Physics IV - Script of the Lecture - qudev.phys.ethz.ch · 3 Solutions to Schr˜odinger’s...

Physics IV - Script of the Lecture

Prof. Simon Lilly

October 21, 2006

Notes from:

Raphael Honegger

$Id: physics4.tex 1507 2006-10-21 15:02:55Z charon $

i

Contents

0 Introduction 1

1 The Limits of Classical Physics & Wave-Particle Duality 21.1 Classical Concepts . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Empirical Problems with classical Physics . . . . . . . . . . . . . 2

1.2.1 Atomic structure & atomic spectra . . . . . . . . . . . . . 21.2.2 Photo-electric effect . . . . . . . . . . . . . . . . . . . . . 31.2.3 Waves behaving as particles . . . . . . . . . . . . . . . . . 31.2.4 Particles acting as waves . . . . . . . . . . . . . . . . . . . 31.2.5 Summay - new ideas . . . . . . . . . . . . . . . . . . . . . 3

1.3 Reminder Interference & Diffraction . . . . . . . . . . . . . . . . 31.3.1 Young’s slits interference . . . . . . . . . . . . . . . . . . 31.3.2 Diffraction . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.3 Fourier Optics . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Thermal radiation and Planck’s constant h . . . . . . . . . . . . 51.4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 51.4.2 Stefan-Bolzmann Law . . . . . . . . . . . . . . . . . . . . 61.4.3 Wien’s Displacement Law . . . . . . . . . . . . . . . . . . 71.4.4 Oscillators . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4.5 Planck’s hypothesis (1900) . . . . . . . . . . . . . . . . . 8

1.5 De Broglie waves & Plank’s constant . . . . . . . . . . . . . . . . 91.6 Measurements & Plank’s constant . . . . . . . . . . . . . . . . . 91.7 Wave paricle duality & quantum reality . . . . . . . . . . . . . . 10

2 The Schrodinger Equation 112.1 Review of waves . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2.1.1 Sinusoided waves . . . . . . . . . . . . . . . . . . . . . . . 112.1.2 Superposition of waves . . . . . . . . . . . . . . . . . . . . 112.1.3 Dispersion relations . . . . . . . . . . . . . . . . . . . . . 11

2.2 Particle wave equation . . . . . . . . . . . . . . . . . . . . . . . . 122.3 A Particle in a potential energy field V (r) . . . . . . . . . . . . . 132.4 The meaning of Ψ(x, t) . . . . . . . . . . . . . . . . . . . . . . . . 132.5 Expectation values and operators . . . . . . . . . . . . . . . . . . 14

3 Solutions to Schrodinger’s Equation 173.1 Separable solutions of definite energy . . . . . . . . . . . . . . . . 173.2 Example: Particle in a box . . . . . . . . . . . . . . . . . . . . . 183.3 States of uncertain E . . . . . . . . . . . . . . . . . . . . . . . . . 193.4 Finite potential wells . . . . . . . . . . . . . . . . . . . . . . . . . 203.5 Barrier penetration = “quantum tunelling” . . . . . . . . . . . . 23

4 The Harmonic Oscillator 264.1 Classical case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.2 Quantum oscillator . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2.1 Starionary states . . . . . . . . . . . . . . . . . . . . . . . 264.2.2 Non-stationary states of uncertain energy . . . . . . . . . 284.2.3 3-dimensional oscillater (degenacy) . . . . . . . . . . . . . 29

5 Observables & Operators - Heisenberg Uncertainty Principle 305.1 Operators & eigenfunctions . . . . . . . . . . . . . . . . . . . . . 305.2 Eigenfunctions for position and momentum . . . . . . . . . . . . 305.3 Compatible observables . . . . . . . . . . . . . . . . . . . . . . . 315.4 Heisenberg uncertainty principle . . . . . . . . . . . . . . . . . . 325.5 Compatibility with H . . . . . . . . . . . . . . . . . . . . . . . . 345.6 Orbital angular momentum . . . . . . . . . . . . . . . . . . . . . 355.7 Angular momentum eigenfunctions . . . . . . . . . . . . . . . . . 355.8 Stern-Gerlach experiment . . . . . . . . . . . . . . . . . . . . . . 375.9 Spin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

ii

6 The Hydrogen Atom 386.1 Motion in “central potentials” . . . . . . . . . . . . . . . . . . . . 386.2 Solutions of Schrodingers equation in a central potential . . . . . 386.3 Hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . 396.4 Zeeman effect - Perturbing the Hamiltonian H . . . . . . . . . . 416.5 Time-dependent perurbations - Radiative transitions . . . . . . . 426.6 Some (small) complicating effects . . . . . . . . . . . . . . . . . . 44

6.6.1 The reduced mass effect . . . . . . . . . . . . . . . . . . . 446.6.2 Spin-orbit coupling . . . . . . . . . . . . . . . . . . . . . . 446.6.3 Relativist effects . . . . . . . . . . . . . . . . . . . . . . . 45

7 Quantum mechanics of identical particles 467.1 Wave functions for identical particles . . . . . . . . . . . . . . . . 467.2 Exchange symmetry with spin . . . . . . . . . . . . . . . . . . . . 47

8 More Complicated Atoms 49

Stichwortverzeichnis 50

iii

0 Introduction 05.04.2006

0 Introduction

Webpage: http://www.exp-astro.phys.ethz.ch/PhysikIV

Topics:

1. Failures of Classical Physics; particles and waves

2. The Schrodinger Equation

3. Position and momentum

4. Energy and Time

5. Particles in potentials

6. Harmonic oscillators

7. Observables and Operators

8. Angular Momentum

9. The Hydrogen Atom

10. Bosons and Fermions

11. Atoms

Books:

• Tipler & Mosca: Physik/Physics for Scientists and Engineers

• Phillips: Introduction to Quantum MechanicsThis will be the book, the professor follows more or less.

• Cohen-Tannoudji, Diu, Laloe: Quantum Mechanics vol 1,2 (E/D/F)A classic book

• Messiah: Quantum Mechanics vol 1,2 (E/D)A classic book

• Schwabl: Quantenmechanik

1

1 The Limits of Classical Physics & Wave-Particle Duality 05.04.2006

1 The Limits of Classical Physics & Wave-Par-ticle Duality

1.1 Classical Concepts

• We have particles

– A particle is a discrete entity

– It has a precise and well defined position and momentum

– It’s obeying Newton’s Laws of Motion

– In principle, the Physics of Particles is completely deterministic

• We also have the electromagnetic fields and waves

– The electromagnetic fields pervade all space

– They’re governed by Maxwell’s equations

– We have wavelike disturbances which propagate through space

The fields and the particles interact via the Lorentz forces

F = q(E + v × B)

and we’ve got the equivalent for gravity.

In 1900-1930 there were two revolutions in Physics: Relativity and QuantumMechanics. The upshot of this was, that Classical Physics is only an approxi-mation and it isn’t valid on everyday scales (Quantum Mechanics) und speeds(Relativity).

1.2 Empirical Problems with classical Physics

1.2.1 Atomic structure & atomic spectra

Atoms consist of positive and neutral nuclei and negative electrons (that can beremoved). In Classical Mechanics we assume, that the electrons do orbit like inthe solar system

£

Fg = GM¯mp

r2Fe =

1

4πε0

q+e−

r2

The orbits are ellipses (sun at a focus) with

r =p

1 − e cos θE =

1

2mv2 −G

Mm

r= −G Mm

2a

P 2 = a3 4π2

G(M +m)

where P is the period. As we know, we can get any energy and any period.

Now, we’ve got two problems:

• Atoms gain or loose energy by absorbing or emitting light at particularfrequencies

• The electrons are continuously accelerated, therefore they should looseenergy through electromagnetic radiation

dE

dt=

2

3

q2a2

4πε0c3=

q2a2

6πε0c3

This means, the orbits should decay and the electrons should spiral intothe nucleus, which doesn’t happen.

2


1.2.2 Photo-electric effect

If we have shining light on a metal surface, there are electrons ejected

£

The incident power per unit area is

P = ε0E2c

The Ejection of the electrons should not depend on ν, but only on E2. What

we see is different. For example, Mg requires ν > 8.9 · 1014Hz.

1.2.3 Waves behaving as particles

Consider an electromagnetic radiation with very high frequency ν like by X-raysand wavelength λ

£

The change of K · (1 − cos θ) in λ and ν of an X-ray radiation is a signature ofparticle-like behaviour.

1.2.4 Particles acting as waves

Interference (constructive or destructive adding of waves) is fundamentally awave phenomena

£

In about 1927, it was seen, that particles also show diffraction patterns. It wasshown, that 54eV acted like a wave with λ = 0.17nm and 40keV like one withλ = 0.006nm. In 1999 it was even shown, that the same effect occurs with C60

molecules.

1.2.5 Summay - new ideas

• Consider waves as particles, with specific energy and momentum

• Consider particles as waves, with specific wavelength and frequency

• The enrgy loss/gain is restricted, so ∆E is not continuous

• We’ve got a propagation as waves and an exchange of energy as particles

1.3 Reminder Interference & Diffraction

The Huygens-Fresnel Principle says: “Every point on the wavefront acts as thesource of a secondary spherical wavefront (wavelet), with same frequence andspeed. The amplitude of the field at any point is the superposition of theirwavelets taking into account amplitude and phase”.

1.3.1 Young’s slits interference

Assume that the slits are very narrow and one-dimensional

£

We have constructive interference if

d sin θ = nλ n ∈ Z

and destructive interference if

d sin θ = (n+ 1/2)λ n ∈ Z

3


For a general θ, the phase difference for two slits is

δ = 2πd sin θ

λ

Place the screen far from the slits (i.e. at a distance LÀ d). Then, the positionon the screen is

y = L tan θ ∼ L sin θ

This approximation is good for small θ and the maxima and minima are equallyspaced

ymax =nλ

dL θmax =

nλ

d

ymin =

(n+

1

2

)λ

dL θmin =

(n+

1

2

)λ

d

For the wave amplitude we get

E = E0 (sin(ωt) + sin (ωt+ δ))

= 2E0 cos

(δ

2

)sin

(ωt+

δ

2

)

The intensity is proportional to E2∝ cos2 (δ/2)

£

What about interference from multiple slits? We need to sum wavelets from allslits, e.g. using a “phasor diagram”

£

The maximum always has the “vectors” aligned, i.e. δ = n2π. We will havesecondary maxima when

mδ = (2n+ 1)π

where m is the number of slits and mδ the total angle rotated in the phasordiagram (in this case, the end of the phasor-sum is at the other side of thecircle).

£

If we have an infinite number of slits, the interference pattern looks like this

£

1.3.2 Diffraction

Consider an interference from a finite sized aperture. Take for example lightpassing through an aperture of finite width and ∞ length

£

We can consider this composed of N finite elements for small θ. The phasedifference between adjacent apertures is

δ = 2πθ

λ

a

N

The phasor diagram will become a circle as N goes to ∞

£

4


For the angle we get

Nδ =2πaθ

λ⇒ Eθ =

E0λ

πaθsin

(πaθ

λ

)

Now we wet φ = πaθλ to get

Eθ =E0

φsinφ

The intensity we then can write as

I0sin2 φ

φ2

This leads us to the diffraction pattern

£

For aÀ λ, we get that θmin → 0 and for a ∼ λ, that θmin → 1.

Now we can construct the actual pattern from 2 finite slits

£

In two dimensions (e.g. by a rectangular aperture), we get something like

£

and the intensity, we can write as

I = I0sin2 φaφ2a

sin2 φbφ2b

φa =πaθaλ

φb =πaθbλ

1.3.3 Fourier Optics

Consider a general aperture

£

Then we have

E(θ, φ) =

∫ ∫dxdy · E0T (x, y)e−i

2πλ (θx+φy)

where T (x, y) is the transmission function of the aperture. Set u = λ−1x,v = λ−1y to get

E(θ, φ) = E0λ2

∫ ∫dudv · T ′(u, v)e−2πi(uθ+vφ)

The diffraction/interference pattern is the Fourier Transform of the aperturescaled by λ and vice versa.

1.4 Thermal radiation and Planck’s constant h

1.4.1 Introduction

All bodies emit electro magnetic radiation because of their temperature. Wecould write

dE(λ,Temperature,material)

dtdλ = rate of energy loss per unit area

per unit time in the interval λ→λ+dλ

Different materials absorb different fractions of incident electro magnetic radi-ation, described by the absorption coefficiant a and the emission coefficiant e.It was found empirically by Ritchie (1833), that the emission properties are thesame as the absorption properties for all materials, i.e.

e

a= const

It was shown with an experiment like this

5


£

where for the emissive power e and die absorptive power a, we have

eαaβ = eβaα ⇒ eαaα

=eβaβ

We’ve got a maximum e, when a = 1, as by a body that absorbs all radiationfalling on it (black body).

Consider a uniform temperature enclosure in equilibrium at a temperature T

£

It turns out, that the energy density u with [u] = Jm−3 depends only on thetemperature T .

To proof this, we imagine a cavity like

£

If we have uA(T ) > uB(T ), then energy flows from A to B, when the shutter isopened.⇒ When we close the shutter, TA has gone down and TB has gone, which is aviolation of the second Law of Thermodynamics.

The radiation escaping from a small hole in a cavity is the same as from anideal black body. But what is the form of u(T ) and more specifically, what isthe spectrum of thermal radiation u(λ, T )dλ, which is the energy density forthe interval λ→ λ+ dλ as f(T )?

1.4.2 Stefan-Bolzmann Law

We want to know, what’s the total energy emitted by a Black Body

£

wheredE

dt=

Acu(T )

4

(see problem sheet 1). The answer is

dE

dt= σAT 4

which Stefan got empirically and Boltzmann proves by theory. σ we call theStefan-Boltzmann constant. Boltzmann did a Thermodynamic argument toprove it: Imagine a reflecting enclosure, containing some thermal radiation.It turns out, that the radiation pressure is equal to 1

3 u, where the 3 comes fromthe three dimensions.

£

The work done is equal to

∆v(p(T ) − p(T − dT )) = ∆vdp =1

3∆v du

The efficiency is1

3

∆v dm43 ∆v u

=dT

T

⇒ dT

T=

1

4

du

u→ u ∝ T 4

The energy density u = aT 4.The emission from the black body is

ac

4T 4 = σT 4

per unit area.

6


1.4.3 Wien’s Displacement Law

What can we say about the spectrum u(λ, T )dλ. Consider again a cavity under-going a reversible expansion. In this case, the wavelengths λ are proportionalto the cavity size.

λ1

λ2=

`1`2

dλ1

dλ2=

`1`2

Using exactly the same argument as before, we get

u(λ1)

T 41

dλ1 =u(λ2)

T 42

dλ2

For an adiabatic process, we have

d[V u(λ) dλ] + p(λ) dV = 0 p(λ) =1

3u(λ) dλ

where the first term is the change in energy and the second is the work.

d[3V p(λ)] + p(λ) dV = 0

3V dp(λ) + 4p(λ) dV = 0

→ p(λ1)V4/31 = p(λ2)V

4/32 (1.1)

→ u(λ1)`41 dλ1 = u(λ2)`

42 dλ2

→ u(λ1)λ51 = u(λ2)λ

52

Back to Stefan’s Lawp(λ1)

T 41

=p(λ2)

T 42

Now we eliminate p(λ), using equation 1.1 to get

T 41 V

4/31 = T 4

2 V4/32 ⇒ T1`1 = T2`2 ⇒ T1λ1 = T2λ2

We haveu(λ1)λ

51

f(λ1, T1)=

u(λ2)λ52

f(λ2, T2)

⇒ The general result for u(λ, T ) is

u(λ, T ) =A

λ5f(λT )

We will also be interested in u(ν) dν which is the energy per unit volume withν → ν + dν and

u(ν, T ) = Bν3g

(T

ν

)νmax ∝ T λmax ∝

1

T

which is the general form of Wien’s Displacement Law.

£

1.4.4 Oscillators

Imagina a cavity surrounded by “oscillators”, each with two degrees of freedom

x = x0 sinωt

x = ωx0 cosωt

x = −ω2x0 sinωt

The energy of it will be constant as it oscillates and equal to

ε =1

2mω2x2

0

7


The emission of electro magnetic waves from an oscillator is instantaneouslyequal to

dE

dt=

q2a2

6πε0c3

(see Physics III). Therefore, the time average of this is

dE

dt=

1

2

q2ω4x20

6πε0c3

where we just put in a = x from above and

dE

dt=

q2ω2

6πε0c3mε

For the absorption we get the analogous result (without proof)

dE

dt=

πq2u(ν)

4πε03m

where ω = 2πν. In equilibrium, the absorption and the emission are equal andso we can then write

q2ω2ε

6πε0mc3=

πq2u(ν)

4πε03m⇒ u(ν) =

8πν2

c3ε

where ε is the energy of the oscillator. If we look at all oscillators on average,we get

u(ν) =8πν2

c3ε

where ε is the average energy of all oscillators with frequency ν. So we want toknow what’s the average energy as a function of T . We would expect it to beε = kT to get

u(ν) =8πν2

c3kT

which is called the Rayleigh Jeans formula. It satisfies

u(ν) = ν3g

(T

ν

)

but it can’t be right, because u → ∞ as , ν → ∞. This is known as theultraviolet catastrophe. If we look, where ε = kT comes from, we see that theanalysis to get it assumes, that we have an infinite range of possible energies,weighted by the Boltzmann factor.

dn(ε) =N

kte−ε/kT dε ⇒ ε =

∫∞

0ε NkT e

−ε/kT dε∫∞

0NkT e

−ε/kT dε= kT

1.4.5 Planck’s hypothesis (1900)

Plank’s idea was to ask, what happened, if we allowed only certain energies

εj = jε0 j = γ = 0, 1, 2, ...

Exactly as before, we had nj = Ae−εj/kT , but

N = A+Ae−ε0/kT +Ae−2ε0/kT +Ae−3ε0/kT + ...

=

∞∑

j=0

Ae−εj/kT =A

1 − e−ε0/kT

E = 0 +Aε0e−ε0/kT +A2ε0e

−2ε0/kT + ...

= A∞∑

j=0

εje−εj/kT =

Aε0e−ε0/kT

(1 − e−ε0/kT )2

8


It follows

ε =E

N=

ε0e−ε0/kT

(1 − e−ε0/kT

) =ε0(

eε0/kT − 1)

and

u(ν) dν =8πν2

3c3ε0(

eε0/kT − 1)

This works of u(ν) = Bν3g(Tν

)if ε0 ∝ ν, i.e.

ε0 = hν h = 6.625 · 10−34Js

where h is Plank’s constant. This then gives the final answer for black bodyradiation density

u(ν) dν =8πhν3

3c31(

ehν/kT − 1) dν

which is called Plank’s formula. It reduces to

8πν2

3c3kT

for hν ¿ kT . At the end we’ve got rid of the ultraviolet catastrophe by cuttingout the continuity of the energy.

1905 Einstein quantised the radiation in a cavity and called this wave packetsphotons with

E = hν

This explained for example the photo-electric effect.

1.5 De Broglie waves & Plank’s constant

The wavelike properties of matter were proposed by de Broglie in 1923 with

λ =h

p

where p is the momentum. In terms of energy we can rewrite this with

ε2 = m20c

4 + p2c2 ⇒ λ =hc√

ε−m0c2√ε+m0c2

In a highly relativistic case, i.e. εÀ m0c2, we can write

λ =hc

ε

which is identical to photons (ε = hν). In the non-relativistic case, i.e for

ε = m0c2 + E, where E = p2

2m0is the kinetic energy, we get

λ =h√

2m0E

For example, an energy of 1.5eV is a λ = 1nm and an energy of 15keV is aλ = 0.01nm.

1.6 Measurements & Plank’s constant

This section is still largely classical and shouldn’t be translated into QuantumMechanical concepts.

We go through the “thought experiment” of Heisenbergs microscope

£

9


We know, that the defraction from the aperture d is of the order ∼ λd . The

uncertainty ∆x is going to be given by

∆x ∼ λ

dz ∼ λ

α

The scattered photon from the electron must have entered the microscope andtherefore, the photon has a momentum with

|px| <h

λsinα <

hα

λ

For the electron, the uncertainty ∆p in the momentum of the electron must beof order ∼ hα

λ . We get to the interesting fact

∆x∆p ≥ h

The correct Quantum Mechanical formulation of this is

∆x∆p ≥ h

4π

1.7 Wave paricle duality & quantum reality

Lets do a little thought experiment

£

The electrons hit the detector with a statistical distribution, so we observea diffraction pattern in the locations of the detected electrons. This implieswave properties through the slits. We could ask, whether we can tell whichslit the electron passed through and indeed we can quite easily, but then, thediffraction pattern dissapears! For example, we could measure the momentumof the electron by the recoil of the screen. If you think about it, near to thecenter of the screen, we would expect

∆pscreen ∼ ped

D∼ h

λe

d

D

From the previous section we know that however we try to measure the momen-tum, we have an uncertainty in position that is

∆x ∼ h

∆p∼ Dλe

d

which is exactly the separation of the interference fringes. So, when we knowwhich slit the electron passed through, we no longer observe wavelike properties.When there is no possibility of knowing this, we observe wavelike behavior. Wecould say: “The wave is covert”. “The act of measurement brings into existencea property”.

10

2 The Schrodinger Equation 26.04.2006

2 The Schrodinger Equation

2.1 Review of waves

2.1.1 Sinusoided waves

The simplest wave with definite wavelength λ, wavenumber k = 2πλ and definite

period τ , angular frequency ω = 2πτ and frequency ν = 1

τ are sinusoided wavesas

Ψ(x, t) = A cos (kx− ωt)

The maxima move at speed ωk . The most general sinusoided wave is

Ψ(x, t) = A cos (kx− ωt) +B sin (kx− ωt) = Aei(kx−ωt)

Note, that in Classical Physics we normally take the real part of the complexwave. In Quantum Mechanics you always consider the complex function.

2.1.2 Superposition of waves

Standing wave: A standing wave we get from a superposition like

Ψ(x, t) = A cos (kx− ωt) +A cos (kx+ ωt)

= 2A cos (kx) cos (ωt)

So we can separate the spacial and the temporal parts.

Wave packets: A wave packet is a superposition with similar k

Ψ(x, t) =

k+∆k∫

k−∆k

A cos (k′x− ω′t) dk′

At t = 0, this simplifies to

Ψ(x, 0) =

k+∆k∫

k−∆k

A cos (k′x) dk′

= cos (kx)S(x)

where

S(x) = 2A∆ksin (∆kx)

∆kx

£

2.1.3 Dispersion relations

Non-dispersive waves are governed by the classical wave-equation

∂2Ψ

∂x2− 1

c2∂2Ψ

∂t2= 0

In three dimensions, this is

∇2Ψ − 1

c2∂2Ψ

∂t2= 0

The solutions of this equation are

Ψ(x, t) = Aei(kx−ωt)

where ω2 = c2k2 and where vphase = ωk = c is the speed of a maximum in

the wave. We call it phase velocity. Because of linearity, the superposition ofsolutions is also a solution.

11


Generally, waves are dispersive, that means they have a more complicated ω, krelation as ω

k 6= const. This is due to a more complicated wave equation. Thephase velocity is equal to the speed of individual crests ω

k and the group velocityis the speed of the point of the maximum constrictive interference in a wavepacket vG = dω

dk

£

To see this, consider 2 sinusoided waves with ω1, ω2, k1, k2. We have constructiveinterference, when the phases are the same

(k1xmax − ω1t) = (k2xmax − ω2t) ⇒ xmax =ω1 − ω2

k1 − k2t =

∆ω

∆kt

e.g. for long wavelength water waves in deep water, we have

ω =√gk vp =

ω

k=

√g

kvG =

dω

dk=

1

2

√g

k=

1

2vp

2.2 Particle wave equation

We want a wave equation for particles by looking at the dispersion relation(don’t worry yet what the wave is). With de Broglie, we had

λ =h

pk =

2π

λ⇒ p =

h

2πk =: ~k

A particle with uncertain p is associated with a wave packet with a range of k.

∆p ∼ ~∆k

The “length” of the wave packet

∆x ∼ 2π

∆k

Lets set the group velocity vG equal to the velocity of the particle

⇒ dω

dk=

p

m=

~k

m

We integrate to find

ω =~k2

2m+ (const)

One could ask, what wave equation this dispersion relation gives

i~∂Ψ

∂t= − ~

2

2m

∂2Ψ

∂x2

or in three dimensions

i~∂Ψ

∂t= − ~

2

2m∇2Ψ

This is the Schrodinger Equation for a “free particle”. Lets take a wave-functionΨ(x, t) = Aei(kx−ωt) and check

i~(−iω) = − ~2

2m− k2 ⇒ ~ω =

~2k2

2m

Note:

• Such wave equations have always complex wave solutions.

• The Schrodinger-equation is linear, that means that any superposition ofsolutions is also a solution.

12


The most general solution then is

Ψ(x, t) =

∞∫

−∞

A(k′)ei(k′x−ωt) dk′

provided we set

~ω′ =~

2k′ 2

2m

A narrow range of k leads to a wave packet moving with speed

dω

dk=

~k

m

The uncertainty in p and in x (∆p and ∆x) are related by

∆p ∼ ~∆k ∆x ∼ 2π

∆k

2.3 A Particle in a potential energy field V (r)

If a particle has a potential energy V (r) as well as a kinetic energy, we modifythe Schrodinger-Equation like

i~dΨ

dt=

(− ~

2

2m∇2 + V (r)

)Ψ

We could try a solution of the form Ψ(x, t) = Aei(kx−ωt) with V (x) = V0, to get

E = ~ω =~

2k2

2m+ V0

2.4 The meaning of Ψ(x, t)

We’ll first do a digression on probabilities and probability distributions.

Consider n discrete outcomes of an experiment xn, each of which have a prob-ability of occuring pn. We know

∑pn = 1

We define the expectation value of x as the average value, if the experiment isrepeated many times

〈x〉 =∑

n

xnpn

With the standard deviation in x, ∆x, we get the variance (∆x)2 and

(∆x)2

=∑

n

(xn − 〈x〉)2 pn =⟨x2⟩− 〈x〉2

Consider a continuous variable x, then p(x)dx is the probability of x between xand x+ dx. Like above, we then have

∫

all x

p(x) dx = 1

and futhermore

〈x〉 =

∫

allx

xp(x) dx⟨x2⟩

=

∫

allx

x2p(x) dx

Now, recall classical waves in the two-slit experiment

13


£

By superposition, we have Ψ = ΨA + ΨB and the intensity is proportional toΨ2. By a Quantum wave function, we also have Ψ = ΨA + ΨB and by analogy,we construct |Ψ|2 = Ψ∗Ψ. But what does this quantity mean?

£

Born’s interpretation of Ψ(x, t) was the following

Ψ∗(r, t)Ψ(r, t) dV or Ψ∗(x, t)Ψ(x, t) dx

is the probability of finding a particle in the volume dV (or within the distancedx) at the time t, when we do some suitable experiment.

As we go on, we can sometimes state some rules about how we have to workwith our wave-functions. Here’s the first one.

Rule 1: We must always normalize Ψ(R, t), such that

∫

all space

Ψ∗ΨdV = 1

¤

If we haveΨ(x, t) =

∑

j

Ajei(kjx−ωjt) pj = ~kj

then we get a probability of measuring the momentum pi which is proportionalto |Ai|2. We can gereralize this with a continuous distribution

Ψ(x, t) =

∞∫

−∞

A(k)ei(kx−ωt) dk

which effectively is a Fourier Transform. We can also construct a Fourier trans-form pair as follows

Ψ(x, t) =1√2π~

∞∫

−∞

Ψ(p, t)eipx

~ dp

Ψ(p, t) =1√2π~

∞∫

−∞

Ψ(x, t)e−ipx

~ dx

We know∞∫

−∞

Ψ∗Ψdx = 1 ⇔∞∫

−∞

Ψ∗Ψ dp = 1

where Ψ∗Ψ is the probability amplitude for x and Ψ∗Ψ the probability amplitudefor p.

2.5 Expectation values and operators

Using the equations from above, we can write down in one dimension

〈x〉 =

∫xΨ∗Ψdx =

∫Ψ∗xΨdx

〈p〉 =

∫p Ψ∗Ψ dp =

∫Ψ∗pΨdp

14


We can also calculate 〈p〉 in the following way

〈p〉 =

∫

allx

Ψ∗(x, t) (−i~)∂

∂x(Ψ(x, t)) dx

which we prove with

Ψ(x, t) =1√2π~

∫Ψ(p, t)e

ipx~ dp

We get

−i~ ∂

∂xΨ(x, t) = −i~ ∂

∂x

1√

2π~

∫Ψ(p, t)e

ipx~ dp

=1√2π~

∫pΨ(p, t)e

ipx~ dp

and therefore it follows

〈p〉 =

∫Ψ∗(x, t)

(−i~ ∂

∂xΨ(x, t)

)dx

=

∫Ψ∗(x, t)

1√

2π~

∫pΨ(p, t)e

ipx~ dp

dx

=

∫ 1√

2π~

∫Ψ∗(x, t)e

ipx~ dx

pΨ(p, t) dp

=

∫Ψ∗pΨ dp

So, we can write

〈x〉 =

∫Ψ∗(x, t)xΨ(x, t) dx

〈p〉 =

∫Ψ∗(x, t)

(−i~ ∂

∂x

)(Ψ(x, t)) dx

Now we introduce the concept of operators. The operator for x is x = x andthe operator for p is p = −i~ ∂

∂x . It turns out, that any obervable quantity canbe represented by an operator. Furthermore, the operators for x2 and p2 aregiven by

x2 = x2 = (x)2

p2 = −~2 ∂2

∂x2= (p)

2

In three dimensions, we have

r = r p = −i~∇

Rule 2: We can apply operators to get any expectation value

〈ξ〉 =

∞∫

−∞

Ψ∗ξΨdx

or with dV in three dimension. ¤

15


Lets look at the operator for the energy E. Classically, we can write it as

E =p2

2m+ V (r)

and we would expect the energy operator as

E =p 2

2m+ V (r) = − ~

2

2m∇2 + V (r) =: H

where H is called the Hamiltonian operator. Now

〈E〉 =

∫Ψ∗HΨdV

Now going back to Schrodingers-Equation, we see that

i~∂Ψ

∂t=

(− ~

2

2m∇2 + V (r)

)Ψ = HΨ

16

3 Solutions to Schrodinger’s Equation 08.05.2006

3 Solutions to Schrodinger’s Equation

3.1 Separable solutions of definite energy

We had the Schrodinger-Equation for a particle in a potential

i~∂Ψ

∂t=

(− ~

2

2m∇2 + V (r)

)Ψ

Now, we’re looking for solutions

Ψ(r, t) = ψ(r)T (t)

We put it in

i~ψ(r)dT

dt=

(− ~

2

2m∇2ψ + V (r)ψ

)T

and separate the variables

i~

T

dT

dt=

1

ψ

(− ~

2

2m∇2ψ + V (r)ψ

)

This equation is true for all t and for all r, so the terms have to be equal to aconstant K. It follows

dT

dt= − i

~KT ⇒ T (t) = Ae−

iKt~

and on the other hand(− ~

2

2m∇2 + V (r)

)ψ = Kψ

So, we’re looking for the eigenfunctions ψn of H associated with the eigenvaluesKn, i.e.

Hψn = Knψn

With Ψn(r, t) = ψn(r)Ae−iKt

~ we then get

〈E〉 =

∫Ψ∗nHΨn d3r =

∫ψ∗nHψn d3r = Kn

So, Kn is equal to the energy of the system and we can write

Hψ(r) = Eψ(r)

which is called the time-independent Schrodinger equation. We also get

⟨E2⟩

=

∫Ψ∗H2Ψd3r = E2

and for the energy uncertainty, we get from before

∆E =

√〈E2〉 − 〈E〉2 = 0

So, if the wave function of a system is an eigenfunction of H, the energy of thesystem is precisely determined and is equal to to eigenvalue En associated tothe eigenfunction. If we make a measurement to determine the energy E, Ψ be-comes the eigenfunction associated with the actual value measured. This sort ofstrange idea is called the collapse of the wave function during the measurement.

Note: The time part T is

T (t) = Ae−iEnt

~ En = hνn = ~ωn

17


3.2 Example: Particle in a box

Consider the potential

V (x) =

{0, 0 < x < a∞, elsewhere

0 a x

We had the 1-dimensional Schrodinger-Equation

i~∂Ψ

∂t=

(− ~

2

2m

d2

dx2+ V (r)

)Ψ

Look for separable solutions of it as

Ψ(x, t) = ψ(x)e−iEnt

~

so we get (− ~

2

2m

d2

dx2+ V (x)

)ψn(x) = Enψ(x)

Inside the box, V = 0 and with En = ~2

2mk2n we get the equation

∂2ψ

∂x2= −k2

nψ

The boundary conditions are ψ(0) = 0, ψ(a) = 0, so that ψ is continuous. Thenthe solutions are

ψn = A sin knx kn =nπ

a

Now, for the n-th energy-state we get

En =n2π2

a2

~2

2mn = 1, 2, 3, ...

and we can writeΨn = A sin

(nπxa

)e−

iEnt~

We can draw out these energy states

n = 2

E n = 3

n = 4

n = 1

Note: The distance En+1 − En goes up as we increase n, but

En+1 − EnE

≈ 2

n

18


goes down as n and therefore the energy E gets bigger. The minimum E 6= 0we have for n = 1, so

E1 =~

2π2

2m

1

a2

For 3-dimensional potential wells

c

a b

Its easy to show, that

ψ(x, y, z) = A sin(nxπx

a

)sin(nyπy

b

)sin(nzπz

c

)

and of course then the energy of this system is

E(nx, ny, nz) =~

2π2

2m

(n2x

a2+n2y

b2+n2z

c2

)

We can see, that we have more complex energy-states, which lead to somethinglike

£

In dependency of a, b, c we can find equal energy-states E for different nx, ny, nz.We call this degeneracy.

3.3 States of uncertain E

We represent more general states by a sum over the eigenstates

Ψ(r, t) =∞∑

n=1

cnψn(r)e−iEnt

~

This is, because the eigenfunctions of H form a complex orthonormal set ofbasis functions. That is

∫ψ∗mψn d3r =

{1, m = n0, m 6= n

It is again easy to prove that

cn =

∫Ψ∗(r, 0)ψn(r) d3r =

∫ψ∗n(r)Ψ(r, 0) d3r

and then〈E〉 =

∑

n

|cn|2En⟨E2⟩

=∑

n

|cn|2E2n

If the Quantum particle is represented by this general Ψ, that satisfies theconditions we’ve set, then the result of an experiment to measure E will beon En with probability |cn|2. Remember, that the eigenfunctions representquantum states, with

• precicely defined energy E

• observable properties that are time independent, like the position proba-bility of the n-th quantum state

Ψ∗nΨn = ψ∗

neiEnt

~ ψne−

iEnt~ 6= f(t)

19


Example 3.1: We want to go through the example of a simple superposition

Ψ(r, t) =1√2ψ1e

−iE1t

~ +1√2ψ2e

−iE2t

~

Then we get

〈E〉 =∑

|cn|2En =1

2(E1 + E2)

⟨E2⟩

=1

2

(E2

1 + E22

)

and therefore

∆E =

√〈E2〉 − 〈E〉2 =

1

2|E1 − E2|

Look at Ψ∗Ψ as function of t

Ψ∗Ψ =

(1

2ψ∗

1ψ1 +1

2ψ∗

2ψ2 +1

2ψ∗

1ψ2ei(E1−E2)t

~ +1

2ψ∗

2ψ1e−

i(E1−E2)t~

)

Therefore Ψ∗Ψ oscillates with frequency

ν =E1 − E2

hor ω =

E1 − E2

~

This of course is a non stationary state. The time scale for change is δt ∝ 1∆E ,

where δt∆E ∼ ~. ¤

3.4 Finite potential wells

Consider a V (x) as

V

0

E

∼ unbound

ε = bound energy

0 a−V0

Now we look for states of definite E solving

− ~2

2m

∂2ψ

∂x2+ V (x)ψ = Eψ

where

V (x) :=

∞, x < 0−V0 0 < x < a

0 x > a

Set ψ(x) and ∂ψ∂x to be continuous at x = 0, x = a (except ∂ψ

∂x at x = 0). Lookfirst at the bound states with −V0 < E < 0. For 0 < x < a, we get

∂2ψ

∂x2= −2m

~2(E + V0)ψ = −k2

0ψ

with k20 = 2m

~2 (E + V0). This has the general solution

ψin = C sin (k0x+ γ)

20


where γ is an arbitrary phase. But γ = 0 satisfies ψ = 0 at x = 0, so

ψin = C sin (k0x)

For x > a, we have∂2ψ

∂x2= −2mE

~2ψ = α2ψ

with α2 = − 2mE~2 and we get solutions

ψout = Ae−αx +A′eαx

We can set A′ = 0, so that ψ doesn’t blow up for x→ ∞ (which wouldn’t haveany Physical meaning). Now stitch together the solutions at x = a. To get acontinuous ψ(x), we need

C sin (k0a) = Ae−αa

and for the first derivative

k0C cos (k0a) = −αAe−αa

Dividing these, we getk0 cot (k0a) = −α

About the energy, we know that

E =~

2

2mk20 − V0 = − ~

2

2mα2 V0 =

~2

2mw2

0

where w0 is a measure of depth of the well. Now there are two things required

1) k20 + α2 = w2

0

2) k0 cot (k0a) = −α

k0

α

w0 = 1

w0 = 2

w0 = 3

We have only one solution for k0, α for π2a < ω0 <

3π2a , two for 3π

2a < ω0 <5π2a

and so on.

21


Note:

• The number of solutions depends on the well depth and it goes to ∞ as−V0 → ∞.

• ψ(x) for the bound states has an exponential tail extending into the clas-sically forbidden region. That was essentially because

ψout ∝ e−αx α2 =2mε

~2

We get a long tail, when ε is small and a small tail, when ε is very large.

−V0

E

Now, look at the unbound states E > 0. We write the energy for 0 < x < a andx > a

E =~

2

2mk20 − V0 =

~2

2mk2

Then we write down the solutions

ψin(x) = C sin (k0x)

ψout(x) = D sin (kx− δ)

Now ψ(x) and ψ′(x) shall be continuous at x = a, so

C sin (k0a) = D sin (ka+ δ)

k0C cos (k0a) = kD cos (ka+ δ)

After dividing them, we get

k0 cot (k0a) = k cot (ka+ δ)

and as before

k20 − k2 = V0

2m

~2

At the end, there’s an infinite number of solutions for k and k0, all with own δ.This, if you like, makes sense.

Note: These are solutions for the stationary states. This does make sense forthe unbound particle, because we can view stationary solutions as a standingwave due to a total reflection of particles at x = 0.

Introduce A0 = −C2i , then

ψin = A0e−ik0x −A0e

ik0x

Now we can write

Ψin = A0e−ik0x−

iEt~ −A0e

ik0x−iEt

~ = ψin(x)e−iEt

~

where the first term describes a travelling wave in negative x-direction andthe second one a travelling wave in positive x-direction. For Ψout, introduceA = −D

2ie−iδ. Applying algebra, we get

Ψout = Ae−ikx−iEt

~ −Ae2iδeikx−iEt

~ = ψout(x)e− iEt

~

Here, the first term describes the incident and the second the reflected wave.

22


£

Note: The reflected wave has a phase shift of 2δ. We’d like to know where thiscomes from

It is due to the time delay in traversing x = a to x = 0 and back and therefore,we would expect δ to be a function of the energy δ(E). In practice, the actualparticles will have a range of E (like an uncertain E) and then, there’ll be awave packet

Ψincident =

∞∫

0

c(E′)e−i“

E′t~

+ p′x~

”

dE′ p′ =√

2mE′

and the reflected wave will be

Ψreflected =

∞∫

0

c(E′)e−i“

E′t~

− p′x~

”

e2iδ(E′) dE′

3.5 Barrier penetration = “quantum tunelling”

V = 0x = 0 x = a

V0

Classically, a particle will be reflected by the barrier. We use the same approachas before, so we solve the time-independent Schroedinger-Equation to get

Ψ(x, t) = ψE(x)e−iEt

~ − ~2

2m

d2ψEdx2

+ V (x)ψE = EψE

For x < 0, we have

d2ψEdx2

= −k2ψE E =~

2k2

2m

because V (x) = 0. Therefore, we get the general solution

⇒ ψE(x) = AIeikx +ARe

−ikx

where the first part is the incident component and the socond part the reflectedone.

For 0 < x < a, we get two cases

a) E > V0:d2ψEdx2

= −k20ψE E =

~2

2mk20 + V0

So we get the same solution as above, but with k0 instead of k.

b) E < V0: This is classically forbidden, but here we get

d2ψEdx2

= α2ψE E = −α2~

2

2m+ V0

which has as general solution

→ ψE(x) = Be−αx +B′eαx

23


For x > a, we get the same as for x < 0, means

ψE(x) = AT eikx

where AT can be complex to account for phasing.

Continuity at x = 0 requires

AI +AR = B +B′ ikAI − ikAR = −αB + αB′

Continuity at x = a requires

Be−αa +B′eαa = AT eika − αBe−αa + αB′eαa = ikAT e

ika

We set B′ = 0, assuming that we have a “wide” barrier. So we get at x = 0

2ikAI ≈ B(ik − α)

and at x = a

AT eika ≈ 2α

(α− ik)Be−αa

Now eliminate B to get

⇒ AT eika ≈ 4ik

(α− ik)2 αe

−αaAI

The transmission probability then is

T =

∣∣∣∣ATAI

∣∣∣∣2

≈ 16k2α2

(α2 + k2)2 e

−2αa

where

k2 =2mE

~2α2 =

2m (V0 − E)

~2

Therefore, we can rewrite this as

T ≈ 16E (V0 − E)

V 20

e−2αa

Remember, this is all right for e−2αa ¿ 1. If we also have E ¿ V0, we can write

T ≈ 16E

V0e−2αa

Example 3.2: Fusion in starsIt turns out, that in the center of the sun, we have H and He gas at temperatureT ∼ 107K. The energy of a proton then is about

Ep = kT ∼ 1keV

As protons approach with E ∼ 1keV, the radius of closest approach is r ∼10−12m. To get to r ∼ 10−15m, which is necessary for fusion, we need anenergy E ∼ 1MeV. The Coulomb-Potential is

V (r) =q2

4πε0r=

ZAZBe2

4πε0r

So, fusion should not occur classically at 107K. We need to solve(− ~

2

2µ∇2 +

ZAZRe2

4πε0r

)ψ = Eψ µ =

mAmB

(mA +mB)

where µ is the reduced mass. ψ(r) for a spherically symmetric wave functionhas the form

ψ(r) =u(r)

rIt turns out, that u(r) satisfies

−~2

2mr

d2u

dr2+ZAZBe

2

4πε0u = Eu

which is the same as in the one-dimensional case for x.

24


rN rC

V ∼ 1r

Now, the tunneling probability, we get by

T =

∣∣∣∣∣∣exp

−

rC∫

rN

β dr

∣∣∣∣∣∣

2

β(r) =

√(ZAZBe2

4πε0r− E

)2m

~2

This leads us to

T ∼ exp(−√EG/E

)EG =

(e2

4πε0~c

)2

2π2µc2

where we call EG the Gamow energy. In numbers, this would be

T ∼ e−22 ∼ 3 · 10−10

by a temperature T ∼ 107K and therefore an E ∼ 1keV. The rate of fusion

dN

dt= e−βE

−1/2

e−αEn2σV dE

where e−βE−

12 is the tunneling probability and e−αE is the Maxwellian distri-

bution, n is the density

N

Note:

1) There is a sharp peak in energy of particles that can fuse e−αEe−βE−1/2

2) The total rate is very strong temperature dependent

¤

25

4 The Harmonic Oscillator 22.05.2006

4 The Harmonic Oscillator

4.1 Classical case

We have the restoring force which is proportional to a displacement

F = −kx

and the potential

V (x) =

∞∫

0

kx′ dx′ =1

2kx2

We can write down the differential equation

md2x

dt2= −kx ↔ x = −ω2x ω =

√k

m

which has solutionsx = A cos (ωt+ α)

The total energy is

Etot =1

2mx2 +

1

2kx2 =

1

2mω2A2

4.2 Quantum oscillator

4.2.1 Starionary states

Lets write down the Hamiltonian in terms of the kinetic and the potential energy

H =p2

2m+

1

2mω2x2 = − ~

2

2m

d2

dx2+

1

2mω2x2

and Schrodinger’s Equation

i~∂Ψ

∂t= HΨ

Now look for stationary states, that is

− ~2

2m

d2

dx2ψn +

1

2mω2x2ψn = Eψn

Change variables, i.e. E = ε~ω, x = q√

~/mω, to get

(− d2

dq2+ q2

)ψ(q) = 2εψ(q)

Remember(q +

d

dq

)(q − d

dq

)f(q) = q2f − d2

dq2f − q

df

dq+

d

dq(qf)

=

(q2 − d2

dq2+ 1

)f(q)

(q − d

dq

)(q +

d

dq

)f(q) =

(q2 − d2

dq2− 1

)f(q)

So we can write(q +

d

dq

)(q − d

dq

)ψ(q) = (2ε+ 1)ψ(q)

and also (q − d

dq

)(q +

d

dq

)ψ(q) = (2ε− 1)ψ(q)

26


We can quickly find two solutions

a) ε = −1/2: (q − d

dq

)Ψ = 0 ψ(q) = Ae

12 q

2

which does not work, because the energy can’t be negative.

b) ε = 1/2: (q +

d

dq

)ψ = 0 ψ(q) = Ae−

12 q

2

This is the ground state of the oscillator, so if ε = 1/2, we have E0 = 12~ω

and

ψ(q) = Ae−12 q

2 ⇒ ψ0(x) = Ae−x2

2a2 a =

√~

mω

Now return to the eigenvalue equation(q +

d

dq

)(q − d

dq

)ψn(q) = (2εn + 1)ψn(q)

(q − d

dq

)(q +

d

dq

)(q − d

dq

)ψn(q)

︸︷︷︸ψm

= (2εn + 1)︸︷︷︸2εm−1

(q − d

dq

)ψn(q)

︸︷︷︸ψm

but we had(q − d

dq

)(q +

d

dq

)ψm(q) = (2ε− 1)ψm(q)

⇒ 2εn + 1 = 2εm − 1

⇒ εn = εm − 1

So the operator (q − d

dq

)

is the operator to raise n by 1, means(q − d

dq

)ψn = ψn+1

Now apply the “raising operator” to ψ0, ψ1, ψ2, ..., so

εn = ε0 + n~ω ⇒ En = ε0 + n~ω

and

ψn(q) = An

(q − d

dq

)ne−

12 q

2

ψn(x) = AnHn

(xa

)e−

x2

2a2 a =

√~

mω

where Hn is a Hermite polynomial of xa . We might write out some of the

normalized functions

ψ0(x) =

(1

a√π

) 12

e−x2

2a2

ψ1(x) =

(1

2a√π

) 12

2(xa

)e−

x2

2a2

ψ2(x) =

(1

8a√π

) 12(

2 − 4(xa

)2)e−

x2

2a2

ψ3(x) =

(1

48a√π

) 12(

12(xa

)− 8

(xa

)3)e−

x2

2a2

...

27


Notes:

1) Hn has either odd or even powers of x, as n is odd or even, so

ψn(−x) = ψn(x) (−1)n

2) The states (of definite E) are time-independent (unlike the classical oscil-lator)

3) ψ∗ψ extends to x = ±∞4) 〈x〉 = 0 for all n, both sides are equally likely and

⟨x2⟩

=

(n+

1

2

)a2 ⇒ ∆x = a

√n+

1

2

5) 〈p〉 = 0⟨p2⟩

=

(n+

1

2

)~

2

a2⇒ ∆p =

~

a

√n+

1

2

It follows

∆x∆p =

(n+

1

2

)~ ≥ ~

2

6) For the energies we get

〈Epot〉 =1

2mω2

⟨x2⟩

=1

2En 〈Ekin〉 =

⟨p2⟩

2m=

1

2En

To be sure, that ψ0(x) is really the “ground state”, we look at the energy

lowering operator(q + d

dq

), that is

(q +

d

dq

)ψn = ψn−1

(q +

d

dq

)ψ0 = 0

where ψ0 has the term e−q2

2 in it.

4.2.2 Non-stationary states of uncertain energy

We write

Ψ(x, t) =∞∑

n=1

cnψn(x)e−iEnt

~

The probability density in x, will oscillate in a complicated way

Ψ∗Ψ =∞∑

n=1

(c∗nψ

∗neiEnt

~

) ∞∑

m=1

(cmψme

−iEmt~

)

=

∞∑

n,m=1

c∗ncmψ∗nψme

−i(Em−En)t

~

where

ωm,n =|Em −En|

~

is an integer multiple of ω, but 〈x〉 oscillates at only one frequency, ω

〈x〉 (t) =

∞∫

−∞

Ψ∗xΨdx

=∑

m,n=1

c∗mcne−

i(En−Em)~

t

∞∫

−∞

ψ∗nxψm dx

which is 0 unless |m − n| ≤ 1. It is called the quasi classical state. In fact, wecan find a set of cn, so that 〈x〉 and 〈p〉 change with t as in the classical case

|cn|2 =nn

n!e−n

with nÀ 1. This is called the Poisson distribution.

28


4.2.3 3-dimensional oscillater (degenacy)

Now we solve the 3-dimensional H, which is seperable in x, y, z

Hx = − ~2

2m

∂2

∂x2+

1

2mω2

xx2

Hy = − ~2

2m

∂2

∂y2+

1

2mω2

yy2 H = Hx + Hy + Hz

Hz = − ~2

2m

∂2

∂z2+

1

2mω2

zz2

From this, we get directly

Enxnynz=

(nx +

1

2

)~ωx +

(ny +

1

2

)~ωy +

(nz +

1

2

)~ωz

or if ωx = ωy = ωz = ω

Enxnynz=

(nx + ny + nz +

3

2

)~ω

∆E

29

5 Observables & Operators - Heisenberg Uncertainty Principle 29.05.2006

5 Observables & Operators - Heisenberg Uncer-tainty Principle

5.1 Operators & eigenfunctions

We had the eigenvalue problem

Hψ = Eψ

Having it resolved, we could decompose

Ψ(r, t) =∑

n

cn(t)ψn(r) +

∫c(E′, t)ψE′(r) dE′

where the sum is for the bound and the integral for the continues case. |cn(t)|2is the probability, that we measure En and |c(E′, t)|2dE′ the probability, thatwe measure an energy E between E ′ and E′ + dE′. We can now generalize thisto any operator A corresponding to an observable A, provided

(1) A must be a linear operator. So, if we have Aψ1 = φ1 and Aψ2 = φ2, then

A (c1ψ1 + c2ψ2) = c1φ1 + c2φ2

We need this because it allows the whole concept of superposition of states.

(2) A must be Hermitian, then

∫Ψ∗

1AΨ2 d3r =

∫ (AΨ1

)∗Ψ2 d3r

It guarantees that the eigenvalues of A are real and that 〈A〉 is real.

(3) The eigenfunctions must form a complete basis set. So any Ψ can berepresented as the sum or the integral of the eigenfunctions, one of whichis “selected” when we make a measurement. We can rewrite generally

Ψ(r, t) =∑

n

can(t)ψan

(r) +

∫c(a′, t)ψa′(r) da′

where |can(t)|2 is the probability of measuring a = an and |c(a′, t)|2da′ is

the probability of measuring a between a′ and a′ + da′.

5.2 Eigenfunctions for position and momentum

For the position, we have the equation

xψx′(x) = x′ψx′(x)

where ψx′(x) is the eigenfunction with eigenvalue x′. Because x = x, we get

xψx′(x) = x′ψx′(x)

This has the Dirac δ-function as solution

ψx′(x) = δ (x− x′)

where by definition

∞∫

−∞

f(x)δ (x− x′) dx = f(x′)

30


The eigenfunctions of position are continuous. Therefore, we can take our gen-eral decomposition equation in one dimension

Ψ(x, t) =

∞∫

−∞

c (x′, t)ψx′(x) dx′

=

∞∫

−∞

c(x′, t)δ (x− x′) dx′

= c(x, t)

where again |c(x, t)|2dx is the probability of finding a value between x andx+ dx.

For the momentum, we get the eigenequation

pψp′(x) = p′ψp′(x)

which is equivalent to

−i~ ∂

∂xψp′(x) = p′ψp′(x)

It has the solutions

ψp′(x) =1√2π~

eip′x

~

where (2π~)−1/2

is an arbitrary constant. We expect, that the function is eithercontinuous in p′ or discrete, so

Ψ(x, t) =

∞∫

−∞

c(p′, t)ψp′(x) dp′

=1√2π~

∞∫

−∞

c(p′, t)eip′x

~ dp′

which is the Fourier integral from before with Ψ(p, t) = c(p, t).

5.3 Compatible observables

There’s an important idea: We can not neccessarily specify a quantum stateusing different observables (c.f. x, px, y, py, z, pz in classical mechanics), but wespecify the position by a Ψ which is a superposition of momentum states andwe specify the momentum by a Ψ which is a superposition of position states.The observables must be “compatible”, but x, p are incompatible. To explorethis, we introduce the commutator of two operators

[A, B

]= AB − BA

If [A, B] = 0, then A,B are comatible. If [A, B] 6= 0, A,B are uncompatible.We take x, p

xpΨ − pxΨ = x

(−i~ ∂

∂xΨ

)+ i~

∂

∂x(xΨ)

= x (−i~)∂

∂xΨ + xi~

∂

∂xΨ + i~Ψ

It follows[x, p] = i~ 6= 0

It is easy to prove that there are no simultaneous eigenfunctions of x and p. Ifthere were, we had

xψx′p′(x) = x′ψx′p′(x) pψx′p′(x) = p′ψx′p′(x)

31


and therefore

[x, p]ψx′p′(x) = (x′p′ − p′x′)ψx′p′(x) = 0

It followedi~ψx′p′(x) = 0 ⇒ ψx′p′(x) = 0

Therefore, if we have a non-zero commutator, we have non identical eigenfunc-tions.

5.4 Heisenberg uncertainty principle

We apply these ideas to get the uncertainties in x and p. The variance in x is

(∆x)2

=⟨x2⟩− 〈x〉2 =

∞∫

−∞

Ψ∗(∆x)2

Ψdx

(∆p)2

=⟨p2⟩− 〈p〉2 =

∞∫

−∞

Ψ∗(∆p)2

Ψdx

Where ∆x = x− 〈x〉 and ∆p = p− 〈p〉. So

[∆x, ∆p

]= (x− 〈x〉) (p− 〈p〉) − (p− 〈p〉) (x− 〈x〉) = [x, p] = i~

To go on, we need to write down some general properties for Hermitian opera-tors:

(1) For the expectation value of A2, where A is Hermitian, we get

⟨A2⟩

=

∞∫

−∞

Ψ∗A(AΨ)dx =

∞∫

−∞

(AΨ)∗(AΨ)dx

(2) With two Hermitian operators, we get A and B

∞∫

−∞

Ψ∗ABΨdx =

∞∫

−∞

(AΨ)∗BΨdx

=

∞∫

−∞

(BAΨ)∗Ψdx

=

∞∫

−∞

Ψ(BAΨ)∗ dx

=

∞∫

−∞

Ψ∗BAΨdx

∗

This implies, that the integral

∞∫

−∞

Ψ∗[A, B

]+

Ψdx =

∞∫

−∞

Ψ∗(AB + BA

)Ψdx

is real and

∞∫

−∞

Ψ∗[A, B

]Ψdx =

∞∫

−∞

Ψ∗(AB − BA

)Ψdx

is purely imaginary.

32


(3) Schwarz’ inequality: Suppose α(x), β(x) are complex functions of x withfinite integrals

∞∫

−∞

α∗α dx,

∞∫

−∞

β∗β dx,

∞∫

−∞

α∗β dx

Construct φ(x) = α+ λβ, λ ∈ C

∞∫

−∞

φ∗φdx =

∞∫

−∞

α∗α dx+ λ∗λ

∞∫

−∞

β∗β dx

+ λ∗∞∫

−∞

β∗α dx+ λ

∞∫

−∞

α∗β dx ≥ 0

This is true for any λ, so true for

λ =

−∞∫

−∞

β∗α dx

∞∫−∞

β∗β dx

It follows

λ∗∞∫

−∞

β∗α dx = −λ∗λ∞∫

−∞

β∗β dx

and therefore

∫φ∗φdx =

∞∫

−∞

α∗α dx−

∞∫−∞

α∗β dx∞∫

−∞

β∗α dx

∞∫−∞

β∗β dx

≥ 0

That is

∫α∗α dx

∫β∗β dx ≥

∫α∗β dx

∫β∗α dx =

∣∣∣∣∣∣

∞∫

−∞

α∗β dx

∣∣∣∣∣∣

2

Now, set α = AΨ and β = BΨ to get

⟨A2⟩ ⟨B2⟩ (1,3)

≥

∣∣∣∣∣∣

∞∫

−∞

(AΨ)∗(BΨ)dx

∣∣∣∣∣∣

2

=

∣∣∣∣∣∣

∞∫

−∞

Ψ∗ABΨdx

∣∣∣∣∣∣

2

=

∣∣∣∣∣∣

∞∫

−∞

Ψ∗ 1

2

([A, B

]+

+[A, B

])Ψdx

∣∣∣∣∣∣

2

(2)=

∣∣∣∣∣∣1

2

∞∫

−∞

Ψ∗[A, B

]+

Ψdx

∣∣∣∣∣∣

2

+

∣∣∣∣∣∣1

2

∫Ψ∗[A, B

]Ψdx

∣∣∣∣∣∣

2

=1

4

∣∣∣∣∣∣

∞∫

−∞

Ψ∗[A, B

]+

Ψdx

∣∣∣∣∣∣

2

+1

4

∣∣∣∣∣∣

∞∫

−∞

Ψ∗[A, B

]Ψdx

∣∣∣∣∣∣

2

33


It follows

〈∆x〉2 〈∆p〉2 ≥ 1

4~

2 +

∣∣∣∣∣∣

∫Ψ∗ [x, p]+ Ψdx

∣∣∣∣∣∣

2

which gives us the Heisenberg uncertainty inequality

∆x∆p ≥ ~

2

If we want equality, we also need

∞∫

−∞

ψ ∗ [x, p]+ ψ dx = 0

Note: In three dimensions, we can “mix” the position and momentum observ-ables to specify a state, like x, y and pz

[x, y] = 0 [x, pz] = 0 [y, pz] = 0

Then we could write any wave function as superposition of functions

ψx′y′p′z=

1√2π~

δ (x− x′) δ (y − y′) e−ip′

zz

~

which are the eigenfunctions of x, also y, also pz.

5.5 Compatibility with H

Consider the expectation value for an observable A for a particle with wavefunction Ψ(r, t)

〈A(t)〉 =

∫

R3

Ψ(r, t)∗AΨ(r, t) d3r

Using Schrodingers Equation

i~∂

∂tΨ(r, t) = HΨ(r, t)

we get

d

dt〈A(t)〉 =

∫

R3

∂Ψ∗∂t

AΨd3r +

∫

R3

Ψ∗A∂Ψ

∂td3r

= − 1

i~

∫

R3

(HΨ)∗AΨd3r +1

i~

∫

R3

Ψ∗AHΨd3r

=1

i~

∫

R3

Ψ∗(AH − HA

)Ψd3r

=1

i~

∫

R3

Ψ∗[A, H

]Ψd3r

where by the first step we assume that A 6= f(t). If A is compatible with H,i.e. [A, H] = 0, then

d 〈A〉dt

= 0

that means, A is a constant of motion.

34


5.6 Orbital angular momentum

Classically, the angular momentum is given by

L = r × p

In Quantum Mechanics, we write

L = r × p = −i~r ×∇or

Lx = ypz − zpy = −i~(y∂

∂z− z

∂

∂y

)

Ly = zpx − xpz = −i~(z∂

∂x− x

∂

∂z

)

Lx = xpy − ypx = −i~(x∂

∂y− y

∂

∂x

)

It is easy to show by expanding, that for i = x, y, z[Lx, Ly

]= Lz [z, pz]

[Ly, Lz

]= Lx [x, px]

[Lz, Lx

]= Ly [y, py]︸︷︷︸

=i~6=0

but that [L

2, Lx

]=

[L

2, Ly

]=

[L

2, Lz

]= 0

We can also verify, that[Li,∇2

]= 0

[Li, V (r)

]= 0

It follows [Li, H

]= 0

for all H associated with a “central potential V (r)”

⇒[L

2, H]

= 0

As expected, L is conserved in a central potential. The most surprising thingis that we can define (or measure) the total L2, plus only one component of it.Conventionally, we take Lz.

5.7 Angular momentum eigenfunctions

We first explore a Cartesian function

ψ(0,0) = R(r)

so a spherically symmetric function. We go on with

ψ(1,0) = R(r)z

rψ(1,+1) = R(r)

x+ iy

rψ(1,−1) = R(r)

x− iy

r

Now we want to know what happens, if we apply Lx, Ly, Lz and L2

• ψ(0,0): We get

Lxψ(0,0) = −i~(y∂R

∂r

∂r

∂z− z

∂R

∂r

∂r

∂y

)

= −i~ ∂R∂r

(yz

r− z

y

r

)= 0

Any spherically symmetric ψ has L2 = Lx = Ly = Lz = 0 and therefore

Lxψ(0,0) = 0ψ(0,0) Lyψ(0,0) = 0ψ(0,0) Lzψ(0,0) = 0ψ(0,0)

L2ψ(0,0) = 0

35


• ψ(1,0): First we write down, that

L

(R(r)

rz

)= zL

(R(r)

r

)+R(r)

rLz =

R(r)

rLz

Furthermore, we get

Lxψ(1,0) =R(r)

rLxz = −i~ R(r)

ry

Lyψ(1,0) = i~R(r)

rx

Lzψ(1,0) = 0

ψ(1,0) is not an eigenfunction of Lx or Ly, but is of Lz, with eigenvalue 0.Going on, we get

L2xψ(1,0) = ~

2 R(r)

rz = ~

2ψ(1,0) = L2yψ(1,0)

and thereforeL

2= L2

x + L2y + L2

z = 2~2ψ(1,0)

so ψ(1,0) is also an eigenfunction of L2

with eigenvalue 2~2.

• ψ(1,1):

ψ(1,1) = R(r)x+ iy

r

ψ(1,1) is an eigenfunction of L2

with

L2ψ(1,1) = 2~

2ψ(1,1)

and an eigenfunction of Lz with

Lzψ(1,1) = ~ψ(1,1)

• ψ(1,−1):

Lzψ(1,−1) = −~ψ(1,−1) L2ψ(1,−1) = 2~2ψ(1,−1)

This can be illustrated in the following way

ψ1,0

ψ1,1

ψ1,−1

√2~

All this is much easier in spherical coordinates (r, θ, φ), then Lz and L2

operateon spherical harmonic functions, where

L2(Y`m(θ, φ)) = ` (`+ 1) ~

2Y`m (θ, φ)

Lz (Y`m (θ, φ)) = m~Y`m (θ, φ)

with m = −`,−`+ 1, ..., 0, ..., `.

L

√l(l + 1)

36


Note that ψ0,0 ∝ Y00, ψ1,−1 ∝ Y1−1, ... and that Y`m are orthonormal

2π∫

0

π∫

0

Y ∗`mY`′m′ sin θ dθ dφ = δmm′δ``′

So we have the two important things

(1) L2 and Lz define orbital angular momentum quantum states

(2) The orbital angular momentum is quantised in units of ~

L =√` (`+ 1) ~ Lz = 0,±1~,±2~, ...,±`~

5.8 Stern-Gerlach experiment

The angular momentum of charged particles lets us expect a magnetic momentµ. e.g. for the electron

µorbitalz =

−e2me

Lz = − e

2mem`~

and for the spin

µspinz =

−2e

2meSz = − e

mems~

Recall that µ in an inhomogeneous B field suffers a force

Fz = µz∂B

∂z

Atoms

Precisely

two

beams

Silver

The result is, that we get exactly two beams, that is Lz is quantised by ± 12~.

5.9 Spin

We have an angular momentum also from the spin of particles, which is notreally meaningful as classical spinning object. There we have eigenvalues

S =√s (s+ 1) ~

and ms = −s,−s + 1..., s − 1, s, but s can now be a half-integer. The electronfor example has s = 1

2 and therefore ms = ± 12 or the W -Boson has s = 1 and

therefore ms = −1, 0, 1.

37

6 The Hydrogen Atom 07.06.2006

6 The Hydrogen Atom

6.1 Motion in “central potentials”

The hydrogen atom is the simplest one, because it is a proton p+ and an orpitingelectron e−. We say e− orbits, because mp À me. There we have a centralpotential

V (r) =1

4πε0

−e2r

which is as known similar to the potential of planetary motions. The centralforce acts radially and therefore, the angular momentum L is conserved

F = −dV

drr L = mr × v = mr × dr

dtN = r × F = 0

where N is the torque.

r

v = dr

dt

A = 12r × dr = 1

2mLdt

We can split the momentum into a radial and a tangential part

pr = mdr

dtpt =

L

r

The kinetic energy then is given by

p2

2m=

p2r

2m+

L2

2mr2

and therefore the total energy by

E =p2r

2m+

L2

2mr2+ V (r) =

p2r

2m+ Ve Ve :=

L2

2mr2+ V (r)

where we call Ve the effective potential with effective force

Fe = −dVedr

=L2

mr3− dV

dr

6.2 Solutions of Schrodingers equation in a central poten-tial

For a wave function, describing a quantum state with sharply defined energy E,we write

Ψ(r, θ, φ, t) = ψ(r, θ, φ)e−iEt

~

where ψ(r, θ, φ) is an energy eigenfunction satisfying the eigenvalue equation

[− ~

2

2m∇2 + V (r)

]ψ = Eψ (6.1)

If we also assume, that the quantum state has as well definite angular momentumproperties, we can write

ψ(r, θ, φ) = R(r)Y`,m`(θ, φ) (6.2)

Note that

∇2ψ =1

r

∂2

∂r2(rψ) +

1

r2

(∂2ψ

∂θ2+

cos θ

sin θ

∂ψ

∂θ+

1

sin2 θ

∂2ψ

∂φ2

)

38


and

L2

= −~2

(∂2

∂θ2+

cos θ

sin θ

∂

∂θ+

1

sin2 θ

∂2

∂φ2

)

So we can also write

∇2ψ =1

r

∂2

∂r2(rψ) − 1

r2~2L

2ψ (6.3)

Substituting into equation (6.1), using equation (6.2), we get

− ~2

2m∇2R(r)Y`,m`

= − ~2

2mr

∂2

∂r2(rR(r)Y`,m`

(θ, φ)) +1

2mr2L

2(R(r)Y`,m`

(θ, φ))

= − ~2

2mr

∂2

∂r2(rR(r)Y`,m`

(θ, φ)) +1

2mr2R(r)` (`+ 1) ~

2Y`,m`(θ, φ)

= ER(r)Y`,m`(θ, φ) − V (r)R(r)Y`,m`

(θ, φ)

Here we can remove the spherical harmonic Y`,m`(θ, φ) and multiply by r to get

− ~2

2m

∂2

∂r2(rR(r)) +

` (`+ 1) ~2

2mr2rR(r) + V (r)rR(r)Y`,m`

= ErR(r)

If we set u(r) = rR(r), we can rewrite

− ~2

2m

∂2

∂r2u(r) +

[` (`+ 1) ~

2

2mr2+ V (r)

]u(r) = Eu(r)

Here we will get eigenfunctions unr,`(r) with energy Enr,` and therefore

ψnr,`,m(r, θ, φ) =unr,`(r)

rY`m(θ, φ)

which must be an eigenfunction of H, Lz, L2.

Introduce the parity of a wave function

ψ(−r) = ±ψ(r)

The eigenvalues are +1 (even parity) and −1 (odd parity). In a central potential,the parity is even if ` is even and it is odd, if ` is odd. The parity is an observableand the parity operator eigenvalues are ±1. H and the parity operator commuteif H(−r) = H(r).

6.3 Hydrogen atom

We would like to solve the radial Schrodinger Equation

− ~2

2me

∂2

∂r2unr,` +

` (`+ 1) ~2

2mer2unr,` −

e2

4πε0runr,` = Enr,`unr,`

with u → 0 as r → 0 (because ψ is finite at r = 0) and as r → ∞. We definethe effective potential

Veff(r) :=` (`+ 1) ~

2

2mer2− e2

4πε0r

` = 0

r

Veff(r)

increasing `

39


The minimum, we get by

dVeff

dr= −` (`+ 1) ~

2

mer3+

e2

4πε0r2= 0

where

r = rmin = ` (`+ 1) ~2 4πε0mee2

=: ` (`+ 1) a0

and a0 is the Bohr radius

a0 =4πε0e2

~2

me= 0.53 · 10−10m

For the energy at rmin we get

Veff(` (`+ 1) a0) = −1

2

e2

eπε0a0

1

` (`+ 1)= − ER

` (`+ 1)

where ER is the Rydberg energy

ER :=e2

8πε0a0= 136eV

For high `, we have a high energy and the probability gets further out. Introducethe fine structure constant

α :=e2

4πε0~c=

1

137.035...

which is a dimensionless constant. Now

a0 =1

α

~

mecER =

e2

8πε0a0=

α2

2mec

2

and we solve

− ~2

2me

∂2

∂r2unr,` +

(` (`+ 1) ~

2

2mer2~

2 − e2

4πε0r

)unr,` = Enr,`unr,`

Setting r = qa0, E = −γ2ER, we can write this as

∂2u

∂q2+

(2

q− ` (`+ 1)

q2

)u = γ2u (6.4)

We have the boundary conditions u(q) = 0 at q = 0 and as q → ∞. Then, atlarge q, we get the equation

d2u

dq2= γ2u

which has the general solutions

u(q) = Ae−γq +Beγq = Ae−γq

where we set B = 0, because u(q) → 0 as q → ∞. For small q, we get

d2u

dq2− ` (`+ 1)u

q2= 0

which has solutions

u(q) = Aq`+1 or u(q) = Bq−`

where the second one wouldn’t satisfy u(q) = 0 at q = 0. So we look for

u(q) = f(q)q`+1e−γq

40


which (by substituting into equation (6.4)) leads to

qd2f

dq2+ 2 [(`+ 1) − γq]

df

dq+ 2 [1 − γ(`+ 1)] f = 0

This has solutionsf(q) = pnr,`(q) =

∑

s

asqs

with

as+1 =

[2γ (s+ `+ 1) − 2

s(s+ 1) + 2(s+ 1)(`+ 1)

]as γ =

1

nr + `+ 1

so that pnr,`(r) terminates at anrqnr . Finally, we get

unr,`(r) = Npnr,`(r)r`+1e

− ra0(nr+`+1) Enr,` = − ER

(nr + `+ 1)2

For each `, there is an infinite number of discrete energy states as nr goes from0 to ∞.

` = 0s

nr = 0

nr = 1

` = 1p

nr = 0

` = 2d

−ER

−ER

4

−ER

9

nr = 0

6.4 Zeeman effect - Perturbing the Hamiltonian H

We introduce the total angular momentum as the sum of the orbital angularmomentum and the spin. We had before

L =√` (`+ 1) ~ m` = −`,−`+ 1, ..., `

S =√s (s+ 1) ~ ms = −s,−s+ 1, ..., s

Now, we add

J =√j (j + 1) ~ mj = −j,−j + 1, ...,+j

The question now is, how to get j from ` and s. The possible values of j rangefrom j = `+ s and j = |`− s| with ∆j = 1. e.g. if we have ` = 2, s = 1

2 , thenj = 5

2 or j = 32 and therefore

mj = −5

2,−3

2,−1

2, ...,

5

2or mj = −3

2,−1

2,1

2,3

2

We also dealt earlier with the magnetic moment, which depends on whether theangular momentum is coming from spin or orbital angular momentum and sowe can write general formulae

µ = IA =q

2mL µz = −g e

2me~mj

This g here is called the Lande g factor, which is given by

g = 1 +j (j + 1) − ` (`+ 1) + s (s+ 1)

2j (j + 1)

41


so g = 1 if s = 0 and g = 2 if ` = 0. Now we can consider an atom placed in amagnetic field. The electron acqires energy

Emag = −µ · B

If we have arranged things, so that |B| = Bz, then

Emag = −µzB = gµBmjB µB =e~

2me

where µB is the Bohr magneton. We had

Hψnr,`,m = Eψnr,`,m H = H0 +mjgµBB︸︷︷︸const

The effect is, that all enery levels are split into 2j+1 components. This is calledthe Zeeman effect.

6.5 Time-dependent perurbations - Radiative transitions

If H changes with time, we get transitions between “stationary states”. Imaginethe unperturbed eigenfunctions Ψ0

k and eigenvalues ε0k. In this case, we havethe time-indipendent form of the Schrodinger equation

i~∂

∂tΨ0k(t) = H0Ψ

0k(t) = ε0kΨ(t)

We’d expect then, that the general solution of this is a superposition

Ψ0(t) =∑

k

a0kΨ

0k(t) =

∑

k

a0ke

−iε0

kt

~ ψ0k

Now we perturb H = H0 + V (t). We can write

i~∂

∂tΨ(t) =

(H0 + V (t)

)Ψ(t)

Assume that

Ψ(t) =∑

k

ak(t)Ψ0k(t) =

∑

k

ak(t)e−

iε0kt

~ ψ0k

and substitute into the Schrodinger equation

i~∑

k

(dak(t)

dt− iε0k

~ak(t)

)e−

iε0kt

~ ψ0k =

∑

k

(ε0k + V (t)

)ak(t)e

−iε0

kt

~ ψ0k

which can be simplified to

i~∑

k

(dak(t)

dt

)e−

iε0kt

~ ψ0k =

∑

k

V (t)ak(t)e−

iε0kt

~ ψ0k

Now we multiply by another state ψ0 ∗n and integrate over d3r

i~∑

k

(dak(t)

dt

)e−

iε0kt

~

∫ψ0 ∗n ψ0

k d3r

︸︷︷︸δnk

=∑

k

ak(t)e−

iε0kt

~

∫ψ0 ∗n V ψ0

k d3r

and it follows

i~d

dtan(t) =

∑

k

ak(t)ei(ε0n−ε0k) t

~

∫ψ0 ∗n V ψ0

k d3r

42


If we assume that the initial state were an eigenfunction ψj , we get

i~d

dtan(t) =

∫ψ0 ∗n V ψ0

j d3r ei(ε0n−ε0j) t

~

and therefore

an(t) = − i

~

t∫

0

∫ψ0 ∗n V ψ0

j d3r ei(ε0n−ε0j) t′

~ dt′

where we define

Vnj :=

∫ψ0 ∗n V ψ0

j d3r εnj := ε0n − ε0j

If the perturbation is constant and applied for some time t, we can write

an(t) = − i

~Vnj

t∫

0

eiεnjt′

~ dt′ = − i

~Vnj

(~

iεnjeiεnj

t′

~

)∣∣∣∣t

0

= −Vnjεnj

(eiεnj

t~ − 1

)

Define

Pjn := |an(t)|2 =|Vnj |2ε2nj

4 sin2

(εnjt

2~

)

where as t→ ∞sin2

(εnjt2~

)

tε2nj

behaves as π2~δ(εnj), that is

→ limt→∞

Pjn(t)

t=

2π

~|Vnj |2 δ(εnj)

If V is time varying, we cannot take outside of the integral∫

dt′, e.g. the E-waveis interacting with the dipole of the atom, e.g.

V = (er)E0 cosωt

where er is the dipole of the atom. So it is exactly as before, but now

δ → δ (εnj ± ~w)

and the transition probability is

limt→∞

Pjn(t)

t=

2π

~|Vnj |2 |A(ωnj)|2

where |A(ω)|2 is the strength of perturbation at ω =εnj

~. We only get transi-

tions, if ~ω = |εnj | and |Vnj |2 6= 0. The interaction with the dipole of the atomis proportional to r which is the operator of r and not the unit vector. Look at

∫ψnrψj d3r

Note that the operator r changes sign when r → −r, therefore Vnj = 0 unlessψn and ψj have opposite parity.

2p

1s

2s

43


6.6 Some (small) complicating effects

So far, we have considered H

• non-relativistic

• only with Coulomb forces

• assuming, that the mass of the proton was ∞.

6.6.1 The reduced mass effect

The mass of the proton mp is not ∞ and therefore the proton and the electronorbit arround a common center of mass. The energy is

E =p2e

2me+

p2p

2mp− e2

4πε0r

In the center of mass frame, we have pe = pp = p and

E =p2

2µ− e2

4πε0rµ =

memp

me +mp

where µ is the reduced mass. The net effect of this is to change the length andenergy scales

a′0 =me

µa0 E′

R =µ

meER E′

n =µ

meEn

For the electron-proton atom (i.e. Hydrogen 1H), we have

µ

me=

1836

1837

or for Deuterium 2H (e− and p+n0) we get

µ

me=

3671

3672

6.6.2 Spin-orbit coupling

This is a modification of H due to the magnetic field produced by the orbitalmotion interacting with µ from the spin. The electron sees an “orbiting” protonwith period

τ =2πr

v= 2π

mer2

Le

This means, that effectively we have a current

I =e

τ⇒ B =

µ0I

2r=

µ0

2r

eLe2πmer2

=e

4πε0mec2r3Le

and we haveµe = −2

e

2meS

and therefore

Emag = −µ · B =e2

4πε0m2ec

2r3L · S

This isn’t completely correct. Correct would be

Emag =e2

8πε0m2ec

2r3L · S

where L ∼ S ∼ ~ and

Emag ∼ e2~2

4πε0m2ec

2a30

= α4mec2

44


6.6.3 Relativist effects

We know the relativistc energy

ε =√m2ec

4 + p2c2 = mec2 +

p2

2me− 1

8

(p

mec

)4

mec2 + ...

where p2 = E2me ∼ α2m2ec

2 and therefore p ∼ αmec. This means that therelativistic correction is of order

ε =α4

8mec

2

45

7 Quantum mechanics of identical particles 19.06.2006

7 Quantum mechanics of identical particles

In deterministic Classical Physics we keep track of even identical particles. InQuantum Physics it is probabilistic, so it’s impossible to keep track of individualparticles.

7.1 Wave functions for identical particles

Consider two particles p, q, then we have a “2-particle wave-function”

Ψ (rp, rq, t)

The probability of finding p at a and q at b is

|Ψ(a, b, t)|2 d3ad3b

If we have two identical particles, we get

|Ψ(a, b, t)|2 = |Ψ(b,a, t)|2

This enables us to say that

Ψ (a, b, t) = eiδΨ(b,a, t)

where e−iδ can be a “phase factor”. Then

Ψ (a, b, t) = e2iδΨ(a, b, t) ⇒ eiδ = ±1

Therefore, the wave function of two identical particles must be symmetric oranti-symmetric under exchange of p and q. Later, the symmetric ones, we willcall Bosons and the anti-symmetric ones Fermions.

Tale the Hamiltonian of the harmonic oscillater, assuming that p and q do notdepend on each other

H(xp,xq) = − ~2

2m

∂2

∂x2p

− ~2

2m

∂2

∂x2q

+1

2mω2x2

p +1

2mω2x2

q

The eigenfuncions are ψn with En =(n+ 1

2

)~ω

1) If p and q are in exactly the same state (nq = np), we get

E = 2En = (2n+ 1) ~ω

and the wave-function

Ψ(xp,xq, t) = ψn(xp)ψn(xq)e−i(2En) t

~

Verify that it is a solution of

HΨ(xp,xq, t) = EΨ(xp,xq, t)

In this case, Ψ is symmetric under the p to q exchange operator. We cannot find an anty-symmetric solution, if p and q are in the same state.

2) If p and q are in different states and also distinguishable (denoted by Dsuperscript), we will have

E = En + En′ = (n+ n′ + 1)~ω

We can construct two solutions

ΨD1 (xp,xq, t) = ψn(xp)ψn′(xq)e

−i(En+En′ ) t~

ΨD2 (xp,xq, t) = ψn(xq)ψn′(xp)e

−i(En+En′ ) t~

and any linear combination will be a solution. In particular, we can write

ΨD(xp,xq, t) = c1ΨD1 (xp,xq, t) + c2Ψ

D2 (xp,xq, t)

where |c1|2 is the probability that p is in state n and q in state n′ and

|c2|2 is the probability that q is in state n and p in state n′.

46


3) If we have n and n′ but p and q are now indistinguishable, then

|c1|2 = |c2|2 =1

2

Now we can construct symmetric and anti-symmetric wave functions

ΨS (xp,xq, t) =1√2

(ψn(xp)ψn′(xq) + ψn(xq)ψn′(xp)) e−i(En+En′ ) t

~

ΨA (xp,xq, t) =1√2

(ψn(xp)ψn′(xq) − ψn(xq)ψn′(xp)) e−i(En+En′ ) t

~

Its interesting to look at ΨD, ΨS , ΨA for xp = xq = x0. We find

ΨD (x0,x0, t) = ψn(x0)ψn′(x0)e−i(En+En′ ) t

~

ΨS (x0,x0, t) =√

2ψn(x0)ψn′(x0)e−i(En+En′ ) t

~

ΨA (x0,x0, t) = 0

This means, that relative to the case of distinguishable particles, we can saythat the symmetric ΨS is twice as likely to have both particles in the sameplace at the same time and the anti-symmetric ΨA never has the two particlesin the same place at the same time. For examples, see the exercice sheet.

7.2 Exchange symmetry with spin

Recall that we had the spin angular momentum S =√s (s+ 1)~ with z-

component Sz = ms~. Lets say, we have a spin wave function χ(p) with

χ(p) =

s∑

ms=−s

cmsχs,ms

(p)

We can combine this with the (normal) spatial wave function ψ, to get

Φn,`,m`,s,ms= ψn,`,m`

χs,ms

Now look at two indistinguishable particles in exactly the same state. We willagain find a wave function

Φ(p, q) = Φn,`,m`,s,ms(p)Φn,`,m`,s,ms

(q)

and again, ΨA for the same states does not exist.

But what about particles in different states? Can we represent the wave functionas product of two single particle states?

ΨS(p, q)

=1√2

(Φn,`,m`,s,ms

(p)Φn′,`′,m′

`,s′,m′

s(q) + Φn,`,m`,s,ms

(q)Φn′,`′,m′

`,s′,m′

s(p))

ΦA(p, q)

=1√2

(Φn,`,m`,s,ms

(p)Φn′,`′,m′

`,s′,m′

s(q) − Φn,`,m`,s,ms

(q)Φn′,`′,m′

`,s′,m′

s(p))

which is exactly the same approach as we had before. But we could also havethe product of two two-particle states as

Φ2,S(p, q) = ψS(p, q)χS(p, q)

Φ2,A(p, q) = ψA(p, q)χS(p, q)

Φ2,A(p, q) = ψS(p, q)χA(p, q)

Φ2,S(p, q) = ψA(p, q)χA(p, q)

47


Example 7.1: Consider two particles with s1, s2 = 12 and remember that if we

have the combination of two spin-numbers s1, s2 then S = |s1 − s2|, ..., s1 + s2.Therefore, we get S = 1 = s1 + s2 with mS = −1, 0, 1 or S = 0 = |s1 − s2| withmS = 0. Lets call χ 1

2 ,+12

= χ+ and χ 12 ,−

12

= χ−, then

χS1,1(p, q) = χ+(p)χ+(q)

χS1,0(p, q) =1√2

(χ+(p)χ−(q) + χ+(q)χ−(p))

χS1,−1(p, q) = χ−(p)χ−(q)

χA0,0(p, q) =1√2

(χ+(p)χ−(q) − χ+(q)χ−(p))

¤

As we already mentioned, the nature gives us then two types of particles, namelyBosons (photon, 4He, ...) and Fermions (e−, p+). Lets simply state the rules:The exchange symmetry depends on the spin of the particle. In particular,particles with integer total spin must have a symmetric wave function underparticle exchange. Particles with half integer spin must have anti-symmetrictwo-particle wave-functions under the action of changing the two particles. TheBosons have integer spin and therefore a symmetric Ψ. The Fermions have halfinteger spin and therefore an anti-symmetric Ψ.

Consequences for Fermions:

1) Two Fermions (e.g. two electrons) can never have absolutely identicalquantum states, because we can only make symmetric ΨS out of identicalwave functions.

2) When the two-particle spin state is anti-symmetric χA, then Ψ must besymmetric ΨS . This means, that the “electrons like to be together”. Wecall it Covalent Bond.

3) When the spin is symmetric χS , then Ψ must be anti-symmetric, whichmeans that the electrons are never together. We call this rigidity of matter.

Consequences for Bosons: We can always make symmetric wave-functions fromtwo or more identical states. This is used by Lasers, the Bose-Einstein Conden-sate and superfluidity.

48

8 More Complicated Atoms 05.07.2006

8 More Complicated Atoms

For more complicated atoms, we’ll have to modify what we’ve got for the H-atom. Multiple e atoms will have a Hamiltonian H that includes the term dueto the electron-electron interaction, e.g. for He we have

H = − ~2

2m

(∇2p + ∇2

q

)+ V (rp, rq)

where

V (rp, rq) = − 2e2

4πε0rp− 2e2

4πε0rq+

e2

4πε0|rp − rq|with two’s in the first two terms, because the nucleus has order Z = 2. Whatwe do, to deal with those more complicated atoms is to modify the centralpotential. Close to the nucleus, the potential is something like

V (r) = − Ze2

4πε0r

Very far away, we could take

V (r) =−e2

4πε0r

We combine these two potentials by

V (r) = z(r)−e2

4πε0rz(r) =

((z − 1)e−

ra + 1

)a =

1

2a0

Therefore, we solve for a single electron to get the eigenfunctions. Once we’vegot the eigenfunctions, we can “populate” them with the electrons. Then thereare two changes:

• At low n, the energy eigenvalues are much lower (by a factor of z2, whereone z is because of the potential V (r) and the other z because of a0 ∝

1z ).

• The degeneracy between ` states is broken

Note: 3d > 4s

More complicated atoms H-Atom

Now assign electrons to these n, `,m states, 2 at a time (↑↓), to get

• 1s: n = 1, ` = 0, m = 0 ⇒ 2 electrons

• 2s: n = 2, ` = 0, m = 0 ⇒ 2 electrons

• 2p: n = 2, ` = 1, m = −1, 0, 1 ⇒ 6 electrons

• 3s: n = 3, ` = 0, m = 0 ⇒ 2 electrons

• 3p: n = 3, ` = 1, m = −1, 0, 1 ⇒ 6 electrons

• 3d: n = 3, ` = 2, m = −2,−1, 0, 1, 2 ⇒ 10 electrons

49

Stichwortverzeichnis

A

absorption coefficiant . . . . . . . . . . . . 5angular frequency . . . . . . . . . . . . . . . 11

B

black body. . . . . . . . . . . . . . . . . . . . . . . 6Bohr radius . . . . . . . . . . . . . . . . . . . . . 40Born. . . . . . . . . . . . . . . . . . . . . . . . . . . .14Boson . . . . . . . . . . . . . . . . . . . . . . . . . . 46

C

collapse of wave function . . . . . . . . 17commutator . . . . . . . . . . . . . . . . . . . . 31Compton Scattering. . . . . . . . . . . . . .3Covalent Bond . . . . . . . . . . . . . . . . . . 48

D

degeneracy. . . . . . . . . . . . . . . . . . . . . .19

E

effective force . . . . . . . . . . . . . . . . . . . 38effective potential . . . . . . . . . . . . . . 38 fenergy density . . . . . . . . . . . . . . . . . . . 6expectation value . . . . . . . . . . . . . . . 13

F

Fermions . . . . . . . . . . . . . . . . . . . . . . . 46fine structure constant . . . . . . . . . . 40frequency . . . . . . . . . . . . . . . . . . . . . . . 11

G

Gamow energy . . . . . . . . . . . . . . . . . . 25

H

Hamiltonian operator . . . . . . . . . . . 16Heisenbergs microscope . . . . . . . . . . 9Hermitian . . . . . . . . . . . . . . . . . . . . . . 30

L

Lande g factor . . . . . . . . . . . . . . . . . . 41

P

parity. . . . . . . . . . . . . . . . . . . . . . . . . . .39period . . . . . . . . . . . . . . . . . . . . . . . . . . 11phase velocity . . . . . . . . . . . . . . . . . . 11photon. . . . . . . . . . . . . . . . . . . . . . . . . . .9Plank’s constant . . . . . . . . . . . . . . . . . 9Plank’s formula . . . . . . . . . . . . . . . . . . 9Poisson Distribution . . . . . . . . . . . . 28

Q

quasi classical state . . . . . . . . . . . . . 28

R

Rayleigh Jeans . . . . . . . . . . . . . . . . . . . 8reduced mass . . . . . . . . . . . . . . . . . . . 44rigidity of matter . . . . . . . . . . . . . . . 48Rydberg energy . . . . . . . . . . . . . . . . . 40

S

Schrodinger equation . . . . . . . . . . . 17Schrodinger Equation . . . . . . . . . . . 12standard deviation . . . . . . . . . . . . . . 13Stefan-Boltzmann constant. . . . . . .6

U

ultraviolet catastrophe . . . . . . . . . . . 8

W

wavelength . . . . . . . . . . . . . . . . . . . . . 11wavenumber . . . . . . . . . . . . . . . . . . . . 11

Z

Zeeman effect . . . . . . . . . . . . . . . . . . . 42

50

Physics IV - Script of the Lecture - qudev.phys.ethz.ch · 3 Solutions to Schr˜odinger’s...

Documents

Transcript of Physics IV - Script of the Lecture - qudev.phys.ethz.ch · 3 Solutions to Schr˜odinger’s...