quantum physics - uni-muenster.de · 1 matter waves 1.1 de broglie’s matter waves in 1924 de...

R. Friedrich

Quantum Physics

July 14, 2010

Springer

Contents

1 Matter Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 De Broglie’s Matter Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Matter Waves: Experiments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2.1 Davisson-Germer Experiment . . . . . . . . . . . . . . . . . . . . . . . 21.2.2 Diffraction Experiment of Mollenstedt and Duker . . . . . . 41.2.3 Double Slit Experiment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.2.4 Tonomura . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.3 Interpretation of the Wave Function . . . . . . . . . . . . . . . . . . . . . . . 41.4 Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.4.1 Gaussian Wave Packet . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.4.2 Initial Condition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.5 Particle in a Box . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2 Schrodinger Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.1 Free particle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.2 Free Particle in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . 162.3 Particle in a Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.3.1 Linear Superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.4 Wave Function: Statistical Interpretation . . . . . . . . . . . . . . . . . . . 17

2.4.1 Probability Density, Current: 1 D . . . . . . . . . . . . . . . . . . . 172.4.2 Probability Density, Current: 3D . . . . . . . . . . . . . . . . . . . . 18

2.5 Expectations, Operators, Brackets . . . . . . . . . . . . . . . . . . . . . . . . . 192.5.1 Expectation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 192.5.2 Momentum Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.5.3 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.5.4 Hamilton Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.5.5 Brackets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202.5.6 Expectations of Operators . . . . . . . . . . . . . . . . . . . . . . . . . . 21

2.6 Quantization of Classical Mechanics . . . . . . . . . . . . . . . . . . . . . . . 212.6.1 Classical Mechanics: Hamilton Formulation . . . . . . . . . . . 212.6.2 Quantization, Jordan Rules . . . . . . . . . . . . . . . . . . . . . . . . . 21

VI Contents

3 Wave Mechanics in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . 233.1 Time Independent Schrodinger Equation . . . . . . . . . . . . . . . . . . . 233.2 Free Particle on a Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.2.1 Expectation of Position . . . . . . . . . . . . . . . . . . . . . . . . . . . . 253.2.2 Expectation of Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 263.2.3 Hamilton Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.3 Free Particle on the Line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263.4 Infinitely Deep Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273.5 Particle in a Finite Potential Well . . . . . . . . . . . . . . . . . . . . . . . . . 28

3.5.1 Bounded States . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293.5.2 Solution with even parity: . . . . . . . . . . . . . . . . . . . . . . . . . . 303.5.3 Solution with odd parity: . . . . . . . . . . . . . . . . . . . . . . . . . . . 323.5.4 Continuous Spectrum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

3.6 Reflection at a Barrier: Tunnel Effect . . . . . . . . . . . . . . . . . . . . . . 353.6.1 E < V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353.6.2 E > V . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.6.3 Wave Packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.7 Wave Mechanics in Two Dimensions . . . . . . . . . . . . . . . . . . . . . . . 383.7.1 Quantum Dots . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383.7.2 Quantum Billards . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

4 Hilbert Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.1 Linear Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394.2 Normed Linear Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

4.2.1 Scalar Product . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.2.2 Schwarz Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 404.2.3 Orthogonal Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.2.4 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.3 Hilbert Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.4 Basis, Dimension, Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

4.5.1 Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 424.5.2 Space of periodic functions . . . . . . . . . . . . . . . . . . . . . . . . . 434.5.3 Square Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . 444.5.4 Square Integrable Functions in N Dimensions . . . . . . . . . 44

4.6 Linear Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.6.1 Example: Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . 454.6.2 Example: Complex Space . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.6.3 Example: Position and Momentum Operator . . . . . . . . . . 45

4.7 Adjoint Operator, Hermitian Operator . . . . . . . . . . . . . . . . . . . . . 454.7.1 Momentum Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.7.2 Operator of Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . 464.7.3 Hamilton Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.8 Eigenvalue Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 464.8.1 Example: Euclidean Space . . . . . . . . . . . . . . . . . . . . . . . . . . 46

Contents VII

4.8.2 Example: Momentum Operator . . . . . . . . . . . . . . . . . . . . . 474.9 Hermitian Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.9.1 Projection Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 474.9.2 Spectral Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

4.10 Functions of Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.11 Commuting, Noncommuting Operators . . . . . . . . . . . . . . . . . . . . . 48

4.11.1 Position, Momentum. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.11.2 Definition: Commutator . . . . . . . . . . . . . . . . . . . . . . . . . . . . 484.11.3 Eigenvectors of Commuting Operators . . . . . . . . . . . . . . . 494.11.4 Functions of Commuting Operators . . . . . . . . . . . . . . . . . . 49

4.12 Uncertainty Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.12.1 Uncertainty Relation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5 Chance and Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.1 Probability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.2 Probability Density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 515.3 Expectations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 525.4 Mathematical Definition of Probability . . . . . . . . . . . . . . . . . . . . . 52

6 Concepts of Quantum Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.1 Axiomatic Quantum Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 536.2 Physical Observables and Hermitian Operators . . . . . . . . . . . . . . 546.3 Probabilistic Interpretation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 546.4 Measurement Process . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

7 Harmonic Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577.1 Classical Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.1.1 Hamilton Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577.1.2 Phase Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 577.1.3 Bohr-Sommerfeld Quantization . . . . . . . . . . . . . . . . . . . . . . 58

7.2 Quantum Mechanical Harmonic Oscillator . . . . . . . . . . . . . . . . . . 587.2.1 Eigenvalue Problem: Hamilton Operator . . . . . . . . . . . . . 597.2.2 Hamilton Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

7.3 Sommerfeld’s Polynom Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.3.1 Ground State . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 607.3.2 Higher Eigenfunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 617.3.3 Properties of Hermite Polynomials . . . . . . . . . . . . . . . . . . . 637.3.4 Spectrum and Eigenfunctions . . . . . . . . . . . . . . . . . . . . . . . 64

7.4 Algebraic Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 647.4.1 Creation, Annihilation Operators . . . . . . . . . . . . . . . . . . . . 647.4.2 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 657.4.3 The Ladder . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 667.4.4 Positivity of Energy Eigenstates . . . . . . . . . . . . . . . . . . . . . 677.4.5 Occupation Number Operator . . . . . . . . . . . . . . . . . . . . . . . 687.4.6 Normalization of Eigenstates . . . . . . . . . . . . . . . . . . . . . . . . 68

VIII Contents

7.4.7 Uncertainty Relationships . . . . . . . . . . . . . . . . . . . . . . . . . . 697.5 Coupled Harmonic Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

7.5.1 Harmonic Oscillator in 3 Dimensions . . . . . . . . . . . . . . . . . 707.5.2 Aplications: Molecular Oscillations . . . . . . . . . . . . . . . . . . 72

8 Hydrogen Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 758.1 Classical Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

8.1.1 Center of Mass, Relative Motion . . . . . . . . . . . . . . . . . . . . 758.1.2 Conservation Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 76

8.2 Quantization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 768.2.1 Separation of Relative and Center of Mass Motion . . . . . 778.2.2 Time Independent Schrodinger Equation . . . . . . . . . . . . . 77

8.3 Hamilton Operator in Spherical Coordinates . . . . . . . . . . . . . . . . 778.3.1 Gradient and Momentum Operator: Spherical

Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 788.3.2 Separation Ansatz . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79

8.4 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808.4.1 Eigenvalue Problem. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 808.4.2 Algebraic Treatment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 828.4.3 Eigenvalue Problem connected with L2, Lz . . . . . . . . . . . 838.4.4 Commutation relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

8.5 Radial Eigenfunctions: Hydrogen Atom. . . . . . . . . . . . . . . . . . . . . 888.5.1 Asymptotic behaviour ρ→ ∞ . . . . . . . . . . . . . . . . . . . . . . . 898.5.2 Sommerfeld’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 898.5.3 Associate Laguerre Polynomials . . . . . . . . . . . . . . . . . . . . . 918.5.4 Summary: Energy Levels and Wave Functions . . . . . . . . . 928.5.5 Associate Laguerre Functions . . . . . . . . . . . . . . . . . . . . . . . 928.5.6 Symmetry and the Spectrum of the Hydrogen Atom . . . 928.5.7 Runge-Lenz-Vector . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

8.6 Alkali-Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 938.7 Appendix: Angular Momentum Operators and Spherical

Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

9 Diatomic Molecule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

10 Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10110.1 Introductory Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10110.2 Perturbation Theory: Nondegenerate Case . . . . . . . . . . . . . . . . . . 10210.3 Perturbation Theory for Degenerate States . . . . . . . . . . . . . . . . . 105

11 Atoms in Electric Field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10711.1 Application of Perturbation Theory . . . . . . . . . . . . . . . . . . . . . . . . 10711.2 Stark Effect of the Ground State n=1,l=0 . . . . . . . . . . . . . . . . . . 108

11.2.1 Induced Dipole Moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10911.3 Linear Stark Effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

1

Matter Waves

1.1 De Broglie’s Matter Waves

In 1924 de Broglie postulated the existence of waves related with matter. Heassumed that the momentum of a free particle is related to the wavelenght λ,or the wave number k = 2π

λ of the wave, according to

p =h

λ= hk (1.1)

The energy of a free particle in classical mechanics is given by

E =p2

2m=

h2

2mλ2

(hk)2

2m(1.2)

Light Matter

E = hν E = hν = (hk)2

2m

p = hνc p = h

λ= hk

Let us calculate the de’Broglie wavelength for some particles:

a) Electrons, accelerated by an electric field

2 1 Matter Waves

p2

2m= eU λ =

h√2eUme

=1, 226nm√

U/V(1.3)

U=1000 V λ = 3.910−11mU=100 V λ = 1.210−10m

b) Macroscopic particles:

m = 10−6g, v = 10m/secλ = 6.610−26m (1.4)

1921 Ramsauer Scattering of slow electronswith gas atoms

1924 De Broglie Postulation of the existenceof matter waves

1926 E. Schrodinger Formulation of theSchrodinger equation

1927 Davisson and Germer Interference patterns inthe intensity of electronsreflected from a crytsal surface

1927 W. Heisenberg Formulation of theuncertainty relation

1931 Stern, Frisch, Estermann Interference Experimentswith Helium atoms

1956 Mollenstedt Fresnel double slit experimentwith electrons

1.2 Matter Waves: Experiments

1.2.1 Davisson-Germer Experiment

The Davisson-Germer Experiment directly verifies the hypotheses of de Broglie’sconcerning matter waves. It uses a beam of electrons, i.e. objects with a certain

1.2 Matter Waves: Experiments 3

energy E and momentum p, in a typical diffraction experiment. The experi-mental arrangement is the one used for Bragg reflection using x-ray’s. First,one investigates whether a kind of Bragg reflection pattern is observed. Fromthe Bragg condition,

λ = 2dsinθ (1.5)

one determines the wavelength of the reflection pattern.The outcomes of the experiments are:

• One detects the typical interference patterns of Bragg reflection.• The Bragg condition for constructive interference allows one to determine

the wave length λ.• The measurement of the momentum of the electrons allows one to calculate

the de Broglie’s wave length λ.• This wavelenght coincides with the one obtained form the hypothesis of

de Broglie.

The Davisson-Germer experiment directly adresses the hypothesis of deBroglie. The electrons are accelerated in a electric field with voltage differenceU. It gains the kinetic energy

E =p2

2m= eU (1.6)

This relationship allows one to calculate the momentum

p =√

2meU (1.7)

According to de Broglie’s hypothesis, this should correspond to a wavelenth

λ =h

p(1.8)

This wavelenght should be related to the scattering angle, in the same wayas it would be for the case of x-ray-scattering. If we denote the lattice spacingby d constructive interference is obtained for the angles

λ = 2ndsinθ (1.9)

where n denotes the various interference maxima. As a consequence, there is adirect relationship between the angle θ and the kinetic energy of the electrons:

sin θ =1

2nd

h

p=

1

2nd

h√2emU

(1.10)

From x-ray scattering experiments using nickel crystals the lattice spacingof

d = 0.091nm (1.11)

is obtained. The potential difference in the experiment was

4 1 Matter Waves

U = 54V (1.12)

This yields a de Broglie wavelength of

λ = 0.165nm (1.13)

The first maximum then, should be, close to 50 degree. This is actually ob-served in the experiment: The Davisson-Germer experiment is a direct verifi-cation of the hypothesis of de Broglie.

C. Davisson, L. H. Germer; Reflections of a electrons by a crystal of Nickel,Nature 119, 558-560 (1927)

1.2.2 Diffraction Experiment of Mollenstedt and Duker

The Fresnel double prism experiment is a fundamental experiment in classicaloptics. The analog of the experiment carried out with electrons has been per-formed by Mollenstedt and Duker. The experimental arrangement is depictedin figure . The double prism has been realized for electrons by a quartz fibrelocated perpendicular to the electron beam. A voltage is induced between thefibre and the counterelectrodes, which depends on the current flowing throughthe voltage. The electrons are deflected by the inhomogenous electrical field.

1.2.3 Double Slit Experiment

C. Jonsson, Zeitschrift fr Physik 161, 454-474 (1961)

1.2.4 Tonomura

One can imagine that one lowers the intensity of the electron beam in adouble slit experiment or another diffraction experiment so that, in fact, onlya single electron passes the slits. This electron hits the screen at a certainpoint reflecting its pointlike behaviour. Such an experiment actually has beenperformed by Tonomura et al. Their results are exhibited in fig. (1.2.4). Thefirst hits of electrons on the screen lead to a kind of random appearanceof the hits. With increasing number of impinging electrons the well-knowninterference patterns emerge. This point to the fact the wave function containsinformation, which has to be interpreted in a statistical sense.

A. Tonomura, J. Enod, T. Matsuda, T. Kawasaki, and H. Exawa, Demon-stration of single-elctron biuldup of an interference pattern,Amer. J. Phys.57, 117 (1989)

1.3 Interpretation of the Wave Function

De Broglie has assigned a wave function to each particle. This leads us to theproblem of the interpretation of the wave function Ψ(x, t).

1.3 Interpretation of the Wave Function 5

Interference pattern in the experiment of Mollenstedt and Duker.

Interference pattern in the experiment of Mollenstedt and Duker.

6 1 Matter Waves

Interference pattern in the double slit exper-iment of Jonsson.

Emergence of interference pattern in thedouble slit experiment of Tanamura.

1.4 Wave Packets 7

If we reconsider the double slit experiment of Tonomura, we recall that theinterference patterns in the double slit experiment only develops gradually ifa whole ensemble of electrons has passed the slits. For a single event, it isnot possible to predict where the particle hits the counter. This introducesa probabilistic element: We relate the intensity of the wave, |ψ(x, t)|2, to theemerging interference pattern. To this end we interpret

|Ψ(x, t)|2dx (1.14)

as the probability to find the particle in the interval [x, x+ dx].Based on this interpretation we can define the expectation of the position

〈x(t)〉 of the particles. This quantity is defined according to

〈x(t)〉 =

∫

x|Ψ(x, t)|2dx (1.15)

Furthermore, we can determine higher order moments of the position,

〈x(t)n〉 =

∫

xn|Ψ(x, t)|2dx (1.16)

1.4 Wave Packets

We shall now consider the wave function, which according to de’Broglie, isassigned to a free particle. To this end we start with a monochromatic wave,i.e. a wave characterized by a single frequency and wavelength, which can berepresented by the complex function Ψ(x, t)

Ψ(x, t) = aei(kx−ωt) (1.17)

If we now take the expressions for matter waves due to de Broglie,

E = hω(k) =h2k2

2mp = hk , (1.18)

we obtainΨ(x, t) = ae

ih

(px−Et) = aei(kx−ω(k)t) (1.19)

However, such a wave is an infinitely extended object. In order to deal withlocalized particles one has to consider wave packets.

Wave packets are linear superpositions of monochromatic waves:

Ψ(x, t) =

∫

dka(k)ei(kx−ω(k)t) (1.20)

We consider the following example of a wave packet:

8 1 Matter Waves

a(k) = a k0 −∆k < k < k0 +∆k0 otherwise

(1.21)

For sufficiently small values of ∆k we can aproximate ω(k) by a Taylorseries:

ω(k) = ω(k0) + (k − k0)ω′(k0) + .... (1.22)

Based on this approximation we evaluate the integral and obtain

Ψ(x, t) = aei(k0x−ω(k0)t)

∫ ∆k

−∆k

dk′ei(k′(x−ω′(k0)t)

= aei(k0x−ω(k0)t)sin∆k(x− ω′(k0)t)

(x − ω′(k0)t)(1.23)

The wave is a product of a carrier wave

aei(k0x−ω(k0)t) (1.24)

as well as the enveloppesin∆k(x− ω′(k0)t)

(x− ω′(k0)t)(1.25)

The maximum of the envelope moves with the group velocity

x = V t = ω′(k0)t (1.26)

For matter waves we can explicitly use the relationship ω(k0) =hk2

0

2m andconclude that the velocity V is just

V =hk0

m=

p

m(1.27)

The width of the moving envelope can be defined as the distance betweenthe two nearest zeros of the envelope, (1.25). It is given by ∆x∆k = 2π or,using the relationship ∆p = h∆k, by

∆x∆p = h (1.28)

If we interpret ∆x as the uncertainty of the location of the particle and ∆pas the uncertainty to determine the momentum of the particle, we encounterthrough equation (??) an example of the Heisenberg uncertainty relation. Thisrelationship demonstrates that it is not possible to sharply localize the particle∆x→ 0 and simultaneously have a sharply defined momentum ∆p→ 0.

1.4.1 Gaussian Wave Packet

We investigate the Gaussian wave packet. This packet is defined by a Gaussiandistribution of the amplitudes:

1.4 Wave Packets 9

Wave packet

Ψ(x, t) =

∫

dke−

k2

2Q

√2πQ

ei(kx−h k2

2mt (1.29)

We can easily evaluate this integral, since

Ψ(x, t) =

∫

dkeikx e−[ 1

2Q+i h

2mt]

√2πQ

(1.30)

is the Fourier-transform of a Gaussian integral, and as a consequence is alsoGaussian.

As can be seen from fig. an initially localized wave function spreads in thecourse of time.

1.4.2 Initial Condition

The amplitude a(k) of a wave packet is determined by the wave packet atinitial time t = 0:

Ψ(x, 0) =

∫

dka(k)eikx (1.31)

This relation shows that the initial wave packet ψ(x, 0) and the amplitudea(k) are related by Fourier transform. The inverse transform yields

10 1 Matter Waves

Gaussian wave packet

a(k) =1

2π

∫

dxψ(x, 0)e−ikx (1.32)

We explicitly see, that the wave function at time t is related to the functionat time t = 0, since

Ψ(x, t) =

∫

dk1

2π

∫

dx′Ψ(x′, 0)eik[(x−x′)−ω(k)t] (1.33)

1.5 Particle in a Box

In this section, we shall treat the wave mechanics of a particle moving in afinite box. This means, that the regions x < 0 and x > L are not allowed.Mathematically, we have to assume that the wave function of the particle hasto become zero at x = 0 and x = L. This problem can be seen as a crudemodel of an electron moving in the potential of a nucleus.

We start from the de’Broglie’s wave

Ψ(x, t) = aei(kx−ω(k)t) (1.34)

In order to take into account the barriers at x = 0, x = L we make a superpo-sition of a wave with positive momentum, p = hk, and negative momentum,

1.5 Particle in a Box 11

Evolution of Gaussian wave packet.

p = −hk,:Ψ(x, t) = a1e

i(kx−ω(k)t) + a2ei(−kx−ω(−k)t) (1.35)

Since

ω(−k) =h(−k)2

2m= ω(k) (1.36)

We can combine this superposition into

Ψ(x, t) = [csin(kx) + dcos(kx)] e−iω(k)t (1.37)

We have to fullfill the boundary conditions

Ψ(0, t) = 0 ; Ψ(L, t) = 0 (1.38)

The first condition leads us to d=0. The second condition leads us to a re-striction of momentum, i.e. the wave number k:

kL = πn n = 1, 2, .... (1.39)

12 1 Matter Waves

Energy Eigenstates and Wave Functionsfor a Particle in a Potential Well.

Quantization of Energy, Momentum

We obtain the following family of wave functions

Ψn(x, t) = cnsinπn

Lxe−i hπ2n2

L2 t (1.40)

Since the number n is allowed to take the values n = 1, ..,∞ there is an infinitenumber of such functions, each characterized by a disrete set of momenta, andenergies:

pn = hkn = hn

L

En =h2n2

2mL2

ΨN (x, t) = cNsinπn

Lxe−i En

ht = ψN (x)e−i En

ht

ψN (x) =

√

2

Lsin

πn

Lx (1.41)

We have normalized the functions ψN (x) in such a way that

∫ L

0

dxψN (x)2 =

∫ L

0

dxψN (x)2 =2

L

∫ L

0

dxsin2πn

Lx (1.42)

1.5 Particle in a Box 13

Summarizing, we see that the momentum hk as well as the energy E =h2k2

2m is not allowed to take any value. Due to the boundaries, the energies andmomenta of the wave functions (1.41 are restricted to an infinite set of disretevalues: Energy and momentum of a particle in a box are quantized. The wavefunctions differ in the number of knots in the interval 0 < x < L. There aren− 1 knots of the wave function ψn(x).

The lowest energy value is

E1 =h2

2m(1.43)

and the smallest absolute value of the momentum is

p1 =h

L(1.44)

Furthermore, we have the following relation

pnL = nh n = 1, ...,∞ (1.45)

Linear Superposition

We can generate other wave functions in terms of a linear superposition of thewave functions (1.41)

Ψ(x, t) =

∞∑

n=1

Cnψn(x)e−i Enh

t (1.46)

with arbitrary coefficients Cn. These coefficients, however, are determinedfrom the initial condition Ψ(x, 0), since

Ψ(x, 0) =

∞∑

n=1

Cnψn(x) =

∞∑

n=1

Cn

√

2

Lsin

nπ

Lx (1.47)

It is possible to invert this relationship by considering the following propertyof the functions ψn(x):

∫ L

0

dxψn(x)ψM (x) = δnm (1.48)

Multiplying equation (??) by ψm(x) and integrating over the interval 0 < x <L, we obtain

∫ L

0

dxψm(x)Ψ(x, 0) =

∞∑

n=1

Cn

∫ L

0

dxψm(x)ψn(x) =

∞∑

n=1

Cnδnm = Cm (1.49)

The reader will have recognized that eq. (1.47) is a Fourier representationof Ψ(x, 0). Therefore, each initial wave function can be represented in theabove manner.

14 1 Matter Waves

Interpretation of Wave Function

We have seen that the function ψn(x) belongs to the energy value En = h2

2mL2 .The wave function is a linear superposition of these wave functions. We havealready seen that the probability to find the particle in a small interval atlocation x is given by

p(x, dx) = |Ψ(x, t)|dx (1.50)

In a similar way, we can assign a statistical interpretation to the amplitudesCn. To this end we do not characterize the particle by its location, but by itsenergy. Energy of the particle is quantized and we can ask for the probabilityp(En) to measure a particle with energy En. It is straightforward to postulatethat the probability to measure the energy value En is given by

p(En) = |Cn|2 (1.51)

For example, the wave function

Ψ(x, t) = C1ψ1(x)e−i

E1h

t + C2ψ2(x)e−i

E2h

t (1.52)

describes an experiment, in which, by measuring the energy, the energy E1

would be obtained with probability p(E1) = |C1|2 and the energy E2 withprobability p(E1) = |C1|2.

2

Schrodinger Equation

The Schrodinger equation is an evolution equation for the wave functionΨ(x, t), which allows one to determine Ψ(x, t) given an initial wave functionΨ(x, 0). The Schrodinger equation was formulated by E. Schrodinger in 1926.

The Schrodinger equation replaces Newton’s law for quantum particles.

2.1 Free particle

We consider the general form of De Broglie’s matter waves for a free particlemoving along a line.

ψ(x, t) =

∫

dka(k)ei(kx− hk2

2mt) (2.1)

Just as electromagnetic waves are described by the wave equation, one maylook for an equation, whose solution yields the matter waves.

One can calculate the temporal derivative of ψ(x, t):

∂

∂tΨ(x, t) =

∫

dk

(

−i hk2

2m

)

a(k)ei(kx− hk2

2mt) (2.2)

However, the right hand side can be expressed as a second derivative withrespect to the coordiante x:

∂2

∂x2Ψ(x, t) =

∫

dk(

−k2)

a(k)ei(kx− hk2

2mt) (2.3)

Combining eqs. (2.2) and (2.3) we obtain the evolution equation for thede Broglie matter wave:

ih∂

∂tΨ(x, t) = − h2

2m

∂2

∂x2Ψ(x, t) (2.4)

This equation has been formulated by E. Schrodinger and is denoted asthe Schrodinger equation for a free particle moving along a line.

16 2 Schrodinger Equation

2.2 Free Particle in Three Dimensions

We generalize our treatment to the case of a free particle in three dimensions.To this end we have to take into account that the momentum of a particle isa vectorial quantity

E =p2

2m=h2k2

2m= hω(k)

p = hk (2.5)

A matter wave, therefore, has the form

ψ(x, t) =

∫

d3ka(k)ei(k·x−ω(k)t) (2.6)

We can formulate the following evolution equation for the wave functionof a free particle in the infinite, three-dimensional space

ih∂

∂tψ(x, t) = − h2

2m∆ψ(x, t) (2.7)

2.3 Particle in a Potential

We generalize our treatment to the motion of a particle in an external potentialU(x). The total energy for a particle in an external potential U(x) is just

E = H(p, x) =p2

2m+ U(x) (2.8)

As a consequence, we extend the Schrodinger equation as follows:

ih∂

∂tψ(x, t) = − h2

2m∆ψ(x, t) + U(x)ψ(x, t) (2.9)

2.3.1 Linear Superposition

The Schrodinger equation is a linear equation. Linearity implies the validityof the superposition principle: Provided ΨI(x, t) and ΨII(x, t) are solutionsthen the superposed solution is a solution of the Schrodinger equation

Ψ(x, t) = N [ΨI(x, t) + ΨII(x, t)] (2.10)

However, the interpretation of the wave function |Ψ(x, t)|2 as a probabilitydensity requires the proper normalization of the sum.

2.4 Wave Function: Statistical Interpretation 17

2.4 Wave Function: Statistical Interpretation

As we have already emphasized the wave function has a statistical interpre-tation. The quantity

ρ(x, t)dx = |ψ(x, t)|2dx = ψ(x, t)∗ψ(x, t)dx (2.11)

denotes the probability to find a particle in the interval [x, dx]. To this end,the wave function has to be normalized:

∫

|ψ(x, t)|2dx =

∫

ψ(x, t)∗ψ(x, t)dx = 1 (2.12)

2.4.1 Probability Density, Current: 1 D

Particle in 1D

Since the probability has to be conserved,∫ ∞

−∞

dxρ(x, t) = 1 (2.13)

the temporal evolution of the probability density has to obey a continuityequation

∂

∂tρ(x, t) +

∂

∂xj(x, t) = 0 (2.14)

This equation implies that the normalization is conserved with respect to time

∂

∂t

∫ ∞

−∞

dxρ(x, t) = −∫ ∞

−∞

dx∂

∂xj(x, t) = −[j(∞, t) − j(−∞, t)] = 0 (2.15)

We have to proof that the existence of the continuity equation is compat-ible with the Schrodinger equation. To this end we determine the temporalevolution of the probability density ρ(x, t)

∂

∂tρ(x, t) = Ψ(x, t)∗

∂

∂tΨ(x, t) + Ψ(x, t)

∂

∂tΨ(x, t)∗ (2.16)

from the Schrodinger equation:

Ψ(x, t)∗∂

∂tΨ(x, t) =

1

ihΨ(x, t)∗

[

− h2

2m

∂2

∂x2Ψ(x, t) + U(x)Ψ(x, t)

]

Ψ(x, t)∂

∂tΨ(x, t)∗ = − 1

ihΨ(x, t)

[

− h2

2m

∂2

∂x2Ψ(x, t)∗ + U(x)Ψ(x, t)∗

]

(2.17)

A straightforward calculation yields

∂

∂tρ(x, t) = − h

2mi

[

Ψ(x, t)∗∂2

∂x2Ψ(x, t) − Ψ(x, t)

∂2

∂x2Ψ(x, t)∗

]

(2.18)


which can easily be transformed into

h

2mi

∂

∂x

[

Ψ(x, t)∗∂

∂xΨ(x, t) − Ψ(x, t)

∂

∂xΨ(x, t)∗

]

(2.19)

Let us summarize: We have shown that the probability density ρ(x, t)obeys a continuity equation (2.14) with the probability density ρ(x, t) andthe probability current j(x, t)

ρ(x, t) = |ψ(x, t)|2

j(x, t) =h2

2mi[ψ(x, t)∗

∂

∂xψ(x, t) − ψ(x, t)

∂

∂xψ(x, t)∗] (2.20)

The continuity equation directly follows from the Schrodinger equation. Thisunderlines the fact that the probabilistic interpretation and the temporal evo-lution of the wave function are compatibel.

2.4.2 Probability Density, Current: 3D

We perform the determination of probability current and density for a particlein 3 dimensional space. The density and current are related by the continuityequation

∂

∂tρ(x, t) + ∇ · j(x, t) = 0 (2.21)

The Schrodinger equation yields

∂

∂tΨ(x, t)∗Ψ(x, t) = Ψ(x, t)∗

∂

∂t+ Ψ(x, t)

∂

∂t

∗

1

ih Ψ(x, t)∗

−h2

2m∆Ψ(x, t) − Ψ(x, t)

−h2

2m∆Ψ(x, t)∗(2.22)

Furthermore, we make use of the identity

∇ · [Ψ(x, t)∗∇Ψ(x, t)] = ∇ · Ψ(x, t)∗ · ∇Ψ(x, t)] + [Ψ(x, t)∗∆Ψ(x, t)] (2.23)

and are led to the continuity equation

∂

∂tρ(x, t) + ∇ · h

2mi[Ψ(x, t)∗∇Ψ(x, t) − Ψ(x, t)∇Ψ(x, t)∗] = 0 (2.24)

The probability current is given by

j =h

2miΨ(x, t)∗∇Ψ(x, t) − Ψ(x, t)∇Ψ(x, t)∗ (2.25)

2.5 Expectations, Operators, Brackets 19

2.5 Expectations, Operators, Brackets

2.5.1 Expectation of Momentum

The expectation value 〈x(t)〉 of the position of a particle is determined by theaverage

〈x(t)〉 =

∫

dxxρ(x, t) (2.26)

The question arises, how one can define the expectation of momentum.In classical mechancis, momentum is defined via the relation

mx = p (2.27)

In quantum mechanics, we can only talk expectation, due to the proba-bilistic nature of single experiments. Therefore, we define the expectation ofmomentum by the relation

m〈x〉 = 〈p〉 (2.28)

It is straightforward to determine the temporal derivative of

d

dt〈x(t)〉 =

∫

dxx∂

∂tρ(x, t)

= −∫

dxx∇ · j(x, t) (2.29)

Thereby, we have made use of the continuity equation for the probabilitydensity ρ(x, t). Partial integration yields

d

dt〈x(t)〉 =

∫

dxj(x, t) (2.30)

Now, we can use the representation of the probability current

d

dt〈x(t)〉 =

h

2mi

∫

dx[Ψ(x, t)∗∇Ψ(x, t) − Ψ(x, t)∇Ψ(x, t)∗] (2.31)

A second partial integration yields

〈p(t)〉 = md

dt〈x(t)〉 =

∫

dxΨ(x, t)∗h

i∇Ψ(x, t) (2.32)

2.5.2 Momentum Operator

We can compare the expectation values of position and momentum:

〈x(t)〉 =

∫

dxΨ(x, t)∗xΨ(x, t) (2.33)


〈p(t)〉 =

∫

dxΨ(x, t)∗h

i∇Ψ(x, t) (2.34)

This comparision suggests to introduce the momentum operator

p =h

i∇ (2.35)

to rewrite the expectation in terms of this momentum operator:

〈p(t)〉 =

∫

dxΨ(x, t)∗pΨ(x, t) (2.36)

2.5.3 Operators

We can view the operation p as a prescription to form the vector hi ∇Ψ(x, t)

from a wave function Ψ(x, t). In this sense, we can also introduce the operatorx, which generates the vector xΨ(x, t) from the wave function Ψ(x, t).

We can use the momentum operator in order to define further operatorslike

(px)2 = (h

i

∂

∂x) = −h2 ∂

2

∂x2(2.37)

or the operator

(p)2 = (px)2 + (py)2 + (pz)2

= = −h2

(

∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)

= −h2∆ (2.38)

Furthermore, we can define operators which are composed of the positionx and momentum p.

2.5.4 Hamilton Operator

An further example of an operator is the so-called Hamilton operator, definedas

H =(p)2

2m+ U(x) (2.39)

Using the representation of the momentum operator it is straightforward toformulate the Schrodinger equation in terms of the Hamilton operator

ih∂

∂tΨ(x, t) = HΨ(x, t) (2.40)

2.5.5 Brackets

It is convenient to introduce a shorthand notation for the integrals occuringin the expressions for the expectations of position and momentum. We define

〈x(t)〉 = 〈Ψ(x, t)|x|Ψ(x, t)〉 (2.41)

as well as〈p(t)〉 = 〈Ψ(x, t)|p|Ψ(x, t)〉 (2.42)

2.6 Quantization of Classical Mechanics 21

2.5.6 Expectations of Operators

In a similar way, we can define expectations for arbitrary operators O:

〈O〉 = 〈Ψ(x, t)|O|Ψ(x, t)〉 (2.43)

2.6 Quantization of Classical Mechanics

2.6.1 Classical Mechanics: Hamilton Formulation

The dynamics of a particle in classical mechanics can be described in termsof Hamilton’s formulation. The Hamilton function is given in terms of the(generalized) positions q and momenta p :

H =p2

2m+ U(q) (2.44)

The equation of motion are given by Hamilton’s equation

qi =∂

∂piH , pi = − ∂

∂qiH (2.45)

Integration of these equations determines the position and momentaq(t),p(t) from the initial conditions q(0),p(0).

2.6.2 Quantization, Jordan Rules

We can perform the quantization of the classical theory due to the followingsteps:

• Introduction of a wave function Ψ(x, t)• Introduction of the position and momentum operators x, p.

q → x

p → p =h

i∇ (2.46)

• Introduction of the Hamilton operator:The replacement of position and momentum by the correponding operatorsin the classical Hamilton function H(p,q) leads to the Hamilton operatorH

H =p2

2m+ U(x) (2.47)


• The temporal evolution of the wave function Ψ(x, t) is given by theSchrodinger equation

ih∂

∂tΨ(x, t) = HΨ(x, t) (2.48)

This equation allows one to determine the wave function Ψ(x, t) from theinitial condition Ψ(x, 0). It replaces Hamilton’s equations (7.92) of classicalmechanics.

• Interpretation of the wave function.Expectations of location and momenta are given by

〈x(t)〉 = < Ψ(x, t)|x|Ψ(x, t)〉〈p(t)〉 = < Ψ(x, t)|p|Ψ(x, t)〉 (2.49)

• Jordan’s rules:a) An operator O is assigned to each observable given by the classicalexpression O(q,p). The outcome of the measurement of this observable isgiven by the expectation value

〈O〉 = 〈Ψ(x, t)|O|Ψ(x, t) > (2.50)

b) The explicit expression of the operator O is obtained by replacing q

and p with the operators x, p:

O = O(x, p) (2.51)

Classical Quantity Quantum Mechanical Operator

q x

p hi ∇

U(q U(x)

T (p) = p2

2m − h2

2m∆

H(p,q) H = − h2

2m∆+ U(x)

L = [q × p] L = hi [x ×∇]

3

Wave Mechanics in One Dimension

In this section we consider solutions of the Schrodinger equation in one di-mension:

ih∂

∂tΨ(x, t) = − h2

2m

∂2

∂x2Ψ(x, t) + U(x)Ψ(x, t) (3.1)

3.1 Time Independent Schrodinger Equation

A first step in the treatment of the Schrodinger equation for a stationarypotential is the separation ansatz:

Ψ(x, t) = e−i Eh

tψ(x) (3.2)

This ansatz immediately leads us to the sc-called time independent Schrodinger

equation

Eψ(x) = − h2

2m

∂2

∂x2ψ(x) + U(x)ψ(x) (3.3)

This is a second order differential equation with respect to x, which hasto be solved under the consideration of certain boundary conditions.

3.2 Free Particle on a Ring

The time independent Schrodinger equation for a free particle reads

Eψ(x) = − h2

2m

∂2

∂x2ψ(x) (3.4)

Thereby, we consider a particle on a ring-like configuration, i.e. we assumethat the wave function Ψ(x, t) and, consequently, ψ(x) obey periodic boundary

conditions

ψ(x) = ψ(x+ L) (3.5)

24 3 Wave Mechanics in One Dimension

We can perform the following ansatz for the solution

ψ(x) = Neikx (3.6)

The periodic boundary condition requires a quantization of the wavenumberk:

k =2π

Ln n = −∞, ..., 0, ...∞ (3.7)

The normalization constant is determined by the conditon

∫ L

0

dxψ(x)∗ψ(x) = N2L (3.8)

Furthermore, we have the relation

∫ L

0

dxψk(x)∗ψl(x) = δkl (3.9)

Our ansatz solves the time independent Schrodinger equation provided theenergy E is related to the wave number by

Ek =(hk)2

2m= (

h

L)2

1

2mn2 (3.10)

We recover explicitly de Broglie’s ansatz, p = hk, for matter waves. However,due to periodic boundary conditions, the energy is quantized.

Since the Schrodinger equation is linear, a general solution is given by alinear superposition

Ψ(x, t) =∑

k

ckψk(x)e−iE(k)

ht (3.11)

Thereby, the sum is over all k-values k = 0,± 2πL , ....

The coefficients are given by the initial condition:

Ψ(x, 0) =∑

k

ckψk(x) (3.12)

and, due to relation (3.16), we are able to determine the coefficients ck ac-cording to

ck =

∫ L

0

dxψk(x)Ψ(x, 0) =

∫ L

9

dx1√Le−ikxΨ(x, 0) (3.13)

Inserting this representation of the coefficients ck into (3.12) we obtain

Ψ(x, 0) =∑

k

ψk(x)

∫ L

0

dx′ψk(x′)∗Ψ(x′, 0) (3.14)

3.2 Free Particle on a Ring 25

This identity shows that the class of functions ψk(x) obeys the so-called com-

pleteness relation∑

k

ψk(x)ψk(x′)∗ = δ(x− x′) (3.15)

Since the class of functions obeys the orthogonality relation

∫ L

0

dxψl(x)∗ψk(x) = δkl (3.16)

the set of functions form an orthonormal basis: Each periodic function can beexpanded into a Fourier-series.

Let us reconsider our results. The wave function is given as a linear super-position of the functions ψk(x):

Ψ(x, t) =∑

k

ckψk(x)e−iEkh

t (3.17)

They are obtained as a solution of the boundary value problem

− h2

2m

∂2

∂x2ψk(x) = Ekψk(x)

ψk(0) = ψk(L) = 0 (3.18)

This boundary value problem simultaneously determines the discrete set of en-ergy values Ek, the so-called energy spectrum, as well as a complete orthonor-mal set of so-called eigenfunctions. The coefficients ck can be determined fromthe initial condition using the orthonormality condition (3.16). Since the set ofeigenfunctions is complete, each wave function obeying the periodic boundaryconditions can be represented as a superposition: The eigenfunctions ψk(x)form a complete orthonormal set of functions.

Let us now adress the question of the interpretation of the wave functionΨ(x, t). We already now that the quantity |Ψ(x, t)|2dx can be interpretedas a probability. However, the representation allows one to make a differentinterpretation. If we consider a wave function of the form

Ψ(x, t) = ψk(x)e−iEkh

tck(0) (3.19)

a single energy value Ek is involved, and, therefore, the corresponding particlehas a definite energy value Ek.

3.2.1 Expectation of Position

The expectation value of the particle position is denoted as

〈x(t)〉 =

∫ L

0

dxρ(x, t)x =

∫ L

0

dxΨ(x, t)∗xΨ(x, t) (3.20)


3.2.2 Expectation of Momentum

We now want to determine the expectation value of momentum. We definethis quantity according to the classical relation

〈p(t)〉 = md

dt〈x(t)〉 =

∫ L

0

dxΨ∗(x, t)

(

h

i

∂

∂x

)

Ψ(x, t) (3.21)

It is instructive to calculate this expectation using the representation ofour wave function:

〈p(t)〉 =∑

k

∑

k′

∫ L

0

dxeiEk−E

k′

htc∗kck′ψ∗

k(x)h

i

∂

∂xψk′(x)

=∑

k

hk|ck|2 (3.22)

However, this formula points out the physical interpretation of the coefficientsck.


The introduction of the momentum operator allows us to introduce the Hamil-ton operator

H =p2

2m= − h2

2m

∂2

∂x2(3.23)

It is interesting to calculate the expectation of the energy in a similar way.We calculate

〈H〉 = 〈Ψ(x, t)|H |Ψ(x, t)〉 =

∫ L

0

dxΨ(x, t)∗HΨ(x, t) (3.24)

A similar calculation as in the previous section yields

〈H〉 =∑

|ck|2(hk)2

2m(3.25)

This representation immediately yields the following interpretation of |ck|2:The quantity |ck|2 has to be interpreted as the probability to find a particlewith the energy Ek.

3.3 Free Particle on the Line

Let us now consider the limit of infinite length L → ∞. In this case theallowed k-values tend to the continuum and it is straightforward to replacethe discrete sum by an integral. To this end we introduce the representation

3.4 Infinitely Deep Potential Well 27

Ψ(x, t) =∑

k

1√Lcke

ikxe−ihk2/2mt

=∑

k

∆kC(k)1√2πeikxe−ihk2/2mt

→∫

dkC(k)1√2πeikxe−ihk2/2mt

(3.26)

where we have defined

∆k =2π

Lck =

C(k)√2πL

(3.27)

The limit L→ ∞ means that we are going from a Fourier-series to a Fourierintegral.

The coefficients C(k) are then given by the inverse Fourier transform

Ψ(x, 0) =

∫

dkC(k)1√2πeikx (3.28)

or,

C(k) =

∫

dxΨ(x, 0)1√2πe−ikx (3.29)

Inserting this relationship back into (3.28) we arrive at the completeness

relation∫

dk1

2πeik(x−x′) = δ(x− x′) (3.30)

which is a well-known representation of the δ-function.

3.4 Infinitely Deep Potential Well

Although we have already discussed the wave mechanics of a particle in aninfinitely deep potential well, we shall reconsider this problem based on thesolution of the Schrodinger equation.

To this end we consider the time independent Schrodinger equation

Eψ(x) = − h2

2m

∂2

∂x2ψ(x) (3.31)

together with the boundary conditions

ψ(x = 0) = ψ(x = L) = 0 (3.32)

These boundary conditions imply that the probability to find a particle closeto the boundary vanishes.


The time independent Schrodinger equation is solved by the ansatz

ψ(x) = asinkx+ bcoskx (3.33)

provided we choose the energy

E =(hk)2

2m(3.34)

The boundary condition ψ(0) = 0 leads us to b = 0, whereas the conditionψ(L) = 0 requires the quantization of the values of k

kL = πn n = 1, 2, ...,∞ (3.35)

and, accordingly, a quantization of energy

En =(hπ)2

2mL2n2 (3.36)

Mathematical remarks:The functions

ϕn(x) =

√

2

Lsinn

π

Lx (3.37)

form a complete function set. This means that each function with the bound-ary conditions

v(x = 0) = v(x = L) = 0 (3.38)

can be expanded in a series according to

f(x) =∑

n

cn

√

2

Lsinn

π

Lx (3.39)

Furthermore, we have the orthogonality relation

∫ L

0

dx

√

2

Lsinn

π

Lx

√

2

Lsinm

π

Lx = δnm (3.40)

The function system is complete.

3.5 Particle in a Finite Potential Well

We consider the motion of a particle in the potential well exhibited in fig. (??)

U(x) = −U0 |x| < L/20 |x| > L/2

(3.41)

Classically, there are bounded states, if the energy of the particle is lessthen zero. In the case of positive energy, the particle can move freely. Thebehaviour of a quantum particle is similar, but we expect that the energyvalues are quantized in the bounded case.

3.5 Particle in a Finite Potential Well 29

Finite potential well.

3.5.1 Bounded States

We restrict the energy of the particle to

−U0 < E < 0 (3.42)

We have to solve the time independent Schrodinger equation

Eψ(x) = − h2

2m

∂2

∂x2ψ(x) + U(x)ψ(x) (3.43)

considering the boundary conditions

ψ(x = −∞) = ψ(x = +∞) = 0 (3.44)

as well as conditions at x = ±L/2, i.e. at positions where the potential ex-hibits the finite discontinuity. We expect to have a continuous probabilitydistribution ρ(X, t) as well as a continous probability current j(x, t) at thesepoints. This leads us to

ψ(±L/2 + 0) = ψ(±L/2 − 0)

ψ′(±L/2 + 0) = ψ′(±L/2 − 0) (3.45)


Since the problem is invariant with respect to reflections x→ −x we expectthat there are symmetric as well as antisymmetric wave functions. These wavefunctions are said to possess even (symmetric) and odd (antiysmmetric) parity.In the following we shall explicitly construct these wave functions.

Graphical solution of the equation determining energy eigenstates for a particle ina finite Potential Well.

3.5.2 Solution with even parity:

We first consider the solutions with even parity:

ψ(x) = ψ(−x) (3.46)

We consider bound states with with energies

−U0 ≤ E ≤ 0 (3.47)

For the following it is convenient to define the quantities κ and k accordingto


κ =

√

2m|E|h2

k =

√

2m|E| − U0

h2

The time independent Schrodinger equation can be written as

d2

dx2ψ = κ2ψ |x| > L

2d2

dx2ψ = −k2ψ |x| < L

2(3.48)

and is solved by

ψ(x) = A′eκx +A′′e−κx

ψ(x) = B′coskx+B′′sinkx (3.49)

The ansatz for the solution with even parity is given by

ψ(x) =Aeκ(x+L/2) x < −L/2Bcoskx |x| ≤ L/2Ae−κ(x−L/2) x > L/2

(3.50)

Particle in a finite potential well: Lowest wavefunctions with even parity.

The wave function ψ(x) as well as its derivative ψ′(x) has to be continuousat x = ±L/2.


A−BcoskL

2= 0

−Aκ+BksinkL

2= 0 (3.51)

This is a homogeneous set of equations, which can be written in matrix form

(

1 coskL2

−sinkL2 1

) (

AB

)

= 0 (3.52)

The solvability condition reads

k

κsin

kL

2− cos

kL

2= 0 (3.53)

Here, κ as well as k depend on the energy E, which then is determined bythe solvability condition. As we shall see from a graphical discussion of thiscondition there is a finite set of energy values Ei for which the conditions isfullfilled.

For our graphical discussion we introduce κ and K according to

κ = κL

2, K = k

L

2(3.54)

Using the definition of κ, k we obtain

K2 + κ2 = R2 = (L

2)2

2mU0

h2 (3.55)

The solvability condition reads:

κ = ktank =√

R2 − k2 (3.56)


R =L

2

√

2mU0

h2 (3.57)

The value of k, and, consequently the energy levels, are given by the crossingsof the two curves of eq. (3.56). Details can be seen in fig. (3.5.1).

3.5.3 Solution with odd parity:

We now consider solutions with odd parity

ψ(x) = −ψ(−x) (3.58)

Again, we consider bound states with energies

−U0 ≤ E ≤ 0 (3.59)


The ansatz for the solution with odd parity is now given by

ψ(x) =

−Aeκx x < −L/2Bsinkx |x| ≤ L/2Ae−κx x > L/2

(3.60)

Thereby the quantities κ and k have been defined above.The wave function ψ(x) as well as its derivative ψ′(x) has to be continuous

at x = ±L/2:

Ae−κL/2 −BsinkL

2= 0

−Aκe−κL/2 −BkcoskL

2= 0 (3.61)

As a result we obtain the solvability condition

k

κcos

kL

2+ sin

kL

2= 0 (3.62)

Again , we can perform a graphical solution of this problem.To this end we introduce κ and K according to

κ = κL

2, K = k

L

2(3.63)

Using the definition of κ, k we obtain

K2 + κ2 = R2 = (L

2)2

2mU0

h2 (3.64)

The solvability condition reads:

−κ = kcotk = −√

R− k2 (3.65)


R =L

2

√

2mU0

h2 (3.66)

Summary: Bound States

Graphically, we can obtain the following results:

• There is at least one solution with even parity ψ(x) = −ψ(−x)• With inreasing value of |U0|, i.e. with increasing depth of the well, the

number of energy levels increase. This can be seen from figure (??), whereit is evident that by increasing U0,i.e. R, the number of solutions of thetranscendental equations increase.

• For large values of U0, i.e. R, the limiting case of an infinitely deep potentialwell is approached.


The wave functions have even and odd symmetry, which is clearly a resultof the reflectional symmetry of the problem. The wave functions differ withrespect to the number of knots. We remind the reader that there is a nonvan-ishing wave function in the classically forbidden regions. Although this wavefunction exhibits an exponential decay in this region

ψ(x) = ae−xd ; d =

1

κ(3.67)

there is a nonvanishing probability to detect particles in this classically for-bidden regions.

3.5.4 Continuous Spectrum

Up to now we have been concerned with particles with energy E < 0. Inthe case of positive energy, we can perform the following ansatz for the wavefunction

Ψ(x, t) =

aeik(x+L/2) + be−ik(x+L/2)

ceik′x + de−ik′x

feik(x−L/2) + ge−ik(x−L/2)

(3.68)

Here,

k =

√

2mE

h2 , k′ =

√

2m(E + |U0|h2 (3.69)

The requirement of continuity of the wave function and the probability currentleads to the following set of equations

a+ b = ce−ik′L/2 + deik′L/2

ceik′L/2 + de−ik′L/2 = f + g

ik(a− b) = ik′[ce−ik′L/2 − deik′L/2]

ik′(ceik′L/2 + de−ik′L/2) = ik(f − g) (3.70)

Continuity leads to four equations for six amplitudes a− g.In order to find a solution, we can fix two constants arbitrarily. One con-

stant can be fixed by, e.g. taking a = 1. A second choice allos us to constructa right-running wave (g=0) or a left-running wave (f=0). Furthermore, wesee that set of equations determining the amplitudes (??) can be fulfilled forarbitrary values of the energy E, in contrast to the case of negative energydiscussed above. In this case, the additional requirement that the wave func-tion decays for x → ±∞ sets two amplitudes to zero, leading to the linearset of four homogeneous equations for four amplitudes (or, two equations fortwo coefficients for symmetric and antisymmetric wave functions). As a con-sequence, a solvability condition has to be fulfilled leading to a discrete set ofenergy values.

3.6 Reflection at a Barrier: Tunnel Effect 35

3.6 Reflection at a Barrier: Tunnel Effect

We consider now the motion of a particle in the potential

V (x) =

(

U0 0 < x < L0 otherwise

)

(3.71)

We assume that a stream of particles with energy E < U0 approaches thepotential barrier from the left. The Schrodinger equation in the three regionsare

∂2

∂x2ψ(x) = −κ2ψ(x) x < 0, x > L

∂2

∂x2ψ(x) = k2ψ(x) x < 0, x > L (3.72)


κ2 =2mE

h2 , k2 =2m|E − V |

h2 (3.73)

3.6.1 E < V

In the following we assume that E < V :Using the solutions of these equations in the different regimes we can

perform the following ansatz

ψ(x) = aeikx + be−ikx x < 0ce−κx + deκx 0 < x < L

feik(x−L)(3.74)

The time dependent solution would be

Ψ(x, t) = [aei(kx−ωt) + be−i(kx+ωt)] x < 0ce−κx−iωt + deκx−iωt 0 < x < Lfei(k(x−L)−ωt)

(3.75)

where the frequency ω of the wave is related to the (yet unspecified) energyE = hω. This wave consists of a right- and left running wave in the spacex < 0, a decaying wave in the classically forbidden regime 0 < x < L and arightrunning wave in the regime x > L. We remind the reader that, in the clas-sical description, a particle would not be able to traverse the potential barrierand would be reflected. A quantum mechanical particle can tunnel throughthe barrier, provided there is a nonvanishing amplitude c of the rightrunningwave.

In order to describe a beam of particles moving from x = −∞ towards thebarrier, we evaluate the probability current


j =h

2mi[Ψ(x, t)∗

∂

∂xΨ(x, t) − Ψ(x, t)

∂

∂xΨ(x, t)∗] (3.76)

We explicitly obtain

j =

hk2mi [(a

∗e−ikx + b∗eikx)ik(aeikx − be−ikx) − c.c.]h

2mi [(c∗e−κx + d∗eκx)(−κ)(ce−κx + d∗eκx) − c.c.]

h2mi [f

∗e−ikxikfeikx − c.c.]

=

hkm (|a|2 − |b|2)κm i[d

∗c− dc∗]hkm |f |2

(3.77)

Continuity of the current yields

|a|2 − |b|2 = |f |2 (3.78)

We define the reflection coefficient R and the transmission coefficient T ac-cording to

R =|b|2|a|2 , T =

|f |2|a|2 (3.79)

with the relationsshipR+ T = 1 (3.80)

In case where T > 0 particles can be detected behind the barriers, also in thecase where E < V . Classically the particles would be reflected in this case,R = 1. A nonvanishing transmission coefficient T indicates that quantumparticles can penetrate through a potential barrier. This effect is denoted astunneling.

Let us now determine the transmission and reflection coefficiennt explic-itly. The amplitudes a, aR as well as the corresponding energy values aredetermined from the boundary conditions. Again, we have to require continu-ity of the wave function, as well as the spatial derivative of the wave functionat the boundaries x = 0, x = L.

Continuity of the wave function ψ(x) at x = 0, L yields the relationships

a+ b = c+ d

ce−κL + deκL = f (3.81)

Continuity of the derivative of the wave function ψ′(x) at x = 0, L yields

ik(a− b) = −κc+ κd

−κce−κL + κdeκL = ikf (3.82)

There are 4 equations for 5 coefficients. We can fix the amplitude of theincoming wave a = 1. Since the amplitudes a, b are the amplitudes of anincoming wave and the reflected wave we can make the ansatz b = ra with afactor r, which describes the reflection. Furthermore, we are able to out a = 1,

3.6 Reflection at a Barrier: Tunnel Effect 37

normalizing the amplitude of the incoming wave to a = 1. As a consequence,we obtain the following inhomogeneous set of linear equations

r − c− d = −1

ce−κL + deκL − f = 0

r +iκ

kc− iκ

kd = 1

ce−κL iκ

k− deκL iκ

k− f = 0 (3.83)

In order to calculate the coefficients, we separate the set of equations into

c(1 + iα) + d(1 − iα) = 2

ce−κL(1 − iα) + deκL(1 + iα) = 0 (3.84)

and

r = (c+ d) − 1

t = ce−κL + deκL (3.85)

We obtain by defining

α =κ

k(3.86)

d = −ce−2κL 1 − iα

1 + iα(3.87)

c = 21 + iα

(1 + iα)2 − (1 − iα)2e−2κL(3.88)

t =4iαe−κL

(1 + iα)2 − (1 − iα)2e−2κL

=2iα

(1 − α2)sinhκL+ 2iαcoshκL

=1

coshκL− 1−α2

2α isinhκL(3.89)

We can now state our result for the transmission coefficient T

T = |t|2 =1

cosh2κL+ (1−α2)2

4α2 sinh2κL=

4α2

4α2 + (1 + α2)2sinh2κL(3.90)

whereas the reflection coefficient is just obtained from

R = T − 1 (3.91)


3.6.2 E > V

If we consider the case of particles with energy larger as V, the potential heightof the barrier, classical particles will all pass the barrier. Quatum particles,however, feel the barrier indicating that there is a nonvanishing transmissioncoefficient.

This transmission coefficient can easily be calculated by noticing that allformulas remain valid, except that we now have to consider that κ is imaginary,since E > V :

κ = iκ α = iα (3.92)

The transmission coefficient reads:

T = |t|2 =4α2

4α2 + (1 − α2)2sin2κL(3.93)

3.6.3 Wave Packets

Up to now we have considered a monochromatic incoming wave. In order todescribe a beam of localized moving towards the barrier, one has to bild upwave packets.

The wave packet on the left side and right hand side of the barrier has theform

Ψ(x, t) =

∫

dka(k)eikx0e−iω(k)t[eikx + r(k)e−ikx∫

dka(k)eikx0e−iω(k)t[c(k)e−κx + d(k)eκx]∫

dka(k)eikx0e−iω(k)tt(k)eikx

(3.94)

where

ω(k) =hk2

2m(3.95)

The amplitude a(k) of the wave packet can be, e.g. taken as a Gaussian

a(k) =1√2πQ

e−(k−k0)2

2Q (3.96)

3.7 Wave Mechanics in Two Dimensions

3.7.1 Quantum Dots

3.7.2 Quantum Billards

4

Hilbert Space

4.1 Linear Space

We consider vectors |ψi >, which can be added, |ψ1 > +|ψ2 > and multipliedby a scalar quantity α, |ψ >= α|ψ1 >. The set of vectors form a linear space,provided the following axioms are fullfilled:

a) Neutral element |0 > |ψ > +|0 >= |ψ >

b) Inverse element |ψ > +| − ψ >= |0 >

c) Associativity |ψ1 > +(|ψ2 > +|ψ3 >) = (|ψ1 > +|ψ2 >) + |ψ3 >

d) Distributivity |α(ψ1 + ψ2) >= α|ψ1 > +α|ψ2 >

e) |(α + β)ψ >= α|ψ > +β|ψ >

f) |(αβ)ψ >= α|βψ >

g) |1ψ >= |ψ >

4.2 Normed Linear Space

A linear space is extended to a normed linear space if to each element ψ thereis a positive real number ||ψ|| obeying the following axioms

40 4 Hilbert Space

a) Positivity ||ψ|| > 0b) Neutral element ||0|| = 0

c) ||αψ|| = |α|||ψ||

d) Triangle inequality ||ψ1 + ψ2|| ≤ ||ψ1|| + ||ψ2||

4.2.1 Scalar Product

A norm in a linear space can be defined via a scalar product. The scalarproduct is denoted as

< ψ1|ψ2 > (4.1)

and can be a complex quantity obeying the following axioms

a) < ψ3|(ψ1 + ψ2) >=< ψ3|ψ1 > + < ψ3|ψ2 >

b) < ψ1|αψ2 >= α < ψ1|ψ2 >

c) < ψ1|ψ2 >=< ψ2|ψ1 >∗

d) < ψ|ψ >≥ 0

e) < ψ|ψ >= 0 implies ψ = 0

The scalar product can be used to define a norm

||ψ|| =√

< ψ|ψ > (4.2)

Frequently, the scalar product is denoted as inner product. A linear space witha scalar product is an inner product space.

4.2.2 Schwarz Inequality

A scalar product obeys the Schwarz inequality:

| < ψ1|ψ2 > | ≤ ||ψ1|| ||ψ2|| (4.3)

For a proof we start from the inequality

4.4 Basis, Dimension, Completeness 41

||(ψ1 − αψ2)||2 ≥ 0 (4.4)

which implies

||ψ1||2 − α < ψ1|ψ2 > −α∗ < ψ2|ψ1 > +|α|2||ψ2||2 ≥ 0 (4.5)

Now we choose

α =< ψ2|ψ1 >

< ψ2|ψ2 >(4.6)

and end up with the inequality of Schwarz.

4.2.3 Orthogonal Vectors

Two vectors are orthogonal with respect to each other if the scalar product

< ψ1|ψ2 >= 0 (4.7)

vanishes

4.2.4 Completeness

A normed linear space is complete, if each convergent Cauchy sequence |ψn >has a limit which belongs to this space. A sequence of vectors ψn > is conver-gent, provided

||(ψn − ψn+N )|| ≤ ǫ(n) (4.8)

4.3 Hilbert Space

A Hilbert space is a complete linear space with a scalar product. It is alsodenoted as inner product space.

4.4 Basis, Dimension, Completeness

In a Hilbert space, we can choose a set of basis vectors denoted by

|ψ1 >, |ψ2 >, .... (4.9)

The vector|ψ >=

∑

i

ci|ψi > (4.10)

yields a new vector, which is an element of the Hilbert space.We can consider an orthogonal set of basis vectors, which obey the orthog-

onality condition< ψi|ψj >= δij (4.11)

42 4 Hilbert Space

An orthogonal basis can be constructed iteratively using the Schmidt or-thogonalization procedure: From the set |ψ1 >, |ψ2 > we form

|ψ1 > = |ψ1 >

|ψ2 > = |ψ2 > − < ψ2|ψ1 > |ψ1 >

|ψ3 > = ..... (4.12)

A basis |ψi > is complete provided each vector of the space can be repre-sented as a linear superposition of basis vectors

|ψ >=∑

i

ci|ψi > (4.13)

the coefficients are the scalar products

ci =< ψi|ψ > (4.14)

Forming the scalar product

< ψi|ψ >=< ψi|∑

j

cjψj >=∑

j

cj < ψi|ψj >= ci (4.15)

Inserting this expression into (??) we obtain

|ψ >=∑

i

|ψi > ci =∑

i

|ψi >< ψi|ψ > (4.16)

which we rewrite in the form

1 =∑

i

|ψi >< ψi| (4.17)

The quantity ψ1 >< ψ2| is denoted as dyadic product.The dimension of a Hilbert space is the number of basis vectors necessary

to expand each vector of the space. The space is separable if the there is acountable set of basis vectors. The number of basis vectors can be infinite.

4.5 Examples

4.5.1 Euclidean Space

The real vectors a = [a1, as, ..aN ] in Rn with the scalar product

< a|b >=

N∑

i=1

aibi (4.18)

form a Hilbert space, the euclidean space.

4.5 Examples 43

4.5.2 Space of periodic functions

Our next example is the set of functions ϕ(x) which are periodic in x withperiod L and are square integrable. This means that the integrals

∫

dx|ψ(x, t)|2 <∞ (4.19)

These functions form a linear space with respect to addition ψ1(x) + ψ2(x)and multiplication with a scalar, αψ(x).

The scalar product is defined by the integral

< ψ1|ψ2 >=

∫ L

0

dxψ1(x)∗ψ2(x) (4.20)

These functions can expanded into a Fourier-series:

f(x) =

∞∑

n=−∞

cn1√Lein 2π

Lx (4.21)

The set of functions

|ϕn >=1√Leink0x n = −∞, ...,∞ k0 =

2π

L(4.22)

form a orthonormal set of basis functions. This follows from the straightfor-ward evaluation of

< ϕn|ϕm >= N2

∫ L

0

dxei(m−n)k0x = N2δnmL (4.23)

The Fourier-coefficients can be calculated by the integral

cn =

∫ L

0

1√Le−ink0xψ(x) (4.24)

The Fourier representation yields

Ψ(x) =∑

n

∫ L

0

dx′ϕn(x′)∗Ψ(x′)ϕn(x) (4.25)

which leads us to the completeness relation

δ(x− x′) =∑

n

∫ L

0

dx′ϕn(x′)∗ϕn(x) (4.26)

We explicitly see that the δ-function can be represented in terms of the basisfunction.

44 4 Hilbert Space

4.5.3 Square Integrable Functions

We consider the space of complex functions ψ(x), for which the integrals

∫ ∞

−∞

dx|ψ(x)|2 <∞ (4.27)

exist. These functions are denoted as square integrable functions.The square integrable functions form a linear space with respect to the

addition and multiplication with a complex variable α:

ψ1(x) + ψ2(x) = ψ(x) (4.28)

The sum ψ(x) is again square integrable, since

|ψ(x)|2 = |ψ1(x) + ψ2(x)|2 ≤ |ψ1(x)|2 + |ψ2(x)|2 (4.29)

so that∫ ∞

−∞

dx|ψ(x)|2 <∫ ∞

−∞

dx|ψ1(x)|2 +

∫ ∞

−∞

dx|ψ2(x)|2 <∞ (4.30)

For square integrable functions the scalar product can be defined by theintegral

< ψ1|ψ2 >=

∫

dxψ(x)∗1ψ2(x) (4.31)

The property e) has to be discussed in some more detail. The space ofsquare integrable functions may contain functions, which are zero on a spaceof Lebesque measure zero. An example is the function

u(x) = 1 forx = 0

u(x) = 0 otherwise (4.32)

The (Lesbeque) integral of this function is zero,

∫ ∞

−∞

dxu(x) = 0 (4.33)

Thus, in principle, the property e) is not fullfilled. However, if we consider theequivalence class of functions which differ from each other by functions whichare different from zero on sets of measure zero, the property e) is fullfilled and(4.31) defines a scalar product in the space of the square integrable functions.

4.5.4 Square Integrable Functions in N Dimensions

Generalization to three, N dimensions:∫

d3x|ψ(x)|2 ≤ ∞ (4.34)

4.7 Adjoint Operator, Hermitian Operator 45

4.6 Linear Operators

Several times we have encountered operators. An operator A transforms anelement of a Hilbert space into another element of this space:

|Ψ2 >= A|Ψ1 > (4.35)

4.6.1 Example: Euclidean Space

In the Euclidean space the operators A are represented by matrices. It turnsa vector |Ψ1 > into a vector |Ψ2 >. If we take a basis |ϕi > we can expand thevectors

|Ψ1 >=

N∑

i=1

ai|ϕi > , |Ψ2 >=

N∑

i=1

bi|ϕi > (4.36)

we obtainN

∑

i=1

bi|ϕi >= A

N∑

i=1

bi|ϕi >=

N∑

i=1

biA|ϕi > (4.37)

Taking the scalar product we obtain the representation of the operator Ain terms of the matrix

bi =

N∑

j=1

Aijaj (4.38)

withAij =< ϕi|A|ϕj > (4.39)

4.6.2 Example: Complex Space

4.6.3 Example: Position and Momentum Operator

Examples of physically relevant opertors are the position and momentumoperator x, p.

4.7 Adjoint Operator, Hermitian Operator

We shall now define the adjoint operator A† by the relation

< Ψ1|A|Ψ2 >=< A†Ψ1|Ψ2 > (4.40)

A selfadjoint operator is defined by the relationship

A = A† (4.41)

The operator need not be defined for all vectors of the Hilbert space. It mayonly be defined on a subset D of H.

A selfadjoint operator is an hermitian opertor if D† = D .

46 4 Hilbert Space

4.7.1 Momentum Operator

The momentum operator is an hermitian operator

< Ψ1(x)|h

i

∂

∂x|Ψ2(x) >=<

h

i

∂

∂xΨ1(x)|Ψ2(x) > (4.42)

Proof: We explicitly write down the scalar products in terms of th integrals

< ψ1(x)|p|ψ2 > =

∫

dxψ1(x)∗ h

i

∂

∂xψ2(x)

= −∫

dx

(

h

iψ1(x)

)∗∂

∂xψ2(x)

=

[

−(

h

iψ1(x)

)∗

ψ2(x)

]∞

−∞

=

∫

dx

(

h

i

∂

∂xψ1(x)

)∗

ψ2(x) =< pψ1|ψ2 > (4.43)

4.7.2 Operator of Kinetic Energy

< Ψ1(x)| −h2

2m∆|Ψ2(x) >=< − h2

2m∆Ψ1(x)|Ψ2(x) > (4.44)


The Hamilton operator is an hermitian operator:

H =(p)2

2m+ U(x) = H† (4.45)

4.8 Eigenvalue Problems

Eigenvalue problemA|ψ >= λ|ψ > (4.46)

The eigenvalue problem connected with a linear operator A in general hasa nontrivial solution only for a certain set of eigenvalues λi.

4.8.1 Example: Euclidean Space

In Euclidean space RN the operators are matrices. Therefore, there are Neigenvalues which are the roots of the so-called characteristic equation

Det[A− λE] = 0 (4.47)

4.9 Hermitian Operators 47

4.8.2 Example: Momentum Operator

h

i

∂

∂xψ(x) = λ|ψ(x) > (4.48)

We consider the space of squar integrable periodic functions. The eigen-functions are given by

|ψk >=1√Leikx (4.49)

and the eigenvalues areλk = hk (4.50)

4.9 Hermitian Operators

The eigenvalues of hermitian operators are real. The set of eigenvalues of anoperator is denoted as the spectrum of AProof:

< Ψ |A|Ψ >= λ < Ψ |Ψ >=< AΨ |Ψ >= λ∗ < Ψ |Ψ > (4.51)

The eigenfunctions ψ1 and ψ2 of a selfadjoint operator belonging to differ-ent eigenvalues are orthogonal:Proof:

< ψ1|A|ψ2 >= λ2 < ψ1|ψ2 >= λ1 < ψ1|ψ2 > (4.52)

For the case λ1 6= λ2 we conclude

< ψ1|ψ2 >= 0 (4.53)

In case of a degeneracy of an eigenvalue, i.e. in case that there are two ormore eigenvectors belonging to a single eigenvalue, we can orthogonalize thesevectors.

The eigenvectors of a hermitian operator form a complete orthogonal set:

< ψi|ψj > = δij∑

i

|ψi >< ψi| = 1 (4.54)

4.9.1 Projection Operator

We can take

P =|ψ1 >< ψ1|

||ψ1||(4.55)

P 2 = P (4.56)

48 4 Hilbert Space

4.9.2 Spectral Decomposition

4.10 Functions of Operators

It is possible to define functions of an operator A, F (A). It is straightforward todefine the operators A2, An. Provided the function F (a) has a representationin terms of a Taylor series with coefficients Fk we are able to define theoperator function according to

F (A) =∑

k

Fk(A)k (4.57)

An example is the matrix exponential of an operator A

eA =∑

k=0

1

k!(A)k (4.58)

4.11 Commuting, Noncommuting Operators

We consider the product of two operators, AB, and compare it to the productBA. In general, these operators are different.

4.11.1 Position, Momentum

As an example we consider the operators A = x, B=p

xp = xh

i

∂

∂x

px =h

i

∂

∂xx (4.59)

In order to compare both expressions we have to consider e.g. the quantity

px|ψ > =h

i

∂

∂x(xψ(x))

= xh

i

∂

∂xψ(x) +

h

iψ(x) = xp+

h

i(4.60)

where we have made use of the product rule of differentiation. We can nowdefine the difference between the operators xp and px

xp− px = ih (4.61)

4.11.2 Definition: Commutator

Let us now generalize: We define the commutator [A,B] of two operators A,B by the difference of the product AB and BA:

[A, B] = AB − BA (4.62)

4.12 Uncertainty Relations 49

4.11.3 Eigenvectors of Commuting Operators

Two commuting operators A, B have common eigenvectors. For the proof weconsider the eigenvalue equation

A|ϕi >= λi|ϕ > (4.63)

From the relationBA|ϕi >= AB|ϕi >= λiB|ϕ > (4.64)

we conclude that B|ϕi > is an eigenvector of A with eigenvalue λi. We assumethat the eigenvalue is simple, then

B|ϕi >= γi|ϕi > (4.65)

which shows that the vector |ϕi > is also an eigenvector of the operator B.

4.11.4 Functions of Commuting Operators

Operator functions of commuting operators commute.

4.12 Uncertainty Relations

We start from Schwarz inequality

< Ψ1|Ψ1 >< Ψ2|Ψ2 > ≥ | < Ψ1|Ψ2 > |2 (4.66)

which is valid for each pair of elements of a Hilbert space. We consider twoHermitian operators A, B. We define by

|Ψ1 >= A|Ψ >

|Ψ2 >= B|Ψ > (4.67)

and obtain using the Schwarz inequality and the property, the A, B are Her-mitian operators

< Ψ |A2|Ψ >< Ψ |B2|Ψ >

≥ | < Ψ |AB|Ψ > |2 (4.68)

We introduce the abbreviation

z =< Ψ |AB|Ψ > (4.69)

and use the estimate

|z|2(Rez)2 + (Imz)2 ≥ (Imz)2 = (z − z∗

2i)2 (4.70)

50 4 Hilbert Space

Sincez =< Ψ |AB|Ψ >=< BA)Ψ |Ψ > (4.71)

and, using the assumption that A , B are hermitian operators, we obtain

z∗ =< Ψ |BA|Ψ > (4.72)

as well as

z − z∗

2i=

1

2i< Ψ |AB − BA|Ψ >

=1

2i< Ψ |[A, B]|Ψ > (4.73)

where we have introduced the commutator [A, B] of A and B.As a consequence, we obtain the general uncertainty relationship between

two operators A, B

∆A∆B ≥ 1

2| < Ψ |[A,B]|Ψ > | (4.74)

Thereby, we have defined the uncertainties

∆A =√

< Ψ |(A− < A >)2|Ψ > , ∆B =√

< Ψ |(B− < B >)2|Ψ >(4.75)

4.12.1 Uncertainty Relation

The commutator between the operators x and p is

[x, p] = ih (4.76)

Consequently, we obtain the uncertainty relation

∆x∆p ≥ h

2(4.77)

5

Chance and Probability

Probability theory has been developed to cope with random events. Diffrac-tion experiments with quantum particles like electrons, neutrons have demon-strated the random outcome of a single event as well as the emergence ofwell-predictable interference patterns.

An operational approach has been developed in order to deal with systemswith random outcomes. It is based on the notion of the probability pi of arandom event denoted as ωi. The elementary events ωi belong to a set Ω,which is denoted as sample space. The probability pi is operationally definedas the number

pi = limN→∞Ni

N(5.1)

Thereby, N denotes a number of experiments, and Ni counts the number ofthese experiments in which the event ωi has been realized. The probability,therefore, can be measured by performing a large number of experiments.

5.1 Probability

Our operational definition of probability immediately implies that the proba-bility pi is normalized:

∑

i

pi =∑

i

limN→∞Ni

N= 1 (5.2)

The sum is over all elementary events.

5.2 Probability Density

For the case of a continuous random variable we expect that the probabilityp(x, dx) is proportional to the width dx of the interval:

52 5 Chance and Probability

p(x, dx) = ρ(x)dx (5.3)

This defines the probability density ρ(x). This probability density is normal-ized

∫

ρ(x)dx = 1 (5.4)

5.3 Expectations

The introduction of the notion of probability allows one to calculate expecta-tion values of random quantities and, in turn, allows one to perform predic-tions about the outcome of experiments with random variables.

The mean value of a quantity is given by

< x >=∑

i

xipi =

∫

ρ(x)xdx (5.5)

Accordingly, we can define expectations of the quantities x2, xn which aredenoted as moments:

< xn >=∑

i

xni pi =

∫

ρ(x)xndx (5.6)

An important quantity is the variance defined as

< x2 > − < x >2 (5.7)

5.4 Mathematical Definition of Probability

The notion of probability requires a mathematical definition. We shall brieflyoutline this definition, which is based on the notion of a sample space Ω. Thesample space consists of sets Ωi including the empty set 0. ......

6

Concepts of Quantum Physics

The behaviour of a quantum system is described by a time dependent statevector

|Ψ(t) > (6.1)

which is element of a Hilbert space H . This state vector is normed, i.e.

||Ψ || =√

〈Ψ(t)|Ψ(t)〉 (6.2)

The wave function contains all information on the quantum system.The spatio-temporal evolution is given by the Schrodinger equation

ih∂

∂t|Ψ(t) >= H |Ψ(t) > (6.3)

Here, H denotes the Hamilton operator.

6.1 Axiomatic Quantum Theory

• Physical states are elements of a complex Hilbert space.• Physical observables are Hermitian operators defined in this Hilbertspace.

Measurements are interpreted in a probabilistic sense. The expectationvalue of a physical variable is given by the scalar product

< A >=< Ψ |A|Ψ > (6.4)

• Temporal EvolutionThe temporal evolution of the state vector is determined by the

Schrodinger equation

ih∂

∂t|Ψ(t) >= H|Ψ(t) > (6.5)

54 6 Concepts of Quantum Physics

For a particle the Hamilton operator is obtained from the classical Hamil-ton function

H =p2

2m+ U(x) (6.6)

• Measurement ProcessA measurement of an observable A yields one of the eigenvalues of the

corresponding operator A.

Observable Operator

Position x = x

Momentum p = hi ∇

Kinetic Energy T = − h2

2m∆

Potential Energy U(x))

6.2 Physical Observables and Hermitian Operators

To each physical observable A we assign an Hermitian operator A. The mea-sured values of the observable are the eigenvalues ai of the operator in Hilbertspace defined by the eigenvalue problem

A|ϕAi >= ai|ϕA

i > (6.7)

Since the operator A is Hermitian, the eigenvalues are real numbers. Thenumber of the eigenvalues equals the dimension of the Hilbert space.

Since the operator A is an Hermitian operator, it has a complete set oforthonormal eigenstates:

< ϕAi |ϕA

j > = δij∑

i

|ϕAi >< ϕA

i | = 1 (6.8)

6.3 Probabilistic Interpretation

The state vector |Ψ(t) > has to be interpreted in a statistical sense.We represent the state vector |Ψ(t) > in terms of the eigenfunctions |ϕA

j >

of A:

6.4 Measurement Process 55

|Ψ(t) >=∑

i

ci(t)|ϕAi > (6.9)

The expectation of the observable A can then be evaluated according to

〈A〉 = < Ψ(t)|A|Ψ(t) >=∑

j

∑

k

c∗jckλAk < ϕA

j |ϕAk >

=∑

j

|cj(t)|2λAk (6.10)

This expression has to be interpreted in a probabilistic sense: The physicalobservable A has the set of eigenvalues λk, which contribute to the expectation< Ψ(t)|A|Ψ(t) > with |cj(t)|2.

A prediction of the outcome of a quantum mechanical experiment, there-fore, has to assign a probability |cj(t)|2 to the measurement of the eigenvalueλA

j .

6.4 Measurement Process

In a double slit experiment it is not possible to predict the location wherea single particle will hit the screen with certainty. Shortly after the particlehas hit the screen, however, it can be definitely localized. Only predictionson probabilities of the outcome of an experiment are possible. The outcomeof a measurement of a physical observable is uncertain, after measurementit is certain. This fact has to be incorporated into a quantum theory and isfrequently denoted as the collapse of the wave function.

We consider a quantum system described by the state vector Ψ(t) >. Weintend to measure a physical quantity represented by the Hermitian operatorA. The possible outcomes of the measurement our the set of eigenvalues ai

with corresponding eigenstates |ϕai >. A decomposition of the state vector

yields

|Ψ(t) >=∑

i

ci(t)|ϕAi > (6.11)

The probability, to measure the value ai is given by

pi = |ci(t)|2 (6.12)

The knowledge of pi allows one to make statistical predictions.After the measurement, where we definitely measure the value aj we know

that the state vector is|Psi(t+ 0) >= |ϕa

j > (6.13)

Therefore, the measurement changes the state vector from

Ψ(t− 0) >=∑

i

ci(t)|ϕai >→ |Ψ(t+ 0) >= |ϕa

j > (6.14)

This property of a quantum system is denoted as the collapse of the state

vector.

7

Harmonic Oscillators

7.1 Classical Harmonic Oscillator

We shall briefly reconsider the treatment of a harmonic oscillator in classicalmechanics.

7.1.1 Hamilton Function

The Hamilton function reads

H =p2

2m+mω2

2q2 (7.1)

and the canonical equations read

q =∂

∂pH(p, q) =

p

m

p = − ∂

∂pH(p, q) = −mω2q (7.2)

Elimination of the momentum p yields a second order differential equation

q + ω2q = 0 (7.3)

The solution reads

q(t) = Asinωt+Bcosωt

p(t) = Aω

msinωt+B

ω

mcosωt (7.4)

7.1.2 Phase Space

One can represent the instantaneous classical state of an oscillator by thespecification of [q(t), p(t)]. Therefore, the temporal evolution can be viewed

58 7 Harmonic Oscillators

by the evolution of this point in phase space spanned by the coordinates q, p.The energy is a constant and the trajectorie of an oscillator is an ellipse

1 =p2

2mE+mω2q2

2E(7.5)

with the semi axis

a =

√

2E

mω2

b =√

2mE (7.6)

7.1.3 Bohr-Sommerfeld Quantization

A simple way to restrict the possible energy values of a harmonic oscillatoris to specify a criterion, which selects specific classical trajectories in phasespace. The Bohr-Sommerfeld quanitization rule consists in resctricting thearea encircled by a certain motion. This quantity has the dimension of anaction, or, the dimension of the Planck’s constant h. The restriction, therefore,reads

∮

pdq = nh (7.7)

The area equals the area of an ellipse with the axis

b =√

2mE , a =

√

2E

mω2(7.8)

The area encircled by an ellipse is

πab =2πE

ω= nh (7.9)

leading to the energy spectrum

En = hω (7.10)

7.2 Quantum Mechanical Harmonic Oscillator

In this section, we shall derive the energy spectrum of the harmonic oscillatorfrom a solution of the Schrodinger equation:

ihΨ(X, t) = [− h2

2m

∂2

∂q2+mω2

2q2]Ψ(x, t) (7.11)

7.2 Quantum Mechanical Harmonic Oscillator 59

7.2.1 Eigenvalue Problem: Hamilton Operator

A separation ansatz of the form

Ψ(x, t) = e−i Eh

tϕ(x) (7.12)

leads us to the linear eigenvalue problem

E|ϕ >= [− h2

2m

∂2

∂q2+mω2

2q2]|ϕ > (7.13)

which determines the energy spectrum and the corresponding eigenvectors.For the following, it is convenient to introduce the dimensionless quantity

x =

√

mω

hq (7.14)

This transformation is taken in the form that the energy is measured in mul-tiples of the quantity hω.

The transformation turns the eigenvalue problem into

Ehωϕ(x) = ǫϕ(x) =1

2[− ∂2

∂x2+ x2]ϕ(x) (7.15)

where E = hωǫ.There are several methods to solve this eigenvalue problem. We shall first

consider a method, which frequently is denoted as Sommerfeld’s polynom

method , which is based on a direct treatment of eq. (7.15) as an exampleof a so-called Sturm-Liouville eigenvalue problem. A second treatment will bebased on an algebraic method using occupation numbers.


As a first step towards the solution of the eigenvalue problem we consider thefollowing representation of the Hamilton operator

H =1

2[− ∂2

∂x2+ x2] (7.16)

We try to split the operator [− ∂2

∂x2 + x2] into a product of two operators. Tothis end we consider the product

[∂

∂x+ x][− ∂

∂x+ x] (7.17)

which would be a valid representation in the case that the opertors ∂∂x and x

would commute. We explicitly calculate


[− ∂

∂x+ x][

∂

∂x+ x] = − ∂2

∂x2− [

∂

∂x, x] + x2 (7.18)

However, the commutator

[∂

∂x, x] = 1 (7.19)

does not vanish. As a consequence,

H =1

2[− ∂

∂x+ x][

∂

∂x+ x] +

1

2(7.20)

7.3 Sommerfeld’s Polynom Method

7.3.1 Ground State

We can easily find a solution of the eigenvalue problem

1

2[− ∂

∂x+ x][

∂

∂x+ x]ϕ(x) = (ǫ− 1

2)ϕ(x) (7.21)

To this end, we seek a solution which obeys

[∂

∂x+ x]ϕ0(x) = 0 (7.22)

with the energy eigenvalue

ǫ0 =1

2(7.23)

The first order differential equation (7.22) is easily solved. It yields

ϕ0(x) = Ne−x2

2 (7.24)

The normalization constant N is determined according to

1 = N2

∫ ∞

−∞

dxe−x2

= N2√π (7.25)

We can now summarize the expression for the so-called ground state

E =hω

2, ϕ0(x) =

1

π1/4e−

x2

2 (7.26)

It will turn out that this state is the one with the lowest energy value, E = hω2 .

7.3 Sommerfeld’s Polynom Method 61

7.3.2 Higher Eigenfunctions

In order to look for other eigenfunctions and eigenvalues we perform the ansatz

ϕ(x) = e−x2

2 H(x) , (7.27)

where H(x) is a function which will be specified in the following.To this end we calculate:

d2

dx2ϕ(x) = e−

x2

2 [d2

dx2H(x) + 2

d

dxH(x)(−x) + x2H(x) −H(x) (7.28)

The linear eigenvalue problem then reads:

λH(x) =1

2(− d2

dx2H(x) + 2x

d

dxH(x) +H(x)) (7.29)

This is the so-called Hermite eigenvalue problem. Possible solutions are theso-called Hermite polynomials. It is an example of a so-called Sturm-Liouville

eigenvalue problem.It will turn out that we can find solutions only for special values of λ,

provided the function ϕ(x) is bounded. To this end, H(x) has to obey in thelimit |x| → ∞

|H(x)| < ex2

2 (7.30)

In this case ϕ(x) → 0 at infinity, which is consistent with the interpretationas a probability density.

Let us consider various types of ansatza)

H0 = c , λ =1

2(7.31)

b)H1 = ax+ b (7.32)

We obtain

λ(ax+ b) =1

2[2ax+ ax+ b] (7.33)

The solvability condition reads

λ =3

2b = 0 (7.34)

This suggests that we can look for an ansatz in terms of a polynomial: Thisis denoted as Sommerfeld’s method:


H(x) =∑

k=0

xkak (7.35)

This yields

λ∑

k=0

akxk =

1

2[−

∑

k=2

akxk−2k(k − 1) + 2

∑

k=1

akkxk +

∑

k=0

akxk

=1

2[−

∑

k=0

ak+2xk(k + 2)(k + 1) + 2

∑

k=1

akkxk +

∑

k=0

akxk(7.36)

Now, we can read off the following recursion relation

ak+2 =[(2k + 1) − 2λ]

(k + 2)(k + 1)ak (7.37)

We can distinguish solutions with even and odd parity. Solutions with even

parity are obtained using a0 6= 0, a1 = 0 H(x) = H(−x), whereas solutionswith odd parity a1 6= 0, a0 = 0 H(x) = −H(−x).

λn = n+1

2(7.38)

The recursion relation yields for the even Hermite polynomials:

Hnk+2 =

2(k − n)

(k + 1)(k + 2)Hn

k k = 0, 2, ..., n = 2l (7.39)

The Taylor coefficients Hnk for the odd Hermite polynomials obey the same

recursion relation, however, with k = 1, 3, ..n = 2l + 1.From this relation we see that the Hn(x) are polynomials of degree n,

since the solvability condition yields an 6= 0, an+2 = an+4... = 0, and an+1 =an+3... = 0. The functions Hn(x) are even or odd polynomials in x, the so-called Hermite polynomials.

Even Hermite polynomials Hn(x):

n = 0 1

n = 2 4x2 − 2

n = 4 16x4 − 48x2 + 12

Odd Hermite polynomials Hn(x):

n = 1 2x

n = 3 8x3 − 12x

n = 5 32x5 − 160x3 + 120x

7.3 Sommerfeld’s Polynom Method 63

7.3.3 Properties of Hermite Polynomials

Orthogonality

The Hamilton-operator H is an Hermitian operator. As a consequence, theeigenfunctions ϕn(x) form a complete orthogonal set. This implies the follow-ing relationship for the Hermite functions

< ϕn|ϕm >=

∫ ∞

−∞

dxe−x2

Hn(x)Hm(x) = 2nn!√πδnm (7.40)

The normalization of the Hermite polynomials are introduced according toAbramowitz, Stegun, Handbook of Mathematical Functions.

Rodriques Formula

Hermite polynomials can be iteratively calculated by subsequent differentia-tion. We obtain

Hn(x) = ex2

(x− d

dx)ne−x2

(7.41)

We shall give a proof of this relation below.

Generating Function

A generating function G(x, t) is a function, whose Taylor expansion in t isrelated with a class of functions fn(x) according to

G(x, t) =∑

n=0

anfn(x)tn (7.42)

The generating function for the Hermite polynomials is

G(x, t) = e2xt−t2 =∑

n=0

1

n!Hn(x)tn (7.43)

Recurrence Relation

One can establish the following relation between the Hermite polynomial oforde n and n+ 1:

Hn+1(x) = 2xHn(x) − ∂

∂xHn(x) (7.44)


7.3.4 Spectrum and Eigenfunctions

We can now summarize: The energy spectrum of the harmonic oscillator is

En = hω(n+1

2) (7.45)

and the eigenfunctions are

ϕn(x) =1

√

2nn!√πe−

x2

2 Hn(x) (7.46)

These eigenfunctions form a complete and orthogonal set.∫ ∞

−∞

ϕn(x)ϕm(x)dξ = N2n

∫ ∞

−∞

dxe−ξ2

Hn(ξ)Hm(ξ) = δnm (7.47)

∞∑

n=0

ϕn(ξ)ϕm(ξ′) =

∞∑

n=0

N2ne

−ξ2/2−ξ′2/2Hn(ξ)Hn(ξ′) = δ(x− x′) (7.48)

We can now express the eigenfunctions in terms of the coordinate x =√

mωh q:

ϕn(q) = (mω

h)1/4 1√

2nn!Hn(

√

mω

hq)e−

mω2h

q2

(7.49)

7.4 Algebraic Treatment

7.4.1 Creation, Annihilation Operators

We start from the Hamilton operator

H = hω1

2[− d2

dx2+ x2] (7.50)

and introduce the operators b, b∗ by the definitions

b =1√2[x+

d

dx]

b∗ =1√2[x− d

dx] (7.51)

It is straightforward to show that b∗ is an operator adjoint to the operatorb. To this end we consider

< ψ1(x, t)|b|ψ2(x, t) =√

12

∫ ∞

−∞

dxψ1(x, t)∗(x +

d

dx)ψ2(x, t)

=√

12

∫ ∞

−∞

dx[(x − d

dx)ψ1(x, t)

∗]ψ2(x, t) (7.52)

7.4 Algebraic Treatment 65

Harmonic oscillator: Eigenstates ϕn(x), andprobability densities |ϕn(x)|2.

We now express the Hamilton operator in terms of the operators b†, b. Tothis end we determine the operator product

b∗b =1

2(x− d

dx)(x+

d

dx) =

1

2[x2 − d2

dx2− d

dxx+ x

d

dx]

=1

2[x2 − d2

dx2− 1] (7.53)

This yields the explicit representation of the Hamilton operator in terms ofthe operators b∗, b:

H = hω(b∗b+1

2) (7.54)

7.4.2 Commutators

In this subsection we determine various commutators, which will be neededin the algabraic treatment of the eigenvalue problem of harmonic oscillators.

First, we evaluate the commutator between b, b†:

[b, b†] =1

2[(x+

d

dx)(x − d

dx) − (x− d

dx)(x+

d

dx)]

=1

2[2d

dxx− x

d

dx] = 1 (7.55)


This allows one to calculate in a straightforward manner the commutators[b, b†b] and [b†, b†b]:

[b, b†b] = bb†b− b†bb = [b, b†]b = b (7.56)

as well as[b†, b†b] = b†b†b− b†bb† = b†[b†, b] = −b† (7.57)

The commutators of the operators b†, b with the Hamilton operator H are

[b,H ] = hωb (7.58)

[b†, H ] = −hωb† (7.59)

7.4.3 The Ladder

We denote the eigenvectors of the Hamiltonian by |n >:

H |n >= En|n > (7.60)

We observe the validity of the following property: If |n > is an eigenstateof the Hamilton operator, then Nb|n > is an eigenstate with energy En − hω.The constant N is determined from the normalization of the eigenstate |n >:

< n|n >= 1 (7.61)

For the proof we consider the relation

bH |n >= Enb|n >= Hb|n > +hωb|n > (7.62)

where we have made use of the commutator [b,H ] = b. We can rearrange thisexpression as follows

Hb|n >= (En − hω)b|n > (7.63)

From this relation we conclude that if |n > is an eigenstate of the Hamiltonoperator, then Nb|n > is an eigenstate with energy En − hω. Therefore, theoperator is denoted as an annihilation operator b.

We can make a similar statement concerning the operator b†. We start bycalculating

b∗H |n >= Enb∗|n >= Hb|n > −hωb∗|n > (7.64)

and rewrite this expression using the commutator as

Hb∗|n >= (En + hω)b∗|n > (7.65)

This indicates that b∗|n > is an eigenstate with eigenvalue En + hω. Theoperator b† is denoted as creation operator.


7.4.4 Positivity of Energy Eigenstates

The expectation value of the Hamilton operator of the harmonic oscillator ispositive:

Proof:

< ψ(x, t)|H |ψ(x, t) > = hω < ψ|b∗b+1

2|ψ >

= hω[< bψ|bψ > +1

2] (7.66)

As a consequence, there exist a ground state, i.e. the state with lowestenergy eigenvalue. We denote this state by |0 >. Inserting |ψ >= |0 > weobtain

e0 =< 0|b†b|0 > +1

2≥ 1

2(7.67)

Furthermore, since there is no eigenstate with E < E0, we conclude onthe basis of the relation

b|0 >= (E0 − hω)b|0 > (7.68)

thatb|0 >= 0 (7.69)

This fixes the energy eigenvalue of the ground state |0 > to

E0 =1

2hω (7.70)

The eigenvalues of the excited state are obtained

E1 = hω(1 +1

2) = E0 + hω

E2 = hω(2 +1

2) = E1 + hω

E3 = hω(3 +1

2) = E2 + hω (7.71)

It is straightforward to determine the ground state |0 > using the explicitdefinition of the operator b,

1√2(x+

d

dx)ϕ0(x) = 0 (7.72)

We have already solved this differential equation of first order.


7.4.5 Occupation Number Operator

Thie operator b†b has the meaning of an occupation number operator n.This becomes obvious from considering the eigenvalue problem

H |n >= hω(b†b+1

2)|n >=

(

n+1

2

)

hω|n > (7.73)

where we have explicitly inserted the obtained expressions for the energyeigenvalues En. This eigenvalue problem defines the eigenvalues and eigenvec-tors of the number operator n

n|n >= b†b|n >= n|n > (7.74)

7.4.6 Normalization of Eigenstates

As we have seen above, we can determine the eigenvector |n + 1 > from theknowledge of the eigenvector |n >:

|n+ 1 >= Nb∗|n > (7.75)

We now calculate the normalization constant N from the condition

< n+ 1|n+ 1 > = N2 < b∗n|b∗n >= N2 < n|bb∗|n >= N2 < n|b∗b|n > +N2 = N2(n+ 1) < n|n > (7.76)

Since we assume |n > to be normalized we explicitly obtain the relationship

|n+ 1 >=1√n+ 1

b∗|n > (7.77)

In a similar way we can determine the eigenstate |n−1 > from the knowldeof |n >:

|n− 1 >= Nb|n > (7.78)

with the normalization constant to be determined from the condition

< n− 1|n− 1 >= N2 < bn|bn >= N2 < n|b∗b|n >= N2 < n|n >= 1 (7.79)

Explicitly, we obtain

|n− 1 >=1√nb|n > (7.80)

Let us summarize:

|n+ 1 > = b∗1√n+ 1

|n >

|n− 1 > = b1√n|n > (7.81)

We can build each eigenvector starting from the ground state by a repeatediteration of (7.81):

|n >=1√n!

(b∗)n|0 > (7.82)


Relation with Hermite’s polynomials:

b|0 >=1√2(x+

d

dx)H0(x)e

− x2

2 = 0 (7.83)

We obtain

|n >= Hn(x)e−x2

2 =1√n!

1√2

n (x − d

dx)nH0e

− x2

2 (7.84)

Which leads us to the following recursion relation for Hermite’s polynomials:

Hn(x) =1√n!

1√2

n ex2

2 (x − d

dx)ne−

x2

2 H0 (7.85)

7.4.7 Uncertainty Relationships

We calculate the expectation of the variables q and p in the excited states.We obtain

q =

√

h

mωx =

√

h

2mω(b+ b∗)

p =h

i

d

dq=h

i

d

dq=

1

i

√hmω

d

dx

=1

i

√

hmω

2(b− b∗) (7.86)

Let us now determine the expectation values

< n|q|n > =

√

h

2mω< n|(b+ b∗)|n >= 0

< n|p|n > =1

i

√

hmω

2< n|(b− b∗)|n >= 0 (7.87)

where we have exploited the fact that

< n|n− 1 >= 0 (7.88)

In order to determine the uncertainties we calculate

< n|q2|n > =h

2mω< n|(b+ b∗)2|n >

=h

2mω< n|(b+ b∗)2|n >=

h

mω(n+

1

2)

< n|p2|n > = −mωh2

< n|(b− b∗)2|n >

= −mωh2

< n|(b)2 − bb∗ − b∗b + (b∗)2|n >

= mωh(n+1

2) (7.89)


We can summarize the uncertainty relation as follows:

√

< n|q2|n >< n|p2|n > = h(n+1

2) (7.90)

7.5 Coupled Harmonic Oscillators

7.5.1 Harmonic Oscillator in 3 Dimensions

In this subsection we consider the quantum mechanical description of theharmonic oscillator in three dimensions. The Hamilton operator reads

H(p,q) =p2

x + p2y + p2

z

2m+mω2

2[q2x + q2y + q2z ] (7.91)

This Hamilton operator can be written as the sum of three operators, eachdescribing a one-dimensional harmonic oscillator:

H(p,q) = Hx(px, qx) +Hy(py, qy) +Hz(pz, qz) (7.92)

In the following we use the Hamiltonian in the dimensionless variable x =√

mωh , i.e.

Hx(px, x) = hω2[− ∂2

∂x2+ x2] (7.93)

and, similarly, for the y,z directions.We will show that the solution of the time independent Schrodinger equa-

tion

[Hx(px, x) +Hy(py, y) +Hz(pz, z)]ϕ(x, y, z) = Eϕ(x, y, z) (7.94)

can be reduced to the eigenvalue problem of the single harmonic oscillator.To this end, we perform a product ansatz for the eigenfunction

ϕ(x, y, z) = ϕx(x)ϕy(y)ϕz(z) (7.95)

Thereby, the functions ϕx(x), ϕy(y), ϕz(z) are only functions of x, y, z, re-spectively. We insert this product ansatz into the eigenvalue problem (7.94)and consider the fact, that the operator Hx(px, x) acts only on the functionϕ(x):

H(px, x)ϕx(x)ϕy(y)ϕz(z) = ϕy(y)ϕz(z)H(px, x)ϕx(x) (7.96)

and similar relations for Hy(py, y)ϕ(x, y, z) and Hz(pz , z)ϕ(x, y, z). This prop-erty, which is based on the special form of the Hamiltonian (7.92), allows oneto rewrite the eigenvalue problem (7.94) in the form

7.5 Coupled Harmonic Oscillators 71

1

ϕx(x)Hx(px, x)ϕx(x) +

1

ϕy(y)Hy(py, y)ϕy(y) +

1

ϕz(z)Hz(pz, z)ϕz(z) = E

(7.97)From the fact that the left hand side is a sum of three functions, the first onedepending on x, the second one on y and the third one on z, whereas the righthand side is independent on x, y, z we conclude that each term on the righthand side has to be a constant, which we denote as Ex, Ey, Ez :

Hx(px, x)ϕx(x) = Exϕx(x)

Hx(py, y)ϕy(y) = Eyϕy(y)

Hz(pz, z)ϕz(z) = Ezϕz(z) (7.98)

We recognize that, for each direction, we obtain the eigenvalue problem of theharmonic oscillator. Each energy eigenvalue Ex, Ey, Ez is then given by

Exnx

= hω[nx +1

2]

Eyny

= hω[ny +1

2]

Eznz

= hω[nz +1

2] (7.99)

with ni = 0, 1, ... and the corresponding eigenfunctions

ϕnx(x) = NHnx

e−x2

2 (7.100)

As a consequence, the energy eigenvalues E of the Hamiltonian of the threedimensional harmonic oscillator are characterized by three quantum numbersnx, ny, nz

Enx,ny,nz= hω[nx +

1

2]

+ hω[ny +1

2] + hω[nz +

1

2] (7.101)

and the corresponding eigenfunction is the product of the eigenfunctions ofthe harmonic oscillators

ϕnx,ny,nz(x, y, z) = ϕnx

(x)ϕny(y)ϕnz

(z) (7.102)

For example, the ground state takes the form

ϕ0,0,0 = Ne−x2+y2+z2

2 (7.103)

These eigenfunctions form an orthornormal product basis of the Hilbertspace of square integrable functions in R3. Each wave function can be repre-sented as a superposition of these basis functions


Ψ(x, y, z, t) =∑

nx,ny,nz

cnx,ny,nz(t)ϕnx,ny,nz

(x, y, z) (7.104)

and the expansion coefficients cnx,ny,nz(t) can be determined from the function

Ψ(x, y, z, t) by the evaluation of the scalar products

cnx,ny,nz(t) =

∫

dx

∫

dy

∫

dz[ϕnx(x)ϕny

(y)ϕnz(z)]∗Ψ(x, y, z, t) (7.105)

We can reformulate the treatment of the three dimensional harmonic oscil-lator in terms of the occupation number formalism. To this end we introduceannihilation and generation operators for each direction. This allows one torewrite the Hamiltonian (7.92) in the form

H = hωx[b†xbx +1

2] + hωy[b

†yby +

1

2] + hωz[b

†zbz +

1

2] (7.106)

Here, we have slightly generalized our treatment by considering different fre-quencies ωx, ωy, ωz for each direction.

It is straightforward to formulate the commutators of the operators b†i , bi.The commutation rules can be summarized as follows

[bi, bj ] = 0 , [b†i , b†j ] = 0 , [bi, b

†j ] = δij (7.107)

The eigenvectors of the Hamilton operator are

|nx, ny, nz >= |nx > |ny > |nz > (7.108)

and obeyH |nx, ny, nz >= Enx,ny,nz

|nx, ny, nz > (7.109)

The occupation number operator acts on the eigenvectors in the followingway

nx|nx, ny, nz >= b†xbx|nx, ny, nz >= nx|nx, ny, nz > (7.110)

The eigenvectors can be determined from the ground state |0, 0, 0 > via

the creation operators b†i :

|nx, ny, nz >=1√nx!

1√

ny!

1√nz!

(b†x)nx(b†y)ny (b†z)

nz |0, 0, 0 > (7.111)

7.5.2 Aplications: Molecular Oscillations

We consider a simple model of a molecule, which we describe by the momentaand the locations of its atoms. The atoms are glued together by electrons. Wedescribe this binding force by an effective interacting between the atoms. Thisforce is derived from the potential V (|Q1, ..Qn). Newton’s law yields

7.5 Coupled Harmonic Oscillators 73

d

dtPi = −∇Qi

V (Q1, ..,Qn) (7.112)

A stable configuration of the molecule, denoted by Q0i is given by the condition

−∇QiV (Q1, ..,Qn) = 0 (7.113)

The Hamilton function is given by

H =∑

i

Q2i

2mi+ V (Q1, ..,Qn) (7.114)

In order to describe oscillations of the molecule around the equilibrium stateQ0

i we introduce the deviations

Qi = Q0i + qi Pi = pi (7.115)

In terms of these quantities, the Hamilton function takes the form

H =∑

j

pi2

2mi+ V (Q0

1 + q1, ...,Q0n + qn) (7.116)

We can now expand the potential energy around the equilibrium positionsand obtain

V (Q01 + q1, ...,Q

0n + qn) = V (Q0

1, ...,Q0n) +

1

2

∑

i,j

∇Q0i∇Q0

jV (Q0

1, ...,Q0n)qiqj

(7.117)The first order term of the Taylor expansion vanishes, since we consider theequilibrium position.

We can lump all coordinates into the vectors p and q and obtain theHamilton function in the harmonic approximation as

H =∑

α

p2α

2mα+

1

2

∑

α,β

Vα,βqαqβ (7.118)

Furthermore, we perform the transformation

p′α = pα/√√

mα (7.119)

The Hamilton function then reads

H =p2

2+

1

2q · V q (7.120)

We introduce the eigenvectors uα of the matrix V

V uα = λαuα (7.121)


and the transformation

q =∑

α

cα(t)uα

p =∑

α

cα(t)uα (7.122)

As a result we obtain the Hamilton function

H =∑

α

[|cα| + λα|cα|2] (7.123)

Since the equilibrium state is stable, all eigenvalues λα ≥ 0.It turns out that due to translational invariance of the molecule there

are at least three eigenvalues which are zero. The corresponding motions arerelated to a pure translation of the molecule.

8

Hydrogen Atom

8.1 Classical Model

The classical treatment of the Hydrogen atom consisting of a positive coresurrounded by an electron is analogous to the Kepler problem of the motion ofplanets around the sun. The classical formulation of this problem is completelyintegrable.

Let us consider the mathematical formulation. The masses and the posi-tions of the core and the electron are denoted by mc, rc and me, re, respec-tively. Newton’s law reads

mere = K(re − rc)

mcrc = −K(re − rc) (8.1)

where we have denoted the force between core and electron by K(r).The Hamilton function reads

H =p2

e

2me+

p2c

2mc+ V (|re − rc|) (8.2)

8.1.1 Center of Mass, Relative Motion

As is well-known, the classical two body problem can be separated into themotion of the center of mass, R = mere+mcrc

me+mc, and the relative coordinate

r = re − rc:

R = 0

mr = K(r) (8.3)

m denotes the reduced mass defined according to

m =memc

me +mc(8.4)

76 8 Hydrogen Atom

The motion of the center of mass and the relative motion decouple.The Hamilton function of the relative motion reads

H =p2

2µ+ V (r) (8.5)

r is the relative coordinate. The potential for the case of the Hydrogen atomis given by

V = −γr

= − 1

4πǫ0

e2

r(8.6)

8.1.2 Conservation Laws

The classical problem (Kepler problem) is integrable due to the existence ofthe conservation of energy and angular momentum.

8.2 Quantization

H = − h2

2mc∆rc

− h2

2me∆re

+ V (|re − rc|) (8.7)

We introduce relative coordinate and center of mass coordinate accordingto

r = re − rc

R =mere +mcrc

me +mc(8.8)

The inverse transform is given by (M = me +mc)

re = R +mc

Mr

rc = R +me

Mr (8.9)

This transformation yields the Hamiltonian

H = − h2

2M∆R − h2

2m∆r + V (|r|) (8.10)

The proof is straightforward. We express the gradients with respect to re

and rc in terms of R and r:

∇re=me

M∇R + ∇r

∇rc=mc

M∇R −∇r (8.11)

Subsequently, we calculate

8.3 Hamilton Operator in Spherical Coordinates 77

8.2.1 Separation of Relative and Center of Mass Motion

The Hamiltonian of the joint motion of core and electron is separated intothe Hamiltonian for the center of mass, HR, and the Hamiltonian Hr of therelative motion:

H = HR +Hr (8.12)

The center of mass part, HR, is the Hamiltonian of the motion of a freeparticle.

For the solution of the time dependent Schrodinger equation we can per-form a separation ansatz

Ψ(R, r, t) = ΨR(R, t)Ψr(r, t) (8.13)

This separation ansatz yields the two Schrodinger equations

ih∂

∂tΨR(R, t) = HRΨR(R, t)

ih∂

∂tΨr(r, t) = HrΨr(r, t) (8.14)

8.2.2 Time Independent Schrodinger Equation

The ansatzΨr(r, t) = e−i E

htψ(r) (8.15)

leads to the time independent Schrodinger equation

Eψ(r) = − h2

2m∆ψ(r) + V (|r|)ψ(r) (8.16)

8.3 Hamilton Operator in Spherical Coordinates

In this section we consider the quantum mechanical treatment of the motionof a particle of mass m in a central force field, which is obtained from thepotential

K(r) = −∇rV (r) = −erV′(r) (8.17)

Due to spherical symmetry, it is convenient to express the wave functionΨ(r, t) in spherical coordinates r, θ, ϕ:

x = rsinθcosϕ

y = rsinθsinϕ

z = rcosθ (8.18)

Our first task consists in expressing the operators ∇, ∆ in spherical coordi-nates.

78 8 Hydrogen Atom

8.3.1 Gradient and Momentum Operator: Spherical Coordinates

In this subsection, we shall derive a representation of the nabla operator, and,in turn, the momentum operator in spherical coordinates. Our starting pointis the decomposition of the gradient of a function f(r) into a radial and atransversal part:

∇f = erer · (∇f) − er × [er × (∇f)] (8.19)

where er denotes the unit vector in radial direction:

[er]i =rir

(8.20)

It is straightforward to proof that in spherical coordinates

er · ∇ =rjr

∂

∂rj=

∂

∂r(8.21)

We consider now the momentum operator

p =h

i∇

= erpr −1

rer × L (8.22)

Here, we have defined the operator of angluar momentum:

L = r × h

i∇ (8.23)

We can now determine the Laplacian in terms of the operator of angularmomentum. We calaculate

∂

∂ri

∂

∂ri=

∂

∂ri[rir

∂

∂r− ǫijk

rjr

i

hLk]

=2

r

∂

∂r+

∂2

∂r2− i

hǫijk

[(

∂

∂ri

rjr

)

+rjr

∂

∂riLk

]

=1

r2∂

∂rr2∂

∂r− i

hǫijk

rjr

∂

∂riLk (8.24)

We can summarize the representation of the Laplacian in terms of theoperator of angular momentum

∆ =1

r2∂

∂rr2∂

∂r− 1

r2h2 L2 (8.25)

In the appendix we shall explicitly specify this operator in spherical coor-dinates:

8.3 Hamilton Operator in Spherical Coordinates 79

∆ =1

r2∂

∂rr2∂

∂r+

1

r2[

1

sinθ

∂

∂θsinθ

∂

∂θ+

1

sin2θ

∂2

∂ϕ2] (8.26)

As a consequence, the Hamiltonian for the motion of a particle in a spher-ically symmetric potential reads

H = − h2

2m

1

r2∂

∂rr2∂

∂r+

L2

2mr2+ V (r) (8.27)

Thereby, we have defined the operator

L2 = −h2[1

sinθ

∂

∂θsinθ

∂

∂θ+

1

sin2θ

∂2

∂ϕ2] (8.28)

As we shall see this operator is related to the operator of angular momentum,L

8.3.2 Separation Ansatz

The time independent Schrodinger equation

Hψ(x) = Eψ(x) (8.29)

is a partial differential equation. It is straightforward to perform the followingseparation ansatz

ψ(x) = R(r)Y (Θ,ϕ) (8.30)

Thereby, we exploit the fact that L2 is an operator acting only on the functionY (Θ,ϕ). Inserting this ansatz into time independent Schrodinger equation weobtain

E = − h2

2m

1

r2∂

∂rr2∂

∂rR(r) +

1

2mr2L2Y (θ, ϕ)

Y (θ, ϕ)(8.31)

In order that this separation ansatz makes sense, the quantity

L2Y (θ, ϕ)

Y (θ, ϕ)= h2Ω (8.32)

has to be a constant, which we take to be h2Ω. We recognize that this condi-tion leads to an eigenvalue problem corresponding to the operator L2:

L2Y (θ, ϕ) = h2ΩY (θ, ϕ) (8.33)

The investigation of this eigenvalue problem will be performed in the nextsection.

The separation ansatz leads us to the following eigenvalue problem for theradial part R(r) of the wave function

[− h2

2m

1

r2∂

∂rr2∂

∂r+

h2Ω

2mr2+ V (r)]R(r) = ER(r) (8.34)

The eigenvalue problem for the motion of a particle in a central potentialis decomposed into two eigenvalue problems, one for the angular part, theother for the radial part of the wave function.

80 8 Hydrogen Atom

8.4 Angular Momentum

8.4.1 Eigenvalue Problem

In this subsection we discuss a direct solution of the eigenvalue problem

L2Y (θ, ϕ) = h2ΩY (θ, ϕ) (8.35)

with the explicit form of the operator L2 in spherical coordinates:

−[1

sinθ

∂

∂θsinθ

∂

∂θ+

1

sin2θ

∂2

∂ϕ2]Y (θ, ϕ) = ΩY (θ, ϕ) (8.36)

We perform the ansatz

Y (θ, ϕ) = Ym(θ)eimϕ (8.37)

In order to fullfill the boundary condition

Y (θ, ϕ+ 2π) = Y (θ, ϕ) (8.38)

we have to assume that m = 0,±1, .... This yields the following equation

−[1

sinθ

∂

∂θsinθ

∂

∂θ− m2

sin2θ]Ym(θ) = ΩYm(θ) (8.39)

We furthermore introduce the coordinate

x = cosθd

dθ=dx

dθ

d

dx= − sin θ

d

dx(8.40)

to obtain the equation

d

dx(1 − x2)

d

dxPm

l − m2

1 − x2Pm

l (x) = −ΩPml (x) (8.41)

This is the associated Legendre differential equation.In the case m = 0 we obtain the Legendre differential equation, which can

be written in the form

(1 − x2)P ′′l (x) − 2xP ′

l (x) = λPl(x) (8.42)

The associated Legendre functions are obtained from the Legendre func-tions by differentiation with respect to x:

Pml (x) = (−1)m(1 − x2)m/2 d

m

dxmPl(x) (8.43)

l Pl(x)0 11 x2 1

2 (3x2 − 1)

3 12 (5x3 − 3x)

4 18 (35x4 − 30x2 + 3)

Legendre Polynomials

8.4 Angular Momentum 81

Sommerfeld’s Polynomial Ansatz

In the following we use Sommerfelds polynom ansatz in order to solve theeigenvalue problem

(1 − x2)P ′′ − 2xP ′ = λP (8.44)

We look for a solution in terms of the Taylor expansion of P (x) at x = 0.

P (x) =∑

k=0

akxk (8.45)

This expansion should converge in the interval −1 ≤ x ≤ 1.Inserting the Taylor expansion into the eigenvalue proble yields

∑

k=0

akk(k − 1)(xk−2 − xk) − 2kakxk =

∑

k=0

λakxk (8.46)

so that we obtain the recursion relation

ak+2(k + 1)(k + 2) − ak(2k + k(k − 1) + λ) = 0 (8.47)

This recursion relation can be cast into the form

ak+2 =(k(k + 1) − λ)

(k + 1)(k + 2)ak (8.48)

Polynomial solutions exist provided the Taylor expansion terminates at k = l.This condition immediately yields the eigenvalues:

λl = −l(l+ 1) (8.49)

The corresponding polynomial eigenfunctions are obtained by the recursionrelation.

Rodrigues Formula

Pl(x) =1

2ll!

dl

dxl(x2 − 1)l (8.50)

Generating Function

1√1 − 2xt+ t2

=∑

l=0

tlPl(x) (8.51)

Bonnet’s Recursion Relation

(l + 1)Pl+1(x) = (2l + 1)xPl(x) − lPl−1(x) (8.52)

82 8 Hydrogen Atom

Orthogonality Property

∫ 1

−1

dxPl(x)Pl′ (x) = δl,l′2

2l+ 1(8.53)

Parity:Pl(x) = (−1)lPl(−x) (8.54)

8.4.2 Algebraic Treatment

The operator of angular momentum is defined by

L = r × p = r × h

i∇ (8.55)

or, explicitly

Li =h

iǫijkrj

∂

∂rk(8.56)

In cartesian ccordinates the x,y,z- components of the operator reads

Lx =h

i[y∂

∂z− z

∂

∂y]Lz =

h

i[z∂

∂x− x

∂

∂z]Lz =

h

i[x∂

∂y− y

∂

∂x] (8.57)

As shown in the appendix, the operators of angular momentum expressedin spherical coordinates are as follows:

Lx =h

i[y∂

∂z− z

∂

∂y]

= − hi[sinϕ

∂

∂θ+ cotθcosϕ

∂

∂ϕ]

Ly =h

i[z∂

∂x− x

∂

∂z]

=h

i[cosϕ

∂

∂θ− cotθsinϕ

∂

∂ϕ]

Lz =h

i[x∂

∂y− y

∂

∂x]

=h

i

∂

∂ϕ(8.58)

Using these representations we can determine L2:

L2 = −h2 1

sinθ

∂

∂θsinθ

∂

∂θ+

1

sin2θ

∂2

∂ϕ2 (8.59)


8.4.3 Eigenvalue Problem connected with L2, Lz

We investigate the eigenvalue problems

L2Y (θ, ϕ) = h2ΩY (θ, ϕ)

LzY (θ, ϕ) = hmΩY (θ, ϕ) (8.60)

Our treatment will be algebraic.

8.4.4 Commutation relations

The various components of the operator of angular momentum do not com-mute. The following commutation rules follow from the definition of the op-erator of angular momentum:

LiLj − LjLi = ǫijkihLk (8.61)

Proof:

LiLj = −h2ǫiklǫjmnxk∂

∂xlxm

∂

∂xn

= −h2ǫiklǫjmnxkxm∂

∂xl

∂

∂xn

− h2ǫiklǫjmnδlmxk∂

∂xn (8.62)

leading to the commutator We interchange the indices i and j and obtainfor the commutator

[Li, Lj ] = −h2ǫiklǫjmnδlmxk∂

∂xn + h2ǫjklǫimnδlmxk

∂

∂xn (8.63)

With the help ofǫiklǫljn = δijδkn − δinδjk (8.64)

we end up with the commutators

[Li, Lj] = −h2[xj∂

∂xi− xi

∂

∂xj] = ihǫijkLk (8.65)

The operators Li are generators of the Lie-group SO(3).

84 8 Hydrogen Atom

The Operator L2

We define tha operator L2

L2 =∑

i

L2i = L2

1 + L22 + L2

3 (8.66)

It is straightforward to determine the commutators between the compo-nents of L and L2:

[Li,L2] = Li

∑

k

L2k −

∑

k

L2kLi (8.67)

We calculate the commutator between

[Li, L2k] = LiL

2k − L2

kLi = +LkLiLk + [Li, Lk]Lk − L2kLi

= L2kLi + Lk[Lk, Li] + [Li, Lk]Lk − L2

kLi =

= ih∑

j

[LkǫkijLj − ǫkijLjLk] = ih∑

j

ǫkij [LkLj − LjLk](8.68)

If we sum over k, we obtain[Li,L

2] = 0 (8.69)

Ladder Operators

It is convenient to define the operators

L+ = Lx + iLy

L− = Lx − iLy = (L+)† (8.70)

They will play a similar role as the operators b† and b for the harmonicoscillator.

The Eigenvalue Problem

We consider the eigenvalue problems

L2|lm > = h2Ω|lm >

Lz|lm > = hm|lm > (8.71)

Thereby, we have denoted the eigenvectors by |lm >. We need to indices toenumerate the eigenvectors since they are eigenvectors to the operators L− zand L2.

a) We first proof the following inequality between the eigenvalues mh andΩh:

Ω ≥ m2 (8.72)


For the proof we explicitly consider

[L2x + L2

y + L2z]|lm >= [L2

x + L2y +m2h2]|lm >= h2Ω|lm > (8.73)

which leads us to

[L2x + L2

y]|lm >= h2(Ω −m2)|lm > (8.74)

Lx, Ly are hermitian operators such that

< lm|L2x + L2

y|lm >=< Lxlm|Lxlm > + < Lxlm|Lxlm >≥ 0 (8.75)

As a result, we obtain the inequality

(Ω −m2) ≥ 0 (8.76)

b) We now show that L+ and L− are ladder operators: Provided that|lm > is an eigenfunction of L2 and Lz with eigenvalue mh, L±|lm > is aneigenfunction of L2 and Lz with eigenvalue (m± 1)h

Proof: The operators L± commute with the operator L2. Therefore, theypossess common eigenvectors. It remains to show that L±|lm > is also aneigenfunction of Lz. To this end we calculate the commutator

[Lz, L±] = [Lz, Lx] ± i[Lz, Ly] = ihLy ± hLx = ±hL± (8.77)

The proof is straightforward:

L±LzYlm = −[Lz, L±]Ylm + LzL±Ylm = hmL±Ylm (8.78)

Using the commutator (8.77) we conclude

LzL±Ylm = h(m± 1)Ylm (8.79)

so that up to a normalization constant Nlm

L±Ylm = NlmYl,m±1 (8.80)

c) We are now in the position to determine the possible eigenvalues Ω andm. Since the quantity

Ω −m2 ≥ 0 (8.81)

there exist minimal and maximal values mmin, mmax. Since the operators L±

are ladder operators we conclude that

L−Ylmmin= 0 , L+Ylmmax

= 0 (8.82)

We now determine the eigenvalues of L2. To this end we consider

L−L+ = (Lx − iLy)(Lx + iLy) = L2x + L2

y + ih[Lx, Ly] (8.83)

86 8 Hydrogen Atom

L−L+ = L2x + L2

y − hLz = L2 − Lz(Lz + h) (8.84)

In an analogous way, we obtain

L+L− = L2x + L2

y + hLz = L2 − Lz(Lz − h) (8.85)

Thus,

L−L+|lmmax >= 0 = h2(Ω −m2max −mmax)|lmmax > (8.86)

L+L−|lmmin >= 0 = h2(Ω −m2min +mmin)|lmmin > (8.87)

However, we have to postulate that

Ω = mmax(mmax + 1) = mmin(mmin − 1) (8.88)

This relation can be combined to yield

0 = mmax(mmax + 1) −mmin(mmin − 1)

= (mmax +mmin)(mmax −mmin + 1) (8.89)

From here, we concludemmax = −mmin (8.90)

Possibilities: mmin = −lAngular momentum: m = −l, .., 0, .., lSpin l = 1/2, l = 3

2 , l = 52

d) The LadderWe specify explicitly the relations between the eigenfunctions lm > and

l,m± 1 > which arise due to the operators L±. We have the relation

L±|l,m >= Nl,±|l,m± 1 > (8.91)

We now determine the normalization constant Nl,±. We obtain

< l,m± 1|l,m± 1 > =1

N2l,±

< L±lm|L±lm >

=1

N2l,±

< lm|L∓L±|lm > (8.92)

We can now express the operators L∓L± in terms of L2, Lz:

L∓L± = L2 − Lz(Lz ± h) (8.93)

Therefore,Nl,± = h

√

l(l+ 1) −m(m± 1) (8.94)

We are now in a position to summarize the relations induced by the ladderoperator:

L+|l,m > = h√

l(l + 1) −m(m+ 1)|l,m+ 1 >

L−|l,m > = h√

l(l + 1) −m(m− 1)|l,m− 1 > (8.95)

These relations especially include

L−|l,−l >= 0 L+|l, l >= 0 (8.96)


Spherical Harmonics

The operators L± can be represented in spherical coordinates

L± = hei±ϕ[± ∂

∂θ+ icotθ

∂

∂ϕ] (8.97)

We can successively obtain the spherical harmonics starting with Yl,−l. Tothis end we notice that

L−Yl,−l = hei±(1−l)ϕ[− ∂

∂θ+ lcotθ]Yl,−l = 0 (8.98)

This is a first order differential equation which can be solved in a straight-forward manner:

Yl,−l = NlsinlΘe−ilϕ (8.99)

The normalization constant is determined by the relationship

∫

sinθdθdϕN2l sin

2lΘ = 2πN2l

∫

sinθdθsin2lΘ (8.100)

The eigenvectors Yl,m can be determined in a straightforward way usingthe ladder operator L±:

Yl,m+1 =1

h

1√

l(l + 1) −m(m+ 1)L+Yl,m (8.101)

In a similar way, we obtain the relationship

Yl,m−1 =1

h

1√

l(l + 1) −m(m− 1)L−Yl,m (8.102)

Spherical Harmonics for Small l

a) l = 0 (s-Orbital)

Y0,0 =1√4π

(8.103)

b) l = 1 (p-Orbital)

Yl,0 =

√

3

4πcosθ =

√

3

4π

z

r(8.104)

Yl,±1 =

√

3

8πsinθei±ϕ =

√

3

8π

x± iy

r(8.105)

c) l = 2 (d-Orbital)

88 8 Hydrogen Atom

Y2,0 =

√

5

16π(3cos2θ − 1) (8.106)

Y2,1 =

√

15

8π3sinθcosθeiϕ (8.107)

Y2,2 =

√

15

32πsin2θe2iϕ (8.108)

Fig. 8.1. Visualization of low order spherical harmonics

8.5 Radial Eigenfunctions: Hydrogen Atom

We adress the eigenvalue problem determining the radial part of the wavefunction of the Hydrogen atom.

− h2

2µ

1

r2∂

∂rr2∂

∂r+h2l(l + 1)

2µr2− γ

rψ(r) = Eψ(r) (8.109)

We perform the ansatz

ψ(r) =u(r)

r(8.110)

since1

r2∂

∂rr2∂

∂r

u(r)

r=

1

r

∂

∂r

∂

∂rru(r) (8.111)

8.5 Radial Eigenfunctions: Hydrogen Atom 89

Furthermore, we introduce a dimensionless radius ρ, where a is Bohr’s radius,

ρ =r

a(8.112)

and obtain

− h2

2µa2

∂

∂ρ

∂

∂ρu(ρ) +

h2l(l + 1)

2µa2− γ

ρau(ρ) = Eu(ρ) (8.113)

− ∂2

∂ρ2u(ρ) +

l(l + 1)

2ρ2− 2

ρu(ρ) = ǫu(ρ) (8.114)

Thereby, we have introduced a dimensionless quantity ǫ related with theenergy

ǫ =E

RH

RH =µγ2

2h2

a =h2

µγ(8.115)

a: Bohr’s radius, RH Rydberg constant

8.5.1 Asymptotic behaviour ρ → ∞

The asymptotic behaviour for ρ→ ∞ is obtained from (8.114):

d2

dρ2u = −ǫu (8.116)

We are interested in bound states: ǫ < 0. The solution of (8.116) is thefunction

u(ρ) = ae√

|ǫ|ρ + be−√

|ǫ|ρ (8.117)

The function u(ρ) should approach 0 for large values of ρ. This results ina = 0.

8.5.2 Sommerfeld’s Method

We determine the radial functions u(ρ) by Sommerfeld’s method. To this endwe perform the ansatz

u(ρ) = e−αρP (ρ) (8.118)

withα =

√−ǫ (8.119)

90 8 Hydrogen Atom

and obtain the following differential equation defining the function P (ρ):

∂2

∂ρ2− 2α

∂

∂ρ− l(l + 1)

2ρ2+

2

ρP (ρ) = 0 (8.120)

This differential equation defines an eigenvalue problem.The asymptotic behaviour for small values of ρ is determined from

∂2

∂ρ2− l(l + 1)

2ρ2P (ρ) = 0 (8.121)

This yields the solution

P (ρ) = cρ(l+1) + dρ−l (8.122)

The second part diverges for ρ→ 0.For the general solution we perform the following polynomial ansatz

P = ρl+1∑

k=0

akρk = ρl+1L(ρ) (8.123)

This ansatz has the correct asymptotic behaviour for ρ → 0. Inserting thisansatz into the differential equation we are led to the relation

∑

k

ak(l + 1 + k)(l + k)ρl+k−1 − 2α(l + 1 + k)akρl+k + 2akρ

l+k

−(l + 1)lakρl+k−1 = 0 (8.124)

The corresponding recursion relation reads

∑

k

ak(l+1+k)(l+k)−l(l+1)]ρl+k−1−2[α(l+1+k)−1]akρl+k = 0 (8.125)

ak+1 =2[α(l + k + 1) − 1]

(l + k + 2)(l + k + 1) − l(l + 1)ak (8.126)

The Taylor expansion (8.123) stops at order K, provided we require

α(l +K + 1) − 1 = 0 (8.127)

This requirement determines the energy eigenvalues

ǫ = −α2 = − 1

(l +K + 1)2(8.128)

We can label the eigenvalues and eigenfunctions introducing the principal

quantum number n

n = l +K + 1 , n = 1, 2, .... (8.129)

8.5 Radial Eigenfunctions: Hydrogen Atom 91

n=1 K=0 l=0

n=2 K=0 l=1K=1 l=0

n=3 K=0 l=2K=1 l=1K=2 l=0

The quantum number of angular momentum l is restricted to the principalquantum number

l = 0, 1, .., n− 1 (8.130)

Magnetic quantum number m

−l,−l+ 1, ..., 0, ...l− 1, l (8.131)

8.5.3 Associate Laguerre Polynomials

The ansatz (8.123) introduces the so-called associate Laguerre polynomialsL(ρ). Inserting this ansatz into the differential equation (8.120) yields

ρL′′ + [2(l + 1) − 2αρ]L′ − [2α(l + 1) − 2]L = 0 (8.132)

The transformationρ =

x

2α(8.133)

The so-called associate Laguerre differential equation is obtained by con-sidering that α = 1

n . Furthermore, we perform the transformation

ρ =α

2x (8.134)

This leads us to the so-called associate Laguerre differential equation

xd2

dx2L(x)+[(2l+1)+1−x] d

dxL(x)+[(n+ l)− (2l+1)]L2l+1

n+l (x) = 0 (8.135)

The solutions of this differential equation are the associate Laguerre polyno-mials L2l+1

n+l (x).Thus, the radial eigenfunctions Rnl(r) can be constructed from the asso-

ciate Laguerre polyonials:

Rnl(r) = Nnl(r

a)lL2l+1

n+l (2r

na) (8.136)

Rodrigues formula

L2l+1n+l =

(

− d

dx

)2l+1

ex

(

− d

dx

)n+l

e−xxn+l (8.137)

92 8 Hydrogen Atom

8.5.4 Summary: Energy Levels and Wave Functions

The energy eigenvalues (energy levels) depend on the principal quantum num-ber n

En = RHǫ = −RH1

n2(8.138)

where RH denotes the Rydberg constant. These energy levels are degeneratewith respect to the quantum number of angular momentum l, whereas l =0, .., n − 1 and furthermore, with respect to the magnetic quantum numberm = −l, .., l

The wave functions depend on the principal quantum number n, the quan-tum number of angular momentum l and the magnetic quantum number m.

ψn,l,m(r, θ, ϕ) = |n, l,m >= Rn,l(r

a)Ylmθ, ϕ (8.139)

R1,0 = 2a−3/2e−ρ

R2,0 = (2a)−3/2(2 − ρ)e−ρ/2

R2,1 = 3−1/2(2a)−3/2ρe−ρ/2

R3,0 = 2(3a)−3/2[1 − 2

3ρ+

2

27ρ2]e−ρ/3

R3,1 =4√

2

9(3a)−3/2ρ[1 − 1

6ρ]e−ρ/3

R3,2 =2√

2

27√

5(3a)−3/2ρ2e−ρ/3 (8.140)

8.5.5 Associate Laguerre Functions

8.5.6 Symmetry and the Spectrum of the Hydrogen Atom

The degeneracy of the energy eigenvalues with respect to the magnetic quan-tum number is the result of the spherical symmetry. This becomes evident bythe form of the Hamiltonian, which only depends on L2, but not on Lz.

8.5.7 Runge-Lenz-Vector

The degeneracy of the energy eigenvalues with respect to the quantum numberof angular momentum is also related to a conservation law. The Runge Lentzvector is defined as

A = r × L− γr

r(8.141)

It is straightforward to show that the Runge Lentz vector A is a constant ofmotion, i.e.

d

dtA = 0 (8.142)

8.7 Appendix: Angular Momentum Operators and Spherical Coordinates 93

8.6 Alkali-Atoms

The Alkali atoms Lithium, Natrium, Kalium, Rubidium, Caesium, Franziumexhibit spectra which are quite similar to the spectrum of the Hydrogen atom.This is related to the fact that they have a single valence electron. A naivepicture of such an atom is one, in which the positively charged core withcharge Z e is surrounded by a cloud of (Z-1) electrons and a single valenceelectron. If the valence electron has orbits which are far away from the coreand the (Z-1) electrons, the behaviour should be quite similar to the hydrogenatom, i.e. the potential energy of the valence electron would be equal to

V (r) = − e2

4πǫ0r(8.143)

However, if this electron comes close to the core, the potential energy changesto

V (r) = − Ze2

4πǫr(8.144)

As a consequence, the effective potential is not proportional to 1/r. The de-generacy of the spectrum of the Hydrogen atom with respect to the quantumnumber l is broken. One expects the following dependency of the energy spec-trum

En,l = −R 1

(n−∆(n, l))2(8.145)

Here, the shift ∆(n, l) is due to the effective potential.

8.7 Appendix: Angular Momentum Operators and

Spherical Coordinates

The coordinates x, y, z are related to spherical coordinates r, θ, ϕ accordingto

x = rsinθcosϕ

y = rsinθsinϕ

z = rcosθ (8.146)

Our aim is to determine the oprator of angular momenta in sphericalcoordinates.

We start by calculating the derivative with respect to ϕ:

∂

∂ϕ=∂x

∂ϕ

∂

∂x+∂y

∂ϕ

∂

∂y+∂z

∂ϕ

∂

∂z

= −y ∂∂x

+ x∂

∂y(8.147)

94 8 Hydrogen Atom

This operator is related to Lz:

Lz =h

i[x∂

∂y− y

∂

∂x] =

h

i

∂

∂ϕ(8.148)

Next we calculate the derivative with respect to θ:

∂

∂θ=∂x

∂θ

∂

∂x+∂y

∂θ

∂

∂y+∂z

∂θ

∂

∂z

= zcosϕ∂

∂x+ zsinϕ

∂

∂y− x

cosϕ

∂

∂z

= zcosϕ∂

∂x+ zsinϕ

∂

∂y− y

sinϕ

∂

∂z(8.149)

Thus, we obtain

cosϕ∂

∂θ= zcos2ϕ

∂

∂x+ zsinϕcosϕ

∂

∂y− x

∂

∂z

sinϕ∂

∂θ= zcosϕsinϕ

∂

∂x+ zsin2ϕ

∂

∂y− y

∂

∂z(8.150)

Now,

cosϕ∂

∂θ= z(cos2ϕ− 1)

∂

∂x+ zsinϕcosϕ

∂

∂y+ z

∂

∂x− x

∂

∂z

sinϕ∂

∂θ= zcosϕsinϕ

∂

∂x+ z(sin2ϕ− 1)

∂

∂y+ z

∂

∂y− y

∂

∂z(8.151)

cosϕ∂

∂θ= −zsin2ϕ

∂

∂x+ zsinϕcosϕ

∂

∂y+ z

∂

∂x− x

∂

∂z

sinϕ∂

∂θ= zcosϕsinϕ

∂

∂x− zcos2ϕ

∂

∂y+ z

∂

∂y− y

∂

∂z(8.152)

cosϕ∂

∂θ= −ycotθsinϕ ∂

∂x+ xcotθsinϕ

∂

∂y+ z

∂

∂x− x

∂

∂z

sinϕ∂

∂θ= ycotθcosϕsinϕ

∂

∂x− xcotθcosϕ

∂

∂y+ z

∂

∂y− y

∂

∂z(8.153)

With the help of (8.147) we obtain

cosϕ∂

∂θ− cotθsinϕ

∂

∂ϕ= z

∂

∂x− x

∂

∂z

sinϕ∂

∂θ+ cotθcosϕ

∂

∂ϕ= z

∂

∂y− y

∂

∂z(8.154)

8.7 Appendix: Angular Momentum Operators and Spherical Coordinates 95

The operators of angular momentum expressed in spherical coordinatesare as follows:

Lx =h

i[y∂

∂z− z

∂

∂y]

= − hi[sinϕ

∂

∂θ+ cotθcosϕ

∂

∂ϕ]

Ly =h

i[z∂

∂x− x

∂

∂z]

=h

i[cosϕ

∂

∂θ− cotθsinϕ

∂

∂ϕ]

Lz =h

i[x∂

∂y− y

∂

∂x]

=h

i

∂

∂ϕ(8.155)

Using these representations we can determine L2:

L2 = −h2 1

sinθ

∂

∂θsinθ

∂

∂θ+

1

sin2θ

∂2

∂ϕ2 (8.156)

9

Diatomic Molecule

We consider a diatomic molecule. The two atoms form a molecule due to thebinding forces resulting from the motion of electrons around the atoms. Wewill take this biding force into account in a phenomenological way postulatingthe existence of an attractive force between the two atoms, which we char-acterize by the potential V (r). This potential will have the form depicted infigure (). Frequently, it can be approximated by the so-called Morse potential

V (r) = V0(1 − e−(r−r0)/a) (9.1)

If we denote the masses and the locations of the atoms by mi and ri weobtain the Hamilton operator

H = − h2

2m1∆1 −

h2

2m2∆2 + V (|r1 − r2|) (9.2)

As described in a previous section we are able to separate the motion ofthe center of mass and the relative motion. The Hamiltonian for the relativemotion reads

H = − h2

2m∆+ V (r) (9.3)

As a result, we have reduced the problem of a molecule consisting of twoatoms by the motion in a central force field given by the potential V (r). TheHamiltonian reads

H = − h2

2m

1

r2∂

∂rr2∂

∂r+

L2

2mr2+ V (r) (9.4)

We assume that the oscillations around the equlibrium positionR are smallso that we can make the transformation

r = R+ x (9.5)

In lowest order we obtain the Hamiltonian

98 9 Diatomic Molecule

Fig. 9.1. Potential energy of the binding force of a diatomic molecule

H = − h2

2m

∂2

∂x2+mω2

2x2 +

L2

2J(9.6)

where we have introduced the momentum of inertia

J = mR2 (9.7)

The Hamiltonian consists of two parts. The first part is an harmonic oscillator,the second part is related to the rotation of the molecule.

The Hamiltonian is obtained under the assumption that the deviation xfrom the equilibrium position is small compared to the equilibrium distanceR. This allows us to perform the following approximations

1

(R + x)2∂

∂x(R+ x)2

∂

∂x≈ ∂2

∂x2

9 Diatomic Molecule 99

V (R + x) = V (R) + x∂

∂RV (R)

+1

2x2 ∂2

∂R2V (R) = V (R) +

mω2

2x2 (9.8)

Introducing creation and annihilation operators for the harmonic oscillatorpart we end up with the Hamiltonian

H = hω[b†b+1

2] +

L2

2J(9.9)

The corresponding energy eigenstates are obtained in a straightforwardmanner. A product ansatz yields the eigenstates

|n, l,m >= |n > |lm > (9.10)

or, in terms

ψ(x,Θ, ϕ) = NnHn(x)e−x2

2 Y ml (Θ,ϕ) (9.11)

The energy spectrum takes the form

En,l = hω(n+1

2) +

h2l(l+ 1)

2J(9.12)

The spectrum is degenerate with respect to the quantum number m.It turns out that the quantity

h2

2J<< hω (9.13)

A schematic picture of the eigenstates are given in fig.

100 9 Diatomic Molecule

Fig. 9.2. Vibrational rotational spectrum of a diatomic molecule

10

Perturbation Theory

Only very few problems in quantum mechanics can be treated exactly. Exam-ples are the harmonic oscillator or the Hydrogen atom. In order to deal withother systems of interest, e.g. the Hydrogen atom in an external electric field,one has to ressort to approximations applying so-called perturbation theory.

In this section we are concerned with time independent perturbation the-ory. The general outline of the problem is as follows:

We have to solve the time independent Schrodinger equation for a Hamil-tonian whih is composed of the operator H0, whose eigenvalue problem canbe solved explicitly, and a perturbation V :

[H0 + ǫV ]|ψ >= E|ψ > (10.1)

A smallness parameter ǫ is introduced in order to enumerate the arising termsof order 1 = ǫ0, ǫ, ǫ2 ....

One assumes that the eigenvalue E = E(ǫ) and the eigenvector |ψ >=|ψ(ǫ) > are smooth functions of ǫ, which are known for ǫ = 0. Perturbationtheory is devoted to the calculation of the corrections, which arise for smallperturbations ǫV

E = E0 + ǫE1 + ǫ2E2 + ...

|ψ > = |ψ0 > +ǫ|ψ1 > +ǫ2|ψ2 > +... (10.2)

If we have , for ǫ = 0, a nondegenerate eigenvalue, we expect a shift of thisvalue. In the case of degen

10.1 Introductory Example

We consider the eigenvalue problem

H |ψ >= [H0 + ǫW ]|ψ >= E|ψ > (10.3)

102 10 Perturbation Theory

with the symmetric 2 × 2 matrices

H0 =

(

E1 00 E2

)

, W =

(

W1 ww W2

)

(10.4)

The eigenvalue problem is then related to the solution of the linear homoge-neous system

(

E1 + ǫW1 − E ǫwǫw E2 + ǫW2 − E

) (

ψ1

ψ2

)

= 0 (10.5)

Nontrivial solutions are obtained provided

Det(H0 + ǫW − EI) = (E1 + ǫW1 − E)(E2 + ǫW2 − E) − ǫ2w2

= E2 − (E1 + E2 + ǫ(W1 +W2))E

+ (E1 + ǫW1)(E2 + ǫW2) − ǫ2w2 = 0 (10.6)

The characteristic equation can be solved

E1,2 =E0

1 + E02 + ǫ(W1 +W2)

2±

√

(E01 − E0

2 + ǫ(W1 −W2))

4+ ǫ2w2 (10.7)

In order to obtain the eigenvalues for small values of ǫ, we can expand intoa Taylor series with respect to the smallness parameter ǫ. To first order in ǫwe obtain

E1 = E01 + ǫW1

E2 = E02 + ǫW2 (10.8)

To second order in ǫ2 the result reads

E1 = E01 + ǫW1 + ǫ2

w2

2(E1 − E2)

E2 = E02 + ǫW2 + ǫ2

w2

2(E2 − E1)(10.9)

10.2 Perturbation Theory: Nondegenerate Case

In this section we shall generalize the above result to the time independentHamilton operator:

[H0 + ǫV ]|ψ >= E|ψ > (10.10)

We assume that the eigenfunctions |n > and the eigenvalues En of theHamiltonian H0 are well-known:

H0|n >= E0n|n > (10.11)

10.2 Perturbation Theory: Nondegenerate Case 103

In order to solve the linear eigenvalue problem we perform the followingexpansion

E = E0 + ǫE1 + ǫ2E2 + .....

|ψ > = |ψ0 > +ǫ|ψ1 > +ǫ2|ψ2 > (10.12)

Inserting this ansatz into the eigenvalue problem and collecting terms ofthe same order in ǫ we find

H0|ψ0 > −E0|ψ0 > = 0

H0|ψ1 > −E0|ψ1 > = −E1|ψ0 > −V |ψ0 >

H0|ψ2 > −E0|ψ2 > = −E2|ψ0 > −E1|ψ1 > −V |ψ1 > (10.13)

In general, one could write down the expansion in n-the order.We shall now solve this set of equations order by order. In zeroth order we

obtain the solution

E0 = E0n , |ψ0 >= |n > (10.14)

where we consider in detail a perturbation of the n-th eigenfunction corre-sponding to the eigenvalue E0

n.Let us now preceed to the solution of the first order equation

(H0 − E0n)|ψ1 > = E1|n > −V |n > (10.15)

We can immediately determine the first order correction E1n to the energy

value E0n by forming the scalar product of this equation with the eigenvector

|n >:

< n|(H0 − E0n)|ψ1 > = E1 < n|n > − < n|V |n > (10.16)

The left hand side, however is zero, since

< n|(H0 − E0n)|ψ1 >=< (H0 − E0

n)n|ψ1 >= (E0n − E0

n) < n|ψ >= 0 (10.17)

As a result, we obtain the first order correction

E1n =< n|V |n > (10.18)

We proceed to determine the first order correction to the eigenvector,which is determined by eq. (10.15). To this end we perform an expansion ofthe first order correction |ψ1 > in terms of the complete set of eigenvectors|n >

|ψ1 >=∑

k

ckn|k > (10.19)

Equation (10.15) yields with the help of H0|k >= E0k |k >

104 10 Perturbation Theory

(H0 − E0n)

∑

k

ckn|k >=∑

k

(E0k − E0

n)ckn|k >= E1n|n > −V |n > (10.20)

Projecting this equation onto the eigenvector |m > yields

∑

k

(E0k − E0

n)Ckn < m|k >= E1δmn− < m|V |n > (10.21)

where we have exploited the fact that < m|n >= δnm. We arrive at

(E0m − E0

n)cmn = E1nδmn− < m|V |n > (10.22)

Again, we see that taking n = m we obtain as solvability condition the firstorder correction E1

n =< n|V |n >. For n 6= m we can determine the coefficientscmn :

cmn =<< m|V |n >E0

n − E0m

(10.23)

Thereby, we have to assume that E0n 6= E0

m, i.e. that we have no degeneracyof the energy spectrum. This case will be treated below.

We can now summarize our results to first order in ǫ

E = E0n + ǫE1

n = E0n + ǫ < n|V |n >

|ψ > = |n > +∑

m

< m|V |n >E0

n − E0m

|m > (10.24)

We shall proceed to the treatment of the second order corrections E2n and

ψ2n >, which are determined by

H0|ψ2 > −E0n|ψ2 > = E2|n > +E1|ψ1 > −V |ψ1 > (10.25)

Again, we can employ a solvability condition by forming the scalar productwith < n|. The left hand side vanishes identically, whereas the right hand sideyields the second order contribution

E2 =< n|V |ψ1n >=

∑

m

< n|V |m > cmn =<| < m|V |n > |2E0

n − E0m

(10.26)

We can now summarize our results to second order in ǫ

E = E0n + ǫ < n|V |n > +ǫ2

∑

m 6=n

<| < m|V |n > |2E0

n − E0m

|ψ > = |n > +∑

m

< m|V |n >E0

n − E0m

|m > (10.27)

The second order contribution to the eigenvector can bed determined in astraightforward manner.

10.3 Perturbation Theory for Degenerate States 105

10.3 Perturbation Theory for Degenerate States

Example:(

E0 ǫwǫw E0

)

(10.28)

Characteristic equation:

(E − E0)2 − ǫ2w2 = 0 (10.29)

E = E0 ± ǫw (10.30)

General: Finite dimensional subspace

H0|n, α >= E0n|n, α > (10.31)

Ansatz:|ψ >=

∑

α

cnα|nα > +ǫ∑

k 6=n

|kβ > +.. (10.32)

ǫW∑

α

cnα|nα >= ǫE1n

∑

α

cnα|nα > (10.33)

Projection∑

α

[< nα|W |nβ > −E1nδα,β ]cnα = 0 (10.34)

11

Atoms in Electric Field

11.1 Application of Perturbation Theory

We consider the Hamilton operator of a Hydrogen atom in an electric field

H = H0 + er · E (11.1)

In order to approximately determine the energy levels we take the directionof the electric field as the z-direction and introduce spherical coordinates. Asa result we obtain

H = H0 + ercosθE (11.2)

As we have seen in the section on time independent perturbation theorythe various corrections are related to the matrix elements

< nlm|V |n′l′m′ >=< nlm|rcosθ|n′l′m′ > (11.3)

where the eigenfunctions of the Hamilton operator H0 are denoted by

|nlm >= Rnl(r)Ym

l (θ, ϕ) (11.4)

The explicit expression for the matrix elements

< nlm|rcosθ|n′l′m′ > =

∫ ∞

0

r2dr

∫ π

0

sinθdθ

∫ 2π

0

dϕRnl(r)rRn′ l′(r)

× Y ml (θ, ϕ)∗cosθY m′

l′ (θ, ϕ) (11.5)

explicitly shows that the integrals factorizes according to

< nlm|V |n′l′m′ >= Mlm,l′m′Knl,n′l′ (11.6)

where the matrix elements Mlm,l′m′ involves integration with respect to thevariables Θ, ϕ

108 11 Atoms in Electric Field

Mlm,l′m′ =

∫ π

0

sinθdθ

∫ 2π

0

dϕY ml (θ, ϕ)∗cosθY m′

l′ (θ, ϕ) (11.7)

and a matrix element in volving the integral

Knl,n′l′ =

∫ ∞

0

r2drRn′l′(r)rRnl(r) (11.8)

It is convenient to represent cosθ as a spherical harmonics according to

cosθ =

√

4π

3Y 0

1 (θ, ϕ) (11.9)

As a consequence, one has to evaluate the integrals

Mlm,l′m′ =

∫ π

0

sinθdθ

∫ 2π

0

dϕY ml (θ, ϕ)∗Y 0

1 (θ, ϕ)Y m′

l′ (θ, ϕ)

= < Y ml |Y m′

l′ Y 01 > (11.10)

We emphasize that this matrix element is an example of a Clebsch-Gordon

coefficient.Due to the fact that

Y ml = eimϕLlm(cosθ) (11.11)

where Llm are polynomial functions of cos θ we immediately obtain the selec-tion rule with respect to the quantum number m:

Mlm;l′m′ = δm,m′Mlm;l′m′ (11.12)

Furthermore, there are selection rules with respect to l,l’. It can be shownthat Mlm,l′m′ is different from zero only in the cases

l = l′ + 1 l = l′ − 1 (11.13)

11.2 Stark Effect of the Ground State n=1,l=0

We remind the reader that

Y 00 (θ, ϕ =

1√4π

(11.14)

As a consequence, the matrix element

Ml,m;00 = δm,0

∫ π

0

sinθdθ

∫ 2π

0

dϕY ml (θ, ϕ)∗

√

4π

3Y 0

1 (θ, ϕ)1√4π

= δl,1δm,01√3

(11.15)

11.2 Stark Effect of the Ground State n=1,l=0 109

can be evaluated using the orthogonality condition of the spherical harmonics:

< Y ml (θ, ϕ)Y m′

1 (θ, ϕ) = δll′δmm′ (11.16)

In order to apply perturbation theory we have to consider the followingmatrix elements which are given by the following integrals:

< nlm|rcosθ|100 >= δl,1δm,01√3

∫ ∞

0

r2drRn1rR10 (11.17)

We are now in the position to determine the corrections to the groundstate |1, 0, 0 > of the hydrogen atom. The first order contribution vanishes

E1100 =< 100|rcosθ|100 >= M00,00K10,10 = 0 (11.18)

since M00,00 = 0, according to (11.17). As a result, the ground state of thehydrogen atom does not exhibit the linear Stark effect.

We proceed to calculate the second order correction to the energy eigen-value E0

1 , which is given by

E2100 = e2E2

∑

n,l

| < nl0|rcosθ|100 > |2E100 − Enl0

(11.19)

Due to the selection rule (11.17) only the matrix elements with l=1 contributeand we obtain explicitly

E2100 = e2E2

∑

n

1

3

|Kn1,10|2E100 − Enl0

(11.20)

with the integrals

Kn1,00 =

∫ ∞

0

r2drrRn1(r)R10(r) (11.21)

involving the (normalized) radial functions of the Hydrogen problem.Taking the contribution with n=2 and neglecting the contirbutions from

the higher order functions Rn1(r) we obtain our final estimate for the secondorder contirbutions

E(2)100 = −9

4a3

BE2e2 (11.22)

We can summarize: The ground state of the Hydrogen atom exhibits thequadratic Stark effect.

11.2.1 Induced Dipole Moment

Energy of a dipol in the electric field:

110 11 Atoms in Electric Field

∆E = −∫

d(E)dE = −α2E2 = −d

2E (11.23)

As a consequence:

< d >=9

2Ea3

B (11.24)

Polarization

P = ρ < d >= χeE (11.25)

11.3 Linear Stark Effect

The linear Stark effect arises for the case of a degenerate energy level. Weconsider the states with principal quantum number n = 2. The energy levelE0

2 = −Ry/4 is fourfold degenerate and we denote the corresponding eigen-states by |nα >, according to our notation in the section on perturbationtheory:

|2, 1 > = |2, 0, 0 >|2, 2 > = |2, 1, 0 >|2, 3 > = |2, 1, 1 >|2, 4 > = |2, 1,−1 > (11.26)

In order to calculate the splitting of the degeneracy by the external electricfield we have to calculate the matrix elements

< 2α|rcosθ|2β > (11.27)

Due to selection rules only the matrix elements

< 22|rcosθ|20 >= W =< 20|rcosθ|22 > (11.28)

is different from zero. It is explicitly given by

W = M10,00K21,20 (11.29)

with

M10,00 =1√3

(11.30)

and

K21,20 =

∫

r2drrR20(r)R21(r) =

∫ ∞

0

r2dr2√

3(2a)3e−r/a(1 − r

a)r

ar (11.31)

The radial wave functions are given by

11.3 Linear Stark Effect 111

R20 =2

(2a)3/2(1 − ρ)e−ρ/2

R21 =1√3

1

(2a)3/2ρe−ρ (11.32)

a: Bohr’s radiusAs a consequence we obtain

W = 3eaE (11.33)

Let us now consider the perturbation theory in first order. We make thefollowing ansatz for the eigenvector

|ψ > =4

∑

α=1

cα2 |2α >

= c1|200 > +c2|210 > +c3|211 > +c4|21 − 1 > (11.34)

We obtain the following eigenvalue problem

4∑

β=1

< 2α|V |2β > cβ2 = E12c

α2 (11.35)

Explicitly,

0 W 0 0W 0 0 00 0 0 00 0 0 0

c1c2c3c4

= E12

c1c2c3c4

(11.36)

This linear set of equations has only nontrivial solutions for eigenvaluesE1

2 determined by the characteristic equation

Det[< 2α|V |2β > −E12δαβ] = 0 (11.37)

The characteristic equation reads

(E12)2(E1

2 −W 2) = 0 (11.38)

with the four roots

E12 = 0 , E1

2 = 0 , E12 = W , E1

2 = −W (11.39)

In first order the energy eigenvalues split into three groups. The states|2, 1, 1 > and |21 − 1 > remain degenerate and there is not change of theeigenvalue in lowest order. The states |210 > and |200 > are coupled by theexternal field and the energy eigenvalues split:

E1 = 3eaE |ψ >=1√2[|210 > +|200 >]

E2 = −3eaE |ψ >=1√2[|210 > −|200 >] (11.40)

The corresponding eigenstates (to lowest) order are linear superpositions.

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