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of 352
Chemistry 351 and 352
Physical Chemistry I and II
Darin J. Ulness
Fall 2006 2007
Contents
I Basic Quantum Mechanics 15
1 Quantum Theory 161.1 The Fall of Classical Physics . . . . . . . . . . . . . . . . . . . . 16
1.2 Bohrs Atomic Theory . . . . . . . . . . . . . . . . . . . . . . . . 17
1.2.1 First Attempts at the Structure of the Atom . . . . . . . . 17
2 The Postulates of Quantum Mechanics 222.1 Postulate I . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.2 How to normalize a wavefunction . . . . . . . . . . . . . . . . . . 23
2.3 Postulates II and II . . . . . . . . . . . . . . . . . . . . . . . . . . 24
3 The Setup of a Quantum Mechanical Problem 273.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3.2 The Quantum Mechanical Problem . . . . . . . . . . . . . . . . . 27
3.3 The Average Value Theorem . . . . . . . . . . . . . . . . . . . . . 29
3.4 The Heisenberg Uncertainty Principle . . . . . . . . . . . . . . . . 30
4 Particle in a Box 314.1 The 1D Particle in a Box Problem . . . . . . . . . . . . . . . . . . 31
4.2 Implications of the Particle in a Box problem . . . . . . . . . . . 34
5 The Harmonic Oscillator 385.1 Interesting Aspects of the Quantum Harmonic Oscillator . . . . . 40
i
5.2 Spectroscopy (An Introduction) . . . . . . . . . . . . . . . . . . . 42
II Quantum Mechanics of Atoms and Molecules 45
6 Hydrogenic Systems 466.1 Hydrogenic systems . . . . . . . . . . . . . . . . . . . . . . . . . . 46
6.2 Discussion of the Wavefunctions . . . . . . . . . . . . . . . . . . . 49
6.3 Spin of the electron . . . . . . . . . . . . . . . . . . . . . . . . . . 51
6.4 Summary: the Complete Hydrogenic Wavefunction . . . . . . . . 52
7 Multi-electron atoms 557.1 Two Electron Atoms: Helium . . . . . . . . . . . . . . . . . . . . 55
7.2 The Pauli Exclusion Principle . . . . . . . . . . . . . . . . . . . . 56
7.3 Many Electron Atoms . . . . . . . . . . . . . . . . . . . . . . . . 58
7.3.1 The Total Hamiltonian . . . . . . . . . . . . . . . . . . . . 59
8 Diatomic Molecules and the Born Oppenheimer Approximation 608.1 Molecular Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
8.1.1 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . 61
8.1.2 The BornOppenheimer Approximation . . . . . . . . . . 62
8.2 Molecular Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 63
8.2.1 The Morse Oscillator . . . . . . . . . . . . . . . . . . . . . 64
8.2.2 Vibrational Spectroscopy . . . . . . . . . . . . . . . . . . . 66
9 Molecular Orbital Theory and Symmetry 679.1 Molecular Orbital Theory . . . . . . . . . . . . . . . . . . . . . . 67
9.2 Symmetry . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
10 Molecular Orbital Diagrams 7210.1 LCAOLinear Combinations of Atomic Orbitals . . . . . . . . . 72
10.1.1 Classification of Molecular Orbitals . . . . . . . . . . . . . 73
10.2 The Hydrogen Molecule . . . . . . . . . . . . . . . . . . . . . . . 74
10.3 Molecular Orbital Diagrams . . . . . . . . . . . . . . . . . . . . . 76
10.4 The Complete Molecular Hamiltonian and Wavefunction . . . . . 78
11 An Aside: Light ScatteringWhy the Sky is Blue 7911.1 The Classical Electrodynamics Treatment of Light Scattering . . . 79
11.2 The Blue Sky . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
11.2.1 Sunsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
11.2.2 White Clouds . . . . . . . . . . . . . . . . . . . . . . . . . 83
III Statistical Mechanics and The Laws of Thermody-namics 88
12 Rudiments of Statistical Mechanics 8912.1 Statistics and Entropy . . . . . . . . . . . . . . . . . . . . . . . . 89
12.1.1 Combinations and Permutations . . . . . . . . . . . . . . . 90
12.2 Fluctuations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
13 The Boltzmann Distribution 9413.1 Partition Functions . . . . . . . . . . . . . . . . . . . . . . . . . . 96
13.1.1 Relation between the Q and W . . . . . . . . . . . . . . . 9713.2 The Molecular Partition Function . . . . . . . . . . . . . . . . . . 99
14 Statistical Thermodynamics 103
15 Work 10715.1 Properties of Partial Derivatives . . . . . . . . . . . . . . . . . . . 107
15.1.1 Summary of Relations . . . . . . . . . . . . . . . . . . . . 107
15.2 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
15.2.1 Types of Systems . . . . . . . . . . . . . . . . . . . . . . . 108
15.2.2 System Parameters . . . . . . . . . . . . . . . . . . . . . . 109
15.3 Work and Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
15.3.1 Generalized Forces and Displacements . . . . . . . . . . . 110
15.3.2 PV work . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
16 Maximum Work and Reversible changes 11316.1 Maximal Work: Reversible versus Irreversible changes . . . . . . . 113
16.2 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115
16.3 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . 116
16.3.1 Example 1: The Ideal Gas Law . . . . . . . . . . . . . . . 116
16.3.2 Example 2: The van der Waals Equation of State . . . . . 117
16.3.3 Other Equations of State . . . . . . . . . . . . . . . . . . . 118
17 The Zeroth and First Laws of Thermodynamics 11917.1 Temperature and the Zeroth Law of Thermodynamics . . . . . . . 119
17.2 The First Law of Thermodynamics . . . . . . . . . . . . . . . . . 121
17.2.1 The internal energy state function . . . . . . . . . . . . . . 121
18 The Second and Third Laws of Thermodynamics 12418.1 Entropy and the Second Law of Thermodynamics . . . . . . . . . 124
18.1.1 Statements of the Second Law . . . . . . . . . . . . . . . . 127
18.2 The Third Law of Thermodynamics . . . . . . . . . . . . . . . . . 127
18.2.1 The Third Law . . . . . . . . . . . . . . . . . . . . . . . . 128
18.2.2 Debyes Law . . . . . . . . . . . . . . . . . . . . . . . . . . 129
18.3 Times Arrow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130
IV Basics of Thermodynamics 134
19 Auxillary Functions and Maxwell Relations 13519.1 The Other Important State Functions of Thermodynamics . . . . 135
19.2 Enthalpy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136
19.2.1 Heuristic definition: . . . . . . . . . . . . . . . . . . . . . . 137
19.3 Helmholtz Free Energy . . . . . . . . . . . . . . . . . . . . . . . . 137
19.3.1 Heuristic definition: . . . . . . . . . . . . . . . . . . . . . . 138
19.4 Gibbs Free Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 138
19.4.1 Heuristic definition: . . . . . . . . . . . . . . . . . . . . . . 139
19.5 Heat Capacity of Gases . . . . . . . . . . . . . . . . . . . . . . . . 139
19.5.1 The Relationship Between CP and CV . . . . . . . . . . . 13919.6 The Maxwell Relations . . . . . . . . . . . . . . . . . . . . . . . . 140
20 Chemical Potential 14220.1 Spontaneity of processes . . . . . . . . . . . . . . . . . . . . . . . 142
20.2 Chemical potential . . . . . . . . . . . . . . . . . . . . . . . . . . 144
20.3 Activity and the Activity coecient . . . . . . . . . . . . . . . . . 14620.3.1 Reference States . . . . . . . . . . . . . . . . . . . . . . . 147
20.3.2 Activity and the Chemical Potential . . . . . . . . . . . . 148
21 Equilibrium 15121.0.3 Equilibrium constants in terms of KC . . . . . . . . . . . . 15321.0.4 The Partition Coecient . . . . . . . . . . . . . . . . . . . 153
22 Chemical Reactions 15622.1 Heats of Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 156
22.1.1 Heats of Formation . . . . . . . . . . . . . . . . . . . . . . 157
22.1.2 Temperature dependence of the heat of reaction . . . . . . 157
22.2 Reversible reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 158
22.3 Temperature Dependence of Ka . . . . . . . . . . . . . . . . . . . 15922.4 Extent of Reaction . . . . . . . . . . . . . . . . . . . . . . . . . . 160
23 Ionics 16123.1 Ionic Activities . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
23.1.1 Ionic activity coecients . . . . . . . . . . . . . . . . . . . 16223.2 Theory of Electrolytic Solutions . . . . . . . . . . . . . . . . . . . 163
23.3 Ion Mobility . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164
23.3.1 Ion mobility . . . . . . . . . . . . . . . . . . . . . . . . . . 165
24 Thermodynamics of Solvation 16924.1 The Born Model . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
24.1.1 Free Energy of Solvation for the Born Model . . . . . . . . 173
24.1.2 Ion Transfer Between Phases . . . . . . . . . . . . . . . . . 174
24.1.3 Enthalpy and Entropy of Solvation . . . . . . . . . . . . . 174
24.2 Corrections to the Born Model . . . . . . . . . . . . . . . . . . . . 175
25 Key Equations for Exam 4 177
V Quantum Mechanics and Dynamics 180
26 Particle in a 3D Box 18126.1 Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . 181
26.2 The 3D Particle in a Box Problem . . . . . . . . . . . . . . . . . . 183
27 Operators 18727.1 Operator Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . 187
27.2 Orthogonality, Completeness, and the Superposition Principle . . 191
28 Angular Momentum 19228.1 Classical Theory of Angular Momentum . . . . . . . . . . . . . . 192
28.2 Quantum theory of Angular Momentum . . . . . . . . . . . . . . 193
28.3 Particle on a Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
28.4 General Theory of Angular Momentum . . . . . . . . . . . . . . . 195
28.5 Quantum Properties of Angular Momentum . . . . . . . . . . . . 199
28.5.1 The rigid rotor . . . . . . . . . . . . . . . . . . . . . . . . 200
29 Addition of Angular Momentum 20129.1 Spin Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . 201
29.2 Addition of Angular Momentum . . . . . . . . . . . . . . . . . . . 202
29.2.1 The Addition of Angular Momentum: General Theory . . 202
29.2.2 An Example: Two Electrons . . . . . . . . . . . . . . . . . 203
29.2.3 Term Symbols . . . . . . . . . . . . . . . . . . . . . . . . . 204
29.2.4 Spin Orbit Coupling . . . . . . . . . . . . . . . . . . . . . 205
30 Approximation Techniques 20730.1 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . 207
30.2 Variational method . . . . . . . . . . . . . . . . . . . . . . . . . . 209
31 The Two Level System and Quantum Dynamics 21131.1 The Two Level System . . . . . . . . . . . . . . . . . . . . . . . . 211
31.2 Quantum Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . 214
VI Symmetry and Spectroscopy 220
32 Symmetry and Group Theory 22132.1 Symmetry Operators . . . . . . . . . . . . . . . . . . . . . . . . . 222
32.2 Mathematical Groups . . . . . . . . . . . . . . . . . . . . . . . . . 222
32.2.1 Example: The C2v Group . . . . . . . . . . . . . . . . . . 223
32.3 Symmetry of Functions . . . . . . . . . . . . . . . . . . . . . . . . 223
32.3.1 Direct Products . . . . . . . . . . . . . . . . . . . . . . . . 225
32.4 Symmetry Breaking and Crystal Field Splitting . . . . . . . . . . 225
33 Molecules and Symmetry 22833.1 Molecular Vibrations . . . . . . . . . . . . . . . . . . . . . . . . . 228
33.1.1 Normal Modes . . . . . . . . . . . . . . . . . . . . . . . . 229
33.1.2 Normal Modes and Group Theory . . . . . . . . . . . . . . 229
34 Vibrational Spectroscopy and Group Theory 23134.1 IR Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231
34.2 Raman Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . . . 233
35 Molecular Rotations 23535.1 Relaxing the rigid rotor . . . . . . . . . . . . . . . . . . . . . . . . 236
35.2 Rotational Spectroscopy . . . . . . . . . . . . . . . . . . . . . . . 236
35.3 Rotation of Polyatomic Molecules . . . . . . . . . . . . . . . . . . 237
36 Electronic Spectroscopy of Molecules 24036.1 The Structure of the Electronic State . . . . . . . . . . . . . . . . 240
36.1.1 Absorption Spectra . . . . . . . . . . . . . . . . . . . . . . 241
36.1.2 Emission Spectra . . . . . . . . . . . . . . . . . . . . . . . 241
36.1.3 Fluorescence Spectra . . . . . . . . . . . . . . . . . . . . . 242
36.2 FranckCondon activity . . . . . . . . . . . . . . . . . . . . . . . 243
36.2.1 The FranckCondon principle . . . . . . . . . . . . . . . . 243
37 Fourier Transforms 24537.1 The Fourier transformation . . . . . . . . . . . . . . . . . . . . . 245
VII Kinetics and Gases 249
38 Physical Kinetics 25038.1 kinetic theory of gases . . . . . . . . . . . . . . . . . . . . . . . . 250
38.2 Molecular Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . 252
39 The Rate Laws of Chemical Kinetics 25439.1 Rate Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 254
39.2 Determination of Rate Laws . . . . . . . . . . . . . . . . . . . . . 258
39.2.1 Dierential methods based on the rate law . . . . . . . . . 25939.2.2 Integrated rate laws . . . . . . . . . . . . . . . . . . . . . . 259
40 Temperature and Chemical Kinetics 26140.1 Temperature Eects on Rate Constants . . . . . . . . . . . . . . . 261
40.1.1 Temperature corrections to the Arrhenious parameters . . 262
40.2 Theory of Reaction Rates . . . . . . . . . . . . . . . . . . . . . . 262
40.3 Multistep Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . 265
40.4 Chain Reactions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 267
41 Gases and the Virial Series 26941.1 Equations of State . . . . . . . . . . . . . . . . . . . . . . . . . . 269
41.2 The Virial Series . . . . . . . . . . . . . . . . . . . . . . . . . . . 270
41.2.1 Relation to the van der Waals Equation of State . . . . . . 271
41.2.2 The Boyle Temperature . . . . . . . . . . . . . . . . . . . 272
41.2.3 The Virial Series in Pressure . . . . . . . . . . . . . . . . . 272
41.2.4 Estimation of Virial Coecients . . . . . . . . . . . . . . . 273
42 Behavior of Gases 27442.1 P, V and T behavior . . . . . . . . . . . . . . . . . . . . . . . . . 274
42.1.1 and T for an ideal gas . . . . . . . . . . . . . . . . . . . 27542.1.2 and T for liquids and solids . . . . . . . . . . . . . . . . 275
42.2 Heat Capacity of Gases Revisited . . . . . . . . . . . . . . . . . . 276
42.2.1 The Relationship Between CP and CV . . . . . . . . . . . 27642.3 Expansion of Gases . . . . . . . . . . . . . . . . . . . . . . . . . . 279
42.3.1 Isothermal and Adiabatic expansions . . . . . . . . . . . . 279
42.3.2 Heat capacity CV for adiabatic expansions . . . . . . . . . 28042.3.3 When P is the more convenient variable . . . . . . . . . . 28142.3.4 Joule expansion . . . . . . . . . . . . . . . . . . . . . . . . 282
42.3.5 Joule-Thomson expansion . . . . . . . . . . . . . . . . . . 283
43 Entropy of Gases 28643.1 Calculation of Entropy . . . . . . . . . . . . . . . . . . . . . . . . 286
43.1.1 Entropy of Real Gases . . . . . . . . . . . . . . . . . . . . 288
VIII More Thermodyanmics 292
44 Critical Phenomena 29344.1 Critical Behavior of fluids . . . . . . . . . . . . . . . . . . . . . . 293
44.1.1 Gas Laws in the Critical Region . . . . . . . . . . . . . . . 294
44.1.2 Gas Constants from Critical Data . . . . . . . . . . . . . . 295
44.2 The Law of Corresponding States . . . . . . . . . . . . . . . . . . 296
44.3 Phase Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . 296
44.3.1 The chemical potential and T and P . . . . . . . . . . . . 29744.3.2 The Clapeyron Equation . . . . . . . . . . . . . . . . . . . 298
44.3.3 Vapor Equilibrium and the Clausius-Clapeyron Equation . 298
44.4 Equilibria of condensed phases . . . . . . . . . . . . . . . . . . . . 299
44.5 Triple Point and Phase Diagrams . . . . . . . . . . . . . . . . . . 300
45 Transport Properties of Fluids 30145.1 Diusion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30145.2 Viscosity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303
45.3 Thermal conductivity . . . . . . . . . . . . . . . . . . . . . . . . . 305
45.3.1 Thermal Conductivity of Gases and Liquids . . . . . . . . 306
45.3.2 Thermal Conductivity of Solids . . . . . . . . . . . . . . . 307
46 Solutions 30846.1 Measures of Composition . . . . . . . . . . . . . . . . . . . . . . . 308
46.2 Partial Molar Quantities . . . . . . . . . . . . . . . . . . . . . . . 308
46.2.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . 309
46.2.2 Partial Molar Volumes . . . . . . . . . . . . . . . . . . . . 310
46.3 Reference states for liquids . . . . . . . . . . . . . . . . . . . . . . 311
46.3.1 Activity (a brief review) . . . . . . . . . . . . . . . . . . . 311
46.3.2 Raoults Law . . . . . . . . . . . . . . . . . . . . . . . . . 312
46.3.3 Ideal Solutions (RL) . . . . . . . . . . . . . . . . . . . . . 314
46.3.4 Henrys Law . . . . . . . . . . . . . . . . . . . . . . . . . . 316
46.4 Colligative Properties . . . . . . . . . . . . . . . . . . . . . . . . . 318
46.4.1 Freezing Point Depression . . . . . . . . . . . . . . . . . . 318
46.4.2 Osmotic Pressure . . . . . . . . . . . . . . . . . . . . . . . 319
47 Entropy Production and Irreverisble Thermodynamics 32247.1 Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322
47.2 The Second Law . . . . . . . . . . . . . . . . . . . . . . . . . . . 324
47.3 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325
47.3.1 Entropy Production due to Heat Flow . . . . . . . . . . . 326
47.3.2 Entropy Production due to Chemical Reactions . . . . . . 328
47.4 Thermodynamic Coupling . . . . . . . . . . . . . . . . . . . . . . 330
47.5 Echo Phenonmena . . . . . . . . . . . . . . . . . . . . . . . . . . 331
Chemistry 351: PhysicalChemistry I
1
1
Solved Problems
I make-up most of the problems on the problems sets, so it might be helpful to
you to see some of these problems worked out.
Even though there arent many book problems assigned during the year, you can
still learn a lot be working these and looking that their solutions in the solution
manual.
Keep in mind this chapter provides some examples of how to solve problems for
both physical chemistry I and physical chemistry II. Consequently early in the
course some of the examples might seem very itimidating. Simply skip those
examples as you scan through this chapter.
Tips for solving problems
Working problem sets is the heart and sole of learning physical chemistry. The
only way that you can be sure that you understand a concept at to be able to
solve the problems associated with it.
This takes time and hard work.
But there are some things that you can do to help yourself with these problems.
Tips
2
2
1. Remember nobody cares if you solve any particular problem on the problem
set. They have all been solved before, so if you solve them you will not
become famous nor will you save the world. The only reason you work them
is to learn.
2. Budget your time so that you dont have to work on an overwhelming number
of problems at a time. Try to whip-o a few on the same day that you getthe problem set. Then work on them consistently during the week. This
will make the problem sets much more ecient at helping you learn.
3. You can do the problem. I dont assign problems that you cannot do. If you
think you cant do the problem then maybe you need try a dierent way ofthinking about it.
4. Part of the trouble is simply understanding what the problem is asking you
to do. There is a tendency to try to start solving the problem before fully
understanding the question.
Read the question carefully Try to think about what topic(s) in lecture and in the notes the problemis dealing with.
Do not worry about not knowing how to solve it yet. Just identify the general ideas that you think you might need. Determine wether you need to approach the problem mathematicallyor conceptually or both.
If the question is long, try to identify subsections of it.
5. For problems that require a mathematical approach...
Do not be afraid. Try to figure out what mathematical techniques youneed to express the solution to the problem.
3
Do the math; either you will be able to do this or you wont. It mighttake some review on your part.
Always check to see if the math makes sense when you are done.
6. For problems that require a conceptual approach...
Make sure that the physical idea that you are using in your argument iscorrect. If you are not sure, start with a related concept that is better
known by you.
Look for self-consistency. Does you final answer jive with what youknow.
Problems Dealing With Quantum Mechanics
Problem: What is the periodicity of the following functions
f(x) = sin2 x
f(x) = cosx
f(x) = e2ix
Solution: For the first function it is easiest to see the periodicity by writing thefunction as f(x) = (sinx)(sinx). We know that this function will repeat zeroswhen ever sinx = 0. This occurs at x = n, n = 0,1,2 . . ., so the periodicityis . The second function we should remember from trig as having a period of 2.Finally for the last function it is best to used Eulers identity and write
e2ix = cos 2x+ i sin 2x (1)
The real part of this function, cos 2x, has a period of as does the imaginarypart, sin 2x. Therefore the entire function has a period of .
4
Problem: Which of the following functions are eigenfunction of the momentumoperator, px = i~ ddx .
(x) = eikx
(x) = ex2
(x) = cos kx
Solution: We need to determine if px(x) = (x) where is a constant. Ifthis equation is true then the function is an eigenfunction with eigenvalue . Forthe case of momentum all we need to do is take the derivative of each function,
multiply by i~ and check to see if the eigenvalue equation holds.For the first function
px(x) = i~d(x)dx
= i~deikx
dx= ~keikx = ~k(x), (2)
so, yes, this function is an eigenfunction of the momentum operator.
For the second function
px(x) = i~d(x)dx
= i~dex2
dx= 2i~xex2 = 2i~x(x), (3)
so, no, this function is not and eigenfunction of the momentum operator.
For the last function
px(x) = i~d(x)dx
= i~d cos kxdx
= i~k6=cos kxz }| {sin kx, (4)
so, no, this function is not an eigenfunction of the momentum operator.
Problem: A quantum object is described by the wavefunction (x) = ex2.What is the probability of finding the object further than away from the origin(x = 0)?
5
Solution: First of all we do not know if this wavefunction is normalized, so weshould assume that it isnt. We could normalize this wavefunction, but we wont.
We are interested in finding the probability that the object is outside of the region
< x < . To do this using an unnormalized wavefunction we must evaluate
P (|x| > ) =R |(x)|2 dx+
R |(x)|2 dxR
|(x)|2 dx. (5)
The first integral in the numerator gives the probability that the object is at a
position x < and the second integral in the numerator gives the probabilityfor x > . The denominator accounts for the fact that the wavefunction is un-normalized. The limits of the integral in the denominator represent all space for
the object. If you were working with a normalized wavefunction the denominator
would be equal to 1 and hence not needed. Plugging in the wavefunctions we have
P (|x| > ) =R e
2x2dx+R e
2x2dxR e
2x2dx. (6)
Mathematica can assist with these integrals to give the final answer of
P (|x| > ) = erfc[2
32 ]. (7)
Problem: A quantum object is described by the wavefunction (x) = ex overthe range 0 x
Which for this case is
N =
sZ 0
|ex|2 dx =sZ
0
e2xdx =r1
2(10)
So finally we get the normalized wavefunction by rearanging unnorm = Nnorm:
norm(x) =p2ex. (11)
Problem: A quantum object is described by the wavefunction (x) = ex overthe range 0 x a0) =Z 20
Z 0
Z a0|1s|2 r2 sin drdd. (14)
Remember the extra r2 sin is needed when integrating in spherical polar coordi-nates. The normalized 1s wavefunction is
1s =1pa30
er/a0 . (15)
7
We can do this integral by hand or have Mathematica help us to give
P (r > a0) =5
e2= 0.677. (16)
So, about 68% of the time the electron would be found at some distance greater
then one Bohr radius from the proton.
Problem: A free particle in three dimensions is described by the Hamiltonian,H = ~
2
2m2. Express the wavefunction (in Cartesian coordinates) as a product
state.
Solution: This problem appears hard at first since we are not studying threedimensional systems, but all it is asking is to express the wavefunction, which is
a function of the three spatial dimensions, (x, y, z) as a product state. We knowthat if the wavefunction is to be a product state then the Hamiltonian must be
made up of a sum of independent terms. To see this we write out the Laplacian
to get
H =~22m
2
x2+
2
y2+
2
z2
. (17)
We see that indeed the Hamiltonian is a sum of term that depends only on x,a term depending only on y and a term that depends only on z. Therefore theappropriate product state is
(x, y, z) = (x)(y)(z). (18)
Problem: Expand the Morse potential in a Taylors series about Req. Verify thatthe coecient for the linear term is zero. What is the force constant associatedwith the Morse potential?
Solution: The Morse potential is
V (x) = De1 e(RReq)
. (19)
8
The Taylor series about Req for this function is
V (x) = V (x)|Req| {z }= 0
+dV (x)dx
Req| {z }
= 0
(RReq) +1
2!
d2V (x)dx2
Req| {z }
= 2De
(RReq)2 + . (20)
So, yes the coecient of the linear term (the term involving (R Req) to thefirst power) is zero. This will always be true when you perform a Taylor series
expansion about a minimum (or maximum). The force constant is given by the
coecient of the quadratic term so in this case k = 2De.
Problem: Without performing any calculations, compare hRi as a function ofthe vibrational quantum number for a diatomic modelled as a harmonic oscillator
versus a Morse oscillator.
Solution: This problem requires the we think qualitatively about the wavefunc-tions and the potentials for the harmonic oscillator and the Morse oscillator. The
potential for the harmonic oscillator is described by a parabola centered about the
equilibrium bond length. Hence no mater what the vibrational quantum number is
there is just as much of the wavefunction on either side equilibrium thus hRi = Reqfor any quantum number. The Morse potential does not have this symmetry. It
is steeper on the short side of equilibrium and softer on the long side of equi-
librium and this softness increases with increasing quantum number. Therefore
without performing any calculations we can at least say that hRi increases as thequantum number increases.
Problems Dealing With Statistical Mechanics and Thermo-
dynamics
Problem: A vial containing 1020 benzene molecules is at 300K. How many mole-cules are in the first excited state of the ring breathing mode (992 cm1)? How
9
many are in the first excited state of the symmetric CH vibrational mode (3063
cm1)?
Solution: This is a problem that deals with the Boltzmann distribution. So,
N rbv=1 =2 sinh
992
2 208e
39922208
1020 = 8.41 1017 (21)
and
NCHv=1 =2 sinh
3063
2 208e
330632208
1020 = 4.02 1013. (22)
We see that about 8.411017
1020 100% = 0.841% of the benzene molecules are in the
first vibrational excited state for the ring breathing mode and 4.021013
1020 100% =
0.0000402% of the benzene molecules are in the first excited state for the CHstretching mode.
Problem: Consider a linear chain of N atoms. Each of the atoms can be in oneof three states A, B or C, except that an atom in state A can not be adjacent toan atom in state C. Find the entropy per atom for this system as N . Tosolve this problem it is useful to define the set of three dimensional column vectors
V (j) such that the three elements are the total number of allowed configurations ofa j-atom chain having the jth atom in state A, B or C. For example,
V (1) =
1
1
1
, V (2) =
2
3
2
, V (3) =
5
7
5
, . (23)
The V (j+1) can be found from the V (j) vector using the matrix equation,
V (j+1) =MV (j), (24)
where for this example
M =
1 1 0
1 1 1
0 1 1
. (25)
10
The matrix M is the so-called transfer matrix for this system. It can be shownthat the number of configurations W = Tr[MN ]. Now for large N, Tr[MN ] Nmax,where max is the largest eigenvalue of M. So
W = limN
Nmax. (26)
1. 1. Use M to find V (4)
2. Verify V (3) explicitly by drawing all the allowed 3-atom configurations.
3. Verify W = Tr[MN ] for N = 1 and N = 2.
4. Use Boltzmanns equation to find the entropy per atom for this chain
as N goes to infinity.
Solution: For part (a) we simply use the transfer matrix as directed in theproblem (we are given V (3)):
V (4) =
1 1 0
1 1 1
0 1 1
5
7
5
=
12
17
12
.
For part (b) we need to list all states for the case of N = 3 and verify the we getthe same result as calculated using the transfer matrix. Remembering that V (3)
gives us the number of sequences that end in a given state we should organize our
list in the same manner
States ending in A States ending in B States ending in C
AAA AAB ABCABA ABB BBCBAA BAB BCCBBA BBB CBCCBA BCB
CBBCCB
5 states
7 states
5 states
.
11
States like AAC are not allowed because A and C are neighbors.For part (c) we evaluate W = Tr[MN ] for N = 1 and 2. For N = 1, W =Tr[M ] = 3 This corresponds to the three distinguishable microstates A, B, andC. For N = 2,
W = Tr[M2] = Tr
1 1 0
1 1 1
0 1 1
1 1 0
1 1 1
0 1 1
= Tr
2 2 1
2 3 2
1 2 2
= 7 (27)
This corresponds to the seven distinguishable microstates AA, AB, BA, BB, BC,CB and CC (Remember C and A cannot be neighbors).For part (d) we use
SN=
kNlnW = lim
N
kNlnNmax = limN
kNN lnmax = k lnmax. (28)
So, we simply need to find the maximum eigenvalue of the Transfer matrix. Using
Mathematica we find max = 1 +2. Therefore the limiting entropy per atom
isSN= k ln
1 +
2. (29)
Problem: Using the classical theory of light scattering, calculate the positions ofthe Rayleigh, Stokes and anti-Stokes spectral lines for benzene. Assume benzene
has only two active modes (992cm1 and 3063cm1) and assume the Laser light
used to do the scattering is at 20000cm1 (this is 500nmgreen light).
Solution: Since there are two vibrational modes we expect two Stokes lines tothe red of 20000cm1, one at 20000cm1 992cm1 = 19008cm1 and one at20000cm1 3063cm1 = 16937cm1. Likewise we expect two anti-Stokes lines,one at 20000cm1 + 992cm1 = 20992cm1 and one at 20000cm1 + 3063cm1 =
23063cm1. There is only one Rayleigh line and it is at the same frequency at the
input laser beam which, in this case, is 20000cm1.
12
Problem: A simple model for a crystal is a gas of harmonic oscillators. De-termine A, S, and U from the partition function for this model.
Solution: For this model the crystal is modelled as a collection of harmonicoscillators so we need the partition function for the harmonic oscillator.
Qcrystal = qNHO =
1
2 sinh ~2
!(30)
From our formulas for statistical thermodynamics
A = kT lnQcrystal = +NKT ln2 sinh
~2
, (31)
where we used properties of logs to pull the N out front and move the sinh termfrom to the numerator,
S = kQcrystal
+ k lnQcrystal (32)
=Nk~2
coth~2 k ln
2 sinh
~2
and
U = Qcrystal
=N~2
coth~2
. (33)
Problem: Express the equation of state for internal energy for a Berthelot gas.
Solution: The equation representing a Berthelot gas is
P =nRTV nb
n2aTV 2
. (34)
We are interesting in an equation of state for U(T, V ). Writing out the totalderivative of U(T, V ) we get
dU =UT
VdT +
UV
TdV. (35)
13
NowUT
V is just heat capacity, CV , but
UV
T is nothing convenient so we must
proceed. We employ the useful relationUV
T= T
PT
V P (36)
to eliminate U in favor of P so that we can use the equation of state for a Berthelotgas. One obtains
TPT
V P = T
nR
V nb +n2aT 2V 2
nRTV nb +
n2aTV 2
=2n2aTV 2
. (37)
Hence the equation of state for internal energy of a Berthelot gas is
dU = CV dT +2n2aTV 2
dV (38)
Problem: Use the identities for partial derivatives to eliminate thePT
V factor
in
Cp = Cv + TVT
P
PT
V
(39)
so that all derivatives are at constant pressure or temperature.
Solution: Here we either remember an identity or turn to our handout of partialderivative identities to employ the cyclic rule to
PT
V :
PT
V=
PV
T
VT
P. (40)
This eliminates the constant V term and so,
Cp = Cv TVT
2P
PV
T. (41)
14
Part I
Basic Quantum Mechanics
15
15
1. Quantum Theory
The goal of science is unification.
Many phenomena described by minimal and general concepts.
1.1. The Fall of Classical Physics
A good theory:
explain known experimental results
self consistent
predictive
minimal number of postulates
Around the turn of the century, experiments were being performed in which the re-
sults defied explanation by means of the current understanding of physics. Among
these experiments were
1. The photoelectric eect
2. Low temperature heat capacity
3. Atomic spectral lines
4. Black body radiation and the ultraviolet catastrophe
16
16
5. The two slit experiment
6. The Stern-Gerlach experiment
See Handouts
1.2. Bohrs Atomic Theory
1.2.1. First Attempts at the Structure of the Atom
The solar system model.
The electron orbits the nucleus with the attractive coulomb force balancedby the repulsive centrifugal force.
Flaws of the solar system model
Newton: OK
Maxwell: problem
17
As the electron orbits the nucleus, the atom acts as an oscillating dipole
The classical theory of electromagnetism states that oscillating dipolesemit radiation and thereby lose energy.
The system is not stable and the electron spirals into the nucleus. Theatom collapses!
Bohrs model: Niels Bohr (18851962)
18
Atoms dont collapse = what are the consequences
Experimental clues
Atomic gases have discrete spectral lines.
If the orbital radius was continuous the gas would have a continuous spec-trum.
Therefore atomic orbitals must be quantized.
r =40N2~2
Zmee2(1.1)
where Z is the atomic number, me and e are the mass and charge of theelectron respectively and 0 is the permittivity of free space. N is a positivereal integer called the quantum number. ~ = h/2 is Plancks constantdivided by 2.
The constant quantity 40~2
mee2appears often and is given the special symbol a0
40~2mee2
= 0.52918 and is called the Bohr radius.
The total energy of the Bohr atom is related to its quantum number
EN = Z2e2
2a0
1
N2. (1.2)
Tests of the Bohr atom
Ionization energy of Hydrogen atoms
The Ionization energy for Hydrogen atoms (Z = 1) is the miniumenergy required to completely remove an electron form it ground state,
i.e., N = 1 N =
Eionize = E E1 =Z2e22a0
1
2 1
12
=
e2
2a0(1.3)
19
Eionize = e2
2a0= 13.606 eV= 109,667 cm1 =R. R is called the Rydberg
constant.
Eionize experimentally observed from spectroscopy is 13.605 eV (verygood agreement)
Spectroscopic lines fromHydrogen represent the dierence in energy betweenthe quantum states
Bohr theory: Dierence energies
Ej Ek =e2
2a0
1
N2j 1N2k
= R
1
N2j 1N2k
(1.4)
Initial state Nk Final States Nj Series Name
1 2,3,4, Lyman2 3,4,5, Balmer3 4,5,6, Pachen4 5,6,7, Brackett5 6,7,8, Pfund
Since the orbitals are quantized, the atom may only change its orbitalradius by discrete amounts.
Doing this results in the emission or absorption of a photon with energy
v =4Ehc
(1.5)
Failure of the Bohr model
No fine structure predicted (electron-electron coupling)
No hyperfine structure predicted (electron-nucleus coupling)
No Zeeman eect predicted (response of spectrum to magnetic field)
20
Spin is not included in theory
The Bohr quantization idea points to a wavelike behavior for the electron.
The wave must satisfy periodic boundary conditions much like a vibrating ring
See Fig. 11.9 Laidler&Meiser
The must be continuous and single valued
Particles have wave-like characteristics
The Bohr atom was an important step towards the formulation of quantum theory
Erwin Schrdinger (18871961): Wave mechanics
Werner Heisenberg (19021976): Matrix mechanics
Paul Dirac (19021984): Abstract vector space approach
21
2. The Postulates of Quantum
Mechanics
2.1. Postulate I
Postulate I: The state of a system is defined by a wavefunction, , which con-tains all the information that can be known about the system.
We will normally take to be a complex valued function of time and coordi-nates: (t, x, y, z) and, in fact, we will most often deal with time independentstationary states (x, y, z)
Note: In general the wavefunction need not be expressed as a function of coordi-
nate. It may, for example, be a function of momentum.
The wavefunction represents a probability amplitude and is not directly observ-able.
However the mod-square of the wavefunction, = ||2 , represents a probabilitydistribution which is directly observable.
That is, the probability of finding a particle which is described by (x, y, z) at theposition between x and x+dx, y and y+dy and z and z+dz is |(x, y, z)|2 dxdydz(or |(r, , )|2 r2 sin drdd in spherical coordinates).
22
22
Properties of the wavefunction
Single valueness
continuous and finite
continuous and finite first derivative
Rspace |(x, y, z)|2 dxdydz
The N is just a constant so it can be pulled out of both the mod-square and theintegral Z
space|unnorm|2 dxdydz = N2
Zspace
|norm|2 dxdydz, (2.3)
but Zspace
|norm|2 dxdydz = 1 (2.4)
because that is the very definition of a normalized wavefunction. Thus wherever
we seeRspace |norm|2 dxdydz we can replace it with 1. So,Z
space|unnorm|2 dxdydz = N2 1 = N2. (2.5)
This gives us an expression for N. Taking the square root of both sides gives.
N =
sZspace
|unnorm(x, y, z)|2 dxdydz. (2.6)
So finally we get the normalized wavefunction by reagranging unnorm = Nnorm:
norm =1
Nunnorm. (2.7)
Notice that no where did we ever specify what unnorm or norm actually were,therefore this is a general procedure that will work for any wavefunction.
To find the probability for the particle to be in a finite region of space we simple
evaluate (here a 1D case)
P (x1 < x < x2) =
R x2x1|(x)|2 dxR
|(x)|2 dxif (x)=
normalized
Z x2x1|(x)|2 dx (2.8)
2.3. Postulates II and II
Postulate II: Every physical observable is represented by a linear (Hermitian)operator.
24
An operator takes a function and turns it into another function
Of(x) = g(x) (2.9)
This is just like how a function takes a number and turns it into another number.
So in quantum mechanics operators act on the wavefunction to produce a new
wavefunction
The two most important operators as far as we are concerned are
x = x
px = i~ xand of course the analogous operators for the other coordinates (y, z) and coordi-nate systems (spherical, cylindrical, etc.).
Nearly all operators we will need are algebraic combinations of the above.
Postulate III: The measurement of a physical observable will give a result thatis one of the eigenvalues of the corresponding operator.
There is a special operator equation called the eigenvalue equation which is
Of(x) = f(x) (2.10)
where is just a number.
For a given operator only a special set of function satisfy this equation. These
functions are called eigenfunctions.
25
The number that goes with each function is called the eigenvalue.
So solution of the eigenvalue equation gives a set of eigenfunctions and a set of
eigenvalues.
Example
Let O in the eignevalue equation be the operator that takes the derivative: O =d = ddx .
So we want a solution to
df(x) = f(x) (2.11)df(x)dx
= f(x)
So, we ask ourselves what function is proportional to its own derivative? f(x) = ex.
So the eigenfunctions are the set of functions f(x) = ex and the eigenvalues arethe numbers
26
3. The Setup of a Quantum
Mechanical Problem
3.1. The Hamiltonian
The most important physical observable is that of the total energy E.
The operator associated with the total energy is called the Hamiltonian operator
(or simply the Hamiltonian) and is given the symbol H.
The eigenvalue equation for the Hamiltonian is
H = E. (3.1)
This equation is the (time independent) Schrdinger equation.
This equation is the most important equation of the course and we will use it many
times throughout our discussion of quantum mechanics and statistical mechanics.
3.2. The Quantum Mechanical Problem
Nearly every problem one is faced with in elementary quantum mechanics is han-
dled by the same procedure as given in the following steps.
1. Define the classical Hamiltonian for the system.
27
27
The total energy for a classical system is
Ecl = T + V, (3.2)
where T is the kinetic energy and V is the potential energy.
The kinetic energy is always of the form
T =1
2mp2x + p
2y + p
2z
(3.3)
The potential energy is almost always a function of coordinates only
V = V (x, y, z) (3.4)
Note: Some quantum systems dont have classical analogs so the Hamil-tonian operator must be hypothesized.
2. Use Postulate II to replace the classical variables, x, px etc., with theirappropriate operators. Thus,
T =~22m
2 = ~2
2m2, (3.5)
where 2 2x2 +2y2 +
2z2 , and
V = V (x, y, z) = V (x, y, z). (3.6)
So,
H = T + V =~22m
2 + V (x, y, z) (3.7)
3. Solve the Schrdinger equation, H = E, which is now a second orderdierential equation of the form
~22m
2 + V (x, y, z) = E
~2
2m2 + (V (x, y, z)E) = 0 (3.8)
28
Note: It is solely the form of V (x, y, z) which determines whether thisis easy or hard to do.
For one-dimensional problems~22m
d2
dx2 + (V (x)E) = 0 (3.9)
3.3. The Average Value Theorem
Postulate III implies that if is an eigenfunction of a particular operator rep-resenting a physical observable, then all measurements of that physical property
will yield the associated eigenvalue.
However, If is not an eigenfunction of a particular operator, then all measure-ments of that physical property will still yield an eigenvalue, but we cannot predict
for certain which one.
We can, however, give an expectation, or average, value for the measurement.
This is given by
hi =Zspace
dxdydz (3.10)
For example,
hxi =Zspace
xdxdydz =Zspace
x ||2 dxdydz (3.11)
and
hpxi =Zspace
pxdxdydz = i~Zspace
x
dxdydz (3.12)
29
3.4. The Heisenberg Uncertainty Principle
In quantum mechanics certain pairs of variables can not, even in principle, be
simultaneously known to arbitrary precision. Such variables are called compli-
mentary.
This idea is the Heisenberg uncertainty principle and is of profound im-portance.
The general statement of the Heisenberg uncertainty principle is
12
Dh,
iE, (3.13)
where the notationh,
imeans the commutator of and . The commutator is
defined as h,
i . (3.14)
The most important example of complimentary variables is position and momen-
tum. We see
pxx 1
2|h[px, x]i| = 1
2|hpxx xpxi| (3.15)
=1
2
Z~i
x
x x x
dx
=
~2i
=~2.
So, at the very best we can only hope to simultaneously know position and momen-
tum such that the product of the uncertainty in each is ~2. (n.b., pxy = 0, we can
know, for example, the y position and the x momentum to arbitrary precision.)
Suppose we know the position of a particle perfectly, what can we say about its
momentum?
30
4. Particle in a Box
We now will apply the general program for solving a quantum mechanical problem
to our first system: the particle in a box.
This system is very simple which is one reason for beginning with it. It also can
be used as a zeroth order model for certain physical systems.
We shall soon see that the particle in a box is a physically unrealistic system and,
as a consequence, we must violate one of our criteria for a good wavefunction.
Nevertheless it is of great pedagogical and practical value.
4.1. The 1D Particle in a Box Problem
Consider the potential, V (x), shown in the figure and given by
V (x) =
x 00 0 < x < a x a
. (4.1)
Because of the infinities at x = 0 and x = a, we need to partition the x-axis intothe three regions shown in the figure.
31
31
Now, in region I and III, where the potential is infinite, the particle can never
exist so, must equal zero in these regions.
The particle must be found only in region II.
The Schrdinger equation in region II is (V (x) = 0)
H = E = ~2
2md2(x)dx2
= E, (4.2)
which can be rearranged into the form
d2(x)dx2
+2mE~2
(x) = 0. (4.3)
The general solution of this dierential equation is
(x) = A sin kx+B cos kx, (4.4)
where k =q
2mE~2 .
Now must be continuous for all x. Therefore it must satisfy the boundaryconditions (b.c.): (0) = 0 and (a) = 0.
32
From the (0) = 0 b.c. we see that the constant B must be zero becausecos kx|x=0 = 1.
So we are left with (x) = A sin kx for our wavefunction.
As can be inferred from the following figure, the second b.c., (a) = 0, placescertain restrictions on k.
In particular,
kn =na, n = 1, 2, 3, . (4.5)
The values of k are quantized. So, now we have
n(x) = A sinnxa
. (4.6)
The constant A is the normalization constant. We obtain A fromZ
n(x)n(x) = 1 =Z a0
A2 sinnxasin
nxa
dx. (4.7)
Letting u = xa , du =adx, this becomes
1 = A2a
Z 0
sin2 nudu = A2a//2=A2a2
. (4.8)
33
Solving for A gives
A =
r2
a. (4.9)
Thus our normalized wavefunctions for a particle in a box are
n(x) =
0 Iq2a sin
nxa II
0 III
. (4.10)
Is this wavefunction OK?
We can get the energy levels from kn =q
2mEn~2 and kn =
na :
En =n22~2
2ma2~= h
2=n2h2
8ma2. (4.11)
4.2. Implications of the Particle in a Box problem
Zero Point Energy
34
The smallest value for n is 1 which corresponds to an energy of
E1 =h2
8ma26= 0. (4.12)
That is, the lowest energy state, or ground state, has nonzero energy. This residual
energy is called the zero point energy and is a consequence of the uncertainty
principle.
If the energy was zero then we would conclude that momentum was exactly zero,
p = 0. But we also know that the particle is located within a finite region ofspace, so x 6=.
Hence, xp = 0 which violates the uncertainty principle.
Features of the Particle in a Box Energy Levels
The energy level spacing is
4E = En+1 En = (n+ 1)2h2
8ma2 n
2h2
8ma2= (n2/ + 2n+ 1 n2/ ) h
2
8ma2
4E = (2n+ 1) h2
8ma2(4.13)
This spacing increases linearly with quantum level n
This spacing decreases with increasing mass
This spacing decreases with increasing a
It is this level spacing that is what is measured experimentally
The Curvature of the Wavefunction
35
The operator for kinetic energy is T = ~2
2md2dx2 . The important part of this is
d2dx2 .
From freshman calculus we know that the second derivative of a function describes
its curvature so, a wavefunction with more curvature will have a larger second
derivative and hence it will posses more kinetic energy.
This is an important concept for the qualitative understanding of wavefunctions
for any quantum system.
Applying this idea to the particle in a box we an anticipate both zero point energy
and the behavior of the energy levels with increasing a.
We know the wavefunction is zero in regions I and III. We also know thatthe wave function is not zero everywhere. Therefore it must do something
between x = 0 and x = a. It must have some curvature and hence some zeropoint energy.
As a is increased, the wavefunction is less confined and so the curvature doesnot need to be as great to satisfy the boundary conditions. Therefore the
energy levels decrease in energy as does their dierence.
The particle in a box problem illustrates some of the many strange features of
quantum mechanics.
We have already seen such nonclassical behavior as quantized energy and zero
point energy.
As another example consider the expectation value of position for a particle in
the second quantum level:
hxi =Z
2(x)x2(x)dx =2
a
Z a0
x sin2[2ax]dx =
a2
(4.14)
36
yet the probability of finding the particle at x = a2is zero: 2(
a2) = 0. There is
a node at x = a2. So even though the particle may be found anywhere else in the
box and it may get from the left side of the node to the right side, it can never
be found at the node.
37
5. The Harmonic Oscillator
The harmonic oscillator model which is simply a mass undergoing simple harmonic
motion. The classical example is a ball on a spring
The harmonic oscillator is arguably the single most important model in all of
physics.
We shall begin by reviewing the classical harmonic oscillator and than we will
turn our attention to the quantum oscillator.
The force exerted by the spring in the above figure is F = k(RReq), where kis the spring constant and Req is the equilibrium position of the ball.
Setting x = R Req we can measure the displacement about the equilibriumposition.
38
38
From Newtons law of motion F = ma = md2xdt2 , we get
md2xdt2
= kx d2xdt2
+kmx = 0 (5.1)
This is second order dierential equation which we already know the solutions to:
x = A sint+B cost, (5.2)
where =q
km and A and B are constants which are determined by the initial
conditions.
For quantum mechanics it is much more convenient to talk about energy rather
than forces, so in going to the quantum oscillator, we need to express the force of
the spring in terms of potential energy V . We know
V = Z
Fdx =1
2kx2 + C. (5.3)
Since energy is on an arbitrary scale we can set C = 0. Thus V = 12kx2.
By postulate III the Schrdinger equation becomes
H = E
~
2
2md2
dx2| {z }K.E.
+1
2kx2| {z }P.E.
= E. (5.4)
This can be rearrange into the form
~22m
d2dx2
+
1
2kx2 E
= 0 (5.5)
This dierential equation is not easy to solve (you can wait to solve it in graduateschool).
39
The equation is very close to the form of a know dierential equation called Her-mites dierential equation the solutions of which are called the Hermite polynom-inals.
As it turns out, the solutions (the eigenfunctions) to the Schrdinger equation for
the harmonic oscillator are
n(y) = AnHn(y)e y
2
2 , y =km~2
14
x, An =1p
2nn!, (5.6)
where An is the normalization constant for the nth eigenfunction and Hn(y) arethe Hermite polynomials.
The eigenvalues (the energy levels) are
En = (n+1
2)~, (5.7)
where again =q
km .
Note the energy levels are often written as
En = (n+1
2)h0, (5.8)
where 0 = 12q
km and is called the vibrational constant.
See Fig. 11.12 Laidler&Meiser
5.1. Interesting Aspects of the Quantum Harmonic Oscilla-
tor
It is interesting to investigate some of the unintuitive properties of the oscillator
as we have gone quantum mechanical
40
1. Consider the ground state (the lowest energy level)
There is residual energy in the ground state because
E0 = (0 +1
2)~.
Just like for the particle in a box, this energy is called the zero pointenergy.
It is a consequence of uncertainty principle If the ground state energy was really zero, then we would concludethat the momentum of the oscillator was zero.
On the other hand, we would conclude the particle was located atthe bottom of the potential well (at x = 0)
Thus we would have p = 0, x = 0, so px = 0 Not allowed!
The uncertainty principle forces there to be some residual zeropoint energy.
2. Consider the wavefunctions.
The wavefunctions penetrate into the region where the classical particleis forbidden to go
The wavefunction is nonzero past the classical turning point.
The probability distribution ||2 becomes more and more like what isexpected for the classical oscillator when v .
This is a manifestation of the correspondence principle whichstates that for large quantum numbers, the quantum system must
behave like a classical system. In other words the quantum me-
chanics must contain classical mechanics as a limit.
3. Interpretation of the wavefunctions and energy levels
41
Remember the wavefunctions are time independent and the energy lev-els are stationary
If a molecule is in a particular vibrational state it is NOT vibrating.
5.2. Spectroscopy (An Introduction)
The primary method of measuring the energy levels of a material is through the
use of electromagnetic radiation.
Experiments involving electromagnetic radiationmatter interaction are called
spectroscopies.
Atoms and molecules absorb or emit light only at specific (quantized) energies.
These specific values correspond to the energy level dierence between the initialand final states.
42
Key Equations for Exam 1
Listed here are some of the key equations for Exam 1. This section should not
substitute for your studying of the rest of this material.
The equations listed here are out of context and it would help you very little to
memorize this section without understanding the context of these equations.
The equations are collected here simply for handy reference for you while working
the problem sets.
Equations
The short cut for getting the normalization constant (1D, see above for 3D).
N =
sZspace
|unnorm(x)|2 dx. (5.9)
The normalized wavefunction:
norm =1
Nunnorm. (5.10)
The Schrdinger equation (which should be posted on your refrigerator),
H = E. (5.11)
43
43
The Schrdinger equation for 1D problems as a dierential equation,~22m
d2
dx2 + (V (x)E) = 0. (5.12)
How to get the average value for some property (1D version),
hi =Zspace
dx. (5.13)
The momentum operatorpx = i~
x
. (5.14)
Normalized wavefunctions for the 1D particle in a box,
n(x) =
r2
asin
nxa
. (5.15)
The energy levels for the 1D particle in a box,
En =n22~2
2ma2~= h
2=n2h2
8ma2. (5.16)
The energy level spacing for the 1D particle in a box,
4E = (2n+ 1) h2
8ma2(5.17)
The wavefunctions for the harmonic oscillator are
n(y) = AnHn(y)ey
2
2 , y =km~2
14
x, An =1p
2nn!, (5.18)
where An is the normalization constant for the nth eigenfunction and Hn(y)are the Hermite polynomials.
The energy levels are
En = (n+1
2)~, =
rkm
(5.19)
44
Part II
Quantum Mechanics of Atomsand Molecules
45
45
6. Hydrogenic Systems
Now that we have developed the formalism of quantum theory and have discussed
several important systems, we move onto the quantum mechanical treatment of
atoms.
Hydrogen is the only atom for which we can exactly solve the Schrdinger equation
for. So this will be the first atomic system we discuss.
The Schrdinger equation for all the other atoms on the periodic table must be
solved by approximate methods.
6.1. Hydrogenic systems
Hydrogenic systems are those atomic systems which consist of a nucleus and one
electron. The Hydrogen atom (one proton and one electron) is the obvious exam-
ple
Ions such as He+ and Li2+ are also hydrogenic systems.
These system are centrosymmetric. That is they are completely symmetric about
the nucleus.
The obvious choice for the coordinate system is to use spherical polar coordinates
46
46
with the origin located on the nucleus.
The classical potential energy for these hydrogenic systems is
V (r) =Ze2(40)r
. (6.1)
So the Hamiltonian is
H =~22me
2 + Ze2
(40)r. (6.2)
Schrdingers equation (in spherical polar coordinates) becomes
E = H (6.3)
E =~22me
2 + Ze2
(40)r
E =~22me
1
r2rr2
r+1
r2
1
sin sin
+
1
sin2 2
2
+Ze2(40)r
The Hamiltonian is (almost) the sum of a radial part (only a function of r) andan angular part (only a function of and ):
H = Hrad +1
r2Hang, (6.4)
Hrad =~22me
1
r2rr2
r Ze
2
(40)r
(6.5)
and
Hang =~22me
1
sin sin
+
1
sin2 2
2
(6.6)
Since the Hamiltonian is the sum of two terms, must be a product state.
(r, , ) = rad(r)ang(, ) (6.7)
It turns out that solving the Schrdinger equation,
Hangang(, ) = Eang(, ), (6.8)
47
yields
ang(, ) = Ylm(, ), (6.9)
where the Ylm(, )s are the spherical harmonic functions characterized by quan-tum numbers l andm. The spherical harmonics are known functions. (Mathematicaknows them and you can use them just like any other built-in function like sine
or cosine.)
We shall use the spherical harmonics more next semester when we develop the
quantum theory of angular momentum.
It also turns out that the energy associated with Hang is found to be
E = El =l(l + 1)~2
2me. (6.10)
So,
Hangang(, ) =l(l + 1)~2
2meang(, ) (6.11)
Now lets denote the radial part of the wavefunction as rad(r) = R(r).
The full Schrdinger equation becomes
H(r, , ) = E(r, , ) (6.12)
HR(r)Ylm(, ) = ER(r)Ylm(, )Hrad +
1
r2Hang
R(r)Ylm(, ) = ER(r)Ylm(, ),
Operating with Hang we getHrad +
l(l + 1)~2
2mer2
R(r)Ylm(, ) = ER(r)Ylm(, ) (6.13)
48
The Ylm(, ) can now be cancelled to leave a one dimensional dierential equation:
~22me
1
r2rr2
r Ze
2
40r l(l + 1)
r2
R(r) = ER(r). (6.14)
This dierential equation is very similar to a known equation called Laguerresdierential equation which has as solutions the Laguerre polynomials Lln(x).
In fact, the solutions to our dierential equation are closely related to the Laguerrepolynomials.
Rnl() = Anl
2n
le/nL2l+1n+1
2n
, (6.15)
where the normalization constant, Anl, depends on the n and l quantum numbersas
Anl =
s2Zna0
3(n l 1)!2n[(n+ l)!]3
(6.16)
The energy eigenvalues, i.e., the energy levels are given by
En = Z2Rn2
(6.17)
Note: The energy levels are determined by n alonel drops out.
Also Note: the energy levels are the same as for the Bohr model.
So, the total wavefunction that describes a hydrogenic system (ignoring the spin
of the electron, which will be briefly discussed later) is
nlm(r, , ) = Rnl(r)Ylm(, ) (6.18)
6.2. Discussion of the Wavefunctions
We are now very close to having the atomic orbitals familiar from freshman chem-
istry.
49
We have explicitly derived the physicists picture of the atomic orbitals
orbital n l m wavefunctions ( = r/a0)
1s 1 0 0 1s = 100 = e
2s 2 0 0 2s = 200 =1
2
e/2
2p 2 1 0 2p0 = 210 = e/2 cos
2 1 1 2p1 = 211 = e/2 sin ei3d 3 2 0 3d0 = 320 =
2e/3 (3 cos2 1)3 2 1 3d1 = 321 = R32(r) cos sin ei3 2 2 3d2 = 322 = R32(r) sin2 ei2
The wavefunctions in the physicists picture are complex (they have real and
imaginary components). The wavefunctions that chemists like are pure real. So
one needs to form linear combinations of these orbitals such that these combina-
tions are pure real.
The atomic orbital you are used to from freshman chemistry are the chemists
picture of atomic orbitals
In the above table 1s, 2s, 2p0, 3d0 are pure real and so these are the same inthe chemists picture as in the physicists picture.
The table below lists the atomic orbitals in the chemists picture as linear com-
binations of the physicists picture wave functions.
50
orbital n l m wavefunctions ( = r/a0)
1s 1 0 0 1s = 1s2s 2 0 0 2s = 2s2p 2 1 0 pz = 2p0
2 1 1 2px = 122p1 + 2p1
2 1 1 2py = 1i2
2p1 2p1
3d 3 2 0 3dz2 = 3d0
3 2 1 3dxz = 123d1 + 3d1
3 2 1 3dyz = 1i2
3d1 3d1
3 2 2 3dxy = 12
3d2 + 3d2
3dx2y2 =
1i2
3d2 3d2
6.3. Spin of the electron
As we know from freshman chemistry, electrons also posses an intrinsic quantity
called spin.
Spin is actually rather peculiar so we will put o a more detailed discussion untilnext semester.
For now we must be satisfied with the following:
There are two quantum numbers associated with spin: s and ms s is the spin quantum number and for an electron s = 1/2 (always).
ms is the spin orientation quantum number and ms = 1/2 for electrons.
The spin wavefunction is a function in spin space not the usual coordinate space,
so we can not write down an explicit function of the coordinate space variables.
51
We simply denote the spin wavefunction generally as s,ms and tack it on asanother factor of the complete wavefunction.
When a particular spin state is needed a further notation is commonly used:
12, 12(the spin-up state) and 1
2,1
2(the spin-down state)
6.4. Summary: the Complete Hydrogenic Wavefunction
We are now in position to fully describe all properties of hydrogenic systems
(except for relativistic eects)
The full wave function is
n,l,m,s,ms = n,l,ms,ms (6.19)
= Rnl(r)Yl,m(, )
The energy is given by
En = Z2Rn2
, (6.20)
where recall. Again note that for a free hydrogenic system the total energy dependsonly on the principle quantum number n.
The quantum numbers of the hydrogenic system
The principle quantum number, n: determines the total energy of the sys-tems and the atomic shells.
The principle quantum number, n, can take on values of 1,2,3. . .
The angular momentum quantum numbers, l: determines the total angularmomentum of the system. It also determines the atomic sub-shells
52
The angular momentum quantum number, l, can take on values of 0,1, . . . (n 1)
For historical reasons l = 0 is called s, l = 1 is called p, l = 2 is calledd, l = 3 is called f etc.
The orientation quantum number, m: determine the projection of the an-gular momentum onto the z-axis. It also determines the orientation of theatomic sub-shells
The magnetic quantum number, m, can take on values of 0, 1, . . . l.
The spin quantum number, s: determines the total spin angular momentum.
For electrons s = 1/2.
The spin orientation quantum number, ms: determines the projection of thespin angular momentum onto the z-axis (i.e., spin-up or spin-down).
For electrons ms = 1/2
We have accomplished quite a bit. We have determined all that we can about the
hydrogen atom within Schrdingers theory of quantum mechanics.
This is not the full story however. The Schrdinger theory is a non-relativistic
one; that is, it can not account for relativistic eects which show up in spectraldata. We also had to add spin in an ad hoc manner to account for what we know
experimentallyspin did not fall out of the theory naturally.
Dirac, in the late 1920s, developed a relativistic quantum theory in which the
well established phenomenon of spin arose naturally. His theory also made the
53
bold prediction of the existence anti-matter that has now been verified time and
again.
The Dirac theory was still not fully complete, because there still existed exper-
imental phenomena that was not properly described. In 1948 Richard Feynman
developed the beginnings of quantum electrodynamics (QED). QED is the best
theory ever developed in terms of matching with experimental data.
Both the relativistic Dirac theory and QED are beyond our reach, so we limit
ourselves to the non-relativistic Schrdinger theory.
54
7. Multi-electron atoms
7.1. Two Electron Atoms: Helium
We now consider a system consisting of two electrons and a nucleus; for example,
helium.
Although the extension from hydrogen to helium seems simple it is actually ex-
tremely complicated. In fact, it is so complicated that it cant be solved exactly.
The helium atom is an example of the three-body-problemdicult to handleeven in classical mechanicsone can not get a closed form solution.
The Hamiltonian for helium is
H = ~2
2me21| {z }
K.E of electron 1
~2
2me22| {z }
K.E of electron 2
Ze2
40r1| {z }P.E of electron 1
Ze2
40r2| {z }P.E of eletcron 2
+e2
40r12| {z }elec.elec. repulsion
, (7.1)
where r12 = |r1 r2| is the distance between the electrons.
The electronelectron repulsion term is responsible for the diculty of the prob-lem. It makes a closed form solution impossible.
The problem must be solved by one of the following methods
Numerical solutions (we will not discuss this)
55
55
Perturbation theory (next semester)
Variational theory (next semester)
Ignore the electronelectron repulsion (good for qualitative work only)
7.2. The Pauli Exclusion Principle
Electron are fundamentally indistinguishable. They can not truly be la-belled.
All physical properties of a system where we have labelled the electrons as, say, 1
and 2 must be exactly the same as when the electrons are labelled 2 and 1.
Now, only ||2 is directly measurablenot itself.
All this implies that
(1, 2) =
+(2, 1), symmetricor
(2, 1) antisymmetric(7.2)
The Pauli exclusion principle states: The total wavefunctions for fermions(e.g., electrons) must be antisymmetric under the exchange of indistinguishable
fermions.
Note: a similar statement exists for bosons (e.g., photons): The total wavefunction
for bosons must be symmetric under exchange of indistinguishable bosons.
Let us consider the two electron atom, helium
56
The total wavefunction is
= (1, 2)(1, 2) (7.3)
Since a complete solution for helium is not possible we must use approximate
wavefunctions. Since we are doing this, we may as well simplify matters and use
product state wavefunctions (products of the hydrogenic wavefunctions).
= (1)(2)| {z }spatial part
(1)(2)| {z }spin part
, (7.4)
where the single particle wavefunctions are that of the hydrogenic system.
The Pauli exclusion principle implies that if the spatial part is even with respect
to exchange then the spin part must be odd. Likewise if the spatial part is odd
then the spin part must be even.
Now lets blindly list all possibilities for the ground state wave function of helium
a = 1s(1)(1)1s(2)(2) (7.5)
b = 1s(1)(1)1s(2)(2)
c = 1s(1)(1)1s(2)(2)
d = 1s(1)(1)1s(2)(2)
These appear to be four reasonable ground state wavefunctions which would im-
ply a four-fold degeneracy. However considering the symmetry with respect to
exchange we see the following
a has symmetric spatial and spin parts and is there for symmetric. It mustbe excluded.
Similarly for d.
b and c have symmetric spatial parts, but the spin part is neither sym-metric or antisymmetric. So, one must make an antisymmetric linear com-
bination of the spin parts.
57
The appropriate linear combination is
(1)(2) (2)(1). (7.6)
So the ground state wave function for helium is
g = 1s(1)1s(2) [(1)(2) (2)(1)] . (7.7)
Consequences of the Pauli exclusion principle
No two electrons can have the same five quantum numbers
Electrons occupying that same subshell must have opposite spins
7.3. Many Electron Atoms
The remaining atoms on the periodic table are handled in a manner similar to
helium.
Namely the wavefunction is product state that must be antisymmeterized in ac-
cordance with the Pauli exclusion principle.
The product wavefunction for the ground state is determined by applying the
aufbau principle. The aufbau principle states that the ground state wavefunction
is built-up of hydrogenic wavefunctions
To arrive at an antisymmetric wavefunction we construct the Slater determinant :
=
1s(1)(1) 1s(1)(1) n(1)(1) n(1)(1)1s(2)(2) 1s(2)(2) n(2)(2) n(2)(2)
......
......
...
1s(N)(N) 1s(N)(N) n(N)(N) n(N)(N)
(7.8)
58
The reason one can be sure that this wavefunction is the antisymmeterized is that
we know from linear algebra that the determinant is antisymmetric under exchange
of rows (corresponds to exchanging two electrons). It is also antisymmetric under
exchange of columns.
Another property of the determinant is that if two rows are the same (corresponds
to two electrons in the same state) the determinant is zero. This agrees with the
Puli exclusion principle.
As an example consider lithium:
There are three electrons so we need three hydrogenic wavefunctions: 1s,1s, and 2s (or 2s).
We construct the Slater determinant as
1 =
1s(1)(1) 1s(1)(1) 2s(1)(1)1s(2)(2) 1s(2)(2) 2s(2)(2)1s(3)(3) 1s(3)(3) 2s(3)(3)
(7.9)
or
2 =
1s(1)(1) 1s(1)(1) 2s(1)(1)1s(2)(2) 1s(2)(2) 2s(2)(2)1s(3)(3) 1s(3)(3) 2s(3)(3)
(7.10)
The short hand notation for these states is (1s)2(2s)1
7.3.1. The Total Hamiltonian
The total Hamiltonian for a many electron (ignoring spin-orbit coupling which
will be discussed next semester) atom is
H =NXi=1
"~22me
2i Ze2
40ri+Xj>i
e2
40rij
#(7.11)
59
8. Diatomic Molecules and the Born
Oppenheimer Approximation
Now that we have applied quantum mechanics to atoms, we are able to begin the
discussion of molecules.
This chapter will be limited to diatomic molecules.
8.1. Molecular Energy
A diatomic molecule with n electrons requires that 3n+6 coordinates be specified.
Three of these describe the center of mass position.
3n of these describe the position of the n electrons.
This leaves three degrees of freedom (R, , ) which describe the position of thenuclei relative to the center of mass. R determines the internuclear separationand and determine the orientation.
60
60
8.1.1. The Hamiltonian
In the center of mass coordinates the Hamiltonian for a diatomic molecule is
H = TN + Te + VNN + VNe + Vee. (8.1)
TN is the nuclear kinetic energy operator and is given by
TN = ~2
22N =
~2
2R2R
R2R
+~2
2J2, (8.2)
where J is angular momentum operator for molecular rotation and = m1m2m1+m2 isthe reduced mass of the diatomic molecule.
Te =P
i ~2
2me2ei is the kinetic energy operator for the electrons.
VNN = ZAZBee2
40Ris the nuclearnuclear potential energy operator.
VNe = P
i
hZAe240rAi
+ ZBe2
40rBi
iis the nuclearelectron potential energy operator.
Vee =P
i>je2
40rjiis the electronelectron potential energy operator.
61
8.1.2. The BornOppenheimer Approximation
The BornOppenheimer approximation: The nuclei move much slower thanthe electrons. (classical picture)
We put the BornOppenheimer approximation to work by first defining an eec-tive Hamiltonian
Heff = Te + VNN + VNe + Vee. (8.3)
The approximation comes in by treating R as a parameter rather than an operator(or variable). So one writes
Heffe(R, {ri}) = Ee(R)e(R, {ri}). (8.4)
e is the so-called electronic wavefunction.
Now the Schrdinger equation for the diatomic molecule isTN + Heff
(R, {ri}) = E(R, {ri}). (8.5)
Since the Hamiltonian is a sum of two terms, one can write the wavefunction
(R, {ri}) as a product wavefunction
= Ne, (8.6)
where N is the so-called nuclear wavefunction.
Substituting the product wavefunction into the Schrdinger equation givesTN + Heff
Ne = ENe (8.7)
TN +Ee(R)Ne/ = ENe/
TN +Ee(R)N = EN .
62
The last equation is exactly like a Schrdinger equation with a potential equal to
Ee(R).
One now models Ee(R) or determines it experimentally.
8.2. Molecular Vibrations
As stated earlier R is the internuclear separation and and determine theorientation. Consequently, R is the variable involved with vibration whereas and are involved with rotation.
Considering only the R part of the Hamiltonian (under the BornOppenheimerapproximation), we have
~
2
22
R2+Ee(R)
vib = Evibvib. (8.8)
It is convenient at this point to expand Ee(R) in a Taylor series about the equi-librium position, Req:
Ee(R) = E0 +ER
Req
(RReq) +1
2!
2ER2
Req
(RReq)2 + . (8.9)
Now E0 is just a constant which, by choice of the zero of energy, can be set to anarbitrary value.
Since we are at a minimum,ER
Req
must be zero, so the linear term vanishes.
One defines2ER2
Req
ke as the force constant.
The remaining terms in the expansion can collective be defined as O[(RReq)3] Vanh, the anharmonic potential.
63
As a first approximation we can neglect the anharmonicity. With this, the Schrdinger
equation becomes ~
2
22
R2+1
2ke(RReq)2
vib = Evibvib. (8.10)
If we let x = (RReq) this becomes ~
2
22
x2+1
2kex2
vib = Evibvib, (8.11)
which is exactly the harmonic oscillator equation. Hence
vib,n = AnHn(x)ex
2/2, (8.12)
where q
ke~ .
And
Evib,n = hce(n+1
2), (8.13)
where e 12q
ke .
8.2.1. The Morse Oscillator
Neglecting anharmonicity and using the harmonic oscillator approximation works
well for low energies. However, it is a poor model for high energies.
For high energies we need a more realistic potentialone that will allow of bond
dissociation.
The Morse potential
Ee(R) = De[1 e(RReq )]2, (8.14)
64
where De is the well depth and = 2ceq
2De
is the Morse parameter. Note:
this expression for the Morse potential has the zero of energy at the bottom of
the well (i.e. R = Req, ;Ee(Req) = 0).
The Morse Potential can also be written as
Ee(R) = De[e2(RReq ) 2e(RReq )]. (8.15)
Now the zero of energy is the dissociated state (i.e. R, ;Ee(R) = 0).
We approach this quantum mechanical problem exactly like all the other.
The Schrdinger equation is ~
2
22
R2+De[1 e(RReq )]2
vib = Evibvib (8.16)
This is another dierential equation that is dicult to solve.
As it turns out, this Schrdinger equation can be transformed into a one of a broad
class of known dierential equations called confluent hypergeometric equationsthe solutions of which are the confluent hypergeometric functions, 1F1.
Doing this yields the wavefunctions of the form
vib,n(z) = zApnez1F1(n, 1 + 2Apn, 2z), (8.17)
z =2Deh
ex,
A =2h
,
pn =pDe +
12 nA
and energy levels of the form
Evib,n = De + hce(n+1
2) hcexe(n+
1
2)2, (8.18)
65
where exe together is the anharmonicity constant, with xe = hce4De .
See Handout
8.2.2. Vibrational Spectroscopy
Infrared (IR) and Raman spectroscopy are the two most widely used techniques
to probe vibrational levels.
The spectral peaks appear at v = 4Ehc (in units of wavenumbers, cm1).
The transition from the n = 0 to the n = 1 state is called the fundamentaltransition.
Transitions from n = 0 to n = 2, 3, 4 are called overtone transitions.
Transitions from n = 1 to 2, 3, 4 , n = 2 to 3, 4, 5 , etc. are called hottransitions (or hot bands)
Since the energy levels depend on mass, isotopes will have a dierent transitionenergy and hence appear in a dierent place in the spectrum. Heavier isotopeshave lower transition energies.
66
9. Molecular Orbital Theory and
Symmetry
9.1. Molecular Orbital Theory
One of the most important concepts in all of chemistry is the chemical bond.
In freshman chemistry we learn of one model for chemical bondingVSEPR (va-
lence shell electron-pair repulsion) theory, where hybridized atomic orbitals deter-
mine the bonding geometry of a given molecule.
We are now prepared to discuss a bonding theory that is more rigorously based
in quantum mechanics.
Basically we will treat the molecules in the same way as all our other quantum
mechanical problems (e.g., particle in a box, harmonic oscillator, etc.)
As you might expect, it is not possible to obtain the exact wavefunctions and
energy levels so, we must settle for approximate solutions.
As a first example, let us consider the molecular hydrogen ion H+2 .
The Hamiltonianfor H+2 is
H = TN + Tel + VNel + VNN (9.1)
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We use the Born-Oppenheimer approximation and treat the nuclear coordinates
as a parameters rather than as variables. So we only worry about parts of the
Hamiltonian that deal with the electron.
The eective Hamiltonian becomes
H = Tel + VNel (9.2)
=~22me
2 e2
40rA e
2
40rB.
The eigenfunctions of this Hamiltonian are called molecular orbitals.
The molecular orbitals are the analogues of the atomic orbitals.
Atomic orbitals: Hydrogen is the prototype and all other atomic orbitalsare built from the hydrogen atomic orbitals.
Molecular orbitals: The hydrogen molecular ion is the prototype and allother molecular orbitals are built from the hydrogen molecular ion molecular
orbitals.
There is one significant dierence between the above, which is the hydrogen atomicorbitals are exact whereas the hydrogen molecular ion molecular orbitals are not
exact.
In fact, we shall see that these molecular orbitals are constructed as linear com-
binations of atomic orbitals.
9.2. Symmetry
Let the atoms of the hydrogen molecular ion lie on the z-axis of the center of masscoordinate system.
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Inversion symmetry
The potential field of the hydrogen molecular ion is cylindrically symmetricabout the z-axis.
Because of the symmetry the electron density at (x, y, z) must equal theelectron density at (x,y,z).
The above symmetry therefore requires that the molecular orbitals be eigen-functions of the inversion operator, . That is
(x, y, z) = (x,y,z) = a(x, y, z). (9.3)
Moreover the eigenvalue a can be either +1 or 1.
If a = +1 the molecular wavefunction is even with respect to inversion andis called gerade and labelled with a g: g = g
If a = 1 the molecular wavefunction is odd with respect to inversion andis called ungerade and labelled with a u: u = u
The terms gerade and ungerade apply only to systems that posses inversionsymmetry.
Cylindrical symmetry
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The cylindrical symmetry implies that the potential energy can not dependon the .
The molecular wavefunction is described by an eigenvalue = 0,1,2, . . .
We use to label the molecular orbitals as shown in the table
0 1 2 label
Mirror plane symmetry
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There is also a symmetry about the x-y plane called horizontal mirror planesymmetry: operator h.
Thus the molecular wavefunction must be an eigenfunction of h with eigen-value 1.
If the eigenvalue is +1 (even with respect to h) the molecular orbitalis called a bonding orbital.
If the eigenvalue is 1 (odd with respect to h) the molecular orbitalis called an antibonding orbital.
There are also vertical mirror plane symmetries, but we will put that dis-cussion o for the time being.
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10. Molecular Orbital Diagrams
10.1. LCAOLinear Combinations of Atomic Orbitals
Now that we know what symmetry the molecular orbitals must posses, we need
to find some useful approximations for them.
Useful can mean qualitatively useful or quantitatively useful.
Unfortunately we cant have both.
We will discuss the approximation which models the molecular orbitals as linear
combinations of atomic orbitals (LCAO).
LCAO is qualitatively very useful but it lacks quantitative precision.
Let us again consider the hydrogen molecular ion H+2 : let one H atom be labelled
A and the other labelled B.
Linear combination of the 1s atomic orbital from each H atom is used for themolecular orbital of H+2 :
(1sA) = kerA/a0 (10.1)
and
(1sB) = kerB/a0 (10.2)
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We construct two molecular orbitals as
+ = C+(1sA + 1sB) (10.3)
and
= C(1sA 1sB) (10.4)
The normalization condition is Zd = 1 (10.5)
As can be seen from the above figure, + represents a situation in which theelectron density is concentrated between the nuclei and thus represents a bonding
orbital.
Conversely represents a situation in which the electron density is very lowbetween the nuclei and thus represents an antibonding orbital
10.1.1. Classification of Molecular Orbitals
With atoms we classified atomic orbitals according to angular momentum.
For molecular orbitals we shall also classify them according to angular momentum.
But we shall also classify them according to their inversion symmetry and wether
or not they are bonding or antibonding.
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The classification according to angular momentum is as follows.
0 1 2 orbital symbol
Atomic orbitals with m = 0 form type molecular orbitals, e.g., s , pz .
Those with m = 1 form type molecular orbitals, e.g., px etc.
The classification according to inversion symmetry is simply a subscript g oru. For example, g or u etc.
The classification according to bonding or antibonding is an asterisk is used to
denote antibonding. For example, g is a bonding orbital and u is an antibondingorbital.
10.2. The Hydrogen Molecule
Let us now consider the hydrogen molecule. This molecules is a homonuclear
diatomic with two electrons.
If the two atoms are infinitely far apart. The ground state of the system would
consist of two separate hydrogen molecules in their ground atomic states: (1s)1
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As the atom are brought closer together, their respective s orbitals begin to over-lap.
It is now more appropriate to speak in terms of molecular orbitals, so one forms
linear combinations of the atomic orbitals.
There are two acceptable linear combinations. These are
g = 1sA + 1sB (10.6)
and
u = 1sA 1sB. (10.7)
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It can be shown mathematically that the energy level associated with g is lowerthan u.
We can intuit this qualitatively however since the u orbital must have a nodewhereas the g does not.
It is also to be expected since we know H2 is a stable molecule.
10.3. Molecular Orbital Diagrams
The energy levels associated with the molecular orbitals are drawn schematically
is what is called a molecular orbital diagram.
The molecular orbital diagram for H2 is shown below
Molecular orbital diagrams can be drawn for any molecule. Some get very compli-
cated. We will focus on the second row homonuclear diatomics and some simple
heteronuclear diatomics.
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The molecular orbital diagrams for the second row homonuclear diatomics are
rather simple.
See Supplement
The supplement that follows this section contains examples for each of
