =0, the contribution is constant (A0) at all f. For N =±1, the functions repeat 
n times around the cycle and clearly have the periodicity of the system. For N =±2 
the functions repeat 2n times, so they obviously still have the periodicity of the system. 
Since all the individual members of the sum have the periodicity of the system, so does 
the sum itself. The specific nature of the resultant function U(f) depends on the choice 
of coefficients AN. 
We have shown that 
exp(ijf)U(f) (15-13) 
with U(f) n-fold periodic is a general mathematical form for bases for irreducible 
representations for the Cn group. But wavefunctions, .(f), for a periodic potential 
must also be bases for such irreducible representations. Therefore, wavefunctions must 
be expressible in this form: 
.j (f)=exp(ijf)U(f), j =0,±1,±2, . . . (15-14) 
We are not claiming that all functions of the form Eq. (15-13) are wavefunctions for the 
n-fold periodic potential. Rather, we are claiming that all such wavefunctions belong 
to the class of functions represented by Eq. (15-13). Only certain choices of U(f) will 
serve to make Eq. (15-13) become a wavefunction for the system, and the choice of an 
appropriate function U(f) is not necessarily the same at different values of j . To make 
this dependence on j explicit, we include it as a subscript on U: 
.j (f)=exp(ijf)Uj (f), j =0,±1,±2, . . . (15-15) 
with Uj (f) n-fold periodic. This completes the proof of Bloch’s theorem for periodic 
cycles. 
Before extending this result to noncyclic systems, a number of comments and clarifications 
should be made. First, the function Uj (f) may still seem to be something 
of a mystery. Think of it this way. For benzene p MOs we know we need to create 
a basis set having two properties: It should be like a 2pp AO near any carbon atom, 
and it should be the same at each carbon atom. Bloch’s theorem does not comment on 
whether U should look like a 2pp AO at a carbon atom, but it does require that it be 
the same at each carbon atom, that it be the same at the midpoints between adjacent 
carbons, that it be the same at a point one bohr above each carbon atom, etc. In short, 
U must be symmetric for the six-fold rotation operation. It is up to us to figure out 
what that six-fold symmetric basis set should look like in detail, and to construct U 
appropriately. 
Second, U depends on j . What does this mean? Suppose we compare the lowestand 
highest-energy p MOs for benzene. The lowest is totally bonding and the highest is 
totally antibonding. Ordinarily, we use the identical set of 2pp AOs as basis functions 
for both MOs, but this is not required. Perhaps the pp AOs for the bonding MO 
would be more appropriately chosen to be slightly larger, so they overlapped better, and 
maybe those for the antibonding MO should be slightly smaller, to reduce antibonding 
interactions. These objectives could be achieved, for example, by making U a mixture 
of 2pp and 3pp AOs on each center and letting the nature of the mixture depend on j . 
This makes for a more involved variational calculation, so it is normally not included in
536 Chapter 15 Molecular Orbital Theory of Periodic Systems 
cases like this one where such effects are expected to be small. (3pp character should 
not mix in to a significant extent because the 3pp AO is considerably higher in energy 
than the 2pp AO.) However, there are many cases where a unit cell contains several 
AOs in the same symmetry class and of similar energy (e.g., 2s, 2ps ), and it is then 
quite important to allow the mix of these to vary with j . A given band, for example, 
could start out being mainly 2s at low energies and end up being mainly 2ps at the 
high-energy end. This means that the appropriate description of the function U must 
be redetermined for each choice of wavenumber j . 
Third, careful comparison of Eqs. (15-15) and (15-4) shows that they are not exactly 
the same. Equation (15-15) instructs us to find a periodic functionUj (f) and multiply it 
by exp(ijf) at every point in f. Think of the sine or cosine related to the exponential and 
imagine what this means as we multiply it times a 2pp on some carbon. Say the cosine 
is increasing in value as it sweeps clockwise past the carbon nucleus at 2:00 on a clock 
face. This produces a product of cosine and 2pp that is unbalanced—smaller toward 
1:00 than toward 3:00, because the cosine wave modulates Uj (f) everywhere. But 
Eq. (15-4) is different. It instructs us to take the value of the cosine at 2:00 and simply 
multiply the 2pp AO on that atom by that number. The 2pp AO is not caused to become 
unbalanced. Only its size in the MO is determined by the cosine. Equation (15-4) is 
called a Bloch sum. Such sums are approximations to Bloch functions, but any errors 
inherent in this form are likely to be quite small if the basis functions and unit cell are 
sensibly chosen. (Using Bloch sums is similar in spirit to the familiar procedure of 
approximating a molecular wavefunction as a linear combination of basis functions.) 
Extending Bloch’s theorem to linear periodic structures requires identifying an 
appropriate substitute for the coordinate f and reconsidering the appropriate values for 
j (which we will call k in noncyclic systems). We require that these substitutions yield 
eigenfunctions and eigenvalues for the linear system that are the same as what results 
when it is treated as a cycle. Let us suppose that we have some very large number, n, 
of unit cells in the cycle, with a repeat distance of a. This means that the cycle has a 
Cn axis and also a circumference of na, so the linear coordinate (call it x) ranges from 
0 to na as the angular coordinate f ranges from 0 to 2p. When we move s steps around 
the cyclic polymer, the exponential’s argument changes by a factor of 2pijs/n, with j 
an integer. The same number of steps should give us the same effect on the exponential 
for the linear polymer. If we choose our exponential’s argument to be of the form ikx, 
then s steps creates the factor iksa, so we require that iksa = 2pijs/n. This gives 
k=2pj/na, with j an integer. We see that, for extremely large values of n, k becomes 
an effectively continuous variable (points separated by 2p/na). Also, we see that k is 
uniformly distributed (j =0,±1,±2, etc.), which affects our DOS calculations, as was 
mentioned in Section 15-2. 
We can also ask about the range of k, which should correspond to the 0–2p range of 
f. We expect “empty wiggling” to occur when the sin or cos equivalent of exp(ikx) has 
a wavelength shorter than 2a. The largest nonredundant value of k should come when 
cos(kx) goes from cos(0) to cos(2p) as x goes from 0 to 2a. Therefore kx=k(2a)=2p, 
so k has a range of 2p/a. Since the convention is to center k about zero, k runs from 
-p/a to p/a. (Note that this range becomes infinite for the free particle in a constant 
potential, for which a is zero.) The range of k over which all unique wavefunctions for 
the periodic system are produced once and only once (-p/a <k =p/a) is called the 
first Brillouin zone (FBZ).
Section 15-7 An Example: Polyacetylene with Uniform Bond Lengths 537 
For noncyclic, one-dimensional systems, then, eigenfunctions have the form 
.k =exp(ikx)Uk(x), -p/a <k =p/a (15-16) 
Just as was true for cyclic cases, we can approximate this Bloch functional form 
with a Bloch sum:
.k =
n-1 

s=0 
exp(iksa)Uk(sa), -p/a <k =p/a (15-17) 
In chemistry, it is Bloch sums that are normally used. 
Two- or three-dimensional periodic systems can be treated similarly, with separate 
k vectors for each of the independent translational directions. 
15-6 A Retrospective Pause 
We have reviewed some familiar systems and presented Bloch functions and sums for 
one-dimensional cyclic and linear periodic systems. The practical lesson to this point is 
that one can generate useful one-electron wavefunctions for some such systems simply 
by choosing a basis set of AOs that is identical from one unit cell to the next and then 
modulating that set with exponential (or sine and cosine) functions. This is most easily 
done by using the exponential or trigonometric functions as “coefficient waves” from 
which coefficient values are plucked at points where a unit cell basis set is centered 
(i.e., by taking Bloch sums). In a case like the p electrons of benzene, each unit cell is 
a carbon atom and the basis set is a 2pp AO at each atom. The coefficients determined 
by exp(ijf) with j =0,±1,±2, 3 serve to define the wavefunctions completely. Once 
these are known, the energy values follow. In cases where several AOs are present in 
each unit cell the situation is complicated because the mix of these AOs may change 
as j (or k) changes. This added complexity is easily handled by computer programs 
working within the paradigm of the variational method. What is important for us to 
understand as chemists is what the computer is doing in such cases, why it is doing it, 
and what it means for the chemical and physical properties of the system. 
15-7 An Example: Polyacetylene with Uniform 
Bond Lengths 
Consider the molecule produced by addition polymerization of acetylene. The standard 
chemical representation for the all trans version of the molecule is shown in Fig. 15-7a. 
The resonance diagram makes it plausible that this polymer, like benzene, will have 
equal C–C bond lengths, intermediate in value between the lengths of single and double 
bonds. For the moment let us assume this to be true and examine the MOs for the 
p orbitals of this system. To keep matters simple, we begin by treating the system at 
the simple H¨uckel level. This means that the hydrogen atoms can be ignored. Also, 
since only nearest-neighbor p interactions are accounted for, we can pretend that the 
carbon framework is linear (Fig. 15-7b).
538 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-7 
 (a) All-trans polyacetylene in its two “resonance structures.” (b) The linear model 
appropriate for a nearest-neighbor, p-only calculation. 
First we must choose our unit cell. The simplest choice is to make it a single carbon 
atom with the characteristic translational distance being a C–C bond length, shown as 
a in Fig. 15-7b. 
Next we must choose the periodic basis set Uk(x) for the p MOs. The normal 
choice for this (and also the choice dictated by the simple H¨uckel method) is a 2pp AO 
on each carbon. Furthermore, it is normal to assume that this basis does not change 
as we go from one MO to another, so we can dispense with the subscript k on U. 
(Hence, no variational calculation will be necessary in finding thewavefunctions for this 
system.) 
Equation (15-17) tells us how to produce wavefunctions. We simply take exp(ikx) 
times our basis set, with k taking on all values between -p/a and p/a, and pick off 
values at discrete points corresponding to carbon atom positions. Since it is more 
convenient towork with the real forms of solutions, we in effect choose a pair of k values 
(e.g., -p/4a and p/4a) so that we can generate a pair of trigonometric coefficient 
waves [cos(px/4a) and sin(px/4a)]. The p MOs for regular polyacetylene produced 
from Bloch sums are shown in Fig. 15-8 for selected values of k. These can be used to 
illustrate some important points: 
1. The k =0 situation merely reproduces the basis set. 
2. As |k| increases, the coefficient wave goes to shorter wavelength and more nodes. 
3. Only one MO results from each of the ±p/a extremes of the k range—these are 
nondegenerate solutions. 
4. The pair of MOs drawn for ±p/4a are degenerate, as are those for ±p/2a. Therefore, 
they can be mixed, either to generate complex functions or else simply to shift 
the phase. For example, the pair at k=±p/2a can be added or subtracted to produce 
the equally valid alternative pair shown in Fig. 15-9. In this way a phase shift of any 
degree could be created. There is no requirement that the MOs be “lined up” in a 
simple way with the atoms, as we have done in Figs. 15-8 and 15-9. (It is, however, 
usually more convenient.)
Section 15-7 An Example: Polyacetylene with Uniform Bond Lengths 539 
Figure 15-8 
 Coefficientwaves times a periodic basis set of one 2pp AOon each carbon. (a) k=0, 
so the coefficient wave is a constant and the periodic basis is unmodulated. (b)–(d) Plus and minus 
k-value exponentials are mixed to give trigonometric coefficient waves. In (d), one of these waves 
has nodes at every carbon so no Bloch sum function exists for this case. 
Once we have the MOs, also sometimes called “crystal orbitals” (COs) for infinite 
systems, we can consider their energies. In general, these would come from 
computations involving nuclear-electronic interactions, kinetic energies, etc. But the 
simple H¨uckel method takes the much easier approach of relating energy to bond order.
540 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-9 
 An alternative to the Bloch sums shown in Fig. 15-8c. These have the same coefficient 
wavelength but the wave is displaced by a/2. 
(See Section 8-9.) We can see at once that the MO at k = 0 is totally bonding. By 
analogy with the totally bonding MO of benzene, its energy is a + 2ß. The pair of 
MOs due to k=±p/2a are nonbonding. This is obvious from Fig. 15-8, but is also 
not hard to see in Fig. 15-9, which shows the MOs as having equal numbers of bonding 
and antibonding interactions. These MOs, then, have E = a. Finally, the totally 
antibonding MO at k =p/a has an energy of a -2ß. (Note that the pairing theorem 
holds. Compare the coefficients of this MO with those at k =0.) 
EXAMPLE 15-2 What is the energy of an MO in Fig. 15-8 having k=±p/4a? 
SOLUTION 
 The figure shows that these are triads of atoms having bonding interactions within 
themselves. Interactions between triads are nonbonding, since only nearest-neighbor interactions 
are taken into account in the HMO method. So E lies between a and a +2ß. Within a triad, the 
coefficient ratios (if we choose the cosine set) go as cos(-p/4), cos(0), cos(p/4) orv2/2, 1,v2/2. 
Normalizing for a triad gives 1/2, 1/v2, 1/2. Eq. (8-58) gives E =a +2ß(1/2v2+1/2v2)= 
a +1.414ß. This is the same as the result we obtained for the allyl radical in Section 8-6, which 
makes sense since, because they are nonbonding to their neighbors in this CO, they behave like 
isolated allyl systems.  
The H¨uckel p MO energies for regular polyacetylene can be plotted versus k. The 
result appears in Fig. 15-10 along with a plot of the density of states (DOS). Each carbon 
atom brings one p electron to the polymer, and these fill the lower-energy half of the 
MOs, so the highest occupied MO (HOMO) for the polymer is at E =a, |k|=p/2a. 
Because the CO energies are identical for |k| and -|k|, we can observe all the unique 
energies by plotting over the range 0=k =p/a. The k-range that produces all unique 
energies (rather than all the independent wavefunctions) is called the reduced first 
Brillouin zone (RFBZ). 
The term Fermi energy is often used in reference to electrons having the highest 
energy in the ground state of the system. Here, if we ignore the effects of thermal 
energy, the Fermi energy is theHOMOenergy. When there is a gap between theHOMO 
and LUMO energies, there is some disagreement in use of this term. Physicists, for 
good theoretical reasons, place the Fermi level midway between HOMO and LUMO 
energies; chemists often continue to equate it to the HOMO energy.
Section 15-7 An Example: Polyacetylene with Uniform Bond Lengths 541 
a – 2ß 
a + 2ß 
p /a 
a
0 0 0 
E E E 
Fermi 
energy
k 
(a) (b) (c) 
DOS 
– + 
COBO 
Figure 15-10 
 (a) Energy versus k for a chain of p AOs as calculated by the simple H¨uckel 
method with Bloch sums. (b) Sketch of the variation of density of states versus energy. (c) Sketch 
of the crystal orbital bond order (COBO) function. Solid line is bond orders between neighbors. 
Dashed line is bond orders between second-nearest neighbors (reduced by an arbitrary factor of five 
to reflect smaller extent of overlap between p AOs on second-nearest-neighbor atoms). (Adapted 
from Hoffmann et al. [1].) 
It is apparent that the energy span of the p band depends on the magnitude of ß and 
that this in turn should depend on a, the distance between the carbon atom unit cells. 
The simple H¨uckel method in fact assumes ß to be roughly proportional to AO overlap. 
(The extended H¨uckel method makes the assumption explicit.) This means that 
the p band of polyacetylene should become “wider” as the distance between carbons 
decreases. Standard terminology is to refer to the span between lowest and highest 
energies in a band as the band width, even though normal graphical representations 
show this as a vertical distance.) 
There is yet another graphical representation of the relationships within these MOs 
that is useful to chemists. This is a plot of the amount of Mulliken overlap population 
between AOs as a function of E. This quantity, called crystal orbital overlap population 
(COOP), allows one to see at a glance how the bonding interactions change in a 
band. The simple H¨uckel analog of COOP is the crystal orbital bond order (COBO). 
Figure 15-10c shows the COBO for the linear polyacetylene p system as a function 
of k. The lower-energy COs have a high degree of net bond order, but this falls off to 
become zero at E =a and then negative at higher energies. This is just what we know 
must be happening. The H¨uckel MO energies are, after all, directly proportional to 
bond order, so low energy must go with large positive net bond order, etc. Also, the 
pairing theorem holds, so the nearest-neighbor COBO curve must behave antisymmetrically 
through E =a. Although this particular COBO curve is not especially subtle, 
it is a good first example because we understand so well what it is telling us. 
A useful feature of COOP or COBO curves is that one has complete freedom as to 
which AOs (or groups of AOs) one can look at in this way. For example, the dashed 
line in Fig. 15-10c shows the overlap populations between p AOs on second-nearest 
neighbors in linear polyacetylene. (Even though all overlaps are formally assumed to 
be zero in the H¨uckel method, once we have the MOs we are at liberty to calculate 
the overlaps that actually exist between the AOs in these MOs.) We see that the 1,3 
interactions start out with positive overlap at low E, decrease to a negative value at 
E =a, and then rise to a positive value at high energy. This behavior is less obvious 
than the 1,2 COOP curve, though examination of the MOs at k = 0,p/2a, and p/a
542 Chapter 15 Molecular Orbital Theory of Periodic Systems 
in Fig. 15-8 makes sense of it. (COOP curves are often shown without an explicit 
numerical overlap scale because they are usually used for qualitative comparisons.) 
EXAMPLE 15-3 Compare Fig. 15-8(c) to Fig. 15-9 with regard to second-nearestneighbor 
COBO values. 
SOLUTION 
 The values should be equal since these are equivalent orbitals. It is easy to see that 
the value should be negative in Fig. 15-8(c) because half of the next-nearest-neighbor coefficient 
products are negative, and the other half are zero. In Fig. 15-9, they are all negative. Because the 
normalization constant in 15-8(c) is larger by a factor of v2, the sum is the same in each case. 
These are nonbonding orbitals, so their energies equal a. Fig. 15-8(c) shows that the corresponding 
second-nearest-neighbor COBO is indeed negative.  
The p system of polyacetylene is a convenient first example. It provides elementary 
examples of the unfamiliar (to chemists) band, DOS, and COOP plots. Also, the MOs 
are trivial to generate because they are identical to the Bloch sums of the basis functions. 
We have already seen similar behavior in the context of diatomic molecules, where the 
analog to a Bloch sum is a symmetry orbital (SO). (See Chapter 7.) In the case of H2, 
we saw that use of a minimal basis set ( 1sA and 1sB) gives two symmetry orbitals 
(1sA ±1sB) and that these are identical to the MOs for H2. However, when we go to 
larger basis sets (1s, 2s, 2px, 2py, 2pz on each center) we find that symmetry orbitals 
may no longer be the same as MOs—some MOs are mixtures of symmetry orbitals of 
like symmetry (e.g., the 2ssg SO mixes with the 2psg SO).We will now explore a case 
where the analogous mixing of Bloch sums occurs for a polymer. 
Consider what happens to our treatment of the linear chain of carbon atoms when 
we use an all-valence basis set (2s, 2px,y,z) on each carbon. Since our chain lies along 
the x-coordinate, the symmetries of the 2s and 2px AOs is s, while that for 2py and 
2pz is p. We can use the extended H¨uckel (EH) method to deal with this basis. At very 
large internuclear distances, interactions betweenAOs on different atoms vanish, so our 
energy diagram becomes the simple two-level picture for isolated atoms (Fig. 15-11a). 
As the atoms move closer together, bands of finite width begin to develop (Fig. 15-11b). 
At first, while the bands are still fairly narrow compared to the distance between E2s and 
E2p, we can describe them as almost pure s and pure p in character. That is, the MOs 
are essentially identical to Bloch sums of the basis set at this point. The 2s band rises 
in energy from a totally bonding set of 2s AOs to a totally antibonding set of 2s AOs. 
The 2py and 2pz bands (degenerate) behave similarly. The 2px band looks peculiar in 
that it drops in energy as k increases. Sketching out the Bloch sums for 2px at k =0 
and p/a (Fig. 15-12) shows why this happens. The 2px AOs in their original basis 
set arrangement are antibonding. The k =0 Bloch sum keeps them that way. At the 
other extreme, k=p/a and the Bloch sum reverses the sign of every second coefficient, 
making all of the interactions bonding. This is just the opposite of what happens for 
the other bands. We say that the 2s, 2px, and 2py bands “run up” as |k| increases and 
that the 2px band “runs down.” The simple rule is that bands run up when unit cell 
basis functions are symmetric across the unit cell (so they have phase agreement on 
the two ends) and run down for antisymmetric basis functions. Note that the 2px band 
is wider than the 2py, 2pz band because the 2px orbitals overlap more strongly due to 
their orientation. Note also how the DOS plot (Fig. 15-11c) reflects the degeneracy of
Section 15-7 An Example: Polyacetylene with Uniform Bond Lengths 543 
Figure 15-11 
 (a) Valence AO energies for an EHMO calculation of a linear chain of carbon 
atoms. (b) EHMO band diagram calculated at an interatomic distance of 220 pm. (c) DOS diagram 
and (d) COOP curve between adjacent carbons, for the same calculation. (Adapted from Hoffmann 
et al. [1].) 
Figure 15-12 
 2px and 2py basis sets at k =0,p/a. 
the 2py, 2pz (or 2pp ) band as a large area. Because the data leading to Fig. 15-11 are 
produced by the extended H¨uckel (EH) method, we see the characteristic magnification 
of antibonding effects (Chapter 10), leading to asymmetries in DOS and COOP curves. 
The variational method underlying EH calculations on this system evaluates interactions 
between various Bloch sums at each k value. Because the Bloch sums (BSs) for 
2s and 2px AOs have the same symmetry (s ), a nonzero hamiltonian matrix element 
may exist between them, permitting mixing. However, the higher energy of the 2px 
BS compared to the 2s BS discourages much mixing as long as the overlap between 
them is small. At large distances, the overlap is indeed small, so we see the relatively
544 Chapter 15 Molecular Orbital Theory of Periodic Systems 
“pure” bands of Fig. 15-11. Since the 2pp band disagrees in symmetry with the s BSs, 
it remains unmixed with them regardless of C–C distance. 
EXAMPLE 15-4 How should the areas compare in the DOS of 2s and 2p regions in 
Fig. 15-11? 
SOLUTION 
 To the extent that these regions remain “pure,” we should have three times more 
area in the p regions because, with three times as many basis functions, there are three times as 
many states. (The integral over DOS gives the number of states.)  
Recalculating the band diagram for all-valence linear carbon at a smaller C–C distance 
gives the band diagram in Fig. 15-13a. Several things have happened: The lowest 
band has dropped significantly in energy at k =0 and risen slightly at k =p/a, so it is 
wider. The p band has gotten slightly wider too. The highest band has shot up to very 
high energies and has undergone a noticeable change of shape: It has a large hump at 
intermediate k values that did not show at large C–C separation. What is responsible 
for these changes? 
Even without BS mixing we should expect band widths to increase as a result of 
increased overlap between AOs on adjacent carbons. Hence, at first glance the lowestenergy 
band could be simply a more spread out 2s band. This cannot be right, though, 
because the EH method on a pure 2s band would give an antibonding-caused energy 
rise at k =p/a that is larger than the bonding-caused energy drop at k =0. Instead, 
we are seeing a smaller rise. Something must be happening to either enhance the drop 
at k = 0 or counteract some of the rise at k = p/a. The DOS plots with 2s and 2ps 
portions shaded (Fig. 15-13b, c) indicate what is happening. We see that the lowest 
Figure 15-13 
 (a) EHMO band diagram for a linear chain of carbon atoms separated by 140 pm. 
(b) DOS plot, with 2s AO contributions shaded. (c) DOS plot with 2ps AO contributions shaded. 
(From Hoffmann et al. [1].)
Section 15-7 An Example: Polyacetylene with Uniform Bond Lengths 545 
Figure 15-14 
 Band diagram for a linear chain of carbon atoms showing some COs and intended 
correlations. Positive and negative phases are indicated by shading or lack of shading. (From 
Hoffmann et al. [1].) 
band is still primarily 2s for small k, but becomes mainly 2ps at larger k. The highest 
band shows opposite behavior. It is mostly 2ps at smaller k and 2s at larger k. Mixing 
is occurring between Bloch sums. 
Another way to see these features is to sketch the MOs that the computer program 
reports for various k values. These are shown for k=0,p/a in Fig. 15-14. It is obvious 
that the lowest band is all 2s at k =0 and all 2ps at k =p/a, whereas the highest band 
is exactly the reverse. The bands have switched character. The pure 2s band we were 
discussing, that should have gone to very high energy, is represented by a dashed line 
in Fig. 15-14, and the pure 2ps band by another, and these dashed lines do just what 
we argued they should. 
The variational program is figuring out, at each value of k, how to get the minimumenergy 
s band, minimum-energy p band, etc. and is not seeking to maintain 2s or 
2ps purity in the bands. Evidently, for the lowest band, it finds pure 2s to be lowest in 
energy at k =0 and pure 2ps to be lowest at k =p/a. At intermediate k values it finds 
a mixture to be best. 
The dashed lines in Fig. 15-14 are the bands for Bloch sums of pure 2s and 2ps AOs. 
They differ from the bands for variational COs, where the Bloch sums are mixed. Even 
though they are not real bands, these dashed lines are nevertheless useful for guessing 
in advance of calculation what a band diagram will look like. The dashed lines are 
sometimes referred to as intended correlations because they show where the band that 
starts out as 2s, for instance, “intends” to be at k =p/a. These two dashed lines are 
forced to cross at some value of k, but we have seen (Section 7-6) that such crossings
546 Chapter 15 Molecular Orbital Theory of Periodic Systems 
are not allowed for wavefunctions of like symmetry (i.e., Bloch sums that have nonzero 
overlap). Because the two dashed lines belong to Bloch sums of the same symmetry, 
the noncrossing rule prevents their crossing, forcing the alternative correlation scheme 
shown by the solid lines. We will see later that an easyway to construct some qualitative 
band diagrams is to draw them first for simple Bloch sums and then to reroute some of 
the lines to remove forbidden crossings.1 
15-8 Electrical Conductivity 
The all-valence band diagram of Fig. 15-14 still leads to the prediction that the p band 
is half filled. Many texts make the point that metallic conductivity results when there 
are empty MOs immediately above the Fermi energy in a bulk solid, so we should 
consider whether or not polyacetylene is a metallic conductor. This leads us into the 
topic of electrical conductivity in one-dimensional periodic systems. 
The essence of electrical conductivity is electron flux, and this requires net electron 
momentum. Consider the “band” for the one-dimensional free particle, shown in 
Fig. 15-1. If we had a band of this kind partially filled with electrons, the levels would 
be occupied from the lowest up to the Fermi level. For each occupied state exp(ikx) 
therewould be an occupied mate exp(-ikx). In short, each electron having momentum 
k ¯h would be balanced by one having momentum -k ¯h. Net momentum and electrical 
current would be zero. But if we apply a voltage, states exp(ikx) and exp(-ikx) lose 
their degeneracy. The state corresponding to motion in the direction of lower potential 
energy for the electron will now have lower energy than that for equal momentum in the 
opposite direction. As a result, some states that were originally above the Fermi level 
will move below it because they correspond to motion in the “right” direction, while 
others that were originally below the Fermi level will move above it. The electrons 
readjust their occupancies to fit the new scheme, and we now have more states occupied 
that correspond to momentum in the right direction and fewer corresponding to motion 
in the wrong direction. Electric current flows. Notice that this mechanism will work 
for very small applied voltages (metallic conductivity) only if there are empty MOs 
just above the Fermi energy. 
If the first empty levels are separated from the Fermi level by a modest energy gap 
(on the order of kBT , where kB is Boltzmann’s constant, 1.38×10-23J K-1), then a 
small population of electrons will exist in the “empty” orbitals at thermal equilibrium. 
This allows electrical conductivity (though typically much lower than that normal for 
metals) which increases with temperature—the situation in intrinsic semiconductors. 
If the gap is large, no conductivity occurs except at extreme voltages and the material 
is an insulator. 
The above discussion is based on a free-particle wavefunction. In real materials 
the potential along a coordinate is not constant, so the band structure becomes more 
complicated than the parabola of Fig. 15-1. Also, charge-density adjustments occur 
that tend to screen out the applied field. However the basic requirement for metallic 
conductivity continues to be the absence of a gap between the highest filled and lowest 
empty MOs. Therefore, if the p band of polyacetylene really is partly filled, pure 
1Apparent symmetry disagreement of bands for reflection through a plane perpendicular to the polymer axis 
and bisecting an atom or a bond does not lead to allowed crossing. This is explained in Section 15-13.
Section 15-9 Polyacetylene with Alternating Bond Lengths—Peierls’ Distortion 547 
polyacetylene should be a good (metallic) conductor. However, we will show next that 
there are other factors operating that always open up a band gap just above the Fermi 
level in one-dimensional systems, thereby preventing metallic conductivity. 
15-9 Polyacetylene with Alternating Bond 
Lengths—Peierls’ Distortion 
We now reconsider our assumption of uniform C–C bond lengths. Since the molecules 
leading up to polyacetylene (butadiene, hexatriene, etc.) have alternating long and 
short C–C bonds, it is reasonable to ask whether this variation disappears completely 
in the limit of the infinite polymer. 
We return to the p-only system at the simpleH¨uckel level. This allows us to continue 
treating the polymer as though it were a straight chain of carbon atoms (Fig. 15-15a). 
Since we are considering the possibility that the bond lengths alternate, we can no 
longer take a single carbon as our unit cell. (Thatwould require two different translation 
distances.) Instead we take two atoms, say bonded through the shorter distance, and 
let the translation distance, a, equal the sum of the two bond lengths (Fig. 15-15b). 
The band diagram undergoes a marked change in appearance when we change from 
a one-atom to a two-atom unit cell, quite aside from the fact that we have two bond 
distances. To show this, we will first work out the two-atom unit cell band diagram for 
equal bond lengths. 
The most obvious change comes in the range of k. Recall that -p/a <k = p/a. 
But our new choice of unit cell gives a new translation distance, a, equal to twice its 
previous value, so the numerical range of k is cut in half. 
Figure 15-15 
 (a) A section of a linear carbon polymer with alternating bond lengths. (b) Unit 
cell and translation distance. (c) Ungerade (u) and gerade (g) basis sets for a unit cell.
548 Chapter 15 Molecular Orbital Theory of Periodic Systems 
The other big change is in the basis set U. Before it was one 2pp AO on each carbon. 
Now, however, we must have two AOs for each unit cell, and so we need to consider 
the most useful combinations to choose as basis. Generally we seek choices that will 
minimize subsequent mixing of Bloch sums, because evaluation of such mixing requires 
variation calculations and a computer. Bloch sums will not mix if they disagree in 
symmetry (for operations for which the polymer hamiltonian is invariant). Clearly, 
inversion through the center of a C–C unit cell is a symmetry operation for the whole 
polymer, so let us choose the g and u combinations of 2pp AOs shown in Fig. 15-15c. 
Since these basis functions cannot mix without producing a polymer MO of mixed 
symmetry (which is not allowed unless the MOs are degenerate), we can take the Bloch 
sums of these bases as identical to the nondegenerate COs at least. 
We can now construct the band diagram. At k =0, our two basis functions give two 
Bloch sums, shown in Fig. 15-16a. The ungerade combination at k =0 is obviously 
totally bonding over the whole molecule, has energy a + 2ß, and is identical to the 
MO we obtained at k = 0 when we chose one carbon atom per unit cell. (Compare 
with Fig. 15-8.) The gerade basis function at k =0 is totally antibonding, has energy 
a-2ß, and is the same as what we previously had at k=p/a (with a equal to one C–C 
bond length). These, then, are the COs we previously indicated to be nondegenerate. 
When we shift to a equal to two C–C bond lengths, we find at k =p/a that the g and 
u bases produce the polymer MOs shown in Fig. 15-16b. These are easily seen to be 
nonbonding MOs (E = a) exactly like those in Fig. 15-9. When we start the band 
diagram at k =0, we find two energies (E =a ±2ß). The u basis has the same phase 
on the ends of the unit cell, so it runs up (from a +2ß) as k increases, ending up at 
E =a when k=p/a. The g basis has opposite phases on the ends of the unit cell, runs 
down from a -2ß as k increases, and finishes at E =a too, as shown in Fig. 15-17. 
The Fermi level is still at E =a, but this point now occurs on the right edge of the band 
diagram instead of at the midpoint. 
In going from a uniform chain of p AOs with one carbon per unit cell to an identical 
chain with two carbons per unit cell, we have not changed anything physically. We have 
merely changed our choice of representation. Therefore we must obtain the same MOs 
and energies either way, and it appears that we do, at least at the edges of the diagram. 
However, the appearance of the band diagram is affected strongly. The relation between 
Figure 15-16 
 Results of translating u and g basis functions through unit cell translation distances 
a, modulated by cos(kx) with k =0 and p/a.
Section 15-9 Polyacetylene with Alternating Bond Lengths—Peierls’ Distortion 549 
Figure 15-17 
 p-band diagram from using two-atom unit cell of Fig. 15-15 with (a) s=l, (b) s =l. 
the two band diagrams turns out to be very simple, though. Comparison of Figs. 15-17a 
and 15-10a shows that the band diagram for the two-atom unit cell is generated from 
that for the one-atom unit cell by simply folding the latter through a vertical line halfway 
across the diagram. There is always a smallest possible entity that can be chosen as 
the unit cell, and it is always possible to select any multiple of this for the unit cell and 
still generate the periodic structure. The band diagrams based on smaller unit cells can 
always be converted to those based on larger cells by the process of folding. 
Note that what was one p band in the earlier band diagram becomes two p bands 
when we double the unit cell size. We shall refer to the lower of these two as the p band 
and the other as the p* band, the asterisk indicating net antibonding character. Whatwas 
previously a degenerate level in the middle of the |k| range is nowat the right edge of the 
diagram, with each member belonging to a different band (though still degenerate). 
There is a subtlety that needs to be addressed at this point. All that has been said up 
to now suggests that no energy changes occur as a result of our doubling the size of the 
unit cell. This is misleading. The Bloch sums resulting at intermediate k values are not 
identical for corresponding states in these two representations. This reflects the fact that, 
as k varies for a single-carbon unit cell, the coefficient of the pp AOcan change from one 
carbon to the next. For the two-carbon unit cell, theAO coefficients are locked together 
in pairs. We cannot get a CO that looks like the one in Fig. 15-8b, for instance, using 
a Bloch sum based on the pu basis set in a two-carbon unit cell. Doubling the unit cell 
forces the Bloch sums to be a more “coarse-grained” approximation. What saves the day 
is that the variational procedure mixes together the p and p* Bloch sums at intermediate 
k values in precisely the manner needed to undo the result of this limitation. The lesson 
we take from this is that, if we seek to estimate band wavefunctions and energies from 
Bloch sums without variational modification, we make the smallest error if we use the 
smallest possible unit cell, since this gives the finest-grained first approximation. 
Now that we understand the band diagram for a two-atom unit cell, we can consider 
the effects of bond-length alternation. This is most easily treated as a perturbation on the 
uniform bond-length system, with half of the bonds getting longer and the intervening
550 Chapter 15 Molecular Orbital Theory of Periodic Systems 
half getting shorter. It will suffice to consider the effects of this perturbation to first 
order and to consider only the four MOs shown in Fig. 15-16. Looking first at the k=0 
MOs, we see that making half of the bonds longer will raise the energy of the bonding 
MO, but making the other half of the bonds shorter will lower the energy. Hence we 
expect little net energy change. The totally antibonding MO will likewise respond 
oppositely to each type of change, so the effects will again cancel. To first order, then, 
we expect the perturbation to have little effect on the left side of the p band diagram. 
On the right side, the situation is much more interesting. The MOs here (Fig. 15-16b) 
are degenerate (both nonbonding, E = a), so we must be wary. There is only one 
pair (among the infinite number of possible mixtures) that is “proper” for evaluating 
this particular perturbation. From Chapter 12 we know that the proper pair is that 
which responds most differently to the perturbation. Consider first the MOs shown in 
Fig. 15-8c. It is easy to see that neither of these would undergo any energy change as 
bond lengths change because there is no p AO overlap at all between nearest neighbors. 
On the other hand, the versions of these MOs shown in Fig. 15-9 or 15-16b respond 
strongly to the perturbation. One of them is bonding across the bonds being shortened 
and antibonding across the bonds being lengthened, and its energy will be lowered 
by both factors. The other is antibonding across the shortening bond, bonding across 
the lengthening bond, so its energy will rise. Clearly, this choice of MOs leads to the 
most different responses to this perturbation, so this is the prediction we use—that the 
endpoints of the curves at the right side of the band diagram will split apart, one going 
up and one going down (Fig. 15-17b). States at slightly smaller |k| are slightly less 
affected, etc., so the curves split apart over a range, not just at their termini. 
Will the polymer elect tomake this bond-length distortion? Yes, because the occupied 
MOs are affected only by energy lowering. All of the MOs of increasing energy are 
empty. This situation is analogous to what we saw in Chapter 8 for molecules having 
partly occupied degenerate MOs. In that context, the molecule undergoes what is 
called Jahn–Teller distortion. For infinite periodic materials, the phenomenon is called 
a Peierls distortion. 
We should consider whether the perturbation affects the other bands (s bands) in a 
manner to counteract the effect on the p bands. Although some splitting does occur 
for those bands too, the effect on overall energy is much smaller because these bands 
are either completely occupied, so that the effects of energy rise and lowering are both 
registered, or else are empty, so that neither effect registers. Thus, it is the partly filled 
band that determines the overall energy change due to the perturbation. 
We have been dealing with a half-filled band, but it is not difficult to show that a 
suitable distortion can be found to produce a gap at the Fermi energy for any fractional 
degree of filling. 
We arrive, then, at the conclusion that a one-dimensional periodic structure having 
a partly filled band is unstable with respect to a distortion in bond lengths that will 
produce a gap at the Fermi level. This means that polyacetylene cannot be a metallic 
conductor, though it might be a semiconductor if the gap is not too large. 
Finally, we must address the question of how any metal can be a conductor. How can 
sodium, say, have a partly filled band without a band gap? Why doesn’t the metal relax 
into a lower-energy structure having a gap at the Fermi energy? The answer is that, 
while one-dimensional systems can always find a distortion that has opposite effects on 
the energies of the MOs above and below the Fermi level, multidimensional systems
Section 15-10 Electronic Structure of All-Trans Polyacetylene 551 
cannot. A distortion that would work to split the MOs around the Fermi level according 
to their behavior in one coordinate does not necessarily work when we examine what is 
happening in the other coordinates. A kind of coincidence is needed that is not usually 
found except in crystals called quasi–one-dimensional. As a result, two- and threedimensional 
crystals with partly filled bands can exist without undergoing spontaneous 
structural reorganization and losing metallic conductivity. 
15-10 Electronic Structure of All-Trans Polyacetylene 
We finish our discussion of this polymer by examining the band structure for the 
nonlinear, all trans structure with alternating bond lengths (Fig. 15-18a) as calculated 
by the extended H¨uckel method. 
The EHMO band structure appears in Fig. 15-18b. There are 10 valence AO basis 
functions per unit cell (two carbons and two hydrogens) so we get 10 lines in the band 
diagram. Note that, since the chain is no longer linear, the 2py and 2pz bands are no 
longer degenerate. One of these is still perpendicular to the molecular plane, hence is 
still of p symmetry, but the other now lies in the plane and has s symmetry. 
60 
–30 
50 
40 
30 
20 
10
0 
–10 
–20 
s
s 
s 
s
s 
s 
s s 
s 
s 
p 
p 
p* 
p* 
H
C 
H 
H 
C 
C 
H
C 
(a) 
136 
144 
–4 
–6 
–8 
–10 
–12 
–14 
–16 
0 p/a 0 p/a 0 
k k DOS 
(b) (c) (d) 
Figure 15-18 
 (a) Unit cell for all-trans polyacetylene with alternating bond lengths of 136pmand 
144 pm. (b)Valence-band diagram calculated by the EHMO method. All bands have gaps at k=p/a. 
(c) Magnified band diagram showing region near the Fermi level. (d) Density of states in region of 
Fermi level. Dashed line shows p-band portion. (Parts a–c adapted from Hoffmann et al. [1].)
552 Chapter 15 Molecular Orbital Theory of Periodic Systems 
A gap resulting from bond-length alternation occurs at the right edge for each of 
the bands, though it is quite small in bands associated with MOs that have small C–C 
overlap. A density-of-states plot for the occupied valence s bands and the p bands 
shows the gap to be just above the highest-energy p-band level. 
With two exceptions, the EHMO band structure shown in Fig. 15-18a, b for alternating 
bond lengths is almost indistinguishable from that for uniform bond lengths 
(not shown). The only significant differences are that (1) the avoided crossing between 
s levels shown in 15-18b becomes a true crossing in the more symmetric, uniform bondlength 
case, and (2) all the gaps at k =p/a disappear for uniform C–C bond lengths. 
15-11 Comparison of EHMO and SCF Results 
on Polyacetylene 
So far our entire discussion of band theory has been in the context of H¨uckel-type models. 
It is possible to do band calculations using a self-consistent-field (SCF) approach 
wherein the contributions to one-electron energies resulting from kinetic energy, 
nuclear-electron attraction energy, and electron–electron repulsion and exchange 
energies are each explicitly calculated as described in Chapter 11. As pointed out 
there, the total energy is not the same as the sum of one-electron energies in such 
calculations, since this would double-count the interelectronic interactions. Also, 
the energies for virtual orbitals are not given the same physical interpretation as for 
occupied orbitals. For a calculation on a neutral n-electron system, the occupied MO 
energies are appropriate for an electron in the neutral system interacting with the n-1 
other electrons. The virtual MO energies are appropriate for an additional electron 
(electron n+1) interacting with the n other electrons (with the proviso that the original 
n electrons have not reacted to the presence of electron n+1). Thus, the virtual MOs 
refer to a different system (the unrelaxed negative ion) than the occupied MOs. We 
will see that this affects the nature of band diagrams. 
In contrast to SCF calculations, H¨uckel methods evaluate orbital energies entirely 
in terms of nodal behavior. Simple H¨uckel theory calculates bond orders and converts 
these to energies. As more and more nodes go into the MO, the bond orders drop 
from net positive through zero (nonbonding) to net negative, and the MO energies rise 
accordingly. Since overlap is neglected, the MO coefficients remain similar in absolute 
value over this range, so the high-energy antibonding MOs are elevated above zero 
energy as much as the bonding MOs are depressed below it, as the pairing theorem 
requires. The extended H¨uckel method is similar in spirit but evaluates the overlaps 
between AOs explicitly in order to arrive at interaction elements that are tailored to 
explicit AO types, the particular kinds of atoms involved, and the explicit distance 
between them. Importantly, this method does includeAOoverlap inMOnormalization, 
which leads to very much larger coefficients in the highly noded MOs and a resulting 
“inflation” on the high-energy end of the energy spectrum. However, because these 
high-energy MOs are usually not occupied with electrons, they do not affect predictions 
and can generally be ignored. In contrast to the SCF situation, there is no qualitative 
difference between the interpretation appropriate for an empty and a filled simple or 
extended H¨uckel MO. The lowest empty MO is simply an MO that is a little more 
antibonding than the highest occupied MO. If twoH¨uckelMOsare equally antibonding,
Section 15-11 Comparison of EHMO and SCF Results on Polyacetylene 553 
Figure 15-19 
 SCF band diagrams computed for all-trans-polyacetylene. (a) All C–C bonds 
1.39 Å. (b) Alternating bond lengths of 1.3636Å and 1.4292 Å. (From Andr´e and Leroy [2].) 
they have equal energy, even if one is occupied and the other is not, whereas this is not 
the case for two ab initio SCF MOs. 
We can see the results of these factors when we compare the band diagram for 
polyacetylene calculated by an SCF procedure with that from an EHMO program. 
Figure 15-19 shows the SCF bands resulting from all-trans polyacetylene with uniform 
and with alternating C–C bond lengths. Comparing these with the EHMO results 
(Fig. 15-18a and discussion in previous section), we see that the occupied valence 
bands agree quite well in the two approaches (the five lowest curves in Fig. 15-18a 
and the third through seventh lowest curves in Fig. 15-19b). This is important if these 
methods are to agree in predictions of relative energies of various polymer structures. 
The highest-energy band in theEHMOdiagram is much higher than in the SCF diagram 
as a result of the inflationary factor described above. This is not a problem as long as we 
do not try to use this band for predictions. The most important difference is that the SCF 
band diagram for uniform C–C distances shows a sizable p–p* band gap, while the 
EHMO method would show no gap there. The SCF gap gets larger when bond-length 
alternation is introduced, while the EHMO gap becomes finite, so the methods agree 
that bond-length alternation increases gap size. 
The apparent disagreement about the existence of a gap in the uniform polymer is 
really not a disagreement in light of what we have seen about the different ways these 
methods define orbital energies. The EHMO method indicates that the highest p and 
lowest p* COs have the same amount of overlap-induced bonding energy (the MOs are 
the same, merely differing in phase). The SCF method indicates that the least stable 
electron in the neutral polymer (at the top of the p band) is at lower energy than an 
electron can achieve in the best CO it can find in the (unrelaxed) negative ion (at the 
bottom of the p* band). The EHMO method is suggesting that the uniform polymer 
should be a conductor (if we could make it stay uniform). The SCF method is telling
554 Chapter 15 Molecular Orbital Theory of Periodic Systems 
us (via Koopmans’ theorem) that the energy needed to remove an electron from neutral 
polyacetylene is larger than the energy released when neutral polyacetylene gains an 
electron (without relaxing). It would be incorrect to use the EHMO band diagram for 
a Koopmans’ theorem analysis and say that the ionization energy and electron affinity 
of polyacetylene have the same absolute value. It would also be improper to argue 
that the large gap seen in this SCF calculation on uniform polyacetylene means such 
a system would not be a conductor. Both types of calculation have advantages and 
disadvantages, and these include their appropriateness for revealing various properties. 
We will continue to use H¨uckel methods in this chapter. 
15-12 Effects of Chemical Substitution on the p Bands2 
We have seen that Peierls’ distortion produces a band gap at the Fermi level in polyacetylene. 
Another type of change we can consider is replacement of one kind of atom 
in the unit cell with another. While this is not a modification that the polymer can make 
spontaneously, it is a way for us to predict what sort of changes in band structure would 
result if substitutional relatives of polyacetylene were synthesized. 
Consider what would happen if one C–H group in the two-carbon unit cell were 
replaced by N. In making this substitution, we lose a hydrogen atom and its 1s AO, 
so the all-valence band diagram would have 9 lines rather than 10. Also, we replace 
one carbon with a more electronegative atom having greater attraction for its electrons. 
At the simple H¨uckel level, this would lead to use of a coulomb integral for nitrogen 
that is lower in energy than that for carbon: aN· =aC +hN·ß. According to Table 8-3, 
hN· is 0.5. At the extended H¨uckel level, valence state ionization energies for nitrogen 
would replace those for carbon. Ab initio methods would account for the change by 
increasing the nuclear charge from 6 a.u. to 7. Because the results of all these methods 
agree qualitatively, we continue the discussion in terms of the simple H¨uckel method 
since all relevant calculations can be done mentally. 
We again start with uniform bond lengths and consider only the p electrons. It is not 
difficult to predict the changes, to first order, that will occur at k =0 when N replaces 
C–H. Referring once more to Fig. 15-16, we note that both the p and p* bands place 
equal amounts of p-electron density on every atom. Therefore, each energy should 
drop at k =0 by an amount reflecting the fact that each electron now spends half of its 
time in a nitrogen 2pAO. At the simple H¨uckel level this is easily seen to be an energy 
drop of 0.25ß. At k = p/a, we once again confront the question as to which of the 
infinite number of linear combinations of degenerate COs is the correct set to use in 
evaluating the substitutional perturbation. Both members of the pair we used to evaluate 
bond-length alternation (shown in Fig. 15-16b) place equal density on every carbon, so 
both COs would drop from a to a+hß/2 and remain degenerate. Maximum difference 
in response comes from using the pair shown in Fig. 15-8c because one of these COs 
places all of its density on the set of atoms that become nitrogen while the other places 
all of its density on the atoms that remain carbon. The first-order change predicted by 
the H¨uckel method, based on the principle of “maximum difference in response” from 
degenerate-level perturbation theory, is that one level drops to a +hß while the other 
remains at a. The results of these considerations are sketched in Fig. 15-20a. 
2This section follows closely the discussion of Lowe and Kafafi [3].
Section 15-13 Poly-Paraphenylene—A Ring Polymer 555 
Figure 15-20 
 (a) p band of polyacetylene (simple H¨uckel method) with alternating C–H groups 
replaced by nitrogen. (b) Same as (a) except that the remaining C–H groups have been replaced by 
an atom for which the correction factor h is half as large as that for nitrogen. 
We can go further and consider the effects of replacing the other C–H groups with 
something else. If we continue the perturbational treatment using the original polyacetylene 
functions, it is easy to see that now it will be the other band that will be 
affected at k =p/a (Fig. 15-20b). 
Since substitution removes the degeneracy at k = p/a, we now find the situation 
changed if we go on to consider alternating bond lengths. Whereas before we found 
that the uniform structurewould distort to open a gap, we nowfind a gap already present 
due to substitution. Furthermore, the COs now describing the nondegenerate states at 
the upper and lower edges of the gap are the ones that we saw earlier are not affected by 
bond-length changes. This means that the substitutional change strongly suppresses the 
tendency of the polymer to have alternating bond lengths. (To first order for degenerate 
states, perturbations will counteract each other unless they share the same set of proper 
zeroth-order wavefunctions. See Section 12-8.) 
There is a lot going on that we are ignoring. The C–N length is different from the 
C–C length, the COs change when we go to higher levels of perturbation theory, and 
H¨uckel methods are very approximate and become less reliable in systems with polar 
bonds. It should nevertheless be clear that, if one were looking for ways to predict the 
effects of chemical substitution on band gap, this sort of approach would provide the 
conceptual framework. 
15-13 Poly-Paraphenylene—A Ring Polymer 
We turn now to the p COs of poly-paraphenylene (PPP). This system serves as an 
excellent illustrator of principles introduced in earlier sections.3 
3This section follows closely the discussion of Lowe et al. [4].
556 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-21 
 (a) Two of the choices for the unit cell of PPP. (b) The p MOs of benzene. Shading 
indicates negative phase. (From Lowe et al. [4].) 
As its name indicates, PPP results when benzene molecules are attached in a planar 
chain, with each benzene being linked to two neighbors at para (opposite) positions 
(Fig. 15-21a). There is an infinity of ways we could select the repeating segment, or 
unit cell, for this polymer. Two of the most obvious are indicated in Fig. 15-21a, but 
any portion of length a (or na) would serve. Since selection of the unit cell determines 
the basis set of MOs we will use to construct the COs, we choose the unit cell on the 
left and start with the familiar p MOs of benzene, sketched in Fig. 15-21b. 
Two of the benzene MOs, f1 and f6, are nondegenerate. There are two degenerate 
pairs: f2,f3 and f4,f5. Our first task is to decide how these MOs will respond to the 
perturbation resulting from linking them to a pair of neighbors in an infinite chain. We 
will assess the interaction energies at the simple H¨uckel level, since the computations 
are so trivial that the reasoning is less obscured. Before the perturbation, then, these 
MOs are at energies Ef1 =a +2ß;Ef2,f3 =a +ß;Ef4,Ef5 =a -ß;Ef6 =a -2ß, 
where ß is a negative quantity.
Section 15-13 Poly-Paraphenylene—A Ring Polymer 557 
Figure 15-22 
 Crystal orbitals for PPP constructed as Bloch sums from f1 at |k|=0,p/2a,p/a. 
(From Lowe et al. [4].) 
We first consider the lowest-energy MO, f1. We want to know the energy that 
results when f1 is a component of COs in which |k| ranges from 0 to p/a. We can 
estimate this to first order by imagining that the MO does not undergo any change 
when it is incorporated into the CO. That is, it remains six 2pp AOs, all with the same 
coefficient (1/v6), just as in the molecule. When k =0, the MO repeats itself [times 
exp(ika)=exp(0)=1] for each translation through the distance a, giving the pattern 
shown at left in Fig. 15-22. An energy of a+2ß results from the unchanged bond order 
within each benzene, but there is now some additional bond order between molecules. 
Recalling that the bond order between AOs j and k in MO i equals cj,ick,i this is 
(1/v6)(1/v6)=1/6. SinceMOenergy involves 2ß times bond order, such newbonds 
are worth |ß|/3. Since there is one such new bonding interaction per ring, the energy 
of the CO at k=0 is lowered by |ß|/3 to a +2.33ß. Qualitatively, the MOs have been 
linked in a bonding manner so the first-order energy is lower than that for an isolatedMO. 
At k=p/a, the MO f1 is multiplied by exp(ika)=exp(ip)=cos(p)+i sin(p)=-1, 
for each translation through the distance a, resulting in the situation sketched at the 
right of Fig. 15-22. When we analyze this to first order, all is as before except that the 
additional interaction is antibonding, so the energy rises by |ß|/3 to a +1.67ß. At the 
halfway point, when k =p/2a,f1 oscillates half as fast, going through zero on every 
second unit cell, giving the middle sketch of Fig. 15-22. Obviously, f1 is not interacting 
at all with a companion on either side, so its energy remains unperturbed at a +2ß. 
The resulting CO energy curve, labeled p1, runs up (by 2ß/3 to first order) because the 
unit cell MO has phase agreement at the two points of attachment to the polymer chain. 
The analysis for f6 is identical except that there is phase disagreement at the points 
of attachment, no matter which opposite set of carbons we choose. This means that the 
band p6 runs down. Since the coefficients all have the same absolute value as in f1, 
the first-order interaction magnitudes are the same, so p6 runs from a -2.33ß at k=0 
through a -2ß at k =p/2a to a -1.67ß at k =p/a. 
Analysis of the degenerate MOs requires that we work with the proper zeroth-order 
versions. We need to ask, for each level, which orthogonal pair responds most differently 
to the perturbation resulting from linking onto para carbons. Alternatively, we 
can select a reflection plane that does not move the linkage sites and then seek the 
orthogonal pair having opposite symmetries for reflection through that plane. Such a 
plane is the one perpendicular to the benzene plane and intersecting both of the linking 
sites. Either of these approaches leads to the conclusion that the proper zeroth-order 
MOs are pairs such that one MO places a node at the linking carbons and the other
558 Chapter 15 Molecular Orbital Theory of Periodic Systems 
places an antinode there. Looking at Fig. 15-21b, we see that we can use these MOs 
as they are sketched provided we assume that the linking occurs at the carbons located 
at the 3 o’clock and 9 o’clock positions (i.e., atoms 1 and 4 in Fig. 15-21a). Once we 
ascertain this starting point for analysis, the rest is easy. COs p2 and p5, resulting from 
f2 and f5, are unaffected by the polymerization perturbation because they are zero at 
the linking sites. p3 runs down because theAO phases disagree at the linking sites, but 
p4 runs up by an equal amount because phases agree. p3 and p4 are wider bands than 
p1 and p6 because the coefficients at the linking sites are larger. In fact, they are v2 
times larger (see Appendix 6 for coefficients), so the bands are twice as wide. All of 
these considerations are brought together in Fig. 15-23. 
The first-order simpleHMOband diagram of Fig. 15-23 must be considered tentative 
because it shows two curve crossings and we have not yet considered whether these are 
allowed by symmetry. The noncrossing rule tells us that such crossings are forbidden 
if the COs agree in symmetry for all operations, so it is necessary to compare the 
symmetries of p2 and p3 and also of p4 and p5. Examination of the COs sketched in 
Fig. 15-23 makes it clear that, in each pair, there is symmetry disagreement for reflection 
through a plane perpendicular to the polymer plane and containing the polymer axis. 
(Call this s1.) Therefore, the crossings are allowed. 
Figure 15-23 
 First-order simple H¨uckel band diagram for PPP showing fragments of COs at 
|k| = 0,p/a. The curves are sketched between points computed only at |k| = 0,p/2a, and p/a. 
(From Lowe et al. [4].)
Section 15-13 Poly-Paraphenylene—A Ring Polymer 559 
One might select a different plane, say the one that is perpendicular to the polymer 
axis and bisects a benzene ring. (Call it s2.) This also appears to be a plane of symmetry 
disagreement between p4 and p5, but, surprisingly, this disagreement alone would not 
allow crossing. This subtlety arises from the existence of two degenerate COs for 
each value of |k| (except at 0 and p/a). For example, p4 at |k|=p/2a is multiplied 
by either +i or -i for each translation by a. For convenience in picturing COs, we 
take appropriate linear combinations to achieve the real forms that correspond to the 
+, 0,-, 0,+, 0,- pattern of a cosine wave and the corresponding 0,+, 0,-, 0,+, 0 
sinewave pattern, just as we did for Figs. 15-8b and c. These real pairs for p4 and p5 are 
shown in Fig. 15-24. Evidently, if one member of a pair is symmetric for the reflection 
s2, the other is antisymmetric. This is simply a manifestation of a general property 
of sine and cosine functions and obviously holds for all the bands at intermediate |k| values. Now we can see that the apparent symmetry disagreement we found between 
p4 and p5 for s2 reflection is only part of the story and that, if one of the real p4 COs 
disagrees with a real p5 CO in symmetry, the other real p4 must agree with it. This in 
turn means that, when we convert back to exponential versions, which are sine-cosine 
mixtures, there will always be some symmetry agreement between p4 and p5, so the 
bands still cannot cross (unless there is some other symmetry disagreement). Therefore, 
symmetry with respect to a reflection perpendicular to the k axis is not useful in bandcrossing 
analyses. (A more general approach to such analyses is to assign symmetry 
p'4 
p4 
k 
k 
–p 
p 
2a 
2a 
s2 
s2 
s2 
s2 
s2 
s2 
s1 
s1 
s1 
s1 
s1 
s1 
s1 
s1 
+
+
+ 
+
– 
–
–
– 
(a) 
(b) 

 
 

 
 
/ 
/ 
= 
= 
p'5 
p5 
k 
k 
–p 
p 
2a 

 
 2a 

 
 
/ 
/ 
= 
= 
Figure 15-24 
 Sine and cosine versions of two crystal orbitals of PPP. (a) p4 at |k| = p/2a. 
(b) p5 at |k|=p/2a. s1 and s2 indicate where reflection planes intersect the molecular plane. (From 
Lowe et al. [4].)
560 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-25 
 EHMO p-energy bands in eV for (a) PPP, (b) PPP-N2. (c) PPP-N2N6. All C–C 
and C–N bond lengths are set at 1.40 Å, all angles at 120.. (From Lowe et al. [4].) 
labels to the COs, which effectively reveals symmetry agreement or disagreement for 
all symmetry operations of the polymer.) 
A variational EHMO band calculation results in the diagram of Fig. 15-25a. There 
is much similarity with the simple H¨uckel first-order band diagram of Fig. 15-23, but 
there are also some differences. The p2 and p5 lines are not exactly horizontal, so 
these bands now have finite width. This results from the ability of AOs separated by 
two or more bonds to interact weakly in EHMO calculations. The p4 and p6 bands are 
much wider than p1 and p3, a result of the inflation at high energies characteristic of 
EHMO calculations. The CO coefficients indicate that some mixing of Bloch sums has 
occurred. For instance, in p1 at k =0 not all carbon coefficients are equal: The ones 
on the linking para carbons are larger. Variational mixing has managed this because it 
produces more overlap population and therefore a lower minimum energy root. Finally, 
the degeneracies at k = p/a of p1 and p3 and also of p4 with p6 are lost. The gap 
between p4 and p6 results from an avoided crossing. We can tell that this is the case 
because the EHMO-computed wavefunction for p6 at k = p/a, when sketched (not 
shown), looks like the second-highest CO sketched at the right of Fig. 15-23. That is, 
it is made up of f4-type monomer MOs, so it is the “intended correlation” point for 
p4. But p4 and p6 are both symmetric for s1, hence cannot cross. On the other hand, 
examination of the CO coefficients at k =p/a for p1 and p3, shows that this gap does 
not result from an avoided crossing. These bands merely fail to meet as they run up or 
down to their intended correlation points. 
The effects of chemical substitution on the band structure of this system can be analyzed 
using perturbation arguments, just aswas done earlier. Consider the effect, to first 
order, of replacing the C–H group at position 2 (see Fig. 15-21a) with a nitrogen atom
Section 15-13 Poly-Paraphenylene—A Ring Polymer 561 
Figure 15-26 
 p1 and p3 at |k|=p/a, shown at left as Bloch sums from the benzene basis, are 
degenerate and mix under the influence of a perturbation at position 2 to produce proper zeroth-order 
functions p
(0) 
a and p
(0) 
b . (From Lowe et al. [4].) 
to form what we will code as PPP-N2. Because of nitrogen’s greater electronegativity 
(more negative valence-state ionization energy) we expect any band’s energy to be 
lowered at each k value by an amount proportional to the amount of electronic charge 
the band places at position 2. All of the COs place some charge at position 2 when 
k =0 (see Fig. 15-23), so all of these band energies should drop on the left side of the 
band diagram. It appears that all the COs place charge at position 2 for k =p/a also, 
but now we have degeneracies (or near-degeneracies) to consider. We need to discover 
the linear combinations of degenerate COs that lead to zeroth-order COs having the 
greatest difference in their responses to the perturbation. Clearly it should be possible 
to mix p3 and p4 to achieve cancellation at position 2. The orthogonal mate for this 
new CO is correspondingly larger at position 2 because the charge placed at any site 
by this pair of COs must not vary with CO mixing. Figure 15-26 shows the results of 
mixing p1 and p3 to produce p(0) 
a and p(0) 
b . Note that p(0) 
a is zero not only at position 2 
but also at position 6, due to the symmetries of p1 and p3 with respect to a s1 reflection. 
We expect, then, that p(0) 
a will be unaffected by substitution at position 2, whereas 
p(0) 
b will be strongly lowered in energy. A similar analysis results for COs p4 and p6. 
Looking at the results of an EHMO variational calculation on PPP-N2 (Fig. 15-25b), 
we find that these expectations are borne out: All band energies at k=0 are lowered, as 
are all but two at k=p/a. These two exceptions (p2 and p6) are essentially unaffected 
by nitrogen substitution at position 2. 
Notice that the p2–p3 and p4–p5 crossings in PPP become avoided crossings in 
PPP-N2. This results from the loss of s1 reflection as a symmetry operation for the 
system when nitrogen is substituted at position 2. Whereas these COs can have different 
symmetries for s1 reflection in PPP, this is not possible for PPP-N2. (There is only 
one point-symmetry operation for PPP-N2, namely a reflection through the molecular 
plane. All p COs are antisymmetric for this. Hence no crossings are possible among 
p bands.) 
When we replace yet another C–H (at position 6) with nitrogen, giving PPP-N2N6, 
we find (Fig. 15-25c) that the band energies all move down some more in energy except 
for the two that did not move before, because, as we saw, the corresponding COs 
for these two are also zero at position 6. This substitution restores s1 reflection as a 
symmetry operation, so the p4–p5 crossing recurs. Accidental degeneracy now occurs 
for p2 and p3 at k =p/a.
562 Chapter 15 Molecular Orbital Theory of Periodic Systems 
15-14 Energy Calculations 
Suppose we wish to compare the calculated energy of regular polyacetylene to that of a 
bond-alternating version, or that of PPP with PPP-N2. Howdo we get energies from the 
band calculations for these two cases? In a H¨uckel-type calculation of a molecule, we 
simply add up the one-electron energies. In a band calculation we are faced with a very 
large set of one-electron energies—one for each k value. How do we deal with this? 
Consider again the simple H¨uckel band diagram for regular polyacetylene 
(Fig. 15-17a). This band diagram indicates that electrons in the p CO at k =0 have an 
energy of a +2ß, those at k =p/a have E =a, and those at k =p/2a have an energy 
somewhat below a +ß. It is fairly obvious that the CO average energy is somewhat 
below a + ß. Each CO has two electrons delocalized over the polymer, but the net 
number of electrons per unit cell is two. Therefore, with an average p CO energy of 
below a +ß and two p electrons per unit cell, we have a p energy per unit cell below 
2a +2ß. To be more precise would require a more precise average energy for the CO. 
[Even without such a calculation, however, we can see that bond alternation causes the 
average energy of the occupied p band to drop (Fig. 15-17b), so the p-electron energy 
per unit cell is lower for the alternating structure.] 
The problem, then, is to calculate an accurate average energy for each of the filled 
bands, the average being over the first Brillouin zone, and then multiply each such 
average energy by the number of electrons in that band, per unit cell. The sum of these 
is the H¨uckel total energy per unit cell. (This procedure assumes that the bands are not 
partially filled.) 
The practical difficulty in doing this is that each variational calculation is carried 
out at a single point in k space. Thus, for the variational bands sketched in Fig. 15-23, 
an EHMO calculation was made at k =0, another was made at k =p/2a, and a third 
was made at k = p/a. Then lines were drawn to connect the energy points found at 
these k values, with consideration being given to the noncrossing rule. Obviously, this 
produces a rather qualitative diagram. To refine it would require making additional 
variational calculations at intermediate points in k space. Thus, we must be concerned 
with the trade-off between accuracy of final average energy and effort needed to achieve 
well-characterized band energies from which to calculate that average. 
Achieving accurate average CO energies from values at a few k points is possible if 
those points are sensibly chosen and if appropriate weighting factors are employed. The 
problem of choosing the correct fewpoints and their weight factors has beenworked out 
by Chadi and Cohen [5] for multidimensional systems of various symmetries. Although 
this is a matter of real practical interest, we will not explore it further here. 
15-15 Two-Dimensional Periodicity and Vectors 
in Reciprocal Space 
In one-dimensional problems there is a single translation direction, a single step size 
a, and a single accompanying variable k. It is a simple matter in such problems to treat 
a and k as scalars and to plot band energies versus k in the range of the first Brillouin 
zone, -p/a<k=p/a. Step size a has dimensions of length and k of reciprocal length 
to give an argument for exp(ikna) that is dimensionless.
Section 15-15 Two-Dimensional Periodicity and Vectors in Reciprocal Space 563 
Figure 15-27 
 (a)Atwo-dimensional rectangular lattice (solid points) showing translation vectors 
a1 and a2 as well as reciprocal vectors b1 and b2. (b) The reciprocal lattice (hollow points) with 
the first Brillouin zone outlined by dashed lines. (c) The first Brillouin zone with the reduced first 
Brillouin zone (RFBZ) shaded. 
In multidimensional systems we must keep track of k values forwaves oriented along 
different directions, and we need to calculate each CO energy at a point corresponding 
to k values for each of these directions. This makes treatment of such problems 
considerably more complicated. It is convenient and conventional to handle all this in 
terms of vectors. We will outline the vector treatment here, using the two-dimensional 
rectangular lattice shown in Fig. 15-27a as an example. 
For a three-dimensional crystal, we label the translation vectors for a unit cell in 
real space a1, a2, and a3. Any point r in the unit cell can then be expressed in terms of 
vectors a. For the vectors in reciprocal space, we use the symbols b1,b2, and b3. The 
position k in reciprocal space can then be expressed in terms of vectors b. Vectors a 
have dimensions of length, and b of reciprocal length. 
The vectors b in reciprocal space are defined according to the formula 
bi =2p(aj ×ak)/(ai · aj ×ak) (15-18)
564 Chapter 15 Molecular Orbital Theory of Periodic Systems 
This makes each vector bi orthogonal to the plane of the “other” two a vectors, aj and 
ak. In the case of a two-dimensional crystal, the missing vector ak is assumed to be a 
unit vector perpendicular to the plane of ai and aj . The denominator of Eq. (15-18) 
is a scalar with the same value in all three cases, so the length of bi is greater if the 
vector product of aj and ak is greater, i.e., if |aj ||ak|sin f is greater, where f is the 
angle between aj and ak. In the example of Fig. 15-17a, all angles are the same, so 
the fact that a1 is twice as long as a2 makes b1 half as “long” (in reciprocal distance) 
as b2. The “length” of bi is 2p/|ai |, which means that these vectors are of appropriate 
dimension to define edges for the first Brillouin zone. However, we will see that they 
need some adjustment in position. 
As is indicated in Fig. 15-27a, moving a lattice point through translations a generates 
the real, or direct lattice. Moving a point in reciprocal space through translations b 
generates the reciprocal lattice. Figure 15-27a and b indicate that a real lattice with 
orthogonal vectors a results in a reciprocal lattice with orthogonal vectors b but that 
the reciprocal lattice is rotated with respect to the real one. When the vectors a are not 
orthogonal, the reciprocal lattice may look quite different from the real one. 
According to Fig. 15-27b, the zone defined by vectors b runs from 0 to 2p/|a|. 
We seek an FBZ that runs from -p/|a| to p/|a| in each direction. We can arrange 
this by drawing lines connecting one reciprocal lattice point to all its nearest neighbors 
and then bisecting all these lines with perpendicular bisectors. This gives the 
rectangle bounded by dashed lines in Fig. 15-27b. In essence, we are finding a cell 
in reciprocal space that fills up that space when translated by steps b and that also 
associates every point in reciprocal space with the reciprocal lattice point to which it 
is closest. This is sometimes called a Wigner–Seitz unit cell. It is also the FBZ for 
the crystal. In the example at hand, the FBZ looks like the zone defined by the b 
vectors, simply shifted, but in more complex systems, these may look quite different. 
Generally, the FBZ continues to have the symmetry of the real crystal. Thus, both the 
direct lattice and the FBZ in Fig. 15-27 are symmetric for reflections in the xz and yz 
planes. 
We have seen that one-dimensional systems have degenerate COs for equal values of 
|k| and so, if we wish to portray only the unique energies of the system, we need consider 
only the range from 0 to p/a. The analogous situation in two or three dimensions 
is that symmetrically equivalent positions k give degenerate COs. Hence we need 
consider only a symmetrically unique portion of the FBZ—the reduced first Brillouin 
zone (RFBZ). Since the FBZ is symmetric for the reflections mentioned above, the 
RFBZ is the quadrant shown in Fig. 15-27c. 
It is standard to use the label 
 for the central point of the FBZ. Certain other unique 
points are often labeled with symbols such as K,L,M, and X. These points often 
generate waves having the same periodicity as the crystal. 
If we choose a basis of one s-typeAO on each real-space lattice point, then the COs 
resulting from the special reciprocal-space lattice points labeled in Fig. 15-27c appear 
as shown in Fig. 15-28. The point 
 has coordinate intercepts (or k1, k2 values) of (0,0), 
so exp(ik ·na)=1 and the sAO is translated without change from point to point, giving 
the totally bonding CO of Fig. 15-28a. Point X corresponds to (0,p/a2), meaning that 
no phase change occurs as we move through the direct lattice by steps of a1 but that 
the phase reverses for each step by a2 (Fig. 15-28b). Point M is for (p/a1,p/a2), so 
phase reversal occurs for a step by either a1 or a2, resulting in the totally antibonding
Section 15-16 Periodicity in Three Dimensions—Graphite 565 
Figure 15-28 
 Bloch sums based on an s-type AO at each lattice point in the two-dimensional 
rectangular lattice. (k1, k2)= (a) (0,0), (b) (0,p/a2), (c) (p/a1,p/a2). 
situation in Fig. 15-28c. The waves for points X and M can be seen to indeed have the 
periodicity of the lattice. 
Comparison of these three COs leads us to expect lowest energy at 
, higher energy 
at X, and highest energy at M. A complete map of the CO energies would require 
a surface lying over the RFBZ. Instead of showing this surface, it is standard to plot 
the energy as a function of various straight-line paths connecting special points—e.g., 
from 
 to X, from X to M, from M to X, from 
 to M. Such line plots are portrayed 
side by side on a common k axis, as in Fig. 15-29, even though the identity of k (k1, k2, 
or some combination) changes from panel to panel. 
15-16 Periodicity in Three Dimensions—Graphite 
Graphite is an important commercial material. Some of its uses derive from the fact that 
it is a very good conductor of electricity—nearly as good as metals. The most stable 
crystalline form of graphite (Bernal graphite) is depicted in Fig. 15-30a. It comprises
566 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-29 
 Energy versus k along various paths in the RFBZ for the system of s-type AOs in 
a rectangular two-dimensional lattice. 
carbon atoms in two-dimensional sheets having hexagonal symmetry. These are stacked 
so that half of the carbons in a sheet lie directly between carbons in adjacent sheets 
while the other half lie between empty hexagon centers. The atoms within a sheet are 
strongly covalently bonded to their neighbors, with nearest-neighbor bond lengths of 
1.42 Å. The sheets are separated by 3.5 Å, indicative that the forces between them are 
very weak. (This explains why they fracture into sheetlike fragments that can slide 
over each other like a pile of playing cards, making ground graphite a good lubricant.) 
A sensible approach to understanding the electronic structure of graphite, then, is to 
first analyze the two-dimensional sheet and then consider the perturbation involved in 
forming layers of sheets. 
A section of a two-dimensional sheet, the translation vectors a1 and a2, and the 
primitive unit cell are sketched in Fig. 15-30b. The primitive unit cell contains two 
carbon atoms. The angle between a1 and a2 is 120.. Notice that the lattice points in real 
space, at the vector origin and at the termini of integral numbers of steps from the origin, 
fall at the centers of hexagons, where there are no atoms. This illustrates that there is no 
particular identification of direct lattice points with atoms or other structural features of 
the molecules that constitute a crystal. In the present instance the direct lattice, shown 
in Fig. 15-30c, looks like a graphite crystal that has been rotated by 30. with an extra 
point in the center of each hexagon. In general, even though the direct lattice need not 
look identical to the crystal, it will have the same symmetry as the crystal, in this case 
hexagonal. In Fig. 15-30d is shown the relation between the vectors a and the reciprocal 
space vectors b. Since b1 must be perpendicular to a2, etc., it follows that the angle 
between b vectors is 60.. The reciprocal lattice generated by the b vectors appears 
in Fig. 15-30e and is again a centered-hexagon pattern, but rotated with respect to the 
direct lattice. The FBZ is constructed by drawing lines from one reciprocal lattice point 
to all its nearest neighbors and then cutting them with perpendicular bisectors. This 
yields a hexagon, as shown in Fig. 15-30f. Also shown is the RFBZ. The entire FBZ 
can be filled by putting the RFBZ through the symmetry operations of the hexagonal 
sheet, so it includes all the points not equivalent by symmetry. 
Points of special interest in the RFBZ are labeled 
,M, and K. The point 
 corresponds 
to the (k1, k2) values (0,0). This is a unique point in the FBZ, so it produces 
one nondegenerate Bloch sum for each basis function in the unit cell. The point M
Section 15-16 Periodicity in Three Dimensions—Graphite 567 
a1 
a2 
a2 
a1 b1 b1 
b b2 2 
(a) 
(c) (d) 
(b) 
(e) 
Figure 15-30 
 (a) The crystal structure of Bernal graphite. Each point represents a carbon atom. 
(b) Top view of a hexagonal sheet of graphite showing the unit cell and translation vectors for the 
two-dimensional structure. The “origin point” of the vectors does not correspond to an atom location. 
(c) The direct lattice produced by translations of the “origin point” of the a vectors. The full lines 
emphasize that this is a lattice of centered hexagons. The graphite structure is indicated with dashed 
lines. (d) Reciprocal lattice vectors b are shown in relation to vectors a. (e) The reciprocal lattice, 
produced by translating the origin point of the b vectors. (f) Construction of the FBZ by bisecting 
lines connecting one reciprocal lattice point to all of its nearest neighbors. The RFBZ is shaded. 
(g) Geometry of the relation between point K and the vectors connecting it to the origin. 
corresponds to (p/a, 0). We might expect this to be a six-fold degenerate point because 
it is one of six symmetrically equivalent points in the FBZ. However, the point diametrically 
opposite this in the FBZ, at (-p/a, 0), does not produce a separate independent 
Bloch sum because, if we combine these complex functions to form a real (cosine)
568 Chapter 15 Molecular Orbital Theory of Periodic Systems 
wave, the other real (sine) wave puts a node at every real lattice point, meaning that 
the basis function is multiplied everywhere by zero, which is not acceptable. This is 
similar to what we saw in the one-dimensional case, which led us to exclude -p/a 
from the range of k. So here again we recognize that two points on opposite sides of 
the FBZ where perpendicular bisectors intersect lines between reciprocal lattice points 
really only account for one independent state. This means that the point M represents 
three independent Bloch sums rather than six. We shall have more to say about this 
later. The point labeled K does not lie on a b vector, so must be expressible as a sum 
of two vectors. Figure 15-30g shows the relevant construction. Since the lowest ray 
has length p/a, the central ray has length p/(a cos 30.). Then the length of k1 or k2 
must be p/a(2 cos2 30.), or 2p/3a. So (k1, k2) = (2p/3a, 2p/3a) at K. As will be 
explained later, K does not have reduced degeneracy as M does. 
Since there are two carbon atoms per unit cell and four valence AOs per carbon, 
we have eight basis functions per unit cell for a minimal valence basis set calculation. 
Therefore, we expect eight bands. Since the system is planar we can distinguish two p 
AOs and six s AOs. We know that these two symmetry types will not interact for any 
k, so we can predict or interpret the s and p band behaviors independently. 
Let us begin by making some predictions about the two p bands. As was the case 
for polyacetylene, the two basis sets that are simplest to work with are the bonding (pu) 
and antibonding (pg) combinations in a unit cell. We shall label the band resulting 
from the bonding function p and the other band p*. At the point 
, the unit cell basis 
functions are simply translated along a1 and a2 with no change in phase, yielding the 
COs sketched in Fig. 15-31a. The p CO is bonding in every bond, p* is antibonding, 
so these will be widely separated in energy. Indeed, there is no way they could become 
more widely separated by intermixing, so we can assume that, if we were to do a 
variational calculation, these Bloch sums would not be modified. 
At point M, translation by each step along a1 is accompanied by phase reversal 
because k1 = p/a1, whereas translation along a2 involves no change because k2 = 0 
(Fig. 15-31b). Nowwe find each carbon to be bonding to two neighbors and antibonding 
to the third in the p CO and the reverse in the p*CO. Once again, it is not hard to see 
that any mixing of these two would decrease their energy difference (Problem 15-24), 
so these Bloch sums should be unchanged in a variational calculation. Therefore, the p 
and p* bands should be separated in energy at M, but only by about one third as much 
as at 
. As we mentioned above, the COs sketched in Fig. 15-31b are each degenerate 
with two other COs corresponding to symmetrically equivalent points in the FBZ, 
for example, the points (0,p/a) and (p/a,-p/a). Sketching these (Problem 15-25) 
shows that they are like the CO already sketched except rotated by ±60.. 
Point K is more complicated to analyze because it has k values that do not produce 
COs having the periodicity of the crystal. This means that the decreased degeneracy 
we find for high-symmetry points like X in Section 15-15, caused because one of the 
real coefficient waves has a node at every unit cell, does not occur at K. (If the wave 
lacks the periodicity of the crystal, it cannot put a node at every unit cell, by definition.) 
That means we will have two p and two p* solutions at K. Also, the fact that these 
waves lack the periodicity of the crystal makes them more complicated to sketch. 
Nevertheless, undaunted, we plunge ahead. Since K corresponds to (2p/3a, 2p/3a), 
we expect steps along either a1 or a2 in real COs to be modulated by either cos(2np/3) 
or sin(2np/3). For n=0, 1, 2, 3, this gives the following repeating series for the cosine
Section 15-16 Periodicity in Three Dimensions—Graphite 569 
Figure 15-31 
 p and p* COs at the point (a) 
, (b) M, (c) K. 
and sine cases, respectively: 1,-0.5,-0.5, 1 and 0, 0.866,-0.866, 0. The results 
of using these factors to modulate translations of the pu and pg basis functions are 
sketched in Fig. 15-31c. Inspection of these sketches shows that the extent of bonding 
and antibonding interactions around any carbon exactly cancels in every case. This 
means that the p and p* bands are degenerate at K. Each of the pairs of COs sketched 
in Fig. 15-31c is degenerate with two other pairs that are rotated by ±60.. Therefore, 
the degeneracy of this energy level is 12. 
The band diagram for two-dimensional graphite is sketched in Fig. 15-32 for a circuit 
around the RFBZ. The p and p* bands behave essentially as we anticipated, being 
widely separated at 
, much less separated at M, and degenerate at K. We note that 
there are three s bands at lower energy and three more at higher energy, corresponding 
to predominantly bonding and antibonding situations, respectively. However, these 
bands are less easily analyzed because the 2s, 2px, and 2py AOs undergo varying extents 
of mixing with each other as k varies. That is, these are bands where the variational 
method causes extensive mixing of Bloch sums. Nevertheless, we can see patterns that 
would not be difficult to investigate. For example, we can see that the p band and one 
of the s bands run up from 
 to M, whereas the other two s bands run down. This is 
consistent with the fact that the 2s and 2pp AOs relate to correspondingAOs in adjacent 
unit cells in one way, while 2ps AOs relate in the opposite way (Problem 15-28).
570 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-32 
 Valence-band diagram for two-dimensional graphite. (Adapted from Painter and 
Ellis [6].) 
There are eight valence electrons per unit cell in neutral graphite. For the band 
diagram of Fig. 15-32 it is simple to figure out the band-population scheme. Eight 
electrons require four bands, and we see that the four lowest-energy bands lie below 
the other four bands for all values of k. This means that the four lowest-energy bands 
are filled and the others are empty. Therefore, the Fermi level comes at the top of the 
p band, which occurs atK. We see that there is an empty band (p*) at the same energy, 
so there is no gap at the Fermi level. Thus the two-dimensional lattice is predicted to be 
a good electrical conductor. We could examine the possibility that lattice deformation 
would split the levels to produce a gap (a Peierls distortion), but, since we have structural 
information indicating that graphite does in fact have hexagonal symmetry, we will not 
pursue that point. 
If there were crossing of one of the s* COs with the p CO over part of the range of 
k, then some fraction of the electronic population would occupy the low-energy portion 
of the s* band at the expense of the high-energy portion of the p band. Analysis of this 
would require more effort, but the result would still predict good conductivity since, 
instead of a full band “touching” the bottom of an empty band, we would have two 
partially filled bands. 
Next we consider the effects of stacking two-dimensional sheets to form the threedimensional 
crystal.4 As Fig. 15-30a indicates, the stacking pattern repeats the orientation 
of a sheet after one intervening layer in an ABABAB stacking pattern. This 
means that the three-dimensional unit cell must contain carbon atoms from two layers. 
The unit cell now has three associated translation vectors. The two “intrasheet” translations 
a1 and a2 are as before, and the “intersheet” translation a3 is perpendicular to 
the planes of the sheets and 7.0Å long. Because a3 is the longest vector in real space, 
b3 is the shortest vector in reciprocal space, leading to a reciprocal lattice where sheets 
4A discussion of effects due to other stacking arrangements can be found in LaFemina and Lowe [7].
Section 15-16 Periodicity in Three Dimensions—Graphite 571 
of centered hexagons are layered with short intersheet distances. The resulting FBZ, 
produced by perpendicular bisecting planes (since we are now working in three dimensions), 
appears in Fig. 15-33a. The RFBZ is similar to that in Fig. 15-30f except that 
it now has a third dimension, over which k3 ranges from 0 to p/a3 (Problem 15-26). 
b1 
b2 
a3 
b3 
a1 
L 
M
H
K 
p 
a1 
3a1 
3a1 
3a2 
3a2 
p 
2p p 
2p 2p 
2p 
a3 
a3 
p 
0
0 0 
A 
a3 
p 
G(0, 0, 0) 
0 0 
0 
RFBZ 
(a) 
(b) 
, , 
, ,
, , 
, , 
, , 
Figure 15-33 
 (a) The FBZ and RFBZ for three-dimensional Bernal graphite. (b) View of the 
unit cell for Bernal graphite, as seen along an axis parallel to the hexagonal sheets. (Compare to 
Fig. 15-30a.)
572 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Let us see if we can predict the modifications in energies of COs that should occur 
at some of the special points in the RFBZ as a result of intersheet interactions. Because 
three-dimensional Bernal graphite is symmetric for reflection through the plane of a 
hexagonal sheet, we continue to have a valid distinction between s and p COs. We 
expect intersheet interactions to be small, so we can take a perturbational approach. 
The interactions involving p COs should be largest since pp AOs project farthest from 
the plane, so we will restrict our attention to the p bands. 
A side view of the unit cell appears in Fig. 15-33b. Obviously, some of the intersheet 
interactions are already included in the unit cell, and the others will occur between unit 
cells. We will first consider the energy changes due to new interactions within the cell. 
These do not depend on k values. Then we will consider energy changes caused by 
stacking the unit cells. These depend on k3. We will ignore all interactions except those 
between carbons directly above or below each other in adjacent sheets. 
We begin with the situation where the p and p* bands are nondegenerate, e.g., at 
 
or M. We know from our two-dimensional treatment that each covalently bonded pair 
of carbons in the unit cell will have a pu MO when we are dealing with the p band and 
a pg MOfor the p* band. (See Fig. 15-31a, b.) But we have a choice of arranging these 
in the unit cell to give long-range bonding or antibonding between layers (Fig. 15-34). 
These two arrangements cause the energies to rise or fall slightly (indicated by E . or 
E .), splitting both the p and p* bands. 
Next we examine what happens when we stack these unit cells. Consider first the 
point 
. Here k3 =0, so we stack the unit cells with no change. If we start with panti 
of Fig. 15-34a, we obtain Fig. 15-35a, which shows that the unit cells interact in an 
antibonding manner. Here, then, is a Bloch sum that is antibonding within and also 
between unit cells. The symbol E .. indicates two destabilizing interactions. Each 
arrow represents the same amount of energy because the interlayer distance within 
a unit cell is the same as that between unit cells and also because there are equal 
numbers of inter- and intralayer interactions in the crystal. Similar sketches easily 
demonstrate how the other three unit-cell function sets of Fig. 15-34 behave at 
 : 
pbond,E ..; p*anti, E ..; p*bond,E ... Therefore, at 
, the p and p* bands are both 
split. 
Now let us consider point A. Here we stack the unit cells together with a reversal 
in sign. The situation for panti is shown in Fig. 15-35b. The intercell interaction has 
reversed from the k3 =0 case. This means that the E .. and E .. cases from 
 now 
are E .. and E ... The levels are not split by interlayer effects at A. 
When we consider points M and L, we again are dealing with the same pu and pg 
functions in the unit cell (see Fig. 15-31b), so we obtain the same results with k3 =0, 
or p/a3. The p and p* bands are split at M by the same amount as at 
, and at L they 
are not split at all. 
At point K, the p and p* bands are degenerate. This means we must ask which are 
the proper zeroth-order two-dimensional COs with which to analyze the perturbation. 
We know that these will be four COs produced by mixing the four COs shown in 
Fig. 15-31c. This mixing is determined by solving a 4 × 4 determinantal equation, 
which, as was pointed out in Chapter 12, is the same equation we would use to discover 
the COs having minimum and maximum energy changes as a result of the perturbation. 
As a shortcut, then, we can simply look for the COs that respond most differently. Our 
analysis at 
 and M tells us that the pu and pg bases will give a splitting of all the
Section 15-16 Periodicity in Three Dimensions—Graphite 573 
Figure 15-34 
 The four independent arrangements of p (or pu) and p* (or pg) functions in a unit 
cell. Labels “anti” and “bond” refer to the interactions betweenAOs directly above/below each other. 
E . and E . indicate whether these interactions raise or lower the energy. 
COs. Not bad, but we can do even better by mixing the COs pictured in Fig. 15-31c. 
Adding and subtracting the cosp* and cosp COs produces two new functions whose 
unit cells (in two dimensions) have a 2pp AO on one carbon and nothing on the other, 
and vice versa. (In effect, we are taking the sum and difference of the pu and pg basis 
functions that we constructed in the first place.) These new basis functions give us 
four new unit cell bases for the three-dimensional case, shown in Fig. 15-36. Two of 
these place all the 2pp AOs on atoms that lie directly above and below each other, so 
these rise or fall in energy due to the intracell interactions. The other two place all 
the 2pp AOs on carbons that do not have neighbors directly above or below them, so 
these do not undergo energy change. The rise and fall here are larger than for the cases 
pictured in Fig. 15-34 because theAO coefficients are larger when the basis set exists on
574 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-35 
 (a) The result of translating a panti unit-cell basis function one step along a3 with 
no change of sign or value (k3 =0). (b) Same as (a) except now sign reverses (k3 =p/a). 
fewer atoms. These, then, are the proper zeroth-order functions with which to analyze 
the stacking perturbation. (The argument here is equivalent to saying that a calculus 
min-max problem will select roots -2, 0, 0,+2 in preference to -1,-1,+1,+1.) 
When we put these unit-cell bases together at K, where k3=0, we find that the functions 
that do not interact within the cell do not interact between cells either (Fig. 15-37a). 
These two cases remain unsplit in energy. The cases that interact within the cell interact 
again and in the same way between cells, so these two cases are split even more 
(Fig. 15-37b). At L, the intercell interactions reverse but the intracell interactions are 
unchanged. The E· cases are unaffected, but E . and E . become E .. and E .. (Fig. 15-37c), just as at A and H, so the splitting disappears at L. 
A sketch of the p and p* bands for a route around the edges of the RFBZ appears 
in Fig. 15-38. In general, no splitting of these bands occurs for points on the top of the 
RFBZ. For points on the bottom, splitting is larger at K than at 
 or M. 
Computing the electronic energy per unit cell for graphite requires knowing the 
average energy of the occupied bands, not only over the paths shown but also over the 
interior points of the RFBZ. The method of Chadi and Cohen [5] can be used to select
Section 15-16 Periodicity in Three Dimensions—Graphite 575 
Figure 15-36 
 The four independent unit-cell functions that result from requiring maximum difference 
in response to the interlayer interaction. E· indicates no energy change. 
an optimized set of weighted points in the RFBZ for purposes of calculating accurate 
average energies. 
One might wonder whether the splittings depicted in Fig. 15-38 have any detectable 
consequences. One change that occurs as a result of this splitting is that the Fermi level, 
EF, now refers to p electrons in COs that are not split along K–H by the perturbation, 
i.e., in COs like the one pictured in Fig. 15-37a. This means that the electrons having the 
highest energy are predicted to be in 2pp AOs at carbons that do not have carbon atoms 
directly above or belowthem in the crystal. This comes about because the electrons that 
do have such neighbors are lowered slightly in energy due to weak bonding between 
layers. This is relevant because there exists an experimental technique called scanning 
tunneling microscopy (STM) that detects the locations of surface electronic charge 
having energy near the Fermi level. When STM maps are made of electron distribution 
at a clean graphite surface, a trigonal pattern is seen, rather than the expected hexagonal 
one. Evidently, not all the atoms are “seen” equally well. The trigonal pattern is
576 Chapter 15 Molecular Orbital Theory of Periodic Systems 
Figure 15-37 
 (a and b) Result of translating two of the functions from Fig. 15-36 along a3 at 
point K (k3 = 0, so no change in function). (c) Same as (b) except now at point L (k3 = p/a, so 
function reverses sign). 
consistent with more charge being detected over every-other atom. If the device is 
tuned to include electronic charge from deeper energies, the familiar hexagonal pattern 
of graphite emerges. This suggests that the former experiment detects charge in the 
unsplit, horizontal band shown between K and H, corresponding to COs on atoms 
without neighbors in the adjacent layer, and that the latter measurement detects charge 
in both that band and the one just below it. (Of course, the situation at the surface is not 
identical to that in the bulk, but surface carbons having carbons directly below them 
will still have their 2pp energies lowered somewhat by the weak bonding interaction.) 
15-17 Summary 
Understanding or predicting the nature of the electronic band structure for a periodic 
material requires that we identify a unit cell and accompanying basis set in real space 
and then move this basis set along translation vectors a, possibly with accompanying
Section 15-17 Summary 577 
Figure 15-38 
 Sketch of p and p* band energies for a circuit around the RFBZ of threedimensional 
graphite. Heavy dots at 
,K,H, and M represent energies for the isolated sheet. 
Splittings at 
 and M should be about half that at K. 
rotations. Each time the basis set is moved to a different cell, it is multiplied by a 
coefficient that can be taken from a formula of the type cos(knp), where n is the 
number of steps taken through a and k is the wavenumber for the wave that determines 
the coefficient. This process generates Bloch sums. Sometimes Bloch sums are the 
same as the COs, sometimes they are not. The reasons they might be different are 
that (1) there may be intended correlations that violate the noncrossing rule and (2) the 
basis set mixture in a unit cell may change with k in order to take better advantage of 
the changing interaction between unit cells, even when there is no problem with band 
crossings. 
Degeneracies at the Fermi level result in Peierls distortions in one-dimensional systems, 
opening a band gap. Degenerate levels, whether at a gap or not, also allow 
one to predict which energies at the edges of the band diagram will be unaffected by 
certain chemical substitutions. Both of these phenomena are easily analyzed using 
the principle of “maximum difference in response” from degenerate-level perturbation 
theory. 
The values of k giving independent COs fall into a zone of reciprocal space called 
the first Brillouin zone. A subspace within it, called the reduced first Brillouin zone 
(RFBZ), contains all the k values that are needed to calculate every energy in the 
band system. Certain points at corners of the RFBZ are of special interest because 
they are easy to use for constructing Bloch sums. They often give waves having the 
same periodicity as the real lattice, which means that the cosine Bloch sum will give 
alternating positive and negative coefficients at each unit cell and the sine Bloch sum 
will give zero at each unit cell, hence not exist. This results in reduced degeneracy. 
Points in the FBZ that are related by symmetry to a point in the RFBZ correspond 
to COs that differ by a symmetry operation (e.g., a rotation) from the CO produced by 
the point in the RFBZ. 
Readers interested in further treatment of either the chemical or physical aspects of 
periodic systems should consult references at the end of this chapter [1-11].
578 Chapter 15 Molecular Orbital Theory of Periodic Systems 
15-17.A Problems 
15-1. Work out the relations in Eq. (15-3) (for a particle of mass m) so that it can be 
written as equalities rather than proportionalities. 
15-2. Derive the functional form for Fig. 15-2b. 
15-3. a) Obtain MOs and energies for benzene using Eqs. (15-4) and (15-5) with 
j =0-5. 
b) Show that the results for j =6 are identical to those for j =0. 
15-4. a) Derive the mathematical formula for the curve of Fig. 15-5b. 
b) What would this function be if DOS/N were constant over the energy range? 
15-5. For benzene, j =1 and j =-1 give different complex degenerate MOs. Produce 
these and find linear combinations that are real and orthogonal. Sketch these 
real MOs and compare them with those tabulated in Appendix 6 for the same 
energy. 
15-6. Demonstrate that Eq. (15-10) is correct if N is an integer. 
15-7. Using sketches, show that the Bloch sum for (C2H2)n at |k|=p/3a is different 
from that for (CH)2n at |k|=p/6a. Which do you think has the lower energy? 
What will happen in a variation calculation on each of these two cases? 
15-8. Suppose an infinite straight chain of atoms existed with a band of s orbitals that 
was one-third filled. Indicate what sort of bond-length pattern should result 
from a Peierls distortion. 
15-9. Figure 15-14 indicates that the 2s and 2ps Bloch functions avoid crossing. Use 
symmetry to argue why this is so. 
15-10. In connection with Fig. 15-18b, the text indicates that the avoided crossing 
between s levels for alternating bond lengths becomes an allowed crossing 
when uniform bond lengths are used. This means that there is a symmetry 
disagreement for uniform bond lengths that does not exist for alternating bond 
lengths. The new symmetry element that comes into existence with uniform 
bond lengths is a two-fold screw axis. (The operation corresponding to this 
element is to rotate the polymer by 180. about the axis and also to translate 
it parallel to the axis.) Show, using sketches, that these s bands disagree in 
symmetry for such a symmetry operation. 
15-11. CrystallineH2(H–H···H–H···H–H·) is an insulator. It is thought that it should 
become a metallic conductor at extremely high pressures. Explain why this is so. 
15-12. A heteronuclear diatomic moleculeA–B crystallizes end to end to form a linear 
chain: . . .A–B···A–B···A–B. The internuclear distance between molecules 
is significantly longer than that within molecules. The monomer has a bonding 
valence s MO and a higher energy s* MO: 
Sketch a band structure diagram for the crystal showing qualitatively the 
energies of the crystal orbitals as a function of |k| in the first Brillouin zone. 
Sketch the appearances of the COs for the s and s* bands at |k|=0 and p/a.
Section 15-17 Summary 579 
15-13. Andr´e and Leroy [2] report an ionization energy of 8.19 eV and an electron 
affinity of -4.09eV (i.e., energy is released in forming the anion) from applying 
Koopmans’ theorem to their ab initio calculations on trans-polyacetylene 
with regular bond lengths. a) What is the energy of the HOMO? b) Of the 
LUMO? c) What is the value of the energy gap? d) What should happen to the 
values of these three numbers when the bond lengths are allowed to alternate? 
15-14. It is stated in Section 15-11 that it is improper to interpret a H¨uckel band 
diagram having zero gap as meaning that the ionization energy (from the 
HOMO energy) equals the electron affinity (from the equal LUMO energy). 
Yet we do use HOMO and LUMO H¨uckel energies to estimate the IE and EA 
in Chapter 8. What is the explanation for this apparent disagreement? 
15-15. In a variational EHMO calculation on PPP, the coefficients for p1 at k = 0 
are 0.37 at linking carbons (1 and 4) and 0.28 at the others. This results from 
mixing the Bloch function p1 at k = 0 (with all coefficients the same) with 
another Bloch function at k =0. Which Bloch function, px, can mix with p1 
at k =0 to accomplish this modification? (See Fig. 15-23 for sketches.) What 
happens to the energy of the px band at k =0 as a result of mixing with p1? 
What happens to the coefficients in px? 
15-16. When PPP-N2 is substituted at position 6, the band energy values of -13.94eV 
and -5.64eV at k =p/a are not affected. Would you expect the same thing to 
happen if substitution occurred at position 5 instead of 6? Why or why not? 
15-17. The PPP diagram of Fig. 15-23 is constructed from Bloch sums built from 
benzene MOs. We could have used the MOs for the other choice of unit cell 
shown in Fig. 15-21a. Obtain the appropriate unit cell simple H¨uckel MOs and 
energies from Appendix 6 and construct the appropriate PPP band diagram. 
Discuss the differences with Fig. 15-23. 
15-18. Predict the first-order H¨uckel-level energy band diagram for 
Show the actual band-edge energies in terms of a and ß. Show how the 
edge energies at k = 0 and p/a would shift if substitution occurred in all 
the nonlinking sites so that a = a + 0.25ß, with no change in ß. What if 
substitution occurred in only one nonlinking site in each monomer? 
15-19. Sketch the CO for point X in Fig. 15-27c for s-type unit-cell basis functions. 
Comparing with Fig. 15-28, how would you expect the energy of this CO to 
compare to the energies of COs at 
,X, and M?
580 Chapter 15 Molecular Orbital Theory of Periodic Systems 
15-20. Figure out the (k1, k2) coordinates and sketch the COs for s-type basis functions 
for the following points in the RFBZ of Fig. 15-27c. a) Halfway between X 
and M. b) Halfway between X and M. () Halfway between 
 and X. d) 
Halfway between 
 and M. Rank these in order of energy. 
15-21. Sketch COs based on a 2px AOat each lattice point for the RFBZ of Fig. 15-27c, 
at points 
,X, and X. Rank according to predicted energy. 
15-22. What degeneracy would you anticipate for energies associated with points 

,X,M, and X of Fig. 15-27c?. 
15-23. The rectangular lattice of Fig. 15-27a leads to a rectangular RFBZ. What shape 
RFBZ would result for a square lattice? 
15-24. It is stated in the text that mixing p(M) and p*(M) of Fig. 15-31b will decrease 
the difference in their energies. Sketch the functions p(M) ± p*(M) and 
comment on their energies. 
15-25. For the FBZ of Fig. 15-30f, locate the points (0,p/a) and (p/a,-p/a). 
Sketch the p CO corresponding to the first of these and compare it with the p 
CO for (p/a, 0), sketched in Fig. 15-31b. How are these COs related? 
15-26. What physical situation allows us to use the half-range of 0 = k3 = p/a3 for 
the RFBZ pictured in Fig. 15-33a? 
15-27. Give a general argument for there being no interlayer-induced splitting of 
p bands for any point at the top of the RFBZ of Fig. 15-33a. What does your 
argument predict for s bands? 
15-28. For two-dimensional graphite, sketch the Bloch sums at 
 and M resulting 
from a bonding pair of 2px AOs in each unit cell. Let x be the bisector of a1 
and a2. Which of these COs would you expect to have higher energy? 
References 
[1] R. Hoffmann, C. Janiak, and C. Kollmar, Macromolecules 24, 3725 (1991). 
[2] J.-M. Andr´e and G. Leroy, Int. J. Quantum Chem. 5, 557 (1971). 
[3] J. P. Lowe and S. A. Kafafi, J. Amer. Chem. Soc. 106, 5837 (1984). 
[4] J. P. Lowe, S. A. Kafafi, and J. P. LaFemina, J. Phys. Chem. 90, 6602 (1986). 
[5] D. J. Chadi and M. L. Cohen, Phys. Rev. B 8, 5747 (1973). 
[6] G. S. Painter and D. E. Ellis, Phys. Rev. B 1, 4747 (1970). 
[7] J. P. LaFemina and J. P. Lowe, Int. J. Quantum Chem. 30, 769 (1986). 
[8] J. C. Slater, Quantum Theory of Molecules and Solids, Vol. 2. McGraw-Hill, 
NewYork, 1965.
Section 15-17 Summary 581 
[9] R. Hoffmann, Solids and Surfaces: A Chemist’s View of Bonding in Extended 
Structures.VCH Publishers, NewYork, 1988. 
[10] N.W.Ashcroft and N. D. Mermin, Solid State Physics. Holt, Rinehart andWinston, 
NewYork, 1976. 
[11] C. Kittel, Introduction to Solid State Physics, 7th ed.Wiley, NewYork, 1996.
Appendix 1 
Useful Integrals 

 xneax dx =(xneax/a)-(n/a) 
 xn-1eax dx 

 8 
0 
xne-ax dx =(n!/an.+1)=
n+1(a), n>-1, a>0 

 8 
0 
e-ax2
dx = 
1
2p/a 

 8 
0 
xe-ax2
dx =1/2a 

 8 
0 
x2e-ax2
dx = 
1
4
p/a3 

 8 
0 
x3e-ax2
dx =1/2a2 

 8 
0 
x2ne-ax2
dx = 
1 · 3 · ··· · (2n-1) 
2n+1 
 p 
a2n+1 

 8 
0 
x2n+1e-ax2
dx = 
n! 
2an+1 

 8 
1 
e-axdx = 
e-a 
a 

 1 
0 
e-axdx =(1/a)(1-e-a) 

 8 
1 
xe-axdx =(e-a/a2)(1+a) 

 1 
0 
xe-axdx =(1/a2)[1-e-a(1+a)] 

 8 
1 
x2e-axdx =(2e-a/a3)(1+a +a2/2) 

 1 
0 
x2e-axdx =(2/a3)[1-e-a(1+a +a2/2)] 

 8 
1 
xne-axdx =(n!e-a/an+1) 
n 

k=0 
ak/k!=An(a) 
582
Appendix 1 583 

 8 
y 
xne-axdx =(n!e-ay/an+1) 
n 

k=0
(ay)k/k! 

 +1 
-1 
e-axdx =(1/a)(ea -e-a) 

 +1 
-1 
xe-axdx =(1/a2)[ea -e-a -a(ea +e-a)] 

 +1 
-1 
xne-axdx =(-1)n+1An(-a)-An(a) 

 +1 
-1 
xndx =
0, n=1, 3, 5, . . . 
2/(n+1), n=0, 2, 4, . . . 

 sinx dx=-cos x 

 cosx dx =sin x 

 sin2 x dx = 
x
2 - 
sin 2x 
4 

 cos2 x dx = 
x
2 + 
sin 2x 
4 

 x sinx dx =sin x -x cos x 

 x cosx dx =cos x +x sin x 

 x sin2 x dx = 
x2 
4 - 
x sin 2x 
4 - 
cos 2x 
8 

 x cos2 x dx = 
x2 
4 + 
x sin 2x 
4 + 
cos 2x 
8
Appendix 2 
Determinants 
A determinant is a scalar calculated from an ordered set of elements according to a 
specific evaluation recipe. The elements are ordered in a square array of rows and 
columns, bounded at left and right by straight vertical lines. For instance, (A2-1) is a 
2×2 determinant: 





 
x -i 
2 y2





 
(A2-1) 
The recipe for evaluating a 2×2 determinant is: From the product of the elements 
on the principal diagonal (upper left to lower right) subtract the product of the other 
two elements. Thus, (A2-1) has the value xy2 +2i. 
Larger determinants are evaluated by a process that reduces them step by step to a 
linear combination of smaller determinants until, finally, they are all 2×2’s, which are 
then evaluated as above. The process of reduction involves the concept of a cofactor. 
As our example, we use the 4×4 determinant (A2-2), symbolized |M|, whereM is the 
array of elements within the vertical bars: 
(A2-2) 
The elements are numbered so that the first index tells which row, and the second 
index which column, the element is in. The cofactor of element a11 is defined as the 
determinant obtained by removing the rowand column containing a11. We see in (A2-2) 
that striking out row 1 and column 1 gives us a 3×3 determinant (dashed outline) as 
cofactor of a11. Symbolize this cofactor as |A11|. 
To evaluate the determinant |M|, we expand in terms of cofactors. We begin by 
choosing any row or column of M. (We will choose row 1.) Then we write a linear 
combination containing every element in this row or column times its cofactor: 
|M|=a11|A11|-a12|A12|+a13|A13|-a14|A14| (A2-3) 
The sign of each term in the linear combination is determined as follows. lf the sum of 
row and column indices is even, the sign is plus. If the sum is odd, the sign is minus. 
Since the indices of a12 and a14 sum to odd numbers, they are minus in (A2-3). 
584
Appendix 2 585 
The method of expanding in cofactors is successively applied until a large determinant 
is reduced to 3×3’s or 2×2’s that can be evaluated directly (see Problem A2-1). 
Thus, a 5 × 5 is first expanded to five 4 × 4’s and each 4 × 4 is expanded to four 
3×3’s giving a total of 20 3×3’s. This method becomes extremely clumsy for large 
determinants. 
Some useful properties of determinants, symbolized |M|, are stated below without 
proof. The reader should verify that these are true using 2×2 or 3×3 examples, or by 
examining Eq. (A2-3). 
1. Multiplying every element in one row or one column of M by the constant c multiplies 
the value of |M| by c. 
2. If every element in a row or column of M is zero, then |M|=0. 
3. lnterchanging two rows or columns of M to produce M
 results in


M



=-|M|; 
i.e., it reverses the sign of |M|. 
4. Adding to any row (column) of M the quantity c times any other row (column) of 
M does not affect the value of the determinant. 
5. If two rows or columns of M differ only by a constant multiplier, then |M|=0. 
A2-1 Use of Determinants in Linear Homogeneous Equations 
Suppose that we seek a nontrivial solution for the following set of linear homogeneous 
equations: 
a1x +b1y +c1z = 0 (A2-4) 
a2x +b2y +c2z = 0 (A2-5) 
a3x +b3y +c3z = 0 (A2-6) 
Here, x,y, and z are unknown and the coefficients ai, bi, ci are given. Let us collect 
the coefficients into a determinant |M|, 
|M|=







 a1 b1 c1 
a2 b2 c2 
a3 b3 c3







 
(A2-7) 
As before, let |A1| be the cofacter of a1, etc. Now, multiply Eq. (A2-4) by |A1| (since |A1| is a determinant, it is just a number, and so this is a scalar multiplication), 
Eq. (A2-5) by -|A2|, and Eq. (A2-6) by |A3| and add the results to get 
x(a1|A1|-a2|A2|+a3|A3|)+y(b1|A1|-b2|A2|+b3|A3|) 
+z(c1|A1|-c2|A2|+c3|A3|)=0 (A2-8) 
The coefficient of x is just |M|. The coefficients of y and z correspond to determinants 
having two identical rows and hence are zero. Therefore, 
|M|x =0 (A2-9)
586 Determinants 
In order for x to be nonzero (i.e., nontrivial) it is necessary that |M|=0. This is a result 
that is very useful. The condition that must be met by the coefficients of a set of linear 
homogeneous equations in order that nontrivial solutions exist is that their determinant 
vanish. 
A2-2 Problems 
A2-1. Expand the 3×3 determinant (Fig. PA2-1), by cofactors and showthat this result 
is equivalent to the direct evaluation of the 3×3 by summing the three products 
parallel to the main diagonal (solid arrows) and subtracting the three products 
parallel to the other diagonal (dashed arrows). 
Figure PA2-1 
 
A2-2. Evaluate (a)





 
0 1 
2 1





 
(b)







 
1 1 0 
0 1 1 
1 1 1







 
(c)









 
1 2 0 1 
3 0 1 4 
1 1 0 1 
0 2 1 1









 
(d)





 
x 2 
1 x






=0 for x 
A2-3. Verify that the coefficient of y in Eq. (A2-8) is zero. 
A2-4. Consider the following set of linear homogeneous equations: 
4x +2y -z=0, 3x -y -2z=0, 2y +z=0 
Do nontrivial roots exist? 
A2-5. Find a value for c that allows nontrivial solutions for the equations 
cx -2y +z=0, 4x +cy -2z=0, -8x +5y -cz=0 
A2-6. Five properties of determinants have been listed in this appendix. (a) Prove 
statement (5) is true assuming statements (1)–(4) are true. (b) Demonstrate 
statements (1)–(4) using simple examples.
Appendix 3 
Evaluation of the Coulomb Repulsion 
Integral Over 1s AOs 
Evaluation of 


 1s(1)1s(2)(1/r12)1s(1)1s(2)dv(1)dv(2) (A3-1) 
where 
1s(1)=. 3/p exp(-.r) (A3-2) 
may be carried out in two closely related ways.1 Each method is instructive and sheds 
light on the other, and so we will give both of them here. 
The first methodworks from a physical model and requires knowledge of two features 
of situations governed by the 1/r2 force law (e.g., electrostatics, gravitation). Suppose 
that there exists a spherical shell in which charge or mass is distributed uniformly, like 
the soap solution in a soap bubble. The first feature is that a point charge or mass 
outside the sphere has a potential due to attraction (or repulsion) by the sphere that 
is identical to the potential produced if the sphere collapsed to a point at its center 
(conserving mass or charge in the process). Thus, the electrostatic interaction between 
two separated spherical charge distributions may be calculated as though all the charge 
were concentrated at their centers. The second feature is that the potential is identical 
for all points inside the spherical shell; that is, if the core of the earth were hollow, a 
person would be weightless there. There would be no tendency for that person to drift 
toward a wall or toward the center. 
Armed with these facts, we can evaluate the integral. First, we remark that all the 
functions in the integrand commute. This enables us to write Eq. (A3-1) as 


 1s2(1)(1/r12)1s2(2)dv(1)dv(2) (A3-3) 
The functions 1s2(1) and 1s2(2) are just charge clouds for electrons 1 and 2, and 
the integral is evidently just the energy of repulsion between the clouds. Suppose 
(see Fig. 3-1) that, at some instant, electron 1 is at a distance r1 from the nucleus. What 
is its energy of repulsion with the charge cloud of electron 2? The charge cloud of 
electron 2 can be divided into two parts: the charge inside a sphere of radius r1 and the 
charge outside that sphere. From what we just said, electron 1 experiences a repulsion 
due to the cloud inside the sphere that is the same as the repulsion it would feel if that 
1Other methods, not discussed here, also exist. See, for example, Margenau and Murphy [1, pp. 382–383]. 
587
588 Evaluation of the Coulomb Repulsion Integral Over 1s AOs 
Charge cloud 1s2 (2) 
r1 
1 
Figure A3-1 
 Sketch of spherical charge cloud 1s2(2) with electron 1 at a distance r1 from the 
nucleus. 
part of the cloud were collapsed to the center. We can calculate the energy due to this 
repulsion (call it the “inner repulsion energy”) by dividing the product of charges by 
the distance between them: 
inner repulsion energy = (fraction of charge cloud 2 inside r1)×1/r1 
= 4p 
 r1 
0 
1s2(2)r2
2 dr2 ×1/r1 (A3-4) 
where the factor 4p comes from integrating over .2,f2. Electron 1 also experiences 
repulsion from charge cloud 2 outside the sphere of radius r1. But, from what we said 
above, electron 1 would experience this same “outer repulsion” no matter where it was 
inside the inner sphere. Therefore we will calculate the energy due to this repulsion 
as though electron 1 were at the center, since this preserves spherical symmetry and 
simplifies the calculation. It follows that all the charge in a thin shell of radius r2 repels 
electron 1 through an effective distance of r2. Integrating over all such shells gives 
outer repulsion energy=4p 
 8 
r1 
(1/r2)1s2(2)r2
2 dr2 (A3-5) 
The total energy of repulsion between charge cloud 2 and electron 1 at r1 is the sum 
of inner and outer repulsive energies. But electron 1 is not always at r1. Therefore, 
we must finally integrate over all positions of electron 1, weighted by the frequency of 
their occurrence: 
repulsive energy=16p2 
 8 
0 
1s2(1) (1/r1) 
 r1 
0 
1s2(2)r2
2 dr2 
+ 
 8 
r1 
1s2(2)r2dr2r2
1 dr1 (A3-6)
Appendix 3 589 
Figure A3-2 
 
The inner integrals contain but one variable, r2, and can be evaluated with the help of 
Appendix 1. After they are performed, the integrand depends only on r1 and this is also 
easily integrated yielding 5./8 as the result. A positive value is necessary since a net 
repulsion exists between two clouds of like charge. 
The second method of evaluation is more mathematical and more general. The function 
1/r12 is expressible2 as a series of terms involving associated Legendre functions: 
1 
r12 = 
8 
l=0 
+l 

m=-l 
(l -|m|)! 
(l +|m|)! 
rl< 
rl+1 
> 
P|m| l (cos .1)P |m| l (cos .2) exp[im(f1 -f2)] (A3-7) 
This infinite series will give the distance between particles 1 and 2 located at positions 
r1, .1,f1 and r2, .2,f2 (Fig. A3-2). All we need to do is pick the larger of r1 and r2 and 
call that r>, the other being r<, and substitute those numbers into the formula. We also 
need all the Legendre functions for cos .1 and cos .2, and values of exp[im(f1 -f2)] for all integral values of m. Putting all this together as indicated by Eq. (A3-7) would 
give a series of numbers whose sum would converge to the value of 1/r12. While this is 
an exceedingly cumbersome way to calculate the distance between two points, it turns 
out that use of the formal expression (A3-7) enables us to integrate Eq. (A3-1). This 
comes about because the Legendre functions satisfy the relation 

 p 
0 
P|m| l (cos .)P|m| l (cos .) sin. d. = 
2 
2l +1 
(l +|m|)! 
(l -|m|)!
dll . (A3-8) 
The first Legendre polynomial P0 is equal to unity. Therefore, 1s(i), which has no 
. dependence, can be written 
1s(i)=(. 3/p)1/2 exp(-.ri)P0(cos .i) (A3-9) 
Thus, in the integral (A3-1) there will be an integration over .1 of the form 

 p 
0 
P0(cos .1)P |m| l (cos .1)P0(cos .1) sin .1 d.1 (A3-10) 
for each term in the sum (A3-7) (and a similar integral over .2). However since 
12 =1, [P0(cos .1)]2 =P0(cos .1), and integral (A3-10) becomes [by Eq. (A3-8)] 

 p 
0 
P0(cos .1)P |m| l (cos .1) sin .1 d.1 =2d0l (A3-11) 
2See Eyring et al. [2, Appendix 5].
590 Evaluation of the Coulomb Repulsion Integral Over 1s AOs 
[m must equal zero here, otherwise the integral over f will vanish.] In other words, all 
terms of the sum over l and m vanish except the first term, for which l =m=0. This 
gives that the 1/r12 operator is equal to 1/r>. Hence, the repulsion integral is 
16p2 
 8 
0 

 8 
0 
1s2(1)1s2(2)r2
1 r2
2 
r> 

dr2dr1 (A3-12) 
where r> is the greater of r1, r2. Suppose that we integrate over r2 first. As r2 changes 
value, it is sometimes smaller, sometimes larger than a particular value of r1. When it 
is larger, r> is r2. When it is smaller, r> is r1. Putting this argument into mathematical 
form gives 
16p2 
 8 
0 

 r1 
0 
1s2(1)1s2(2)r2
1 r2
2 dr2 
r1 +
 8 
r1 
1s2(1)1s2(2)r2
1 r2
2 dr2 
r2 
	
dr1 (A3-13) 
Since the variable of integration in the two inner integrals is r2, the quantities 1s2(1), r2
1 , 
and r1 may be brought outside these inner integrals, giving us the same equation (A3-6) 
that we obtained by the first method. This second method is more generally useful 
because it can be used when repulsions involving p, d, etc. charge clouds are calculated. 
In these cases, terms involving l = 1, 2, etc., become nonvanishing, but the series 
generally truncates after a few terms. 
References 
[1] H. Margenau and D. M. Murphy, The Mathematics of Physics and Chemistry. Van 
Nostrand-Reinhold, Princeton, New Jersey, 1956. 
[2] H. Eyring, J. Walter, and G. E. Kimball, Quantum Chemistry. Wiley, New York, 
1944.
Appendix 4 
Angular Momentum Rules 
A4-1 Introduction 
In Chapters 4, 5, and 7, we deal with the angular momentum of electrons due to 
their orbital motions and spins and also with angular momentum due to molecular 
rotation. The conclusions we arrive at are based partly on physical arguments related 
to experience with macroscopic bodies. While this is the most natural way to introduce 
such concepts, it is not the most rigorous. In this appendix we shall demonstrate how 
use of the postulates discussed in Chapter 6 leads to all of the relationships we have 
introduced earlier. The approach is based entirely on the mathematical properties of 
the relevant operators, making no appeal to physical models. 
A4-2 The Classical Expressions for Angular Momentum 
According to postulate II (Section 6-3), we must first find the classical physical expressions 
for the quantity of interest in terms of x, y, z,px,py,pz , and t . We know that the 
angular momentum for an object of mass m moving with velocity v in a circular orbit 
of radius r is 
L=mr×v=r×p (A4-1) 
where the × symbol means we are taking a cross-product of vectors. To examine this 
in detail, we must resolve r and p into x,y, and z components: 
r =ix +jy +kz (A4-2) 
p=mv=midx 
dt +mjdy 
dt +mkdz 
dt =ipx +jpy +kpz (A4-3) 
Here i, j,k are unit vectors pointing, respectively, along the x, y, z Cartesian axes. 
A cross product is taken by expanding the following determinant: 
L=r×p=







 
i j k 
x y z 
px py pz







 
(A4-4) 
to give 
L=i(ypz -zpy)+j(zpx -xpz)+k(xpy -ypx) (A4-5) 
591
592 Angular Momentum Rules 
The coefficient for i is the magnitude of the x-component of angular momentum, etc. 
That is, 
L=Lxi+Lyj+Lzk (A4-6) 
where we now see that 
Lx =ypz -zpy (A4-7) 
Ly =zpx -xpz (A4-8) 
Lz =xpy -ypx (A4-9) 
(The simple cyclic x, y, z relationships among these formulas can be used as a memory 
aid.) We will also be interested in the square of the angular momentum, L2: 
L2 =L ·L=(Lxi+Lyj+Lzk)(Lxi+Lyj+Lzk)=L2x 
+L2y 
+L2z
(A4-10) 
where we have used the orthonormality of the unit vectors (i · i=1, i · j=0, etc). Note 
that L2 is a magnitude, not a vector. 
This gives the classical expressions we need for the x,y, and z components of angular 
momentum and for the square of its magnitude. 
A4-3 The Quantum-Mechanical Operators 
Postulate II tells us what to do next: Replace px with ( ¯h/i)./.x, and similarly for 
py,pz . The resulting operators are: 
Lˆx =-ih¯ y 
. 
.z -z 
. 
.y 
 (A4-11) 
Lˆy =-ih¯ z 
. 
.x -x 
. 
.z
 (A4-12) 
Lˆz =-ih¯ x 
. 
.y -y 
. 
.x 
 (A4-13) 
and 
Lˆ2 =Lˆ2x 
+Lˆ2y 
+Lˆ2z
(A4-14) 
All of these operators are hermitian. 
Now that we have the quantum-mechanical operators, we are free to transform them 
to other coordinate systems. In spherical coordinates, they are1 
Lˆx =ih¯ sin f 
. 
.. +cot . cosf 
. 
.f
 (A4-15) 
Lˆy =-ih¯ cosf 
. 
.. -cot . sin f 
. 
.f
 (A4-16) 
1See Eyting et al. [1, p. 40.]
Appendix 4 593 
Lˆz =-ih¯ 
. 
.f 
(A4-17) 
Lˆ2=-¯h2  .2 
..2 +cot . 
. 
.. + 
1 
sin2 . 
.2 
.f2
 
=-¯h2  1 
sin . 
. 
.. 
sin . 
. 
.. + 
1 
sin2 . 
.2 
.f2
 (A4-18) 
In atomic units, the quantity ¯h becomes unity and does not appear. We will use 
atomic units henceforth in this appendix. 
A4-4 Commutation of Angular Momentum Operators with 
Hamiltonian Operators and with Each Other 
We have indicated (Chapter 4) that a rotating classical system experiencing torque 
maintains E,Lz , and |L| (or, equivalently, L2) as constants of motion, but not Lx or Ly. 
We might anticipate a similar situation in quantum mechanics. This would mean that 
a state function . would be an eigenfunction for Hˆ ,Lˆ2, and Lˆz but not for Lˆx or Lˆy. 
This in turn requires that Hˆ ,Lˆ2, and Lˆz commute with each other, but that Lˆx and Lˆy 
do not commute with all of them. 
We first consider Hˆ and Lˆ2. Note that Lˆ2 appears in .2 [compare Eq. (A4-18) 
with (4-7)]: 
.2 = 
1 
r2 
. 
.r 
r2 . 
.r - 
1 
r2 ˆ L2 (A4-19) 
Also, since L2 does not contain the variable r,Lˆ2 commutes with any function depending 
only on r. Since Lˆ2 must also commute with itself, it follows that Lˆ2 and .2 
commute, i.e., that 
Lˆ2,.2=0 (A4-20) 
If V in a hamiltonian operator is a function of r only, then 
Lˆ2,Hˆ =0 (A4-21) 
This proves that Hˆ for a hydrogenlike ion commutes with Lˆ2. 
From the expressions for Lˆ2 and Lˆz in spherical polar coordinates, it is obvious that 
Lˆz ,Lˆ2=0 (A4-22) 
Lˆz does not operate on functions of r, and so Lˆz commutes with .2. For a spherically 
symmetric system, V =V (r) and we have that 
Lˆz ,Hˆ =0 (A4-23) 
Thus, we have shown that Hˆ ,Lˆ2,Lˆz all commute in a system having a spherically 
symmetric potential.
594 Angular Momentum Rules 
Now let us test Lˆx and Lˆy with each other and also with Lˆ2. 
Lˆx,Lˆy=LˆxLˆy -LˆyLˆx (A4-24) 
We will examine this using the Cartesian system. We write out each operator product 
and subtract:
LˆxLˆy =-y 
. 
.z -z 
. 
.y
z 
. 
.x -x 
. 
.z
 
=-y 
. 
.x +yz 
.2 
.z.x -z2 .2 
.y.x -yx 
.2 
.z2 +zx 
.2 
.y.z
 (A4-25) 
LˆyLˆx =-zy 
.2 
.x.z -xy 
.2 
.z2 -z2 .2 
.x.y +x 
. 
.y +xz 
.2 
.z.y 
 (A4-26) 
Subtracting (A4-26) from (A4-25): 
Lˆx,Lˆy=-y 
. 
.x -x 
. 
.y 
=x 
. 
.y -y 
. 
.x 
=iLˆz (A4-27) 
A similar approach to other operator combinations gives 
Lˆy,Lˆz=iLˆx (A4-28) 
Lˆz ,Lˆx=iLˆy (A4-29) 
(Notice the x, y, z cyclic relation.) These commutation relations do not depend on 
choice of coordinate system. Use of the r, ., f coordinate system would give the 
same results. Evidently, these operators do not commute with each other since their 
commutators are unequal to zero. 
From this point we will dispense with the carat symbol, since the context of the 
discussion makes it obvious that we are referring to operators. 
We still have not checked L2 with Lx and Ly. We will now show that 
Lx,L2=Lx, (L2x 
+L2y 
+L2z
)=0 (A4-30) 
We proceed by finding the commutator of Lx with L2x
,L2y
, and L2z
individually. It is 
obvious that the first of these, Lx,L2x
equals zero. The second can be evaluated as 
follows. 
LxLy -LyLx =iLz (from A4-27) (A4-31) 
We multiply (A4-31) from the left by Ly: 
LyLxLy -L2y
Lx =iLyLz (A4-32) 
We multiply (A4-31) from the right by Ly: 
LxL2y 
-LyLxLy =iLzLy (A4-33)
Appendix 4 595 
We sum (A4-32) and (A4-33): 
LxL2y 
-L2y
Lx =Lx,L2y
=i(LyLz +LzLy) (A4-34) 
A similar strategy, starting with (A4-29), multiplying from left and right by Lz , and 
summing, yields 
L2z
Lx -LxL2z 
=-Lx,L2z
=i(LzLy +LyLz) (A4-35) 
Equation A4-34 minus A4-35 plus zero (from Lx,L2x
) is equal to (A4-30) and is easily 
seen to equal zero. 
A similar proof gives 
Ly,L2=0 (A4-36) 
It follows at once that, since Lx and Ly operate only on . and f and since .2 contains 
all . and f terms in the form of L2: 
Lx,.2=Ly,.2=0 (A4-37) 
If the potential V for a system is independent of . and f, then 
Lx,H=Ly,H=0 (A4-38) 
We have shown that H for an atom and L2 commute with each other and also with 
Lx,Ly, and Lz , but that the latter three operators do not commute with each other. 
Therefore, we can say that there exists a set of simultaneous eigenfunctions for H,L2, 
and one ofLx,Ly,Lz , but not the other two. We chooseLz to be the privileged operator. 
This means that an atom can exist in states having “sharp” values of energy, magnitude 
of angular momentum (hence square of L), and z component of angular momentum, 
but not x or y components. 
H and L2 will commute if V in H is independent of . and f (central field potential). 
H and Lz will commute if V is independent of f, even if it is dependent on .. This 
is the case for any linear system. Therefore, ML continues to be a sharp quantity for 
linear molecules like H2 or C2H2, but total angular momentum value does not, because 
L2 does not commute with H, so L is not a good quantum number. This is why the 
main term symbol for a linear molecule is based onML, whereas the main term symbol 
for an atom is based on L. 
A4-5 Determining Eigenvalues for L2 and Lz 
We will now make use of our operator commutation relations to determine the nature 
of the eigenvalues for L2 and Lz . We begin by defining two new operators: 
L+ =Lx +iLy (A4-39) 
L- =Lx -iLy (A4-40) 
These are called “step-up” and “step-down” operators, respectively, or “raising” and 
“lowering” operators. They correspond to no observable property and are not hermitian.
596 Angular Momentum Rules 
They have been devised solely because they are useful in formal analysis of the sort we 
are doing here. The reason for their names will become apparent soon. 
We now prove the following theorem for the step-up operator L+: 
LzL+ =L+(Lz +1) (A4-41) 
Expanding L+ gives 
LzL+ =Lz(Lx +iLy)=LzLx +iLzLy (A4-42) 
We now add zero in the form [see Eq. A4-29] 
-(LzLx -LxLz -iLy)=0 (A4-43) 
to obtain 
LzL+ =LxLz +iLy +iLzLy (A4-44) 
We again add zero, this time in the form [see Eq. A4-28] 
iLyLz -iLzLy +Lx =0 (A4-45) 
This results in 
LzL+ =LxLz +iLy +iLyLz +Lx (A4-46) 
Rearranging, 
LzL+ =(Lx +iLy)Lz +Lx +iLy =(Lx +iLy)(Lz +1)=L+(Lz +1) (A4-47) 
This proves Eq. A4-41. An analogous procedure proves the analogous relation for the 
step-down operator L-: 
LzL- =L-(Lz -1) (A4-48) 
Our ultimate goal is to discover the nature of the eigenvalues of L2 and Lz . Since 
these two operators commute, there must exist a common set of eigenfunctions for them. 
Let us symbolize these simultaneous eigenfunctions of L2 and Lz with the symbol Y . 
A given eigenfunction Y will be associated with an eigenvalue for L2 and a (possibly) 
different eigenvalue for Lz . To keep track of these, we use subscript labels l and m, 
defined in the following manner: 
L2Yl,m =klYl,m (A4-49) 
LzYl,m =kmYl,m (A4-50) 
Nowwe make use of our “ladder” operators. It is not difficult to show(ProblemA4-1) 
that 
LzL+Yl,m =(km +1)L+Yl,m (A4-51) 
This shows that operating on Yl,m with the step-up operator L+ produces a new function 
and that this new function is also an eigenfunction of Lz . Furthermore, the new
Appendix 4 597 
eigenfunction has an eigenvalue that is greater by one than the eigenvalue (km) of 
the original eigenfunction. This is the reason L+ is called a step-up operator. The 
analogous relation for L- is
LzL-Yl,m =(km -1)L-Yl,m (A4-52) 
The import of Eqs. (A4-51) and (52) is that a set of eigenfunctions for Lz exists with 
eigenvalues separated by unity. 
We now use the ladder operators in another pair of useful operator relations 
(Problem A4-2): 
L2 =L+L- +L2z 
-Lz (A4-53) 
L2 =L-L+ +L2z 
+Lz (A4-54) 
It is also possible to show (Problem A4-3) that 
L+Yl,m =C+Yl,m+1 (A4-55) 
L-Yl,m =C-Yl,m-1 (A4-56) 
where C+ and C- are constants. Equations A4-55 and A4-56 tell us that the ladder 
operators change the functions Y in a manner such that their eigenvalues for L2 do not 
change. 
We will now show that 
kl =k2m
(A4-57) 
We begin with the obvious relation 
(L2 -L2z
)Yl,m =(kl -k2m
)Yl,m (A4-58) 
But, from Eq. A4-14, 
L2 -L2z 
=L2x 
+L2y
(A4-59) 
and so 
(L2x 
+L2y
)Yl,m =(kl -k2m
)Yl,m (A4-60) 
Equation A4-60 is remarkable because it indicates that the functions Yl,m (which are 
not eigenfunctions of Lx or Ly, since these do not commute with Lz ) are nevertheless 
eigenfunctions for the combination L2x +L2y
. 
We will now show that the eigenvalues of Eq. A4-60 must be positive definite, which 
suffices to prove that kl =k2m
. We proceed by recognizing that the operators Lx and Ly 
do possess eigenfunctions, though they are different from the functions Yl,m. Let us 
symbolize them f and g, respectively. Then 
Lxfi =xifi (A4-61) 
Lygi =yigi (A4-62) 
Because Lx and Ly are hermitian operators, x and y are real numbers and the function 
sets {f } and {g} are complete and can be assumed to be orthonormal. We can therefore 
expand the functions Y in terms of either set: 
Yl,m =
i 
cl,m 
i fi =
j 
dl,m 
j gj (A4-63)
598 Angular Momentum Rules 
Substituting Eqs. A4-63 into A4-60 and operating: 
(kl -k2m
)Yl,m =(L2x 
+L2y
)Yl,m =L2x
Yl,m +L2y
Yl,m 
=L2x

i 
cl,m 
i fi +Ly2
j 
dl,m 
j gj 
=
i 
cl,m 
i x2
i fi +
j 
dl,m 
j y2
j gj (A4-64) 
We can isolate kl -k2m
by multiplying from the left by (Yl,m)* and integrating. On the 
left: 
	 (Yl,m)*(kl -k2m
)Yl,mdv =(kl -k2m
) 	 (Yl,m)*Yl,mdv =(kl -k2m
) (A4-65) 
On the right: 
	 
p 
(cl,m 
p fp)*
i 
cl,m 
i x2
i fi dv +	 
q 
(dl,m 
q gq )*
j 
dl,m 
j y2
j gjdv 
=
p 

i 
(cl,m 
p )*cl,m 
i x2
i 
	 (fp)*fi dv +
q 

j 
(dl,m 
q )*dl,m 
j y2
j 
	 (gq )*gj dv 
(A4-66) 
But {f } and {g} are orthonormal sets, and so Eq. (A4-65-66) becomes 
(kl -k2m
)=
i 
|cl,m 
i |2x2
i +
j 
|dl,m 
j |2y2
i (A4-67) 
Since x and y are real, neither sum has negative terms, so (kl -k2m
)=0. (If we resort 
to a physical argument, we see the reasonableness of this since L2x 
+L2y
is the square 
of the angular momentum projection in the x,y plane.) 
Knowing that kl = k2m
, we now consider the implications of applying the step-up 
operator many times to Yl,m and then operating with Lz : 
LzL+L+L+ ···L+Yl,m =(km +1+1+1 ···+1)Yl,m+1+1+1···+1 
=kmYl,m (A4-68) 
Eventually we will get a km that is too large to satisfy k2
m =kl . At that point the series 
must terminate, which means that L+Yl,m = 0. Let us call the maximum-value km 
reached in this way kmax. Use of L- likewise gives us a lowest possible value of km 
which we call kmin. The corresponding eigenfunctions are labeled Yl,max and Yl,min. 
Since 
L+Yl,max =0 (A4-69) 
it follows that 
L-L+Yl,max =0 (A4-70) 
But, substituting for L+L- from Eq. A4-54, this can be written 
(L2 -L2z 
-Lz)Yl,max =0 (A4-71)
Appendix 4 599 
so 
k2
l -k2
max -kmax =0 (A4-72) 
or 
k2
l =k2
max +kmax (A4-73) 
A similar treatment on L-L+Yl,min yields 
k2
l =k2
min -kmin (A4-74) 
From these two expressions for kl, we have 
k2
min -k2
max =kmin +kmax (A4-75) 
This relation can be satisfied only if
kmax=-kmin (A4-76) 
What has been shown is that, for functions Yl,?, the eigenvalue of L2, kl , determines 
the maximum value of km through kl =k2
max +kmax, that the minimum value of km is 
-kmax, and that intermediate values of km are given by kmax -1, kmax -2, etc. If we 
let the value of kmax be symbolized by l, then we have that kl =l2 +l =l(l +1), so that 
L2Yl,m =l(l +1)Yl,m (A4-77) 
If we let the value of km be symbolized by m, then 
LzYl,m =mYl,m (A4-78) 
For a given value of l,m can take on the values 
m=l, l -1, l -2, . . . ,-l +1,-l (A4-79) 
There are only two possible scenarios for such a set of numbers m. One set is the 
integers; e.g., if l = 3,m = 3, 2, 1, 0,-1,-2,-3. The other set is the half-integers; 
e.g., if l =3/2,m=3/2, 1/2,-1/2,-3/2. Either way, there are 2l +1 allowed values 
of m. 
All of the above relations have been derived from the commutation relations for the 
angular momentum operators. They hold for: 
Electron orbital angular momentum L2, Lz 
Electron spin angular momentum S2, Sz 
Resultant of orbital and spin momenta J 2, Jz 
Molecular rotational angular momentum J 2, Jz 
Nuclear spin angular momentum I 2, Iz 
If we seek analytical expressions for the functions Yl,m, we can start with the assumption 
that they are separable into products of two kinds of functions, one type depending 
only on ., the other only on f: 
Yl,m =(.)(f) (A4-80)
600 Angular Momentum Rules 
Since Lz =-i./.., since LzYl,m =mYl,m, and furthermore since Yl,m must be single 
valued, it follows that (f)=(1/v2p)exp(imf), with m=0,±1,±2, . . . . Note that 
m must be an integer. We have the curious result that, if Yl,m is separable into . and f 
parts, the eigenvalues for Lz cannot belong to the half-integer set mentioned above. 
The separable, analytic functions Yl,m with l an integer are the spherical harmonics 
described in Chapter 4 and long known to classical physics. Indeed, the symbol Yl,m 
has come to stand for those functions. For systems involving half-integer l and m 
values, no analytical functions of the usual sort can be written. Instead, matrices and 
vectors are used that manifest the correct relationships. Thus, for the spin of a single 
electron, Pauli used the following representation: 
a =

1
0

, ß=

0
1

, a|a=1 0
1
0

=1 
Sz = 
1
2 

1 0 
0 -1

, Sx = 
1
2 

0 1 
1 0

, Sy = 
1
2 

0 -i 
i 0 
 
A4-5.1 Problems 
A4-1. Prove Equation A4-51. 
A4-2. Prove Equation A4-53. 
A4-3. Prove Equation A4-55. What does this equation imply about L2 and L+? 
A4-4. Evaluate Yl,m|Lx|Yl,m using only operator relations from this appendix. 
A4-5. Demonstrate that the Pauli spin matrices and vectors satisfy the following 
relations: 
a) a|ß=0 
b) S+ß =a 
c) S+a =0 
d) [Sx, Sy]=iSz 
A4-6. Use the Pauli spin matrices to evaluate S2a. 
Reference 
[1] H. Eyring, J. Walter, and G. E. Kimball, Quantum Chemistry. Wiley, New York, 
1944.
Appendix 5 
The Pairing Theorem1 
A5-1 Pairing of Roots and Relation Between Coefficients 
for Alternant Systems 
The HMO assumptions are 
Hij =
... 
.. 
a if i =j 
ß if i 
=j are bonded together 
0 ifi 
=j are not bonded together 
(A5-1) 
Sij = dij (A5-2) 
fk =
i 
cik.i (A5-3) 
where .i is a 2pp AO on carbon i. 
An alternant hydrocarbon can be labeled with asterisks to demonstrate the existence 
of two sets of carbon centers in the molecule such that no two atoms in the same set 
are nearest neighbors (see Section 8-9). The following discussion pertains to alternant 
systems. 
The simultaneous equations leading to H¨uckel energies and coefficients are of the 
form 
cix +cj +ck +cl +···=0 (A5-4) 
where x = (a -E)/ß. Atoms j, k, l must be bonded to atom i if cj, ck, cl are to be 
unequal to zero. Hence, atoms j, k, and l belong to one set of atoms, and atom i belongs 
to the other set. 
If we have already found a value of x and a set of coefficients satisfying the simultaneous 
equations (A5-4), it is easy to show that these equations will also be satisfied if 
we insert -x and also reverse the signs of the coefficients for one set of centers or the 
other. If we reverse the coefficient signs for the set j, k, l, we obtain, on the left-hand 
side 
ci(-x)-cj -ck -cl -··· (A5-5) 
which is the negative of Eq. A5-4 and hence still equals zero. If we reverse the sign of 
ci, we have 
-ci(-x)+cj +ck +cl +··· (A5-6) 
which is identical to Eq. A5-4. 
1See Coulson and Rushbrooke [1]. 
601
602 The Pairing Theorem 
This proves that each root of an alternant hydrocarbon at x(
=0) has a mate at -x 
and that their associated coefficients differ only in sign between one or the other sets of 
atoms. Note that, if x =0, the ci term vanishes, leaving coefficients for only one set of 
centers. Reversing all signs in this case corresponds to multiplying the entire MO by 
-1, which does not generate a new (linearly independent) function. Thus, it is possible 
for an alternant system to have a single, unpaired root at x =0. It is necessary for odd 
alternants to have such a root. An even alternant may have a root at x =0, but, if it has 
one such root, it must have another since, in the end, there must be an even number of 
roots. 
A5-2 Demonstration That Electron Densities Are Unity in 
Ground States of Neutral Alternant Hydrocarbons 
From n AOs result n MOs. The AOs as well as the MOs can be assumed normalized 
and (in the HMO method) orthogonal. If each AO contains one electron, the electron 
density at eachAO is unity. If theseAOs are combined to form MOs, and each MO has 
one electron, each AO electron density would still be unity, since the set of all singly 
occupied MOs is just a unitary transformation of the set of all singly occupied AOs. 
(The matrix equivalent of these statements is 
C†C=1=CC† (A5-7) 
The left equality means that the sum of squares (absolute) of coefficients over all atoms 
in one MO is unity, so the MO is normalized. The right equality means that the sum of 
squares of coefficients over one atom in all MOs is unity.) 
For an alternant hydrocarbon, however, the squares of coefficients in an MO at 
E =a +kß are identical to those in the MO at E =a -kß. Therefore, no change in 
electron density will result if each electron in the upper half of ourMOenergy spectrum 
is shifted to its lower-energy mate. Thus, the resulting state, which is the neutral ground 
state, still has unit electron density at each AO. 
A5-3 A Simple Method for Generating Nonbonding MOs 
An immediate consequence of Eq. A5-4 is that the coefficients for any nonbondingMO, 
for which x =0 by definition, satisfy the following simple rule: The coefficients on all 
the atoms attached to any common atom sum to zero. Consider the nonbonding MOs 
below: 
0
0 0 
0 
0 
– 
– 
– – 
It is clear that, no matter which atom one chooses as reference, the sum of coefficients 
of attached atoms is zero.
Appendix 5 603 
It is possible to use this observation to generate nonbonding MOs without the aid 
of a computer or tabulation. For odd-alternant systems such an MO is guaranteed to 
exist, and the recipe is especially easy, so we start with these. The procedure is: 
1. Divide the centers into asterisked and unasterisked sets, as described in Chapter 8. 
One set will have fewer centers. Set all of the coefficients in this set equal to zero. 
2. Choose one of the nonzero sites and set its coefficient to be x. Then work around 
the molecule, setting other coefficients to values needed to satisfy the “sum to zero” 
rule. 
3. When finished, evaluate x by requiring that the sum of squares of coefficients equal 
unity. 
For example, consider the naphthyl system: 
8 
7 
5 
6 
4 
3
2 
9 1 
10 
11 
* 
* * 
* 
* 
* 
The asterisked centers are more numerous, so we set the others to zero. 
0 0 
0 0 
0 
Now we set one of the nonzero sites equal to x. Let us use the one marked with an 
arrow. Then we move around the molecule, setting the other values. 
0 
0 
0 
0 
0 
0 
0 0 
0 0 
0 
0 0 
0 0 
0 
0 0 
0 0 
x x
–x 
–x 
–x 
–x 
–x 
–x 
x x x x 
2x 2x –3x 
Now we set the sum of squares to unity: 17x2 =1;x =0.242. The final nonbonding 
MO is 
–0.242 –0.727 
–0.242 
0.242 0.242 
0.484 
0 
0 
0 
0 
0 
If an even alternant has a nonbonding MO, then it must have a pair of them. Each 
one corresponds to setting a different subset of coefficients equal to zero. 
This simple recipe would be little more than a parlor trick were it not for the fact 
that the nonbonding MO is often very important in determining a molecule’s chemical 
or physical properties. For example, the neutral molecule used above as an example 
is alternant, hence has p-electron densities of one at every carbon. However the spin 
density is controlled by the nonbonding MO, since that is where the unpaired electron 
is, so the ESR splitting pattern should correspond to a series of coupling constants 
proportional to the squares of these nonbonding MO coefficients (if we ignore negative 
spin density). 
If the neutral radical is ionized to the cation, the electron is lost from the nonbonding 
MO, and so the deficiency of electronic density (i.e., the positive charge) appears on
604 The Pairing Theorem 
carbons with nonzero coefficients, in proportion to the squares of the coefficients. In 
the case of our previous example, the cation charge distribution is predicted to be 
+0.06 +0.23 +0.53 
+0.06 
+0.06 
+0.06 
0 
0 
0 
0 
0 
If we were to modify the molecule chemically by attaching a methyl group (which 
donates p electron charge via hyperconjugation) or by substituting for carbon a more 
electronegative nitrogen atom, we could hope to influence the ease of ionization. But 
it is apparent that such modifications will have greatest effect if they occur at sites 
where the largest changes in electron density occur upon ionization, i.e., at positions 
11 and 1. Modifying the molecule at a position where a zero coefficient exists in 
the nonbonding MO (i.e., positions 4, 5, or 7) should have little effect on the ease of 
carbocation formation. 
Reference 
[1] C. A. Coulson and G. S. Rushbrooke, Proc. Cambridge Phil. Soc. 36, 193 (1940).
Appendix 6 
H¨uckel Molecular Orbital Energies, 
Coefficients, Electron Densities, 
and Bond Orders for Some Simple 
Molecules 
Each molecule is labeled as alternant or nonalternant. For alternants, only the occupied 
MO data are tabulated since the remainder may be generated by use of the pairing 
theorem (see Appendix 5). Bond orders are tabulated for only one bond from each 
symmetry-equivalent set in a molecule. 
x HMO root =(a -E)/ß 
n number of electrons in MO when molecule is in neutral ground state 
ci LCAO-MO coefficient of AO at atom i 
qi p-electron density on atom i 
pij p-bond order between atoms i and j 
Ep total p energy of the molecule=sum of p-electron energies 
Molecules in this tabulation are grouped according to the number of centers in the 
conjugated system. (In all cases, it is assumed that the system is planar and undistorted 
which, in some cases, is not correct.) 
605
606 H¨uckel Molecular Orbitals 
Index of System Tabulated 
Butadiene-2,3-bimethyl 
Benzene 
Fulvene 
Heptatrienyl 
Benzyl 
Cycloheptatrienyl 
Cyclooctatetraene 
Octatetraene 
Benz-cyclopentadienyl 
Azulene 
Hexatriene Naphthalene 
Cyclobutadienyl methyl 
Cyclopentadienyl 
Pentadienyl 
Methylene cyclopropene 
2-Allylmethyl 
Cyclobutadiene 
Butadiene 
Cyclopropenyl 
Allyl 
Ethylene
Appendix 6 607 
Two Centers 
Ethylene (alternant) C2H2 
all q =1.0,p12 =1.0, 
Ep =2a +2ß 
n x c1 c2 
2 -1.000 0.7071 0.7071 
Three Centers 
Allyl radical (alternant) C3H5 
all q =1.0,p12 =0.707, 
Ep =3a +2.8284ß 
n x c1
c2 c3 
2 -1.4142 0.5000 0.7071 0.5000 
1 0.0000 0.7071 0.0000 -0.7071 
Cyclopropenyl radical (nonalternant) C3H3 
all q =1.0,p12 =0.5, 
Ep =3a +3.0000ß 
n x c1
c2 c3 
2 -2.0000 0.5774 0.5774 0.5774 
1
2 1.0000 -0.8165 0.4082 0.4082 
1
2 1.0000 0.0000 0.7071 -0.7071
608 H¨uckel Molecular Orbitals 
Four Centers 
Butadiene (alternant) C4H6 
all q =1.0,p12 =0.8944,p23 =0.4472, 
Ep =4a +4.4721ß 
n x c1
c2 c3 c4 
2 -1.6180 0.3718 0.6015 0.6015 0.3718 
2 -0.6180 0.6015 0.3718 -0.3718 -0.6015 
Cyclobutadiene (alternant) C4H4 
all q =1.0,p12 =0.5,Ep =4a +4.000ß 
n x c1
c2 c3 c4 
2 -2.0000 0.5000 0.5000 0.5000 0.5000 
1 0.0000 0.5000 0.5000 -0.5000 -0.5000 
1 0.0000 0.5000 -0.5000 -0.5000 0.5000 
2-Allylmethyl (alternant) C4H6 
all q =1.0,p12 =0.5774,Ep =4a +3.4641ß 
n x c1
c2 c3 c4 
2 -1.7320 0.7071 0.4082 0.4082 0.4082 
1 0.0 0.0000 0.7071 -0.7071 0.0000 
1 0.0 0.0000 0.4082 0.4082 -0.8165 
Methylene cyclopropene C4H4 (nonalternant) 
Ep =4a +4.9624ß,p12 =0.4527, 
p23 =0.8176,p14 =0.7583 
n x c1
c2 c3 c4 
2 -2.1701 0.6116 0.5227 0.5227 0.2818 
2 -0.3111 0.2536 -0.3682 -0.3682 0.8152 
0 1.0000 0.0000 0.7071 -0.7071 0.0000 
0 1.4812 0.7494 -0.3020 -0.3020 -0.5059 
qi = 0.8768 0.8176 0.8176 1.4881
Appendix 6 609 
Five Centers 
Pentadienyl radical (alternant) C5H7 
all q =1.0,p12 =0.7887,p23 =0.5774, 
Ep =5a +5.4641ß 
n x c1
c2 c3 c4 c5 
2 -1.7320 0.2887 0.5000 0.5774 0.5000 0.2887 
2 -1.0000 0.5000 0.5000 0.0000 -0.5000 -0.5000 
1 0.0000 0.5774 0.0000 -0.5774 0.0000 0.5774 
Cyclopentadienyl radical (nonalternant) C5H5 
all q =1.0,p12 =0.5854,Ep =5a +5.8541ß 
n x c1
c2 c3 c4 c5 
2 -2.0000 0.4472 0.4472 0.4472 0.4472 0.4472 
3/2 -0.6180 0.6325 0.1954 -0.5117 -0.5117 0.1954 
3/2 -0.6180 0.0000 -0.6015 -0.3718 0.3718 0.6015 
0 1.6180 0.6325 -0.5117 0.1954 0.1954 -0.5117 
0 1.6180 0.0000 0.3718 -0.6015 0.6015 -0.3718 
Cyclobutadienylmethyl radical (alternant) C5H5 
all q =1.0,p12 =0.3574,p23 =0.6101 
p15 =0.8628,Ep =5a +5.5959ß 
n x c1
c2 c3 c4 c5 
2 -2.1358 0.5573 0.4647 0.4351 0.4647 0.2610 
2 -0.6622 -0.4351 0.1845 0.5573 0.1845 -0.6572 
1 0.0000 0.0000 -0.7071 0.0000 0.7071 0.0000
610 H¨uckel Molecular Orbitals 
Six Centers 
Hexatriene (alternant) C6H8 
all q =1.0,p12 =0.8711,p23 =0.4834, 
p34 =0.7848,Ep =6a +6.9879ß 
n x c1
c2 c3 c4 c5 c6 
2 -1.8019 0.2319 0.4179 0.5211 0.5211 0.4179 0.2319 
2 -1.2470 0.4179 0.5211 0.2319 -0.2319 -0.5211 -0.4179 
2 -0.4450 0.5211 0.2319 -0.4179 -0.4179 0.2319 0.5211 
Butadiene-2,3-bimethyl (alternant) C6H8 
all q =1.0,p12 =0.6667,p23 =0.3333,Ep =6a +6ß 
n x c1
c2 c3 c4 c5 c6 
2 -2.0000 0.2887 0.5774 0.5774 0.2887 0.2887 0.2887 
2 -1.0000 0.4082 0.4082 -0.4082 -0.4082 0.4082 -0.4082 
2 0.0000 -0.5000 0.0000 0.0000 -0.5000 0.5000 0.5000 
Benzene (alternant) C6H6 
all q =1.0,p12 =0.6667,Ep =6a +8ß 
n x c1
c2 c3 c4 c5 c6 
2 -2.0000 0.4082 0.4082 0.4082 0.4082 0.4082 0.4082 
2 -1.0000 0.0000 0.5000 0.5000 0.0000 -0.5000 -0.5000 
2 -1.0000 0.5774 0.2887 -0.2887 -0.5774 -0.2887 0.2887 
Fulvene (nonalternant) C6H6 
Ep =6a +7.4659ß,p12 =0.7779,p23 =0.5202,p45 =0.4491, 
p56 =0.7586 
n x c1
c2 c3 c4 c5 c6 
2 -2.1149 0.4294 0.3851 0.3851 0.4294 0.5230 0.2473 
2 -1.0000 0.0000 0.5000 0.5000 0.0000 -0.5000 -0.5000 
2 -0.6180 0.6015 0.3718 -0.3718 -0.6015 0.0000 0.0000 
0 0.2541 -0.3505 0.2795 0.2795 -0.3505 -0.1904 0.7495 
0 1.6180 -0.3718 0.6015 -0.6015 0.3718 0.0000 0.0000 
0 1.8608 -0.4390 0.1535 0.1535 -0.4390 0.6635 -0.3566 
qi = 1.0923 1.0730 1.0730 1.0923 1.0470 0.6223
Appendix 6 611 
Seven Centers 
Heptatrienyl radical (alternant) C7H9 
all q =1.0,Ep =7a +8.0547ß 
p12 =0.8155,p23 =0.5449,p34 =0.6533 
n x c1
c2 c3 c4 c5 c6 c7 
2 -1.8478 0.1913 0.3536 0.4619 0.5000 0.4619 0.3536 0.1913 
2 -1.4142 0.3536 0.5000 0.3536 0.0000 -0.3536 -0.5000 -0.3536 
2 -0.7654 0.4619 0.3536 -0.1913 -0.5000 -0.1913 0.3536 0.4619 
1 0.0000 -0.5000 0.0000 0.5000 0.0000 -0.5000 0.0000 0.5000 
Benzyl radical (alternant) C7H7 
all q =1.0,p12 =0.5226,p23 =0.7050, 
p34 =0.6350,p17 =0.6350,Ep =7a +8.7206ß 
n x c1
c2 c3 c4 c5 c6 c7 
2 -2.1010 0.5000 0.4063 0.3536 0.3366 0.3536 0.4063 0.2380 
2 -1.2593 -0.5000 -0.1163 0.3536 0.5615 0.3536 -0.1163 -0.3970 
2 -1.0000 0.0000 0.5000 0.5000 0.0000 -0.5000 -0.5000 0.0000 
1 0.0000 0.0000 -0.3780 0.0000 0.3780 0.0000 -0.3780 0.7560 
Cycloheptatrienyl radical (nonalternant) C7H7 
all q =1.0,p12 =0.6102,Ep =7a +8.5429ß 
n x c1
c2 c3 c4 c5 c6 c7 
2 -2.0000 0.3780 0.3780 0.3780 0.3780 0.3780 0.3780 0.3780 
2 -1.2470 -0.5345 -0.3333 0.1189 0.4816 0.4816 0.1189 -0.3333 
2 -1.2470 0.0000 -0.4179 -0.5211 -0.2319 0.2319 0.5211 0.4179 
1
2 0.4450 0.5345 -0.1189 -0.4816 0.3333 0.3333 -0.4816 -0.1189 
1
2 0.4450 0.0000 0.5211 -0.2319 -0.4180 0.4180 0.2319 -0.5211 
0 1.8019 -0.5345 0.4816 -0.3333 0.1189 0.1189 -0.3333 0.4816 
0 1.8019 0.0000 0.2319 -0.4179 0.5211 -0.5211 0.4179 -0.2319
612 H¨uckel Molecular Orbitals 
Eight Centers 
Octatetraene (alternant) C8H10 
all q =1.0,p12 =0.8621,p23 =0.4948, 
p34 =0.7581,p45 =0.5288,Ep =8a +9.5175ß 
n x c1
c2 c3 c4 c5 c6 c7 c8 
2 -1.8794 0.1612 0.3030 0.4082 0.4642 0.4642 0.4082 0.3030 0.1612 
2 -1.5321 0.3030 0.4642 0.4082 0.1612 -0.1612 -0.4082 -0.4642 -0.3030 
2 -1.0000 -0.4082 -0.4082 0.0000 0.4082 0.4082 0.0000 -0.4082 -0.4082 
2 -0.3473 0.4642 0.1612 -0.4082 -0.3030 0.3030 0.4082 -0.1612 -0.4642 
Cyclooctatetraene (alternant) C8H8 
all q =1.0,p12 =0.6035,Ep =8a +9.6568ß 
n x c1
c2 c3 c4 c5 c6 c7 c8 
2 -2.0000 0.3536 0.3536 0.3536 0.3536 0.3536 0.3536 0.3536 0.3536 
2 -1.4142 0.3536 0.0000 -0.3536 -0.5000 -0.3536 0.0000 0.3536 0.5000 
2 -1.4142 0.3536 0.5000 0.3536 0.0000 -0.3536 -0.5000 -0.3536 0.0000 
1 0.0000 0.3536 0.3536 -0.3536 -0.3536 0.3536 0.3536 -0.3536 -0.3536 
1 0.0000 0.3536 -0.3536 -0.3536 0.3536 0.3536 -0.3536 -0.3536 0.3536 
Nine Centers 
Benzcyclopentadienyl radical (nonalternant) C9H7 
p12 =0.6592,p49 =0.6071,p56 =0.6630, 
p18 =0.4790,p45 =0.6363,p89 =0.5118, 
Ep =9a +11.8757ß 
n x c1 c2 c3 c4 c5 c6 c7 c8 c9 
2 -2.3226 0.3203 0.2758 0.3203 0.2988 0.2259 0.2259 0.2988 0.4681 0.4681 
2 -1.5450 -0.3114 -0.4031 -0.3114 0.2689 0.4934 0.4934 0.2689 -0.0780 -0.0780 
2 -1.1935 0.2992 0.0000 -0.2992 -0.4841 -0.2207 0.2207 0.4841 0.3571 -0.3571 
2 -0.7293 -0.2054 -0.5634 -0.2054 0.0935 -0.3454 -0.3454 0.0935 0.4136 0.4136 
1 -0.2950 0.5428 0.0000 -0.5428 0.3355 0.2591 -0.2591 -0.3355 0.1601 -0.1601 
0 0.9016 0.1548 -0.3434 0.1548 -0.5424 0.2852 0.2852 -0.5424 0.2038 0.2038 
0 1.2950 -0.2591 0.0000 0.2591 -0.1601 0.5428 -0.5428 0.1601 0.3355 -0.3355 
0 1.6952 0.4840 -0.5711 0.4840 0.1884 -0.0699 -0.0699 0.1884 -0.2495 -0.2495 
0 2.1935 -0.2207 0.0000 0.2207 0.3571 -0.2992 0.2992 -0.3571 0.4841 -0.4841 
qi = 0.9571 1.1119 0.9571 0.9218 0.9920 0.9920 0.9218 1.0730 1.0730
Appendix 6 613 
Ten Centers 
Azulene (nonalternant) C10H8 
p12 =0.6560, p4,10 =0.5858, p56 =0.6389, 
p19 =0.5956, p45 =0.6640, p9,10 =0.4009, 
Ep =10a +13.3635ß 
n x c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 
2 -2.3103 0.3233 0.2799 0.3233 0.2886 0.1998 0.1730 0.1998 0.2886 0.4670 0.4670 
2 -1.6516 -0.2678 -0.3243 -0.2678 0.1909 0.4333 0.5247 0.4333 0.1909 -0.1180 -0.1180 
2 -1.3557 0.2207 0.0000 -0.2207 -0.4841 -0.3571 0.0000 0.3571 0.4841 0.2992 -0.2992 
2 -0.8870 -0.2585 -0.5829 -0.2585 0.2186 -0.1598 -0.3603 -0.1598 0.2186 0.3536 0.3536 
2 -0.4773 0.5428 0.0000 -0.5428 0.1601 0.3355 0.0000 -0.3355 -0.1601 0.2591 -0.2591 
0 0.4004 -0.0632 0.3158 -0.0632 0.4699 0.1023 -0.5109 0.1023 0.4699 -0.2904 -0.2904 
0 0.7376 -0.2992 0.0000 0.2992 -0.3571 0.4841 0.0000 -0.4841 0.3571 0.2207 -0.2207 
0 1.5792 0.4364 -0.5527 0.4364 -0.0844 0.2697 -0.3416 0.2697 -0.0844 -0.1365 -0.1365 
0 1.8692 -0.2500 0.2675 -0.2500 -0.3233 0.4045 -0.4328 0.4045 -0.3233 0.1998 0.1998 
0 2.0953 -0.2591 0.0000 0.2591 0.3355 -0.1601 0.0000 0.1601 -0.3355 0.5428 -0.5428 
qi = 1.1729 1.0466 1.1729 0.8550 0.9864 0.8700 0.9864 0.8550 1.0274 1.0274 
Naphthalene (alternant) C10H8 
all q =1.0, Ep =10a +13.6832ß 
p12 =0.7246, p23 =0.6032, p19 =0.5547, p9,10 =0.5182 
n x c1 c2 c3 c4 c5 c6 c7 c8 c9 c10 
2 -2.3028 0.3006 0.2307 0.2307 0.3006 0.3006 0.2307 0.2307 0.3006 0.4614 0.4614 
2 -1.6180 0.2629 0.4253 0.4253 0.2629 -0.2629 -0.4253 -0.4253 -0.2629 0.0000 0.0000 
2 -1.3028 0.3996 0.1735 -0.1735 -0.3996 -0.3996 -0.1735 0.1735 0.3996 0.3470 -0.3470 
2 -1.0000 0.0000 -0.4082 -0.4082 0.0000 0.0000 -0.4082 -0.4082 0.0000 0.4082 0.4082 
2 -0.6180 0.4253 0.2629 -0.2629 -0.4253 0.4253 0.2629 -0.2629 -0.4253 0.0000 0.0000
Appendix 7 
Derivation of the Hartree–Fock 
Equation 
This appendix is divided into two parts. In the first section we develop the formula for 
the expectation value E¯ =
.|H|. for the case in which . is a single determinantal 
wavefunction over MOs. In the second section we derive the Hartree–Fock equation 
by requiring E¯ to be stationary with respect to variations in .. 
A7-1 The Expansion of E¯ in Terms of Integrals over MOs 
We limit discussion to the case in which . is a single, closed-shell determinant. We 
will develop our arguments by referring to a four-electron example: 
.4 =(4!)-1/2 


f1(1) ¯ f1(2)f2(3) ¯ f2(4)


(A7-1) 
Recall that this is the shorthand formula for a Slater determinant. Each f is a normalized 
MO,theMOsare assumed to be orthogonal, and a bar signifies that an electron possesses 
ß spin. As we develop our arguments within the context of ., we will generalize them 
to apply to the general 2n electron closed-shell wavefunction 
.2n =[(2n!)]-1/2|f1(1)f¯1(2)f2(3)f¯2(4) ···fn(2n-1)f¯n(2n)| (A7-2) 
When .4 is expanded according to the rule for determinants (Appendix 2), we obtain 
4! products. We note the following features of the expanded form. 
1. There is one product, occurring with coefficient+1, which is identical to the product 
appearing in the shorthand form of Eq. (A7-1). We refer to this as the “leading 
term.” 
2. An equivalent way of expressing a Slater determinant is via the expression (for .4) 
.4 =(4!)-1/2
P 
(-1)pP f1(1)f¯1(2)f2(3)f¯2(4) (A7-3) 
Here P stands for all the sequences of permutations of electron labels that lead 
to different products (i.e., P is a permutation operator), and p is the number of 
pairwise permutations in a given sequence. For .4 there are 4! sequences P, the 
simplest being “no permutations” (hence, p =0) which produces the leading term. 
Then there are single permutations, such as P1,2 (with p =1), which produces the 
term -f1(2)f¯1(1)f2(3)f¯2(4). There are also double permutations, etc. According 
to Eq. (A7-3), any term differing from the leading term by an odd number of 
614
Appendix 7 615 
permutations will appear with coefficient -1. We will be particularly concerned 
with products that differ from the leading term by a single permutation. 
3. A single permutation may be made to occur between electrons in MO’s with the 
same spins or different spins. In the latter case, two electrons in the singly permuted 
product will disagree in spin with their counterparts in the leading term. In the 
former case, no such spin disagreement will exist. 
4. Terms also appear in .4 corresponding to more than a single permutation of electron 
indices. 
It is useful to pick a representative example of each type of product mentioned above. 
For .4, we have the following: 
Leading term: f1(1)f¯1(2)f2(3)f¯2(4) 
Singly permuted term: f1(1)f¯1(4)f2(3)f¯2(2); Spin agreement: (P2,4) 
Singly permuted term: f1(2)f¯1(1)f2(3)f¯2(4); Spin disagreement: (P1,2) 
Doubly permuted term: f1(2)f¯1(4)f2(3)f¯2(1); (P1,2P1,4) 
Upon expanding 
.4|Hˆ |.4, we obtain a set of 4! products on both the left and righthand 
sides of Hˆ : 
E¯ = (4!)-1  {f1*(1)f¯1*(2)f2*(3)f¯2*(4)-f1*(1)f¯1*(4)f2*(3)f¯2*(2)-···} 
×Hˆ (1, 2, 3, 4){f1(1)f¯1(2)f2(3)f¯2(4)-f1(1)f¯1(4)f2(3)f¯2(2)-···}dt (A7-4) 
This can be expanded into a set of integrals, one for each term on the left: 
E¯ = (4!)-1  f1*(1)f¯1*(2)f2*(3)f¯2*(4)Hˆ (1, 2, 3, 4) f1(1)f¯1(2)f2(3)f¯2(4) 
- f1(1)f¯1(4)f2(3)f¯2(2)-f1(2)f¯1(1)f2(3)f¯2(4)-··· dt 
- f1*(1)f¯1*(4)f2*(3)f¯2*(2)Hˆ (1, 2, 3, 4) f1(1)f¯1(2)f2(3)f¯2(4) 
- f1(1)f¯1(4)f2(3)f¯2(2)-f1(2)f¯1(1)f2(3)f¯2(4)-··· dt	 (A7-5) 
etc. In Eq. (A7-5), E¯ is a sum of 4! integrals, each containing one term from the set on 
the left of Hˆ and all 4! from the set on the right. 
At first, it might seem that we must evaluate all of the 4! integrals in Eq. (A7-5). But 
it can be shown that these integrals, times their +1 or -1 coefficients, are all equal to 
each other, enabling us to write E¯ as 4! times the first integral: 
E¯ =  f1*(1)f¯1*(2)f2*(3)f¯2*(4)Hˆ (1, 2, 3, 4) 

P 
(-1)pP(f1(1)f¯1(2)f2(3)f¯2(4)) dt (A7-6)
616 Derivation of the Hartree–Fock Equation 
The demonstration that the various integrals in Eq. (A7-5), times their coefficients, 
are equal to each other is as follows. Consider the second integral in Eq. (A7-5). 
Note that, if we permute electrons 2 and 4 in that integral, we restore the term on the 
left of Hˆ to its original “leading term” order, thereby making that product identical 
to its counterpart in the first integral. Furthermore, if we carry out this permutation 
throughout the whole of the integrand of integral number 2 (i.e., in Hˆ , in all 4! products 
to the right of Hˆ , and in dt ), we will not affect the value of the integral. (Recall that, 
for example, 
 1 
0 
x2dx  2 
1 
y3dy = 1 
0 
y2dy  2 
1 
x3dx 
The result of this permutation P2,4 on the second integral in Eq. (A7-5) is 
 f1*(1)f¯1*(2)f2*(3)f¯2*(4)Hˆ (1, 4, 3, 2) f1(1)f¯1(4)f2(3)f¯2(2) 
-f1(1)f¯1(2)f2(3)f¯2(4)-··· dt (A7-7) 
Now Hˆ is invariant under exchange of electron indices, and the set of products to the 
right of Hˆ in Eq. (A7-7) is the same set we had before, but their order is changed, and 
the whole product evidently differs by a factor of -1 from the set in the first integral. 
Therefore, we can say that the first and second integrals of Eq. (A7-5) have the same 
absolute value but different signs. However, the fact that these integrals contribute to 
E¯ with opposite signs cancels the sign disagreement. In this way, every integral in 
Eq. (A7-5) can be compared to the leading integral and Eq. (A7-6) verified. This much 
simplified expression for E¯ is, for the 2n-electron case 
E¯ = f1*(1)f¯1*(2) ···f*n(2n-1)f¯*n(2n)Hˆ (1, 2, . . . , 2n) 



P 
(-1)pP(f1(1)f¯1(2) ···fn(2n-1)f¯n(2n))

dt (A7-8) 
At this point we write out Hˆ more explicitly. It is, in atomic units, 
Hˆ (1, 2, . . . , 2n) = 
2n 

i=1 

-
1
2.2
i -
nuclei 

µ 
Zµ/rµi


+
2n-1 

i=1 
2n 
 
j=i+1 
1/rij (A7-9) 
= 
2n 

i=1 
Hcore 
(i) + 1/rij (A7-10) 
Here,  is a shorthand symbol for the double sum in Eq. (A7-9). We see that Hˆ is 
composed of one-electron operators, Hcore 
(i) , which deal with the kinetic and nuclearelectron 
attraction energies for electron i, and two-electron operators for interelectronic 
repulsion. The internuclear repulsion is omitted since, for a given nuclear configuration, 
it is simply a constant that can be added to the electronic energy. Note for future 
reference that Hˆ has no dependence on electron spin coordinates.
Appendix 7 617 
If we insert the expression (A7-10) for Hˆ into Eq. (A7-6) for E¯, it is easy to to see 
that we can expand the result into separate sets of integrals over one-and two-electron 
operators. We consider first the one-electron integrals. For .4, these are 
 f1*(1)f¯1*(2)f2*(3)f¯2*(4) 
× 
4 

i=1 
Hcore 
(i) f1(1)f¯1(2)f2(3)f¯2(4)-f1(1)f¯1(4)f2(3)f2(2)-··· dt (A7-11) 
Upon expanding this, the first set of integrals we obtain is 
 f1*(1)f¯1*(2)f2*(3)f¯2*(4) Hcore 
(1) +Hcore 
(2) +Hcore 
(3) +Hcore 
(4) f1(1)f¯1(2)f2(3)f¯2(4)dt 
(A7-12) 
This can be expanded again. The first integral contains the operator Hcore 
(1) , which does 
not act on electrons 2-4. This allows a separation into a product of integrals as follows: 
 f1*(1)f¯1*(2)f2*(3)f¯2*(4)Hcore 
(1) f1(1)f¯1(2)f2(3)f¯2(4) dt (A7-13) 
= f*1 (1)Hcore 
(1) f1(1)dt1  f¯1*(2)f¯1(2) dt2  f*2 (3)f2(3)dt3  f¯2*(4)f¯2(4) dt4 
(A7-14) 
The last three integrals are overlap integrals and are all unity by virtue of normality of 
the MOs. The first integral, a “core integral,” is normally symbolized H11. Here the 
subscripts refer to the MO index, not the electron index: 
Hii = f*i (1)Hcore 
(1) fi(1)dt1 (A7-15) 
Therefore, Eq. (A7-13) equalsH11. By continued expansion, Eq. (A7-12) can be shown 
to be equal to H11 + H11 + H22 + H22 = 2(H11 + H22). In this case, we have been 
dealing with identical MO products on the two sides of the operator. As we continue 
evaluating the expansion of Eq. (A7-12), we next encounter an integral in which the 
products differ by a permutation, namely, 
- f1*(1)f¯1*(2)f2*(3)f¯2*(4)Hcore 
(1) +Hcore 
(2) +Hcore 
(3) +Hcore 
(4) f1(1)f¯1(4)f2(3)f¯2(2) dt 
(A7-16) 
Again, for Hcore(1), this may be written 
- f*1 (1)Hcore 
(1) f1(1)dt1  f¯1*(2)f¯2(2) dt2  f*2 (3)f2(3)dt3  f¯2*(4)f¯1(4) dt4 
(A7-17) 
Orbital orthogonality will cause the second and fourth integrals to vanish. In general, 
if the two products differ by one or more permutations, they will have two or more sites 
of disagreement. Upon expansion, at least one disagreement will occur in an overlap
618 Derivation of the Hartree–Fock Equation 
integral, causing the integral to vanish. Thus, except for the integral involving identical 
products [Eq. (A7-12)], all the integrals obtained by expansion of Eq. (A7-11) vanish. 
Our result, generalized to the 2n-electron case, is (seeAppendix 11 for bra-ket notation) 

.2n





 
2n 

i=1 
Hcore 
(1) 





 
.2n

= 
n 

i=1 
2Hii (A7-18) 
We now turn to integrals containing two-electron operators. Consider, for example, 
an integral that has products differing in two places, 
-f1(1)f¯1(2)f2(3)f¯2(4) |1/r13|f1(3)f¯1(2)f2(1)f¯2(4) (A7-19) 
This can be partly separated into a product of integrals over different electron 
coordinates. 
(A7-19)=-
f1(1)f2(3)|1/r13|f1(3)f2(1)
f¯1(2)|f¯1(2)
f¯2(4)|f¯2(4) (A7-20) 
Observe that the two disagreements are inside the two-electron integral, and that the 
overlap terms both show complete internal agreement and are therefore equal to unity. 
It is clear that, if our two products differed in more than two places, at least one such 
disagreement would appear in an overlap integral, causing the whole integral to vanish. 
Therefore, two-electron integrals need be considered only if they involve products 
differing by zero or one permutations. Let us consider these two possibilities separately. 
If there are no disagreements, we have for .4, 
f1(1) ¯ f1(2)f2(3) ¯ f2(4)



 1/rtj 



f1(1)f¯1(2)f2(3)f¯2(4) 
=
f1(1)f¯1(2)|1/r12|f1(1)f¯1(2)+
f1(1)f2(3)|1/r13|f1(1)f2(3) 
+
f1(1)f¯2(4)|1/r14|f1(1)f¯2(4)+
f¯1(2)f2(3)|1/r23|f¯1(2)f2(3) 
+
f¯1(2)f¯2(4)|1/r24|f¯1(2)f¯2(4)+
f2(3)f¯2(4)|1/r34|f2(3)f¯2(4) (A7-21) 
These integrals give the coulombic repulsion between electrons in MOs. They are 
symbolized Jij , where 
Jij =fi(1)fj (2) |1/r12|fi(1)fj.(2)=
ij |ij  (A7-22) 
Here fi and fj may be associated with either spin. Because the operator and MOs 
commute, the integrand can be rearranged to give 
Jij = f*i (1)fi(1)(1/r12)f*j (2)fj (2)dt1dt2 =(ii|jj ) (A7-23) 
Some people prefer this form because it places the two mutually repelling charge 
clouds on the two sides of the operator. It is important to realize that the parenthetical 
expression (ii|jj ) and the bra-ket shorthand 
ij |ij  correspond to different conventions 
for electron index order, and are really the same integral. Returning to Eq. (A7-21), we 
see that it is equal to 
J11 +J12 +J12 +J12 +J12 +J22 = 
2 

i=1 
.
.
Jii +
j =i 
2Jij
.
. 
(A7-24)
Appendix 7 619 
We must now consider the case where the products differ by a single permutation, 
hence in two places. Anexample has been provided in Eq. (A7-19). We noted that, when 
the operator 1/rij corresponds to electrons i and j in the positions of disagreement, 
the overlap integrals are all unity. Otherwise, at least one overlap integral vanishes. An 
integral like that in Eq. (A7-20) is called an exchange integral. 
Exchange integrals can occur only when the product on the right of the operator 
differs from the leading term by a single permutation. Hence, exchange integrals always 
enter with a coefficient of -1. 
We noted earlier that two classes of singly permuted products exist. One class 
involves permutations between electrons of like spin. In such a case, fi and fj appear 
throughout the integral Kij with spin agreement. For cases where electrons of different 
spin have been permuted, spin disagreement forces the exchange integral to vanish. 
(Since 1/rij is not a spin operator, the integration over spin coordinates factors out and 
produces a vanishing integral if spins disagree.) 
The result of all this is that each singly permuted product can give -Kij if the 
permutation is between electrons of like spin in MOs fi and fj., and zero otherwise. 
For .4, the acceptable permutations can be seen to be electron 1 with 3 and electron 2 
with 4, both of these occurring between f1 and f2 space MOs. Hence, the contribution 
to E is -2K12. Combining this with J terms gives 
.4



 1/rij 



.4=J11 +4J12 -2K12 +J22 (A7-25) 
From the definitions of J and K, it is apparent that 
Jij =Jji, Kij =Kji, Kii =Jii 
This allows us to rewrite Eq. (A7-25) as 
2J11 -K11 +2J12 -K12 +2J21 -K21 +2J22 -K22 
= 
2 

i=1 
2 

j=1
(2Jij -Kij ) (A7-26) 
Generalizing to the 2n-electron, closed-shell case and adding in our one-electron 
contribution, 
E¯ =
.2n|Hˆ |.2n=2 
n 

i=1 
Hii + 
n 

i=1 
n 

j=1
(2Jij -Kij ) (A7-27) 
This is the desired expression for E¯ in terms of integrals over MOs fi for a singledeterminantal, 
closed-shell wavefunction. 
A7-2 Derivation of the Hartree–Fock Equations 
To find the “best”MOs, we seek those that minimize E¯, that is, thoseMOs f for which 
E¯ is stationary to small variations df. But there is a restriction in the variations df. 
The MOs can only be varied in ways that do not destroy their orthonormality since this 
property was assumed in deriving Eq. (A7-27). This means that, for proper variations
620 Derivation of the Hartree–Fock Equation 
df at the minimum E¯, both E¯ and all the MO overlap integrals Sij = 
fi |fj  must 
remain constant. (Sij must equal unity when i =j , zero otherwise.) If E¯ and Sij are 
constant, any linear combination of them is constant too. Thus, we may write that, at 
the minimum E¯, 
c0E¯ +
i 

j 
cijSij =constant (A7-28) 
for our restricted type of df. This equation will hold for any set of coefficients c as 
long as df is of the proper restricted nature. However, it is possible to show that, for 
a particular set of coefficients, Eq. (A7-28) is satisfied at minimum E¯ for any small 
variations df. The particular coefficients are called Lagrangian multipliers. They are 
of undetermined value thus far, but their values will become known in the course of 
solving the problem. The technique, known as “Lagrange’s method of undetermined 
multipliers” is from the calculus of variations.1 The Lagrangian multipliers will ultimately 
turn out to be essentially the MO energies. For future convenience we write 
Eq. (A7-28) in the form 
E¯ -2
i 

j 
ijSij =constant for df (A7-29) 
where we now understand df to be unrestricted and ij to be some unknown special 
set of constants. The stability of the quantity on the left-hand side of Eq. (A7-29) may 
be expressed as follows: 
dE -2d
i 

j 
ijSij =0 (A7-30) 
or, expanding E¯, 
2 
n 

i=1 
dHii + 
n 

i=1 
n 

j=1
(2dJij -dKij )-2 
n 

i=1 
n 

j=1 
ij dSij =0 (A7-31) 
The variations occur in the MOs f, and so 
dSij =  df*i (1)fj (1)dt1 + f*i (1) dfj (1)dt1 (A7-32) 
dHii =  df*i (1)Hcore 
(1) fi(1)dt1 + f*i (1)Hcore 
(1) dfi(1)dt1 (A7-33) 
dJij =  df*i (1)f*j (2)(1/r12)fi(1)fj (2)dt1dt2 
+  f*i (1)df*j (2)(1/r12)fi(1)fj (2)dt1dt2 +complex conjugates (A7-34) 
It is convenient to define a coulomb operator Jˆi(1) as 
Jˆi(1)= f*i (2)(1/r12)fi(2)dt2 (A7-35) 
1For an introduction to this topic, see Margenau and Murphy [1].
Appendix 7 621 
Using this definition we can rewrite Eq. (A7-34) as 
dJij = dfi*(1)Jˆj (1)fi(1) dt1 + df*j (1)Jˆi(1)fj (1) dt1 +complex conjugates 
(A7-36) 
In the same spirit, we define an exchange operator Kˆi , which, because it involves an 
orbital exchange, must be written in the context of an orbital being operated on: 
Kˆi(1)fj (1)= f*i (2)(1/r12)fj (2)dt2fi(1) (A7-37) 
This enables us to write dKij 
dKij =  df*i (1)Kˆj (1)fi(1) dt1 + df*j (1)Kˆi(1)fj (1) dt1 
+ complex conjugates 
Employing the operators Jˆ and Kˆ , Eq. (A7-31) can be written as follows: 
2
i 
 df*i (1)Hcore 
(1) fi(1)+
j 
(2Jˆj (1)-Kˆj (1))fi(1)-
j 
ijfj (1)dt1 
+2
i 
 dfi(1)Hcore* (1) f*i (1)+
j 
(2Jˆ* j (1)-Kˆ *j (1))f*i (1) 
-
j 
*ijf*j (1)dt1 =0 (A7-38) 
Here we have made use of the hermitian properties of Hcore, Jˆ, and Kˆ , and also the 
relation ji  dfj (1)f*i (1)dt1=ij  dfi(1)f*j (1)dt1, which is merely an index interchange. 
Since the variations df*i and dfi are independent, each half of Eq. (A7-38) must 
independently equal zero. Hence, we can select either half for further development. We 
will select the first half. This equation indicates that the sum of integrals equals zero. 
Either the integrals are all individually equal to zero or else they are finite but cancel. 
However the latter possibility is ruled out because the variations df*i are arbitrary. By 
appropriately picking df*i , we could always spoil cancellation if the various integrals 
for different i were nonzero. But the equation states that the sum vanishes for every 
df*i . Therefore, we are forced to conclude that each integral vanishes. 
Continuing in the same spirit, we can conclude that the term in brackets in the 
integrand is zero. For the integral to vanish requires the integrand either to be identically 
zero or else to have equal positive and negative parts. If the latter were true for some 
choice of df*i , it would be possible to change df*i so as to unbalance the cancellation 
and produce a nonzero integral. Since the integral is zero for all df*i , it must be that 
the bracketed term vanishes identically. Thus, 
Hcore 
(1) +
j 
(2Jˆj (1)-Kˆj (1))fi(1)=
j 
ijfj (1) (A7-39) 
for all i =1 to n and for a certain set of constants ij .
622 Derivation of the Hartree–Fock Equation 
The original development of SCF equations was performed by Hartree for simple 
product wavefunctions. Fock later extended the approach to apply to antisymmetrized 
wavefunctions. For this reason, the collection of operators in brackets in Eq. (A7-39) 
is called the Fock operator, symbolized Fˆ, and Eq. (A7-39) becomes 
Fˆ(1)fi(1)=
j 
ijfj (1) (A7-40) 
Equation A7-40 is a differential equation for each MO fi . But as it stands it is 
not an eigenvalue equation because, instead of regenerating fi , we obtain a sum of 
functions fj times the various unknown constants ij . However, there remains a degree 
of freedom in the problem that can be used to throw Eq. (A7-40) into eigenvalue form. 
It is pointed out in Appendix 2 that the value of a determinant is unchanged if any row 
or column, multiplied by a constant, is added to any other row or column. This means 
that a Slater determinant of “best” MOs is unaffected by such internal rearrangements. 
In other words, if we were to solve Eq. (A7-40) for a set of “best” MOs, fb
i , we could 
form various new orthonormal MOs, (e.g., fb
i +.ikfb
k,fb
k -.kifb
i ) by mixing them 
together, and our wavefunction ., and all values of observables predicted from ., 
including E¯, would be precisely the same. 
A transformation that mixes the MOs f without affecting the property of orthonormality 
is called a unitary transformation (see Chapter 9). Letting U stand for such a 
transformation, we have that a transformed set of f’s, called f, is given by 
fi =
j 
Ujifj, i=1, . . . ,n (A7-41) 
In matrix notation, this is 
˜  =˜ U (A7-42) 
where ˜  and ˜ are row vectors, viz. 
˜  =(12 ···n), (A7-43) 
and U is an n×n matrix, with 
UU† =U†U=1 (A7-44) 
In terms of these matrices, Eq. (A7-40) is 
Fˆ˜ =˜ E (A7-45) 
where E is an n×n matrix. If we multiply this from the right by U, we obtain 
Fˆ˜ U=˜ EU (A7-46) 
Inserting 1 (in the form UU†) between ˜ and E gives 
Fˆ˜ U=˜ UU†EU (A7-47) 
or 
Fˆ˜  =˜ U†EU (A7-48)
Appendix 7 623 
We can now require that the matrix U be such that U†EU is a diagonal matrix E. 
(This requires that E be a hermitian matrix, which can be shown to be the case.)2 This 
requirement defines U, and we have 
Fˆ˜  =˜ E (A7-49) 
which corresponds to 
Fˆfi =ifi, i=1, 2, . . . ,n (A7-50) 
This equation has the desired eigenvalue form, and is commonly referred to as the 
Hartree–Fock equation. It is discussed at length in Chapter 11. 
It is important to bear in mind that our transformation U is for mathematical convenience 
and has no physical effect. We may imagine that our original basis set spans a 
certain function space, and that solution of Eq. (A7-50) produces a set of occupied MOs 
fi that span a “best” subspace. Transformations by unitary matrices produce new sets 
of MOs f but these still span the same subspace as f. However, they are generally 
not eigenfunctions of Fˆ, and satisfy the less convenient Eq. (A7-40). Nonetheless, 
there are occasions when it is useful to use some set of MOs other than f, and we can 
always do this without having to worry about introducing physical changes as long as 
our converted MO’s are related to f by a unitary transformation. 
The Hartree–Fock equation is ordinarily used in quantum chemistry in connection 
with a basis set of AOs, and it is possible to carry through a derivation of the Hartree– 
Fock equation for this type of basis. Detailed treatments of this derivation may be 
found in the paper by Roothaan [2] and in the book by Pople and Beveridge [3]. 
References 
[1] H. Margenau and G. M. Murphy, The Mathematics of Physics and Chemistry. 
Van Nostrand-Reinhold, Princeton, New Jersey, 1956. 
[2] C. C. J. Roothaan, Rev. Mod. Phys. 23, 69 (1951). 
[3] J. A. Pople and D. L. Beveridge, Approximate Molecular Orbital Theory. 
McGraw-Hill, NewYork, 1970. 
2See Roothaan [2].
Appendix 8 
The Virial Theorem for Atoms 
and Diatomic Molecules 
A8-1 Atoms 
In Chapter 3 itwas shownthat, for the ground state of the quantum-mechanical harmonic 
oscillator, the average value of the kinetic energy is equal to the average value of the 
potential energy. We now consider how the average electronic kinetic and potential 
energies are related in an atom. We begin by deriving a rather general expression, 
and then we discuss how it applies to different levels of calculation. As our first step, 
we examine the effects of coordinate scaling on average values. In order to follow 
this discussion, it is useful to recall that one can manipulate variables and limits in an 
integral as follows: 

 b 
a 
f (x)dx =
 b 
a 
f (y)dy =
 .x=b 
.x=a 
f (.x)d(.x)=. 
 x=b/. 
x=a/. 
f (.x)dx (A8-1) 
Let .(r1, r2, . . . , rn) be a normalized function of the space coordinates of n electrons. 
Let 
T¯ = 
.|Tˆ|. (A8-2) 
V¯ = 
.|Vˆ |. (A8-3) 
where Tˆ and Vˆ are, respectively, the kinetic and potential energy operators for some 
system, and are independent of spin. 
We introduce a scale factor . into .. This factor affects the lengths of the vectors 
ri but not their directions. That is, 
.. =.(.r1,.r2, . . . ,.rn) (A8-4) 
lf .>1,.n is more contracted in 3n-dimensional space than .. For .<1,.n is 
more diffuse. 
We must check to see if our scaled function .n is normalized. We know that 
1 = 
 .*(r1, . . . , rn).(r1, . . . , rn)dv 
= 
 .*(.r1, . . . ,.rn).(.r1, . . . ,.rn)d(.v) (A8-5) 
because we have simply relabeled all variables r by .r, including the volume element, 
just as in Eq. A8-1. We now factor . out of the volume element and divide the limits of 
624
Appendix 8 625 
integration by ., just as in Eq. A8-1. However, the limits are zero and infinity, so they 
are unaffected. The volume element d(.v) is given by 
d(.v)=(.r1)2 sin .1d(.r1)d.1df1(.r2)2 sin .2d(.r2)d.2df2 ··· (A8-6) 
and so .3 appears for each electron. Thus, we are led to 
1=.3n 
 .*... dv (A8-7) 
Therefore, our normalization constant for .. is .3n/2, and our normalized, scaled 
function is 
.. =.3n/2.(.r1,.r2, . . . ,.rn) (A8-8) 
[Compare this with the specific example encountered in Eq. (7-7).] We now inquire as 
to the values of V¯. and T¯., where 
T¯. = 
..|Tˆ|.. (A8-9) 
V¯. = 
..|Vˆ |.. (A8-10) 
For an n-electron atom, 
Tˆ = -
1
2 
n 

i=1 
.2
i (A8-11) 
Vˆ = - 
n 

i=1
(Z/ri)+
n-1 

i=1 
n 
 
j=i+1 
1/rij (A8-12) 
Therefore 
V¯. =.3n 
 .*(.r1, . . . )
.
. 
n 

i=1 
-Z 
ri 
+
n-1 

i=1 
n 
 
j=i+1 
1 
rij 
.
.
.(.r1, . . . ) dv (A8-13) 
We couldmake the integral equal to V¯ ifwe could get the scale factor into all the r terms 
in the operator and also into dv. The volume element dv requires .3n, which is already 
present in Eq. A8-13 from the normalization constants. To get . into the operator, we 
need to multiply the operator by .-1. Multiplying Eq. A8-13 by ..-1 gives, then 
V¯. =. 
 .*(.r1, . . . )
.
. 
n 

t=1 
-Z 
.ri 
+
n-1 

t=1 
n 
 
j=i+1 
1 
.rij 
.
.
.(.r1, . . . ) d(.v) (A8-14) 
or 
V¯. =.V¯ (A8-15) 
The same approach to T¯. gives 
T¯. =.2 T¯ (A8-16)
626 The Virial Theorem for Atoms and Diatomic Molecules 
This arises from the fact that 
.2 = 
1 
r2 
. 
.r 
r2 . 
.r 
+ 
1 
r2 sin . 
. 
.. 
sin . 
. 
.. 
+ 
1 
r2 sin2 . 
.2 
.f2 (A8-17) 
and scaling the r terms here requires multiplying by .-2. Hence, the integral is multiplied 
by .2.-2 in the final step. 
The general result is that, for any quantum-mechanical system where 
Vˆ =f (r-v) (A8-18) 
scaling results in 
T¯. =.2 T¯, V¯. =.v V¯ (A8-19) 
For atoms, Vˆ contains r as r-1. For all systems, Tˆ involves .2, which contains r to 
the net power of -2. We can now write the expression for the total energy of the atom 
as given by the scaled function
E¯. =T¯. +V¯. =.2 T¯ +.V¯ (A8-20) 
Now we can seek the best value of the scale factor. We do this by minimizing E¯. 
with respect to variations in .:
.E¯./..=2.T¯ +V¯ =0 (A8-21) 
(T¯ and V¯ are independent of ..) 
We are now in a position to make some statements about the average values of Tˆ 
and Vˆ for certain wavefunctions. Let us consider first the exact values of T¯ and V¯ . We 
know that, if . were an exact eigenfunction, no further energy lowering would result 
from rescaling. That is, . equals unity in Eq. A8-21. As a result, 
2T¯ +V¯ =0 (A8-22) 
or 
V¯ =-2T¯ (A8-23) 
or, since T¯ +V¯ =E¯, 
E =-T¯ = 
1
2 ¯ V (A8-24) 
Thus, for an atom, we know that the exact nonrelativistic energy is equal to minus 
the exact average kinetic energy and is equal to one half the exact potential energy. 
Knowing that the exact energy of the ground-state neon atom is -128.925 a.u. enables 
us to say that T¯ =+128.925 a.u. and V¯ =-257.850 a.u. without actually knowing .. 
Moreover, the same relation holds for any stable state of an atom. 
This same argument holds, not only for exact solutions, but for any trial function that 
has already been energy-optimized with respect to a scale factor, for then a new scaling 
parameter . gives no improvement, . = 1, and all is as above. Thus, any nonlinear 
variation scheme consistent with uniform scaling should ultimately lead to the relations
Appendix 8 627 
[Eqs. A8-22–A8-24]. Satisfying these relations is frequently referred to as satisfying 
the virial relation. Completely optimized single-. and double-. functions satisfy the 
virial relation. 
It follows that Hartree–Fock atomic wavefunctions must satisfy the virial relation. 
Such solutions are, by definition, the best (lowest energy) attainable in a single determinantal 
form. “Best” includes all conceivable variation, linear or nonlinear, so all 
improvements achievable by scale factor variation are already present at the Hartree– 
Fock level, and .=1. 
In the event that E¯ can be lowered by scaling, it is possible to evaluate the optimum. 
from Eq. A8-21, which gives 
.=-V¯ /2T¯ (A8-25) 
One of the useful applications of the virial theorem is as an indicator of closeness of 
approach to the Hartree–Fock solution for an atom. If the calculation involves nonlinear 
variation (uniformly applied to all r coordinates), then the resulting wavefunction will 
satisfy the virial relations no matter how deficient it is as an approximation to the 
true eigenfunction. However, if the calculation involves only linear variation, as for 
example, when a linear combination of gaussian functions is used to approximate an 
AO, then there is no guarantee that the virial relation will be satisfied. If the basis set 
is extensive enough, however, the Hartree–Fock limit will be approached, and V¯ /T¯ 
will approach -2. Strictly speaking, a linear variation calculation on an atom that 
gives V¯ /T¯ =-2 is simply one that cannot be improved by uniform scaling. Therefore, 
approach to-2 is not a guarantee of approach to the Hartree-Fock limit. It is a necessary 
but not a sufficient condition. 
A8-2 Diatomic Molecules 
The treatment here is very similar to that for atoms. We make the Born–Oppenheimer 
approximation by assuming that . depends parametrically on the internuclear separation 
R: 
. =.(r1, r2, . . . , rn,R) (A8-26) 
When we scale ri , we scale R as well: 
.. =.(.r1,.r2, . . . ,.rn,.R) (A8-27) 
Henceforth, we let .R = .. Performing the same variable manipulations as in 
Section A8-1, we find 
T¯. = T¯(., .)=.2 T¯(1, .) (A8-28) 
V¯. = V¯ (., .)=.V¯ (1, .) (A8-29) 
Here, Vˆ may or may not include the internuclear repulsion term. This gives, for the 
total energy, 
E¯. =.2 T¯(1, .)+.V¯ (1, .) (A8-30)
628 The Virial Theorem for Atoms and Diatomic Molecules 
Upon taking the derivative with respect to ., we obtain 
.E¯n 
.. =2.T¯(1, .)+V¯ (1, .)+.2 .T¯(1, .) 
.. +. 
.V¯ (1, .) 
.. =0 (A8-31) 
This differs fromthe atomic case in that V¯ (1, .) and T¯(1, .) depend on . through ..But 
. 
.. = . 
..
.. 
..
= . 
.. 
R (A8-32) 
and so Eq. A8-31 becomes 
.E¯. 
.. =0=2.T¯(1, .)+.V¯ (1, .)+.2R
.T¯(1, .) 
.. +.R
.V¯ (1, .) 
.. 
(A8-33) 
If we assume that . is the exact eigenfunction, then .=1, and 
2T¯ +V¯ +R .E¯ 
.R 
=0 (A8-34) 
Indeed, this relation holds for any case in which all improvement in the nature of 
a scale factor variation has been made, such as, for example, the Hartree-Fock limit. 
Note that, if V¯ contains internuclear repulsion, E¯ is the total energy. If not, E¯ is the 
electronic energy. 
A8-2.1 Problems 
A8-1. Use the methods outlined in this appendix to show that V¯ =T¯ for any stationary 
state of the quantum mechanical harmonic oscillator. 
A8-2. Evaluate V¯ and T¯ with . = 1/vp exp(-r) for the Li2+ ion. From these, 
establish the optimum scale factor . and write down the expression for the 
normalized optimized .. and the optimized energy E.. Compare these results 
with the eigenfunction for Li2+.
Appendix 9 
Bra-ket Notation 
Bra-ket, or Dirac, notation is frequently used in the literature because of its economical 
form. Perhaps the best way to learn how this notation is used is by studying its use in 
a few familiar relations and proofs. Accordingly, we have outlined a few of these uses. 
The applications and subtleties of this notation go considerably beyond the treatment 
summarized here.1 
“Usual” notation Dirac notation 

 f*m
fndt = fm 
bra
| fn 
ket
=m|n (A9-1) 

 f*m
Afndt = fm|A|fn=m|A|n=Amn (A9-2) 

 f*m
fndt* =
 f*nfmdt fm|fn* =fn|fm 
or (A9-3) 
m|n* =n|m 
For hermitian operator A: 

 f*m
Afndt = 
 fn(Afm)*dt m|A|n=n|A|m* 
= 
 f*nAfmdT * (A9-4) 
Any function. can be written as a sum of a complete set of orthonormal functions f: 
. =m cmfm |. =m cm|fm=m cm|m 

 f*n.dt =m cm 
 f*nfmdt =cn n|. =m cmn|m=cn 
cn =
 f*n.dt cn =n|. 
. =m cmfm =m 
 f*m
.dtfm |. =mm|.|m 
=m |mm|. 
1Strictly speaking, for instance, fm|fn and m|n are not identical in meaning. The former refers to specific 
functions, fm and fn, which represent state vectors in a specific representation. The latter refers to the state vectors 
in any representation and hence is a more general expression. Distinctions such as this will not be necessary at 
the level of this text. 
629
630 Bra-ket Notation 
Example of Use: Proof that eigenvalues of A (hermitian) are real. 
A|m = am|m (A9-5) 
m|A|m = amm|m 
   =0,=8
, (A9-6) 
m|A|m* = a*m 
m|m (A9-7) 
Combining Eqs. A9-4, A9-6, and A9-7, we have 
(am -a*m
)=0
Appendix 10 
Values of Some Useful Constants 
and Conversion Factors 
631
Values of Some Useful Constants a 
Value 
Quantity 
Symbol 
and/or 
formulac Atomic units SI units Other units 
Uncertainty 
(ppm) 
a) Fundamental constants 
Planck’s constant h 2p 6.626176×10-34 J sec 4.135701×10-15 eV sec 5.4 
Planck’s constant h/2p ¯h 1 1.0545887×10-34 J sec 6.582173×10-16 eV sec 5.4 
Rest mass of electron me 1 9.109534×10-31 kg 9.109534×10-28 gm 5.1 
Charge of electron -e -1 -1.602189×10-19 C -4.803242×10-10 esu 2.9 
Rest mass of proton mp 1.83615×103 1.6726485×10-27 kg 1.6726485×10-24 gm 5.1 
Rest mass of neutron mn 1.83868×103 1.6749543×10-27 kg 1.6749543×10-24 gm 5.1 
Speed of light in vacuum c 137.039 2.99792458×108m sec-1 2.99792458×1010 cm sec-1 0.004 
Avogadro’s numberb NA — 6.0220943×1023 mol-1 — 1.05 
Bohr magneton ße 
1
2 9.274078×10-24 JT-1 
Electron g value ge — 2.00232 
b) Derived quantities 
Bohr radius a0 1 5.2917706×10-11m 0.52917706Å 0.82 
Vacuum permittivity 4p0 1 1.11265×10-10 J-1 C2 m-1 
Twice the ionization potential 
of the hydrogen atom with 
infinite nuclear mass 
Ea = e2/4p0a0 1 4.359814×10-18 J 27.21161 eV; 2 rydbergs 6.6 
Electric field strength one 
bohr radius from proton e/4p0a2
0 1 5.1423×1011V m-1 1.715270×1010 esu cm-2 4.4 
632
Polarizability (of a 
molecule) a =e2a0
2/Ea 1 1.648776×10-41 C2 m2 J-1 1.481846×10-25 esu2 cm2 erg-1 1.4 
Bohr magneton µB =e ¯h/2me 
1
2 9.274078×10-24 J T-1 5.788378×10-9 eV G-1 3.9 
Nuclear magneton µN =e ¯h/2mp 2.723087×10-4 5.050824×10-27 J T-1 3.1524515×10-12 eV G-1 3.9 
Electric dipole moment 
of electron–proton 
separated by one Bohr 
radius ea0 1 8.478418×10-30C m 
2.541765×10-20 esu m 
=2.541765 debyes 3.8 
Time for 1s electron in 
hydrogen atom to travel 
one bohr radius t = ¯h/Ea 1 2.41888×10-17 sec — 12 
Atomic unit of velocity a0/t 1 2.18769×106m sec-1 2.18767×108 cm sec-1 13 
Atomic unit of volume a0
3 1 1.481846×10-31m3 0.14818 Å3 2.0 
Atomic unit of probability 
density a0-3 1 6.748340×1030 m-3 6.748340 Å-3 3.0 
aFrom Cohen and Taylor [1]. C=coulomb, J=joule, V=volt, T=tesla, G=gauss, Å=angstrom. 
bSee Ref. [2]. 
cFormula appropriate for atomic units. 
633
Energy Conversion Factorsa 
eV jouleb kcal/mole Hz m-1 .K a.u. 
eV 1 1.6021892×10-19 23.060362 2.4179696×1014 8.065479×105 1.160450×104 3.674901×10-2 
jouleb 6.2414601×1018 1 1.4393033×1020 1.5091661×1033 5.034037×1024 7.242902×1022 2.293675×1017 
kcal/mole 4.336445×10-2 6.947805×10-21 1 1.0485393×1013 3.497551×104 5.032223×102 1.593601×10-3 
Hz 4.1357012×10-15 6.6261759×10-34 9.5370770×10-14 1 3.3356412×10-9 4.799274×10-11 1.519829×10-16 
m-1 1.239852×10-6 1.986477×10-25 2.859144×10-5 2.9979243×108 1 1.438786×10-2 4.556333×10-8 
.K 8.617347×10-5 1.380662×10-23 1.987191×10-3 2.083648×1010 6.950303×101 1 3.166790×10-6 
a.u. 27.21161 4.359816×10-18 6.275098×102 6.579686×1015 2.194747×107 3.157772×105 1 
aTo convert from units in the left hand column to units in the top row, multiply by the factor in the row-column position, e.g., 1 kcal/mole= 1.0485393×1013 Hz. 
b1 joule=107 erg. 
634
Appendix 10 635 
References 
[1] E. R. Cohen and B. N. Taylor, J. Phys. Chem. Ref. Data 2, No. 4, 663 (1973). 
[2] A. L. Robinson, Science, 185, 1037 (1974). 
See also P. J. Mohr and B. N. Taylor, Rev. Mod. Phys. 72 (2000) for revised values.
Appendix 11 
Group Theoretical Charts and Tables 
A11-1 Flow Scheme for Group Symbols 
Is the molecule linear? 
No 
No 
Is there a reflection plane 
perpendicular to the axis? 
T, T h, T d, O, or O h 
(Examine character tables 
to choose among these) 
Is there a 
reflection plane? 
Is there a point 
of conversion? 
C 1 
C I 
C s 
I 
or 
I h 
C v D h 
Are there one or more 
proper rotation axes? 
Does the molecule 
possess six C5 axes? 
Does the molecule 
possess four C3 axes? 
Yes 
No 
No 
No 
No 
No 
Yes Yes 
Yes 
Yes 
Yes 
Yes 
636
Appendix 11 637 
Selection of reference axis 
Cn. (Defines vertical direction.) 
Is there an axis is having 
higher order than any 
other proper axis? 
No 
No 
No 
Yes 
Yes 
Yes 
Yes 
Yes
Yes 
Yes 
Yes 
No 
No 
No 
No 
No 
Of the several axes having 
highest order, is there one 
which is geometrically unique? 
Choose any of the 
equivalent axes as Cn 
(Cn and n now defined) 
Select this as 
reference axis Cn 
Is there an S2n axis 
collinear with Cn? 
S2n 
Cn h
Cn v 
Cn 
Dnh
Dnd 
Dn 
Are there n twofold axes 
perpendicular to Cn? 
Are there any symmetry 
elements present other 
than Cn, S2n, or i ? 
Is there a reflection 
plane (s h) perpendicular 
to Cn ? 
Is there a set of 
n reflection planes 
(sv) containing Cn ? 
Is there a set of 
n reflection planes 
containing Cn and 
bisecting the n 
twofold axes? 
Is there a reflection 
plane (sh) perpendicular 
to Cn? 
Yes (D branch) 
No (C branch) 
A11-2 Meaning of Labels for Representations 
Symbol Interpretation 
Main Symbol 
A One-dimensional representation symmetric for rotation by 2p/n 
about principal axis. (For c1, cs, ci , which have no principal axis, 
this symbol merely means a one-dimensional representation.) 
B One-dimensional representation but antisymmetric for rotation by 
2p/n about principal axis 
E Two-dimensional representation 
T (or F) Three-dimensional representation 
G Four-dimensional representation 
Subscripts 
1 Symmetric for perpendicular C2 rotations (or else sv or sd reflections) 
2 Antisymmetric for perpendicular C2 rotations (or else sv or sd 
reflections)
638 Group Theoretical Charts and Tables 
g Symmetric for inversion 
u Antisymmetric for inversion 
Superscripts 

 Symmetric for sh reflection 


 Antisymmetric for sh reflection 
A11-3 Character Tables 
A11-3.A Special High-Symmetry Groups: C8v,D8h,I,Ih,T,Th, 
Td,O,Oh 
C8v E 2C8
 ··· 8sv 
A1 =+ 1 1 ··· 1 z x2 +y2, z2 
A2 =- 1 1 ··· -1 Rz 
E1 = 2 2 cos ··· 0 (x, y);(Rx,Ry) (xz, yz) 
E2 = 2 2 cos 2 ··· 0 (x2 -y2, xy) 
E3 = 2 2 cos 3 ··· 0 
··· ··· ··· ··· ··· 
D8h E 2C8
 ··· 8sv i 2S8
 ··· 8C2 
g+ 1 1 ··· 1 1 1 ··· 1 x2 +y2, z2 
g- 1 1 ··· -1 1 1 ··· -1 Rz 
g 2 2 cos ··· 0 2 -2 cos ··· 0 (Rx,Ry) (xz, yz) 
g 2 2 cos 2 ··· 0 2 2 cos 2 ··· 0 (x2 -y2, xy) 
··· ··· ··· ··· ··· ··· ··· ··· ··· 
u+ 1 1 ··· 1 -1 -1 ··· -1 z 
u- 1 1 ··· -1 -1 -1 ··· 1 
u 2 2 cos ··· 0 -2 2 cos ··· 0 (x, y) 
u 2 2 cos 2 ··· 0 -2 -2 cos 2 ··· 0 
··· ··· ··· ··· ··· ··· ··· ··· ··· 
I E 12C5 12C5
2 20C3 15C2 
A 1 1 1 1 1 x2 +y2 +z2 
T1 3 1
2 (1+v5) 1
2 (1-v5) 0 -1 (x, y, z);(Rx,Ry,Rz) 
T2 3 1
2 (1-v5) 1
2 (1+v5) 0 -1 
G 4 -1 -1 1 0 
H 5 0 0 -1 1 (2z2 -x2 -y2, 
x2 -y2,xy, yz, zx)
Ih E 12C5 12C5
2 20C3 15C2 i 12S10 12S10
3 20S6 15s 
Ag 1 1 1 1 1 1 1 1 1 1 x2 +y2 +z2 
T1g 3 1
2 (1+v5) 1
2 (1-v5) 0 -1 3 1
2 (1-v5) 1
2 (1+v5) 0 -1 (Rx,Ry,Rz) 
T2g 3 1
2 (1-v5) 1
2 (1+v5) 0 -1 3 1
2 (1+v5) 1
2 (1-v5) 0 -1 
Gg 4 -1 -1 1 0 4 -1 -1 1 0 
Hg 5 0 0 -1 1 5 0 0 -1 1 (2z2 -x2 -y2, 
x2 -y2, 
xy, yz, zx) 
Au 1 1 1 1 1 -1 -1 -1 -1 -1 
T1u 3 1
2 (1+v5) 1
2 (1-v5) 0 -1 -3 -1
2 (1-v5) -1
2 (1+v5) 0 1 (x, y, z) 
T2u 3 1
2 (1-v5) 1
2 (1+v5) 0 -1 -3 -1
2 (1+v5) -1
2 (1-v5) 0 1 
Gu 4 -1 -1 1 0 -4 1 1 -1 0 
Hu 5 0 0 -1 1 -5 0 0 1 -1 
639
640 Group Theoretical Charts and Tables 
T E 4C3 4C3
2 3C2  =exp(2pi/3) 
A 1 1 1 1 x2 +y2 +z2 
E 
1  * 1 (2z2 -x2 -y2, x2 -y2 1 ) *  1 
T 3 0 0 -1 (Rx,Ry,Rz);(x, y, z) (xy, xz, yz) 
Th E 4C3 4C3
2 3C2 i 4S6 4S6
5 3sh  =exp(2pi/3) 
Ag 1 1 1 1 1 1 1 1 x2 +y2 +z2 
Au 1 1 1 1 -1 -1 -1 -1 
Eg 
1  * 1 1  * 1 (2z2 -x2 -y2, 


1 *  1 1 *  1 x2 -y2) 
Eu 
1  * 1 -1 - -* -1 
1 *  1 -1 -* - -1 
Tg 3 0 0 -1 1 0 0 -1 (Rx,Ry,Rz) (xz, yz,xy) 
Tu 3 0 0 -1 -1 0 0 1 (x, y, z) 
Td E 8C3 3C2 6S4 6sd 
A1 1 1 1 1 1 x2 +y2 +z2 
A2 1 1 1 -1 -1 
E 2 -1 2 0 0 (2z2 -x2 -y2, x2 -y2) 
T1 3 0 -1 1 -1 (Rx,Ry,Rz) 
T2 3 0 -1 -1 1 (x, y, z) (xy, xz, yz) 
O E 6C4 3C2(=C4
2) 8C3 6C2 
A1 1 1 1 1 1 x2 +y2 +z2 
A2 1 -1 1 1 -1 
E 2 0 2 -1 0 
(2z2 -x2 -y2, 
x2 -y2) 
T1 3 1 -1 0 -1 (Rx,Ry,Rz); 
(x, y, z) 
T2 3 -1 -1 0 1 (xy, xz, yz)
Appendix 11 641 
Oh E 8C3 6C2 6C4 
3C2 
(=C4
2) i 6S4 8S6 3sh 6sd 
A1g 1 1 1 1 1 1 1 1 1 1 x2 +y2 +z2 
A2g 1 1 -1 -1 1 1 -1 1 1 -1 
Eg 2 -1 0 0 2 2 0 -1 2 0 (2z2 -x2 -y2, 
x2 -y2) 
T1g 3 0 -1 1 -1 3 1 0 -1 -1 (Rx,Ry,Rz) 
T2g 3 0 1 -1 -1 3 -1 0 -1 1 (xz, yz,xy) 
A1u 1 1 1 1 1 -1 -1 -1 -1 -1 
A2u 1 1 -1 -1 1 -1 1 -1 -1 1 
Eu 2 -1 0 0 2 -2 0 1 -2 0 
T1u 3 0 -1 1 -1 -3 -1 0 1 1 (x, y, z) 
T2u 3 0 1 -1 -1 -3 1 0 1 -1 
A11-3.B Groups with No Axis of Symmetry: C1,Ci,Cs 
C1 E 
A 1 
Cs E sh 
A
 1 1 x, y,Rz x2, y2, 
z2, xy 
A

 1 -1 z,Rx,Ry yz, xz 
Ci E i 
Ag 1 1 Rx,Ry,Rz x2, y2, z2, 
xy, xz, yz 
Au 1 -1 x, y, z 
A11-3.C The S2n Groups 
S4 E S4 C2 S4
3 
A 1 1 1 1 Rz x2 +y2, z2 
B 1 -1 1 -1 z x2 -y2, xy 
E 
1 i -1 -i 
1 -i -1 i 
(x, y);(Rx,Ry) (xz, yz) 
S6 E C3 C3
2 i S6
5 S6  =exp(2pi/3) 
Ag 1 1 1 1 1 1 Rz x2 +y2, z2 
Eg 
1  * 1  * 
1 *  1 *  
(Rx,Ry) (x2 -y2, xy); (xz, yz) 
Au 1 1 1 -1 -1 -1 z 
Eu 
1  * -1 - -* 
1 *  -1 -* - 
(x, y)
642 Group Theoretical Charts and Tables 
S8 E S8 C4 S8
3 C2 S8
5 C4
3 S8
7  =exp(2pi/8) 
A 1 1 1 1 1 1 1 1 Rz x2 +y2, z2 
B 1 -1 1 -1 1 -1 1 -1 z 
E1 
1  i -* -1 - -i * 
1 * -i - -1 -* i  
(x, y); 
(Rx,Ry) 
E2 
1 i -1 -i 1 i -1 -i  (x2-y2 1 - , xy) i -1 i 1 -i -1 i 
E3 
11 --* -ii * --11 * -ii --* (xz, yz) 
A11-3.D The Cn Groups 
C2 E C2 
A 1 1 z,Rz x2, y2, z2, xy 
B 1 -1 x, y,Rx,Ry yz, xz 
C3 E C3 C3
2  =exp(2pi/3) 
A 1 1 1 z,Rz x2 +y2, z2 
E 
1  * 
1 *  
(x, y);(Rx,Ry) (x2-y2, xy);(yz, xz) 
C4 E C1 C2 C4
3 
A 1 1 1 1 z,Rz x2 +y2, z2 
B 1 -1 1 -1 x2 -y2, xy 
E 
1 i -1 -i 
1 -i -1 i 
(x, y);(Rx,Ry) (yz, xz) 
C5 E C5 C5
2 C5
3 C5
4  =exp(2pi/5) 
A 1 1 1 1 1 z,Rz x2 +y2, z2 
E1 
1  2 2* *  
1 * 2* 2  
(x, y);(Rx,Ry) (yz, xz) 
E2 
1 2 *  2* 
1 2*  * 2 (x2 -y2, xy)
Appendix 11 643 
C6 E C6 C3 C2 C3
2 C6
5  =exp(2pi/6) 
A 1 1 1 1 1 1 z,Rz x2 +y2, z2 
B 1 -1 1 -1 1 -1 
E1 
1  -* -1 - *  
1 * - -1 -*  
(x, y) 
(Rx,Ry) 
(xz, yz) 
E2 
1 -* - 1 -* -  
1 - -* 1 - -* 
(x2 -y2, xy) 
C7 E C7 C7
2 C7
3 C7
4 C7
5 C7
6  =exp(2pi/7) 
A 1 1 1 1 1 1 1 z,Rz x2 +y2, z2 
E1 
1  2 3 3* 2* *  1 * 2* 3* 3 2  
(x, y) 
(Rx,Ry) 
(xz, yz) 
E2 
1 2 3* *  3 2* 
1 2* 3  * 3* 2 (x2 -y2, xy) 
E3 
1 3 * 2 2*  3* 
1 3*  2* 2 * 3 
C8 E C8 C4 C2 C4
3 C8
3 C8
5 C8
7  =exp(2pi/8) 
A 1 1 1 1 1 1 1 1 z,Rz x2 +y2, z2 
B 1 -1 1 1 1 -1 -1 1 
E1 
1  i -1 -i -* - * 
1 * -i -1 i - -*  
(x, y);(Rx,Ry) (xz, yz) 
E2 
1 i -1 1 -1 -i i -i  
1 -i -1 1 -1 i -i i 
(x2 -y2, xy) 
E3 
1 - i -1 -i *  -* 
1 -* -i -1 i  * - 
A11-3.E The Cnv Groups 
C2v E C2 sv(xz) sv
(yz) 
A1 1 1 1 1 z x2, y2, z2 
A2 1 1 -1 -1 Rz xy 
B1 1 -1 1 -1 x,Ry xz 
B2 1 -1 -1 1 y,Rx yz 
C3v E 3sv 2C3 
A1 1 1 1 z x2 +y2, z2 
A2 1 -1 1 Rz 
E 2 0 -1 (x, y);(Rx,Ry) (x2 -y2, xy);(xz, yz)
644 Group Theoretical Charts and Tables 
C4v E 2C4 C2 2sv 2sd 
A1 1 1 1 1 1 z x2 +y2, z2 
A2 1 1 1 -1 -1 Rz 
B1 1 -1 1 1 -1 x2 -y2 
B2 1 -1 1 -1 1 xy 
E 2 0 -2 0 0 (x, y);(Rx,Ry) (xz, yz) 
C5v E 2C5 2C5
2 5sv 
A1 1 1 1 1 z x2 +y2, z2 
A2 1 1 1 -1 Rz 
E1 2 2 cos 72. 2 cos 144. 0 (x, y);(Rx,Ry) (xz, yz) 
E2 2 2 cos 144. 2 cos 72. 0 (x2 -y2, xy) 
C6v E 2C6 2C3 C2 3sv 3sd 
A1 1 1 1 1 1 1 z x2 +y2, z2 
A2 1 1 1 1 -1 -1 Rz 
B1 1 -1 1 -1 1 -1 
B2 1 -1 1 -1 -1 1 
E1 2 1 -1 -2 0 0 (x, y);(Rx,Ry) (xz, yz) 
E2 2 -1 -1 2 0 0 (x2 -y2, xy) 
A11-3.F The Cnh Groups 
C2h E C2 i sh 
Ag 1 1 1 1 Rz x2, y2, z2, xy 
Bg 1 -1 1 -1 Rx,Ry xz, yz 
Au 1 1 -1 -1 z 
Bu 1 -1 -1 1 x,y 
C3h E C3 C3
2 sh S3 S3
5  =exp(2pi/3) 
A
 1 1 1 1 1 1 Rz x2 +y2, z2 
E
 
1  * 1  * 
1 *  1 *  
(x, y) (x2 -y2, xy) 
A

 1 1 1 -1 -1 -1 z 
E

 
1  * -1 - -* (Rx,Ry) (xz, yz) 
1 *  -1 -* -
Appendix 11 645 
C4h E C4 C2 C4
3 i S4
3 sh S4 
Ag 1 1 1 1 1 1 1 1 Rz x2 +y2, z2 
Bg 1 -1 1 -1 1 -1 1 -1 x2 -y2, xy 
Eg 
1 i -1 -i 1 i -1 -i 
1 -i -1 i 1 -i -1 i 
(Rx,Ry) (xz, yz) 
Au 1 1 1 1 -1 -1 -1 -1 z 
Bu 1 -1 1 -1 -1 1 -1 1 
Eu 
1 i -1 -i -1 -i 1 i 1 - (x, y) i -1 i -1 i 1 -i 
C5h E C5 C5
2 C5
3 C5
4 sh S5 S5
7 S5
3 S5
9  =exp(2pi/5) 
A
 1 1 1 1 1 1 1 1 1 1 Rz x2 +y2, z2 
E1
 
 1  2 2* * 1  2 2* *  
1 * 2* 2  1 * 2* 2  
(x, y) 
E2
 
1 2 *  2* 1 2 *  2* 
1 2*  * 2 1 2*  * 2 (x2 -y2, xy) 
A

 1 1 1 1 1 -1 -1 -1 -1 -1 z 
E1

 
1  2 2* * -1 - -2 -2* -*  
1 * 2* 2  -1 -* -2* -2 - 
(Rx,Ry ) (xz, yz) 
E2

 
1 2 *  2* -1 -2 -* - -2* 
1 2*  * 2 -1 -2* - -* -2 
C6h E C6 C3 C2 C3
2 C6
5 i S3
5 S6
5 sh S6 S3  =exp(2pi/6) 
Ag 1 1 1 1 1 1 1 1 1 1 1 1 Rz x2 +y2, z2 
Bg 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 
E1g 
1  -*-1 - * 1  -* -1 - * 
1 *- -1 -*  1 * - -1 -*  
(Rx,Ry ) (xz, yz) 
E2g 
1 -*- 1 -* - 1 -* - 1 -* -  1 - -* 1 - -* 1 - -* 1 - -* (x2 -y2, xy) 
Au 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 z 
Bu 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 
E1u 
1  -*-1 - * -1 - * 1  -* 
1 *- -1 -*  -1 -*  1 * - 
(x, y) 
E2u 
1 -*- 1 -* - -1 *  -1 *   
1 - -* 1 - -* -1  * -1  *
646 Group Theoretical Charts and Tables 
A11-3.G The Dn Groups 
D2 E C2(z) C2(y) C2(x) 
A 1 1 1 1 x2, y2, z2 
B1 1 1 -1 -1 z,Rz xy 
B2 1 -1 1 -1 y,Ry xz 
B3 1 -1 -1 1 x,Rx yz 
D3 E 2C3 3C2 
A1 1 1 1 x2 +y2, z2 
A2 1 1 -1 z,Rz 
E 2 -1 0 (x, y);(Rx,Ry) (x2 -y2, xy);(xz, yz) 
D4 E 2C4 C2(=C4
2) 2C2
 2C2

 
A1 1 1 1 1 1 x2 +y2, z2 
A2 1 1 1 -1 -1 z,Rz 
B1 1 -1 1 1 -1 x2 -y2 
B2 1 -1 1 -1 1 xy 
E 2 0 -2 0 0 (x, y);(Rx,Ry) (xz, yz) 
D5 E 2C5 2C5
2 5C2 
A1 1 1 1 1 x2 +y2, z2 
A2 1 1 1 -1 z,Rz 
E1 2 2 cos 72. 2 cos 144. 0 (x, y);(Rx,Ry) (xz, yz) 
E2 2 2 cos 144. 2 cos 72. 0 (x2 -y2, xy) 
D6 E 2C6 2C3 C2 3C2
 3C2

 
A1 1 1 1 1 1 1 x2 +y2, z2 
A2 1 1 1 1 -1 -1 z,Rz 
B1 1 -1 1 -1 1 -1 
B2 1 -1 1 -1 -1 1 
E1 2 1 -1 -2 0 0 (x, y);(Rx,Ry) (xz, yz) 
E2 2 -1 -1 2 0 0 (x2 -y2, xy)
Appendix 11 647 
A11-3.H The Dnd Groups 
D2d E 2S4 C2 2C2
 2sd 
A1 1 1 1 1 1 x2 +y2, z2 
A2 1 1 1 -1 -1 Rz 
B1 1 -1 1 1 -1 x2 -y2 
B2 1 -1 1 -1 1 z xy 
E 2 0 -2 0 0 (x, y);(Rx,Ry) (xz, yz) 
D3d E 2C3 3C2 i 2S6 3sd 
A1g 1 1 1 1 1 1 x2 +y2, z2 
A2g 1 1 -1 1 1 -1 Rz 
Eg 2 -1 0 2 -1 0 (Rx,Ry) (x2 -y2, xy), (xz, yz) 
A1u 1 1 1 -1 -1 -1 
A2u 1 1 -1 -1 -1 1 z 
Eu 2 -1 0 -2 1 0 (x, y) 
D4d E 2S8 2C4 2S8
3 C2 4C2
 4sd 
A1 1 1 1 1 1 1 1 x2 +y2, z2 
A2 1 1 1 1 1 -1 -1 Rz 
B1 1 -1 1 -1 1 1 -1 
B2 1 -1 1 -1 1 -1 1 z 
E1 2 v2 0 -v2 -2 0 0 (x, y) 
E2 2 0 -2 0 2 0 0 (x2 -y2, xy) 
E3 2 -v2 0 v2 -2 0 0 (Rx,Ry) (xz, yz) 
D5d E 2C5 2C5
2 5C2 i 2S3
10 2S10 5sd 
A1g 1 1 1 1 1 1 1 1 x2 +y2, z2 
A2g 1 1 1 -1 1 1 1 -1 Rz 
E1g 2 2 cos 72. 2 cos 144. 0 2 2 cos 72. 2 cos 144. 0 (Rx,Ry ) (xz, yz) 
E2g 2 2 cos 144. 2 cos 72. 0 2 2 cos 144. 2 cos 72. 0 (x2 -y2, xy) 
A1u 1 1 1 1 -1 -1 -1 -1 
A2u 1 1 1 -1 -1 -1 -1 1 z 
E1u 2 2 cos 72. 2 cos 144. 0 -2 -2 cos 72. -2 cos 144. 0 (x, y) 
E2u 2 2 cos 144. 2 cos 72. 0 -2 -2 cos 144.-2 cos 72. 0
648 Group Theoretical Charts and Tables 
D6d E 2S12 2C6 2S4 2C3 2S5
12 C2 6C2
 6sd 
A1 1 1 1 1 1 1 1 1 1 x2 +y2, z2 
A2 1 1 1 1 1 1 1 -1 -1 Rz 
B1 1 -1 1 -1 1 -1 1 1 -1 
B2 1 -1 1 -1 1 -1 1 -1 1 z 
E1 2 v3 1 0 -1 -v3 -2 0 0 (x, y) 
E2 2 1 -1 -2 -1 1 2 0 0 (x2 -y2, xy) 
E3 2 0 -2 0 2 0 -2 0 0 
E4 2 -1 -1 2 -1 -1 2 0 0 
E5 2 -v3 1 0 -1 v3 -2 0 0 (Rx,Ry) (xz, yz) 
A11-3.I The Dnh Groups 
D2h E C2(z) C2(y) C2(x) i s(xy) s(xz) s(yz) 
Ag 1 1 1 1 1 1 1 1 x2, y2, z2 
B1g 1 1 -1 -1 1 1 -1 -1 Rz xy 
B2g 1 -1 1 -1 1 -1 1 -1 Ry xz 
B3g 1 -1 -1 1 1 -1 -1 1 Rx yz 
Au 1 1 1 1 -1 -1 -1 -1 
B1u 1 1 -1 -1 -1 -1 1 1 z 
B2u 1 -1 1 -1 -1 1 -1 1 y 
B3u 1 -1 -1 1 -1 1 1 -1 x 
D3h E 2C3 3C2 sh 2S3 3sv 
A1
 1 1 1 1 1 1 x2 +y2, z2 
A2
 1 1 -1 1 1 -1 Rz 
E
 2 -1 0 2 -1 0 (x, y) (x2 -y2, xy) 
A1

 1 1 1 -1 -1 -1 
A2

 1 1 -1 -1 -1 1 z 
E

 2 -1 0 -2 1 0 (Rx,Ry) (xz, yz)
Appendix 11 649 
D4h E 2C4 C2 2C2
 2C2

 i 2S4 sh 2sv 2sd 
A1g 1 1 1 1 1 1 1 1 1 1 x2 +y2, z2 
A2g 1 1 1 -1 -1 1 1 1 -1 -1 Rz 
B1g 1 -1 1 1 -1 1 -1 1 1 -1 x2 -y2 
B2g 1 -1 1 -1 1 1 -1 1 -1 1 xy 
Eg 2 0 -2 0 0 2 0 -2 0 0 (Rx,Ry) (xz, yz) 
A1u 1 1 1 1 1 -1 -1 -1 -1 -1 
A2u 1 1 1 -1 -1 -1 -1 -1 1 1 z 
B1u 1 -1 1 1 -1 -1 1 -1 -1 1 
B2u 1 -1 1 -1 1 -1 1 -1 1 -1 
Eu 2 0 -2 0 0 -2 0 2 0 0 (x, y) 
D5h E 2C5 2C5
2 5C2 sh 2S5 2S5
3 5sv 
A1
 1 1 1 1 1 1 1 1 x2 +y2, z2 
A2
 1 1 1 -1 1 1 1 -1 Rz 
E1
 2 2 cos 72. 2 cos 144. 0 2 2 cos 72. 2 cos 144. 0 (x, y) 
E2
 2 2 cos 144. 2 cos 72. 0 2 2 cos 144. 2 cos 72. 0 (x2 -y2, xy) 
A1

 1 1 1 1 -1 -1 -1 -1 
A2

 1 1 1 -1 -1 -1 -1 1 z 
E1

 2 2 cos 72. 2 cos 144. 0 -2 -2 cos 72. -2 cos 144. 0 (Rx,Ry ) (xz, yz) 
E2

 2 2 cos 144. 2 cos 72. 0 -2 -2 cos 144. -2 cos 72. 0 
D6h E 2C6 2C3 C2 3C2
 3C2

 i 2S3 2S6 sh 3sd 3sv 
A1g 1 1 1 1 1 1 1 1 1 1 1 1 x2 +y2, z2 
A2g 1 1 1 1 -1 -1 1 1 1 1 -1 -1 Rz 
B1g 1 -1 1 -1 1 -1 1 -1 1 -1 1 -1 
B2g 1 -1 1 -1 -1 1 1 -1 1 -1 -1 1 
E1g 2 1 -1 -2 0 0 2 1 -1 -2 0 0 (Rx,Ry ) (xz, yz) 
E2g 2 -1 -1 2 0 0 2 -1 -1 2 0 0 (x2 -y2, xy) 
A1u 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 
A2u 1 1 1 1 -1 -1 -1 -1 -1 -1 1 1 z 
B1u 1 -1 1 -1 1 -1 -1 1 -1 1 -1 1 
B2u 1 -1 1 -1 -1 1 -1 1 -1 1 1 -1 
E1u 2 1 -1 -2 0 0 -2 -1 1 2 0 0 (x, y) 
E2u 2 -1 -1 2 0 0 -2 1 1 -2 0 0
D8h E 2C8 2C8
3 2C4 C2 4C2
 4C2

 i 2S8 2S3
8 2S4 sh 4sd 4sv 
A1g 1 1 1 1 1 1 1 1 1 1 1 1 1 1 x2 +y2, z2 
A2g 1 1 1 1 1 -1 -1 1 1 1 1 1 -1 -1 Rz 
B1g 1 -1 -1 1 1 1 -1 1 -1 -1 1 1 1 -1 
B2g 1 -1 -1 1 1 -1 1 1 -1 -1 1 1 -1 1 
E1g 2 v2 -v2 0 -2 0 0 2 v2 -v2 0 -2 0 0 (Rx,Ry) (xz, yz) 
E2g 2 0 0 -2 2 0 0 2 0 0 -2 2 0 0 (x2 -y2, xy) 
E3g 2 -v2 v2 0 -2 0 0 2 -v2 v2 0 -2 0 0 
A1u 1 1 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 
A2u 1 1 1 1 1 -1 -1 -1 -1 -1 -1 -1 1 1 z 
B1u 1 -1 -1 1 1 1 -1 -1 1 1 -1 -1 -1 1 
B2u 1 -1 -1 1 1 -1 1 -1 1 1 -1 -1 1 -1 
E1u 2 v2 -v2 0 -2 0 0 -2 -v2 v2 0 2 0 0 (x, y) 
E2u 2 0 0 -2 2 0 0 -2 0 0 2 -2 0 0 
E3u 2 -v2 v2 0 -2 0 0 -2 v2 -v2 0 2 0 0 
650
Appendix 12 
Hints for Solving Selected Problems 
Chapter 1 
1-1. Use Eq. (1-25). 
1-7. 
P.E.(t)=-
 .(x,t) 
0 
m.2(x, t)/.t2 d(x, t)= 
1
2
m.22(x, t ). 
Next integrate P.E.(t) over one complete cycle (0-t 
). 
Chapter 2 
2-9. sin x sin y = 1
2 [cos(x -y)-cos(x +y)]. 
2-10. What kind of function has .II=8? When could such a function join smoothly 
onto a sine function in region I? 
Chapter 3 
3-1. Imagine an auto runs from A to B at 30 mph and from B to C at 60 mph. Sketch 
the distribution function for the auto. Then reason howyou arrived at this function 
and apply similar reasoning to the harmonic oscillator. 
3-6. a) Seek points where H2(y) equals zero. 
b) Seek places where d.2/dy=0; then evaluate .2 at these points and compare. 
3-15. 

 8 
1 
exp(-y2)dy ~0.10 exp(-1.052)+exp(-1.152)+exp(-1.252)+··· 
to convergence. 
3-19. Consider howthe solutions for the harmonic oscillatorwould meet the conditions 
imposed by this new potential. 
651
652 Hints for Solving Selected Problems 
Chapter 4 
4-3. The classical turning point occurs when the total energy equals the potential 
energy. 
4-11. E =V (r) when r equals classical turning point. 
4-12. Ignore all but the . and f dependences in Eq. (4-30). Do not forget to square 
these dependences, and do not forget to include the 31/2 term of 3d2
z . 
4-17. x =r sin . cosf. 
4-18. a) Do not forget to include . dependence of dv. 
4-26. Refer to Problem 2-11. 
4-37. µ should be in units of kg/molecule, and you have masses in a.m.u. These can 
be taken directly as g/mole and then converted. 
Chapter 5 
5-4. Square . and integrate, using the fact that 1s, 2s are orthonormal. 
Chapter 6 
6-9. b) * must be same at t =0 and t =1/.. 
6-11. Use the Schmidt orthogonalization method. 
6-13. a) Means: is . an eigenfunction of the momentum operator? 
6-18. Wavefunction is not normalized. 
6-24. Use that exp(ikx)=cos(kx)+i sin(kx). 
Chapter 7 
7-4. a) Note that, for symmetric .p ,  .fdx = 2  L/2 
0 .nf dx. For antisymmetric 
.n, the integral can be evaluated by inspection. A useful integral is 
 x sinx dx =sin x -x cos x. 
c) Use the fact that E = 
nc2n
En. The series can be estimated with a small 
calculator (tedious) or else by summing the first few terms and integrating 
over a function that envelopes the higher terms. 
7-11. Use the fact that f =
ici.i and E =
ic*i ciEi . 
7-13. b) Take limit as F .0 rather than simply evaluating at F = 0.1. Note that 
(1+x)m =1+mx +[m(m-1)/2!] x2+···. 
7-16. Do not forget that overlap between fa and fb must enter normality condition: 
c2
a +c2
b +2cacbS =1 
7-17.  f2(-1/ra)dv=-1+(. +1) exp(-2.) using the method of Appendix 3.
Appendix 12 653 
7-19. Note that .+ and .- are degenerate at R=8. 
7-20. In the limit of R.0,HAA is indeterminate. Use l’Hospital’s rule (i.e., take 
d/dR on the numerator and denominator and evaluate at R=0). Alternatively, 
you can expand exp(-2R) in powers of R and evaluate at R =0. 
7-27. Note that (f) and (g) have bothAOs on center a. When Hˆ is present in the integral, 
you are restricted to considering symmetry operations that do not affectHˆ . 
7-34. Use the Schmidt orthogonalization procedure (Chapter 6) to construct 2s
. 
Chapter 8 
8-15. b) Simply note where HOMO is bonding, antibonding, nonbonding, and recognize 
that some of the effect of this MO will be lost upon ionization. 
Chapter 12 
12-2. Recall that perturbation should be greatest for states with .2 largest in region 
of perturbation. 
12-7. To evaluate the first-order correction to the energy, you can recognize that 
1s| - 1/r|1s is identical to the potential energy of the H atom. The virial 
theorem tells you the value of this quantity at once. 
12-16. Use the fact that Ep =a
MOs 
i ni +2ß
MOs 
i 

neighbors 
k<l pkl . 
Chapter 13 
13-9. Notice which MOs are degenerate when assigning symmetries. For the final 
part of the question, notice that the molecular x axis corresponds to the group 
theoretic z axis. 
Chapter 14 
14-13. Do not forget that the cyclobutadiene molecule differs from two acetylenes in 
both p and s systems. There are a total of eight orbitals to be sketched for each 
side of this reaction. 
Chapter 15 
15-2. The curve in Fig. 15-2b is proportional to the reciprocal of the slope of the curve 
in Fig. 15-2a. 
15-3. It is helpful to use that exp(inp) = 1 for even n,-1 for odd n, and that 
exp [i(a +b)]=exp(ia) exp(ib). 
15-4. Remember that E is degenerate due to states at negative j .
Appendix 13 
Answers to Problems1 
Chapter 1 
1-1. [See Hint.] (A+D +C)cos(kx)+(B +iC -iD) sin(kx). 
1-2. .(x)= same as Eq. (1-32) except cos instead of sin when n is odd. 
1-3. a2 =ß2 =(2p/.)2. 
1-4. Work functions: Cs=1.9eV, Zn=3.7eV.h=4.13×10-13 eVs. 
1-5. a) 0.055 nm. b) 3.31×10-25 nm. 
1-6. Argument of cosine must change by 2p. Cycle time=2p/.. 
1-7. [See Hint.] Integrating PE(t)= 1
2m.22(x, t) over a cycle and dividing by t 
 
to give average PE per unit time gives PE=m.2.2(x)/4.[m is really .dx.] An 
identical result comes from integrating KE(t)= 1
2m.2.2(x) sin2(.t) over the 
same cycle. 
1-8. a) No. Becomes infinite at x =±8. b) Same as a). c) Yes. d) No. Becomes 
infinite at x=-8. e)Yes. 
1-9. . =sin x or cos x are examples. 
1-10. We need establish only one of the extreme profiles for the string. Then, as cos(.t) 
oscillates between+1 and-1, the string oscillates between the two extremes. In 
other words, sin(x) and -sin(x) are the same solution at different times. (They 
differ by a phase factor.) 
1-11. Only c), d) and f). [The latter is most easily seen after recognizing that the 
function equals exp(4ix).] For c), the eigenvalue is zero. 
Multiple Choice (MC): d c e e b 
Chapter 2 
2-1. J2s2kg-1m-2 =(kgm2s-2)2 sec2 kg-1m-2 =kgm2s-2 =J. 
2-2. A=
 L 
np  L 
0 sin2( npx 
L )d( npx 
L )-1/2 
= L 
np  np 
0 sin2 ydy-1/2 
= L 
np · np 
2 -1/2 =2
L 
1Hints for some of these are given in Appendix 12. 
654
Appendix 13 655 
2-3. Recognize that, since sin2+cos2=1,  L 
0 (sin2+cos2)dx is the area of a rectangle 
of height 1 and width L, i.e., has area L. Half of this area goes with sin2. For 
cases where n>1, there are more wiggles, but the value of the integral over sin2 
still equals L/2. 
2-4. 0.6090, 0.1955, 13
, 13
. 
2-5. a) Use height times width of this narrow rectangle in .2. Height = 
(v2/L) sin(0.5pL/L)
2=(2/L) sin2(p/2)=2/L. Width=0.01L. Area= 0.02. This is 2% of the probability density in 1% of the box width. That is 
twice the classical probability, which is uniform in the box. 
b) (2/L) sin2(p/3) times 0.01L=0.015. 1.5 times classical. 
2-6. a) 1/5. (There are five equal hills in .2
5 , and 0=x =L/5 contains one of them. 
b) Smaller. .2
1 (Fig. 2-5) is small in this region (less than 1/L in value) and 
integrates to less than 1/5. 
2-7. a) S. b) -A. c) SS. d) AA. e) -AS. f) AASASSA. g) -AASASAA. 
Rule: Product antisymmetric when odd number of antisymmetric functions is 
present. 
2-8. All are zero by symmetry except d), e), h). i) is integral of 2 sin2 x cos x which is 
SA. j) is integral of sin2 x cos x which is SS from -p to p, but SA in each half 
of the range (i.e., like i). If each half must give zero, so must the whole. 
2-9. [See Hint.] (2/L)  L 
0 sin(npx) sin(mpx)dx = (using Hint) (1/p)  p 
0 {cos[(n- 
m)y]-cos[(n+m)y]}dy =0, for n and m integers and n =m. 
2-10. [See Hint.] .II=8, and so .II is a constant. A constant has a zero derivative, 
and .1 must arrive at x =L with zero derivative if successful junction is to be 
made. This requires that an odd number of quarter-waves fit between 0 and L 
(so wave arrives with either a peak or a valley at x = L). This requires that 
[(2n + 1)/4].I = L; .I = 4L/(2n + 1) = h/v2mU; U = (2n + 1)2h2/32mL2 
is the relation between U and L that is required for a state to exist at E = U. 
(Strictly speaking, one can only approach .=8as a limit, and this problem is 
physically meaningless. However, it makes a good exercise.) 
2-11. Given that H.1 = E1.1,H.2 = E2.2,E1 = E2 = E and f = c1.1 + c2.2. 
Then Hf =c1H.1 +c2H.2 =c1E1.1 +c2E2.2 =E(c1.1 +c2.2)=Ef. 
Q.E.D. 
2-12. a) . should oscillate on right with same . as on left. b) . should be symmetric 
or antisymmetric (and . should be same in each side). c) . should be smooth 
(i.e., have no cusp) at finite barrier. d) . should be a decaying exponential at 
right. e) Same as d). . should not become infinite. 
2-13. a) .4 is two sine waves. 
b) E4 is the same as E2 in a half-box of width L:E4=22h2/8mL2=h2/8mL2. 
Or, one can also calculate it as the n=4 solution with width =2L.
656 Answers to Problems 
2-14. For exponential eigenfunctions, .*. = exp(ijf)]*[exp(ijf) = exp(-ijf) 
exp(ijf)=exp(0)=1. This gives a “rectangle” of height 1 and width 2p, hence 
area 2p and normalizing constant 1/v2p. For sine or cosine functions, we get 
a squared function that oscillates from 0 to 1, gives us half the area, namely p, 
and yields normalizing constant 1/vp. 
2-15. exp(iv2f) does not join onto itself when f.f +2p. 
2-16. a)H =-(h2/8p2I)d2/df2. b)H.=(9h2/8p2I).. c)[(h/2pi)d/df]. = constant times .. Not a constant of motion. 
2-17. The barrier forces solutions to vanish at f = 0. Of our four choices sin(kf), 
cos(kf), exp(±ikf) only the sin(kf) set has this property. Therefore a) E=0 is 
lost. b) All degenerate levels become nondegenerate. c) No, only sine solutions 
exist. d) No, sine solutions are not eigenfunctions of the angular momentum 
operator. 
2-18. a) Enx,ny =(h2/8m)(n2x
/L2x 
+n2y
/L2y
)]=(n2x 
+4n2y
)h2/8mL2x
. 
b) Zero point energy =E1,1 =5h2/8mL2x 
c) nx 1 2 3 1 4 2 3 5 4 
ny 1 1 1 2 1 2 2 1 2 
E/(h2/8mL2x
) 5 8 13 17 20 20 
 	 
 
25 29 32 
degeneracy 1 1 1 1 2 1 1 1 
d) 
e) i) All eigenvalues increase by 10J . (ii) No effect on eigenfunctions. 
2-19. Both 3h2/4mL2x
. Accidental degeneracy results when nodes that are not equivalent 
by symmetry nevertheless give the same energy. 
2-20. a) Yes. The two eigenfunctions being mixed are degenerate (52 + 12 + 12 = 32 +32 +32), so the linear combination remains an eigenfunction. (See Problem 
2-11.) b) Instead of integrating, we can take the value of .2
V because 

V is small enough to make .2 essentially constant in it. .2 = (v2/L)6 
sin2(p/2) sin2(p/2) sin2(p/2) = 8/L3;
V = 0.001L3;.2
V = 0.008. The 
classical value (uniform distribution) is 0.001, so the quantum-mechanical probability 
for finding the particle at the center is 8 times greater than classical. (Not 
coincidentally, this is the cube of the answer for the one-dimensional analog in 
Problem 2-5.)
Appendix 13 657 
2-21. 
E =(6+n)2 -(5+n)2h2/8mL2 
. = (8mcl2/h)(2n+10)2/(2n+11)=637(2n+10)2/(2n+11)Å 
n : 0 1 2 3 
. : 5791 7056 8323 9592 
2-22. Momentum is a constant of motion if [(h/2pi)d/dx]. = constant times .. 
This applies for b) and d), and the momentum equals 3h/2p and -3h/2p, 
respectively. Kinetic energy is equal to k2h2/8p2m, or 9h2/8p2m for all four 
cases. 
2-23. The z component of angular momentum is a constant of motion if 
[(h/2pi)d/df]. = constant times .. This applies for b) and d), and the 
angular momentum equals -3h/2p and 3h/2p, respectively. Kinetic energy 
equals 9h2/8p2l in all four cases. 
2-24. . and d./dx are real at x = a. (. = .*)x=a;Aexp(ika) + B exp(-ika) = 
A* exp(-ika)+B* exp(ika); so Aexp(ika)-A* exp(-ika)=B* exp(ika)- 
B exp(-ika). (d./dx)=(d./dx)*x=a; Aexp(ika) + A* exp(-ika) = 
B* exp(ika) + B exp(-ika); adding these two equations gives 
Aexp(ika)=B* exp(ika), so A = B*. Subtracting the two equations gives 
A* exp(-ika)=B exp(-ika), soA*=B. (Or we could take the complex conjugate 
of the previous equation.) Then, |A|2/|B|2 =A*A/B*B =A*A/A*A=1, 
so |A|=|B|. 
2-25. Matching values at x =0 gives A+B =C. Matching slopes gives kA-kB = 
k
C. Using the first of these to eliminate C from the second gives the first part 
of (2-73). Using the first to eliminate B from the second gives the second part 
of (2-73). 
2-26. NowAis zero, B =C +D,-kB =k
C -k
D,C/D=(k-k
)/(k+k
),B/D= 2k
/(k +k
). 
2-27. a) |C|2/|A|2 gives the relative spatial densities of transmitted to impinging particles, 
but we need relative fluxes, which depend on both particle densities 
and velocities. The relative velocities are the same as the relative momenta 
(since masses are the same), and these are in turn the same as the relative 
k values. So k
/k gives the relative velocities of impinging and transmitting 
particles. For reflecting particles, the analogous ratio is k/k, so it does not 
appear. 
b) k
|C|2/k|A|2 + |B|2/|A|2 = 4k
k/(k + k
)2 + (k - k
)2/(k + k
)2 = 
(k +k
)2/(k +k
)2 =1. 
2-28. 100% transmission occurs when an integral number of de Broglie half-waves fit in 

x =d;n./2=d; also, .=h/p=h/v2T =h/v2m(E -U). Equating these 
expressions for . : 4d2/n2 = h2/ [2m(E -U)] ;(E/U)-1 = n2h2/(8d2mU). 
Substituting for U on the right: (E/U)-1 = n2p2/16. When n = 1,E/U = 1.617. When n=2,E/U =3.467. (Compare with Fig. 2-18b.) 
2-29. 
E1 0.805cm-1,
E2 36cm-1.
658 Answers to Problems 
MC: c d e a e b d b d a c d d b a 
Chapter 3 
3-1. [See Hint.] P(x) is proportional to 1/v(x), which is [dx(t)/dt ]-1, which is 
-vk/mLsin(vk/mt)-1. This is proportional to 

sin2(k/mt)-1/2 
=
1-cos2(k/mt)-1/2 
=1-x2/L2-1/2 
The normalized probability distribution function is (pvL2 -x2)-1. 
3-2. a) x(t)=0.100 cos(v2t), t in seconds, x in meters. b) -0.0453m. c) 1.00× 10-2 J. d) 2.00 × 10-3 J. (e, f) 5.00 × 10-3 J. g) 0.126ms-1. h) ±0.100m. 
i) 0.225 s-1 
3-3. x=±
(n+1/2)h/p)/vkm
1/2
. 
3-4. a) (vß/p)1/(2nn!)
1/2 is the normalizing factor. It keeps the total probability 
density equal to one. Hn(y) is a Hermite polynomial. It provides the nodes 
in the wavefunction. Exp(-y2/2) forces . to decay at large values of x. It 
gives the correct asymptotic behavior. 
b) .0(y) = 4 ß/p exp(-y2/2) 
.1(y) = 4 ß/p(1/v2)2y exp(-y2/2) 
.2(y) = 4 ß/p(1/v8)(4y2 -2) exp(-y2/2) 
3-5. H.0 = -h2/(8p2m) d2/dx2 +kx2/2
 (ß/p)1/4 exp(-ßx2/2) 
= h2ß/(8p2m)-ß2h2x2/(8p2m)+kx2/2.0 
= 
(h/4p)k/m-kx2/2+kx2/2.0 
= (h/2)(1/2p)k/m).0 =(hv/2).0. 
3-6. [See Hint.] a) y=±v1/2. b) y=±v5/2. 
3-7. The first function is asymmetric; the second becomes infinite in both limits of y. 
3-8. a) Approaches zero. Decaying exponential overwhelms polynomial. b) Antisymmetric. 
Polynomial is antisymmetric and exponential is symmetric. 
c) Value =0, slope =120. 
3-9. a) c0 =7. b) c1 =0. 
3-10. a) Not, because missing y3 term should cause the polynomial to terminate, not 
permit y5 term. 
b) Not, because of mixed symmetry. (Powers 5, 3, 1, 0.) 
c) OK. 
3-11. 
3-12. a) Zero (integrand antisymmetric). b) vp/2 (symmetric).
Appendix 13 659 
3-13. Zero (integrand antisymmetric). 
3-14. a) xtp =±
3h/(2pvmk)
1/2 
=±v3/ß. b) .2 is maximum at y =±1, or 
x = ±1/vß = ±h/(2pvmk). c) 1/2. d) 0.0144 [from (2/e)vß/p times 
0.02v3h/(2pvkm)1/2]. 
3-15. [See Hint.] 2  8 a vß/p exp(-ßx2)dx = (2/vp) 8 1 exp(-y2)dy ~ 0.156, 
where a =1/vß. 
3-16, 17. H2(y)=4y2 -2. 
3-18. Zero point energy = 3
2h.; 10; 3. 
3-19. [See Hint.] The barrier requires . = 0 at x = 0 but does not affect Hˆ at 
x > 0. Therefore, antisymmetric solutions of harmonic oscillator are still 
good. Results are .x>0 =v2.harm,osc,n=1, 3, 5, . . . , and E =(n+ 1
2 )h., n= 1, 3, 5, . . . .x=0 =0. 
3-20. a) 40. b) -1. 
3-21. V 5 =E5/2=(11/4)h. =T 5. 
3-22. For each unit of energy going into vibration, half goes into kinetic energy, which 
registers as a rise in temperature, and half is “hidden” as potential energy. None 
is hidden as potential energy in rotation or translation. 
3-23. Using nominal nuclear masses gives k values of 958, 512, 408, 311Nm-1, 
respectively. Decreasing bond stiffness implies decreasing bond strengths. 
(Observed D0 values are, in kJ mol-1, 564, 428, 363, 295, respectively.) 
MC: e e d, c c 
Chapter 4 
4-1. E2 - E1 = 2.46737 × 1015 Hz using me. Using µ gives 0.999455 times this 
value, a difference of 545 ppm. 
4-2. The radial distribution function vanishes at r =0 because 4pr2 is zero there and 
again at r =8 because the decaying exponential of .2 overwhelms the finite 
power term 4pr2. 
4-3. [See Hint.] E1s=-1
2 a.u. V (r)=-1/r a.u. Equal when r =2 a.u. 8 2 .2dv= 0.238, and so 23.8%. 
4-4. a) 1/Z a.u. b) 3/2Z a.u. c) 0 (by inspection of .2). d) -Z2 a.u. Differs 
from -Z times the reciprocal of b) because average of 1/r =1/(average of r).
660 Answers to Problems 
d) is lower because 1/r blows up at small r, contributing large (negative) amounts 
of potential energy. 
4-5. (1/v2) 8 0 exp(-r)(2-r) exp(-r/2)r2dr =. . . 0. (Do not forget r2 from dv.) 
4-6. p-1/2 is the normalizing factor. 
4-7. ¯x = +8 
-8 .2x dx. .2 is symmetric, x is antisymmetric, and ¯x =0. 
4-8. a)  L 
0 .n(x - L/2)2.n dx. b) Expect it to increase as n increases, approaching 
a limiting value. This because .2 favors the box center for lower states 
and approaches classical (uniform) distribution as n increases. c) L2(1/12 - 1/2n2p2), n=1 : 0.03267L2, n=2 : 0.070668L2. Approaching L2/12, which 
is the classical value achieved if .2 in a) is replaced by 1/L. 
4-9. For reflection in the xy plane, 2pz is antisymmetric, and 3dxy is symmetric. 
4-10. a) 4/Z a.u. b) 5/Z a.u. c) 2/Z a.u. d) -Z2/4 a.u. 
4-11. [See Hint.] rtp =2n2/Z. 
4-12. [See Hint.] The sum of squares of angular dependencies equals 4
3 . Use trigonometric 
identities to remove angle dependence. 
4-13. a) -1/2 a.u. b) -2 a.u., -1/2 a.u. c) 25 d) 2, 1 e) -2 a.u. 
4-14. a) l =m=0. (No angular dependence: must be s.) 
b) 2 (quadratic in r). 
c) -1/18 a.u. (2 nodes means it is 3s, so n=3.) 
d) 18 a.u. 
4-15. a) Yes, at r =6 a.u. b) 3py(r sin . sin f =y, so it is py, and there is one radial 
node). 
4-16. a) Looks OK. b) No. Blows up at large r. c) No. Lacks exponential decay 
function in r. 
4-17. [See Hint.]  1s2r sin . cosf dv = 0 (because cosf is antisymmetric in each 
subrange 0–p, p–2p). The average value of x should be zero because the electron 
is equally likely to be found at equal ±x positions due to the spherical symmetry 
of .2. 
4-18. [See Hint.] a) .mp = 35.15
, 144.45
. b) Angular nodes come where 
(3 cos2 . -1)=0. . =54.74., 125.26.. 
4-19. xy =(r sin . cosf)(r sin . sin f)=r2 sin2 . cosf sin f =(r2 sin2 . sin 2f)/2. 
4-20. Equation (4-45) predicts 2/3 and 0 for these integrals. Actual integration over 
x2 and over 5x4/2-3x2/2 gives 2/3 and 0. Integral over all space for 3p03d-1 
involves integral from 0 to p of P0 
1 (cos .)P-2 
2 (cos .) sin. d., which is same as 
integral from 1 to -1 of P0 
1 (x)P -1 
2 (x) dx, so the integral vanishes. 
4-21. 
15/(2v10p (1-cos2 .) cos . exp(-2if), f-2. 
4-22. LˆzYl,m(.,f) = (h¯/i)(d/df)
l,m(.) exp(imf) = mh¯
l.m(.) exp(imf) = mh¯ 
Yl,m(.,f).
Appendix 13 661 
4-23. ˆ Lx2p0 = ˆ LxR(r) cos . =-iR(r) sin . sin f =constant ×2p0 
ˆ Ly2p0 = iR(r) sin . cosf =constant ×2p0 
Lˆx1s = 0=Lˆy1s=Lˆz1s=Lˆ2ls ···=0×ls 
Since the vector has zero length, its x, y, z components must also have zero 
length. The question of vector orientation becomes meaningless. 
4-24. [-1/(sin .)(d/d.) sin .d/d.]R(r) cos . =2R(r) cos ., l(l +1)=2, l =1. 
4-25. 
Same for 6f 
4-26. [See Hint.] Will be eigenfunctions in cases b) and e). Only these correspond to 
mixing degenerate eigenfunctions. 
4-27. a) v6¯h or v6 a.u. b) d-2. 
4-28. a) 0. b) -3¯h or -3 a.u. 
4-29. a) 0. b) 2¯h2 or 2 a.u. c) 0 (same as if all x’s were z’s). d) 0 (by symmetry). e) 0. 
4-30. a) 6.3,2,1. b) 2.2px. c) (-1/18).3px. d) -.2p-1 . 
4-31. a)Yes, -1/18. e)Yes, 0 (same as Lˆz.3pz ). f)Yes, 2. All others, No. 
4-32. 1.44×10-3 a.u. 
4-33. 2.80×1010 Hz, 11.4 ppm. 
4-34. a) -1/32. b) 16. c) 0, v2,v6,v12. d) 7. e) 4. 
4-35. µr2 =?=m1r2
1 +m2r2
2 . Chug, chug, Bingo. 
4-36. Using Eq. (4-56), we write Eq. (4-68) as (Lˆ2/h¯2). =(-2IE/h¯2).; (Lˆ2/2I )= 
E.; since there is no potential term, E is all kinetic, so this completes the 
demonstration. 
4-37. [See Hint.] First find µ in a.m.u., then divide by 6.0221 × 1026 to convert to 
kg/molecule. µ=1.6529×10-27 kg. r =vI/µ=1.4144×10-10 m. 
4-38. 10.976cm-1, 21.953cm-1, 32.929cm-1. 
4-39. 1.129×10-10m(µ=1.1385×10-26 kg, I =1.4504×10-46 kg m2). 
4-40. J =3, so mJ =3, 2, 1, 0,-1,-2,-3. The states split into seven equally spaced 
levels. 
4-41. v3/2, or 0.866 a.u.
662 Answers to Problems 
MC: e b b e d e b c c 
Chapter 5 
5-1. -1
2 (.2
1 +.2
2 +.2
3 )-3/r1 -3/r2 -3/r3 +1/r12 +1/r23 +1/r13. 
5-2. ¯r1s =3/2Z =(for He+) 3
4 a.u., ¯r2s =6/Z =(for He+) 3 a.u. 
5-3. E = 2.343 × 105 eV compared with an IE of 13.6 eV. It shows that . small 
enough to locate the electron with useful precision involves photons with energy 
sufficient to excite the electron completely out of the system. 
5-4. [See Hint.]  .2dv = 1
2  [1s(1)22s(2)2 + 2(1s(1)2s(1)2s(2)1s(2)) + 
2s(1)21s(2)2]dv(1)dv(2)= 1
2 (1 · 1+2 · 0 · 0+1 · 1)=1. 
5-5. .a(2.1)=(1/v2)[1s(2)2s(1)-2s(2)1s(1)]=-.a. 
5-6. Upon substitution and expansion, complete cancellation occurs. 
5-7. (1/v6)[1s2p1s(aßß -ßßa)+1s1s2p(ßaß -aßß)+2p1s1s(ßßa -ßaß)]. 
5-8.  1s*2sdv  1s*1sdv  2s*1sdv  a*ad. ß*ad. a*ßd. = 0 · 1 · 0 · 1 · 0 · 0=0. 
5-9. For r1 =1, r2 =2, r3 =0 get 
.(1, 2, 0) = (1/v6)[1¯s(r =1)2pz(r =2)1s(r =0)+2pz(r =1)1s(r =2)1¯s(r =0) 
-2pz(r = 1)1¯s(r =2)1s(r =0)-1s(r =1)2pz(r =2)1¯s(r =0)] 
The other cases are the same except for factor of -1. Thus, .2 is identical for 
all three cases, and no physical distinction exists. 
5-10. Replace each a in the lowest row with ß. 
5-11. a) aa, ßß, .., aß +ßa, a. +.a, ß. +.ß 
b) aß -ßa, a. -.a, ß. -.ß 
5-12. a) Yes. Antisymmetric for exchange of any two electrons. 
b) -1/2(.2
1 +.2
2 +.2
3 )-3/r1 -3/r2 -3/r3 +1/r1,2 +1/r2,3 +1/r1,3. 
c) No. It is a product of one-electron orbitals, hence an independent-electron 
solution, but H is not separable. 
d) (-9/2)(1+1/4+1/9)=-6.125 a.u. e) -3/2 a.u. 
5-13. Equation (5-41) with U1 =1s, U2 =1¯s, U3 =2s, U4 =2¯s. 
5-14. F, 1s22s22p5, Z =9, .1s =8.7, .2s =.2p =2.6. 
5-15. LetAˆf=af and  f*fdt =1. (Aˆ)av = f*Aˆfdt = f*af dt =a f*f dt =a. 
5-16. Li2+ is a hydrogenlike ion, and hence should have all states of same n degenerate. 
Li differs in that potential seen by electron is not of form -Z/r, due to screening 
of nucleus by other electrons. Hence, degeneracy is lost. The 2s AO of Li is 
lower than the 2p due to the fact that the 2s electron spends a larger fraction of 
time near nucleus where it experiences full nuclear charge. 
5-17. 
1
2 
1
2 +1=3
4
a.u.
Appendix 13 663 
5-18. n electrons of a spin give one state. Each time one a is changed to a ß we get a 
different state. There are na spins available to change. 
5-19. Symmetric combination gives e2 +8e either way. Antisymmetric gives e2 -8e 
one way and -(e2 -8e) the other. 
5-20. a) 2s1/2( 2 states). b) 2p3/2,2p3/2,2s1/2( 8 states). 
5-21. a) Not satisfactory since electron 1 is identified with the 1s AO, etc. 
[1s(1)3d2(2)-3d2(1)1s(2)]a(1)a(2). 
b) 3D3. 
5-22. 120. 
5-23. a) 2P3/2, 2p1/2. b) 2. c) 4, 2. 
5-24. a) a(1)ß(2) - ß(1)a(2). b) -22.5 a.u. c) -22.5 a.u. + J1s,2P1 + K1s,2P1 
. 
d) Eigenvalue for S2 is 0. 
5-25. All in a.u.: a) 6. b) 2. c) 12. d) 2, 1, 0,-1,-2. e) 1, 0,-1. f) 3, 2, 1, 0,-1, 
-2,-3. 
5-26. a) In both cases, maximum net z components are: spin =1, orbital=1. 
b) 3P0 below 3P1 below 3P2 for p2 case, reverse for p4. 
c) Each pairing reduces multiplicity to m+1, where m is number of unpaired 
electrons, which equals the number of holes. Each pair’s contribution to ML 
due to their orbital’s ml value is either totally canceled by a pair at -ml or 
else half-canceled by an unpaired electron at -ml . Therefore, the uncanceled 
portions reflect the presence of unpaired electrons, hence holes. Therefore, 
the amount of uncanceled ml is equal to minus that which we could assign to 
the holes. 
5-27. a) 2S1/2. b) 4S3/2. c) 1S0. d) 3F2. 
5-28. a) 12. b) 3P2, 3P1, 3P0, 1P1. 
5-29. a) 2P1/2 below 2P3/2. b) 4S3/2 below 2D3/2, 2D5/2 below 2P1/2, 2P3/2. (Rules 
do not allow us to sort by J in this case since shell is exactly half filled.) 
5-30. Striving for maximum multiplicity means avoiding pairing electrons. 
5-31. a) 15. b) 35. 
5-32. No. Impossible for doublet and singlet combinations to arise from same number 
of electrons. Each pairing reduces number of unpaired electrons by two, so 
allowed multiplicities are all even or all odd. 
5-33. a) 20. b) 12. c) 6. d) 60. e) 100. 
5-34. a) 28. b) 4F9/2, 4F7/2, 4F5/2, 4F3/2. 
5-35. 3D3. Also 3D2, 3D1. 
5-36. 2g2 -g, 15, 45. 
5-37. Splittings equal 9.274×10-24 J times g, with g equal to 4/3, 7/6, 1/2, 1, respectively.

664 Answers to Problems 
MC: d e e d 
Chapter 6 
6-1.  8 
-8 .*(d2/dx2)fdv 
?=
 8 
-8 f(d2/dx2).*dv. Use 8 
-8 vdu = uv|8-8 - 
 8 
-8 udv. On the left, let v=.*, u=df/dx, du=d2f/dx2, dv=d.*/dx. On 
the right, let v =f, u=d.*/dx, etc. The uv term vanishes since .* and f each 
vanish at limits. The remaining integrals are identical. 
6-2. Each equals -4v8/27. 
6-3.  .*.dv =1,  .*i .jdv =di,j , . =ici.i . Then 
 .*.dv =1= ic*i .*i j cj.jdv =ij c*i cj dij =ic*i ci . 
Q.E.D. 
6-4. . =iciµi ,  µ*i
µjdv =di.j , want ck: 
 µ*k
.dv = µ*k
iciµidv =ici  µ*k
µidv =icidk,i =ck 
6-5. a) (1/p)  2p 
0 
cos 2f(¯h/i)(d/df) cos 2fdf = -(2/¯hpi)  2p 
0 
cos 2f sin 2fdf 
. 2p 
0 
sym · antisym = 0 
b) . =(1/v2) 
(1/v2p)exp(2if)+(1/v2) 
(1/v2p)exp(-2if) 
Terms in [] are normalized eigenfunctions of Lz with eigenvalues of +2¯h and 
-2¯h. So (Lz)av =(1/v2)2(2¯h)+(1/v2)2(-2¯h)=0. 
6-6. [x,px] = [x(¯h/i)(d/dx) - ( ¯h/i)(d/dx)x]f (x)=( ¯h/i) xf 
 -f -xf 
=-( ¯h/i)f 

x · 
px = 
1
2

 .*(-¯h/i).dt
=|-¯h/2i|= ¯h/2 
6-7. f must be identical to the eigenfunction .0. 
6-8. No. The existence of some real eigenvalues does not guarantee that the operator 
satisfies the definition of hermiticity: 
(d/dr) exp(-ar)=-a exp(-ar) 
but 
 8 
0 
exp(-ar)(d/dr) exp(-br)r2dr =  8 
0 
exp(-br)(d/dr) exp(-ar)r2dr 
if a = b 
6-9. a) Each side equals -(1/2v2).1s exp(it/2)-(1/8v2).2p0 exp(it/8). 
b) [See Hint.] (*)t=0 = (1/2)(1s2 + 21s2p0 + 2p20
);(*)t=1/. = 
(1/2)(1s2 + 2 cos(3/8.)1s2p0 + 2p20
). Equal when 3/8. =2p;. =3/16p. 

E = (1/2)(1 - 1/4) = 3/8 a.u.; 
E = h. = 2ph./2p = 2p. ¯h = 2p. in 
a.u. . =
E/2p =(3/8)/2p =3/16p.
Appendix 13 665 
6-10. (1-S2)-1/2. 
6-11. [See Hint.] S =v3/2, and so f =(2/v3p)(r - 3
2 ) exp(-r). 
6-12. a) 8v8/27=0.838. b) 0 (by symmetry). 
6-13. [See Hint.] a) ˆ px.0 =-(ß ¯hx/i).0 =constant times .0. Momentum is hence 
not a constant of motion. 
b)  .0*pˆx.0dx=-(ßh¯/i)  .0*x.0dx= (sym)(anti)(sym)=0 (Must be zero 
since otherwise motion in one direction would involve greater momentum than 
motion in the other.) 
6-14. First pair: No. x2(d2/dx2) + x3(d3/dx3) = 2x(d/dx) + 4x2(d2/dx2) + 
x3(d3/dx3). Second pair: Yes. Both arrangements give 2x2(d2/dx2) + 
x3(d3/dx3). 
6-15. a) 6 a.u. (since l =2). b) 0 (since real form of . involves equal mix of ±ml ). 
6-16. a) Because these operators commute, yes, there must be a set of simultaneous 
eigenfunctions. 
b) Mixing degenerate-energy cases gives functions that remain eigenfunctions 
for the energy operator but not for the momentum operator. (These are the 
sine and cosine versions.) 
c) Exactly knowable because knowledge of momentum gives knowledge of an 
eigenfunction which in turn gives knowledge of energy. 
6-17. E =(1/3)(-1/2-1/8-1/18)=-0.2268 a.u. 
6-18. [See Hint]. Normalized . = 0.26726[1s + 2(2p1) + 3(3d2)]. Lˆz = 0.267262[0+4(1)+9(2)]=1.571 a.u. 
6-19. Proofs are in Sections a) 6-8, b) 6-9, c) 6-11. 
6-20. ||2 =(1/2) |.1|2 +|.2|2 +2|.1.2|cos[(E2 -E1)t/¯h]
. 
6-21. Cycle time =mL2/h=mL2/2p a.u. 
6-22. cn =(2v2/L)  L/2 
0 sin(2px/L) sin(npx/L)dx. 
a) v2.2 and  are same function in range 0=x =L/2, guaranteeing c2 =v2. 
No other c can be larger or the resulting function will give a total probability 
density greater than 1. 
b) .4 and  have opposite symmetry for 0=x =L/2. 
c) As .i becomes more oscillatory, the positive and negative portions of its 
product with  will cancel more effectively. 
6-23. a) c1 =0.838, c2 =0.2048. 
b) Whereas only s-type AOs appear in Eq. (6-41), explicit account of changing 
potentialwould yield a nonspherically symmetric potential so that p-typeAOs 
would enter too. 
6-24. [See Hint.] ck is small if there is effective cancellation between positive and 
negative portions of the product of exp(-ax2) and cos(kx). Larger k makes 
cos(kx) more oscillatory and makes cancellation more effective. The broader 
exp(-ax2) is, the more effective is this cancellation for a given value of k.
666 Answers to Problems 
6-25. a) Real V means realHˆ . Hˆ(x, y, z, t)=-(h¯/i)(./.t)(x, y, z, t). Complex 
conjugate of equation gives Hˆ*(x, y, z, t) = (h¯/i)(./.t)*(x, y, z, t). 
Transform t .-t throughout (does not affect equality), then recognize 
./.(-t) = -./.t , so Hˆ*(x, y, z,-t) = -(h¯/i)(./.t)*(x, y, z,-t). 
Q.E.D. 
b) (x, y, z, t) becomes .(x, y, z) exp(-iEt/¯h). *(x, y, z,-t) becomes 
.*(x, y, z) exp[iE(-t)/¯h] = .*(x, y, z) exp(-iEt/¯h). Carrying these 
through Eqs. (6-3,4,5) shows that Hˆ .* =E.*. 
c) If . and .* are independent, we have two solutions with the same energy, 
i.e., degeneracy. If E is not degenerate, . and .* are not different, so . =.* 
and is real. 
d) 2p-1 becomes -2p+1. No change for 2p0. 
e) No. While all complex eigenfunctions must be degenerate, real eigenfunctions 
can be degenerate too. Cases we have seen are (1) accidental degeneracies 
such as .3,3,3 and .5,1,1 in the 3-dimensional cubic box and (2) real eigenfunctions 
constructed as linear combinations of complex ones, as 2px and 2py 
from 2p+1 and 2p-1. 
MC: c a d 
Chapter 7 
7-1. a) Yes. c1 = v2/L L 
0 f (x) sin(px/L)dx, c2 = v2/L L 
0 f (x) sin(2px/L) 
dx. c1 =0 by symmetry. c2 is positive. 
b) No. Both a) and b) are continuous, smooth, single-valued functions, but b) 
does not go to zero at x =0,L as do all the box eigenfunctions. 
7-2. a) 
b) 1.3h2/8mL2. 
7-3. c1 =v0.4=0.632, c3 =0 by symmetry. 
7-4. a) [See Hint.] For n=odd, 
cn =2  L/2 
0 
f.dx=± 
4v6 
n2p2 
+for n = 1, 5, 9, 13, . . . 
-for n = 3, 7, 11, 15, . . . 
For n=even, cn=0 (by symmetry) (i.e., f is symmetric and so contains only 
symmetric .n). 
b) fapprox(x =L/2) = 
m 

n=1 
cn.n(x =L/2) 
= (4v6/p2)2/L 
m 
 
n=1(odd)
(1/n2)
Appendix 13 667 
m: 1 3 5 7 9 . . . 135 vLfapprox(x =L/2): 1.40395 1.55994 1.61609 1.64475 1.66208 1.72689 vLf(x =L/2)=v3=1.73205. 
c) [See Hint.] 
E¯ = 
odd n
(4v6/n2p2)2(n2h2/8mL2)=(12h2/p4mL2)
odd n
(1/n2) 
= (12h2/p4mL2)
 m 
 
odd n=1
(1/n2)+ 
1
2 
 8 
m+1
(1/x2)dx
 
= [(h2/8mL2)(96/p4)(1.23386)]m=9 
= [1.21602(h2/8mL2)]m=9;[1.21432(h2/8mL2)]m=135 
7-5. Normalizing factor = (2a/p)3/4; E¯ = (3a/2) - 2v2a/vp; a = 8/9p; 
E¯(min)=-4/3p =-0.4244 a.u.; r¯ =1.5 a.u.; rmp =v9p/4=1.329 a.u. (For 
.exact, E=-0.5 a.u., ¯r =1.5 a.u., rmp =1.0 a.u.) 
7-6. a) and b) See text and Eqs. (7-16)–(7-20). c) a = 53
,E¯ =-0.370 a.u. 
7-7. a) 1/v8. b) -0.292 a.u. 
7-8. a) Zero, because the function is symmetric for xy reflection whereas 2pz is 
antisymmetric. b) -0.4215 a.u., (by assuming c3s is v0.05 and all higher terms 
vanish). 
7-9. a) . =5/3,E¯ =-0.2777 a.u. 
b) c1 =0.897. 
c) . =2.26(f -0.897.1s). 
d) No. . cannot have E¯ lower than the n=2 value, which is the lowest-energy 
case orthogonal to n=1 and which has E=-0.125 a.u. 
7-10. -0.75 a.u. 
7-11. [See Hint.] The energy E¯ =-0.375 a.u.=(0.9775)2(-1
2 a.u.)+higher-energy 
contributions. But this leading term equals -0.478 a.u., and so the net value of 
the higher energy terms must be positive. Therefore, at least one of them must 
correspond to a state with positive energy—a continuum state. 
7-12. Slater’s rules give . = 1.7, whereas the variation method gives . = 27/16 = 1.6875. 
7-13. a) S11 = 1, S12 = 0, S22 = 1, H11 =-1
2 , H12 = -F, H22 = 0. E¯ = -1
4 - 1
4
v1+16F2; for F = 0.1,E¯ = -0.51926 a.u. This trial form is superior 
because z · .1s is more contracted than .2pz , closer in size to .1s, hence 
interferes constructively and destructively with .1s more effectively. 
b) [See Hint.] lim(F .0) of v1+16F2 =1+8F2, in lim, E¯ =-1
2 -2F2, 
-1
2aF2=-2F2; a =4. E(e2/a0).aF2 (a units) ·(e/a2
0 )2, a units =a3
0. 
(See Appendix 10.) 
7-14. Since 2s is isoenergetic with 2p states, these should mix freely in response to 
field. Hence, 2s is more polarizable.
668 Answers to Problems 
7-15. E¯(min)=-12/5=-2.4 a.u. . =v2/5(fa +fb). 
7-16. [See Hint.] E¯(lowest)=-2.030 a.u. . =1.045fa -0.179fb. 
7-17. [See Hint.] E¯elec = (. 2/2) - 2 + 2(. + 1) exp(-2. ); .best = 0.9118, E¯elec = 
-0.9668 a.u.; E¯elec + 1/R = E¯tot = -0.4668 a.u. Since this energy exceeds 
that of H +H+ (-0.5 a.u. at R=8), this function does not demonstrate the 
existence of a bound state. 
7-18. 1su becomes a 2p AO of He+, so E=-0.5 a.u. 
7-19. [See Hint.] Since they are degenerate, .+ and .- may be mixed. The sum gives 
1sa, describing the case in which the electron is at a. The difference gives 1sb. 
7-20. [See Hint.] Both equal -3
2 a.u. This is higher than the lowest He+ eigenvalue 
because these are hydrogen atom 1s functions instead of He+ functions. 
7-21. k must be greater than S. 
7-22. a) Hˆ =-1
2.2 -(1/rH)-2/rHe. 
b) Separated atoms: lowest energy for H+ +He+(1s)=-2 a.u. For the united 
atom; Li2+(1s)=-4.5 a.u. 
7-23. a) pu bonding. b) su antibonding. c) sg bonding. d) dg bonding. e) pu bonding. 
7-24. 1ssg 2pzsu 3pypu 3dxydg 
2ssg 2pxpu 3dz2sg 3dxzpg 
7-25. a) Antibonding. b) Bonding. c) Bonding. 
7-26. 2px, 2py, 3px, 3py, 3dxy, 3dyz. 
7-27. [See Hint.] The integrals that vanish by symmetry are b), c), e), and f); g) does 
not vanish. The AOs are orthogonal due to different symmetry for reflection in 
the xy plane at a. But Hˆ is not invariant to this reflection. In f), the relevant 
reflection is through the yz plane; Hˆ is invariant to this one. 
7-28. Sketches show that several planes qualify, but symmetries are opposite no matter 
which is chosen. For instance, if the xz plane is selected, dx2-y2 is symmetric, 
dxy is antisymmetric, pxz is symmetric, pyz antisymmetric. 
7-29. 0. (It is a s MO.) 
7-30. a) su antibonding. b) pu bonding. c) pg antibonding. d) dg bonding. 
7-31. a) 4. b) 3 (a triplet). c) (1) increase, (2) decrease. d) 2 (a doublet). 
7-32. a) 1s2 
g 1s2 
u 2s2 
g 2s2 
u 3s2 
g 1p4 
u 1p1 
g. b) 5. c) O+2
has larger D0. d) 2g. 
e) All s MOs. 
7-33. For He2, the second MO (su1s) correlates with third united atomAO (2ps ). For 
LiH, the second MO (s ) correlates with second united atomAO (2s). Thus, this 
MO is less antibonding in heteronuclear case. 
7-34. [See Hint.] 2s
 = 1.0295 2s - 0.2447 1s; 2sg = 0.0136 1sA - 0.6523 2s
A - 0.0854 2ps,A and similarly for B. 
7-35. Eelec = Esepatoms - Vnn - De; De =-Eelec - Vnn + Esepatoms = (1.1026 - 0.500-0.500)a.u.=0.1026 a.u.
Appendix 13 669 
MC: a a d b d 
Chapter 8 
8-1. For .prod, E=E1+E2+E3. For .det, E= 1
6(E1+E2+E3)six times. Energies 
of products in . are E1 +E2 +E3 and E1 +E2 +E4. These can be factored out 
to give Hˆ . =E. only if E3 =E4. 
8-2. a) 
b) 
c) 
 
x 1 
1 x 
 
(only 2 unsaturated carbons, so the same as ethylene). 
d) Same as c). Same as two ethylenes since the two p systems are noninteracting 
due to spatial separation. 
e) Same as c). Same as two ethylenes since the two p systems are orthogonal 
and noninteracting. 
8-3. 
 
x 1 1 1 
1 x 0 0 
1 0 x 0 
1 0 0 x 
 
=0, x4 -3x2 =0 See Appendix 6 for results. 
8-4. 13
, since only .3 is common to both MOs. 
8-5. S =1/v2, 1/v1-S2 =v2, f
2(normalized)=v2(f2 -f1/v2). This results 
in all |c| = 1/2, with negative values on same side. That is, the node for f
2
is 
vertical, while that for f1 is horizontal. 
8-6. a) a -2ß (using octagon in circle). 
b) 
(since totally antibonding). 
8-7. a) E =a +2ß(v2/3v3)+0+0+0)=a +0.544ß. 
b) p12 =0.272. All others zero.
670 Answers to Problems 
c) E = a + ß. (It is an ethylene pi bond, distributed over three locations, or 
three-thirds of a double bond. Or one can use a =1/v6 and equations for p 
and E.) 
8-8. Highest two levels have same magnitudes as lowest two, with coefficients 
multiplied by -1 on alternating atoms (here taken to be second and fourth). 
8-9. a) No. This is an odd alternant, so expect paired energy levels with an MO at 
E = a. The unpaired electron is therefore not in a degenerate level. b) Yes. 
Pentagon in circle forces degeneracies. Second level has three electrons. c)Yes. 
Square in circle shows that second level is degenerate. It contains one electron. 
d)Yes. The degenerate level of benzene now has an odd number of electrons. 
8-12. Bond orders: CH2–CH, 0.8944.0.6708; CH–CH, 0.4472.0.5854. Bond 
lengths in A (Eq. 8-61): CH2–CH, 1.354.1.392; CH–CH, 1.436.1.408; 

CH2–CH=+0.038A, 
CH–CH=-0.028Å. 
8-13. For benzene, cµi should be taken as 1/v6 since all carbons are equivalent. 
8-14. Only fluoranthene deviates markedly because it is nonalternant. Hence, its 
LUMO and HOMO energies are not symmetrically disposed about E =a. 
8-15. a) Oxidation potential ~ 0.97 V, reduction potential ~ 1.41 V. b) [See Hint]. 
To shorten: 4–10, 9–10, 8–9; to lengthen: 3–10, 1–9, 4–5, 7–8; otherwise no 
change. 
8-16. E =18a +21.877ß. Error =0.0015ß per p electron. 
8-17. a) Naphthalene; Ep (from Table 8-2) = 10a +13.128ß, from HMO = 10a + 13.6832ß. The difference = 0.055ß per p electron, aromatic. Perylene; 
Ep (Table 8-2) =20a +27.2796ß, Ep (HMO)=20a +28.2453ß; 
Ep = 0.048ß per p electron, aromatic. 
b) RE for perylene is slightly less than double that for naphthalene. The central 
ring does not appear to be contributing. 
c) These two bonds are single in all formal (nonpolar) structures. 
d) The calculated length =1.433Å. The HMO length is too short. The actual 
length is more consistent with these being “truly” single bonds. 
8-18. The fourth molecule in Fig. 8-24 should strive for six electrons in each ring. This 
would make the left side (i.e., the seven-membered ring) net positive. The other
Appendix 13 671 
two molecules become net negative on the left. Electron densities corroborate 
this. For the fourth molecule, charge densities exceed unity in the five-membered 
ring and are less elsewhere. The fifth molecule has only one p electron density 
less than unity, and this is for the methylene carbon. 
8-19. a) 9. b) 10. c) 4. d) 6. e) 10. 
8-20. Since the formal structure always shows C–O single bonds, C1=C2, and C3=C4 
double bonds, and C2–C3 as single, we can use the single-, double-bond distinctions 
of Table 8-3. These give 
 
x 1.1 0 0 0.8 0 0 0 
1.1 x 0.9 0 0 0.3 0 0 
0 0.9 x -0.1 1.1 0 0 0.8 0 
0 0 1.1 x 0.8 0 0 0 
0.8 0 0 0.8 x +2.0 0 0 0 
0 0.3 0 0 0 x +1.5 0 0 
0 0 0.8 0 0 0 x -0.1 3.0 
0 0 0 0 0 0 3.0 x -0.5
 Otherwise, the positions with 1.1 and 0.9 become 1.0. 
8-21. a) Left. b) Left. c) Right. 
8-22. qr = 1 at all centers, and so it does not predict some centers best for nucleophilic, 
and hence worst for electrophilic substitution, so it doesn’t apply to 
this question. Since the HOMO and LUMO have identical absolute coefficients 
(by the pairing theorem), the same site is most favored for both nucleophilic 
and electrophilic substitution. Lr must be identical for nucleophilic, radical, 
or electrophilic substitution because an interrupted even alternant produces an 
odd alternant, for which cationic, neutral, and anionic p energies (ß part) are 
the same. Thus, HOMO and LUMO indices are consistent with coincidence of 
active sites for nucleophilic and electrophilic substitutions, and Lr is consistent 
with the coincidence of these with active sites for radical addition. 
8-23. No. Both types should prefer the most polarizable site, since that is the site to 
which charge is most easily attracted or from which it is most easily repelled. 
8-24. a) F1 =0.0684, F2 =0.4618, F4 =0.9737. 
b) Index Values Preferred site 
qr q2 =0.818 q4 =1.488 4 
HOMO c2
2 =0.1356 c2
4 =0.6646 4 
L+r
L+2 
=2.134ß L+4 
:= 0.962ß 4 
prr p22=-0.4340 p44=-0.4019 2 
c) Only protons on C2 and C3 will produce ESR splitting in simplest theory, 
since the singly occupied MO of the radical anion is zero elsewhere. d) Net
672 Answers to Problems 
bonding, because energy is below E=a and this happens only when bonding 
interactions dominate. 
8-25. The correlation is fairly good except for styrene, which is way off. But styrene 
is the only member of the set where addition is not occurring at a ring position. 
Because the geometric constraints are so different, the relation between 
free valence and transition-state energy is presumably rather different for 
styrene. 
8-26. The five-membered ring because, as it strives for six electrons in order to satisfy 
the 4n+2 rule, it becomes negatively charged. (The seven-membered ring also 
strives for six electrons, becoming positive.) 
8-27. Fulvene a) should experience greater change. It is not an alternant hydrocarbon, 
whereas benzyl b) is. TheMOinto which the electron goes in benzyl is necessarily 
nonbonding, with zero coefficients at positions 1, 3, and 5. (See Appendix 5 
for discussion, Appendix 6 for coefficient values.) 
8-28. a) 11. b) 2 and 4. c) 3, 6, and 8 (1, 9, and 10 have no attached hydrogen). 
8-29. a) is easier to ionize because the HOMO is zero at the nitrogen position (in the 
all-carbon analog). This means ionization does not remove electronic charge 
from the more electronegative atom in case a) but does in case b). 
MC: c b 
Chapter 9 
9-1. a) 212 b) 6a 6b 6c 
7a 7b 7c
 c)  25 13 
i -7 18
 d) 1 0 
0 1
 e) 
.
.. 
3i +16 2i +28 
31 51 
-12 -21 
.
.. 
f) product not defined g) 1 
9-2. H is defined to be hermitian if Hij = Hj*i . Hj i =  .j*Hˆ .idt , and so H*ji = 
 .jHˆ *.*i dt . But if Hˆ is hermitian, this must equal  .*i Hˆ .j dt =Hij . Therefore, 
H*ji =Hij and H is hermitian. 
9-4. AC = "
CA. But 
"
CA = A˜C˜ , so AC = A˜C˜ . This must be true in this example 
because A and C are symmetric. That is, A=A˜ , C=C˜ . 
9-5. a) |B - .i1| = |T-1AT - .i1| = |T-1AT - .iT-11T| = |T-1(A - .i1)T| = 
|T-1||A-.i1||T| = |TT-1
(A-.i1)| = |A-.i1| b) For diagonal B, value of |B-.11| is product of diagonal elements. For this 
to vanish, at least one such element must vanish. This will occur whenever .i 
equals a diagonal element of B. Therefore, the latent roots are the diagonal 
values. 
9-6. If a latent root is zero, then the product of latent roots is zero. But this product 
is the value of the determinant of the matrix. If the determinant of the matrix is 
zero, there is no inverse.
Appendix 13 673 
9-7. a)
tr(ABC) = 
n 

i=1
(ABC)ii =
i 

j 

k 
aij bjkcki 
=
i 

j 

k 
ckiaij bjk which is 
k 
(CAB)kk =tr(CAB) 
=
i 

j 

k 
bjkckiaij which is 
j 
(BCA)jj =tr(BCA) 
=
i 

j 

k 
(ckibjkaij ) which is not
i 
(CBA)ii , hence =tr(CBA). 
b) tr(T-1AT)=tr(TT-1A)=tr(A). 
9-8. (norm ˜TAT)2 =
i,j 
(#˜TAT)ij ( ˜TAT)ji =
i,j 
(T˜A˜ T)ij (T˜AT)j i 
= 
i,j,k,l
(T˜)ik(A˜ )kl(T)lj (T˜)j l(A)lk(T)ki 
=
k,l 
.
..... 
(A˜ )kl(A)lk
i 
( ˜T)ik(T)ki 
 1	 


j 
(T)lj ( ˜T)jl 
 1	 

.
..... 
=
k,l 
(A˜ )kl(A)lk =(norm A)2 
9-9. a) tr=0, det=2, norm=v6; therefore, a+b+c=0, abc=2, a2+b2+c2=6; 
solutions: 2,-1,-1. 
b) Solutions 1, 1,-1. c) 0, 1+v3, 1-v3. 
9-10. Both vectors transform to  3 cos . 
-3 sin .
. Hence, reversal is not possible and transformation 
is singular. This is verified by fact that the determinant vanishes. 
9-11. The matrix is already diagonalized. This means the eigenvector matrix is the 
3×3 unit matrix. 
9-12. Let T-1AT=DA (diagonal) and T-1BT=DB (diagonal). Then DADB =DBDA 
(diagonal matrices commute); T-1ATT-1BT = T-1BTT-1AT; T-1ABT = T-1BAT;TT-1ABTT-1 =TT-1BATT-1;AB=BA. 
9-13. In the second case, C is not unitary, since C†SC=1. The ordinary procedures 
for diagonalizing H have built in the requirement that C†C = 1. The problem 
would be to find a matrix C that simultaneously diagonalizes H and satisfies 
C†SC=1. 
 a†ß d. . (1 0)
0
1

=0;  a†ad..(1 0)
1
0

=1 
Sˆza = 
1
2 
1 0 
0 -1
1
0

= 
1
2 
1
0

= 
1
2
a
674 Answers to Problems 
Chapter 10 
10-1. 
10-2. AO no. Atom AO type AO no. Atom AO type 
1 H1 1s 6 C3 2py 
2 H2 1s 7 O4 2s 
3 C3 2s 8 O4 2px 
4 C3 2px 9 O4 2px 
5 C3 2px 10 O4 2py 
10-3. E=-0.756 a.u.; MO 9 
f9=-0.27 1s1 -0.27 1s2 -0.49 2sC +0.22 2pxC +0.31 2s0 +0.33 2px0 
A s MO, mainly C–H2 bonding and lone pair (nonbonding) on oxygen. Shows 
some C–O antibonding character [see I). 
E=-0.611 a.u.; MO 8 
f8=-0.21 1s1 +0.21 1s2 -0.32 2pyC -0.76 2pyO 
A s MO, CH2 and C–O bonding [see (II)]. 
E=-0.597 a.u.; MO no. 7 
f7 =0.24 2pzC +0.92 2pzO
Appendix 13 675 
A p MO, mostly on oxygen, but somewhat delocalized to give some C–O 
bonding [see (III)]. 
10-4. The p MOs are 4 and 7. All others are s. 
10-5. C and O 2pp AOs are 4 and 8. The 4, 8 overlap population is seen from the data 
to be 0.1936. Since MO 7 is C–O bonding, loss of an electron should cause the 
C–O bond to lengthen. 
10-6. E7 = (0.2456)2(-10.67eV)+(0.9181)2(-15.85eV) 
+2(0.2456)(0.9181)(1.75)(0.2146)(-10.67eV-15.85eV)/2 
= -16.25eV=-0.5972 a.u. 
10-7. The sum of the elements in the upper triangle =12. (Use of all elements would 
count overlap populations twice.) 
10-8. If column 7 is the gross populations of MO no. 7, then it should turn out 
that 0.2175 = 2 c2
47 +(0.5)(2)c47c87S48: 2(0.2456)2 +(0.2456)(0.9181) 
(0.2146)]=0.2175. 
10-9. These must beAOs because MO charges must be 0, 1, or 2. AO 4, for example, 
gets its charge from MO 7. We have just seen (previous problem) that this is 
0.2175. (AO 4 also appears in MO 4, and the “charge matrix” gives a value of 
1.7825 for this. But this does not appear in the gross population because MO 
no. 4 is unoccupied in the ground state configuration.) 
10-10. The net charges are theAO charges plus the nuclear charges (after cancellation 
of some nuclear charge by inner-shell electrons). These results indicate high 
polarity, with oxygen being the negative end of the dipole. The predicted 
polarity is unrealistically high becauseEHMOneglects interelectronic repulsion 
which would tend to counteract such extreme charge imbalance. 
10-11. Number MOs=number AOs=1 on each H and 4 (valence) on each C=22. 
Chapter 11 
11-1. Koopmans 
SCF Experiment 
2s-. 2p 1.0800 1.0830 0.989 
1s-. 2p 31.9220 31.1921 31.19 
11-2. Koopmans–SCF (eV) SCF – observed (eV) 
(electron relaxation) (electron correlation) 
2B1 2.71 -1.54 
2A1 2.52 -1.40 
2B2 1.86 -0.90 
11-3. For electron affinities, these errors should reinforce, rather than cancel, because 
adding an electron should increase electron correlation. 
11-4. ad -cb+.(af -be)=ad +.af -bc-.be.
676 Answers to Problems 
11-5. Neither .1 nor .2 is already the best function in our function space. Hence, 
we cannot argue that mixing will bring no improvement. 
11-6. This must be true to enable a1 to be factored from the expanded form of Aˆ.. 
11-7. For a given choice of basis functions, there are two integrals: 
.a(1).b(2)|.c(1).d (2) and .a(1).b(2)|.d (1).c(2) 
There are five ways to choose a function for each position. Thus, the number 
of integrals is 2×54 =1, 250. 
11-8. a) and c) would be prevented from contributing. 
11-9. Hˆ =-
1
2 
10 

i=1 
.2
i - 
10 

i=1
( 
1 
ri,H1 + 
1 
ri,H2 + 
8 
ri,0 
)+ 
9 

i=1 
10 
 
j=i+1 
1 
rij 
11-10. a) (1/v2)|sg(1)sg(2)|. b) Eelec = -1.804 a.u. c) Etot = -1.090 a.u. 
d) De =0.090 a.u. e) IE (Koopmans)=0.619 a.u. f) KE+Vne=-1.185 a.u. 
Chapter 12 
12-1. E0 +W(1) 
0 =.|H0|.+.|H
|.=.|H|.=W0. 
12-2. [See Hint.] E(1) 
1 >E(1) 
3 >E(1) 
2 . 
12-3. a) d. b) d. c) -d/2. d) d/2. e) -d/2. f) 0. 
12-4. a) W(1) 
2 =3d/4. 
(b-1) Expect c(1) 
21 to cause .1 to shift to right. Since .2 is positive on left of 
box, negative on right, c(1) 
21 should be negative. Since .2 is above .1, it should 
cause energy to be depressed, and c(1) 
21 leads to a negative contribution to W(2) 
1 . 
(b-2) Because c(1) 
ij =-c(1) 
ji , c(1) 
12 must be positive, giving a f(1) 
2 that causes charge 
to shift left and a contribution to W(2) 
2 that is positive. 
12-5. c(1) 
41 = .1|H
|.4 
E1 -E4 = 
(2/L)  L 
0 sin(px/L)(Ux/L) sin(4px/L)dx 
-15p2/2L2 = 
64UL2 
153p4 
c(1) 
21 = 
32UL2 
27p4 , 
c(1) 
41 
c(1) 
21 
= 
2 
125 =1.6% 
12-6. The effect is zero, to first order, because .2 is symmetric for every state and 
the perturbation is antisymmetric. 
12-7. [See Hint.] E=E0+E1=-1
2 a.u.+1s|-1/r|1s=-3
2 a.u. Correction seeks 
to make . less diffuse, yet still spherical. c(1) 
2s,1s should be positive (augments 
hydrogen 1s at small r and cancels at large r) and c2P0 
, 1s(1) should be zero 
(wrong symmetry). 
12-8. W(0) +W(1) =-2 a.u.+2 a.u. = 0. Reasonable, since must be above -0.5 
a.u. (See Problem 12-1.)
Appendix 13 677 
12-9. a) Zero, because the perturbation is antisymmetric for x, y reflection, while 
the 1s function is symmetric. b) c(1) 
2s,1s vanishes. The relevant integral suffers 
the same symmetry disagreement as in part a). Or, in another view, the 2s AO 
will affect only the diffuseness of ., whereas the perturbation affects polarity. 
c(1) 
2pz,1s should be positive. The perturbation lowers the potential for positive z, 
the wavefunction should skew that way, so 2pz should enter with its positive 
lobe reinforcing 1s at positive z. c) It is negative, reflecting the lower energy of 
the electron distribution as a result of first-order polarization. 
12-10. Ep =4a +4.9624ß +(0.1)(q4); q4 =1.4881; Ep =4a +5.1112ß. 
12-11. The effect is least atC6 since q6 is smallest. First-order result: E=6a+7.777ß. 
Computed result: E =6a +7.8546ß. 
12-12. kk = 4*m
j=1*n
i=m+1 c2
kj c2
ki/(Ej - Ei). This must be negative. .qk must 
be negative if .ak is positive. This means that making atom k less attractive 
causes electron density to decrease there. This makes sense. 
12-13. Since the total charge must be conserved, the sum of all atom-atom polarizabilities, 
including p1,1 must be zero. Since the sum over the polarizabilities in 
Example 12-5 is 0.44277, p1,1=-0.44277. 
12-14. a) W(1) =hß (because all q =1). 
b) The lowest-energy one, which is the MO where the central atoms find the 
largest |c|. 
12-15. a) Ep =9a +12.1118ß. 
b) Atom 1 seems likely to have the greater self-atom polarizability because, in 
the cation, the product of squares of HOMO, LUMO coefficients vanishes 
at atom 2. 
12-16. [See Hint.] For butadiene: 
Ep =4ß[(0.3718)2 -(0.6015)2]=-0.894ß. 
For hexatriene: 
Ep =4ß[(0.2319)2-(0.4179)2+(0.5211)2]=0.603ß. The 
energy of hexatriene is lowered, that of butadiene is raised, and so hexatriene 
benefits. 
For cyclobutadiene: E0 +E(1) =4a +3.5778ß, E (H¨uckel) =4a +4.000ß. 
For benzene: E0 +E(1) =6a +7.591ß, E (H¨uckel) =6a +8.000ß. 
12-17. a) 
b) f(1) 
1 = .2|H
|.1 
E1 -E2 
.2 + .3|H
|.1 
E1 -E3 
.3 
Since H
 depends only on density at C2, it comes out of the integral. [See 
(IV).] 
f(1) 
1 = 0+(-1
2 cß/2v2ß).3=-0.1768c.3, f(1) 
2 =0 
f(1) 
3 =+0.1768c.1
678 Answers to Problems 
Thus, .1 +f(1) 
1 has more density at C2 than did .1, .2 +f(1) 
2 is identical 
to .2, and .3 +f(1) 
3 has lost density at C2. 
12-18. p1,2=-0.1768ß-1; p1,3=-0.265ß-1. Both atoms 2 and 3 lose charge, but 
atom 3 loses more than atom 2. 
12-19. a) Nondegenerate MO, f1 = (1/v3)(.1 + .2 + .3) is already correct. The 
correct zeroth-order degenerate MOs must give zero interaction element with 
H
. For this H
, this means that the overlap between these MOs must be zero 
at C2. Thus, one MO must have a node at C2 
f2 =(1/v2)(.1 -.3) and the 
other must be orthogonal to it 
f3 =(1/v6)(2.2 -.1 -.3). 
If one uses data from Appendix 6, one obtains, for degenerate MOs, .1 = 
-0.8165.1 + 0.4082.2 + 0.4082.3, and .2 = 0.7071.2 - 0.7071.3. These 
give H
 11 = 0.1666cß, H
 22 = 0.5cß, H
 12 = 0.2886cß. Since H
 12 = 0, one 
knows .1 and .2 are not correct zeroth-order wavefunctions. Solving the 
determinantal equation gives E1 = 0, E2 = 0.6666cß. These are the firstorder 
corrections to the energies of the two upper levels. Solving for coefficients 
gives, for E = 0, c1 =-0.866, c2 = 0.500, and so the proper zerothorder 
wavefunction having a zero first-order correction is f(0) 
1 =-0.866.1 + 0.500.2 =0.7071.1 -0.7071.3. For E =0.6666cß, c1 =0.500, c2 =0.866, 
and sof(0) 
2 =-0.4082.1+0.8165.2-0.4082.3. These are the same functions 
arrived at intuitively above. b) For the above zeroth-order MOs, the densities 
at C2 are respectively 13, 0, 2
3 . Therefore, the energy of the lowest level drops 
by cß/3, that for one of the originally degenerate levels drops by 2cß/3, and 
the other level is unaffected (to first order). 
12-20. a +2.1ß, a +0.818ß, a +0.618ß, a -1.418ß, a -1.618ß. 
12-21. Answers to part a) and b) follow answers for parts c) and d). 
c)E=a±2ß(0.5)2(2)=a±ß. These agree exactly with benzene orbital energies, 
which is reasonable since the unperturbed fragment nonbonding orbitals 
combine to give exactly the benzene MOs. 
d) a + v2ß + 2ß(0.353)2(2) = a + 1.913ß compared to a + 2ß for 
benzene.
Appendix 13 679 
The allyl coefficients are renormalized by dividing by v2. 
12-22. a) Only p3.p2 is dipole allowed. 
b) It is polarized along the outer C–C bonds. If we pretend the molecule is 
linear, this is the line connecting the carbons. If we adopt a more realistic 
structure, it depends on whether we choose a cis or trans structure:
680 Answers to Problems 
12-23. f(0) 
± |-z|f(0) 
± =±2s|r cos .|2pZ=±3 a.u. For f =cos(a)2s+sin(a)2pZ, 
maximum dipole occurs when cosa = 1/v2, sin a =±1/v2, that is, when 
f = f(0) 
± . This is reasonable since the mixing of degenerate states like these 
requires no energy “expense” in terms of the unperturbed hamiltonian. As soon 
as the slightest external field appears, the mixing will occur to the above extent 
to produce maximum dipoles. 
12-24. Since, for MO 4, c3 =c7, this MO is symmetric for reflection through the yz 
plane. The operator x is antisymmetric for this reflection. Therefore, if integral 
f4|x|f? is to be nonzero, f? must be antisymmetric for this reflection. Of 
the empty MOs shown, only f5 satisfies this requirement. There is no other 
symmetry operation present that will independently cause the integral to vanish, 
and f4.f5 is likely to be the observed transition. 
12-25. a) The HOMO is antisymmetric for reflection through the yz plane. (Assume 
that the coordinate origin is in the center of the 5–10 bond.) The x operator 
is also antisymmetric. Therefore, the allowed transition should be to an MO 
that is symmetric for this reflection. There are three such MOs, at energies of 
+1.000, +1.303, +2.303 in units of -ß. But the middle of these disagrees 
in symmetry with theHOMOfor reflection through the xz plane. Therefore, 
only the other two transitions are allowed. 
b) Here, the integrals containing y will vanish by symmetry except for transitions 
to +0.618 and +1.303, but the latter state disagrees in symmetry with 
the HOMO for the yz plane reflection. Hence, only -0.618.+0.618 is y 
allowed. 
c) No p–p transition is z allowed. 
d) Transitions to 1.303 and 1.618 are not allowed for any polarization. 
12-26. The field polarizes the atom, which means that the 2s state acquires 2p character. 
But the 2p.ls transition is allowed, and so the atom now relaxes to the 1s 
state. 
12-27. 4.7 allowed, polarized along C–O axis. 3.7 forbidden. 2.7 and 1.7 
allowed, polarized perpendicular to molecular plane. 
12-28. a) a, a.a, a +0.5ß. 
b) a, a.a+0.25ß, a+0.25ß. If we viewthis as a sequence of substitutions, 
then case a) is clearly a case of cooperative perturbations since the proper 
zeroth-order orbital that is strongly lowered by the first substitution is lowered 
again by the second. For case b), the second substitution affects the other 
one of the proper zeroth-order orbital energies. To this order, a) should be 
more stable since the lowering of energy due to the two electrons in this level 
is ß, whereas the most that can come from b) is ß/2. [Analysis of character 
tables (Chapter 13) enables us to see that the degeneracy for case b) must 
become split at higher levels of perturbation because aC2v molecule has only 
one-dimensional representations, hence cannot have degenerate orbitals.] 
12-29. Lowest energy rises by 0.0667ß. One of the next pair rises by 0.1ß and the 
other drops by 0.0333ß. 
12-30. x,y components =0. z component = 0.293 
Z a.u.= 0.745 
Z D= 2.485×10-30 
Z Cm.
Appendix 13 681 
MC: c a c b 
Chapter 13 
13-1. No. “Come about 180.” is needed for closure. 
13-2. It means that for each operation there is a right inverse as well as a left inverse. 
If AB =E, then A is the left inverse of B and B is the right inverse of A. 
13-3. a) C4. b) 4. c) Probably only three, since one tends to think of left face and right 
face as being in same class. However, in the C4 group they are not, because 
there is no operation in the group to interchange them (such as reflection through 
a plane containing the z axis). 
13-4. There are many possible arrangements counted in 4! that are not physically 
achievable through operations in the group. For example, 
1 — 2 
| | 
4 — 3 
.
 
1 — 4 
| | 
3 — 2 
For C3v, all possible arrangements are accessible. 
13-5. a) C2v. b) D2h. c) D4h. d) Td. 
13-6. a) U†U=UU† =1 (upon explicit multiplication). 
b) Upon explicit multiplication, U†AU=
1 0 
0 -1
. 
13-7. a) D6h (yes). b) C2v (no). c) D2h (no). d) D3d (yes). e) C3v (yes). 
13-8. a) D3h. b) (1) a

 2 ; (2)e

; (3)a
2
; (4)a
1
. 
13-9. [See Hint.] D2d, 
MO number : 1 2 3 4 5 6 
	
 
7 8 
	
 
9 10 
 	 
 
11 12 13 
 	 
 
14 15 16 
MO symmetry : b2 a1 a1 b2 e e e b2
e a1 b2 a1 
The highest occupied MO is 9,10; the lowest empty MO is 8,7, and so the 
transition is e.e. 
f9|x or y|f8 =  e.e.e= e.e.e.e=0 
f9|z|f8 =  e.b2 .e= a1 .a2 .b1 .b2 =0 
Transition is allowed for (group theory) z polarized light. This means x polarized 
for the coordinate system shown. 
13-10. a) and b) can be checked against C2v character table. 
c) Hydrogens generate characters 2 0 2 0, which is a1 .b1. 
d) The unnormalized symmetry combinations are: a1, 1sA + 1sB;b1, 1sA - 1sB.
682 Answers to Problems 
13-11. a) and b) can be checked against the C4v character table. 
c) Hydrogens generate characters 4 0 0 2 0 or 4 0 0 0 2, depending on howsv and 
sd are selected. Assuming that sv contains corner ammonia molecules gives the 
former set. This resolves to a1 .b1 .e. (The other choice gives b2 instead of 
b1, but reversal of choice of sv and sd has the effect of interchanging the symbols 
b1 and b2, and there is no real difference involved in this choice.) For a situation 
where the nitrogens are numbered as shown in the figure, the unnormalized 
symmetry orbitals are: a1, 2s1 + 2s2 + 2s3 + 2s4; b1, 2s1 - 2s2 + 2s3 - 2s4; 
e, 2s1 - 2s3 and 2s2 - 2s4; or 2s1 
+ 2s2 - 2s3 - 2s4 and 2s1 - 2s2 - 2s3 + 2s4. (Other combinations are also possible, but these two sets are the most 
convenient.) 
13-12. a) Four operations means order 4. Four classes means four representations. 
Hence, each representation must be one dimensional, therefore having character 
+1 or -1 for every operation. One of these must be +1 everywhere 
(A1). The others must all have +1 in the first column and -1 in two of the 
other three columns. The result is as given in Appendix 13 for C2v. 
b) Twelve operations gives order 12. Six classes means six representations. 
This must mean two 2×2 and four 1×1. There is but one unique set of 
orthonormal character vectors that fit this framework. Check against the 
D3h character table. 
13-13. Two two-dimensional representations and two one-dimensional representations. 
13-14. Check against C2v. 
13-15. Order =12. There are six classes. 
13-16. 
1
2 
.
.... 
1 1 -1 1 
1 -1 1 1 
1 -1 -1 -1 
1 1 1 -1
.
.... 
.
.... 
s 
px 
py 
pz
.
....
=sp3 set 
.
.... 
1/v3 0 v2/v3 0 
1/v3 1/v2 -1/v6 0 
1/v3 -1/v2 -1/v6 0 
0 0 0 1
.
.... 
.
.... 
s 
px 
py 
pz
.
....
=sp2 set 
For each matrix T,T†T=1. 
13-17. Yes. Since various hybridized sets are equivalent, the final result of the calculation 
is independent of one’s choice. 
13-18. a) A2 .2B1 .E. Yes. 
b) Ag .2B3u. No. 
c) A1g .A2g .Eg .A1u .A2u .Eu. No. 
13-19. a) e.b2 .e=a1 .a2 .b1 .b2. Need not vanish. 
b) e.e.e=e.e.e.e. Must vanish. 
c) a1 .b2 .b2 =a1. Need not vanish.
Appendix 13 683 
d) a1 .b2 .a1 =b2. Must vanish. 
e) a1 .e.b2 =e. Must vanish. 
13-20. eg.a2u, eu are x,y allowed. eg.eu is z allowed. 
13-21. The symmetry of the molecule is C2v. All p MOs are antisymmetric for C2 so 
must be bases for b1 or b2 representations. The function y (for the molecule) 
is z (in the table), and so it is a basis for the a1 representation. For the product 
f4yfn to contain a1 symmetry, it is necessary, then, that fn have the same 
symmetry as f4. f6 and f7 do have the same symmetry; therefore, f4.f6,f7 
are y allowed (where y is coincident with the symmetry axis). 
13-22. a) A nonspecial point above the top face has seven equivalent positions above 
that face (into which it can be sent by various symmetry operations, so there are 
eight equivalent positions above that face. Because the two sides of the square 
are identical, there are another eight equivalent positions below. Therefore, we 
expect a group of order 16. (See D4h.) b) Like the square, each face of the cube 
has eight equivalent points. The inside of a face is not equivalent to the outside, 
so each face is limited to eight equivalent points, not 16. Six faces times eight 
points per face yields a predicted group order of 48. (Oh) 
MC: d c a e d a d a b 
Chapter 14 
14-1. 1sA|Hˆhyd - 1/rB|1sA = -1
2 - (1/R) + [(1/R)+1] exp(-2R) where R is 
distance between nuclei. For R =2 a.u., Eelec=-0.9725 a.u., Etot=-0.4725 
a.u. For R =1 a.u., Eelec=-1.2293 a.u., Etot =-0.2293 a.u. For R =3 a.u., 
Eelec=-0.8300 a.u., Etot =-0.4967 a.u. 
14-2. Expected QMOT behavior is reversed for very low HAA. The “antibonding” 
MO is lower than the “bonding” for HAA =-20. This happens because the 
loss of energy involved in removing charge from the atoms is not compensated 
by that gained by putting the charge into the bond region. 
14-3. The molecular electron affinities tend to be greater than for the atoms when the 
extra molecular electron goes into a bonding MO, smaller if into an antibonding 
MO. QMOT leads us to expect the MO energy to be lowered (raised) from 
separated atom levels if the interaction is bonding (antibonding). Koopmans’ 
theorem then leads to predictions in qualitative agreement with these data. 
14-4. The observed ground-state angles are more consistent with the triplet-state configuration. 
The first excited state should then be a singlet and have a smaller 
angle. (Accurate calculations give 103..) 
14-5. It should become more bent. (Also, the O–H bonds should lengthen, although 
this is not the kind of geometric change that the figure explicitly treats.) 
14-6. The computed results will largely agree with QMOT ideas. However, it is 
possible that the 1sg - 1a1 level will rise where Walsh’s rules predict it will 
fall. [Whether this occurs depends upon details of EHMO parameter choices.] 
Upon analysis, this turns out to result from a situation similar to that examined
684 Answers to Problems 
in Problem 14-2. That is, the increase in H–H overlap population comes at the 
expense of population elsewhere. If the energy associated with this “population 
elsewhere” is low (i.e., if the original MO energy is low enough) the energy 
cost is greater than the gain due to increased H–H overlap population. This 
inversion of behavior for low levels is ignored in theWalsh approach. However, 
it does not occur for a given MO until it is fairly deeply buried under higher 
filled MOs. Also, since the higher MOs dominate the behavior of the molecule 
anyway, the inversion does not affect our prediction. 
14-7. Since the exact answer depends on details of your EHMO program, allow us to 
take this opportunity to toast your good health. 
14-8. If S = S0 cos ., dS/d. =-S0 sin .. 
S for . = 0 to . = 30. equals -(1 - 0.866)S0=-0.134S0; for 90.60.,
S=-S0(0-0.5)=0.5S0. 
14-9. For a diagram and discussion, see Gimarc [1]. Those with six valence electrons 
(the first three in the list) are planar. Those with eight electrons (the last four) 
are pyramidal. 
14-10. Sketches and discussions of AB2 molecules may be found in the literature [2,3]. 
Molecules with 16 or fewer valence electrons (BeCl2,C3,CO2,N-3 ) should be 
linear. Those with more than 16 should be bent (NO2,O3,F2O). 
14-11. Bending should increase end-to-end antibonding. These MOs favor the linear 
form [see (V)]. 
Bending should increase end-to-end bonding. These MOs favor the bent form 
[see (VI)]. 
Therefore, pg.pu should make molecule more bent. 
14-12. For allyl, draw the three p MOs. For the cyclic molecule, draw a C–C bond 
(lowest in energy), a single p-p AOon the negative carbon (intermediate energy) 
and the C–C antibond (high energy). Conrotary motion preserves a C2 axis, 
disrotatory preserves a reflection plane. For the C2 axis, the allyl MO symmetries 
are (in order of increasing energy) A, S, A. For the cyclopropenyl anion, 
they are S, A, A. For the reflection plane, the same MOs have symmetries S, 
A, S; and S, S, A. The resulting predictions for the four-electron anion are that 
thermal closure goes conrotatory, photochemical goes disrotatory. 
14-13. [See Hint.] The diagram is given in (VII). s and p refer to symmetry (S or A) 
for reflection through the molecular plane. The second symmetry symbol refers 
to reflections through a plane between the two acetylenes. The third symbol
Appendix 13 685 
refers to a reflection orthogonal to the first two (i.e., bisecting both acetylenes 
at their bond midpoints). The reaction does not appear favorable for a thermal 
mode. Photochemically it is less unfavorable, but still corresponds to a rather 
highly excited product. Asquare planar intermediate is unlikely for either mode. 
14-14. No states can differ in symmetry in this case, and no crossings should occur. 
14-15. SeeWoodward and Hoffmann [4, pp. 23, 24]. 
14-16. SeeWoodward and Hoffmann [4, pp. 70, 117]. 
Chapter 15 
15-1. (h/2p)k =px =v2mE =v2mT =h(1/.). 
15-2. [See Hint.] dk/dE = (dE/dk)-1 = 4p2m/kh2 . 1/k . 1/vE so plot E vs 
1/vE. 
15-3. [See Hint.] 
a) For example, f2=(1/v6) pp1 +exp(2pi/3)pp2 -exp(pi/3)pp3 +pp4+ exp(2pi/3)pp5 -exp(pi/3)pp6 .E2 =a -ß. 
b) j = 0 gives exp(0) = l at every atom, j = 6 gives exp(2(n - 1)pi) = 1 at 
every atom. 
15-4. [See Hint.] 
a) DOS, “normalized” by dividing byN, and in units of |ß|-1, is1/(p sin(hp)), 
where h=j/N and 0=h=1/2. E =a-2|ß|cos(hp). Therefore, DOS = 1/{p sin(arc cos [(E -a)/2|ß|])}. 
b) 2N states spread uniformly over an energy range of 4ß gives a “normalized” 
DOS of 0.5|ß|-1. 
15-5. For example, j =-1.(1/v6) pp1 +exp(-pi/3)pp2 +exp(-2pi/3)pp3- pp4 -exp(-pi/3)pp5 -exp(-2pi/3)pp6. Sum of j =±1 functions divided 
by v2 gives one real function, and difference divided by iv2 gives the other.
686 Answers to Problems 
15-6. Requires that exp(-2piqnN/n) equal 1. This requires that the argument equal 
pi times an even number, which is the case if qN is an integer. Since q and N 
are integers, so is qN. 
15-7. (CH)2n has lower energy. Variation would not affect (CH)2n, but would mix p 
and p* versions of (C2H2)n to produce COs like those for (CH)2n. (Note that, 
for convenience, we have arranged phases so that the node is identically placed 
in these diagrams.) 
15-8. The HOMO and LUMO at |k| = p/3a respond oppositely to distortion that 
lengthens every third bond and/or shortens the others. 
15-9. Symmetry for reflection through a plane perpendicular to the polymer axis is 
not useful for this analysis. Symmetry for reflection through a plane containing 
the atoms divides the COs into s and p types. Only s–p crossings are allowed. 
15-10. The s bands involve 2s and 2p AOs. One band pair includes 2s AOs with 2p 
AOs that are oriented parallel to the screw axis. The other band includes no 2s 
AOs and only 2p AOs perpendicular to the screw axis. The former functions 
form a basis that is symmetric for the symmetry operation, whereas the latter 
form a basis that is antisymmetric: 
15-11. At high pressures, intermolecular distance becomes the same as intramolecular 
distance. This yields a zero “gap” at the Fermi level (same as polyacetylene 
with uniform bond lengths). 
15-12.
Appendix 13 687 
15-13. a) HOMO =-8.19 eV. b) LUMO =-4.09 eV. c) Egap = 4.10 eV. d) Gap 
should open (both up and down) so HOMO becomes lower and IE is higher. 
LUMO is higher and EA is less negative. Computed values: IE=8.39eV,EA= 
-2.26eV,Egap =6.13 eV. 
15-14. In Chapter 8 we show correlations between HOMO and IE and between LUMO 
and EA, with different values for ß in each case. The difference in interelectronic 
repulsion and exchange for neutral and anionic species is handled implicitly in 
this manner by simple H¨uckel theory. 
15-15. Only p4 has the proper symmetry to combine with p1 to make c1=c4 =c2=c3 
=c5 =c6. Mixing causes the bands to split apart. p1 is pushed down by p4, 
and p4 is pushed up by p1. The coefficients decrease on carbons 1 and 4 in p4, 
consistent with a rise in energy due to less bonding. 
15-16. No. Atoms 2 and 6 have zero coefficients in the appropriate COs (see p(0) 
a of 
Fig. 15-26), but atom 5 has a nonzero coefficient. 
15-17. The band diagram is identical to the one in Fig. 15-23 except that the a + ß 
level at k=0 has an intended correlation with a-ß at k=p/a, and vice versa. 
However, these have the same symmetry (anti for s2), hence avoid crossing. 
The resulting diagram is identical to that in Fig. 15-23 except that band p2 has 
an intermediate hill and p5 an intermediate valley—height and depth unknown 
until a variational calculation is performed. 
15-18. Edge energies without substitution at k =0 : a +2.5ß, a,a -ß,a -1.5ß. At 
k=p/a :a+1.5ß,a+ß, a,a-2.5ß. (Proper zeroth-order MOs for theE=a 
level of cyclobutadiene are the versions having nodes through atoms rather than 
through bonds.) The lines at a,a - ß for k = 0 cross. This is allowed by 
symmetry disagreement for reflection through the plane containing the polymer 
axis and perpendicular to the molecular plane. Edge energies are lowered by 
substitution by the following amounts as we move up the diagram on the k =0 
side: ß/8,ß/4, 0,ß/8. On the k = p/a side: ß/8, 0,ß/4,ß/8. Symmetry 
is maintained so the crossing is still allowed. For only one substitution, the 
energy lowerings are half as great. Symmetry is lost, and the crossing becomes 
avoided. 
15-19. Like Fig. 15-28a with the middle horizontal row shaded. Higher in energy than 
, but lower than X or M. 
15-20.
688 Answers to Problems 
15-21. 
The energy level for X is lower than those for  and X
. For X, all interactions 
are bonding. For  and X
, nearest-neighbor interactions are antibonding and 
longer-range interactions of both bonding and anti bonding types occur. Judging 
the relative energies of  and X
 is complicated since we need to evaluate 
relative magnitudes of interactions between p AOs on the same long edge of 
the rectangle and also on opposite corners. Since p AO interactions depend on 
distance and angle, this is not trivial, as it is for s-type AOs. 
15-22. All are nondegenerate. 
15-23. The RFBZ is triangular. Referring to Fig. 15-27c, the RFBZ shown there 
becomes square, but symmetry nowmakes points on opposite sides of the-M 
line equivalent (e.g., X and X
 become equivalent), so only the -X-M- line 
is needed. 
15-24. 
Both are nonbonding and would have the same energy. 
15-25. 
COs at (0,p/a) and (p/a, 0) are the same except that the nodal planes are 
rotated by 60o.
Appendix 13 689 
15-26. The crystal is symmetrical for reflection through a hexagonal sheet. Planewaves 
experiencing the same potential and having the same |k| are degenerate. 
15-27. At the top of the RFBZ, unit cell functions are stacked with sign reversal. If the 
upper and lower phases of the unit cell functions agree, this means that we must 
have a bonding interaction in the cell, but we will also get antibonding between 
cells, hence no splitting. On the other hand, if the upper and lower phases are 
opposite, we must have antibonding in the cell but we now get bonding between 
them, again for no splitting. The same is true for sigma bonds. 
15-28. 
Appendix 2 
A2-2. a) -2. b) 1. c) 0. d) x=±v2. 
A2-4. Determinant of coefficients vanishes, and so nontrivial roots do exist. 
A2-5. c=2, -1±v7 are the roots. 
A2-6. a) If two rows or columns differ by factor c, then multiplication of the smaller 
row or column gives |M
| = c|M| (by rule 1), and M
 has two identical rows 
or columns. Interchanging these gives M

, and |M
|=-|M

|. But M

 =M
 
because the interchanged rows or columns are identical. Therefore, |M
| = 
-|M
|, and |M
|=0. Therefore, |M|=c|M
|=0. 
Appendix 4 
A4-1. From Eq. (A4-47), LzL+ =L+(Lz +1). Then LzL+Yl,m =L+(Lz +1)Yl,m = 
L+(km +1)Yl,m =(km +1)L+Yl,m. 
A4-2. L+L- =(Lx +iLy)(Lx -iLy)=L2x 
+L2y 
+iLyLx -iLxLy. Equation (A3- 
31) tells us that iLyLx -iLxLy =-i2Lz =Lz, so L+L- =L2x 
+L2y 
+Lz = 
L2x 
+L2y 
+L2z -L2z 
+Lz =L2 -L2z 
+Lz . Therefore, L2 =L+L- +L2z 
-Lz . 
A4-3. Equation (A4-55) indicates that the result of L+ on Yl,m remains an eigenfunction 
of L2 with the same eigenvalue, i.e., that L2L+Yl,m =L2C+Yl,m+1 = 
klC+Yl,m+1. But also, L+L2Yl,m = L+klYl,m = C+klYl,m+1, so Eq. (A4-55) 
indicates that L+ and L2 commute. This is easily verified since L+ is a linear 
combination of Lx and Ly, both of which commute with L2. We can establish 
Eq. (A4-55) by evaluating the result of the reverse order of operations and 
using the fact that L+ and L2 commute: L+L2Yl,m =L+klYl,m =klL+Yl,m = 
klC+Yl,m+1. But L+L2Yl,m=L2L+Yl,m, so L2L+Yl,m=klC+Yl,m+1. Q.E.D.
690 Answers to Problems 
A4-4. Lx =(1/2)(L+ +L-), (1/2)Yl,m|L+ +L-|Yl,m=(1/2)[C+Yl,m|Yl,m+1+ 
C-Yl,m|Yl,m-1=0+0. 
A4-5. a) (10)
0
1

=0. 
b) S+ß =(Sx +iSy)ß =
0+0 1
2 - i2 
2 
1
2 + i2 
2 0+0 
0
1

=
1
0

=a 
c) S+a =
0 1 
0 0
1
0

=
0
0
 
d) Sx,Sy= 
1
4 
0 1 
1 0
0 -i 
i 0

-
0 -i 
i 0
0 1 
1 0
+
= 
1
4 
2i 0 
0 -2i 

=iSz . 
A4-6. S2 =S2
x +S2
y +S2
z = 
1
4 
0 1 
1 0
0 1 
1 0

+
0 -i 
i 0 
0 -i 
i 0 

+
1 0 
0 -1 
1 0 
0 -1
+ 
= 
3
4 
1 0 
0 1

;S2a = 
3
4 
1 0 
0 1
1
0

= 
3
4 
1
0

= 
3
4
a. 
Appendix 8 
A8-1. Tˆ = -1
2d2/dx2, Vˆ = 1
2 kx2. Equation (A8-19) gives E¯. = .2T + .-2V . 
.E¯/..=0=2.T -2.-3V . For an exact solution, .=1, and T =V . 
A8-2. V = -3 a.u., T = 1
2 a.u., . = -V /2T = 3, .. = .3/p exp(-.r) = v27/p exp(-3r), E. =.2T +.V = 9
2 -9=-4.5 a.u. 
References 
[1] B. M. Gimarc, Accounts Chem. Res. 7, 384 (1974). 
[2] R. S. Mulliken, Rev. Mod. Phys. 4, 1 (1932). 
[3] A. D.Walsh, J. Chem. Soc. 2260 (1953). 
[4] R. B. Woodward and R. Hoffmann, The Conservation of Orbital Symmetry. 
Academic Press, NewYork, 1970.
Index 
A
A–A bonding, 502 
ab initio calculations 
basis sets for, 353–357 
description of, 348 
examples of, 370–384 
for molecules, 374–382 
Abelian groups, 430 
Absolute squares, 21, 461 
Acceptable functions, 22 
Acetylene, 244 
Acyclic polyenes, 281–282 
Additive constant in H
, 401 
All-trans polyacetylene, 551–552 
Allyl, 606 
Allyl radical, 607 
2-Allylmethyl, 606, 608 
AM1, 386t 
Ammonia, 47, 432f 
Amplitude function, 4 
Amplitude of wave, 2 
Angle-dependent functions, 97 
Angular momentum 
as pseudovector, 111 
description of, 51t 
electron, in atoms, 149–159 
electron spin, 599 
expressions for, 591–592 
gyroscope with, 112f 
magnetic moment and, 115–117 
magnitude of, 595 
in molecular rotation, 117–118 
nuclear spin, 599 
quantum-mechanical operators, 
592–593 
spherical harmonics and, 110–115 
spin-orbital 
for equivalent electrons, 156–159 
for many-electron atoms, 152–159 
for nonequivalent electrons, 
153–154 
for one-electron ions, 150–152 
Zeeman effect, 154–156 
total, 149–150 
total orbital, 152 
vectors, 143 
Angular momentum operators, 52, 
593–595, 599 
Angular momentum–angular position, 18 
Angular velocity, 51t 
Antibonding molecular orbitals, 217, 226, 258 
Antinodes, 4 
Approximate density function, 369 
Approximations 
s-p separability, 245 
Born–Oppenheimer, 207–208 
generalized gradient, 370 
independent electron, 127–129 
local density, 370 
orbital, 233–235 
Aromatic properties, 281 
Aromaticity, 281 
Associative law, 430 
Asymptotic behavior, 74–75, 78–79 
Atom(s) 
helium
description of, 134 
nonlinear variation for, 194–197 
1s2s configuration of, 138–144 
hydrogen 
nonlinear variation for, 191–194 
polarizability of, 197–206, 
410–412 
s-type states of, 418 
virial theorem of, 624–627 
Atom self-polarizability, 291 
Atom–atom polarizability, 407 
Atomic p-electron densities, 257 
Atomic ionization energies, 372 
Atomic orbitals 
in acetylene, 244 
basis, for Hückel determinant, 247–248 
decay of, 385 
definition of, 208 
description of, 128, 143, 145 
differential overlap between, 385 
frozen, 486 
gross population, 337 
Hartree–Fock equation and, 623 
linear combination of, 206–220 
in molecular orbitals, 465–467 
net population, 335 
691
692 Index 
Atomic orbitals (Continued) 
p-type, 219 
1s, 498, 587–590 
united-atom, 227f–228f 
Atomic spectral line splitting, 133 
Atomic units, 109t, 109–110, 632 
“Atom-in-molecule” energy, 488 
Aufbau process, 149 
Austin Model 1, 386t 
Average dipole moment, 171 
Average values, postulate for, 171 
Avogadro’s number, 631 
Azimuthal angle, 91 
Azulene, 290–291, 294–295, 606, 613 
B
Band diagrams, 528 
Band width, 541 
Band-crossing analyses, 559 
Basis atomic orbitals, 247–248 
Basis functions, representations generated 
from, 446–451 
Basis sets 
for ab initio calculations, 353–357 
description of, 231–233 
descriptors for, 356–357 
Gaussian, 353 
Slater-type-orbital, 353–354 
Benz-cyclopentadienyl, 606 
Benz-cyclopentadienyl radical, 612 
Benzene, 606, 610 
description of, 526, 530–533 
energies for, 531f 
Hückel molecular orbital method, 260, 
267–268, 530 
molecular orbitals of, 530–531, 
535, 556 
poly-paraphenylene, 555–561 
Benzyl, 606 
Benzyl radical, 611 
Binding energy, 234 
Bloch functions, 534 
Bloch sums 
description of, 536–537, 540f, 
542–543, 545 
without variational modification, 549 
Bloch’s theorem, 533–537 
Bohr magneton, 115, 155, 631 
Bohr radius, 93, 631 
Boltzmann’s constant, 546 
Bond integrals, 249 
Bond length, 269–270 
Bond order, 269–270, 605 
p-bond order, 259 
Bond-angle change, 484 
Bond–atom polarizability, 408 
Bond–bond polarizability, 408 
Bonding molecular orbital, 217 
Born–Oppenheimer approximation, 
207–208, 349 
Boson, 136 
Boundary conditions, 7 
Bra-ket notation, 629–630 
Brillouin’s theorem, 364–365 
Butadiene, 264–265, 279, 507, 608 
Butadiene-2,3-bimethyl, 606, 610 
C
Cartesian coordinates, 90, 324, 449 
C–C length, 555 
C–C separation, 544, 553 
CCSD, 379 
CCSD(T), 379–380 
Character(s) 
absolute squares of, 461 
conditions for, 460–461 
definition of, 458 
description of, 458–462 
reducible representations resolved 
using, 462–463 
Character tables, 459, 637–648 
Charge density index, 291 
Circular motion, 50 
cis-1,3-butadiene, 511f 
Class, 434–436 
Classical wave equation, 4–7 
Closed shells, 132, 226, 349–350 
Closed subshell, 349 
C–N length, 555 
CNDO/1, 386t 
CNDO/2, 386t 
CNDO/BW, 386t 
Coefficient waves, 532f 
Cofactor, 584 
Column vector, 309 
Commutators, 178–179 
Commuting operators, 175–176 
Complete neglect of differential overlap, 386t 
Complex conjugate of a matrix, 309, 311–312 
Compression waves, 5 
Concerted process, 509 
Configuration correlation diagram, 519f 
Configuration interaction 
calculation, 365 
description of, 360–365 
electron motion and, 378 
size consistency of, 366 
truncated, 366 
Conjugate variables, 18 
Conjugative model, 287 
Conrotatory closure, 508, 513 
Constant of motion 
angular momentum as, 111 
description of, 30, 43 
Constant potential, particle in a ring of, 50–52 
Constants, 631–633 
Continuous function, 532 
Contracted Gaussian function, 356 
Coordinate transformation, 312
Index 693 
Correlation diagrams 
configuration, 519f 
description of, 46 
orbital, 519 
state, 519f 
symmetry in, 514 
symmetry-forbidden, 517 
Correlation energy, 357–358 
Correspondence principle, 31 
Coulomb integral, 249, 413, 587–590 
Coulomb operator, 350–351, 620 
Coulomb terms, 369 
Coupled cluster theory, 366–367 
Covalent character, 363 
Covalent terms, 363 
Crystal orbital bond order, 541 
Crystal orbital overlap population, 541 
Crystal orbitals, 539, 559f 
Cyclic groups, 453–456 
Cyclic polyenes, 281–282 
Cycloaddition reaction, 517 
Cyclobutadiene, 265–267, 283–284, 608 
Cyclobutadienyl methyl, 606 
Cyclobutadienyl methyl radical, 608 
Cyclobutene, 511f, 513 
Cycloheptadienyl, 606 
Cycloheptadienyl radical, 611 
Cyclooctatetraene, 606, 612 
Cyclopentadienyl, 606 
Cyclopentadienyl radical, 6089 
Cyclopropenyl radical, 607 
Cyclopropenyl system, 253–256, 606 
D
D3d–D3h, 504f 
De Broglie, 14–15 
Degenerate state, perturbation theory for, 
409–410 
Degenerate-level perturbation theory, 
412–414 
Delocalization energy, 279 
Delocalized effect, 281 
Delocalized molecular orbitals, 472 
Density functional theory, 368–370 
Density of states, 528 
Determinant(s) 
cofactors, 584 
definition of, 584 
4×4, 584 
Hückel molecular orbital method 
a quantity, 248–249 
ß quantity, 249 
basis atomic orbitals, 247–248 
constructing of, 247–248 
generalizations, 259–263 
manipulation of, 249–250 
overlap integrals, 249 
topological, 250 
in linear homogeneous equations, 585 
secular, 200–201, 211 
Slater, 137–139, 349, 369, 622 
topological, 250 
2×2, 584 
Determinantal equation, 248 
Diagonal matrix, 311 
Diatomic molecules 
description of, 84–85 
homonuclear 
closed shells of, 226 
electronic states of, 229 
molecular orbitals of, 220–231 
properties of, 225t 
qualitative molecular orbital 
theory rules applied to, 
490–494 
symmetry orbitals of, 227f 
virial theorem for, 627–628 
Diels–Alder reaction, 517, 518f 
Differential equation for q(x), 75–76 
Diffraction experiment with electrons, 16–19 
Dihedral planes, 438–439 
Dimethylacetylene, 506, 507f 
Dipole moment, 171 
Dipole transition, 419 
Dirac delta function, 170, 178 
Dirac notation, 629–630 
Direct lattice, 564 
Direction of rotation, 451 
Disrotatory closure, 508 
Dissociation energy, 234 
Distance matrix, 343 
Distribution function, 70 
E
Eigenfunctions 
Bloch’s theorem, 533–534 
commuting operators have simultaneous 
eigenfunctions, 175–176 
description of, 20, 32 
gerade, 215 
for hydrogenlike ion in atomic units, 
110t 
lowest-energy, 190 
nondegenerate, 131, 218 
1s, 192 
simultaneous, 175–176 
symmetry of, 244 
ungerade, 215 
unperturbed, 395 
Eigenvalues 
description of, 20 
extended Hückel method, 328–331 
formaldehyde, 343 
of Hermitian operators 
completeness of, 176–177 
degenerate, 174 
expressed as an orthonormal set, 
174–175
694 Index 
Eigenvalues (Continued) 
nondegenerate, 173–174 
orthogonal set formed from, 
173–174 
proof of, 172–173 
Kohn–Sham, 370 
for L2, 595–600 
linear combination of atomic 
orbitals–molecular 
orbitals–self-consistent field, 
351–352 
for Lz , 595–600 
matrix of, 316 
negative, 93 
postulate relating measured values to, 
169–170 
potential function of, 92 
Schrödinger equation, 92–93 
Eigenvectors 
description of, 316 
extended Hückel method, 328–331 
Electrical conductivity, 546–547 
Electrocyclic reaction, 508 
Electromagnetic radiation, 9 
Electromagnetic wave 
description of, 9–10, 12–13 
square of, 21 
Electron(s) 
charge of, 631 
diffraction experiment with, 16–19 
equivalent, 156–159 
independent approximation of, 127–129 
nonequivalent, 153–154 
orbital motions of, 143 
potential energy of, 397 
resonance energy per, 281, 292f 
rest mass of, 631 
spin states for, 136 
uniform electrostatic perturbation of, 
396–403 
p-electron assumption, 246–247 
Electron densities, 602 
p-electron densities, 271–275, 290 
p-electron energy, 279–284 
Electron exchange symmetry, 129–132, 
159–160 
Electron flux, 546 
Electron index order, 618 
Electron motion, 378 
Electron orbital angular momentum, 599 
Electron probability density function, 93 
p-electron repulsion energy, 277 
Electron spin, 132–136 
Electron spin angular momentum, 599 
Electron spin resonance hyperfine splitting 
constants, 271–275 
Electrophilic aromatic substitution, 289 
Elliptical coordinates, 217 
Energies 
for benzene, 531f 
description of, 30–32 
extended Hückel 
experimental energies and, 
340–342 
Mulliken populations and, 
338–340 
ground state 
description of, 368, 371t 
to first-order of heliumlike 
systems, 403–405 
Energy level splitting, 47 
Energy to first order, 392, 397–399 
Energy to zeroth order, 392 
Equation of motion, 69–70 
Equivalent electrons, 156–159 
Equivalent orbitals, 472 
Equivalent representations, 445, 458 
Ethane, 440 
Ethylene, 263–264, 606–607 
Ethylene molecular orbitals, 268 
Exchange integrals, 142, 619 
Exchange operator, 351 
Exclusion principle, 136, 146 
Experimental energies, 340–342 
Explicit integration, 397–398 
Exponentials, 39 
Extended Hückel energies 
experimental energies and, 340–342 
Mulliken populations and, 338–340 
Extended Hückel method 
band calculations, 560 
basis set, 324–325 
description of, 324, 541 
eigenvalues, 328–331 
eigenvectors, 328–331 
hamiltonian matrix, 326–328 
K parameter, 333–335 
Mulliken populations, 335–340 
nuclear coordinates, 324 
overlap matrix, 325–326, 344 
for polyacetylene, 552–554 
total energy, 331–332 
External potential, 368 
F
Fermi contact interaction, 193 
Fermi energy, 540, 550, 577 
Fermion, 136 
Filled molecular orbitals, 472 
Finite central barrier, particle in an infinite 
“box” with, 44–47 
First Brillouin zone 
description of, 536, 564 
reduced, 540, 564, 566, 577 
First-order corrections 
to .1, 399–401 
description of, 392–394
Index 695 
First-order Stark effect, 411 
First-order structure, 485 
Fock operator, 350–351 
Force constant, 69 
Formaldehyde eigenvalues, 343 
Formaldehyde orbital numbering, 343 
Free particle in one dimension 
description of, 47–50, 526–529 
particle in a ring problem and, 
similarities between, 52 
Free radical reactions, 292 
Free valence, 292 
Frequency factor, 4 
Frontier orbitals, 291, 505–508 
Frozen atomic orbitals, 486 
Fulvene, 606, 610 
Functions 
acceptable, 22 
angle-dependent, 97 
Bloch, 534 
continuous, 532 
Dirac delta, 170 
distribution, 70 
electron probability density, 93 
Gaussian, 355–356 
Kronecker delta, 38 
Legendre, 108, 589 
linear combination of, 7 
linearly independent, 77 
orthogonal, 385 
orthonormal, 38 
polarization, 355 
single-valued, 21 
square-integrable, 22 
G
Gauss error function, 80 
Gaussian basis set, 353 
Gaussian functions, 183, 355–356 
Gaussian wave packets, 184 
Generalized gradient approximation, 370 
Gerade, 213, 215, 548 
Gimarc’s diagram, 502 
Givens–Householder–Wilkinson method, 320 
Graphite, 565–576 
Gross atomic orbital population, 337 
Ground state energy 
description of, 368, 371t 
to first-order of heliumlike systems, 
403–405 
Group
abelian, 430 
definition of, 430 
point, 441–443 
representations for 
from basis functions, 446–451 
cyclic groups, 453–456 
description of, 443–446 
irreducible inequivalent, 456–458 
labels, 636–637 
labels for, 451–452 
molecular orbitals and, 452–453 
one-dimensional, 444, 454, 473 
two-dimensional, 444, 454 
symbols for, 635–636 
symmetry point, 431–434 
Group theory 
elementary example of, 429–430 
overview of, 429 
H
H2, 488–490 
H
, 401 
H2+ molecule–ion 
antibonding states of, 487 
bonding states of, 487 
description of, 206–220, 485–488 
H2 vs., 488–490 
H3AAH3 system, 501 
HAB, 499, 500f 
HAH, 497, 499 
Hamiltonian, 20, 349, 404 
Hamiltonian matrix 
description of, 319, 326–328 
integrals of, 473 
partitioned, 469f 
Hamiltonian operators, 20, 53, 131–132, 
180, 593–595 
Harmonic electric-field wave, 9f 
Harmonic motion, 70 
Harmonic oscillation, 2 
Harmonic oscillator 
one-dimensional, 69–72 
quantum-mechanical, 72–74 
Schrödinger equation for 
asymptotic behavior, 74–75, 
78–79 
description of, 72–73 
differential equation for q(x), 
75–76 
energy spectrum, 79–80 
f 
as a power series, 76–77 
recursion relation for, 77–78 
simplifying of, 74 
wavefunctions, 80–81 
wavefunctions 
description of, 73–74, 80–81 
normalization of, 81–82 
orthogonality of, 81–82 
zero-point energy of, 74 
Harmonic wave, 2, 3f 
Hartree–Fock energy 
correlation energy, 357–358 
density functional theory and, 381–382 
estimating of, 371 
restricted, 357, 360 
unrestricted, 357
696 Index 
Hartree–Fock energy curve, 381 
Hartree–Fock equation 
atomic orbitals used with, 623 
definition of, 623 
derivation of, 614–623 
description of, 350 
Hartree–Fock limit, 357, 627 
Hartree–Fock wavefunctions, 375t, 627 
Heat of hydrogenation, 279 
Heisenberg uncertainty principle, 18–19, 31 
Helium atom 
description of, 134 
nonlinear variation for, 194–197 
1s2s configuration of, 138–144 
Heptatrienyl, 606 
Heptatrienyl radical, 611 
Hermite polynomials, 82 
Hermitian adjoint of a matrix, 310 
Hermitian matrix, 317 
Hermitian operators 
description of, 171–172 
eigenvalues of 
completeness of, 176–177 
degenerate, 174 
expressed as an orthonormal set, 
174–175 
nondegenerate, 173–174 
orthogonal set formed from, 
173–174 
proof of, 172–173 
Hermiticity, 174, 213 
Heteroatomic molecules, 284–287, 285t 
Hexatriene, 510–511, 606, 610 
Highest occupied molecular orbitals, 275, 
291, 493, 505, 507f, 519, 540 
Homogeneous magnetic field, 133 
Homonuclear diatomic molecules 
closed shells of, 226 
electronic states of, 229 
molecular orbitals of, 220–231 
properties of, 225t 
qualitative molecular orbital theory 
rules applied to, 490–494 
symmetry orbitals of, 227f 
Hooke’s law, 69 
Hückel molecular orbital method 
assumptions, 601 
benzene, 260, 267–268 
bond length, 269–270 
bond order, 269–270 
butadiene, 264–265, 279 
charge distributions from, 256–259 
cyclobutadiene, 265–267 
determinant 
a quantity, 248–249 
ß quantity, 249 
basis atomic orbitals, 247–248 
constructing of, 247–248 
generalizations, 259–263 
manipulation of, 249–250 
overlap integrals, 249 
topological, 250 
determinantal equation 
for cyclopropenyl system, 
253–256 
description of, 250 
solving for molecular orbitals, 
251–253 
solving for orbital energies, 
250–251 
ethylene, 263–264 
heteroatomic molecules, 284–287, 285t 
for hydrocarbons, 268–269 
independent p-electron assumption, 
246–247 
overview of, 244 
perturbation at an atom in, 406–409 
reaction indices, 289–295 
self-consistent variations of a and ß, 
287–289 
s-p separability, 244–246 
summary of, 295–296 
Hund’s rule, 148, 159–160 
Hybrid orbitals, 470–471 
Hydrocarbons 
Hückel molecular orbital method for, 
268–269 
Mobius conjugated, 283 
neutral alternant, 602 
Hydrogen atom 
nonlinear variation for, 191–194 
polarizability of, 197–206, 410–412 
s-type states of, 418 
Hydrogenlike orbitals, 146, 208 
Hyperfine splitting constant, 272 
I
Identity operation, 459 
Improper axis, 436 
Improper rotation, 436 
Independent electron approximation, 
127–129 
Independent p-electron assumption, 246–247 
INDO, 386t 
Induced dipole, 411–412 
Inductive model, 286 
Inhomogeneous magnetic field, 133 
Inner repulsion energy, 588 
Instantaneous dipole moment, 171 
Integrals 
bond, 249 
Coulomb, 249, 413, 587–590 
exchange, 142 
list of, 582–583 
overlap, 249 
resonance, 249 
Integrand, 472 
Intended correlations, 545
Index 697 
Interaction element, 204 
Intermediate neglect of differential overlap, 
386t 
Internal rotation, 485 
Internuclear repulsion energy, 209 
Inverse matrix, 311–312 
Inverse operation, 430 
Ionization energy 
description of, 278–279 
of neon, 372t 
valence state, 327 
Irreducible inequivalent representations, 445, 
456–458 
J
Jacobi method, 319 
Jahn–Teller theorem, 258 
K
K parameter, 333–335 
Kinetic energy 
description of, 43 
equation for, 11 
of photoelectrons, 11, 12f, 278 
Kinetic energy operators, 140 
Kohn–Sham orbitals, 369–370 
Koopmans’ theorem, 358–360, 372, 472 
Kronecker delta, 82, 170 
Kronecker delta function, 38 
L
L2, 595–600 
Lagrangian multipliers, 620 
LCAO–MO–SCF equation. See Linear 
combination of atomic 
orbitals–molecular 
orbitals–self-consistent field 
equation 
Legendre functions, 108, 589 
Legendre polynomials, 107 
Light
as electromagnetic wave, 9–10 
electromagnetic field theory of, 12 
Light intensity, 17f 
Linear combination, 7 
Linear combination of atomic 
orbitals–molecular orbital 
calculation, 208 
Linear combination of atomic 
orbitals–molecular 
orbitals–self-consistent field 
equation 
density functional theory methods, 
368–370 
eigenvalues, 351–352 
ground state wavefunction, 361 
multideterminant methods, 367 
overview of, 350–351 
Linear harmonics, 110 
Linear homogeneous equations, 585 
Linear momentum, 18 
Linear polyenes, 262 
Linear position, 18 
Linear variation method 
matrix
formulation, 315–317 
hamiltonian, 319 
overview of, 308 
Linearly dependent functions, 7, 76 
Linearly independent functions, 77 
Lithium, 135 
Local density approximation, 370 
Localization energy, 294 
Localized orbitals, 472 
Lowest unfilled molecular orbitals, 275, 291, 
505, 509 
Lowest-energy eigenfunction, 190 
Lowest-energy molecular orbitals, 222, 252, 
262, 327, 520 
Lowest-energy wavefunction, 93–97 
L–S coupling, 152 
Lz , 595–600 
M
Magnetic moment 
angular momentum and, 115–117 
definition of, 115 
Magnitude, 592 
Mass density, 95 
Matrix
addition of, 310–311 
complex conjugate of, 309, 311–312 
definition of, 308 
diagonal, 311 
of eigenvalues, 316 
of eigenvectors, 316 
expression of, 308–309 
geometric model, 312–314 
hamiltonian, 319, 326–328 
hermitian, 317 
hermitian adjoint of, 310 
inverse, 311–312 
multiplication of, 310–311 
nonsingular, 314 
orthogonal, 314 
product of, 311–312 
rotation, 313 
similarity transformation, 314 
singular, 314 
square, 311 
symmetric, 310 
transformation, 314 
transpose of, 309–312 
unit, 311 
Matrix equation, 317–320 
Matter, wave nature of, 14–16 
Maxwell’s differential equations, 10 
McConnell relation, 272
698 Index 
Metastable states, 501 
Methane, 339 
2-Methoxyethanol, 277 
Methylene cyclopropene, 606, 608 
Microscopic systems, 18 
MINDO/3, 386t 
Mirror A universe, 434–435 
Møller–Plesset perturbation theory, 366 
MNDO, 386t 
Mobile bond order, 259 
Mobius conjugated hydrocarbons, 283 
Modified intermediate neglect of differential 
overlap, 386t 
Modified neglect of diatomic overlap, 386t 
Molecular orbitals 
p, 416, 418 
antibonding, 217, 226, 258 
atomic orbitals in, 465–467 
benzene, 530–531, 535, 556 
bonding, 217 
for cyclopropenyl system, 255 
definition of, 208 
delocalized, 472 
ethylene, 268 
filled, 472 
frontier, 505–508 
hierarchy in, 484–485 
higher-energy, 490, 498 
highest occupied, 275, 291, 493, 505, 
507f, 519, 540 
of homonuclear diatomic molecules, 
220–231 
Hückel molecular orbital determinantal 
equation solved for, 251–253 
for linear polyenes, 262 
lowest unfilled, 275, 291, 505, 509 
lowest-energy, 222, 252, 262, 327, 520 
nonbonding, 252, 602–604 
nondegenerate, 329, 467 
occupied, 362 
representation table and, 452–453 
self-consistent field–molecular orbitals, 
384–386 
subjacent, 506 
superjacent, 506 
symmetry of, 296, 463–465 
virtual, 362 
Molecular rotation, angular momentum in, 
117–118 
Molecules 
ab initio calculations for, 374–382 
homonuclear diatomic. See 
Homonuclear diatomic 
molecules 
point group of, 441–443 
Mulliken populations 
description of, 335 
extended Hückel energies and, 338–340 
net atomic orbital, 335 
overlap, 335–336, 345, 541 
Multiconfigurational self-consistent field 
calculation, 367 
N
Naphthalene, 606, 613 
NDDO, 386t 
Nearest-neighbor interaction, 484–485, 496 
Negative eigenvalues, 93 
Neglect of diatomic differential overlap, 386t 
Neutral alternant hydrocarbons, 602 
Newton’s laws of motion, 4–5 
Nodes, 3 
Nonbonding molecular orbitals, 252, 
602–604 
Nonconcerted process, 509 
Noncrossing rule, 230 
Nondegenerate eigenfunctions, 131, 218 
Nondegenerate state, 6, 31 
Nondegenerate wavefunctions, 448 
Nonequivalent electrons, 153–154 
Nonsingular matrix, 314 
Norbornadiene, 416–417 
Nuclear motion, potential energy for, 207 
Nuclear spin angular momentum, 599 
Nucleophilic aromatic substitution, 289 
O
Occupied molecular orbitals, 362 
Occupied orbitals, 353 
Octatetraene, 606, 612 
Off-diagonal determinantal element, 204 
One-dimensional harmonic oscillator, 69–72 
One-dimensional representations, 444, 
454, 473 
One-electron density function, 256 
One-electron energy, 128, 352 
Open-shell systems, 373 
Operators 
angular momentum, 52, 593–595, 599 
commuting, 175–176 
constructing of, 167–168 
coulomb, 350–351, 620 
definition of, 20 
exchange, 351 
Fock, 350–351 
hamiltonian, 20, 53, 180, 593–595 
permutation, 614 
postulate for constructing, 167–168 
quantum-mechanical, 592–593 
Orbital(s) 
atomic. See Atomic orbitals 
crystal, 539, 559f 
description of, 101–102 
equivalent, 472 
frontier, 291, 505–508 
hybrid, 470–471 
hydrogenlike, 146, 208
Index 699 
interaction between, 414–417 
Kohn–Sham, 369–370 
localized, 472 
molecular. See Molecular orbitals 
occupied, 353 
separated-atom, 228f 
Slater-type 
basis sets, 353–354 
contracted Gaussian functions for, 
356 
description of, 146–147, 324 
minimal valence, 471 
p-type, 354 
symmetry, 220–222, 227f 
Orbital approximation, 233–235 
Orbital energies 
definition of, 352 
description of, 128 
internuclear separation for He2, 489f 
ionization energies and, 278–279 
oxidation-reduction potentials and, 
275–277 
solving of Hückel molecular orbital 
determinantal equation for, 
250–251 
Orbital orthogonality, 617 
Orbital symmetries, 515 
Oriented dipole, 133 
Orthogonal functions, 385 
Orthogonal matrix, 314 
Orthogonal transformation, 314 
Orthogonality 
in irreducible inequivalent 
representations, 456–458 
of vectors, 457 
of wavefunctions, 37–38, 81–82 
Orthonormal functions, 38 
Orthonormal set, 174–175 
Oscillating electric charge, 10f 
Oscillating particle distributions, 527 
Oscillation, harmonic, 2 
Outer repulsion energy, 588 
Overlap integrals, 249 
Overlap matrix, 325–326, 344 
Overlap population, 335–336, 345, 541 
Oxidation-reduction potentials, 275–277 
Pp
orbitals, 149 
Pairing of roots, 601–602 
Pairing theorem, 262 
Particle(s) 
in an infinite “box” with a finite central 
barrier, 44–47 
in a one-dimensional box 
definition of, 27 
energies, 30–32 
with one finite wall, 38–43 
overview of, 27–30 
wavefunctions, 32–38 
in a ring 
of constant potential, 50–52 
description of, 529–530 
scattering of, in one dimension, 56–59 
in a square well, 27 
in a three-dimensional box, 53–56 
Partitioned hamiltonian matrix, 469f 
Pauli principle, 137–138, 148, 156, 178, 268 
Peierls distortion, 550, 570, 577 
Pentadienyl, 606 
Pentadienyl radical, 609 
Periodic potential, 526 
Periodic structures, 528 
Periodicity 
in three dimensions, 565–576 
two-dimensional, 562–565 
Permanent dipole, 411 
Permutation operator, 614 
Perturbation 
definition of, 391–392 
uniform electrostatic, 396–403 
Perturbation theory 
degenerate-level, 412–414 
Møller–Plesset, 366 
overview of, 391–392 
Rayleigh–Schrödinger 
.2 effects on, 402–403 
additive constant in H
, 401 
at an atom in the simple Hückel 
molecular orbital 
method, 406–409 
for degenerate state, 409–410 
first-order corrections, 392–394, 
399–401 
formal development of, 391–396 
ground-state energy to first-order 
of heliumlike systems, 
403–405 
overview of, 391 
W1(2), 401–402 
spectroscopic selection rules and, 
417–420 
Phase factor, 3 
 equation, 106 
Photoelectric effect, 10–14 
Photoelectrons, 11, 12f, 278 
Photons 
characteristics of, 14 
definition of, 12 
Einstein’s relation for, 14 
rest mass of, 14 
p bands, 554–555 
Piecewise continuous, 28 
Planck, 10 
Planck’s constant, 18, 631 
Point group of a molecule, 441–443 
Point of inversion, 436 
Polarizability of the hydrogen atom, 
197–206, 410–412
700 Index 
Polarization functions, 355 
Polyacetylene 
all-trans, 551–552 
with alternating bond lengths, 547–551 
energy calculations, 562 
extended Hückel molecular orbital 
method results for, 552–554 
self-consistent field results for, 552–554 
with uniform bond lengths, 537–546 
Polyatomic molecules, 495–505 
Poly-paraphenylene, 555–561 
Population density, 95 
Postulates 
for average values, 171 
for constructing operators, 167–168 
measured values related to eigenvalues, 
169–170 
Schrödinger equation, 168–169 
wavefunction, 166–167 
Potential energy 
determination of, 383 
of electrons, 397 
for nuclear motion, 207 
quantum-mechanical average value of, 
83–84 
Probability density 
description of, 13 
volume-weighted, 95f 
Probability waves, 13 
. 
as stationary state, 180 
conditions on, 21–22 
triple value of, 22f 
.2, 402–403 
p-type atomic orbital, 219 
p-type Slater-type orbitals, 354 
2px orbital, 101–102 
2py orbital, 101–102 
Pyrrole molecule, 286 
2pz orbital, 101–102 
Q
Qualitative molecular orbital theory 
H2+, 485–488 
molecular orbitals, 484–485 
molecular structure, 484–485 
need for, 484 
of reactions, 508–521 
rules for 
description of, 490 
homonuclear diatomic molecule 
applications, 490–494 
Quantum numbers, 30, 97–98 
Quantum-mechanical average, 96–97 
Quantum-mechanical average value of 
potential energy, 83–84 
Quantum-mechanical harmonic oscillator, 
72–74 
Quantum-mechanical operators, 592–593 
Quantum-mechanical tunneling, 41 
q(x), 75–76 
R
R equation, 108–109 
Radial node, 98 
Radical addition, 289 
Rayleigh–Ritz variation principle 
description of, 178 
proof of, 190 
Rayleigh–Schrödinger perturbation theory 
.2 effects on, 402–403 
additive constant in H
, 401 
at an atom in the simple Hückel 
molecular orbital method, 
406–409 
for degenerate state, 409–410 
first-order corrections, 392–394, 
399–401 
formal development of, 391–396 
ground-state energy to first-order of 
heliumlike systems, 403–405 
overview of, 391 
W1(2), 401–402 
Reaction indices, 289–295 
Reactions 
electrocyclic, 508 
qualitative molecular orbital theory of, 
508–521 
stereospecific, 509 
Reciprocal lattice, 564 
Reciprocal operation, 430 
Reciprocal space, 562–565 
Recursion relation, 77–78 
Reduced first Brillouin zone, 540, 564, 
566, 577 
Reducible representations, 445, 462–463 
Redundancies, 432 
Reflection plane, 436 
Reflection symmetry, 205 
Representations 
from basis functions, 446–451 
cyclic groups, 453–456 
description of, 443–446 
irreducible inequivalent, 456–458 
labels, 636–637 
labels for, 451–452 
molecular orbitals and, 452–453 
one-dimensional, 444, 454, 473 
two-dimensional, 444, 454 
Repulsive energy, 588 
Resonance energy, 281 
Resonance energy per electron, 281, 282f 
Resonance integrals, 249 
Rest mass 
description of, 14 
of electron, 631 
of neutron, 631 
of proton, 631
Index 701 
Restricted Hartree–Fock energy, 357, 360 
Rigid-rotor model, 117–118 
Ring of constant potential, particle in, 50–52 
Rotation axis, 436 
Rotation matrix, 313 
Row vector, 309 
Russell–Saunders coupling, 152 
Rydberg series, 232 
Rydberg state, 232 
S
1s atomic orbitals, 498, 587–590 
2s orbital, 101 
Scalar, 309–310, 312 
Scanning tunneling microscopy, 575 
Schmidt orthogonalization, 175, 468 
Schrödinger equation 
in atomic units, 110 
center-of-mass coordinates, 90 
for circular motion, 50 
description of, 19–20, 22–23, 89–91 
eigenvalues, 92–93 
in free particle in one dimension, 47–48 
harmonic oscillator 
asymptotic behavior, 74–75, 
78–79 
description of, 72–73 
differential equation for q(x), 
75–76 
energy spectrum, 79–80 
f 
as a power series, 76–77 
recursion relation for, 77–78 
simplifying of, 74 
wavefunctions, 80–81 
higher-energy solutions for, 98–105 
lowest-energy wavefunction, 93–97 
for one-dimensional harmonic 
oscillator, 72–73 
for particle in the one-dimensional 
square wall, 54 
for periodic structures, 528 
postulate for, 168–169 
quantum numbers, 97–98 
separation of variables, 105–106 
time-dependent, 168–169, 180 
Second-order Stark effect, 411 
Second-order structure, 485 
Secular determinant, 200–201, 211 
Secular equation, 200 
Self-consistent field 
acronyms, 386 
atomic orbitals, 147 
configuration interaction, 360–365 
definition of, 348 
description of, 146–147 
equations, 622 
hamiltonian, 349 
Hartree–Fock limit, 357 
Koopmans’ theorem, 358–360 
multiconfigurational, 367 
polyacetylene results, 552–554 
total electronic energy, 352–353 
wavefunction for, 349–350 
Self-consistent field–molecular orbitals, 
384–386 
Separated-atom basis, 208 
Separated-atom energy, 487–488 
Separated-atom limits, 208 
Separated-atom orbitals, 228f 
s bond network, 295 
s-p separability, 244–246 
Sigmatropic shift, 519 
Similarity transformations, 314, 459 
Singlet, 142 
Single-valued function, 21 
Singular matrix, 314 
Slater determinants, 137–139, 349, 369, 622 
Slater-type orbitals 
basis sets, 353–354 
contracted Gaussian functions for, 356 
description of, 146–147, 324 
minimal valence, 471 
p-type, 354 
Spectroscopic selection rules, 417–420 
Spectroscopists, 14 
Spherical harmonics 
angular momentum and, 110–115 
definition of, 111 
Spherical polar coordinates, 90, 217, 454 
Spherically symmetric potential, 91, 116 
Spin forbidden, 418 
Spin states, 136 
Spin-free density function, 166 
“Split shell” wavefunction, 195 
Splitting, 154 
Splitting constants, 271–275 
Square
absolute, 21, 461 
of electromagnetic wave, 21 
rotation effects on, 431f 
Square matrix, 311 
Square symmetry, 266 
Square-integrable function, 22 
1s2s configuration of helium, 138–144 
Standing waves 
in clamped string, 7–9 
description of, 3–4 
Stark effect, 411 
State correlation diagram, 519f 
State energy, 501 
Stater-type orbitals, 208 
Stereospecific reactions, 509 
Stilbene, 283 
Strain energy, 284 
Subjacent molecular orbitals, 506 
Sum of squares of dimensions, 457–458 
Superjacent molecular orbitals, 506
702 Index 
Symmetric matrix, 310 
Symmetry 
in correlation diagrams, 514 
integration and, 472–476 
reflection, 205 
Symmetry elements, 436–441 
Symmetry operations, 431, 439 
Symmetry orbitals 
definition of, 468 
description of, 220–222, 515 
generating of, 467–470 
for homonuclear diatomic molecules, 
227f 
unnormalized, 468 
Symmetry point groups, 431–434 
Symmetry-forbidden reaction, 513, 517 
T
Term symbols, 152 
 equation, 107–108 
Third-order structure, 485 
Three-dimensional Bernal graphite, 571f, 572 
Three-dimensional box, particle in, 53–56 
Time-dependent differential equation, 6 
Time-dependent Schrödinger equation, 
168–169 
Time-dependent states, 180–185 
Time-independent wave equation 
description of, 6, 8f 
Schrödinger. See Schrödinger equation 
Topological determinant, 250 
Torsional angle change, 485 
Total electronic energy, self-consistent field, 
352–353 
Total extended Hückel energy, 331–332 
Transformation matrix, 314 
Transposing of a matrix, 309–312 
Transverse waves, 5 
Traveling waves, 1–3 
Trimethylenemethane, 292 
Triplet, 142 
Tunneling, quantum-mechanical, 41 
Two-dimensional graphite, 569, 570f 
Two-dimensional periodicity, 562–565 
Two-dimensional representations, 444, 454 
U
Uncertainty principle of Heisenberg, 18–19, 
31, 114, 179 
Ungerade, 213, 215, 548 
Uniform electrostatic perturbation, 396–403 
Unit matrix, 311 
Unitary transformation, 314, 445–446 
United-atom atomic orbitals, 227f-228f 
United-atom limits, 208 
Unperturbed eigenfunctions, 395 
Unperturbed energy, 392 
Unrestricted Hartree–Fock energy, 357 
V
Vacuum permittivity, 631 
Valence state ionization energy, 327 
Variables 
conjugate, 18 
separation of, 105–106 
Variation method 
linear, 197–206 
nonlinear 
calculations, 196–197 
for helium atom, 194–197 
for hydrogen atom, 191–194 
orbital approximation, 233–235 
spirit of, 190 
Variation principle, 178 
Variational wavefunction, 231–233 
Vectors 
addition of, 310–311 
column, 309 
matrix multiplication of, 311 
multiplication of, 310–311 
orthogonality of, 457 
in reciprocal space, 562–565 
representation, 457 
two-dimensional, 312 
Vertical planes, 438–439 
Vibrations of diatomic molecules, 84–85 
Virial relation, 373 
Virial theorem 
for atoms, 624–627 
for diatomic molecules, 627–628 
Virtual molecular orbitals, 362 
Volume-weighted probability density, 95f 
W
W1(2), 401–402 
Walden inversion, 520 
Walsh diagrams, 495–505, 497f, 503f-504f 
Wave(s) 
amplitude of, 2 
classical equation for, 4–7 
compression, 5 
electromagnetic, 12–13 
frequency of, 3 
harmonic, 2, 3f 
light as, 10 
probability, 13 
standing, 3–4, 7–9 
transverse, 5 
traveling, 1–3 
wavelength of, 2 
Wave profile, 1 
Wavefunctions 
antisymmetric, 136, 213 
description of, 32–35 
determinantal, 349–350 
Dirac delta function as, 178 
gerade, 213, 215 
hamiltonian, 144
Index 703 
of harmonic oscillator, 73–74 
Hartree–Fock, 358, 375t, 627 
lowest-energy, 93–97 
nondegenerate, 448 
normalization of, 81–82 
orthogonality of, 37–38, 81–82 
postulate, 166–167 
Schrödinger equation for harmonic 
oscillator, 80–81 
self-consistent field, 349–350 
Slater determinantal, 137, 139 
“split shell,” 195 
symmetry of, 35–37, 47, 136, 213 
ungerade, 213, 215 
variational, 231–233 
zeroth-order, 393 
Wavelength 
de Broglie, 15 
description of, 2 
Wavenumber, 14, 528 
Wigner–Seitz unit cell, 564 
Wolfsberg–Helmholtz relation, 327 
Woodward–Hoffman rule, 510–511 
Work function, 12 
Z
Zeeman effect, 154–156 
Zeeman splitting, 116 
Zero average momentum, 184–185 
Zero differential overlap, 385 
Zero-point energy, 31 
Zeroth-order wavefunction, 393