8.04: quantum physics i - mit - massachusetts institute of...

8.04: Quantum Physics I

Brice Huang

Spring 2017

These are my lecture notes for the Spring 2017 iteration of 8.04, QuantumPhysics I, taught by Prof. Raymond Ashoori.

These notes are written in LATEX during lectures in real time, and may containerrors. If you find an error, or would otherwise like to suggest improvements,please contact me at [email protected].

Special thanks to Evan Chen and Tony Zhang for the help with formatting,without which this project would not be possible.

As of February 22, 2017, these notes have been publicly shared with theclass.

As of May 26, 2017, these notes have been finalized. The permalink to thesenotes is http://web.mit.edu/bmhuang/www/notes/804-notes.pdf.

1

http://web.mit.edu/bmhuang/www/notes/804-notes.pdf

Brice Huang Contents

Contents

1 February 8, 2016 9

1.1 Sampling of Course . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.2 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.3 Determining particle properties . . . . . . . . . . . . . . . . . . . 10

1.3.1 Charge/Mass of Ionizing Water . . . . . . . . . . . . . . . 10

1.3.2 Thompson: Charge/Mass of an Electron . . . . . . . . . . 10

1.3.3 Millikan: Charge of an Electron . . . . . . . . . . . . . . . 10

1.4 History Lesson: the Thompson Model . . . . . . . . . . . . . . . 11

1.5 Why the Thompson Model is Bad . . . . . . . . . . . . . . . . . 11

1.5.1 α-Particle and Electron . . . . . . . . . . . . . . . . . . . 12

1.5.2 α-Particle and Atom . . . . . . . . . . . . . . . . . . . . . 12

1.5.3 Adding Interactions . . . . . . . . . . . . . . . . . . . . . 12

1.5.4 What was Actually Observed . . . . . . . . . . . . . . . . 12

2 February 15, 2017 13

2.1 Rutherford Scattering . . . . . . . . . . . . . . . . . . . . . . . . 13

2.1.1 Differential Scattering Cross-Section . . . . . . . . . . . . 13

2.1.2 The Distribution of Scattering . . . . . . . . . . . . . . . 14

2.2 Classical Models for Blackbody Radiation are Bad . . . . . . . . 14

2.2.1 The Classical Derivation . . . . . . . . . . . . . . . . . . . 15

2.2.2 The 1D Case . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2.3 The 2D Case . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2.4 The 3D Case . . . . . . . . . . . . . . . . . . . . . . . . . 16

2.2.5 Things Go Really, Really, Wrong . . . . . . . . . . . . . . 16

3 February 21, 2017 18

3.1 Blackbody Radiation – Planck Hypothesis . . . . . . . . . . . . . 18

3.1.1 Last time . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.1.2 Planck’s Hypothesis . . . . . . . . . . . . . . . . . . . . . 18

3.1.3 Comparison with Classical Model . . . . . . . . . . . . . . 18

3.1.4 The Quantum Model . . . . . . . . . . . . . . . . . . . . . 19

3.2 Photoelectric Effect . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.2.1 Photoelectric Emission . . . . . . . . . . . . . . . . . . . . 19

3.2.2 Einstein’s Resolution . . . . . . . . . . . . . . . . . . . . . 20

3.3 Short Wavelength Limit of X-ray Generation . . . . . . . . . . . 20

3.4 Compton Scattering . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.4.1 The Experiment . . . . . . . . . . . . . . . . . . . . . . . 21

3.4.2 What was Observed . . . . . . . . . . . . . . . . . . . . . 22

3.5 Epilogue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22

2


4 February 22, 2017 23

4.1 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.2 Matter Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

4.3 The Bohr Atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.4 Bohr Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

4.4.1 Energy and Radius of Bohr Orbits . . . . . . . . . . . . . 24

4.4.2 Emission spectra . . . . . . . . . . . . . . . . . . . . . . . 25

4.5 Wilson-Sommerfield Quantization . . . . . . . . . . . . . . . . . . 26

4.5.1 Quantize All the Things! . . . . . . . . . . . . . . . . . . 26

4.5.2 Case Study: Position and Momentum . . . . . . . . . . . 26

4.5.3 Case Study: Angular Momentum and Angle . . . . . . . . 27

4.6 Moseley’s Experiment; Shielding . . . . . . . . . . . . . . . . . . 27

4.6.1 What Moseley Saw . . . . . . . . . . . . . . . . . . . . . . 28

4.6.2 Moseley’s Conclusions . . . . . . . . . . . . . . . . . . . . 28

5 February 27, 2017 29

5.1 Double-Slit Experiment . . . . . . . . . . . . . . . . . . . . . . . 29

5.1.1 The Gun Model . . . . . . . . . . . . . . . . . . . . . . . 29

5.1.2 The Wave Model . . . . . . . . . . . . . . . . . . . . . . . 29

5.1.3 Electrons are Waves! . . . . . . . . . . . . . . . . . . . . . 30

5.1.4 Where did the electron really go? . . . . . . . . . . . . . . 30

5.2 Heisenberg’s Uncertainty Principle . . . . . . . . . . . . . . . . . 30

5.2.1 Heisenberg’s Microscope . . . . . . . . . . . . . . . . . . . 30

5.2.2 Derivation of the Uncertainty Principle . . . . . . . . . . 31

5.2.3 The Uncertainty Principle: Energy and Time . . . . . . . 32

5.2.4 The Uncertiain Principle: A Wave . . . . . . . . . . . . . 32

5.3 Applications of the Uncertianty Principle . . . . . . . . . . . . . 33

5.3.1 Application: Particle in a Box . . . . . . . . . . . . . . . 33

5.3.2 Application: Hydrogen Atom . . . . . . . . . . . . . . . . 33

6 March 1, 2017 34

6.1 Administrivia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34

6.2 Developing the Wave Equation . . . . . . . . . . . . . . . . . . . 34

6.2.1 Non-Quantum Waves . . . . . . . . . . . . . . . . . . . . 34

6.2.2 Quantum Waves . . . . . . . . . . . . . . . . . . . . . . . 34

6.3 Understanding Ψ and probabilities . . . . . . . . . . . . . . . . . 36

6.3.1 Probability Density Functions . . . . . . . . . . . . . . . . 36

6.3.2 Interpretations of Quantum Mechanics . . . . . . . . . . . 36

6.3.3 Probability Distributions: An Example . . . . . . . . . . 36

6.3.4 Standard Deviation . . . . . . . . . . . . . . . . . . . . . 37

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7 March 8, 2017 39

7.1 Administrivia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.2 Probability Densities . . . . . . . . . . . . . . . . . . . . . . . . . 39

7.3 Wave Function Normalization . . . . . . . . . . . . . . . . . . . . 39

7.4 Operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

7.4.1 Operators for Observarbles . . . . . . . . . . . . . . . . . 41

7.4.2 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . 42

7.4.3 Generalized Uncertainty Principle . . . . . . . . . . . . . 43

7.5 The Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

7.6 Time-Independent Schrodinger Equation . . . . . . . . . . . . . . 44

8 March 13, 2017 45

8.1 Generalized Uncertainty Relation . . . . . . . . . . . . . . . . . . 45

8.2 Erhenfest’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 45

8.3 More on Separable Solutions . . . . . . . . . . . . . . . . . . . . 46

8.3.1 Separable Solutions are Stationary States . . . . . . . . . 46

8.3.2 Stationary States have a Definite Total Energy . . . . . . 47

8.3.3 General Solutions are Sums of Separable Solutions . . . . 47

8.3.4 Digression: Operators and Eigenvectors . . . . . . . . . . 48

8.4 The Infinite Square Well . . . . . . . . . . . . . . . . . . . . . . . 48

9 March 15, 2017 50

9.1 Solving the Time-Independent Schrodinger Equation . . . . . . . 50

9.2 Complete Sets of Eigenfunctions . . . . . . . . . . . . . . . . . . 50

9.3 Eigenfunction Expansion for Square Well . . . . . . . . . . . . . 50

9.4 Fundamental Postulates of QM . . . . . . . . . . . . . . . . . . . 51

9.5 Example: Infinite Square Well with Starting State . . . . . . . . 52

9.5.1 Computing Coefficients . . . . . . . . . . . . . . . . . . . 52

9.5.2 Expected Value Computations . . . . . . . . . . . . . . . 53

9.5.3 What Happens After We Measure? . . . . . . . . . . . . . 54

9.5.4 Time Evolution . . . . . . . . . . . . . . . . . . . . . . . . 54

10 March 20, 2017 55

10.1 More on Evolution of Probabilities . . . . . . . . . . . . . . . . . 55

10.2 Dirac Delta Function . . . . . . . . . . . . . . . . . . . . . . . . . 56

10.3 Bound States in a General Potential . . . . . . . . . . . . . . . . 56

10.3.1 Geometric Properties of Wavefunctions . . . . . . . . . . 57

10.3.2 Derivatives of Wavefunctions . . . . . . . . . . . . . . . . 58

10.3.3 Drawing Wavefunctions . . . . . . . . . . . . . . . . . . . 59

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11 March 22, 2017 60

11.1 Drawing wave functions . . . . . . . . . . . . . . . . . . . . . . . 60

11.2 Finite Square Well . . . . . . . . . . . . . . . . . . . . . . . . . . 61

11.2.1 Region I . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

11.2.2 Region II . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

11.2.3 Region III . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

11.2.4 Putting Everything Together . . . . . . . . . . . . . . . . 62

11.2.5 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . 63

11.3 Harmonic Oscillator (Power Series Method) . . . . . . . . . . . . 63

12 April 3, 2017 66

12.1 Administrivia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

12.2 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . . . . . 66

12.3 Harmonic Oscillator – Power Series method . . . . . . . . . . . . 66

12.3.1 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . 66

12.3.2 Convergence and Truncation . . . . . . . . . . . . . . . . 67

12.3.3 Small Energy States . . . . . . . . . . . . . . . . . . . . . 68

12.3.4 Aside: Classical and Quantum Probability Distributions . 68

12.4 Harmonic Oscillator – Ladder Operator method . . . . . . . . . . 70

12.4.1 The operators a+ and a− . . . . . . . . . . . . . . . . . . 70

12.4.2 Laddering . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

12.4.3 Solving the Harmonic Oscillator . . . . . . . . . . . . . . 71

13 April 5, 2017 73

13.1 Delta Function Potential . . . . . . . . . . . . . . . . . . . . . . . 73

13.1.1 Boundary Conditions . . . . . . . . . . . . . . . . . . . . 73

13.1.2 Inside Regions I and II . . . . . . . . . . . . . . . . . . . . 73

13.1.3 Putting It Together . . . . . . . . . . . . . . . . . . . . . 74

13.2 Creating a Wave Packet from Free-Particle Wave Functions . . . 74

13.2.1 Creating a Wave Packet . . . . . . . . . . . . . . . . . . . 75

13.3 Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . 75

13.3.1 Fourier Transform Pairs . . . . . . . . . . . . . . . . . . . 75

13.4 Time Dependencies of Wave Packets . . . . . . . . . . . . . . . . 76

13.5 Velocities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

13.6 Back to the Infinite Square Well . . . . . . . . . . . . . . . . . . 78

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14 April 10, 2017 80

14.1 Infinite Square Well with Time Evolution . . . . . . . . . . . . . 80

14.2 Dimensional Analysis . . . . . . . . . . . . . . . . . . . . . . . . . 81

14.2.1 Harmonic Oscillator . . . . . . . . . . . . . . . . . . . . . 81

14.2.2 Infinite Square Well . . . . . . . . . . . . . . . . . . . . . 81

14.3 Parity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

14.3.1 Parity Operator . . . . . . . . . . . . . . . . . . . . . . . 82

14.3.2 P and H . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

14.3.3 Hyperbolic Functions . . . . . . . . . . . . . . . . . . . . 82

14.4 Infinite Square Well with δ-Function . . . . . . . . . . . . . . . . 83

14.4.1 E > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

14.4.2 E < 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

15 April 19, 2017 85

15.1 Exam Statistics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

15.2 When Waves Meet Barriers . . . . . . . . . . . . . . . . . . . . . 85

15.3 Step-Function Barrier . . . . . . . . . . . . . . . . . . . . . . . . 85

15.4 Probability Current . . . . . . . . . . . . . . . . . . . . . . . . . 86

15.4.1 Probability Current: Intuition . . . . . . . . . . . . . . . . 86

15.4.2 Computing Probability Current . . . . . . . . . . . . . . . 86

15.5 Step-Function Barrier II . . . . . . . . . . . . . . . . . . . . . . . 87

15.6 Tunneling . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

15.6.1 Aside: Tunnel Ionization . . . . . . . . . . . . . . . . . . . 88

16 April 24, 2017 89

16.1 Grade Cutoffs from Last Exam . . . . . . . . . . . . . . . . . . . 89

16.2 WKB Approximation . . . . . . . . . . . . . . . . . . . . . . . . 89

16.3 WKB in Classicaly-Allowed Regions . . . . . . . . . . . . . . . . 89

16.4 An Example: Gradually Varying Infinite Well . . . . . . . . . . . 91

16.5 WKB in Classically-Forbidden Regions . . . . . . . . . . . . . . . 91

16.6 An Example: Gradually Varying Tunneling Barrier . . . . . . . . 91

16.7 Tunneling Across a Barrier . . . . . . . . . . . . . . . . . . . . . 92

16.7.1 Application: Nuclear α-Decay . . . . . . . . . . . . . . . . 92

17 April 26, 2017 95

17.1 Extending QM to 3 Dimensions . . . . . . . . . . . . . . . . . . . 95

17.2 Example: Particle in a 3-D Box . . . . . . . . . . . . . . . . . . . 96

17.2.1 Special Case: The Cubical Box . . . . . . . . . . . . . . . 97

17.3 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . . . . . 97

17.3.1 Schrodinger Equation in Spherical Coordinates . . . . . . 97

17.4 Separability of Eigenfunctions in Spherical Coordinates . . . . . 98

17.4.1 Special case: l = 0 . . . . . . . . . . . . . . . . . . . . . . 99

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18 May 1, 2017 100

18.1 Last Time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

18.2 Example: Infinite Spherical Well . . . . . . . . . . . . . . . . . . 100

18.3 Example: Nuclear Potential . . . . . . . . . . . . . . . . . . . . . 102

18.3.1 Power Series Method, Revisited . . . . . . . . . . . . . . . 102

18.3.2 Writing the s-Wavefunctions . . . . . . . . . . . . . . . . 104

19 May 3, 2017 105

19.1 Gyromagnetic Effect . . . . . . . . . . . . . . . . . . . . . . . . . 105

19.2 Magnetic Moment and Angular Momentum . . . . . . . . . . . . 105

19.3 Stern-Gerlach Experiment . . . . . . . . . . . . . . . . . . . . . . 106

19.4 Bohr Magneton . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

19.5 Angular Momentum in QM . . . . . . . . . . . . . . . . . . . . . 108

19.6 Commuting Operators . . . . . . . . . . . . . . . . . . . . . . . . 109

19.6.1 Commutators . . . . . . . . . . . . . . . . . . . . . . . . . 109

20 May 8, 2017 110

20.1 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . . . 110

20.1.1 Aside: Clarification from Last Time . . . . . . . . . . . . 110

20.1.2 Ladder Operators . . . . . . . . . . . . . . . . . . . . . . 110

20.1.3 Top and Bottom Rungs of the Ladder . . . . . . . . . . . 111

20.1.4 Quantization of Angular Momentum . . . . . . . . . . . . 112

20.2 Angular Part of Spherical Schrodinger Equation . . . . . . . . . 113

20.2.1 Separation of θ, φ . . . . . . . . . . . . . . . . . . . . . . . 113

20.2.2 Solving the θ, φ-Equations . . . . . . . . . . . . . . . . . . 113

21 May 10, 2017 115

21.1 Spherical Schrodinger Equation - The θ-equation . . . . . . . . . 115

21.1.1 Last Time . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

21.1.2 Case m = 0: Power Series, Again . . . . . . . . . . . . . . 115

21.2 Solution to θ-Equation for General m . . . . . . . . . . . . . . . 116

21.3 Spherical Schrodinger Equation - Putting Everything Together . 116

21.3.1 Completing the Angular Part . . . . . . . . . . . . . . . . 116

21.3.2 The Radial Part . . . . . . . . . . . . . . . . . . . . . . . 117

21.3.3 Normalization . . . . . . . . . . . . . . . . . . . . . . . . . 117

21.4 Solving the Hydrogen Atom – The Radial Part . . . . . . . . . . 117

21.4.1 Finding Asymptotic Solutions . . . . . . . . . . . . . . . . 118

21.4.2 Power Series, Again; Laguerre Polynomials . . . . . . . . 118

21.4.3 Conclusion . . . . . . . . . . . . . . . . . . . . . . . . . . 119

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22 May 15, 2017 120

22.1 Finishing the Hydrogen Atom . . . . . . . . . . . . . . . . . . . . 120

22.1.1 The Radial Wavefunctions . . . . . . . . . . . . . . . . . . 120

22.1.2 Electron orbitals, Periodic Table . . . . . . . . . . . . . . 120

22.1.3 Spherical Wavefunctions . . . . . . . . . . . . . . . . . . . 121

22.2 Rydberg Atoms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

22.2.1 Size of a Rydberg Atom . . . . . . . . . . . . . . . . . . . 123

22.2.2 Shapes of Rydberg Orbitals . . . . . . . . . . . . . . . . . 123

22.3 EPR Paradox . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124

23 May 17, 2017 125

23.1 EPR Paradox, Bell’s Inequality . . . . . . . . . . . . . . . . . . . 125

23.1.1 Spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

23.1.2 What Results do we Predict? . . . . . . . . . . . . . . . 126

23.1.3 Bell’s Inequality . . . . . . . . . . . . . . . . . . . . . . . 127

23.1.4 Philosophy . . . . . . . . . . . . . . . . . . . . . . . . . . 128

23.2 Quantum Transport in 1-D . . . . . . . . . . . . . . . . . . . . . 128

8

Brice Huang 1 February 8, 2016

1 February 8, 2016

1.1 Sampling of Course

Two-slit experiment: light shined through two slits forms a wave on thewall. It turns out electons do the same – even if we shoot them one at a time.This doesn’t seem to make sense, because the electron is a particle – unless theelectron went through both slits?

What if we put a detector on one of the slits, to see where the electron went?Then we force the electron to choose a slit, and we just get a superposition oftwo one-slit patterns.

This is very counterintuitive – somehow the act of measuring somethingchanges its behavior.

We’ll touch on wave particle duality: electrons can be described as waves,and EM waves can be described as particles.

Quantum interactions are governed by a constant called Planck’s constant,defined by

h = 6.626× 10−27 erg · sec .

Dimensional analysis: this has dimensions

energy · time =g cm2

s2s =

g cm

s· cm = momentum · length.

Just an idea of scale: a car has momentum times length

(106g · 103cm/s) · 300g = 3 · 1011g · cm2/s = 1011erg · s.

An atom has momentum times length around 2 · 10−27erg · s, which is on theorder of Planck’s constant. Wavelike properties show up when things are on thisorder.

1.2 Units

We use the cgs system throughout the course. Differences between cgs and mks:

We define charge units in a way to make the 4πε0 go away. In particular,F = q1q2

r2 , with units defined:

1 dyne =1 esu2

1 cm2.

Energy is also defined in electronvolts: 1 eV is energy gained by electronwhen accelerated across 1 V potential difference. We can convert this to ergs:

1 eV = 1.602× 10−19 J = 1.602× 10−12 erg

Some energy quantities, in eV:

• Ionization energy of hydrogen 13.6 eV;

• Mass rest energy of electron 511 KeV;

• Mass rest energy of proton 938 MeV.

9


1.3 Determining particle properties

1.3.1 Charge/Mass of Ionizing Water

We can ionize water by putting two electrodes in water and applying a potentialdifference across them. We can measure that 2.89 · 1014 esu of charge is neededto collect 1 g of hydrogen (and thus, 8 g of oxygen).

Joke. “If you were like me, you collected them in the same container and tooka match...” - Ashoori

By comparing quantities of gases collected we can infer the relative massesof hydrogen and oxygen atoms. But we can’t get the mass of the atoms.

1.3.2 Thompson: Charge/Mass of an Electron

Thompson did the following experiment. He passed an electron beam throughperpendicular E and B fields, and tuneed the fields until the beam goes straight.Then qvB = qE, so v = E/B gives the velocity. He then turned off the Efield and measured the curvature of the resulting beam. This (details omitted)

determines |qe|me= 5.28× 1017 esu / g.

Comparing this with the ratio of mass to charge in ionizing water, we inferthat electrons are very light.

1.3.3 Millikan: Charge of an Electron

Millikan’s experiment determined the charge of an electron. He used an ionizerto produce oil droplets with very small charges and suspended them in an Efield, tuned so the electric force balanced gravity.

F = QE0 = mg

Q = mg/E0

F = (Q− e)E1mg

Q(E1 − E0)

E1= e = 4.8× 10−10esu.

In fact, this gives us Avogadro’s number:

NA =2.89× 1014esu

4.8× 10−10esu= 6.02× 1023.

We also get the mass of an electron and hydrogen. By looking at density, wecan also conclude that a hydrogen atom has radius about 1A = 10−10 m.

10


1.4 History Lesson: the Thompson Model

Initially people thought electrons couldn’t possibly orbit atoms – they wouldradiate, lose energy, and spiral into the atom.

So, when it was found that electrons were small, Thompson proposed theplum pudding model – the atom is a mass of positive charge, with negativeelectrons stuck on the surface.

In a sphere concentric with the atom, we could compute the electric fieldwith the (cgs) Gauss’s law: ∮

E · da = 4πQenc

4πr2E = 4π( rR

)3

ze

E =ze

R3r.

This is proportional to r, so an electron rattling around this atom will undergosimple harmonic motion, with force

F = −ze2

R3r.

Thus we expect a spring constant k = ze2

R3 , and angular frequency

ω =

√k

m=

√ze2

mR3.

Plugging in constants for hydrogen yields a frequency

f =ω

2π= 2.5× 1015 Hz .

Then, from fλ = c, we get a wavelength λ = 1184A

People compared this with the the emission spectrum of hydrogen, whichhad wavelengths from 1000A to 10000A, and decided this checked out.

This theory stuck for a while.

1.5 Why the Thompson Model is Bad

An experiment analogous to Thompson’s experiment determining |qe|/me, wasdone for alpha-particles emitted from U93. They discovered that its charge:massratio is half that of hydrogen. In an experiment, they aimed the beam of alphaparticles at a gold foil.

If the Thompson model is correct, what do we expect to happen? We analyzethe possible interactions.

11


1.5.1 α-Particle and Electron

What happens when an α-particle, with velocity v0, collides with a stationary e−?Recall that electrons are tiny compared to alpha particles – me ≈ 1

2000mH ≈1

8000mα. From the reference frame of the α-particle, the electron comes inwith velocity v0 and bounces elastically back with velocity about v0. So, inthe original frame, the electron exits the collision with velocity 2v0, with thismomentum coming from the α-particle.

The α-particle loses about 18000 · 2 = 1

4000 of its momentum. Thus we expectits direction (in the lab frame) to deflect by at most 1

4000 rad ≈ 0.014. This issmall.

1.5.2 α-Particle and Atom

Now let’s consider the other possible interaction. What if the α-particle hitsthe positive-sphere part of the atom? We assume a grazing impact because theelectrical force is greatest far from the center. The angle of deflection is

θ ≈ tan θ =∆p

p=F∆t

p.

Moreover, F is

F =qQ

R2=

(2e)(ze)

R2=

2ze2

R2

and ∆t ≤ 2Rv . So,

θ =2ze2

R2· 2R

v· 1

mv=

2ze2

R· 2

mv2.

The kinetic energy can be measured, so we can conlcude 2mv2 ≈

15 MeV . Plugging

in, we get θ ≈ 4.5× 10−4 ≈ 0.03 radians. This is also small.

1.5.3 Adding Interactions

Maybe deflections can add up. If each α particle goes through 100 deflections, theoverall deflection angle should be Gaussian with standard deviation

√100·0.03 =

0.3. Gaussians have really, really small tails. This means that we expect, forexample,

Pr(θ > 10) ≈ 10−483.

1.5.4 What was Actually Observed

Instead, what was found was a distribution of deflection angles θ with probability1

sin4 θ2

. About one in every 8000 α particles bounced backwards.

This completely invalidated the Thompson model. People concluded thatthat there’s something really small in an atom, with really strong positive charge– the nucleus!

12


2 February 15, 2017

2.1 Rutherford Scattering

In scattering situations, we define the scattering cross-section σ as

σ =scattered flux

incident flux per area.

So, σ has units of area.

Suppose we shoot a beam of N0 particles at a piece of gold foil of depth ∆x,such that the target region has area a. Then, the number of scattered particlesis

N0

(σa

)per target. If n is the (per-volume) density of targets, then na∆x is the numberof targets, so the total number of scattered particles is

N0

(σa

)na∆x = N0σn∆x.

2.1.1 Differential Scattering Cross-Section

We define the Differential Scattering Cross-Section as

dσ(θ, φ)

dΩ=

flux scattered into spherical angle dΩ(θ, φ)

incidentfluxperarea.

This measures the differential scattering in a given spherical angle.

We can write

σ =

∫ 2π

0

dφ

∫ π

0

sin θdσ(θ, φ)

dΩdθ.

Suppose an alpha particle with charge q and velocity v0 is approaching agold nucleus with charge Q. The impact parameter is b. Because of their mutualrepulsion the gold nucleus will scatter the α-particle.

Then, for each scatterer (gold atom):

Iscat(θ) =

(I0a

)dσ(θ)

dΩdΩ.

Here, dσdΩ doesn’t depend on φ because of azimuthal symmetry. dΩ is the

spherical angle intercepted by the detector. If the detector has area A and is raway from the scattering center, then dΩ = A

r2 .

So, for a foil target:

Iscat(θ) =

(I0a

)(an∆x)

dσ(θ)

dΩdΩ.

13


2.1.2 The Distribution of Scattering

In the homework, we will find the relation between the impact parameter b andthe ultimate deflection angle θ. We will show, given the mapping b 7→ θ, how tocompute dσ

dΩ .

Note that deflection angle is a function only of b. A circular ribbon ofpre-impact area dσ = 2πbdb gets scattered to a ring of spherical-angle

dΩ =A

r2=

(2πr sin θ)rdθ

r2= 2π sin θdθ.

Thus,dσ

dΩ=

b

sin θ

∣∣∣∣dbdθ∣∣∣∣

Joke. Now I can write down the answer, since you’ll do the hard work.

After doing the computations, we get

dσ

dΩ=

1

16

r20

sin4(θ2

) ,where

r0 =qαQAuEk

and Ek = 12mv

20 .

This prediction for how gold foil will deflect alpha particles is much differentthan the Thompson model’s prediction – and this prediction fit with the data.This led physicists to conclude that the nuecleus was a point charge.

This means that atoms are mostly empty space! Electrons orbit at 10−8 cmfrom the nucleus, while the nucleus itself has diameter 10−13 cm.

2.2 Classical Models for Blackbody Radiation are Bad

When a cavity is held at constant temperature and allowed to reach thermalequilibrium, its spectrum will become the same no matter where in the cavity itis observed.

So, physicists measured the output power per area per frequency of thespectrum emitted by a cavity (denoted Rν). They found that the peak frequencyincreased linearly with temperature; the area of the curve increased by therather simple relation

R(T ) =

∫ ∞0

Rν(ν, T )dν = σT 4,

where σ is the Stephan-Boltzmann constant, equal to 5.67×10−5erg ·K−4 ·cm−2 · sec−1.

Physicists tried to derive the curve Rν(ν, T ) from classical statistical me-chanics.

14


2.2.1 The Classical Derivation

For simplicity’s sake, assume our cavity is in the shape of a box.

What standing EM waves can we have in the cavity? For each n there arestanding waves with n nodes. The energy of a particular standing wave is

En =

∫1

8π(E2

n +B2n)dV.

Remark 2.1. This is analogous to the equation of a harmonic oscillator, U =12kx

2 + 12mp

2.

The probability that a harmonic oscillator has a particular energy E isproportional to e−E/kBT .1 We can divide by the integral to get a normalizedprobability density:

p(E) =e−E/kBT∫∞

0e−E/kBT

=e−E/kBT

kBT.

Once we have this probability distribution, we can ask, what is the meanenergy in this oscillator? We do this by just computing the expected value ofthis probability distribution:∫ ∞

0

Ep(E)dE = kBT.

It turns out that at room temperature, kBT is about 0.03eV .2

Now we ask: what is the distribution of these energies with respect tofrequency? We’ll work through the 1-D and 2-D cases before doing the 3-D case,which we care about.

2.2.2 The 1D Case

Let’s think about standing waves in 1D, on an interval of length L. The possiblestanding waves are sin kx, where

k =π

L,

2π

L,

3π

L, . . . .

So, if N(k) is the number of modes with up to k nodes, then

N(k) =k

π/L=L

πk.

We will work in terms of frequnecy. Recall that k = 2πλ , and λ = c

ν , so k = 2πc ν.

Thus, if N(ν) is the number of modes with frequency up to ν, then

N(ν) =L

π

2π

cν =

2L

cν.

So, dNdν = 2L

c .

1Proof: 8.044. We don’t have to know this, but it’s pretty.2Hydrogen ionizes with 13.6eV. This is why things don’t spontaneously ionize at room

temperature.

15


2.2.3 The 2D Case

Now, consider standing waves in a 2D box of side length L and area A = L2.The standing waves are E = E0 sin(kxx) sin(kyy), where kx, ky are among thefrequencies

k =π

L,

2π

L,

3π

L, . . . .

If we plot (kx, ky) we get a lattice. For each mode we frequency k equals√k2x + k2

y. We do the same computation.

Let N(k) be the number of modes with frequency up to k. Then

N(k) =1

4π

(kL

π

)2

.

Since k = 2πν/c, we have

N(ν) =1

4π

(2πνL

cπ

)2

=Aπν2

c2

anddN

dν=

2Aπ

c2ν.

Note that in this case, the density of modes does depend on frequency.

2.2.4 The 3D Case

We do the analogous computation. Let N(k) be the number of nodes with

frequency k =√k2x + k2

y + k2z up to k. Then,

N(k) =1

8· 4

3π

(kL

π

)3

· 2

where the last multiplicative factor of 2 is due to polarization. Thus, usingk = 2πν

c , we get

N(ν) =8πL3ν3

3c3,

anddN

dν=

8πV

c3ν2.

2.2.5 Things Go Really, Really, Wrong

Classically we expect kBT of energy in each mode. The energy density we findper frequency is

uT (dν)dν =(# of modes in range ν to ν + dν) · (energy per mode)

volume of cavity.

So,

uT (ν)dν =dNdν dν · kBT

V=

8πkBT

c3ν2dν.

This exactly fit the data for low frequencies, and completely missed for higherfrequencies, in what was known as the ultraviolet catastrophe. (See Figure 1)

Nothing in classical mechanics can explain this. Tune in next time for thequantum resolution...

16


Figure 1: Ultraviolet Catastrophe. Image source: University of Colorado

17


3 February 21, 2017

3.1 Blackbody Radiation – Planck Hypothesis

3.1.1 Last time

Last time we talked about blackbody radiation. We talked about the function

uT (ν)dν =dN(ν)

dν· kBT ·

1

V,

the energy density of modes in frequency space. dN(ν)dν is the number of modes

in the frequency range [ν, ν + dν] and kBT is the energy per mode.

We found that this function grows as ν2 because N(ν) grows as ν3 by ageometric argument. Concretely,

uT (ν)dν =8πkBT

c3ν2dν.

We find that this follows the experiemntal curve very well at low frequenciesbut deviates at higher frequencies.

So, the natural question is: what’s going wrong here? The mode-countingpart has to be right. The dispersion relation ν = ck has to be right. So, MaxPlanck hypothesized, the kBT part is wrong.

3.1.2 Planck’s Hypothesis

Planck’s Hypothesis was that energy comes in lumps, ∆E = hν, for some h.At small frequencies, hν << kBT , so we can approximate energy as continuous,like in the classical model. When hν ≈ kBT , then there isn’t enough thermalenergy to create a lump, so oscillators with these modes don’t vibrate.

Then n lumps of energy is En = nhν, and the probability that a lump hasenergy En is

P (En) ∝ e−nhν/kBT .

So, normalizing yields

P (En) =e−nhν/kBT∑

m≥0 e−mhν/kBT

= e−nhν/kBT(

1− e−hν/kBT).

The average energy is

< En >=∑n≥0

EnP (En).

3.1.3 Comparison with Classical Model

For hν << kBT , 1− e−hν/kBT ≈ hνkBT

, so this becomes

P (En) =hν

kBTe−nhνkBT .

18


Recall that in the classical model, when we assumed continuous energy, we hadthe probability density function

p(E) =e−E/kBT

kBT.

The probability that E falls in an interval of width hν << kBT is then

hνp(E) =hν

kBTe−E/kBT ,

which matches the quantized result. Thus, when hν << kBT , then

< En >=∑n≥0

EnP (En) ≈∫ ∞

0

Ep(E)dE = kBT

is a good approximation – the sum is a Riemann sum for the integral. But,when hν > kBT the first term in the sum is exponentially suppressed, so theclassical approximation fails.

3.1.4 The Quantum Model

So, (computation omitted) the average energy of a mode with frequency ν is

< E >=hν

ehν/kBT − 1.

Note that if hν << kBT , ehν/kBT ≈ 1 + hνkBT

, so

< E >≈ hν

hν/kBT= kBT,

while if hν >> kBT the exponential term dominates the demoninator, and

< E >≈ hνe−hν/kBT .

The natural question then, was, what value of h makes this work? By fittingthe data, it was found that

h = 6.626× 10−27 erg · sec .

This value is now known as Planck’s Constant.

3.2 Photoelectric Effect

3.2.1 Photoelectric Emission

In an experiment, a cathode and anode, at opposite ends of a tube, are connectedby a battery. Light was shined at the cathode. The following observations werecollected:

• When red light was used, no current flowed.

• When green light was used, current flowed, according to the followingpattern:

19


– There was some threshold voltage −V0, below which no currentflowed. Above this threshold, as voltage increased current increased,eventually tapering at some maximum.

– If the light intensity increased, the same V to I curve was observed,scaled upward along the I axis.

This doesn’t match classical predictions.

3.2.2 Einstein’s Resolution

Einstein explained this as follows. Electrons in a material are bound by someamount of energy. For each electron, a “work function” W is needed to releasethe electron from the material; the least tightly-held electrons have work functionW0. Moreover, light comes in bundles called photons, each with energy hν.

Thus, if hν < W0, we get no current because the photons don’t have enoughenergy to release any electrons. Moreover, we have the relation

hν = W0 + eV0.

This is because the electron, after being ejected from the metal, has kineticenergy K = hν − W0. For current to flow, this energy must be enough toovercome the repulsion eV0 of the anode.

In summary, Einstein’s postulates:

• Light comes in bundles of energy hν called photons.

• Electrons are bound to metals with some energy W0. This is the energyneeded to eject an electron.

• Each photoelectron (electron ejected by light) is ejected by one photon.

• The kinetic energy is given by K = hν −W (W ≥W0).

• The maximum observed kinetic energy is K = hν −W0.

An interesting consequence of this: two metals (e.g. gold and silver) havedifferent work functions, so when they are connected by a wire electrons willtend to move from one to the other. So, we expect an electric field between thetwo metals – and this has been observed!

Amazingly, the same h Einstein found was the one Planck computed forblackbody radiation! This was a huge triumph for theoretical physics.

3.3 Short Wavelength Limit of X-ray Generation

In another experiment, physicists put a filament on one end of a vacuum tube.The filament was heated enough that the temperature is comparable to the workfunction W0, so electrons will “boil” off the filament. On the other end of thetube was a metal plate.

The filament and plate were connected by a huge voltage V0 – on the orderof 10000 V. As expected, the electrons from the filament hit the plate at highspeed, and their sudden deceleration produced light.

20


Figure 2: Short Wavelength Limit. Image source: ammrf.org.au

If we plot wavelength vs. intensity, we get a graph that looks like Figure 2.What we notice is that there’s a “cutoff” wavelength λmin, below which no lightis created. Moreover, it was found that

1

λmin=

e

hcV0.

This is once again consistent with the quantized model of light: each electronhas energy eV0 and can create photons of at most this energy. Thus:

eV0 = hνmax =hc

λmin⇒ 1

λmin=

e

hcV0.

3.4 Compton Scattering

3.4.1 The Experiment

The sky is blue because air molecules in the upper atmosphere preferentiallyscatter blue light. The air molecules aren’t “producing” blue light – what’sgoing in is what’s coming out.

It turns out that this isn’t true at the quantum scale.

In Bragg scattering, light is shined at a multilayered crystal so that onlycertain wavelengths experience constructive interference. Shining an x-ray ata Bragg scattering crystal is one way to select a particular wavelength from abeam.

Using an X-ray source and a Bragg scattering crystal, we get a monochromaticX-ray beam with wavelength λ0. We can shoot this beam at a target. We canuse a detector to find and plot wavelength vs. intensity of the scattered rays atvarious angles around the target.

21


3.4.2 What was Observed

In the classical model, we expect wavelength in to equal wavelength out. If theclassical model holds, the wavelength vs. intensity graph of the scattered raysshould have a peak at λ0, and nowhere else.

What was actually observed:

• At θ = 0, if we plot wavelength with intensity at the detector, we see amonochromatic peak with wavelength λ.

• At θ = 45, we see two peaks with wavelengths λ0 and λ1, with λ1 > λ0.

• At θ = 90, we again see two peaks with wavelengths λ0 and λ1. λ1 islarger than for θ = 45.

• At θ = 135, we again see two peaks. λ1 is even larger.

What is the quantum explanation for this? Why do we see an increasedwavelength?

Joke. “I’ll explain some of this. I’m not going to explain all of it, because it’sa homework problem.” -Ashoori

When a photon with energy and momentum E0, p0 hits a stationary electron,the electron scatters with angle φ and energy and momentum K, p, and thescattered photon scatters with angle θ and energy and momentum E1, p1. So,in the x and y directions, conservation of momentum gives:

p0 = p1 cos θ + p cosφ

p1 sin θ = p sinφ.

Using this and conservation of energy, we get (computation in homework):

∆λ = λ1 − λ0 =h

mec(1− cos θ).

The value hmec

is called the Compton wavelength. For an electron, this value

is about 0.024A.

Why don’t we see this effect for visible light? Visible light has wavelengthabout 5000A, so the Compton shift is negligible.

It turns out that like energy, momentum comes in lumps! For photons,E = hν, and

p =E

c=h

λ.

3.5 Epilogue

It turns out that once we accept this, the rest of quantum follows.

De Broglie hypothesized: if light is a particle and a wave, can electrons bewaves too? He hypothesized that an electron with momentum p has wavelength

λdeBroglie =h

p.

This has been observed! More on this... next time!

22


4 February 22, 2017

4.1 Dimensional Analysis

So far in this class, we’ve learned these constants, in some sense parameters ofthe universe:

• e, the electron charge;

• me, the electron mass;

• h, Planck’s constant.

Can we derive a length scale from these units? Let’s say

lnatural = mαe (e2)βhγ .

Remember that e2 has units

F · l2 =ml3

t2

and h has units

h = E · t =ml2

t.

We can plug these in:

lnatural = mα ·(ml3t−2

)β · (ml2t−1)γ.

We want the units on the right to multiply out to l, so:

α+ β + γ = 0 (4.1)

3β + 2γ = 1 (4.2)

−2β − γ = 0 (4.3)

(4.4)

Solving this (details omitted) gives us (α, β, γ) = (−1,−1, 2). Thus, the “natural”unit length derived from these constants is

lnatural = m−1e (e2)−1h2 =

h2

mee2= (2π)2 · 0.528A.

It turns out that 0.528A is the radius of a hydrogen atom. This isn’t a coinci-dence! We’ll see the constant

0.528A =

(h2π

)2mee2

again later.

4.2 Matter Waves

De Broglie theorized that all particles have an associated wavelength. A particlewith momentum p has de Broglie wavelength

λ =h

p.

23


Example 4.1

Let’s find the de Broglie wavelength of a 100 g stone, moving at 103 cm / s:

λ =h

p=

6.6× 10−27 erg · s102 g ·103 cm / s

= 6.6× 10−24A.

This is far too small to care about.

Now let’s find the de Broglie wavelength of an electron with energyK = 100 eV.

Since K = p2

2m , we have p =√

2mK, and

λ =h

p=

h√2mK

= 1.2A.

This is big enough to be resolved in scattering experiments!

4.3 The Bohr Atom

4.4 Bohr Orbits

We know electrons orbit an atom. How can this be, if electrons are a wave? Wedon’t want the wave to destructively interfere with itself, so the wave has toclose in on itself.

Say the wave has n nodes, orbits at radius rn, and has wavelength λn. Then,

nλn = 2πrn.

But, de Broglie says that

λn =h

pn=

h

mvn,

sonh

mvn= 2πrn ⇒ mvrn =

nh

2π.

In fact, the term h2π appears often enough in quantum that we’ll give it a name:

~.

Then we havemvrn = n~.

In other words: the angular momentum L = mvrn = n~ is quantized!

It was found that when electrons stay in Bohr orbits, they don’t radiate.Only when electrons jump between orbits do they radiate. When an electronjumps from an orbit with energy Ei to a lower-energy orbit with energy Ef , itemits a photon with energy

Ephoton = hν = Ei − Ef .

4.4.1 Energy and Radius of Bohr Orbits

What is the radius of Bohr orbits? We get this by plugging into classical physics:

mv2

r=ze2

r2⇒ mv2r2 = ze2r.

24


Since L = mvr = n~, we have v = n~mr . Plugging this in gives

m

(n~mr

)2

r2 = ze2r.

Solving (computation omitted) gives us

rn =n2

z

(~2

me2

).

We see the constant a0 = ~2

me2 = 0.528A again! This is why a0 is the radius of ahydrogen atom.

What’s the energy of this electron? Well,

E = K + V =1

2mv2 − ze2

r= −ze

2

2r.

This is negative in the sense that we take potential energy to be zero infinitelyfar away.

In the nth orbital, we have

En = −1

2ze2 zme

2

n2~2= − 1

n2

z2e2

2a0.

We can write this another way (computation omitted):

En = −1

2α2mc2

(z2

n2

).

Here, mc2 = 511 KeV is the rest energy of the electron. α is the fine structureconstant, a famous dimensionless constant in quantum:

α =e2

~c≈ 1

137.

If we take n = z = 1 in the formula for En, we get E1 = −13.6 eV. This isprecisely the ionization energy of hydrogen! Quantum theory just works.

4.4.2 Emission spectra

If we look at the emission spectrum of hydrogen in a gas discharge tube, wesee a bunch of discrete spectral lines, organized in several series, as shown inFigure 3.

One series, called the Lyman series, corresponds to transitions down to then = 1 energy state. Likweise, the Balmer series, corresponds to transitionsdown to the n = 2 energy state, and the Paschen series corresponds totransitions down to the n = 3 energy state.

What frequencies ν do we expect from these transitions? Well, we havehν = Ei − Ef , so

ν =Ei − Ef

h=

e2

2ha0

(1

n2i

− 1

n2f

).

This was an exact match for Hydrogen!

...unfortunately it completely failed for Heilum, for reasons we’ll discussanother day.

25


Figure 3: Emission Spectrum of Hydrogen. Image Source: Wikipedia.

4.5 Wilson-Sommerfield Quantization

4.5.1 Quantize All the Things!

So far we’ve discussed two kinds of quantization:

• Quantization of EM energy for a harmonic oscillator into lumps hν led tounderstanding of blackbody radiation;

• Quantization of angular momentum L led to the Bohr atom.

The natural question arising from this is: what else can we quantize?

Wilson and Sommerfield proposed a general rule: if we integrate variableswith respect to their conjugate variables in a closed action, we get a multiple ofh: ∮

p · dq = nh,

if p, q are conjugate, and the integral is over one cycle.

4.5.2 Case Study: Position and Momentum

Position x and momentum px are conjugate. Consider a harmonic oscillator,oscillating in the x direction. The potential energy of this oscillator V (x) isquadratic in x.

Let’s say the oscillator has energy E and amplitude xmax. Then

E =p2

2m+

1

2kx =

1

2kx2

max.

The oscillator has angular frequency

ω =

√k

m,

26


so k = mω2. Thus:

p =

√1

2mω2(x2

max − x2) · 2m = mω√x2max − x2.

We want to integrate∮p · dx along one oscillation.

We do this by integrating this on a quarter-cycle (x = 0 to x = xmax) andmultiplying by 4 (details somewhat omitted):

nh = mω

∮one period

√x2max − x2dx

= 4mω

∫ xmax

0

√x2max − x2dx

= 4mω(√π4x2

max

).

Putting this together gives us

nh = πmωx2max

2π

ω

(1

2mω2x2

max

).

But, 12mω

2x2max is energy E, so

2π

ωE = nh⇒ E =

nhω

2π= nhν.

This tells us that the energy levels of this oscillator are quantized, and equallyspaced.

Using this method, we can find the energy levels of any oscillator defined bya potential energy function V (x).

4.5.3 Case Study: Angular Momentum and Angle

Applying Wilson-Sommerfield to conjugate variables L, θ gives us∮L · dθ = 2πL = nh⇒ L =

nh

2π= n~.

4.6 Moseley’s Experiment; Shielding

Initially physicists didn’t understand why atoms’ atomic mass were usuallytwice their charge; atoms were ordered by atomic mass, not charge. This led tosome pairs of consecutive elements (e.g. nickel, cobalt) being swapped in earlyorderings of the elements.

In an experiment by Moseley, electrons from a filament were acceleratedby a big potential into a target containing some element. Moseley knew thefollowing:

• Bohr orbits define “shells,” which may have more than one electron. Eachshell has a single energy level.

27


• When an incident electron kicks out an electron from (say) n = 1, anelectron from (say) n = 2 drops down and emits a photon. The emittedphoton has energy

E = hν = Q2E1

(1

12− 1

22

)where E1 is the ground-state energy for hydrogen. More generally, photonsemitted when electrons drop from n = ni to n = nf have energy

E = hν = Q2E1

(1

n2f

− 1

n2i

).

4.6.1 What Moseley Saw

In the wavelength vs. intensity plots of the emitted light, Moseley saw a“background” intensity curve from Bremstrahhlung, and peaks from electronsbeing kicked out of orbitals and replaced.

4.6.2 Moseley’s Conclusions

The shortest-wavelength peak corresponds to the highest-energy photons, whichresult from electrons falling into the n = 1 layer. Likewise, the second shortest-wavelength peak corresponds to electrons falling into the n = 2 layer, and soon.

By the formula

E = hν = Q2E1

(1

n2f

− 1

n2i

),

we expect√ν ∝ Q. So, Moseley plotted

√ν, where ν is the frequency of

the shortest-wavelength peak, against atomic number z, and found a linearrelationship: √

ν ∝ z − 1.

3 Why the −1? There’s two electrons in the n = 1 layer, and one of them shieldsthe nucleus, so the “effective” charge of the nucleus is z − 1.

When this experiment was done for the second shortest-wavelength peaks(corresponding to the n = 2 layer), it was found that

√ν ∝ z − 7.5.

Again, this is due to shielding! We see a higher degree of shielding here becausethe n = 2 layer is more shielded than the n = 1 layer.

3This is how, for example, the reversing of nickel and cobalt was detected. The values of√ν were in the wrong order.

28


5 February 27, 2017

5.1 Double-Slit Experiment

5.1.1 The Gun Model

Consider having a gun that shoots bullets in every direction. The bullets areaimed at a wall with two slits, behind which is another wall.

Joke (Professor Ashoori draws a smiley face at the target of the gun).

We can make an empirical probability distribution function for the likelihoodthat a bullet lands at a location x on the rear wall:

p(x) =# of bullets hitting x

Total # of bullets.

What distributions do we expect?

If only the first hole is open, we expect to see a unimodal distribution p1(x)centered behind the first slit. Likewise, if only the second hole is open, we expectto see a unimodal distribution p2(x) centered behind the second slit.

When both slits are open, we’ll expect to see distribution p12(x) = p1(x) +p2(x), which will be another unimodal distribution with mode in the rear of themiddle wall.

5.1.2 The Wave Model

Now, let’s replace the gun with a wave source. Then, by Huygens’ principle, thetwo slits each act as wave sources. When only the first slit is open, the intensityI1(x) of the wave on the rear wall is a unimodal distribution centered behindthe first slit. Similarly I2(x) is is unimodal, centered behind the second slit.

But, when both slits are open, the resulting intensity on the rear wall I12 isa diffraction pattern. In other words, I12 6= I1 + I2.

Let’s work with this analytically. Let the waves be

Ψi = Ai cos(ωt− kl1 + φi),

for i = 1, 2. Consider a point x on the rear wall that is l1 from the nearer slit,and l2 > l1 from the far slit.

The difference of the phases is

δ = k(l1 − l2)− (φ1 − φ2).

The intensity of the individual waves are

I1 = |A1|2, I2 = |A2|2.

But, the intensity of the sum is

I12 = |A1 +A2|2 = |A1|2 + |A2|2 + 2|A1||A2| cos δ,

and the last cross-term is responsible for diffraction patterns.

29


5.1.3 Electrons are Waves!

Physicists did the two-slit experiment with electrons. To everyone’s surprise, thefrequencies with which the electrons hit the rear wall looked like a diffractionpattern! The electrons behaved like waves. And the wavelength that fit thedata was precisely λ = h

p , where p is the electron’s momentum.

But, the electron also has to be a particle – the receiving wall’s detectorsclick audibly when an electron arrives, so it has to be receiving whole electrons.

So, what gives? Does the electron split into two, go through both holes, andthen somehow hit only once? This doesn’t explain how at some places on thereceiving wall, the probability of an electron is zero when both slits are open,but nonzero when one of the slits is covered. Somehow reducing the number ofpaths from the gun to the location increases the probability of an electron there.

No. The only possible conclusion is that electrons have both particle-likeand wave-like properties.

In summary:

• The receiving wall only receives whole electrons.

• When one slit is open, we get the probability distributions p1 and p2.

• p12 is a diffraction pattern, not the sum p1 + p2.

• The electrons aren’t behaving like bullets!

In fact, even bullets have wavelike properties. Their wavelength is just sotiny that the crests of the resulting diffraction pattern are too fine to resolve,making the resulting distribution look like p1 + p2.

Later, we’ll see that the wavelike properties of electrons are determined bya wavefunction Ψ – this is the thing that’s diffracting. We’ll see that theprobability an electron is at x is proportional to |Ψ(x)|2.

5.1.4 Where did the electron really go?

What if we put a bright light by one of the slits, so we can see which slit theelectron really went through? Then the distribution p12 we get is p1 + p2 – thediffraction pattern goes away! Somehow, observing the electrons forces them tochoose a hole to go through.

If we gradually dim the light until we start missing some electrons, then theelectrons our light detected follow the distribution p1 + p2, while the ones wemissed follow the diffraction pattern. Weird.

The same thing happens if we gradually increase the wavelength of the light.When the wavelength is too big, the light can’t resolve the electrons anymore,and the electrons show the diffraction pattern.

5.2 Heisenberg’s Uncertainty Principle

5.2.1 Heisenberg’s Microscope

Suppose we have a point source of light, a front wall with a slit of width D, anda back wall. The light from the source produces a single-slit diffraction patternon the back wall.

30


[diagram]

Suppose we have two different point sources in front of the front wall. TheRayleigh Criterion for resolving the sources says that in order to resolve thetwo sources, the two peaks of the diffraction patterns must be separated by atleast the maximum-to-first-minimum distance of each pattern.

[diagram]

The minimum angle ∆θ needed for this to occur obeys (derivation omitted)

sin(∆θ) ≈ 1.22λ

D.

Let f be the distance from a light source to the aperature. Then

f sin(∆θ) = ∆xmin.

Let the acceptance angle θ′ be the angular width of the aperature from oneof the sources.

f sinθ′

2=D

2⇒ f sin θ′ = D ⇒ sin θ′ =

D

f,

where we use the small angle approximation. So:

∆xminf

= sin(∆θ) ≈ λ

D≈ λ

f sin θ′,

where the ≈ drops the 1.22 factor. So, as a crude approximation, the minimumdistance our microscope can resolve is

∆xmin ≈λ

sin θ′,

provided θ′ is small.

5.2.2 Derivation of the Uncertainty Principle

Heisenberg’s Uncertainty Principle states:

∆p∆x ≥ ~2.

Let’s say we have an electron in Heisenberg’s microscope, as described above.From the above derivation, the minimum distance to which we can resolve theelectron’s position is

∆xmin ≈λ

sin θ′.

When the light strikes the electron, by Compton scattering it will give theelectron a “kick,” changing its momentum. We’ll show that the uncertainty inthe momentum ∆p satisfies the inequality above.

So, if we see a flash of light from the electron scattering the photon, we knowthe photon was deflected within an angle φ ∈ [θ′,−θ′].4 We ask, how much recoilcould the electron have gotten from this interaction?

4We drop factors of 2 because we’ll be off by a bit here. The derivation works in spirit.

31


The photon’s x-momentum range is [−p sin θ′, p sin θ′]. So:

∆px = 2p sin θ′ =2h

λsin θ′.

By conservation of momentum the electron’s x-momentum uncertainty is alsothis.

Then,

∆p∆x ≈ 2h

λsin θ′ · λ

sinθ′= 2h.

5.2.3 The Uncertainty Principle: Energy and Time

Recall that the “x-direction energy” of a particle is Ex =p2x2m . So,

∆E =∂E

∂px∆px =

pxm

∆px = vx∆px.

If we observe a particle with velocity vx at a location with uncertainty ∆x, thenthe uncertainty in when it was measured is

∆t =∆x

vx.

So, we also get∆E∆t = ∆px∆x ≈ 2h.

If we’re a bit more careful with constants, we get that the best we can do is

∆p∆x ≥ ~2, ∆E∆t ≥ ~

2.

5.2.4 The Uncertiain Principle: A Wave

Let’s consider a wavey(x, t) = A cos(k0x− ω0t).

Then, k0 = 2πλ0

for wavelength λ0 and w0 = 2πν0 = 2πT0

for frequency ν0 andperiod T0.

Let’s say we really know k0. Then

p =h

λ=

2π~λ

= ~k.

Then we have no idea where the particle is. This matches our intuition that asuncertainty in p decreases, uncertainty in x increases.

We can do the same for ω and t. Since ∆ω∆t ∼ 1, we have (~∆ω)∆t =∆E∆t ∼ ~.

It turns out the best we can do is when x, k each have Gaussian distribution.Then, if x is known to standard deviation σ, k is known to standard deviation1

2σ , so

∆x∆k =1

2,

and

∆x∆p = ∆x (~∆k) =~2.

32


5.3 Applications of the Uncertianty Principle

5.3.1 Application: Particle in a Box

Let’s say we have a particle in a box of width a. We want to know what theminimum kinetic energy of the particle is.

By the Uncertainty Principle, the best we can know x, p is

∆x∆p =~2.

So, ∆p is best known when x = a, and

∆p =~2a

is an order-of-magnitude estimate for pmin, the smallest momentum. Theminimum kinetic energy is then, approximately,

Emin ≈p2min

2m≈ ~2

8ma2.

5.3.2 Application: Hydrogen Atom

We have the energy

E =p2

2m− e2

r.

We have ∆x ∼ r, and ∆p · r ∼ ~⇒ ∆p ∼ ~r . So,

E =~2

2mr2− e2

r.

This function goes to +∞ as r → 0, becomes negative at some intermediatevalue of r, reaches a minimum, and then tapers to 0 as r →∞. [diagram] Bysetting ∂E

∂r = 0, we find that E is minimized when

rmin =~2

me2

Emin =me4

2~2.

This is the real reason the electron can’t fall into the nucleus. It would haveto have too much kinetic energy to do that, and to do so would require it tolocalize itself, in violation of the Uncertainty Principle.

33

Brice Huang 6 March 1, 2017

6 March 1, 2017

6.1 Administrivia

There will be a quiz on Monday, on the material up to, but not including, thislecture.

6.2 Developing the Wave Equation

6.2.1 Non-Quantum Waves

In 8.03 we had an equation for waves on a string:

∂2y

∂x2=

1

v2

∂2y

∂t2.

It turns out that all solutions of this equation are of the form

f(x− vt) + g(x+ vt).

The first term is a waveform that just moves to the right, and the second awaveform that just moves to the left.

It’s also nice that the solutions of this wave equation are linear, in the sensethat linear combinations of solutions are also solutions.

6.2.2 Quantum Waves

Let’s try to derive a quantum version of the wave equation.

What properties do we demand of our wave equation? Well:

• De Broglie said that matter behaves like waves, with wavelength λ = hp .

Einstein said that energy is given by E = hν = ~ω.

• In non-relativistic environments, energy is given by

E =p2

2m+ V (x),

where V is potential energy. When we add in relativity this becomes

Erel =√p2c2 +m2c4 + V.

It turns out the wave function in the relativistic case is a lot harder toanalyze (involving something called Dirac functions), so we’ll focus on thenon-relativistic case.

• We want our set of wave functions Ψ to be linear, i.e. closed under takinglinear combinations.

• In general the potential energy is V = V (x, t). When V is constant weexpect k, ω to be constant.

• We want to be able to take constants out of derivatives, i.e. ∂(cΨ)∂x = c∂Ψ

∂x .

34


Then:p2

2m+ V (x, t) = hν = ~ω,

so~2k2

2m+ V (x, t) = ~ω.

We want an equation like this:

α∂2Ψ(x, t)

∂x2+ V (x, t)Ψ(x, t) = β

∂Ψ(x, t)

∂t,

for some constants α, β.

In the non-quantum version, taking two derivatives on a cosine on eitherside returns cosines on both sides, so cosine waves are a solution. This time, theright-hand side has a single t-derivative. This makes the solutions necessarilycomplex.

We try a solution of the form:

Ψ(x, t) = ei(kx−ωt).

Taking derivatives, we find:

∂2Ψ

∂x2= −k2Aei(kx−ωt) = −k2Ψ,

and∂Ψ

∂t= −iωAei(kx−ωt) = −iωΨ.

Plugging in:−αk2Ψ + V0Ψ = −iωβΨ,

so−αk2 + V0 = −iωβ.

From the equation~2k2

2m+ V (x, t) = ~ω,

we get

α = − ~2

2m, β =

~−i

= i~.

So, we get the Schrodinger Equation

− ~2

2m

∂2Ψ

∂x2+ V (x, t)Ψ = i~

∂Ψ

∂t.

Keep in mind this isn’t a derivation! This is a postulated wave equation – thesimplest one that makes sense and has the properties of a wave equation. Nobodyknew what physical implications this wave had!

Later, it was Bohr and Heisenberg who came up with the interpretationthat |Ψ|2 is the probability of a particle-wave being at a given position. Ψ is aprobability-amplitude wave.

35


6.3 Understanding Ψ and probabilities

Consider a mass m in a potential well V (x, t). How do we analyze its movement?

Classically, we can determine the location x(t) from mechanics equations:F = ma = −∂V∂x , K = 1

2mv2, p = mv, etc.

In the quantum universe, we determine Ψ using the Schrodinger equation.In the one-dimensional case, the probability if finding the particle in [x, x+ dx]at time t is |Ψ(x, t)|2dx. We say that Ψ∗Ψ = |Ψ|2 is a probability density

function. Ψ has units distance−1/2.

In three dimensions the picture is similar. |Ψ|2dxdydz is the probabilityof finding the particle in [x, x + dx] × [y, y + dy] × [z, z + dz]. Ψ has units

distance−3/2.

6.3.1 Probability Density Functions

Back to the one-dimensional case. If p(x) is the probability density function of

a location,∫ bap(x)dx is the probability that the particle is in [a, b]. Since the

particle has to be somewhere, integrating along the real line tells us that∫ +∞

−∞p(x)dx = 1.

But, measurement collapses the wave function, localizing it to a particularlocation. After measurement, an immediate subsequent measurement finds theparticle in the same place.

What constitutes a measurement? This isn’t really understood yet.

6.3.2 Interpretations of Quantum Mechanics

• (Realist View): The particle really is somewhere at all times; the locationjust isn’t known to the experimenter.

• (Orthodox View): The particle isn’t really anywhere until it is measured.Somehow the act of measuring forces it to be somewhere.5

• (Agnostic View): This is all very ill-defined, so shut up and calculate!

6.3.3 Probability Distributions: An Example

Let’s say we have a room of 14 people, with various ages. The distribution ofages is as follows.

Age Number14 115 116 322 224 225 5

5Here, “measurement” isn’t just the process of detecting; the ability to detect is enough.For example, in the two-slit experiment with a detector on one of the slits, it doesn’t matterwhether we’re listening to the detector. The presence of the detector is enough to affect theelectron’s behavior.

36


Let N(j) be the number of poeple with age j. Then we can plot j and N(j)on a histogram.

We can think of the age j as a random variable, so the probability P (j)of selecting someone with age j is

P (j) =N(j)

Ntot.

The most likely age (mode) is 25. The middle age (median) is 23. The average(mean) age is 〈j〉, defined by:

〈j〉 =

∑j jN(j)

Ntot= 21.

We can also get the mean of, say, j2:

〈j2〉 =

∑j j

2N(j)

Ntot.

In general, the mean of any function f(j) is

〈f(j)〉 =∑j

f(j)P (j).

For continuously-distributed random variables x, we can find the mean value(or, expectation value) of x as

〈x〉 =

∫xp(x)dx.

Expectation doesn’t commute with (nonlinear) functions! If the randomvariable x obeys

x =

1 p = 0.5

3 p = 0.5,

then 〈x〉2 = 4, while 〈x2〉 = 5.

6.3.4 Standard Deviation

We want a notion of width for probability distributions. For example, both ofthe following random variables have mean, median, and mode 5:

x1 =

4 p = 1/8

5 p = 3/4

6 p = 1/8

, x1 =

2 p = 1/8

3 p = 1/8

4 p = 1/8

5 p = 1/4

6 p = 1/8

7 p = 1/8

8 p = 1/8

.

We can measure the width of the distribution by the mean square distribution:

〈 (j − 〈j〉)2 〉.

37


This is called the variance. Its square root, the standard devaition, isdenoted σ:

σ =

√〈 (j − 〈j〉)2 〉.

A useful computation trick (derivation omitted but not hard): for any randomvariable j, the standard deviation σ is given by

σ2 = 〈j2〉 − 〈j〉2.

38


7 March 8, 2017

7.1 Administrivia

Exam 1: µ = 77, σ = 16. Rough grade cutoffs:

• A: 85-100

• B: 70-85

• C: 60-70

7.2 Probability Densities

If I have a wave function, I can ask, what is the probability that the particle isin a range of places? This is defined in terms of a probability density functionp(x), such that p(x)dx is the probability that the location x lies in [x, x+ dx]. 6

Then, we have the following properties:

P (a ≤ x ≤ b) =

∫ b

a

p(x)dx P (−∞ < x <∞) =

∫ ∞−∞

p(x)dx = 1

We can define the expectation of x, denoted 〈x〉, as

〈x〉 =

∫ ∞−∞

xp(x)dx.

Similarly, we can define the expectation of any function f(x) by

〈f(x)〉 =

∫ ∞−∞

f(x)p(x)dx.

7.3 Wave Function Normalization

Given a wave function Ψ, the probability density of a particle being at x isΨ∗Ψ(x), so in terms of the wave function we have

P (a ≤ x ≤ b) =

∫ b

a

Ψ∗(x, t)Ψ(x, t)dx

∫ ∞−∞

Ψ∗(x)Ψ(x)dx = 1

Since any multiple of a Ψ satisfying the Schrodinger equation also satisfiesthe Schrodinger equation, the relation

∫∞−∞Ψ∗(x)Ψ(x)dx = 1 doesn’t hold for

all Ψ. But, we can always scale Ψ to make this relation hold. This is callednormalization.

6Generally we don’t ask, “what is the probability that the particle is at a specific location,”because for usual probability densities this is zero. The exceptions are when the pdf is a deltafunction.

39


Example 7.1

Suppose we have the wavefunction

Ψ(x, t) = Ae−λ|x|e−iωt.

First, we want to find what A is, so the wavefunction is normalized. Wewant that

1 =

∫ ∞−∞

Ψ∗(x)Ψ(x)dx.

So,

1 = A∗A

∫ ∞−∞

e2−λ|x|dx = |A|2 · 2∫ ∞

0

e−2λxdx =|A|2

λ.

So, |A|2 = λ, from which we infer A =√λeiφ for some φ.

Therefore, the normalized wavefunction is

Ψ(x, t) =√λeiφe−λ|x|e−iωt.

In general, the wavefunction is known to an arbitrary phase factor – wecan set eiφ to whatever we want.

To make life easy, we just set eiφ = 1. So, our wavefunction is

Ψ(x, t) =√λe−λ|x|e−iωt.

Example 7.2

Let’s take the wavefunction from the previous example and find the meanand standard deviation. Recall that

|Ψ|2 = λe−2λ|x|.

So,

µ = 〈x〉 =

∫ ∞−∞

x|Ψ|2 = λ

∫ ∞−∞

xe−2λ|x|dx = 0

because the integrand is an odd function. Moreover,

〈x2〉 =

∫ ∞−∞

x2|Ψ|2dx = 2λ

∫ ∞0

x2e−2λxdx =1

2λ2.

To get the standard deviation, recall that

σ2 = 〈x2〉 − 〈x〉2 =1

2λ2,

so σ = 1λ√

2.

7.4 Operators

If 〈x〉 is the expectation of position x, then

d〈x〉dt

=d

dt

∫ ∞−∞

x|Ψ|2dx =

∫ ∞−∞

∂x

∂t|Ψ|2dx+

∫ ∞−∞

x∂

∂t|Ψ|2dx.

40


Note that the integral∫∞−∞

dxdt |Ψ|

2dx is zero, because the coordinate x isn’tchanging! Therefore,

d〈x〉dt

=

∫ ∞−∞

x∂

∂t|Ψ|2dx

=

∫ ∞−∞

x∂

∂tΨ∗Ψdx

=

∫ ∞−∞

x

(Ψ∗

∂Ψ

∂t+ Ψ

∂Ψ∗

∂t

)dx.

But, we know the Schrodinger equation has a ∂Ψ∂t term! Recall that the

Schrodinger equation says

∂Ψ

∂t=

i~2m

∂2Ψ

∂x2− i

~VΨ.

By taking the complex conjugate of this equation we get

∂Ψ∗

∂t= − i~

2m

∂2Ψ∗

∂x2+i

~VΨ∗.

We plug this into the above equation and get (computation omitted)

d〈x〉dt

=i~2m

∫ ∞−∞

x∂

∂x

(Ψ∗

∂Ψ

∂x− ∂Ψ∗

∂xΨ

).

What we’ve done is change the time-derivatives, which we didn’t know what todo with, into space-derivatives! In fact, after integrating by parts twice:

d〈x〉dt

= − i~m

∫ ∞−∞

Ψ∗∂Ψ

∂xdx.

For now, we’ll take as a given that7

〈v〉 =d〈x〉dt

Then, the expected momentum 〈p〉 = m〈v〉 is

〈p〉 = md〈x〉dt

= −i~∫ ∞−∞

Ψ∗dΨ

dxdx.

Intuitively: Things that wiggle fast have to have big momenta.

7.4.1 Operators for Observarbles

We write

〈x(t)〉 =

∫ ∞−∞

Ψ∗(x, t)xΨ(x, t)dx.

We write x in the middle because we’re thinking of x as an operator thatoperates on Ψ by multiplication. Analogously, we write

〈px(t)〉 =

∫ ∞−∞

Ψ∗(x, t)

(~i

∂

∂x

)Ψ(x, t)dx.

7It’s not clear that this is well-defined – how do we define velocity when position isuncertain? It turns out this is fine, for reasons we won’t go into.

41


Here, p = ~i∂∂x is the “momentum operator.”

More generally, for an operator A, we can write

〈A〉 =

∫ ∞−∞

Ψ∗AΨ.

As a rule, operators for observables always goive real expectation values. Inthe langauge of abstract algebra, operators for observables are Hermitian:∫ ∞

−∞(AΨ)∗Ψdx =

∫ ∞−∞

Ψ∗(AΨ)dx.

This is why we write operators in the middle – it doesn’t matter which waywe operate.

7.4.2 Commutators

In general, operators don’t commute. For example.

xp 6= px.

This motivates us to define the commutator. For any two operators A, B, wedefine the commutator

[A, B] = AB − BA.

Observe that [A, B] = 0 iff A, B commute. Also, the commutator is itself anoperator!

Example 7.3

Let’s compute [p, x]. Note that:

[p, x]Ψ = pxΨ− xpΨ

= ~i∂

∂x(xΨ)− x~

i

∂Ψ

∂x

=

(x~i

∂Ψ

∂x+

~i

)− x~

i

∂Ψ

∂x

=~i

Ψ.

Thus, we write:

[p, x] =~i.

In multiple dimensions, we have, for example:

[x, y] = xy − yx = 0.

Moreover, because the ∂∂x operator doesn’t care about y,

[y, px] = 0.

Likewise,[px, py] = 0.

42


7.4.3 Generalized Uncertainty Principle

This is all related to the Uncertainty Principle – somehow, when two operatorsdon’t commute, knowing more about one of them means we know less about theother. This can be formalized as the Generalized Uncertainty Principle:

σ2Aσ

2B ≥

(1

2i[A, B]

)2

,

where σA, σB are the standard deviations of the random variables associatedwith A,B. In particular, for (A, B) = (x, px), this yields

∆p∆x ≥ ~2.

So, from this we can recover the original Uncertainty Principle.

7.5 The Hamiltonian

Recall the Schrodinger Equation

− ~2

2m

∂2Ψ

∂x2+ V (x, t)Ψ = i~

∂Ψ

∂t.

We can write this as

HΨ = i~∂Ψ

∂t,

where the operator H, the Hamiltonian, is

H = − ~2

2m

∂2

∂x2+ V (x, t).

Why do we care? It turns out H is an energy operator. If you’ve taken classicalmechanics (8.223):

H = T + V =p2

2m+ V = total energy.

In the quantum world, we have

p =~i

∂

dx= −i~ ∂

∂x,

so

p2 = −~2 ∂2

∂x2.

Plugging this into the relation H = p2

2m + V gives us back exactly the quantumdefinition of H:

H =p2

2m+ V = − ~2

2m

∂2

∂x2+ V.

43


7.6 Time-Independent Schrodinger Equation

In classical mechanics, we decoupled space and time to find standing waves ofthe classical wave function. We’ll try to do the same in quantum.

We take the Schrodinger equation

− ~2

2m

∂2Ψ

∂x2+ V (x, t)Ψ = i~

∂Ψ

∂t.

and try a separable solution Ψ(x, t) = ψ(x)f(t). We stipulate that V is time-independent, so V (x, t) = V (x). Then, we get

− ~2

2mψ′′f + V (x)ψf = i~ψf.

We divide by ψf to get

− ~2

2m

ψ′′

ψ+ V (x) = i~

f

f.

The left-hand side is only a function of x, and the right-hand side is only afunction of t – so both sides are independent of both x and t, and are in factconstant! We’ll call this constant E (laughter from class), the “separationconstant.”

Now we have two equations:

− ~2

2mψ′′ + V ψ = Eψ ⇒ Hψ = Eψ

i~f = Ef ⇒ df

dt= − iE

~f.

The first equation depends on V , so let’s solve the second equation first. Weknow the second equation has solution

f(t) = Ce−iE~ t.

The quantity Et~ must be dimensionless, so E must have units of energy... hence

the naming choice E (laughter from class).

The other equation,Hψ = Eψ

is called the Time-Independent Schoridnger Equation.

44


8 March 13, 2017

8.1 Generalized Uncertainty Relation

A correction to what was said last time: the Generalized Uncertainty Relationsays that

σAσB ≥∣∣∣∣ 1

2i〈[A, B]〉

∣∣∣∣ =1

2

∣∣∣〈[A, B]〉∣∣∣ .

8.2 Erhenfest’s Theorem

Theorem 8.1 (Erhenfest)

For any operator A:d

dt〈A〉 =

1

i~〈[A, H]〉

Why is this? We assume A itself has no time dependence, i.e. ∂A∂t = 0. Then,

d

dt〈A〉 =

d

dt

∫ ∞−∞

Ψ∗AΨdx

=

∫ ∞−∞

∂Ψ∗

∂tAΨdx+

∫ ∞−∞

Ψ∗A∂Ψ

∂tdx.

By the Schrodinger Equation,

∂Ψ

∂t=

1

i~HΨ,

∂Ψ∗

∂t= − 1

i~

(HΨ

)∗.

Hence,d

dt〈A〉 = − 1

i~

[∫ ∞−∞

(HΨ

)∗AΨdx−

∫ ∞−∞

Ψ∗AHΨdx.

]But, because H is Hermitian, this means

d

dt〈A〉 = − 1

i~

[∫ ∞−∞

Ψ∗HAΨdx−∫ ∞−∞

Ψ∗AHΨdx.

]=

1

i~〈[A, H]〉,

as desired.

This is useful because we can explicitly compute the commutator [A, H]:recall that

H =p2

2m+ V (x).

45


Example 8.2

Let’s compute d〈x〉dt = 1

i~ 〈[x, H]〉. Since xH = p2

2m + V (x), and V (x) com-mutes with x, the commutator we care about is [x, p2]. But:

[x, p2] = xpp− ppx.

But, from last time we know [x, p] = i~, so px = xp− i~. Thus,

[x, p2] = xpp− ppx= xpp− p(xp− i~)

= xpp− pxp+ i~p= (xp− px)p+ i~p= 2i~p.

Therefore,

[x, H] =1

2m[x, p2] =

i~mp,

andd〈x〉dt

=1

i~〈[x, H]〉 =

p

m.

8.3 More on Separable Solutions

When we ended class last time, we talked about finding separable solutions.Recall that the time-dependent Schrodinger equation is

− ~2

2m

∂2Ψ

∂x2+ V (x)Ψ = HΨ = i~

∂Ψ

∂t.

We found the separable solutions of the form Ψ(x, t) = ψ(x)f(t) by pluggingthis in, which, after computation, yields

− ~2

2m

ψ′′(x)

ψ(x)+ V (x) = i~

f(t)

f(t)= E,

for constant E. Solving the differential equation in time yields

f(t) = Ce−iEt/~.

The differential equation in space is called the Time-Independent SchrodingerEquation, and can be written

Hψ = Eψ.

Why do we care about separable solutions? It turns out we can representany non-separable solution by a sum of separable solutions. Thus, we can reducesolving the general Schrodinger Equation to just solving the time-independentform.

8.3.1 Separable Solutions are Stationary States

The separable solutions are the stationary states of the Shrodinger equation, inthe sense that if

Ψ(x, t) = ψ(x)e−iEt/~,

46


then the probability density

p(x, t) = Ψ∗(x, t)Ψ(x, t) = ψ∗(x)ψ(x)

does not depend on t. In fact, we can show that no observable depends on t!This is because if we write the observable as A(x, p), then

〈A(x, p)〉 =

∫ ∞−∞

Ψ∗(x, t)A

(x,

~i

∂d

∂x

)Ψ(x, t)dx.

When we plug in Ψ(x, t) = ψ(x)e−iEt/~, the time-dependent terms cancel.

So: nothing ever happens in a stationary state!

8.3.2 Stationary States have a Definite Total Energy

Recall that

H =p2

2m+ V (x) = − ~2

2m

∂2

∂x2+ V,

and HΨ = EΨ, so

〈H〉 =

∫ ∞−∞

Ψ∗HΨdx = E

∫ ∞−∞

Ψ∗Ψdx = E.

Moreover,

〈H2〉 =

∫ ∞−∞

Ψ∗HHΨdx = E2

∫ ∞−∞

Ψ∗Ψdx = E2.

Thus,

σH =

√〈H2〉 − 〈H〉2 =

√E2 − E2 = 0.

Therefore, the energy is exactly known!

Remark 8.3. Why isn’t this a violation of the Uncertainty Principle ∆t∆E ≥~2 ? The ∆t is infinite, so the ∆E may be zero.

8.3.3 General Solutions are Sums of Separable Solutions

For integer k, let Ψk(x, t) = ψk(x)e−iEkt/~ be the separable solutions of theSchrodinger Equation for a particular V .

It turns out we can write any solution to the Schrodinger equation in theform

Ψ(x, t) =

∞∑n=1

cnψn(x)e−iEnt/~.

We’ll see this in action on the infinite square well.

What’s really happening here is that the space of solutions Ψ forms an infinite-dimensional vector space, and the stationary solutions form an orthonormalbasis of eigenvalues of this space. (Don’t worry if you don’t understand this)

47


8.3.4 Digression: Operators and Eigenvectors

Let’s consider vectors in (x, y, z)-space. Let the operator Rz(θ) rotate a vectoran angle θ about the z-axis. We consider the operator Rz(180).

Note that if ~r points in the z direction, then Rz(180)~r = ~r, so ~r is a vectorof eigenvalue 1.

Similarly, if ~r is in the (x, y)-plane, then Rz(180)~r = −~r, so ~r is a vector ofeigenvalue −1.

In general, for an operator A, if AΨ = aΨ, then Ψ is an eigenvector of A.

8.4 The Infinite Square Well

We’re going to solve Schrodinger’s Equation for the potential function

V (x) =

0 0 < x < L

∞ elsewhere.

Think of this as a particle trapped in a box of width L, with infinitely highwalls. First, we’ll find the stationary states Ψ(x) = ψ(x)f(t).

Outside the range (0, L), ψ must be zero. In the range (0, L), ψ has to satisfythe time-independent Schrodinger Equation

Hψ = Eψ ⇒ − ~2

2mψ′′(x) = Eψ.

Rearranging this gives

ψ′′(x) = −2mE

~2ψ = −k2ψ,

where k =√

2mE~ . This has solutions

ψ(x) = A sin kx+B cos kx.

This, by itself, doesn’t give us enough information. We need to apply someQM rules:

• ψ(x) must be continuous always (because otherwise we get an infinitederivative);

• ψ′(x) must be continuous except where V →∞ (because otherwise we getan infinite second derivative);

•∫∞−∞|ψ(x)|2dx = 1, because normalization.

By continuity, ψ satisfies the boundary conditions ψ(0) = ψ(L) = 0. Sinceψ(0) = 0,

ψ(0) = A sin 0 +B cos 0 = 0⇒ B = 0.

Since ψ(L) = 0,A sin kL = 0⇒ kL = nπ.

If n = 0, then k = 0 and ψ(x) = 0 everywhere, which isn’t interesting (ornormalizable). The solutions corresponding to negative n are just the opposites

48


of the solutions corresponding to positive n, so we can assume n is positive.Therefore, we can set

ψn(x) = An sin(nπLx),

for n ∈ N. We can get the corresponding energies by computing

kn =nπ

L=

√2mEn~

⇒ En =~2π2

2mL2n2.

Our next job is to find An. We do this by normalization:

1 =

∫ ∞−∞|ψ(x)|2dx =

∫ L

0

|A|2 sin2 (kx) dx =L

2|A|2,

because sin2(kx) has average value 12 in (0, L). Therefore

|A|2 =2

L.

By fiat, we can take A to be positive real, so A =√

2L . So, the energy eigenstates

are

ψn(x) =

√2

Lsin(nπxL

).

Next time, we’ll add up different stationary solutions and see what happens!

49


9 March 15, 2017

9.1 Solving the Time-Independent Schrodinger Equation

The main equation that we’ve been solving is

Hψ(x) = Eψ(x).

As we said last time, the solutions ψ to this equation are called stationarystates.

We saw that we can write the expectation value of an observable A, as afunction of time:

〈A(t)〉 =

∫ ∞−∞

Ψ∗(x, t)AΨ(x, t)dx.

In a stationary state, we saw last time that this also has no time-dependence.

We also argued last time that in stationary states, energy is definite.

We say that a state is an eigenfunction of position if its position is definite;similarly a state is an eigenfunction of momentum of its momentum is definite.Solutions to the time-independent Schrodinger equation are eigenfunctions ofenergy.

9.2 Complete Sets of Eigenfunctions

We argued last time that any solution to the time-independent Schrodingerequation is of the form

Ψ(x, t) =∑n

cnψn(x)e−iEnt/~.

This means that the solutions to Ψ(x, t) form an infinite-dimensional vectorspace, and the eigenfunctions of the Hamiltonian operator form a completebasis.

In general: the eigenfunctions of any operator for an observable form acomplete set.

9.3 Eigenfunction Expansion for Square Well

We return to the infinite square well from last lecture:

V (x) =

0 0 < x < L

∞ elsewhere.

Recall that the solutions are of the form

ψn(x) =

√2

Lsin(nπxL

),

with energies

En(x) =π2~2

2mL2n2.

Note that the functions ψn have the following properties:

50


• The ψn alternate between even and odd about x = 12L.

• ψn has n− 1 nodes, points where the wavefunction is zero.

This turns out to always be true – when we count eigenfunctions of boundstates, the nth eigenfunction has n− 1 nodes.

• The ψn are mutually orthogonal, with respect to the dot product (ψ, φ) =∫∞−∞ψ

∗(x)φ(x)dx. That is,∫ ∞−∞

ψ∗m(x)ψn(x)dx = δmn,

the Kronecker delta function

δmn =

1 m = n

0 m 6= n.

• The eigenfunctions form a complete set – any function f(x) defined for0 < x < L can be defined as a linear combination of ψn’s. That is, thereis a sequence c1, c2, . . . such that

f(x) =∑n

cnψn(x).

How do we find the cn’s? By orthonormality we have

(ψm, f(x)) =

(ψm,

∑n

cnψn

)=∑n

cn (ψm, ψn)

=∑n

cnδmn

= cm.

Therefore, we have the relation

cm = (ψm, f) =

∫ ∞−∞

ψ∗m(x)f(x)dx.

Moreover, if we are given the initial state Ψ(x, 0), such that

Ψ(x, 0) =∑n

cnψn(x),

then the time-evolution of the wavefunction is governed by

Ψ(x, t) =∑n

cnψn(x)e−iEnt/~.

9.4 Fundamental Postulates of QM

The postulates are:

51


• The state of a system is completely described by its wavefunction Ψ, whichis both continuous and differentiable (except at infinities)

• For any observable A, there is a Hermitian operator A.

• Measuring with A always yields an eigenvalue of A; if that eigenvalue isan, then the system is left in the corresponding state φn, an eigenfunctionof A. 8

(It’s possible that there are mutliple eigenfunctions with the same eigen-value, which causes things to go wrong. This is called degeneracy; fornow we’ll assume no degeneracy.)

• The probability of measuring a particular eigenvalue an (at any giventime) is:

P (an) = |(φn,Ψ(x, t))|2 =

∣∣∣∣∫ ∞−∞

φ∗n(x)Ψ(x, t)dx

∣∣∣∣2 = |cn|2.

Since these probabilities must add to 1, this means: if we write Ψ(x, t)as an infinite dimensional vector in the orthonormal basis φn, then thisvector has length 1. This can also be seen as a result of the normalizationcondition (Ψ(x, t),Ψ(x, t)) = 1.

• The Schrodinger Equation governs time evolution.

9.5 Example: Infinite Square Well with Starting State

9.5.1 Computing Coefficients

We return to the infinite square well

V (x) =

0 0 < x < L

∞ elsewhere.

This time we’ll start in the non-stationary state

Ψ(x, 0) = A sin3(πxL

).

We want to write this as a sum of stationary states. Recall again that thestationary states are

ψn =

√2

Lsin(nπxL

).

First, we’ll find the normalization constant A. By omitted computation,

1 =

∫ L

0

|A|2 sin6(πxL

)dx =

5L

16|A|2.

Thus, A = 4√5L

. We write

Ψ(x, 0) =∑n

cnψn(x).

8This is the weird part. Basically, if we ask where the particle is, it’s somewhere.

52


We then compute cn with

cn = (ψn,Ψ(x, 0))

=

∫ L

0

ψ∗n(x)Ψ(x, 0)dx

=

∫ L

0

√2

Lsin(nπxL

)· 4√

5Lsin3

(πxL

)=

√32

5

1

L

∫ L

0

sin(nπxL

)sin3

(πxL

).

It turns out (integration by parts omitted) that this integral is only nonzerowhen n = 1 or n = 3. Moreover, (more computations omitted)

c1 =

√32

5

1

L

∫ L

0

sin4(πxL

)dx =

√9

10

and

c3 =

√32

5

1

L

∫ L

0

sin

(3πx

L

)sin3

(πxL

)dx = −

√1

10.

Therefore, when we measure the particle’s energy we get the energies

E1 =π2~2

2mL2E3 =

9π2~2

2mL2

with probabilities

P (E1) = |c1|2 =9

10P (E3) = |c3|2 =

1

10.

9.5.2 Expected Value Computations

We’ll show how to find 〈x〉 at t = 0. We have

〈x〉 =

∫ ∞−∞

Ψ∗(x, 0)xΨ(x, 0)dx

=

∫ L

0

[c∗1ψ∗1(x) + c∗3ψ

∗3(x)]x [c1ψ1(x) + c3ψ3(x)]

= c∗1c1

∫ ∞−∞

L

0

ψ∗1xψ1dx+ c∗1c3

∫ L

0

ψ∗1xψ3dx+ c∗3c1

∫ L

0

ψ1xψ33dx+ c∗3c3

∫ L

0

ψ∗3xψ3dx.

The cross-terms vanish, leaving

〈x〉 = |c1|2∫ ∞−∞

L

0

ψ∗1xψ1dx+ |c3|2∫ ∞−∞

L

0

ψ∗3xψ3dx =(|c1|2 + |c3|2

) L2

=L

2.

At any time t, we can do the same computation, with

Ψ(x, t) = c1ψ1e−iE1t/~ + c3ψ1e

−iE3t/~.

We can also ask: what is 〈E〉 at t = 0? This is

〈E〉 = c∗1c1E1 + c∗3c3E3 =9

10E1 +

1

10(9E1) =

9

5E1.

53


If we do this computation by expanding∫ ∞−∞

Ψ∗(x, 0)HΨ(x, 0)dx,

we’ll get the same thing – by orthonormality all the cross terms cancel, and theterms that remain are precisely c∗1c1E1 + c∗3c3E3.

9.5.3 What Happens After We Measure?

Let’s suppose we measure the energy at time t = 0, and find that we’re in stateE3. Then, we’ve collapsed into the stationary state ψ3, so future measurementswill return that we’re in state E3.

9.5.4 Time Evolution

Suppose we didn’t collapse the wavefunction at t = 0. Then, at time t we have

Ψ(x, t) = c1ψ1e−iE1t/~ + c3ψ1e

−iE3t/~

=

√2

L

[√9

10sin(πxL

)e−iE1t/~ −

√1

10sin

(3πx

L

)e−iE3t/~

].

Important. the phase factors here matter! The only time they don’t matteris when we’re in a stationary state. In non-stationary states, phase factorsyield nontrivial cross terms that affect the evolution of the probability function|Ψ(x, t)|2.

54


10 March 20, 2017

10.1 More on Evolution of Probabilities

Last time we talked about decomposing wavefunctions into energy eigenstates:

Ψ(x, t) =

∞∑n=1

cnψn(x)e−iEnt/~,

where ψn(x) are the stationary states. The expectation of the energy is

〈E〉 =

∫ ∞−∞

Ψ∗HΨdx =

∫ ∞−∞

∞∑j=1

c∗jψ∗j (x)eiEjt/~

H [ ∞∑k=1

ckψk(x)e−iEkt/~

]dx.

But, because Hψk = Ekψk,

H

[ ∞∑k=1

ckψk(x)e−iEkt/~

]=

∞∑k=1

ckEkψk(x)e−iEkt/~.

Therefore, the expression for 〈E〉 collapses into the double sum

〈E〉 =∑j,k

c∗jckEke−i(Ek−Ej)t/~

∫ ∞−∞

ψ∗j (x)ψk(x)dx =∑j,k

c∗jckEke−i(Ek−Ej)t/~δj,k,

where the last equality is from the orthonormality relation of the ψn.

All the terms where j 6= k are zero, so

〈E〉 =

∞∑k=1

c∗kckEk =

∞∑k=1

|ck|2Ek.

So, the expectation of the energy doesn’t change with time! In fact, even thedistribution of energy doesn’t change with time; the probability of measuringenergy En is the probability of measuring the eigenstate ψn, which is

P (En) = |(ψn,Ψ)|2

=

∣∣∣∣∫ ∞−∞

ψ∗nΨdx

∣∣∣∣2=

∣∣∣∣∣∞∑m=1

∫ ∞−∞

ψ∗ncmψme−iEmt/~dx

∣∣∣∣∣2

=

∣∣∣∣∣∞∑m=1

cme−iEmt/~

∫ ∞−∞

ψ∗nψmdx

∣∣∣∣∣2

= |cn|2.

Now, let’s suppose we have another operator A with eigenvalue an – sothere’s an eigenfunction φn such that Aφn = anφn. Then, by the postulates of

55


QM, the probability of measuring this eigenvalue is

P (an) = |(φn,Ψ)|2

=

∣∣∣∣∫ ∞−∞

φ∗nΨdx

∣∣∣∣2=

∣∣∣∣∣∞∑m=1

∫ ∞−∞

φ∗ncnψm(x)e−iEmt/~

∣∣∣∣∣2

.

This probability does have a time dependency – since the φn are not necessarilyorthonormal with the ψn,9 the sum will yield multiple terms, and squaring yieldscross-terms which have time dependency.

10.2 Dirac Delta Function

The Dirac delta function is defined as

δ(x) =

0 x 6= 0

∞ x = 0,

normalized such that∫∞−∞δ(x)dx = 1.

We can think about this as an infinitely high and narrow spike – we startwith the rectangular function

f(x) =

1w −w/2 ≤ x ≤ w/20 elsewhere

and take the limit as w → 0. Alternatively, we can think of it as the Gaussian

p(x) =1√2πσ

e−x2

2σ2 ,

in the limit as σ → 0.

As an example, if we have a point charge at a, we can write the distributionof the charge as e(x) = qδ(x− a).

We can also take integrals of delta functions. For any (ordinary) function f ,we have ∫ ∞

−∞f(x)δ(x− a)dx = f(a).

10.3 Bound States in a General Potential

In much of this course we’ll deal with “trapped” particles – for example, an atomtraps electrons. We’re particularly interested in two cases: the finite potentialwell

V (x) =

0 −x0 ≤ x ≤ x0

V0 elsewhere

(parametrized by positive values V0, x0) and the harmonic oscillator

V (x) ∝ x2.

9In fact, if A doesn’t commute with H, orthonormality doesn’t hold.

56


[diagram. Regions I, II, III]

In classical mechanics, a particle with energy E can only be found in regionswhere V (x) ≤ E – the classically allowed region. But, in QM there’s anonzero probability of finding the particle in classically disallowed regions.

One special case is when we have an energy valley, bounded by a barrier, butwith another valley on the other side. Even if a particle doesn’t have enoughenergy to surmount the barrier, there is some probability it will just appear onthe other side.

[diagram]

Joke. “I stand against this wall for the lifetime of the universe, I might justend up on the other side!” - Ashoori

10.3.1 Geometric Properties of Wavefunctions

Let’s try to solve the time-independent Schrodinger equation

~2

2m

d2ψ

dx2+ V ψ = Eψ.

In regions where E > V , we can write this as

d2ψ

dx2= α2ψ,

where

α =

√2m(V − E)

~2.

This yields solutions of the form

ψ = Ae−αx +Beαx.

In regions where E < V , we get

ψ′′ =2m

~2(V − E)ψ.

This yields solutions of the form A sin(kx) +B cos(kx), where

k =

√2m(E − V )

~2.

When E − V is large, k is large and the wavefunction is more squiggly.

In summary:

• In regions where E > V we ge exponentially damped solutions, with

damping factor α =√

2m(V−E)~2 ;

• In regions where E < V we get sinusoidal solutions with angular frequency

k =√

2m(E−V )~2 .

57


10.3.2 Derivatives of Wavefunctions

V (x) =

VL x < a

VH x > a,

where VL < E < VH . We analyze the behavior of ψ′ on either side of thediscontinuity:

ψ′(a+)− ψ′(a−) = lim∆→0

ψ′(a+ ∆)− ψ′(a−∆)

= lim∆→0

∫ a+∆

a−∆

d

dxψ′(x)dx

= lim∆→0

∫ a+∆

a−∆

ψ′′(x)dx

= lim∆→0

∫ a+∆

a−∆

2m

~2[V (x)− E]ψ(x)dx.

We consider cases here.

Case 1: if V (x) is finite near a, then 2m~2 [V (x)− E]ψ(x) is bounded, so we

must have ψ′(a+) = ψ′(a−).

Case 2a: If V (x) is infinite near a and there is a hard wall at x = a: soVH =∞, then:

ψ′(a+)− ψ′(a−) = lim∆→0

∫ a+∆

a−∆

2m

~2[V (x)− E]ψ(x)dx

= lim∆→0

[∫ a

a−∆

2m

~2[V (x)− E]ψ(x)dx+

∫ a+∆

a

2m

~2[V (x)− E]ψ(x)dx

]

= lim∆→0

2m

~2

[(∞− E)ψ(a+)− (V (a−)− E)ψ(a−)

]∆.

We can work out what this means, but for our purposes it’s enough to just knowthat the hard wall forces ψ(x) = 0 beyond x = a.

Case 2b: If V (x) is infinite near a and there is a δ-function barrier at a, soV (x) = V0δ(x− a): then

ψ′(a+)− ψ′(a−) = lim∆→0

∫ a+∆

a−∆

2m

~2[V0δ(x− a)− E]ψ(x)dx

= lim∆→0

2m

~2

[V0

∫ a+∆

a−∆

δ(x− a)ψ(x)dx−∫ a+∆

a−∆

Eψ(x)dx

]

= lim∆→0

2m

~2

[V0

∫ a+∆

a−∆

δ(x− a)ψ(x)dx−∫ a+∆

a−∆

Eψ(x)dx

]

=2m

~2V0ψ(a)

where the second integral goes to zero.

This tells us the exact amount of discontinuity in the slope.

What we got from this is two rules:

• ψ is continuous;

58


• ψ′ is continuous where the potential is finite; for the specific case of aδ-function barrier, the magnitude of the discontinuity is ψ′(a+)−ψ′(a−) =2m~2 V0ψ(a).

10.3.3 Drawing Wavefunctions

We look at the time-independent Schrodinger equation

ψ′′(x) =2m

~2[V (x)− E]ψ(x).

Suppose we’re in a classically-allowed region – that is, E > V . Then, thecoefficient of ψ(x) above is negative, so ψ is sinusoidal. Geometrically, thecurvature of ψ is always towards the x-axis.

If E < V , then the coefficient of ψ(x) is positive, so ψ is exponentiallydamped. Geometrically, the curvature of ψ is always away from the x-axis.

Recall that with the infinite square well, the nth energy state has n − 1nodes. This turns out to be true in general.

59


11 March 22, 2017

Sorry, some diagrams are missing here.

11.1 Drawing wave functions

Suppose we have a potential well with energies V1, V2 with V1 < V2. Call theseregions 1 and 2. We want to draw the fifth bound energy E5, with E5 > V2:

[diagram]

What do we know about this wavefunction?

• It has 4 notes;

• It wiggles faster in region 1 than in region 2;

• It has a bigger amplitude in region 2 than in region 1.

So, a qualitative sketch is:

[diagram]

Why does it have a bigger amplitude in region 2? In region i (i ∈ 1, 2), let

ψi(x) = Ai sin(kix+ φi).

Then, by taking derivatives we have

1

kiψ′i(x) = Ai cos(kix+ φi).

Squaring and adding yields:

A2i = ψ2

i (x) +

[ψ′i(x)

ki

]2

.

But, at the junction a of the two regions, φ1 and φ2 match in value and firstderivative. So, for i = 1, 2:

ψ2(a) +

[ψ′(a)

ki

]2

= A2i .

Therefore, if the junction is not at a peak (so ψ′(a) 6= 0), then k2 < k1 impliesA2 > A1.

As another example, let’s draw the ninth bound energy state of this potential:

[diagram]

The energy state has these properties:

• There are 8 nodes;

• The potential increases faster to the left, so the decay is faster to the left;

• At the maximum depth, the wavelength and amplitude are shortest;

• ψ,ψ′ are continuous.

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[Lots of diagrams here. Diagrams: n=1,2,3 infinite well, 1,2,3 for finite well,0,1,large for harmonic oscillator. ]

In the infinite square well, the energy states are sine waves that meet theboundaries at 0, at sharp corners.

In the finite square well, the nth energy state is more spread out than thecorresponding state of the infinite well, because it doesn’t have to meet a hardzero. A longer wavelength also means that the wave energies are lower than thecorresponding states of the infinite well.

In the harmonic oscillator, we label the energy states n = 0, 1, 2, . . . insteadof n− 1, 2, 3, . . . because of reasons relating to photonics. It turns out that in aharmonic oscillator, the lowest state is a Gaussian. We’ll prove this later.

11.2 Finite Square Well

Let’s consider a potential well that looks like:

V (x) =

−V0 |x| < a

0 elsewhere.

Let E be an energy satisfying −V0 < E < 0.

Let Region I be where x < −a, Region II be where |x| < a, and Region IIIbe where x > a. We’ll solve the Schrodinger Equation for each of these regions.

11.2.1 Region I

In Region I,

− ~2

2m

d2ψ

dx2= Eψ ⇒ d2ψ

dx2=−2mE

~2ψ.

We define α =√−2mE

~2 ,so

d2ψ

dx2= α2ψ.

This has solution of the form

ψI(x) = Ae−αx +Beαx.

Since ψI(x)→ 0 as x→ −∞, A = 0. Thus,

ψI(x) = Beαx

for x < −a.

11.2.2 Region II

The Schrodinger Equation gives

− ~2

2m

d2ψ

dx2− V0ψ = Eψ ⇒ d2ψ

dx2= −2m(E + V0)

~2ψ.

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We define k =√

2m(E+V0)~2 , so

d2ψ

dx2= −k2ψ,

which has solutionsψII(x) = C sin kx+D cos kx,

for −a < x < a.

11.2.3 Region III

By similar reasoning as Region I, we get solutions of the form

ψIII(x) = Fe−αx +Geαx.

Since ψIII(x)→ 0 as x→ −∞, G = 0. Thus,

ψIII(x) = Fe−αx

for x > a.

11.2.4 Putting Everything Together

Because V (x) = −V (x), we can assume all the solutions have even or odd parity(cf. Pset 4). First, let’s find the even parity solutions.

Since the solutions are even, the Region II solution is just ψII(x) = D cos kx.Moreover, the Region I and III solutions are symmetric, so F = B. Thus:

ψ(x) =

Fe−αx x > a

D cos kx −a < x < a

ψ(−x) x < −a.

By matching the evaluations and first derivatives of ψII and ψIII at a, we get:

Fe−αa = D cos ka,

−αFe−αa = −kD sin ka.

We can divide these to getα = k tan ka.

We can define z = ka, z0 = a~√

2mV0. (Intuition: z0 is what ka would be ifE = 0)

This produces the equation:

αa

z= tan z.

Moreover, by definition

k2 + α2 =2mV0

~2=z2

0

a2⇒ αa =

√z2

0 − z2.

Therefore: √(z0

z

)2

− 1 = tan z.

62


This is precisely the boundary condition! Note that z0 is completely parametrizedby the problem; every solution z to this equation yields a k, which yields a validwavefunction.

We can find the valid z by plotting the functions

f(z) = tan z

g(z) =

√(z0

z

)2

− 1.

[diagram]

On the intervals (0, π2), (π/2, 3π/2), (3π/2, 5π/2), . . . , the function f in-creases. The function g is large near z = 0 and decreases until it reaches 0 atz = z0. Until z = z0, each of the intervals (0, π2), (π/2, 3π/2), (3π/2, 5π/2), . . .has exactly one solution z. As the depth of the well increases, z0 increases andwe get more bound solutions, which makes sense.

Somewhat surprisingly, there’s always at least one bound solution! This isbecause for small z, f → 0 and g →∞, so f and g always intersect at least oncebefore g reaches 0.

Let’se see what happens when z0 is huge. Then, the intersections zn occurat

zn =nπ

2,

for odd n. Then, kn = nπ2a , and

En =~2n2π2

2m(2a)2,

which agrees with the analysis for the infinite square well.

To complete the analysis, we do the same thing for the odd case. Details areleft as an exercise to the reader... on the homework! ;)

11.2.5 Normalization

Finally, we normalize ψ. We want:

1 =

∫ ∞−∞|ψ(x)|2dx = 2

∫ ∞a

F 2e−2αxdx+ 2

∫ a

0

D2 cos2 kxdx.

After some computations (details omitted), we get:

D =1

a+ 1α

, F =eαa cos ka√

a+ 1α

.

11.3 Harmonic Oscillator (Power Series Method)

Some motivation: we care about the harmonic oscillator because any localminimum of a potential function looks, locally, like a harmonic oscillator. Also,electrons going in circles in a magnetic field behave like they’re in a harmonicoscillator.

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Recall the defining equation for a harmonic oscillator:

F = −kx = md2x

dt2⇒ x(t) = A sinωt+B cosωt,

where ω =√

km . The associated potential is

V (x) =1

2kx2 =

1

2mω2x2.

This is the form we’ll use in the Schrodinger equation:

− ~2

2m

d2ψ

dx2+

1

2mω2x2ψ = Eψ.

There are multiple ways of solving this. Today we’ll start the brute forceway of solving this with power series. Next time we’ll see the nicer ladderoperator approach.

As we saw in the finite square well example, defining the problem in termsof dimensionless constants is useful. So, we define:

y =

√mω

~x.

(Intuition:√

~mω is a length scale for the lowest energy solution)

We also define:

E =2E

~ω=

E12~ω

(Intuition: 12~ω is the scale of the smallest allowed energy)

Plugging into the Schrodinger Equation gives (details omitted)

d2ψ

dy2= (y2 − E)ψ.

When y is big, we have the asymptotic behavior

d2ψ

dy2≈ y2ψ.

This isn’t hard to solve. The functions ey2/2 and e−y

2/2 solve this, so

ψ(y) ≈ Ae−y2/2 +Bey

2/2.

By normalization, B = 0. So,

ψ(y) ≈ Ae−y2/2.

This governs behavior at large y, so in sense the Gaussian Ae−y2/2 is an envelope

for our solutions. We write:

ψ(y) = h(y)e−y2/2,

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for some function h. Then, we can compute:

dψ

dy=

(dh

dy− yh

)e−y

2/2

d2ψ

dy2=

[d2h

dy2− 2y

dh

dy+ (y2 − 1)h

]e−y

2/2.

Plugging this all back into the Schrodinger Equation gives us[d2h

dy2− 2y

dh

dy+ (y2 − 1)h

]e−y

2/2 = (y2 − E)he−y2/2 = 0.

After simplifying, we get the much nicer differential equation

d2h

dy2− 2y

dh

dy+ (E − 1)h = 0.

We solve this with the Method of Frobenius – we define h as a power seriesof y. So:

h(y) =

∞∑j=0

ajyj .

Next time, we’ll finish this by plugging this back in. We’ll end up witha recursion rule defining each aj in terms of previous aj ; once we have this,the entire solution will be parametrized by a0, and from this we can extractsolutions.

65

Brice Huang 12 April 3, 2017

12 April 3, 2017

12.1 Administrivia

The second midterm will be next Wednesday, April 12, in Walker during classhours. It covers homeworks 5 and 6, today, and this Wednesday’s lecture.

12.2 Harmonic Oscillator

Today we find the wavefunctions of the harmonic oscillator. Recall that theharmonic oscillator has potential

V (x) =1

2kx2 =

1

2mω2x2.

So, we can write the Schrodinger Equation as

− h2

2m

d2ψ

dx2+

1

2mω2x2ψ = Eψ.

Today, we’ll solve this in two ways – we’ll finish the power series method fromlast time, and then we’ll see a more elegant solution with ladder operators,which are ubiquitously used in QM.

12.3 Harmonic Oscillator – Power Series method

12.3.1 Power Series

Recall from last time, we rewrote the Schrodinger equation in terms of thedimensionless distance and energy units

y =

√mω

~x E =

E12~ω

,

to get the reparametrized equation

d2ψ

dy2= (y2 − E)ψ.

If y2 >> E , then d2ψdy2 ≈ y

2ψ, and from this we get the approximate solution

ψ(y) ≈ Ae−y2/2 +Bey

2/2.

By normalization B = 0, so in fact ψ(y) ≈ Ae−y2/2.

This solves the problem far away, but we want to know the solution close-up.To do this, we write

ψ(y) = h(y)e−y2/2,

for some polynomial h, and solve for h. After some computations (omitted, orsee last lecture’s notes), we get

d2h

dy2− 2y

dh

dy+ (E − 1)h = 0.

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We write h as a power series of y:

h(y) = a0 + a1y + a2y2 + · · · =

∞∑j=0

ajyj .

By taking derivatives we get

h′(y) =

∞∑j=0

jajyj−1 ⇒ 2yh′(y) =

∞∑j=0

2jajyj .

h′′(y) =

∞∑j=0

j(j − 1)ajyj−2 =

∞∑j=0

(j + 1)(j + 2)aj+2yj .

Substituting this all back gives:∞∑j=0

[(j + 1)(j + 2)aj+2 − 2jaj + (E − 1)aj ] yj = 0.

For this to be zero for all y, all the coefficients must be zero! Therefore

(j + 1)(j + 2)aj+2 − 2jaj + (E − 1)aj = 0

for all j, yielding the recurrence relation

aj+2 =2j + 1− E

(j + 1)(j + 2)aj .

So, the entire function h(y) is determined by a0 and a1.10.

12.3.2 Convergence and Truncation

Let’s analyze the convergence properties of this series. Notice that for large j:

aj+2 ≈2

jaj ⇒ aj ≈

C

(j/2)!

for some constant C. Therefore:

h(y) ≈ C∑j

1

(j/2)!yj = C

∑k

1

k!y2k = Cey

2

.

Uh-oh. This means that

ψ(y) ≈ Cey2

· e−y2/2 = Cey

2/2

blows up at large y, and isn’t normalizable.

How do we prevent this? In order for this blowup to not happen, the tailterms of the series have to be zero. In other words, the value of E has to be suchthat

2j + 1− E = 0

for some j. This will force, for this value of j, aj+2 and every subsequent termof the same parity to be zero. This only truncates the terms aj of one paritythough – in order for the terms of the other parity to not go on forever (andmake the solution non-normalizable), they must all be zero.

Let j = n be where the series stops, so E = 2n+ 1. Then, for all j:

aj+2 =(2j + 1)− (2n+ 1)

(j + 1)(j + 2)aj = − 2(n− j)

(j + 1)(j + 2)aj .

10This makes sense, because two parameters describe the solutions of a second-orderdifferential equation.

67


12.3.3 Small Energy States

What do the small energy states look like?

For n = 0, a0 6= 0, but aj = 0 for all larger even j; all the odd aj are zero, so

ψ0(y) = a0e−y2/2.

The ground state is a Gaussian! For n = 1, a1 6= 0 but aj = 0 for all larger oddj; all the even aj are zero, so

ψ1(y) = a1ye−y2/2.

For n = 2, we have a2 = −2a0, aj = 0 for all larger even j, and all the odd ajare zero, so

ψ2(y) = a0(1− 2y2)ey2/2.

The solutions for h are multiples of the Hermite Polynomials

H0(y) = 1

H1(y) = 2y

H2(y) = 4y2 − 2

H3(y) = 8y3 − 12y

H4(y) = 16y4 − 48y2 + 12

... =...

which can be recursively generated.11

Properly normalized, the solutions ψn(x) are (derivation omitted):

ψn(x) =(mω~π

)1/4 1√2nn!

Hn(y)e−y2/2,

where y =√

mω~ x.

Figure 4 shows the wavefunctions ψ0 through ψ3.

12.3.4 Aside: Classical and Quantum Probability Distributions

Classically, the pdf of a particle in a harmonic oscillator is U-shaped in the regionwhere the particle has positive energy; the probability density is higher towardsthe sides, where the particle has smaller velocity. The probability density is zerooutside the classically-allowed region.

For large n, the graph of |ψn|2 has a U-shaped envelope in the classically-allowed region, where the polynomial h(y) is dominant; outside this region

it exponentially tapers to zero because of the e−y2/2 term. For large n, the

squiggles in the graph of |ψn|2 are too fine to resolve, and the exponentialtapering at the sides is very rapid, so the quantum model looks like the classicalmodel. See Figure 5.

11Traditionally, Hn is scaled such that its leading coefficient is 2n, though this doesn’tmatter.

68


Figure 4: Wavefunctions ψ0 through ψ3 for the Harmonic Oscillator. ImageSource: Lecture.

Figure 5: Comparison of Classical and Quantum pdfs of Harmonic Oscillator.Image Source: Lecture.

69


12.4 Harmonic Oscillator – Ladder Operator method

12.4.1 The operators a+ and a−

We go back to the Schrodinger Equation:

− h2

2m

d2ψ

dx2+

1

2mω2x2ψ = Eψ.

Recall that the momentum operator is defined by

p =~i

d

dx.

With this observation, we rewrite the time-independent Schrodinger equation as

1

2m

[(~i

d

dx

)2

+ (mωx)2

]ψ = Eψ.

We have a sum of two squares! This motivates us to write

u2 + v2 = (u− iv)(u+ iv).

...unfortunately this doesn’t work, because in the operator-universe u and vdon’t necessarily commute. Darn.

Oh well. We’ll go ahead with this anyway, and correct for it later!12 Weconsider the operators a+ and a−, defined by

a± =1√2m

(~i

d

dx± imωx

).

We consider the operator a−a+. On a test function f :

(a−a+) f(x) =1

2m

(~i

d

dx− imωx

)(~i

d

dx+ imωx

)f(x)

=1

2m

(~i

d

dx− imωx

)(~i

df

dx+ imωxf

)=

1

2m

[−~2 d

2f

dx2+ ~mω

d

dx(xf)− ~mωx

df

dx+ (mωx)2f

]=

1

2m

[(~i

d

dx

)2

+ (mωx)2 + ~mω

]f(x).

Since f was an arbitrary test function, this means:

a−a+ =1

2m

[(~i

d

dx

)2

+ (mωx)2

]+

1

2~ω.

The first term on the right is just the Hamiltonian! So, in fact we can write theTime-Independent Schrodinger Equation as:(

a−a+ −1

2~ω)ψ = Eψ.

12This is soooo physics in a nutshell. This math major is amused :)

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It turns out, if we do the same computation with a+a−, we get:

a+a− =1

2m

[(~i

d

dx

)2

+ (mωx)2

]− 1

2~ω.

So, the Time-Independent Schrodinger Equation can also be written as(a+a− +

1

2~ω)ψ = Eψ.

And, the commutator [a−, a+] is:

[a−, a+] = ~ω.

12.4.2 Laddering

Now comes the cool part. We claim:

Theorem 12.1

If ψ is an eigenstate of H, then so are a+ψ and a−ψ.

Proof. Let Hψ = Eψ. We compute:

H(a+ψ) =

(a+a− +

1

2~ω)

(a+ψ)

=

(a+a−a+ +

1

2~ωa+

)ψ

= a+

(a−a+ +

1

2~ω)ψ

= a+

[(a−a+ −

1

2~ω)

+ ~ω]ψ

= a+

[H+ ~ω

]ψ

= a+ [E + ~ω]ψ

= [E + ~ω] a+ψ.

Similarly, we can derive:

H(a−ψ) = [E − ~ω] a−ψ.

This is also the reason for the name “ladder operator” – operating by a+ stepsup the energy by ~ω, while operating by a− steps down the energy by ~ω.

12.4.3 Solving the Harmonic Oscillator

We know that we can’t step down the energy forever, since no energy statehas energy lower than the minimum potential. Therefore, a− applied to the

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lowest-energy wavefunction (the “bottom rung” of the ladder) must kill thewavefunction. That is,

a−ψ0 = 0.

This gives us a nice differential equation to solve for ψ0! We have:(~i

d

dx− imωx

)ψ0 = 0,

which yieldsdψ0

dx= −mω

~xψ0 ⇒

dψ0

ψ0= −mω

hxdx.

Integrating yields:

logψ0 = −mω2~

x2 + C ⇒ ψ0(x) = A0e−(mω/2~)x2

for some normalization constant A0. What is the energy of this state? Well: weknow (

a+a− +1

2~ω)ψ0 = E0ψ0,

and a−ψ0 = 0, so E0 = 12~ω. Moreover, by properties of the raising ladder

operator,

ψn(x) = An(a+)ne−(mω/2~)x2

for some normalization constants An, and this state has energy 2n+12 ~ω.

72


13 April 5, 2017

13.1 Delta Function Potential

Recall that the δ-function is defined as

δ(x) =

0 x 6= 0

∞ x = 0,

normalized such that∫∞−∞δ(x)dx = 1. We consider the (infinitely deep, infinitely

narrow) potential wellV (x) = −wδ(x).

Note that δ(x) has units 1length , so w has units energy · length.

Let’s find the bound states of this potential. First, clearly E < 0 for anybound state.

13.1.1 Boundary Conditions

Say Region I is where x < 0, and Region II is where x > 0. We look at theboundary conditions where Region I meets Region II.

By continuity of ψ, ψ(0−) = ψ(0+). What about the first derivative ψ′? Weintegrate the Schrodinger equation

− ~2

2mψ′′(x)− wδ(x)ψ(x) = Eψ

on an interval (−ε, ε) for small ε, to get

− ~2

2m[ψ′(x)]

∣∣∣ε−ε− wψ(0) = E

∫ ε

−εψ(x)dx.

Taking the limit as ε→ 0, the right-hand integral vanishes, so we get:

− ~2

2m

[ψ′(0+)− ψ′(0−)

]− wψ(0) = 0,

which we write as

ψ′(0+)− ψ′(0−) = −2mw

~2ψ(0).

13.1.2 Inside Regions I and II

In Regions I and II, the Schrodinger Equation gives

ψ′′(x) = −2mE

~2ψ = α2ψ,

where α =√− 2mE

~2 . This has general solution

ψ(x) = Aeαx +Be−αx.

In Region I, the second term must be 0 so ψ doesn’t explode as x→ −∞, so

ψI(x) = Aeαx.

Similarly, ψII(x) = Be−αx. By continuity at x = 0, ψI(0) = ψII(0), so A = B.Thus actually

ψII(x) = Ae−αx.

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13.1.3 Putting It Together

We plug our values for ψI(x) and ψII(x) into

ψ′(0+)− ψ′(0−) = −2mw

~2ψ(0)

to get

−αA− αA = −2mw

~2A⇒ α =

mw

~2.

From the relation α =√

2mE~2 we get

E = −mw2

2~2.

Therefore, there is exactly one bound state, given by this E! Finally we can findnormalization constant A:

1 = 2

∫ ∞0

|A|2e−2αxdx =|A|2

α⇒ A =

√α,

where we take A to be positive real by fiat. Therefore, the unique wavefunctionfor this potential is

ψ(x) =

√mw

~e−

mw~2 |x|.

13.2 Creating a Wave Packet from Free-Particle Wave Func-tions

Let’s say we have a free particle, so V = 0 everywhere. Then the Time-Independent Schrodinger Equation is

− ~2

2m

d2ψ

dx2= Eψ → d2ψ

dx2= −k2ψ,

where k =√

2mE~2 .

This is a “free” particle that has uniform probability of being anywhere from−∞ to ∞ – this means its wavefunction isn’t normalizable, which is a bit weird.But it turns out we can still make wave packets from this wave function thatare localizable.

We can write the solutions as

ψ(x) = C sin kx+D cos kx,

and asψ(x) = Aeikx +Be−ikx.

Unlike in the bound case, k now is a continuous variable – there’s no boundaryconditions that force k to be discrete.

There’s some redundancy here – the solutions eikx and e−ikx have the sameenergy. In fact, if we allow k to be negative, we can write the general solutionjust as Aeikx because we can take linear combinations.

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Recall that if Ψ(x, t) is a particle with energy E, then

Ψ(x, t) = Ψ(x, t = 0)e−iωt,

where ω = E~ . Thus we can write:

Ψ(x, t) = Aei(kx−ωt) +Be−i(kx+ωt).

These two terms represent a waveform traveling right and a waveform travelingleft.

As we said before, if we allow k > 0 and k < 0, we can write our generalsolution simply as

Ψk(x, t) = Aei(kx−ωt).

Unfortunately this isn’t normalizable, because∫ ∞−∞|Ψ|2dx = |A|2

∫ ∞−∞

dx

diverges.

13.2.1 Creating a Wave Packet

Let φ(k) be some continuous function of k. We write the wave packet

Ψ(x, t) =

∫ ∞−∞

φ(k)ei(kx−~k22m t)dk,

as an integral (i.e. sum) of many eigenstates ei(kx−~k22m t). Here, ~k2

2m = ω. Then.

Ψ(x, 0) =

∫ ∞−∞

φ(k)eikxdk.

By properties of Fourier transforms, given any desired wave Ψ(x, 0) we caninvent a φ(k) such that the above equation holds.

13.3 Fourier Transforms

13.3.1 Fourier Transform Pairs

By 18.03, pairs of functions ψ(x) and ψ′(k) related by

ψ(k) =1√2π

∫ ∞−∞

ψ(x)e−ikxdx

are also related by

ψ(x) =1√2π

∫ ∞−∞

ψ(k)eikxdk

and are Fourier Transform Pairs. Parseval’s Theorem says that∫ ∞−∞|ψ(x)|2dx = 1⇔

∫ ∞−∞|ψ(k)|2dk = 1.

75


Therefore, if a wavefunction is properly normalized in x-space, its Fouriertransform is properly normalized in k-space. Because we transform back andforth between ψ and ψ, these are really two alternative descriptions of the samefunction.

In fact, each x-space operator has a corresponding k-space operator; forexample:

x-space k-space

x = x x = i ddkp = ~

iddx p = ~k

This means that we can write, for example,

〈x〉 =

∫ ∞−∞

ψ∗(k)

(id

dk

)ψ(k)dk.

Recall the relation

Ψ(x, 0) =

∫ ∞−∞

φ(k)eikxdk

from before. Properties of Fourier transforms tell us that the φ(k) we want isjust φ(k) = 1√

2πψ(k).

13.4 Time Dependencies of Wave Packets

Given Ψ(x, 0), how do we get Ψ(x, t)? By Fourier Transform:

Ψ(x, t) =1√2π

∫ ∞−∞

ψ(k)ei(kx−ωt)dk

where ω = ~k22m . We get ψ(k) by

ψ(k) =1√2π

∫ ∞−∞

Ψ(x, 0)e−ikxdx.

76


Example 13.1

Suppose we have a free particle, with starting wavefunction

Ψ(x, 0) =

A |x| < a

0 |x| > a.

By normalization we can get A = 1√2a

. We compute for ψ(k):

ψ(k) =1√2π

∫ ∞−∞

Ψ(x, 0)e−ikxdx

=1√2π

1√2a

∫ a

−ae−ikxdx

=1

2√πa

[e−ikx

−ik

]∣∣∣a−a

=1

k√πa

[e−ika − eika

−2i

]=

sin ka

k√πa.

[Sidenote: this means that ψ(k) looks like a diffraction pattern! So, whenwe were doing diffraction in 8.03 we were doing Fourier transforms all along.Cool.]

Now we can get the time dependence:

Ψ(x, t) =1√2π

∫ ∞−∞

sin ka

k√πaei(kx−

~k22m t)dk.

We can actually compute this numerically for various t and watch it evolve:https://www.youtube.com/watch?v=xlcTsLPe0mA.

[Sidenote: Sharp edges require large values of k. This is why in the time-evolution, the edges of the box distort first. But after a while, everythingtime-evolves and turns to mush.]

Moreover, note that the central maximum of ψ(k) has width ∆k = 2πa .

k is related to the momentum p by p = ~k, so ∆p = 2π~a . The original

wavefunction has ∆x = 2a, so

∆p∆x =2π~a

2a = 4π~.

This is another statement of the Uncertainty Principle.

13.5 Velocities

The eigenstate corresponding to k is

Ψk(x, t) = Aei(kx−ωt) = Aei(kx−~k22m t).

This has phase velocity

|Vphase =∣∣∣ωk

∣∣∣ =~|k|2m

√E

2m.

77

https://www.youtube.com/watch?v=xlcTsLPe0mA


Figure 6: Envelope of a Waveform. Image Source: Lecture.

Classically, E = 12mv

2 implies v =√

2Em . This isn’t good – somehow the

quantum and classical velocities differ.

It turns out this isn’t a problem, because the plane wave Ψk(x, t) is aninfinitely big, repeating thing whose movement doesn’t really have meaning.Every meaningful wavefunction is contained in an envelope (See Figure 6).The envelope’s velocity, called the group velocity Vg, doesn’t necessarily equalthe phase velocity of the squiggles it contains.

The group velocity satisfies (proof omitted in lecture)

Vg =dω

dk=hk

m=

√2E

m,

which agrees with the classical result.

13.6 Back to the Infinite Square Well

Recall from before that the infinite square well

V (x) =

0 0 < x < a

∞ elsewhere

has normalized wavefunctions

ψn(x) =

√2

asin(nπx

a

),

with associated energies

En =n2π2~2

2ma2.

Properties of Fourier series tell us that for any ψ(x), we can invent a sequenceof coefficients cn such that

ψ(x) =

∞∑n=1

cnψn(x).

78


The coefficients cn are given by

cn = (ψn,Ψ(x, 0)) =

∫ ∞−∞

ψ∗n(x)Ψ(x, 0)dx.

And the time-evolution is governed by

Ψ(x, t)

∞∑n=1

cnψn(x)e−iEnt/~.

Next time: we’ll see how for a particle defined by Ψ(x, 0) = 1√2

[ψ1(x) + ψ2(x)],

with time-evolution

Ψ(x, t) =1√2

[ψ1(x)e−iE1t/~ + ψ2(x)e−iE2t/~

],

the particle will rattle back and forth in the well.

79


14 April 10, 2017

14.1 Infinite Square Well with Time Evolution

Last time we left off with an infinite square well

V (x) =

0 0 < x < a

∞ elsewhere

with wavefunctions and energy levels

ψn(x) =

√2

asin(nπx

a

), En =

n2π2~2

2ma2.

We consider the wavefunction Ψ consisting of an equal superposition of ψ1 andψ2:

Ψ(x, t) =1√2

[ψ1(x)e−iω1t + ψ2(x)e−iω2t

],

where ωi = Ei~ . Expanding, we get

Ψ(x, t) =1√2

[√2

asin(πxa

)e−iω1t +

√2

asin

(2πx

a

)e−iω2t

].

Moreover, note that ω2 = E2

~ = 4E1

~ = 4ω1. Therefore, the wavefunctionoscillates with angular frequency ω, with period treturn satisfying

ω1treturn =E1

~t = 2π ⇒ treturn =

h

E1.

What does thie time-evolution of this system look like? At t = 0, ψ1 and ψ2

interfere constructively on the left and destructively on the right, yielding thewavefunction and probability density:

[diagram]

At t = 12 treturn = h

2E1, ψ1 and ψ2 interfere destructively on the left and

constructively on the right, yielding the wavefunction and probability density:

[diagram]

Therefore, the particle, in some sense, sloshes back and forth over time.

Actually, the sloshing is faster than the above calculation indicates, becausewe don’t care about an overall phase factor in Ψ(x, t). The probability densityof the particle is:

p(x, t) =

∣∣∣∣∣ 1√2

[√2

asin(πxa

)ei(ω2−ω1)t +

√2

asin

(2πx

a

)]e−iω2t

∣∣∣∣∣2

.

The e−iω2t vanishes when we take the absolute value, so the probability densityis perodic with period Tp satisfying

i(ω2 − ω1)Tp = 2π ⇒ Tp =h

3E1.

So, even though the wavefunction’s value oscillates with period hE1

, the proba-bility density oscillates three times as fast!

80


14.2 Dimensional Analysis

14.2.1 Harmonic Oscillator

We will use dimensional analysis to get an order-of-magnitude estimate for theharmonic oscillator’s ground energy. Energy has units:

E = [M ][L][T ]−2.

13 The harmonic oscillator is parametrized by m,ω, ~, with units:

m = [M ]

ω = [T ]−1

~ = [M ][L]2[T ]−1.

SettingE = mαωβ~γ

yields the system:

α+ γ = 1

2γ = 2

β + γ = 2.

Solving (computation omitted) gives (α, β, γ) = (0, 1, 1). So, an order-of-magnitude estimate for energy is

E = ~ω.

This is the same order of magnitude as the actual answer, E = 12~ω.

14.2.2 Infinite Square Well

We do the same thing for the infinite square well. This time our parameters arem, a, ~, with units

m = [M ]

a = [L]

~ = [M ][L]2[T ]−1.

We setE = mαaβ~γ

to get the system

α+ γ = 1

β + 2γ = 2

−γ = −2.

Solving gives (α, β, γ) = (−1,−2, 2), which yields the estimate

E =~2

ma2.

This is the same order of magnitude as the actual answer, E = π2~2

2ma2 .

13To be as general as possible, we use generic units [M ], [L], [T ], as opposed to, say, g, cm, s.

81


14.3 Parity

A wavefunction ψ is even if ψ(−x) = ψ(x), and odd if ψ(−x) = −ψ(x). Forexample, the infinite square well

V (x) =

0 −a2 < x < a

2

∞ elsewhere,

the wavefunctions ψn(x) = cos(nπa x)

(n odd) are the even eigenstates, and the

wavefunctions ψn(x) = sin(nπa x)

(n even) are odd eigenstates.

In general, an even wavefunction Ψ(x) is a sum of only even eigenstates, andan odd wavefunction is a sum of only odd eigenstates.

14.3.1 Parity Operator

The parity operator P is:Pψ(x) = ψ(−x).

Note that for all even ψeven, Pψeven(x) = ψeven(x), while for all odd ψodd,Pψodd(x) = −ψodd(x).

So, even functions are eigenstates of P with eigenvalues 1, while odd functionsare eigenstates of P with eigenvalues −1.

Are there any other eigenstates? Suppose for a function ψ, Pψ(x) = λψ(x)for some λ. Then,

P 2ψ(x) = P λψ(x) = λ2ψ(x).

But, P 2ψ(x) = Pψ(−x) = ψ(x), so λ2 = 1. Thus no other eigenstates arepossible.

In fact, every wavefunction ψ can be written as a sum of even and oddwavefunctions:

ψ(x) =1

2[ψ(x) + ψ(−x)] +

1

2[ψ(x)− ψ(−x)] .

14.3.2 P and H

Suppose V (x) is symmetric, i.e. V (x) = V (−x). Then, the Schrodinger equationis unchanged when we substitute −x for x. So, for symmetric potentials, Pcommutes with the Hamiltonian, and [P , H] = 0.

Moreover, this means that for any eigenstate ψ of H,

H(Pψ)

= P(Hψ)

= E(Pψ).

This says that for symmetric potentials, any eigenfunction of H is also aneigenfunction of P – that is, all eigenfunctions of H are even or odd.

14.3.3 Hyperbolic Functions

The hyperbolic functions are defined by

sinhx =ex − e−x

2coshx =

ex + e−x

2tanhx =

sinhx

coshx.

Note that sinh, tanh are odd, while cosh is even. We will solve this to solve theδ-function infinite square well.

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14.4 Infinite Square Well with δ-Function

We consider the potential

V (x) =

η2 δ(x) |x| < a

∞ |x| > a.

Here, η is a unit of energy× length. For |x| < a, the Schrodinger Equation says

− ~2

2m

d2

dx2ψ(x) +

η

2δ(x)ψ(x) = Eψ(x).

Like before, we make our variables dimensionless by setting u = xa . Recall that

δ(au) = 1aδ(u). So, the Schrodinger Equation becomes

− 1

a2

~2

2m

d2

du2ψ(u) +

η

2

1

aδ(u)ψ(u) = Eψ(u)

⇒ d2

du2ψ +

mηa

~2δ(u)ψ(u) =

2ma2E

~2ψ(u).

We make the substitutions

w =mηa

~2E =

2ma2E

~2,

where w and E are parameters describing the strength of the δ-function and theenergy, to get

− d2

du2ψ + wδ(u)ψ = Eψ.

What are the boundary conditions here? Since potential is infinite at theouter boundary, ψ(1) = ψ(−1) = 0. By continuity of ψ, ψ(0+) = ψ(0−).

We find the behavior of ψ′ near 0 by integrating the above equation on aninterval (−ε, ε): ∫ ε

−ε[ψ′′ + wδ(u)ψ]du =

∫ ε

−εEψdu.

Taking the limit as ε→ 0, we get

−ψ′(0+) + ψ′(0−) + wψ(0) = 0⇒ ψ′(0+) = ψ′(0−) + wψ(0).

14.4.1 E > 0

We first examine E > 0. Since the potential is symmetric, all the eigenfunctionsare even or odd.

For odd solutions ψ(0) = 0, so in fact ψ′(0+) = ψ′(0−). Then ψ′ is actuallycontinuous, so we just have ψ(u) = A sin ku, where k2 = E = n2π2.

For even solutions, on the intervals (−1, 0) and (0, 1), ψ is sinusoidal, so wesay ψ(u) = A sin(ku+ b). This has to be zero at u = ±1, so b = −k. Since thesolution is even,

ψ(u) =

A sin(ku− k) u > 0

A sin(ku+ k) u < 0.

Plugging this into ψ′(0+) = ψ′(0−) + wψ(0) gives us

kA cos(−k) = −kA cos(−k) + wA sin(−k)⇒ tan k =−2k

w.

This is a transcendental equation that we can solve graphically, like for the finitesquare well.

83


14.4.2 E < 0

For this to have a solution at all, we must have w < 0.

Recall that with a δ-function potential, there’s exactly one bound state, andit is even. When we place the δ-function in a well, the wavefunction ψ must haveadditional curvature, so that it meets the boundaries of the well at ψ(±a) = 0.

It turns out that we’ll only get bound solutions when the well is sufficientlydeep. We’ll discuss the condition needed for a bound state to exist.

We set α2 = |E|, so (many details omitted, but similar to previous section)

ψ(u) =

A sinh(αu− α) u > 0

A sinh(−αu− α) u < 0.

Plugging this into ψ′(0+) = ψ′(0−)+wψ(0) gives tanhα = −2αw . For this to have

a nonzero solution α, the graphs of tanhα and −2w α must have an intersection

other than α = 0. This happens iff 0 < −2w < 1.

Thus the condition for a bound state to exist is w < −2.

84


15 April 19, 2017

15.1 Exam Statistics

µ = 70.78, σ = 20.98. Exams will be given back during recitation tomorrow.

15.2 When Waves Meet Barriers

A plane wave is a solution to the Schrodinger equation, but it isn’t normalizable.We can get normalizable solutions to the Schrodinger Equation by summingplane waves into wave packets.

Today we want to understand: if a wave packet Ψ(x, t) runs into a barrier

V (x) =

0 |x| > a

V0 |x| < a,

part of the wave packet gets transmitted, and part gets reflected. What governshow much is transmitted, and how much reflected?

15.3 Step-Function Barrier

Wave packets contain a lot of frequencies, which makes analyzing hard. To makethings easier, we just send in a plane wave with constant frequency. We considerthe step function potential

V (x) =

0 x < 0

V0 x > 0.

We send in the plane wave A0eik1x and energy E = ~2k2

2m > V0. Classical physicspredicts that the entire wave will pass through.

The Schrodinger Equation, on Regions I (x < 0) and II (x > 0), is:

− ~2

2m

d2ψ

dx2= Eψ, x < 0,

− ~2

2m

d2ψ

dx2+ V0 = Eψ, x > 0.

On Region I, the solutions are

ψI(x) = A0eik1x +Ae−ik1x,

where k1 =√

2mE~ . These terms are, respectively, the (right-traveling) wave we

sent in and the (left-traveling) reflection. On Region II, the solution is

ψII(x) = Beik2x,

where k2 =

√2m(E−V0)

~ . We don’t include the e−ik2x term because the wavetransmitted into Region II must be going to the right.

Now, we solve the boundary condition:

A0 +A = B

ik1A0 − ik1A = ik2B.

85


Solving this (computation omitted) yields B = 2k1k1+k2

A0, A = k1−k2k1+k2

A0.

We can define the transmission and reflection coefficients:

R =

∣∣∣∣ AA0

∣∣∣∣2 =(k1 − k2)2

(k1 + k2)2T = 1−R =

4k1k2

(k1 + k2)2=k2

k1

∣∣∣∣ BA0

∣∣∣∣2In macroscopic conditions, we have E >> V0, so k1 ≈ k2 and R ≈ 0, as

expected.

Classically we expect T =∣∣∣ BA0

∣∣∣2. To see why the quantum model predicts

an additional k2k1

, we first need the notion of probability current.

15.4 Probability Current

15.4.1 Probability Current: Intuition

In an earlier homework problem, we defined a probability current density.The values of T and R should equal:

T =probability current of transmission

probability current of original

R =probability current of reflection

probability current of original.

For a stream of particles, the current is proportional to particle density×particle velocity. Thus, we expect the probability current to be proportional toprobability density× probability wave velocity.

A plane wave with angular frequency k has velocity

v =~km.

∣∣∣ BA0

∣∣∣2 is the ratio of probability densities, and~k2m

~k1m

is the ratio of velocities. This

is the reason for the k2k1

factor in T .

15.4.2 Computing Probability Current

We can make the handwavy reasoning in the previous section rigorous byexplicitly computing the probability current. Recall that the probability offinding a particle in an interval [x1, x2] is

P (x1, x2) =

∫ x2

x1

Ψ∗Ψdx.

In a homework problem, we showed that

∂P (x1, x2)

∂t= J(x1, t)− J(x2, t),

where J is the probability current, defined by

J =i~2m

(Ψ∂Ψ∗

∂x−Ψ∗

∂Ψ

∂x

).

86


For a plane wave, Ψ(x, t) = Aei(kx−ωt). Performing the computation, we get

J(x) =i~2m

(Ψ∂Ψ∗

∂x−Ψ∗

∂Ψ

∂x

)=

i~2m

(−ikAA∗ − ikA∗A)

=~km|A|2.

Thus,

T =~k2m |B|

2

~k1m |A0|2

=k2

k1

∣∣∣∣ BA0

∣∣∣∣2 .15.5 Step-Function Barrier II

Suppose now that the energy of the plane wave is less than the potential, so~2k2

2m < V0. Then, we get the solutions

ψI(x) = A0eikx +Ae−ikx

ψII(x) = Be−αx,

where α =

√2m(V0−E)

~ . We don’t include the eαx term because it will explodeas x→∞.

Solving the boundary condition

A0 +A = B

ik(A0 −A) = −αB

gives (computation omitted) B = 2ikik−αA0, A = ik+α

ik−αA0.

What are the corresponding T and R? Well:

R =|A|2

|A0|2=k2 + α2

k2 + α2

A20

A20

= 1

and so T = 0.

This is strange. We know that there exists probability on the other side,so why is T = 0? This is because the probability density is positive, but theprobability velocity is zero – the wave past the barrier doesn’t travel. Thus theprobability current through the barrier is zero.14

Computationally, we can verify that J(x) = 0 for the wavefunction ψ(x) =βe−αx.

15.6 Tunneling

Suppose we have a potential

V (x) =

0 x < 0

V0 0 < x < L

0 x > L

,

14This is analogous to evanescent waves in optics.

87


and send in a plane wave from the left with energy E < V0, it’ll exponentiallydecay through the barrier but emerge, with small amplitude, on the other side.This is called tunneling, a purely quantum phenomenon. Let’s compute thesolutions of this system.

For the same values of k, α as in the previous Section, our solutions onRegions I (x < 0), II (0 < x < L), and III (x > L) are:

ψI(x) = A0eikx +Ae−ikx

ψII(x) = Be−αx + Ceαx

ψIII(x) = Deikx.

Our boundary conditions are:

A0 +A = B + C

ik(A0 −A) = −αB + αC

Be−αL + CeαL = DeikL

−αBe−αL + αCeαL = ikDeikL.

We can solve this to get (computation omitted):

D

A0=

4ikαe−ikL

e−αL(α+ ik)2 − eαL(α− ik)2.

The transmission coefficient is:

T =

∣∣∣∣ DA0

∣∣∣∣2 =4E(V0 − E)

4E(V0 − E) + V 20 sinh2(αL))

.

We can sanity-check this by analyzing what happens when L is big. Then thebarrier looks like a step function, so we expect zero transmission. Indeed, forαL >> 1, the value of sinh2(αL) is very large, so T → 0.

15.6.1 Aside: Tunnel Ionization

Ordinarily, an electron gets ionized when it absorbs a photon, which lifts the

electron out of the well V (x) = − ze2

r2 .

But, if the EM radiation passing through the atom has big enough electricfield, the electric field will actually tilt the potential by providing an additionallinear potential V (x) = Ex.

[diagram]

This is called a tunnel ionization. It’s possible, for example, for 1.5 eVphotons to tunnel-ionize hydrogen, which requires 13.6 eV to ionize normally.

88


16 April 24, 2017

16.1 Grade Cutoffs from Last Exam

Roughly:

• A: 80-100

• B: 65-80

• C: <65

16.2 WKB Approximation

Named after Wentzel, Kramers, Brillouin.

Joke. “In France, it’s called BWK because Brillouin was French; In Dutch,KWB because Kramers was Dutch; In Britain, JWKB because Jeffreys, a Britonalso contributed.” -Ashoori

Recall that for a free particle, the eigenstates are given by the sinusoids

ψ(x) = Aeikx,

where k =

√2m(E−V )

~ . The WKB approximation concerns potentials that varygradually enough that eigenstates are, locally, approximately sine waves.

Similarly, in classically forbidden regions the eigenstates of a free particleare given by the exponentials

ψ(x) = Ae±αx,

where α =

√2m(V−E)

~ . For gradually-varying potentials, the WKB approxi-mation also approximates eigenstates in classically-forbidden regions by theseexponentials.

16.3 WKB in Classicaly-Allowed Regions

For now, suppose we are in a region where E > V (x). The Schrodinger Equationgives

− ~2

2m

d2ψ

dx2+ V (x)ψ = Eψ.

We rewrite this asd2ψ

dx2= −p

2(x)

~2ψ,

where we definep(x) =

√2m[E − V (x)].

15

15Intuition: p looks like a momentum. If V is constant p is a constant momentum.

89


Remark 16.1. Warning: the WKB approximation doesn’t work near turningpoints, where the classically-allowed and classically-forbidden regions meet. Thisis because near turning points, λ ∝ 2π

k → ∞. Because of this, the WKBapproximation works best in wells with infinite walls, where we don’t have todeal with turning points.

To do the approximation, we write:

ψ(x) = A(x)eiφ(x),

where we assume A and φ are both real. Note that if V is constant, we can takeφ(x) = kx. Taking derivatives, we get (computation omitted)

d2ψ

dx2=[A′′ + 2iA′φ′ + iAφ′′ − (φ′)2A

]eiφ.

Plugging this into d2ψdx2 = −p

2(x)~2 ψ yields

A′′ + 2iA′φ′ + iAφ′′ − (φ′)2A = −p2

~2A.

Since we demand that A, φ both be real, we can separate the real and imaginaryparts to get two equations:

A′′ − (φ′)2A = −p2

~2A ⇒ A′′

A= (φ′)2 − p2

~2

2A′φ′ +Aφ′′ = 0 ⇒ d

dx(A2φ′) = 0.

The second equation implies that A2ψ′ is constant. Set A2ψ′ = C2, so

A(x) =C√φ′(x)

.

Now, we will use approximation to solve the first equation. We assume that∣∣∣∣A′′A∣∣∣∣ << (φ′)2,

∣∣∣∣A′′A∣∣∣∣ << p2

~2.

Then, locally we have (φ′)2 ≈ p2

~2 , so φ′ ≈ ± p~ . Hence, we have

φ(x) ≈ ±1

~

∫p(x)dx.

This implies

A(x) ≈ C√φ′(x)

= ±C

√~

p(x)

and

ψ(x) ≈ C√p(x)

e±i~∫p(x)dx,

where we absorb a√~ into the C. More generally, taking both the positive and

negative solutions,

ψ(x) ≈ 1√p(x)

(C+e

i~∫p(x)dx + C−e

− i~∫p(x)dx

).

90


16.4 An Example: Gradually Varying Infinite Well

We consider an infinite well potential V (x), where V (x) varies gradually for0 < x < a and V (x) =∞ outside this interval.

From the discussion above, we have

ψ(x) ≈ 1√p(x)

[C+eiφ(x) + C−e

−iφ(x)],

where φ(x) = 1~∫ x

0p(x′)dx′. Writing the exponentials trigonometrically yields

ψ(x) ≈ 1√p(x)

[c1 sin[φ(x)] + c2 cos[φ(x)]].

Our boundary conditions are ψ(0) = ψ(a) = 0.

Since φ(0) = 0, ψ(0) = 0 gives c2 = 0.

ψ(a) = 0 givesc1√p(a)

sin[φ(a)] = 0,

so φ(a) = nπ. Therefore, ∫ a

0

p(x′)dx′ = nπ~.

Because p(x) =√

2m(E − V ), this equation yields the allowed energies E.

Finally, note that if V = 0, the equation defining the allowed energiesbecomes ∫ a

0

√2mEdx =

√2mEa = nπ~⇒ E =

n2π2~2

2ma2,

in agreement with our analysis of the finite square well.

16.5 WKB in Classically-Forbidden Regions

Suppose E < V . Then, p(x) =√

2m(E − V ) is imaginary. The result weeventually get is that

ψ(x) ≈ C√|p(x)|

e±1~∫|p(x′)|dx′

.

Remark 16.2. The earlier derivation doesn’t work, because the separationof real and imaginary parts fails when p is nonreal. There are alternativederivations. We’re just taking this result for granted.

16.6 An Example: Gradually Varying Tunneling Barrier

.

We consider a potential V (x), where V (x) > 0 varies gradually for 0 < x < a,and V (x) = 0 outside this interval. Suppose we send in a plane wave ψ(x) =Aeikx from the left. We’ll get out a reflected wave Be−ikx on the left, and atransmitted wave Feikx on the right. Thus, we have

ψ(x) = Aeikx +Be−ikx

91


Figure 7: Tunneling Across a Barrier. Image source: Lecture.

for x < 0, andψ(x) = Feikx

for x > a. The transmission coefficient T is T = |F |2|A|2 . By the discussion above,

on the interval 0 < x < a, we have

ψ(x) ≈ 1√|p(x)|

[Ce

1~∫ x0|p(x′)|dx′

+De−1~∫ x0|p(x′)|dx′

].

We make another approximation: we assume the barrier is very high or verywide, which forces C = 0.

Then,|F ||A|≈ e−γ ,

where γ = 1~∫ x

0|p(x′)|dx′, and the transmission coefficient is

T =|F |2

|A|2= e−2γ .

16.7 Tunneling Across a Barrier

We consider a particle trapped in the region (0, a) by a barrier in the potentialshown in Figure 7.

The particle has an energy E, relative to the potential on the right side.Suppose the particle rattles around the well with velocity v. Then, bounces thewell once in time ∆t = 2a

v , and thus bangs against the barrier with the attemptfrequency 1

∆t = v2a .

Each attempt, it escapes with probability e−2γ , so the probability of escapingper unit time is v

2ae−2γ . This gives it a lifetime of 2a

v e2γ .

16.7.1 Application: Nuclear α-Decay

We consider an α-particle trapped in a nucleus. The nuclear potential is shownin Figure 8.

92


Figure 8: Nuclear α-decay. Image source: Lecture.

Let the nucleus have size r1. Let the particle have energy E (horizontal line),such that the nuclear potential equals E again at distance r2.

The Coulomb potential of the α-particle is

V (r) =(2e)(ze)

r=

2ze2

r.

Since V (r2) = E, we can solve to get r2 = 2ze2

E . Then, γ, the parameter thatdetermines tunneling probability, is given by

γ =1

~

∫ r2

r1

√2m(V (r)− E)dr

=1

~

∫ r2

r1

√2m

(2ze2

r− E

)dr

=1

~

∫ r2

r1

√2mE

(r2

r− 1)dr

=2mE

~

∫ r2

r1

√r2

r− 1dr

=

√2mE

~

[cos−1

(√r1

r2

)−√r1(r2 − r1)

].

Moreover, because r1 << r2,√r1

r2≈ sin

√r1

r2= cos

(π

2−√r1

r2

).

Hence,

cos−1

(√r1

r2

)≈ π

2−√r1

r2,

and

γ ≈√

2mE

~

[r2

(π

2−√r1

r2

)−√r1r2 − r2

1

]≈√

2mE

~

[π2r2 − 2

√r1r2

]= K1

z√E−K2

√zr1,

for constants

K1 =πe2√

2m

~K2 =

4√me2

~.

93


Why is this all interesting? The parameter γ governs the tunneling probability.When E is larger, γ is smaller, and the tunneling probability is larger. Theattempt frequency is v

2r1, and the probability of escape per unit time is v

2r1e−2γ ,

so the lifetime16 is 2r1v e

2γ .

This formula really works! For various isotopes of Uranium, we can plotlog10(half life) against energy1/2 of emitted α-aparticles.

Over about 20 orders of magnitude of half-life, the plot we get is linear, withslope −K1z, exactly as predicted!

16Not quite the half-life, but within a small constant factor

94


17 April 26, 2017

Today we’ll talk about 3-dimensional QM. We’ll see that often, we’ll be able toreduce to 1-dimensional QM by decomposing into x, y, z components.

17.1 Extending QM to 3 Dimensions

We’ll generalize 1-dimensional QM by extending operators in the natural way:

px = −i~ ∂

∂x, py = −i~ ∂

∂y, py = −i~ ∂

∂z.

Since these operators are orthogonal, they commute, and (e.g.) [px, py] = 0.

The Hamiltonian generalizes to:

H =px

2

2m+py

2

2m+py

2

2m+ V (x, y, z, t),

and the (time-dependent) Schrodinger Equation becomes:

− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)Ψ + V (x, y, z, t)Ψ = i~

∂Ψ

∂t.

The operator ∇2 = ∂2

∂x2 + ∂2

∂y2 + ∂2

∂z2 is called the Laplacian, and measures thecurvature of a function. Thus, like in the 1-dimensional case, the more squigglya wavefunction is, the bigger its kinetic energy.

In this notation, we can write the 3-dimensional Schrodinger Equation as:

−~2

2m∇2Ψ + V (~r, t)Ψ = i~

∂Ψ

∂t.

Note that |Ψ(x, y, z, t)|2 is now a volume probability density, so

|Ψ(x, y, z, t)|2dxdydz

is the probability if finding a particle in the region [x, x + dx] × [y, y + dy] ×[z, z + dz].

As before, if potential is unchanging in time, i.e. V (~r, t) = V (~r), we (by thesame argument as for the 1-D case) get a complete set of stationary states

Ψn(~r, t) = ψn(~r)e−iEnt/~,

and a general wavefunction is given by

Ψ(~r, t) =∑n

cnψne−iEnt/~.

Since the particle has to be somewhere, normalization is given by

1 =

∫ ∞−∞

∫ ∞−∞

∫ ∞−∞|Ψ(x, y, z, t)|2dxdydz =

∫d3r|Ψ(~r, t)|2.

95


17.2 Example: Particle in a 3-D Box

Consider a particle trapped in a box [0, a], [0, b], [0, c]. This box has potential:

V (x, y, z) =

0 0 < x < a, 0 < y < b, 0 < z < c

∞ elsewhere.

Since ψ(x, y, z) = 0 outside the box, we get the boundary condition thatψ(x, y, z) = 0 when x ∈ 0, a, y ∈ 0, b, or z ∈ 0, c.

We want to separate the x, y, z-dimensions and deal with each separately.The key observation is:

Fact 17.1. If we can write V (x, y, z) = Vx(x) + Vy(y) + Vz(z), then we can getseparable solutions ψ(x, y, z) = ψx(x)ψy(y)ψz(z).

Thus we write ψ(x, y, z) = ψx(x)ψy(y)ψz(z) and plug into the SchrodingerEquation:

− ~2

2m

(d2ψxdx2

ψyψz + ψxd2ψydy2

ψz + ψxψyd2ψzdz2

)= Eψxψyψz,

which is equivalent to

− ~2

2m

(1

ψx

d2ψxdx2

+1

ψy

d2ψydy2

+1

ψz

d2ψzdz2

)= E.

But, the functions − ~2

2m1ψx

d2ψxdx2 , − ~2

2m1ψy

d2ψydy2 , − ~2

2m1ψz

d2ψzdz2 , are, respectively,

functions of just x, y, and z, so if their sum is constant, they are all constant!Thus, we can write:

− ~2

2m

1

ψx

d2ψxdx2

= Ex, − ~2

2m

1

ψy

d2ψydy2

= Ey, − ~2

2m

1

ψz

d2ψzdz2

= Ez,

where Ex + Ey + Ez = E.

In the x-direction, we write

ψx(x) = Ax sin kxx+Bx cos kxx,

where kx =√

2mEx~ . Applying the boundary condition ψx(0) = ψx(a) = 0 gives

(details omitted; this is exactly the same as before)

ψx = Ax sin kxx,

where kx = nxπa and Ex =

n2xπ

2~2

2ma2 . The computations in the y, z-directions areidentical.

Putting everything together:

ψnx,ny,nz (x, y, z) = A sin(nxπx

a

)sin(nyπy

b

)sin(nzπz

c

),

where we index the eigenstates by (nx, ny, nz). By normalization:

1 = |A|2∫ a

0

sin2(nxπx

a

)dx

∫ b

0

sin2(nyπy

b

)dy

∫ c

0

sin2(ncπz

c

)dz

= |A|2(

1

2a

)(1

2b

)(1

2c

),

96


so A =√

8abc , and

Enx,ny,nz =π2~2

2m

(n2x

a2+n2y

b2+n2z

c2

).

17.2.1 Special Case: The Cubical Box

Suppose that a = b = c. Then,

Enx,ny,nz =π2~2

2ma2(n2x + n2

y + n2z).

For example, E1,1,1 = 3π2~2

2ma2 .

We also have E2,1,1 = E1,2,1 = E1,2,2 – this means that we have somedegeneracy in our space of solutions, in the sense that any linear combination ofψ2,1,1, ψ1,2,1, ψ1,1,2 is also an eigenfunction, with energy E2,1,1 = E1,2,1 = E1,2,2.This mucks up our notion of completeness of our eigenfunction system.

17.3 Spherical Coordinates

17.3.1 Schrodinger Equation in Spherical Coordinates

We consider spherical coordinates, where each point is described by a triple(r, θ, φ). We’d like the write the Schrodinger Equation in spherical coordinates.

What is the Laplacian, in spherical coordinates? Recall that for a functionf , the Laplacian equals

∇2f = ~∇ · ~∇f,

the divergence of the gradient of f .

First, note that

(~∇ψ)r =∂ψ

∂r.

So far, so good. We want to take the divergence of this.

Recall that for a vector-valued function ~F , divergence is defined by

Div~F =

∫small box

~F d~a

Volume.

The numerator is the flux coming out of the box.

Suppose ~F is spherically symmetric. Then, we can take the box to a shell ofradius r and width dr. If Φ(r) is the flux coming out of the sphere of radius r,then Φ(r + dr)− Φ(r) is the net flux coming out of this shell. So:

Div~F =Φ(r + dr)− Φ(r)

4πr2dr

=∂∂r (4πr2Fr)dr

4πr2dr

=1

r2

∂

∂r(r2Fr)

=∂Frdr

+2

rFr.

97


With Fr = ∂ψ∂r , this yields

∇2ψ =∂2ψ

dr2+

2

r

∂ψ

∂r

=1

r2

∂

∂r

(r2 ∂ψ

∂r

).

If ψ isn’t spherically-symmetric, we have (derivation omitted) the more generalresult:

∇2ψ =1

r2

∂

∂r

(r2 ∂ψ

∂r

)+

1

r2 sin θ

∂

∂θ

(sin θ

∂ψ

∂θ

)+

1

r2 sin2 θ

∂2ψ

∂φ2.

With this, the Schrodinger Equation is

− ~2

2m∇2ψ + V (r)ψ = Eψ,

for the above value of ∇2ψ.

17.4 Separability of Eigenfunctions in Spherical Coordinates

We attempt separability by writing ψ(r, θ, φ) = R(r)Y (θ, φ), as a product ofradial and angular components R and Y . The Schrodinger Equation

− ~2

2m∇2φ+ V (r)φ = Eφ

gives

− ~2

2m

[Y

r2

d

dr

(r2 dR

dr

)+

R

r2 sin θ

d

dθ

(sin θ

∂Y

∂θ

)+

R

r2 sin2 θ

∂2Y

∂φ2

]+V (r)RY = ERY.

This rearranges to:

1

R

d

dr

(r2 dR

dr

)− 2mr2

~2[V (r)−E] = − 1

Y

(1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

∂2Y

∂θ2

).

The left side is a function of r, while the right side is a function of θ, φ, so infact both sides equal a constant C! By fiat, we write C = l(l + 1) for some l.17

Looking at only the radial component:

d

dr

(r2 dR

dr

)− 2mr2

~2[V (r)− E]R = l(l + 1)R.

Note that ddr

(r2 dR

dr

)= r d

2(rR)dr2 . Therefore, we have

d2(rR)

dr2− 2m

~2[V (r)− E](rR) =

l(l + 1)

r2(rR).

This looks like the Schrodinger equation, with u(r) = rR(r) in place of ψ!

17Eventually we’ll see that l is an integer that parametrizes the shape of the resulting orbit.For example l = 0 is the spherical s-orbital, and l = 1 is the p-orbital. For now, we just knowl is some (not necessarily even real) number.

98


17.4.1 Special case: l = 0

If l = 0, the equation

− 1

Y

(1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

∂2Y

∂θ2

)= 0

is solved when Y is constant. So, the solutions with l = 0 are the spherically-symmetric solutions!

99

Brice Huang 18 May 1, 2017

18 May 1, 2017

Today, we will continue with the Schrodinger Equation in spherical coordinates.

18.1 Last Time

Last time, we derived the Schrodinger Equation in spherical coordinates:

− ~2

2m

[1

r2

∂

∂r

(r2 dψ

dr

)+

1

r2 sin θ

∂

∂θ

(sin(θ)

∂ψ

∂θ

)+

1

r2 sin2 θ

∂ψ

∂φ2

]+ V ψ = Eψ.

We looked for separable solutions ψ(r, θ, φ) = R(r)Y (θ, φ), which gave us:

1

R

d

dr

(r2 dR

dr

)− 2mr2

~2[V (r)− E] = F (θ, φ) = C = l(l + 1),

for some ugly function F . We stated, but didn’t prove, that if l = 0, Y (θ, φ) =constant satisfies this equation, so the resulting solutions are spherically symetric.

We’ll explore this some more. For today, we set l = 0. We’ll set Y (θ, φ) = 1,so that ψ(r, θ, φ) = R(r). When we plug this into the Schrodinger Equation, weare left with:

− ~2

2m

1

r

d2

dr2(rψ) + V (r)ψ = Eψ.

We can write this as

− ~2

2m

d2

dr2(rψ) + V (r)(rψ) = E(rψ).

Note that this looks like the Schrodinger Equation, with rψ in place of ψ! Wemake the substitution u(r) = rψ to get:

− ~2

2m

d2u

dr2+ V (r)u(r) = Eu(r).

Observe that the units here check out. ψ has to have dimensions r−3/2 becauseits square is a volume probability density; rψ thus has dimensions r−1/2, whichmakes sense because rψ looks like a solution to the 1-D Schrodinger equation.

18.2 Example: Infinite Spherical Well

Consider the spherical well

V (r) =

∞ r > a

0 r < a.

This well admits non-spherically-symmetric solutions, both in the classical andquantum models. For today we’ll focus only on the spherically-symmetricsolutions, which are also the solutions with zero angular momentum.

For r > a, we must have u = ψ = 0. Inside the well, the SchrodingerEquation implies

− ~2

2m

d2u

dr2= Eu(r).

100


Figure 9: Graphs of ψ1, ψ2 and Associated Probability Densities. Image source:Lecture.

Just like in the 1-D square well, this implies

d2u

dr2= −k2u,

for k =√

2mE~ , with boundary condition u(a) = 0. We know that this differential

equation has solutions

u(r) = A sin kr +B cos kr,

and hence

ψ(r) =A sin kr

r+B cos kr

r.

The second term blows up near the origin, so it doens’t solve the Time-Independent Schrodinger Equation, and we get rid of it. 18 Therefore,

ψ(r) =A sin kr

r,

and the boundary condition ψ(a) = 0 implies ka = nπ. Graphs of ψ for thelowest two energies, and their corresponding probability densites, are shown inFigure 9.

Like in the 1-D square well, we get En = ~2n2π2

2ma2 . Finally, we can normalize.The normalization condition is

1 =

∫ ∞−∞

∫ ∞−∞

∫ ∞−∞|Ψ|2dxdydz =

∫ ∞0

|Ψ|2(4πr2dr) = 4π|A|2∫ a

0

sin2(kr)dr = 2πa|A|2.

Thus, A = 1√2πa

.

18Actually, it’s not immediate that this isn’t OK; maybe this term can blow up because wedon’t have a boundary condition near the origin. But, recall from electrostatics that

∇2φ = −ρ

ε0,

where φ is a Coulomb ( 1r

) potential. Thus, if the second term is alowed to blow up at r = 0,then

∇2ψ = ∇2

(1

r

)= δ3(r),

a 3-dimensional delta function. But, our potential doesn’t have a delta-function spike, so thisdoesn’t solve the Time-Independent Schrodinger Equation.

101


18.3 Example: Nuclear Potential

We consider the nuclear Coulomb potential

V (r) = −ze2

r.

The Schrodinger Equation implies

− ~2

2m

d2u

dr2− ze2

ru = Eu⇒ d2u

dr2+

2mze2

~2

u

r+

2mE

~2u = 0.

To solve this, we reparametrize in terms of the dimensionless length

ρ =

√−8mE

~2r

(note that E < 0 for any bound state, so this length is well-defined) to get:

−8mE

~2

d2u

dρ2+

2mze2

~2

(−8mE

~2

)1/2u

ρ+

2mE

~2u = 0,

which simplifies to

d2u

dρ2+

2mze2

~2

(~2

−8mE

)1/2u

ρ− 1

4u = 0.

We then define the dimensionless constant

λ =2mze2

~2

(~2

−8mE

)1/2

=ze2

~2

√m

−2E,

to simplify our equation to

d2u

dρ2+ λ

u

ρ− 1

4u = 0.

18.3.1 Power Series Method, Revisited

To solve the last equation, we will use the power-series approach we saw forthe Simple Harmonic Oscillator. We first find an asymptotic solution awayfrom zero, which will be an envelope for our solutions; then, we write our exactsolutions as the asymptotic solution times a power series.

In the limit as ρ→∞, our equation becomes

d2u

dρ2− 1

4u ∼ 0.

This has solutions u(ρ) ∼ e±ρ/2; of these, we can reject u(ρ) ∼ eρ/2 because itisn’t normalizable. Thus u(ρ) ∼ e−ρ/2.

We now write our exact solutions as u(ρ) = h(ρ)e−ρ/2. We compute:

du

dρ= e−ρ/2

[1

2h− dh

dρ

]d2u

dρ2= e−ρ/2

[1

4h− dh

dρ+d2h

dρ2

].

102


Plugging this into the equation we are solving yields[1

4h− dh

dρ+d2h

dρ2

]+λ

ρh− 1

4h = 0⇒ d2h

dρ2− dh

dρ+λ

ρh = 0.

We attempt to solve this with a power-series solution

h(ρ) =

∞∑j=1

bjρj .

Note that we there is no j = 0 term, because otherwise ψ ∼ h(ρ)ρ e−ρ/2 blows up

at the origin.

We plug this in for h and expand:

∞∑j=1

bjj(j − 1)ρj−2 −∞∑j=1

bjjρj−1 + λ

∞∑j=1

bjρj−1 = 0.

After some re-indexing, this implies:

∞∑j=1

[bj+1(j + 1)j − bjj + λbj ] ρj−1 = 0,

and the coefficients for each ρj−1 must be zero. Therefore, for all j, we get:

bj+1(j + 1)j − bjj + λbj = 0⇒ bj+1 =j − λj(j + 1)

bj .

Finally, we find what values of λ are permissible. When j >> λ, we havebj+1 ∼ 1

j+1bj , so bj ∼ 1j! . This is, actually, pretty bad – the bj don’t die fast

enough. Indeed, this implies

h(ρ) ∼∞∑j=1

1

j!ρj = eρ,

and thus u(ρ) = e−ρ/2h(ρ) =∼ eρ/2. This blows up at large ρ, and we’re sad.

In order for this blowup to not happen, the power series for h has to truncate– some coefficient j−λ

j(j+1) must be zero, causing all larger bj to be zero. Thus, we

must have λ = n for some integer n, causing the power series to terminate atj = n.

From the quantization of λ, we get

λ =ze2

~

√m

−2En= n⇒ En = −z

2me4

2~2

1

n2,

which are the Bohr energies we derived in the beginning of the course! This isone of the triumphs of QM.

We just derived all of the s-orbitals. We know there’s also p, d, f, . . . orbitals– these correspond to orbitals with nonzero angular momentum, which we’ll seelater.

103


18.3.2 Writing the s-Wavefunctions

We’ll write the s-wavefunctions for n = 1, 2.

From r =(

~2

−8mE

)ρ, we get r = na0

2z ρ, where a0 = ~2

me2 was defined in the

beginning of the course. Thus ρ = 2zn

ra0

.

We’ll first write ψ1. Note that h1(ρ) = b1ρ, so u1(ρ) = b1ρe−ρ/2, and thus

ψ1(r) =u1

r=

constant

rρe−ρ/2 = A1e

−zr/a0 ,

wher we use ρ = 2zn

ra0

. We compute the normalization constant:

1 = |A1|2∫ ∞

0

e−2zr/a0(4πr2dr)⇒ |A1| =1√π

(z

a0

)3/2

.

Next, we write ψ2. We have that b2 = 1−22 b1 = − 1

2b1, so

h2(ρ) = b1

(ρ− 1

2ρ2

).

Then,

ψ2(r) =u2

r=

constant

r

[ρ− 1

2ρ2

]e−ρ/2 =

A2z

a0

[1− z

2a0r

]e−zr/2a0 .

Note that node at r = 2a0z . Moreover, because of ρ’s n-dependence, this decays

twice as slowly as ψ1.

104


19 May 3, 2017

Today, we’ll step away from the radial equation and think about angular mo-mentum – remember that so far, we’ve ignored the equation involving Y (θ, φ)by assuming Y is constant!

19.1 Gyromagnetic Effect

There’s an experiment we can do to show that something interesting happensto angular momentum in an atom. We take a solid bar, made of some mag-netic material, and suspend it by a string. We heat the bar above its Curietemperature.19 We then put the bar in a constant magnetic field and lower thetemperature below the Curie temperature.

As we expect, the material’s magnetic moments align with the externalmagnetic field. Surprisingly, something else happens: the bar spins, withoutthe application of a torque! This means that magnetic moments must have anassociated angular momentum. This is known as the gyromagnetic effect.

It turns out electrons also have an associated magnetic moment. It’s as ifthese point charges are spinning. We call this property the electron’s spin.

Nucleons, like electrons, have associated magnetic moments. But, sinceangular momentum is on the order of ~, and nucleons are about 1000 timesas massive as electrons, we expect the magnetic moments of nucleons to beabout 1

1000 times those of electrons. And, this is exactly what we experimentallyobserve.

Properties of nucleons’ magnetic moments have many applications, includingMRI.

Joke. “People used to call then Nuclear Magnetic Resonance Imaging, and thenpeople got scared of the ‘nuclear’ part, so now it’s called Magnetic ResonanceImaging.” - Ashoori

19.2 Magnetic Moment and Angular Momentum

In an electric field not constant across space, an electric dipole feels a forcebecause its two ends experience slightly different electric fields. The same holdsfor magnetic dipoles in non-constant magnetic fields.

We consider a magnetic field B, changing in the z-direction. Let µz be thecomponent of the magnetic dipole moment pointing along the z direction. Then,the dipole experiences a z-direction force of

Fz = µz∂Bz∂z

= µ cos θ∂Bz∂z

.

For a magnetic dipole, the angular momentum is proportional to magneticmoment, with the relation

~µ = γ ~Ls,

where γ is the gyromagnetic moment.

19Recall that above the Curie temperature, the material’s magnetic domains un-align, andthe material isn’t magnetic

105


Figure 10: Stern-Gerlach Experiment Setup. Image source: Lecture.

Figure 11: Lengths in the Stern-Gerlach Experiment. Image source: Lecture.

For a current loop with current I surrounding area A producing a magneticfield, the magnetic moment is

~µ =I ~A

c.

In an external magnetic field ~B, the torque τ experienced by a magneticmoment ~µ is

τ = ~µ× ~B.

19.3 Stern-Gerlach Experiment

Suppose we have an oven spewing Cs atoms. We pass the output through acollinator, giving us a beam of Cs atoms. We pass the Cs beam through acurved magnet, designed so dBz

dz points upwards (see Figure 10), and detect thedeflection of the Cs atoms on the far side.

Suppose the magnet has width d, and the free-flight region between themagnet and the detector has width D. Suppose the Cs atoms deflect by avertical distance z1 in the magnetic region and z2 in the free-flight region; saythe final trajectory of the atoms deflected by an angle α.

We can calculate the acceleration:

|a| = Fzm

=µzm

dBzdz

.

Thus,

z1 =1

2at2 =

1

2

F2

m

(d

v

)2

=µzd

2

2mv2

∂Bzdz

,

so the z-momentum is

Pz = Fzt =d

vµz∂Bz∂z

,

106


Figure 12: Stern-Gerlach Experiment Results. Image source: Lecture.

and

tanα =Pzmv

=µzd

mv2

∂Bz∂z

.

So, z2 = D tanα and the deflection is

z1 + z2 =d

mv2

(D +

d

2

)µz∂Bz∂z

.

What do we expect to observe for the deflections z1 + z2? Classically, thespin can point in any direction, so µz is a smear between |µ| and −|µ|. So, weexpect z1 + z2 to be a smear centered at 0.

When the magnetic field is off, we do get a smear of deflections near 0 –which can be caused by, for example, imperfect collination. So far, so good.

When the magnetic field is on, we get two distinct peaks on opposite sides of0. This means that, somehow, by making a measurement of µz

20 we are forcingthe Cs atoms to choose between going full spin-up and full spin-down.

From the data, we get µz = ±8.7× 10−21 erg /Gauss.

19.4 Bohr Magneton

We consider a Bohr magneton, an electron moving in a circular loop of radiusr. Then, the electron’s period is T = 2πr

v , and the current of the loop isI = e

T = ev2πr . As we said before, the magnetic moment of the Bohr magneton is

µ =IA

c=

ve

2πr· πr

2

c=evr

2c.

But, by Bohr Quantization of angular momentum, L = mvr = ~, so thefundamental quantum unit of magnetic moment is

µB =

(e

2mec

)(mevr) =

e~2mec

= 9.27× 10−21 erg /Gauss .

We can write quantum magnetic moments as multiples of the Bohr magneton:

µz = gµB

(Lz~

),

where g is a dimensionless factor. For atomic orbitals, g = 1, and Lz =0,±h,±2h, . . . .

20This is, effectively, what this apparatus is doing

107


The spin of electrons turns out to (a result of Relativistic Quantum FieldTheory) have angular momentum

Lz = Sz = ±~2,

with g-factor ≈ 2.21 So,

µSz ≈ 2µB

( 12~~

)= µB .

That is, an electron spin is very close to a Bohr magneton!

19.5 Angular Momentum in QM

By definition of angular momentum, ~L = ~r × ~p. So,

Lx = ypz − zpyLy = zpx − xpzLz = xpy − ypx.

But, we know px = −i~ ∂∂x (and likewise for py, pz), so:

Lx = −i~(y∂

∂z− z ∂

∂y

)Ly = −i~

(z∂

∂x− x ∂

∂z

)Lz = −i~

(x∂

∂y− y ∂

∂x

).

But, because x = r sin θ cosφ and y = r sin θ sinφ, we have

∂

∂φ=∂y

∂φ

∂

∂y+∂x

∂φ

∂

∂x= x

∂

∂y− y ∂

∂x.

So, actually,

Lz = −i~ ∂

∂φ.

Suppose ψ is a eigenfunction of Lz, with angular momentum Lz (the number,as opposed to Lz, the operator). Since Lzψ = Lzψ, this implies

Lzψ(r, θ, φ) = −i~ ∂

∂φψ(r, θ, φ) = Lzψ(r, θ, φ).

This implies∂ψ

∂φ= − 1

i~Lzφ,

soφ(r, θ, φ) = eiLzφ/~f(r, θ).

But, φ is an angle! For this function to be single-valued, we must haveeiLz(2π)/~ = 1, i.e.

Lz2π

~= 2πm⇒ Lz = m~.

21Actually, ge ≈ 2.0023318416, a result of Quantum Electrodynamics.

108


Hence, we have the solution

φ(r, θ, φ) = eimφf(r, θ).

Caution: the operators Lx, LyLz don’t commute, so we can never measuretwo of these simultaneously.

19.6 Commuting Operators

Suppose the operators A, B commute. Let ψ be an eigenfunction of both A andB, so that

Aψ = aψ Bψ = bψ.

Then,

BAψ = B(aψ) = abψ

ABψ = A(bψ) = abψ,

as expected. So, it’s possible for commuting operators to share eigenfunctions.

19.6.1 Commutators

We’ve shown before that:

[x, y] = 0 [x, px] = i~ [x, py] = 0.

It turns out that[Lx, Ly] = i~Lz,

so Lx, Ly, Lz don’t commute. Moreover, recall that the Generalized Uncertainty

Principle says that for any operators A, B,

σ2Aσ

2B ≥

(1

2i〈[A, B]〉

)2

.

So,

σLxσLy ≥~2|〈Lz〉|.

Finally, it turns out the operator

L2 = Lx2

+ Ly2

+ Lz2

commutes with all of Lx, Ly, Lz. This means that, for example, eigenfunctions

of Lx are also eigenstates of definite angular momentum.

109


20 May 8, 2017

Today we will finish up the general discussion about angular momentum, anddiscuss the raising and lowering operators for Lz.

We will then revisit and solve the angular part of the spherical SchrodingerEquation. This equation will turn out to also be separable.

Next lecture we’ll put the radial and angular parts together, to get all theorbitals of the hydrogen atom.

20.1 Angular Momentum

Last time we said that

[Lx, Ly] = i~Lz, [Ly, Lz] = i~Lx, [Lz, Lx] = i~Ly.

This can be written compactly as

[Li, Lj ] = i~εijkLk,

where εijk is the sign of the permutation (i, j, k) of (1, 2, 3).

20.1.1 Aside: Clarification from Last Time

Last time, we derived from the Generalized Uncertainty Principle that

σLxσLy ≥~2|〈Lz〉|.

Suppose we know Lz = 0. This doesn’t mean that the uncertainty product ofLx, Ly is zero – the Generalized Uncertainty Principle only gives a lower bound,and in this scenario the bound isn’t tight.

20.1.2 Ladder Operators

We consider the operator L2, defined by

L2 = Lx2

+ Ly2

+ Lz2.

If we look at its commutator with Lx, Ly, Lz, we’ll find that actually

[L2, Lx] = [L2, Ly] = [L2, Lz] = 0.

This means that L2 has simultaneous eigenstates with Lx, Ly, Lz, but theseoperators don’t have simultaneous eigenstates with each other. It’ll turn outthat eigenstates of Lx are eigenstates of L2 but not vice versa, and similarly forLy, Lz.

Let’s consider a simultaneous eigenstate ψ of L2 and Lz. We define theoperators

L+ = Lx + iLy, L− = Lx − iLy.

110


These will turn out to be our raising and lowering operators – like with theharmonic oscillator, L+ will raise Lz while L− will lower Lz.

From the known commutator relations among L2, Lx, Ly, Lz, we can showthat:

[Lz, L±] = ±~L± [L2, L±] = 0.

We claim:

Theorem 20.1

If ψ is an eigenstate of L2 and Lz, then so is(L±ψ

).

Proof. Suppose that

L2ψ = aψ, Lzψ = bψ.

Since [L2, L±] = 0, we have

L2(L±ψ

)= L±

(L2ψ

)= L±(aψ) = a

(L±ψ

).

Therefore,(L±ψ

)is an eigenstate of L2 with the same eigenvalue.

Moreover,

Lz

(L±ψ

)=(LzL± − L±Lz

)ψ + L±Lzψ

= [Lz, L±]ψ + L±Lzψ

= [Lz, L±]ψ + L±(bψ)

= ±~L±ψ + bL±ψ

= (b± ~)L±ψ.

So,(L±ψ

)is an eigenstate of Lz with Lz-eigenvalue b± ~.

20.1.3 Top and Bottom Rungs of the Ladder

This computation shows that the raising and lowering operators preserve thetotal angular momentum, and raise or lower the z-angular momentum by ~.

But, the z-angular momentum can’t exceed the total angular momentum,so there are finitely many rungs. This means that, if ψtop, ψbot are the top andbottom rungs of the ladder, then22

L+ψtop = 0, L−ψbot = 0.

We’ll work with the first relation. Let’s let l~ be the Lz eigenvalue of ψtop.Then, we have

L2ψtop = aψtop, Lzψtop = l~ψtop.22Note the similarity to the Simple Harmonic Oscillator ladder-operator solution.

111


We compute:

L±L∓ =(Lx ± iLy

)(Lx ∓ iLy

)= L2 − L2

z ∓ i(i~Lz

),

soL2 = L±L∓ + L2

z ∓ ~Lz.Therefore,

L2ψtop =(L−L+ + L2

z + ~Lz)ψtop

=(0 + l2~2 + l~2

)ψtop

= l(l + 1)~2ψtop.

So, ψtop, and thus all eigenstates in the ladder, have L2-eigenvalue l(l + 1).23

Similarly, if l~ is the Lz eigenvalue of ψbot, then from the relations

L−ψbot = 0 L2ψbot = aψbot Lzψbot = l~ψbot

we getL2ψbot = l(l − 1)~2ψbot.

Thus, all eigenstates also have L2-eigenvalue l(l − 1). So,

l(l + 1) = l(l − 1).

This, with the relation l ≤ l (by maximality of l) implies l = −l.

20.1.4 Quantization of Angular Momentum

Each application of L+ increases the z-angular momentum by ~, and we go from−l~ to l~ in some integer (say, N) applications of L+. So, l = −l +N , whichimplies l = N

2 ! This means angular momentum is quantized with

l = 0,1

2, 1,

3

2, 2, . . . .

By application of ladder operators to ψtop, ψbot, we get that for each l, there are

simultaneous eigenstates ψl,m of L2 and Lz, where

m = −l,−l + 1, . . . , l − 1, l,

obeying

L2ψl,m = l(l + 1)~2ψl,m

Lzψl,m = m~ψl,m.

It turns out that for orbital states, l is equal to an integer. The intrinsicangular momenta of particles can have non-integer l. We’ll deal with this morein 8.05.

We can notice that the overall angular momentum vector has length√l(l + 1),

whose magnitude is larger than l, the maximum possible z-angular momenutum.This means that no matter how we measure, we can never get all the angularmomentum to point in one direction. Qualitatively, this makes sense – if all theangular momentum points in the z-direction, then we simultaneously know thatthe x, y-angular momenta are zero; but, the non-commutativity of Lx, Ly, Lzprohibits this simultaneous measurement.

23This is the motivation for defining C = l(l + 1) in Lecture 17.

112


20.2 Angular Part of Spherical Schrodinger Equation

Recall that for a spherically-symmetric potential V , the Spherical SchrodingerEquation is

−f~22m∇2ψ(~r) + V (r)ψ(~r) = Eψ(~r).

We separated ψ(r, θ, φ) = R(r)Y (θ, φ), and (by computations) got the radialand angular parts

1

R

d

dr

(r2 dR

dr

)− 2mr2

~2[V (r)− E] = l(l + 1),

1

Y

[1

sin θ

∂

∂θ

(sin θ

∂Y

∂θ

)+

1

sin2 θ

∂2Y

∂φ2

]= −l(l + 1).

20.2.1 Separation of θ, φ

We’ll separate the angular part by writing Y (θ, φ) = Θ(θ)Φ(φ). Multiplying theangular equation through by sin2 θ gives

sin θ∂

∂θ

(sin θ

∂Y

∂θ

)+∂2Y

∂φ2= −l(l + 1)Y sin2 θ

⇒[

1

Θ

[sin θ +

∂

∂θ

(sin θ

dΘ

dθ

)]+ l(l + 1) sin2 θ

]+

[1

Φ

d2Φ

dφ2

]= 0.

The terms on the left-hand side of the last equation are a pure function of θand a pure function of φ, so these two terms are... (class in unison:) constant!By fiat, we can set

1

Θ

[sin θ +

∂

∂θ

(sin θ

dΘ

dθ

)]+ l(l + 1) sin2 θ = m2

1

Φ

d2Φ

dφ2= −m2.

for some complex number m.24

20.2.2 Solving the θ, φ-Equations

The resulting differential equation in φ is

d2Φ

dφ2= −m2Φ,

which has solutionsΦ(φ) = eimφ.

Since Φ is periodic with period 2π, this forces m to be an integer.

We now tackle the equation in θ:[sin θ

d

dθ

(sin θ

dΘ

dθ

)]+[l(l + 1) sin2 θ −m2

]= 0.

24This m turns out to be an integer, and equal to the Lz-eigenvalue, hence the notation.

113


We let x = cos θ, so

d

dθ=dx

dθ

d

dx= − sin θ

d

dx⇒ sin θ

d

dθ= − sin2 θ

d

dx= −(1− x2)

d

dx.

Moreover, we write Θ = Pl,m, where Pl,m is a function of x. This allows us torewrite the θ-equation (computation omitted) as

d

dx

[(1− x2)

dPl,mdx

]+

[l(l + 1)− m2

1− x2

]Pl,m(x) = 0.

Let’s first solve this when m = 0. We write Pl,0(x) = Pl(x), so

d

dx

[(1− x2)

dPldx

]+ l(l + 1)Pl(x) = 0.

We’ll try a power series solution

Pl(x) =

∞∑k=0

akxk.

When we plug this in, we get a recursion relation, and we have to (like be-fore) terminate the power series to prevent it from blowing up. The resultingpolynomials are called Legendre polynomials.

We’ll finish this next time, and then put the radial and angular componentstogether.

114


21 May 10, 2017

21.1 Spherical Schrodinger Equation - The θ-equation

21.1.1 Last Time

Last time we separated the angular part of the 3-D Schrodinger Equation intocomponents Y (θ, φ) = Θ(θ)Φ(φ). The Φ-equation had a simple solution

Φ(φ) ∼ eimφ.

The θ-equation was[sin θ

d

dθ

(sin θ

dΘ

dθ

)]+[l(l + 1) sin2 θ −m2

]Θ = 0.

We set x = cos θ, and let Pl,m(x) be the function Θ, parametrized by x. Theθ-equation reduces to

d

dx

[(1− x2)

dPl,mdx

]+

[l(l + 1)− m2

(1− x2)

]Pl,m(x) = 0.

We’ll derive the m = 0 case first; the solution for general m depends on them = 0 solution. We set Pl,0(x) = Pl(x), so:

d

dx

[(1− x2)

dPl,mdx

]+ l(l + 1)Pl(x) = 0.

21.1.2 Case m = 0: Power Series, Again

We try the power series solution

Pl(x) =

∞∑k=0

akxk.

Plugging in this solution and setting corresponding coefficients equal gives usthe relation (computations omitted):

(k + 1)(k + 2)ak+2 + [l(l + 1)− k(l + 1)] ak = 0,

which implies the recursion relation

ak+2 = − l(l + 1)− k(k + 1)

(k + 1)(k + 2).

For large k, this means ak+2 ≈ −ak, which is a problem because the series∑∞k=0 x

k diverges at ±1, which are in the range of cos θ. So, the series has toterminate, which means k = l for some l. But this only stops the series for theeven or odd terms; for the series to stop, the terms of the other parity have toall be zero. This means that the Legendre polynomials are even for all even land odd for all odd l.

This produces one polynomial Pl for each l; these polynomials are called theLegendre polynomials. The first few Legendre polynomials are below:

P0(x) = 1

P1(x) = x

P2(x) =1

2(3x2 − 1).

115


Incidentally, the Rodrigues Formula generating all the Legendre polynomialsis:

Pl(x) =1

2ll!

(d

dx

)l(x2 − 1)l.

21.2 Solution to θ-Equation for General m

For general m, it turns out that the solution is

Pl,m(x) = (1− x2)|m|/2(d

dx

)|m|Pl(x).

We will not derive this result.

From this formula, we can see that Pl,m(x) 6= 0 only when −l ≤ m ≤ l.Given l, these are the only m for which we have a perissible solution to theθ-equation.

Note that Pl,m is a polynomial in x, multiplied by (1− x2)|m|/2. For m odd,this is an integer power of 1− x2 times

√1− x2 = sin θ. So:

Fact 21.1. Pl,m is a polynomial in cos θ when m is even; when m is odd, it isa polynomial in cos θ multiplied by sin θ.

For sake of concreteness, some values of Pl,m for small l,m:

P1,±1(cos θ) = sin θ

P1,0(cos θ) = cos θ

P2,±2(cos θ) = 3 sin2 θ

P2,±1(cos θ) = 3 sin θ cos θ

P2,0(cos θ) = 3(3 cos2 θ − 1).

21.3 Spherical Schrodinger Equation - Putting Everything To-gether

21.3.1 Completing the Angular Part

Combining the θ and φ-components, we get the angular solutions to theSchrodinger Equation:

Yl,m(θ, φ) = Θ(θ)Φ(φ) = Al,mPl,m(cos θ)eimφ.

These are eigenstates of L2 and Lz. Some examples:

Y0,0 =1√4π

Y1,±1 =

√3

8πsin θe±iφ

Y1,0 =

√3

4πcos θ

116


21.3.2 The Radial Part

We said the radial part to the Sphrerical Schrodinger Equation was

d

dr

(r2 dR

dr

)− 2mr2

~2[V (r)− E]R = l(l + 1)R.

Now, we actually know that l has to do with the angular momentum. When wewrite u(r) = rR(r), we get

− ~2

2m

d2u

dr2+

[V +

~2

2m

l(l + 1)

r2

]u = Eu.

This looks like the 1-D Schrodinger Equation, but now we have an extra term~2

2ml(l+1)r2 . But, we’re treating l as a constant, so can just lump this term into

Veff , the effective potential:

Veff = V +~2

2m

l(l + 1)

r2= V +

L2

2mr2.

The additional term is called the centrifugal term – the need to maintain aconstant angular momentum means the particle needs additional kinetic energyto get close to the center, which acts as a repulsive force. Thus, we can rewritethe radial component as

− ~2

2m

d2u

dr2+ Veffu = Eu.

21.3.3 Normalization

Since Φ(r, θ, φ) = R(r)Y (θ, φ), the normalization condition is

1 =

∫all space

|Ψ|2r2 sin θdθdφdr =

∫ ∞0

r2|R|2dr∫ π

0

∫ 2π

0

|Y |2 sin θdφdθ.

Since we can shift constant factors between R and Y arbitrarily, to normalizewe’ll demand that

1 =

∫ ∞0

r2|R|2dr

1 =

∫ π

0

∫ 2π

0

|Y |2 sin θdφdθ

21.4 Solving the Hydrogen Atom – The Radial Part

We’ll work with the potential V (r) = − ze2

r .25

The radial part is

− ~2

2m

d2u

dr2+

[−ze

2

r+

~2

2m

l(l + 1)

r2

]u = Eu

⇒ 1

α2

d2u

dr2=

[1− 2mz2e2

~2α

1

αr+l(l + 1)

(αr)2

]u,

25This computation doesn’t work for atoms larger than hydrogen, because we assumethere’s only one electron. So, this computation represents hydrogen, once-ionized helium,twice-ionized lithium, and so on. Atoms with more electrons are harder to analyze because ofshielding effects.

117


where α =√

2mE~ .

We set ρ = αr and ρ0 = 2zαa0

(recall that a0 = ~2

me2 ). This gives us (compu-tation omitted) the simplified equation

d2u

dρ2=

[1− ρ0

ρ+l(l + 1)

ρ2

]u.

21.4.1 Finding Asymptotic Solutions

As we’ve done before, we’ll first find the asymptotic solution, and then write theactual solutions as the asymptotic solution times a power series. In the limit asρ→∞, we have

d2u

dρ2≈ u⇒ u(ρ) ≈ Ae−ρ +Beρ.

We discard the eρ solution because it blows up as ρ→∞. Thus u(ρ) ≈ Ae−ρ.In the limit as ρ→ 0, we have

d2u

dρ2=l(l + 1)

ρ2u⇒ u(ρ) = Cρl+1 +Dρ−l.

We discard the ρ−l solution because it blows up as ρ→ 0. Thus u(ρ) ≈ Cρl+1.

Note that this is fine even when l = 0, which makes l(l+1)ρ2 = 0. In this case,

u(ρ) = Cρ, which solves the above equation when C = 1.

21.4.2 Power Series, Again; Laguerre Polynomials

OK, we’ve found our asymptotic solutions. We write the general solution as

u(ρ) = ρl+1e−ρv(ρ).

The first two derivatives of u are (messy computation omitted):

du

dρ= ρle−ρ

[(l + 1− ρ)v + ρ

dv

dl

]d2u

dρ2= ρl; e−ρ

[(−2l − 2 + ρ+

l(l + 1)

ρ

)v + 2(l + 1− ρ)

dv

dρ+ ρ

d2v

dρ2

]When we substitute d2u

dρ2 into our original equation

d2u

dρ2=

[1− ρ0

ρ+l(l + 1)

ρ2

]u,

we get the following equation in v:

ρd2v

dρ2+ 2(l + 1− ρ)

dv

dρ+ [ρ0 − 2(l + 1)] v = 0.

We set v(ρ) =∑∞j=0 ajρ

j , so

dv

dρ=

∞∑j=0

jajρj−1 =

∞∑j=0

(j + 1)aj+1ρj ,

d2v

dρ2=

∞∑j=0

j(j + 1)aj+1ρj−1.

118


We plug all this back in to the v-equation and get:

∞∑j=0

j(j+1)aj+1ρj+2(l+1)

∞∑j=0

(j+1)aj+1ρj−2

∞∑j=0

jajρj+[ρ0 − 2(l + 1)]

∞∑j=0

ajρj = 0.

This yields the recursion relation

aj+1 =2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)aj .

Like before, if this series doesn’t truncate we get an explosion at ρ→∞. This

is because for large j, aj+1 ≈ 2aj , so aj ∼ 2j

j! . This means

v(ρ) ∼∞∑j=0

2j

j!ρj = e2ρ

which causesu(ρ) ∼ ρl+1e−ρe2ρ = ρl+1eρ

to blow up.

So, the series truncates. This forces ρ0 = 2(j + l + 1) for some j. Theresulting polynomials are called the Laguerre polynomials.

21.4.3 Conclusion

If jmax is the order of the polynomial v(ρ) (i.e. the largest j for which aj 6= 0),then

2(jmax + l + 1)− ρ0 = 0.

We set jmax + l + 1 = n, so ρ0 = 2n. Then,

ρ0 =2z

αa0=

2mze2

~2α, α =

2z

ρ0a0=

z

na0.

So,

E = −~2α2

2m= − ~2z2

2n2a20m

= −z2e2

2a0· 1

n2.

We can define the Rydberg energy ERydberg = − e2

2a0≈ −13.6 eV, so the Bohr

energies are

z2 · ERydberg ·1

n2.

Note that unlike in the original Bohr model, this n is compound, defined by

n = jmax + l + 1.

This n (called the Principal Quantum Number) has radial and angularcomponents; jmax is the order of the radial polynomial, which is the number ofradial notes, while the l + 1 is a function of the angular momentum.

This says that we can make the same energy in multiple ways – having a bigradial component and little angular momentum, or vice versa.

In fact, the Principal Quantum Number is what determines the rows ofthe Periodic Table – not the orbital radius size, but this sum of radial andangular numbers! The orbital types (s, p, d, . . . ) are determined by l. The msadd degeneracy within each orbital type.

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22 May 15, 2017

22.1 Finishing the Hydrogen Atom

22.1.1 The Radial Wavefunctions

Last time, we said that if R(r) solves the spherical Schrodinger equation withangular momentum number l, then:

u(ρ) = ρl+1e−ρv(ρ),

where u(r) = rR(r), ρ = zrna0

, and v(ρ) = vn,l(ρ) is a polynomial of order N ,and

n = N + l + 1.

Here, n is the principal quantum number. For each principal quantum numbern, the energy is

En = −z2e2

2a0· 1

n2.

From this, we can back out the solutions R(r):

R(r) ∼ rle−zr/na0vn,l(r).

22.1.2 Electron orbitals, Periodic Table

Figure 13 shows the allowed quantum states in the (N, l) plane. Some things toobserve:

• The diagonal lines shown are lines of constant n.

• For constant n, the l-number determines the orbital type. For example,the states (n, l) = (3, 0) through (3, 3) are the 3s, 3p, 3d, 3f orbitals.

• The state (n, l) is degenerate with multiplicity 2(2l + 1). This is becausethere are 2l + 1 possible values of m, and 2 possible spins.

• The nth diagonal has 2n2 states, because

n−1∑l=0

2(2l + 1) =

n−1∑l=0

2[(l + 1)2 − l2

]= 2n2.

Note that the energy levels depend only on n, and not on l,m.26

26This does not mean that electrons fill orbitals in exactly this energy order, becauseelectron-electron interactions change these energies. For example, electrons preferentially fill2s before 2p, because 2s electrons are better shielded from the nucleus. Moreover, 4s fillsbefore 3d, even though 3d naively has a strictly lower energy.

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Figure 13: Quantum states in (N, l)-plane. Image source: Lecture.

22.1.3 Spherical Wavefunctions

We can finally put together the spherical wavefunctions. Since

u(ρ) = ρl+1e−ρv(ρ),

and ρ = zrna0

, we have

Rn,l ∼ A1

rρl+1e−lvn,l(ρ),

which implies

ψn,l,m(r, θ, φ) = N

(zr

a0

)l(polynomial in zr

na0of

degree N = n− l − 1

)e−zr/na0Yl,m(θ, φ),

where Yl,m was computed in Lecture 21.

Some sample wavefunctions for small n, l,m:

ψ1,0,0 =

√1

πa30

e−r/a0

ψ2,0,0 =1

8

√2

πa30

(2− r

a0

)e−r/2a0

ψ2,1,0 =1

8

√2

πa30

r

a0e−r/2a0 cos θ

ψ2,1,±1 =1

8

√1

πa30

r

a0e−r/2a0 sin θe±iφ.

By integrating out the angular parts, we can find the probability density offinding the electron at a radius r:

p(r)dr = r2[Rn,l(r)]2dr.

Figure 14 shows the graphs of p(r) for small orbitals: Things to note:

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Figure 14: Probability density p(r) for 1s, 2s, 2p orbitals. Image source: Lecture.

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• N = n − l − 1 controls the number of radial nodes; for fixed principalnumber n and increasing angular momentum l, the number of radial nodesdecreases.

• For fixed n, the size of the orbital is roughly the same.

22.2 Rydberg Atoms

22.2.1 Size of a Rydberg Atom

Suppose we have a very excited hydrogen atom – a proton nucleus and anelectron in an orbital with very large n.

We ask: for given n, how big is this atom? Recall that

ψn,l,m = Arlvn,l(r)e−r/na0Yl,m = fn,l(r)Yl,m,

where

fn,l(r) = Arlvn,l(r)e−r/na0

∼ rlrn−l−1e−r/na0

= rn−1e−r/na0 .

Thus, the probability density p(r) of finding the electron at radius r is:

p(r) = r2|fn,l(r)|2dr∼ r2r2n−2e−2r/na0

= r2ne−2r/na0 .

Note that this probability doesn’t depend on l. We can estimate the size of thisatom by finding when p(r) is maximized:

d

drp(r) =

(2n

r− 2

na0

)r2ne−2r/na0 = 0⇒ r = n2a0.

This agrees with the Bohr model!

Note that for a Rydberg atom with n = 350, this yields

r = 0.53A× 3502 = 6.5µm.

22.2.2 Shapes of Rydberg Orbitals

Recall that for given angular momentum l, the effective potential is

Veff (r) =~2l(l + 1)

2mr2− e2

r.

This increases as l increases. So, for given energy E, the classically-allowedregion decreases when l increases, and the resulting wavefunction has fewerwiggles.

Say that the region of classically-allowed radii is [r−, r+]. We can solve forr± from the equation:

~2l(l + 1)

2mr2±− e2

r±= − e2

2a0

1

n2,

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which is quadratic in r±.

For example:

(n, l) = (100, 60) ⇒ (r−, r+) = (2000a0, 18000a0)

(n, l) = (100, 99) ⇒ (r−, r+) = (9000a0, 11000a0).

Observe that for fixed n, as l increases the interval (r−, r+) becomes narrower,and the orbit becomes less spherical and more circular.

22.3 EPR Paradox

Einstein totally hated the idea that the act of observing something changes it.He and Schrodinger tried to draw bizarre conclusions for this – for example,Schrodinger’s Cat. He and his nemesis, Bohr, would go back and forth on theirinterpretations of QM. Finally, Einstein came up with a paradox that stumpedeveryone:

Consider a pion π decaying into an electron-positron pair e+, e−. The pionhas spin 0, so by conservation of angular momentum the e+, e− must haveopposite spin – but aside from this, we don’t know anything about the spindirections.

Suppose we take the e+, e− pair and transport them miles away from eachother. We orient a spin-detector in any direction and measure one member of thepair with it. Once we do this, we instantaneously know that the other memberof the pair has the opposite spin. From this, no matter what interpertation ofQM we use, only two interpretions are possible:

• The collapse of the first particle’s wavefunction instantaneously collapsesthe second’s, which is weird;

• The universe already knows which way you’re going to set the detector,which is weirder.

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23 May 17, 2017

We’ve made it to the last lecture! Whew. Today: fun stuff!

23.1 EPR Paradox, Bell’s Inequality

Recall from last time:

Einstein didn’t believe the probabilistic interpretation of QM, and maintainedthat at all times there is a local reality. He devised a thought experiment wherea pion decays into a positron-electron pair with opposite spin. After separatingthe particles, we measure the spin of one of the particles by a detector, alongsome axis (that we don’t have to set until after separating the particles); oncewe measure it we immediately force the spin of the other particle to also bealong this axis. This allows some kind of instantaneous communication. Spooky.

Einstein believed that QM is an incomplete model – that there must be localhidden variables that allow some form of coordination between the two particles,so information doesn’t actually travel instantaneously. Bell’s Ienquality rulesthis out – it states that there might be global hidden variables, but not localhidden variables.

Let’s write the wavefunction of the two particles’ spins as the superpositionof the two possibilities: up/down, and down/up:

ψspin(s1, s2) =1√2

(↑↓ − ↓↑) .

This is called the singlet state, and has total angular momentum stot = 0.Then, there are the triplet states, which have nonzero angular momentum:

ψspin = ↑↑,

ψspin =1√2

(↑↓ + ↓↑) ,

ψspin = ↓↓ .

What values can we get for the product of the spins?

positron electron product1 -1 -1-1 1 -11 1 1-1 -1 1

If we measure the spins along the same axis, only the first two measurementsare possible.

23.1.1 Spinors

We can define the spin-up and spin-down states by the spinors:[10

] [01

].

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A spin in a general direction (θ, φ) is given by

[cos θ2

eiφ sin θ2

]. The z-spin operator

is given by

sz =

[1 00 −1

].

So, for example, for the spin-up state:

〈sz〉 =[1 0

] [1 00 −1

] [10

]= 1,

and for a general spin

[cos θ2

eiφ sin θ2

],

〈sz〉 =[cos θ2 eiφ sin θ

2

] [1 00 −1

] [cos θ2

eiφ sin θ2

]= cos2 θ

2− sin2 θ

2= cos θ.

Likewise, the x, y-spin operators are:

sx =

[0 11 0

]sy =

[0 −ii 0

].

And, for a general direction, given by a unit vector ~r:

~r = sin θ cosφ~x+ sin θ sinφ~y + cos θ~z,

the spin operator s~r is given by:

S~r = ~S · ~r =

[cos θ eiφ sin θ

eiφ sin θ − cos θ

],

where ~S = (sx, sy, sz).

23.1.2 What Results do we Predict?

We orient the detectors in a direction r in the xz plane offset by an angle θ′.We define the operator

sr =

[cos θ′ sin θ′

sin θ′ − cos θ′

].

Then:

〈srsz〉 =[cos θ2 sin θ

2

] [cos θ′ sin θ′

sin θ′ − cos θ′

] [1 00 −1

] [cos θ2sin θ

2

]= cos θ′ cos2 θ

2+ cos θ′ sin2 θ

2.

The particles come out at all angles, so the average of this over all θ is

〈〈srsz〉〉 = cos θ′.

But, 〈〈srsz〉〉 is the mean product of the r, z-spins, which is correlation betweenthe measurements of the two detectors.27 Thus, if the two detectors are pointed

27In the sense that each detector outputs spin measurements +1,−1

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Figure 15: Correlation Prediction of Local State Model.

in directions ~a,~b, separated by an angle θ′, we expect the average correlation ofour measurements to be

p = − cos θ′ = −~a ·~b.

Suppose the particles had some local state. We’ll start with a brain-deadmodel: the electron comes in with some spin, and the positron comes in withthe opposite spin. Each detector records spin +1 if the particle’s spin is in some180 angle, and −1 if the particle’s spin is in the opposite 180 angle. We canshow that under this model, the correlation p is linear in θ′. This is pictured inFigure 15.

This is measureably different from the quantum prediction that p = − cos θ′.

Joke. “Who believes this? (few people raise their hands) Well, you’ve boughtall this shit so far!” - Ashoori

23.1.3 Bell’s Inequality

Suppose we have three measureable properties A,B,C; particles either haveor don’t have each of these properties. Let N(A,B) be the particles that haveproperty A but not B, and likewise for N(B,C), N(C,A). By staring at a venndiagram, we get:

N(A,B) +N(B,C) ≥ N(A,C).

For example, in a group of students, A,B,C can be “wearing green,” “heightat least 5 foot 6,” “having blue eyes.”

But, at the quantum level, the above inequality fails! For example, if A,B,Care the spins of particles along axes ~a,~b,~c, then the above inequality, translatedinto quantum correlations, is

1 + p(~b,~c) ≥ |p(~a,~b)− p(~a,~c)| ⇒ 1−~b · ~c ≥ | − ~a ·~b+ ~a · ~c|.

This fails when ~a,~b are orthogonal, and ~c is 45 between ~a,~b, because

1− 1√2≥∣∣∣∣0 +

1√2

∣∣∣∣ .Joke. “This means that, when you look at whether someone’s wearing green,you turn their eyes from blue to brown!” - Ashoori

This rules out hidden variables.

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23.1.4 Philosophy

We’ve shown is that there’s no local hidden variables before the particle hits thedetector.

One interpretation of this is that maybe everything is preordained – after all,the universe started at a point, so perhaps everything is entangled. Free will, aswe know it, might not exist. We just don’t know.

23.2 Quantum Transport in 1-D

We’ve talked about the infinite square well, with width L and angular frequenciesk = nπ

L .

We can glue the ends of the infinite square well together – this models,for example, electrons in a circular wire. Our solutions are periodic functionseikx with angular frequency k = 2π

L n. In k-space, the solutions correspond tomultiples of 2π

L . When we fill the wire with electrons, they fill these statesin increasing order of energy (each state is filled by a spin-up electron and aspin-down electron): ± 2π

L ,±4πL ,±

6πL , . . . .

Suppose we’ve filled up to states ±kF . The total energy of the electronswe’ve added is

EF =~2k2

F

2m.

Now we can ask, how dense are the states EF ? Well, we have

kF =N

4

2π

L=nπ

2,

where n = NL is the electron density. Then,

EF =~2k2

F

2m=

~2n2π2

8m⇒ n =

√8EFm

~π.

So, the density of the states is

dn

dEF=

d

dEF

√8EFm

~π=

1

2√EF

√8m

~π.

Finally, we can ask, how fast are the electrons going? The velocity is

Vg =dω

dk=

1

~dE

dk=

~kFm

.

This is called the Fermi velocity.

Now, let’s put a voltage V across this wire. We think of this voltage as apoint on the wire, where there is difference in EF from one side of the point tothe other. The potential difference produced across this point is eV .

What is the current across this barrier? The number of states that haveenough energy to cross the barrier is

eV · 2m

~2kFπ.

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Of these, half of these states travel in the appropriate direction; moreover, thevelocity of the electrons os ~kF

m , and each has charge e. This yields a current of

I = eV · 2m

~2kFπ· 1

2· ~kFm· e =

e2V

~π=

2e2

hV.

The quantity e2

h is called the quantum of conductance. The 2 is becausethere are two spin-states.

This is something we can experimentally measure! We can put a gate in awire narrow enough to allow only one quantum state through, and then use abattery to send a current through this gate. We’ll actually see the current go

up in steps of 2e2hV as we increase the energy.

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