Part A Introductory Quantum Mechanics

2010, 1st edition

Abstract

These notes encompass the Part A (second year) physics course on introductory quan-tum mechanics lectured at the University of Oxford in the academic year 2006-2007.The Michaelmas Term part is concerned with enunciating the quantum postulates aswell as giving simple examples of 1-dimensional quantum mechanics. The Hilary Termpart takes us into full-blown 3D quantum mechanics which will be used to study suchsystems as the Hydrogen and Helium atoms. The course was lectured by John March-Russell and the notes were typeset by Simon Ellersgaard Nielsen.

Contents

1 Part I: Michaelmas Term 21.1 Lecture 1: The Genesis of Quantum Mechanics . . . . . . . . . . . . . . . . . . 2

1.1.1 The problems of classical physics . . . . . . . . . . . . . . . . . . . . . . 21.1.2 The photoelectric effect . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.1.3 The Bohr atom . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.4 Particles and waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.2 Lecture 2: The Schrodinger equation . . . . . . . . . . . . . . . . . . . . . . . . 61.2.1 Dynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2.2 What does ψ(x, t) mean? . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.3 Lecture 3: ψ(x, t) and probability . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.1 The Born postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3.2 The Infinite Square Well . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

1.4 Lecture 4: Energy eigenstates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4.1 Eigenstates of H . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4.2 On superpositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.5 Lecture 5: Hermitian operators and completeness . . . . . . . . . . . . . . . . 121.5.1 Operators and observables . . . . . . . . . . . . . . . . . . . . . . . . . 121.5.2 Proof of reality of eigenvalues . . . . . . . . . . . . . . . . . . . . . . . . 141.5.3 The expansion theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

1.6 Lecture 6: Dirac formulation of QM . . . . . . . . . . . . . . . . . . . . . . . . . 161.7 Lecture 7: Fundamental postulates of QM . . . . . . . . . . . . . . . . . . . . . 19

1.7.1 The postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.7.2 How do we recover wavefunctions? . . . . . . . . . . . . . . . . . . . . 211.7.3 Rules for amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

1

1.8 Lecture 8: Conserved quantities and the potential step . . . . . . . . . . . . . . 231.8.1 Conserved quantities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231.8.2 Probability current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.8.3 The potential step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

1.9 Lecture 9: The conservation equation and compatible observables . . . . . . . 271.9.1 The Ehrenfest theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271.9.2 More on commutators . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.9.3 Complete sets of quantum numbers . . . . . . . . . . . . . . . . . . . . 30

1.10 Lecture 10: The uncertainty principle . . . . . . . . . . . . . . . . . . . . . . . . 311.10.1 The collapse postulate . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311.10.2 The uncertainty principle . . . . . . . . . . . . . . . . . . . . . . . . . . 331.10.3 Proof of general uncertainty principle . . . . . . . . . . . . . . . . . . . 35

1.11 Lecture 11: The simple harmonic oscillator . . . . . . . . . . . . . . . . . . . . 361.11.1 Two approaches to the SHO . . . . . . . . . . . . . . . . . . . . . . . . . 361.11.2 The spectrum of the SHO: . . . . . . . . . . . . . . . . . . . . . . . . . . 38

1.12 Lecture 12: Scattering, tunneling, and finite potential wells . . . . . . . . . . . 401.12.1 The potential step . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401.12.2 The finite square well: . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

2 Part II: Hilary Term 452.1 Lecture 13: Three dimensional TDSE . . . . . . . . . . . . . . . . . . . . . . . . 45

2.1.1 The cubical box and degeneracy . . . . . . . . . . . . . . . . . . . . . . 452.1.2 The 3D SHO and degeneracy . . . . . . . . . . . . . . . . . . . . . . . . 47

2.2 Lecture 14: Spherical symmetry and angular momentum . . . . . . . . . . . . 482.2.1 L2 in classical and quantum mechanics . . . . . . . . . . . . . . . . . . 482.2.2 The angular momentum quantum numbers . . . . . . . . . . . . . . . 50

2.3 Lecture 15: Quantum systems with V = V (r) . . . . . . . . . . . . . . . . . . . 512.3.1 The angular momentum eigenvalue equations . . . . . . . . . . . . . . 512.3.2 Appendix: The spherical harmonics . . . . . . . . . . . . . . . . . . . . 54

2.4 Lecture 16: Angular momentum: a better method . . . . . . . . . . . . . . . . 542.5 Lecture 17: Spin and the radial equation . . . . . . . . . . . . . . . . . . . . . . 59

2.5.1 The meaning of the angular momentum operator results . . . . . . . . 592.5.2 The radial equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

2.6 Lecture 18: The Hydrogen atom . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.7 Lecture 19: Hydrogen continued and spin . . . . . . . . . . . . . . . . . . . . . 68

2.7.1 Normalization of the Hydrogen atom . . . . . . . . . . . . . . . . . . . 682.7.2 Spin 1/2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

2.8 Lecture 20: Dipole moments and the Stern-Gerlach experiment . . . . . . . . . 712.8.1 Spin dipole moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 712.8.2 The Stern-Gerlach experiment . . . . . . . . . . . . . . . . . . . . . . . 73

2.9 Lecture 21: Multi-particle systems . . . . . . . . . . . . . . . . . . . . . . . . . 752.9.1 Two-particle systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 752.9.2 The addition of classical angular momentum . . . . . . . . . . . . . . . 77

2.10 Lecture 22: The addition of angular momentum in QM . . . . . . . . . . . . . 782.10.1 The ang. mom. quantum numbers of 2-particle systems . . . . . . . . 782.10.2 The triangle rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

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1 Part I: Michaelmas Term

1.1 Lecture 1: The Genesis of Quantum Mechanics

1.1.1 The problems of classical physics

At the start of the 20th century there were a number of outstanding problems to do withthe structure of matter. Broadly speaking, there were two areas:

One area concerned atoms - in particular with respect to stability, structure and radiation.Now it was known that atoms have both positive and negative charges in bound states, butif one orbits around the other then classical electromagnetic theory predicts that the systemradiates energy (EM radiation due to accelerating charges) which would cause a collapsein a very short time, so why are atoms stable?Moreover, experiments with discharge tubes by Balmer, Lyman and others showed that theradiation (IR, visible, UV, etc.) emitted by excited atoms cam in discrete spectral lines. TheBalmer series from Hydrogen epitomizes this, and a well-known formula for the frequencyof the absorption/emission lines is:

ν

c= R

(1− 1

n2

)where R is the Rydberg constant and n = 2, 3, 4, 5 etc. Several other series had also beenidentified together with the ”rule” for the frequencies. But why is the spectrum discrete?And why do we have these simple rules for ν for Hydrogen?

The second area concerned black body radiation. All bodies at a temperature T > 0 emitelectromagnetic radiation (and absorb it if in an equilibrium). Schematically, a black bodycan be construed by from a box surrounded by heating coils s.t. radiation exists in equi-librium inside the box. Now a small aperture is made in one of the walls enabling radia-tion to emerge (and be detected). According to classical physics (statistical mechanics andMaxwell’s theory of EM) the amount of electromagnetic radiation per unit volume per unitfrequency is

δE ∼ ν2δν

in the range ν to ν + δν. This is the Rayleigh-Jeans law. but this implies the nonsensical resultthat the total energy density is

Etot ∼∫ ∞

0

ν2dν →∞

(Infinite energy density at high ν for arbitrarily small temperature). Experimentally, thespectrum looks like this

3

where the position of the peak moves to higher µ as T increases.Max Planck’s great contribution was to realize that the problem was due to the continuousnature of radiation energy in classical theory and that the correct spectrum1:

ρ(ν) = (constants)ν3dν

exp(hνkT

)− 1

arose from the assumption that EM radiation of frequency ν requires a minimum excitationenergy

∆E = hν (1)

where h def= Planck’s constant = 6.626 · 10−34Js.This is completely different from classical theory - it is, if you will, the birth of the quantumand h is a new fundamental constant of nature2.

1.1.2 The photoelectric effect

• An equally important contribution was Einstein’s analysis of The photoelectric effect. Theimportance of this experiment is that it showed how Planck’s quanta are real irreducibleentities and not just something that exist in a statistical sense.The basic set-up of the experiment is as follows: radiation incident on a cathode knocks outelectrons, which are then attracted to an anode by an electric field, thence giving rise to adetectable current.

1Planck derived this perfect fit to the blackbody spectrum.2It might be interesting to read Parisi’s article about subtleties in Planck’s argument.

4

Classical theory says that the energy of light is determined by the intensity of light. Thisenergy is delivered continuously by the EM waves and the energy of light is independentof frequency.Thus, classical theory predicts that once light is turned on, the electrons in the cathode startabsorbing energy until eventually they have gained enough to free themselves from themetal surface (this minimum amount is the work function, w, of the metal). Thus

1. There should be a delay between switching on the light and observing the current.

2. If the intensity of light is decreased the delay should increase.

3. The frequency of the light should not matter at all, 1. and 2. should still be true.

These are the predictions of the very well verified wave theory of light.In fact what happens is

I There is no delay in the current flow (delay < 10−9 sec.).

II Decreasing intensity decreases the size of the current, but no effect otherwise.

III The frequency nu is crucial: if nu is too small no current flows no matter how long we wayor how high the intensity is.

Einstein (1905) explained these results by hypothesizing that

Light should be regarded as consisting of irreducible, particle like objects (pho-tons) each with energy E = hν.

This appears to be in total contradiction to 150 years of experiments proving light was awave (Young, Huygens,..., Maxwell, Hertz, Fraunhofer, ...) and even in 1915 it was re-garded as a tremendous mistake by Einstein.

Anyhow, Einstein’s hypothesis had the following consequence:

If a photon (γ) hits an electron (e−) in the cathode, the electron absorb an energyof hν. For e− to escape the surface, its energy must be greater than w − ν.

Thus, Einstein said

1. If ν falls below (w − ν)/h no current can flow (ignoring the unlikely event of two γshitting the same e−).

2. There is no delay - we only need one photon to hit some electron on the cathode fora current.

3. Decreasing the intensity decreases the number of γs per unit time, hence the rate atwhich the e−s are ejected, hence the current.

These all agreed. Furthermore, Einstein predicted that the kinetic energy of the electrons is

KEe− = 12mev

2 = hν − (w − ν)

which was later verified in an experiment by Millikan (1916).

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1.1.3 The Bohr atom

Bohr (1913) was able to derive the Balmer series etc. for Hydrogen by supposing that theelectron is in an orbit around the nucleus with quantized angular momentum

J =nh

2π, n = 1, 2, 3, . . .

(note: [J ] = [h]) and the electron can make discontinuous transitions from one orbit to another,with change in energy E − E′ appearing as radiation with frequency

ν =E − E′

h.

This was a very important breakthrough as it showed that the quantum hypothesis was usefulfor atoms, not just light. But on closer inspection there are a number of things physicallywrong with the Bohr atom - so we’ll wait to do Hydrogen properly via the Schrodingerequation.

1.1.4 Particles and waves

Einstein and Planck argued that light has particle-like properties even though it of coursealso is wave-like, as shown by interference effects (Young’s slits, diffraction) and the successof Maxwell’s equations.Really:

Light is neither a particle nor a wave, but some kind of new entity that has as-pects of both (Feynman called these entities ’wavicles’ as a serious joke).

De Broglie (1923) had the insight to propose that every object - matter as well as light - wasreally one of these new entities. Thus, he was motivated to associate a wavelength, λ, withmatter (the so-called de Broglie wavelength).What should λ be? Recall for γs: E = hν = hc/λ. But for photons E = pc (see problem set)so equally λ = hc/E = h/p and de Broglie took for matter the definition

λde Brogliedef=

h

p(2)

(p is the momentum of the particle) and suggested that diffraction effects for matter shouldbe searched for.

6

Evidence for this hypothesis ultimately came from an experiment performed by Davissonand Gerner. It is not so easy to see diffraction effects for, e.g., e−1s as λ is much shorter (e.g.a 10 eV electron has λ = 2.9 A, where 1 A= 10−10 m). Davisson and Gerner (1925) used aNickel crystal as a diffraction grating

Thus, we do get diffraction of matter!

Finally, the Compton effect confirmed the particle nature of light (see problems).

1.2 Lecture 2: The Schrodinger equation

1.2.1 Dynamics

Heisenberg/Schrodinger in 1925/1926 developed two totally equivalent formulations ofnon-relativis-tic quantum mechanics. We’ll start with Schrodinger: de Broglie had postu-lated wave associated with every particle, so Schrodinger introduced the wave function,ψ(x, t), and took ψ(x, t) to satisfy the partial partial differential equation

Hψ(x, t) = i~∂ψ(x, t)∂t

(3)

- the so-called Schrodinger equation, where ~ = h/2π. H is the Hamiltonian - the partialdifferential operator for the total energy. For a particle moving in 1D:

H =p2

2m+ V (x) (4)

where in QM the momentum operator is

p = −i~ ∂

∂x. (5)

These are hypotheses of QM verified by experiment, with p as given in the Sch. eqn. (in 1D)reads:

−~2

2m∂2ψ

∂x2+ V (x)ψ = i~

∂ψ

∂t. (6)

Try solution via separation of variables method,ψ(x, t) = φ(x)T (t) so[−~2

2m∂2φ(x)∂x2

+ V (x)φ(x)]T (t) =

[i~∂T (t)∂t

]φ(x)

7

Divide both sides by φT to get

1φ(x)

[−~2

2m∂2φ(x)∂x2

+ V (x)φ(x)]

︸ ︷︷ ︸only a function of x

=i~T (t)

∂T (t)∂t︸ ︷︷ ︸

only a function of t

.

The only way that LHS = f(x) and RJS = g(t) is if actually both equal a constant (E).Thus we get two eqns:

~2

2md2φ

dx2+ V (x)φ = Eφ, (7)

i~dT

dt= ET. (8)

(8) is easy to solve: T (t) = constant× e−iEt/~. (7) is called the time independent Schrodingerequation (TISE). It can be very hard to solve as it depends on V (x).Consider the easy case of the free particle (V = 0). For V = 0 TISE is

−~2

2md2φ

dx2= Eφ

with solutions φ ∼ eikx, e−ikx where ~2k2

2m = E. Thus putting T (t) and φ(x) together wehave

ψ(x, t) ∼ ei(kx−Et/~) −wave in +ve x-direction.

ψ(x, t) ∼ e−i(kx+Et/~) −wave in -ve x-direction.

So the angular frequency is ω = E/~ whence E = ~ω = hv (so asymptotically incorporatesEinstein’s relation). Also for classical free particle E = p2

2m so get E = ~2k2

2m = p2

2m whereforep = ~k = ~ 2π

λ = hλ (so automatically incorporates de Broglie’s ’matter wave’ relation).

1.2.2 What does ψ(x, t) mean?

ψ(x, t) cannot be a physical wave like oscillating string, or EM wave of Maxwell theory.TDSE is a complex equation (the i) - solutions are inherently complex (real and imag. parts of ψdo not separately solve equation). Something which is complex cannot be directly measured (allexperiments return real results!). Also note that for a physical wave eik(x−vpt), where vp isthe phase velocity, but for ψ(x, t):

ψ(x, t) = ei(kx−Et/~) = ei~ (px−Et)

so vp = Ep = (p2/2m)

p = p2m = v

2 - i.e. half the speed of particle with momentum p and massm. On the other hand, the group velocity vg = dω

dk = dEdp = p

m = v. This is very different toe.g. EM waves where oscillating E and B have vp = vg = c. So we know what ψ(x, t) isnot ... next lecture we’ll see what it is.

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1.3 Lecture 3: ψ(x, t) and probability

1.3.1 The Born postulate

To understand ψ(x, t) we must introduce

Probability density = P (x, t) (9)

The basic idea is that we can no longer be certain of exact position of particle - only thatprobability of measuring it between x and x+ δx is Prob(in x to x+ δx) = P (x, t)δx and (Born,1926) a postulate of QM is that

P (x, t) = ψ∗(x, t)ψ(x, t) = |ψ(x, t)|2 (10)

If we know P (x, t) (from a solution ψ(x, t) of TDSE) then we can calculate averages or expec-tation values. Discretely,

x ≈∑i=1

xiP (xi)δx.

In the limit as δx→ 0 and the number of slices→∞, the sum becomes an integral:

xdef= 〈x〉 =

∫ +∞

−∞xP (x, t)dx =

∫ +∞

−∞x|ψ(x, t)|2dx (11)

Similary,

〈x2〉 def=∫ +∞

−∞x2|ψ(x, t)|2dx

etc. Of course, the total probability of finding the particle anywhere must be 1, so we have anormalization condition on ψ(x, t):

1 =∫ +∞

−∞P (x, t)dx =

∫ +∞

−∞|ψ(x, t)|2dx (12)

This must be imposed on ψ(x, t) as the Schrodinger eqn is (exactly - not an approximation)linear in ψ.

If ψ solves Hψ = i~∂ψ∂t so does ψ × constant.

Thus, the value of undetermined constant is fixed (up to irrelevant const. phase factor - ignore)by the normalization condition.Complications arise for plane waves, ψ(x, t) = Ge

i~ (px−Et), so∫ +∞

−∞|ψ|2dx =

∫ +∞

−∞|G|2 →∞

regardless of G. Plabe wave is not renormalizable! Reason: plane waves are not physicallyrealizable (they exist for all time and are spread over all space). A more physical situationis a ’plane wave’ confined to some finite region (for example, our experimental apparatus),then

∫ L−L dx|G|

2 = 1⇒ G = 1√2L

- ok!

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1.3.2 The Infinite Square Well

. Let us now investigate solutions to TDSE in the following potential

V (x) =

+∞, x ≤ 00, 0 ≤ x ≤ a+∞, x ≥ a.

Since V 6= V (t) solution is of the form ψ(x, t) = φ(x, t)e−itE~ where

~2

2md2φ

dx2+ V (x)φ = Eφ, (13)

In regions where V = ∞, φ mus t be zero, φ = 0, for finite E solutions (as KE ≥ 0). Insidethe well, since V = 0,

− ~2

2md2φ

dx2= Eφ

so

φ = Aeikx +Be−ikx

withE = ~2k2

2m . Now we have to match inside to outside solutions via the boundary condition:φ must be continuous.

• At x=0

0 =(Aeikx +Be−ikx

)x=0

= A+B

• At x=a

0 = Aeika +Be−ika.

Since 1st boundary condition gives B = −A we get

0 = A(eika − e−ika

)= 2iA sin(ka)

⇒ Either A = 0 (reject) or sin(ka) = 0 whence

k =nπ

an = 1, 2, 3, ... (14)

Substituting back into the expression for E we get that the possible energies of the particleare quantized:

E =~2

2mπ2

a2n2 n = 1, 2, 3, ... (15)

Typical for particles bound in a potential well (atoms, ...). We also have to normalize thesolution φn(x) = A′ sin

(nπxa

).

1 =∫ +∞

−∞|φ(x)|2dx = A′2

∫ a

0

sin2(πnxa

)dx = A′2 a2

10

so A′ =√

2/a. So

φn(x) =

√2a

sinnπx

a(16)

Note that the φn are orthogonal:∫ +∞

−∞φ∗n(x)φm(x)dx = 2

a

∫ a

0

sin(nπxa

)sin(mπxa

)dx = δmn

What do the φn look like? For n = 1 (the ground state) the wave has no nodes and isstrongly peaked in the middle.

For n = 2, the first excited state, there is one node:

The physics of low n states is very different from classical expectations. Classically we expectthat P (x) = 1

a (”ball bouncing back and forward”) s.t.

〈x〉 =∫ a

0

xP (x)dx =1a

∫ a

0

xdx =a

2and 〈x2〉 =

∫ a

0

x2P (x)dx =1a

∫ a

0

x2dx =a2

3

But at the quantum level (see problem set), using P (x) = |φn|2, we get

〈x〉n=1 =a

2(same) and 〈x2〉n=1 =

a2

3− a2

2π2(differ).

As n→∞we find that 〈x2〉 → ’the classical limit’, and the probability distribution becomesmore spread out (example of general behaviour).

1.4 Lecture 4: Energy eigenstates

1.4.1 Eigenstates of H

A wave function of the form ψn(x, t) = φn(x)e−iEnt/~ is special in the sense that it is anenergy eigenstate or equivalently a stationary state. To see this notice that

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Hψn(x, t) =(

~2

2m∂2

∂x2 + V)ψn(x, t)

= i~ ∂∂tψn(x, t) (by TDSE)

= Enψn(x, t) (substituting the expression for ψn(x, t))

So psin(x, t) is an eigenfunction of H with eigenvalue En. Physically,

Energy eigenstate⇔ particle is in a state of definite energy (here En)⇔ if enegyis measured, the result is En with unitary probability (i.e. P (En) = 1).

1.4.2 On superpositions

The wave ψn(x, t) = φn(x)e−iEnt/~ is not the most general solution to TDSE. Since TDSE isan exact linear equation, the general solution is sum of the above:

ψ(x, t) =∑n

anφn(x)e−iEnt/~ (17)

where the an are arbitrary complex coefficients. E.g. superposition of two energy eigen-states:

ψ(x, t) = a1φ1(x)e−iE1t~ + a2φ2(x)e−iE2t~

Act with energy operator to get

Hψ(x, t) = a1E1φ1(x)e−iE1t~ + a2E2φ2(x)e−iE2t~ 6= const.× ψ(x, t)

But ψ is a solution of TDSE, just not an energy E-state as E1 6= E2.For ψ to be a valid wavefunction it must be normalized, i.e.

1 =∫|ψ|2dx =

∫ a

0

a∗1φ1(x)eiE1t~ + a∗2φ2(x)eiE2t~ × a1φ1(x)e−iE1t~ + a2φ2(x)e−iE2t~

=∫ a

0

|a1|2φ21 + |a2|2φ2

2 + a1a∗2ei(E2−E1)t~φ1φ2 + a∗1a2e

i(E1−E2)t~φ1φ2

But φ1 and φ2 are orthogonal and normalized so we find

1 = |a1|2 + |a2|2 + 0 + 0

A big hint: looks like the sum of probabilities. Thus, interpretation of superposition: ifmeasurement made of energy of particle with this wavefunction will get result

E1 with amplitude a1, probability |a1|2.

E2 with amplitude a2, probability |a2|2.

The expectation value of energy should be

〈H〉ψ = E1|a1|2 + E2|a2|2

12

Can check this by calculating 〈H〉 directly ...

〈H〉 def=∫ a

0

dxψ∗Hψ =∫ a

0

dxψ∗(Hψ)

=∫ a

0

dxa∗1φ1(x)eiE1t~ + a∗2φ2(x)eiE2t/~a1E1φ1(x)e−iE1t~ + a2E2φ2(x)e−iE2t~

= E1|a1|2 + E2|a2|2

using the orthogonality of the φis. Note that 〈H〉 is time independent. However, not allexpectation values are time independent for energy superpositions e.g.

〈px〉def=∫ a

0

dxψ∗(−i~ ∂

∂xψ

)=∫ a

0

a∗1 sin πxa e

iE1t/~ + a∗2 sin 2πxa eiE2t/~(−i~)

2aa1

π

acos πxa e

−iE1t/~ + a22πa

cos 2πxa e−iE2t/~

Using the integrals ∫ a

0

dx sin πxa cos πxa =

∫ a

0

dx sin 2πxa cos 2πx

a = 0

∫ a

0

dx sin 2πxa cos πxa = 4a

3π∫ a

0

dx sin πxa cos 2πx

a = − 2a3π

we get

〈px〉 = −i~ 2aa1a

∗2ei(E2−E1)t/~ π

a4a3π + a∗1a2e

i(E1−E2)t/~ 2πa

(−2a3π

)

= − i~83a

(a1a∗2ei(E2−E1)t/~ − c.c.

)In the simple case where a1, a2 ∈ R, 〈p〉 = 16~

3a a1a2 sin (E2−E1)t~ so momentum oscillates in

time.

1.5 Lecture 5: Hermitian operators and completeness

1.5.1 Operators and observables

We have already seen how numerical values of classical quantites (e.g. momentum, energy)become differential operators in QM.

x-momentum: px = −i~ ∂∂x .

Energy: H = p2

2m + V (x) = − ~2

2m∂2

∂x2 + V (x).

13

This is a general principle of QM: every physical observable corresponds to an operatorQ.How are the numbers measured in expectations related to operator? For energy, we foundHφn = Enφn. This is an eigenvalue equation with En the eigenvalue and φn the eigen-function (just like eigenvalue eqn for a matrix). In general, for operatorQ:

Qχn(x) = qnχn(x) (18)

where χn are the eigenfunctions and qn are the eigenvalues (labeled by n). The set ofall qn is called the spectrum ofQ.A general principle of QM is:

Measurements of a physical observable always give a result which is one of theeigenvalues of the correspondingQ.

Since expectations always return real numbers as results, class of allowable operators mustbe such as to have only real eigenvalues: these are Hermitian operators. So principle of QM:

Every Hermitian operator representing a physical observable must be a Hermi-tian operator.

Definition of Hermitian operator,Q: for 1D QM an op. Q is Hermitian if∫ +∞

−∞χ∗Qψdx =

∫ +∞

−∞(Qχ)∗ψdx (19)

for any functions χ(x), ψ(x) which are normalizable and vanish at x = ±∞.What kind of operators are Hermitian?

I The position operator

∫ +∞

−∞χ∗xψdx =

∫ +∞

−∞(x ∗ χ)∗ψdx =

∫ +∞

−∞(xχ)∗ψdx

so since x ∈ R,

x† = x (notation for Hermitian op.)

and relation trivially true.

II The potential energy V (x)

∫ +∞

−∞χ∗V (x)ψdx =

∫ +∞

−∞(V ∗ χ)∗ψdx

So, as long as coefficients in V (x) are real, have

V (x)† = V (x)

what we expect as complex potential doesn’t make sense.

14

III Momentum operator∫ +∞

−∞χ∗(−i~ ∂

∂x

)ψdx = −i~

∫ +∞

−∞χ∗ ∂ψ∂x dx

We need to get op. acting in χ so integrate by parts:

RHS = −i~(

[χ∗ψ]+∞−∞ −∫ +∞

−∞

∂χ∗

∂x ψdx

)Since x ∈ R: ∂

∂x ∈ R so ∂f∗

∂x = ∂f∗

∂x∗ =(∂f∂x

)∗. So

RHS =∫ +∞

−∞

(−i~∂χ∂x

)∗ψdx

Thus, p is Hermitian. Note, i in definition of p is vital ( ∂∂x is not Hermitian).

IV Kinetic energy operator p2

2m = −~2

2m∂2

∂x2

See problem set (hint: integrate by parts twice).

V Hamiltonian operator

Since H = T + V and T and V are Hermitian, H†

= H . Also, sums of Hermitianoperators are Hermitian.

1.5.2 Proof of reality of eigenvalues

LetQχn = qnχn, thus∫χ∗m implies∫ +∞

−∞χ∗mQχndx =

∫ +∞

−∞χ∗mqnχndx

But by assumptionQ is Hermitian so

LHS =∫ +∞

−∞(Qχm)∗χndx

=∫ +∞

−∞(qmχm)∗χndx

=∫ +∞

−∞q∗mχ

∗mχndx

Comparing this with the RHS we immediately get 0 = (qn−q∗m)∫ +∞−∞ χ∗mχndx. Now choose

n = m then

0 = (qn − q∗n)∫ +∞

−∞|χn|2︸ ︷︷ ︸

=1 by norm.

dx ⇒ qn = q∗n

which establishes that qn is real. QED.

15

But more: choose n 6= m and use reality of q’s then 0 = (qn − qm)∫ +∞−∞ χ∗mχndx. If qm 6= qn

then we must have

0 =∫ +∞

−∞χ∗mχndx

and thus χm and χn are orthogonal functions. This should remind you of the infinite poten-

tial well where φm =√

2a sin πnx

a . Indeed we know by direct calculation∫ +∞−∞ φ∗nφmdx = 0.

1.5.3 The expansion theorem

But yet more: we know for φm’s (sin functions) that have Fourier series. Any function can beexpanded as

f(x) =∞∑n=1

anφn(x)

This is true in general using eigenfunctions of Hermitian operator.

Theorem: let χn(x) be the eigenfunctions of any Hermitian operator. Then anynormalizable function f(x) can be written as f(x) =

∑∞n=1 anχn(x).

•We say that the χn(x) form a complete set of functions (or states).• It is this expansion theorem that enables connection between operators and probabilities. Let’s seehow: suppose at a given time (say t = 0) we have wavefunction ψ(x, 0), and we measureQ. We can expand ψ(x, 0) is eigenfunctions ofQ:

ψ(x, 0) =∑n

anχn(x) (20)

Finding an’s is simple: multiply (20) by χ∗m and integrate:

RHS =∫ +∞

−∞

∑n

χ∗manχn =∑n

∫ +∞

−∞anχ

∗mχn

=∑n

anδnm = am

using orthogonality. Since the LHS is∫ +∞−∞ χ∗m(x)ψ(x, 0)dx it follows that

am =∫ +∞

−∞χ∗mψ(x, 0)dx (21)

Now substitute expanded ψ(x, 0) in eqn for 〈Q〉:

〈Q〉ψ =∫ (∑

n

anχn(x)

)∗Q

(∑m

amχm(x)

)

=∫ (∑

n

anχn(x)

)∑m

amqmχm(x)

=∑m,n

a∗namqm

∫χ∗nχm =

∑n

|an|2qn

16

where |an|2 is the probability that result of single measurement of Q gives qn, and the qnare the possible results of single masurement ofQ. Note that by settingQ = 1 in the abovewe get

∫|ψ|2 =

∑n |an|2 = 1 as desired.

Example:Consider the case of the infinite square well from before. We know that Hφn = Enφn with

φn =√

2a sin πnx

a , En = n2π2v~2

2ma2

Note that the φns are not momentum eigenstates since

−i~∂φn∂x = (−i~)√

2anπa cos nπxa 6= const.× φn

But upon solving the differential equation −i~∂φp∂x = pφp we see that

φp =1√aeipx/~ (22)

Suppose a particle is in an energy eigenstate φn and we decide to measure its momentum.To find the probabilities of the possible outcomes we rewrite the energy eigenfunction interms of φp whence

φn =

√2a

(einπx~/a~ − e−inπx~/a~

)=

1i√

2φpn −

1i√

2φ−pn

where pn by direct comparison is pn = nπ~a . Hence, we find the values

+pn = nπ~a with probability P (pn) = | 1

i√

2|2 = 1

2 .

−pn = −nπ~a with probability P (−pn) = | −1

i√

2|2 = 1

2 .

1.6 Lecture 6: Dirac formulation of QMMore powerful and more general than wavefunctions. All serious physicists use it.

Let’s abstract general features of QM:

•Wavefunctions can be linearly superposed with complex coefficients. This mathematicalstructure is tantamount to a complex linear vector space. The basic object is the state ket,|ψ〉, representing the state of the system. A particular set of states are energy eigenstatesH|n〉 = En|n〉. We also have complex conjugates of wavefunctions and eigenstates; werepresent these by state bras, 〈ψ| or 〈n|.

• The overlap of wavefunctions∫dxφ(x)∗ψ(x) generalises to unitary inner products for bras

and kets:

〈φ|ψ〉 = complex number giving overlap of |ψ〉 and |φ〉. (23)

〈φ|ψ〉 satisfies rules explained in maths methods course, e.g. 〈φ|ψ〉 = 〈ψ|φ〉∗.

17

A ket is normalized if 〈ψ|ψ〉 = 1 and |ψ〉 and |φ〉 are orthogonal if 〈φ|ψ〉 = 〈ψ|φ〉 = 0.

The unitary3 complex vector space describing the possible states of the system is called theHilbert space of the system and is sometimes denoted by curlyH.

• Like any linear vector space H has a dimension determined by the maximal number oflinearly independent vectors:

c1|u1〉+ c2|u2〉+ ...+ cN |uN 〉 = 0

satisfies only if c1 = c2 = ... = cN = 0.A difference with vector spaces you are familiar with is that the dimension of H typicallyis∞! E.g. we know particle in 1D square well has an infinite number of energy eigenstates,φn(x)↔ |n〉, which are linearly independent. However, later in course we’ll meet very im-portant physical systems where H is finite dimensional (spins,...) and cannot be describedby wavefunctions.

• Existence of inner product 〈 | 〉 allows us to construct complete sets of orthonormal kets(a basis forH) via Schmidt procedure: |um〉with 〈un|um〉 = δnm.If we have such a basis then any state |ψ〉 of system can be expanded

|ψ〉 =dimH∑n=1

cn|un〉 (24)

cns determined by taking inner product of this eqn with 〈um|:

〈um|ψ〉 =dimH∑n=1

cn 〈um|un〉︸ ︷︷ ︸δmn

= cm

Thus expansion for |ψ〉 is equivalent to

|ψ〉 =dimH∑n=1

|un〉〈un|ψ〉. (25)

The bras and kets are related by taking the adjoint (complex conjugate transpose for usual

vectors) so eqn |ψ〉 =∑dimHn=1 cn|un〉 implies (|ψ〉)† =

(∑dimHn=1 cn|un〉

)†whence

〈ψ| =dimH∑n=1

c∗n〈un| (26)

Note the c∗n. Thus normalization implies:

3Means: with inner product 〈...|...〉.

18

1 = 〈ψ|ψ〉 =dimH∑n=1

c∗n〈un|dimH∑m=1

cm|um〉

=dimH∑n,m=1

c∗ncm 〈un|um〉︸ ︷︷ ︸δmn

=dimH∑n=1

|cn|2.

• In QM we also have Hermitian operators which represent observables.

A key principle of QM is that all operators are linear, i.e., if |ψ〉 = c1|u1〉 + c2|u2〉then

O|ψ〉 = c1(O|u1〉) + c2(O|u2〉).

To be truthful there is one, and only one exception to this, the ’time reversal’ operator Twhich is anti-linear - this is an advanced (and interesting!) topic.

Operators also act on the vector space of bras linearly (O can act ’left’ or ’right’).

〈ψ|O = (c∗1〈u1|+ c∗2〈u2|) O= c∗1(〈u1|O) + c∗2(〈u2|O).

Since bras and kets exchanged by taking adjoint, it is useful to define action of adjoint onoperators too:

Definition: the adjoint A†

of an operator A is defined by

〈ψ|A†|φ〉 def= 〈φ|A|ψ〉∗

Since Hermitian operators are defined by the criterion A†

= A it follows that for Hermitianoperators only:

A is Hermitian↔ 〈φ|A|ψ〉∗ = 〈ψ|A|φ〉∗ (27)

The quantity 〈φ|A|ψ〉 is called the matrix element of A between states |φ〉 and |ψ〉. If |un〉 isan orthonormal basis then 〈um|A|un〉 ≡ Amn is a matrix element in the conventional sense.

• A surprisingly useful operator secretly appears in the expansion of general |ψ〉: equa-tion (25). since this holds for any ket |ψ〉 we can cancel it from both LHS and RHS in theexpansion, so

I =dimH∑n=1

|un〉〈un| (28)

where I is the identity operator, which leaves all states unchanged:

19

I|φ〉 =dimH∑n=1

|un〉〈un|φ〉 = |φ〉

by expansion theorem for |φ〉 and the orthonormality of the basis. The matrix element of Iis (in a basis):

〈up|I|uq〉 =dimH∑n=1

〈up|un〉︸ ︷︷ ︸δpn

〈un|uq〉︸ ︷︷ ︸δnq

= δpq

Thus written explicitly as a matrix I is represented as

I ↔

111 0 . . . 00 122 . . . 0...

.... . .

...0 0 . . . 1HH

The unit matrix!

• Another useful example is the matrix element of a product of operators

〈up|AB|uq〉︸ ︷︷ ︸(AB)pq

= 〈up|AIB|uq〉

=∑n

〈up|A|un〉〈un|B|uq〉

=∑n

ApnBnq.

which is just the matrix multiplication rule of matrices representing A and B.

Also, our earlier definition of A†

agrees with adjoint of matrices: A†≡ 〈um|A|un〉 ≡

〈un|A|um〉∗ ≡ (A)∗nm ≡ (Amn)∗T.

1.7 Lecture 7: Fundamental postulates of QM

1.7.1 The postulates

I States of a system are represented by normalized kets |ψ〉 or bras 〈φ| in a Hilbert space(which varies from system to system).

II Observables are represented by linear Hermitian operators acting on kets or bras.

III All such Hermitian operators A are assumed to possess a complete set of orthonormaleigenstates:

|1〉with eigenvalue a1.|2〉with eigenvalue a2.

...|n〉with eigenvalue an.

20

Complete set means that |1〉, |2〉, ... forms a basis for the space, so any |ψ〉 can be ex-panded in eigenstates of A (for any A)

|ψ〉 =∑n

|n〉〈n|ψ〉

IV The fundamental probability postulate for measurements is

a Possible results of measurement of A are eigenvalues of A only.b Prob(A = an) = |〈n|ψ〉|2.c After measurement of A with result an the state |ψ〉 is reduced to (’collapses to’)|ψafter〉 = |n〉where the coefficient is 1 and |n〉 is satisfies A|n〉 = an|n〉.

Note: if an is a degenerate eigenvalue (i.e. if more than one linearly independentket has same an eigenvalue) the procedure of reduction is slightly more involved- this is an advanced topic.

Note: subsequent measurements of same A on |ψafter〉 = |n〉 return an with prob-ability = 1 as long as no measurements of other operators B are made at interme-diate times (more on this later).

V In the absence of a measurement the time evolution of the ket |ψ(t)〉 describing the stateof the system at time t changes smoothly in time according to the TDSE:

i~∂|ψ(t)〉∂t

= H|ψ(t)〉. (29)

a The TDSE is a linear equation, which is also deterministic, i,e, given the state att = 0 |ψ(0)〉 the state at a later time t is uniquely determined as long as no mea-surements are performed!b Thus the probabilistic, non-deterministic aspects of QM are purely due to thecollapse of the state upon measurement!c H is the Hamiltonian - the operator corresponds to the energy of the system.d We can formally integrate the TDSE from time = t0 to tf :

|ψ(tf )〉 = e−iH(tf−ti)/~|ψ(t0)〉

where the exponential is defined by its power series, e.g.

eO = 1 + O +O

2

2!+ ...

Since H = H†

(Hermitian) the operator U def= e−iH(tf−ti)/~ is unitary:

U†U = eiH

†(tf−ti)/~e−iH(tf−ti)/~

= eiH(tf−ti)/~e−iH(tf−ti)/~

= 1.

So time evolution is ”unitary evolution”.

21

1.7.2 How do we recover wavefunctions?

Consider the position operator x. This has a continuous spectrum of eigenvalues:

x|x〉 = x|x〉.

The eigenkets |x〉 are normalized as 〈x|x′〉 = δ(x− x′) (which is the analogue 〈n|m〉 = δnmin the discrete case).The expansion of a normalized state |ψ〉 of the particle in terms of position eigenkets reads

|ψ〉 =∫dx|x〉〈x|ψ〉

(Compare this with |ψ〉 =∑n |n〉〈n|ψ〉 from before). Now the general rules of the Dirac

formalism tells us that the interpretation of 〈x|ψ〉 is

〈x|ψ〉 = Probability amplitude that particle in state |ψ〉 is located at x.

I.e., 〈x|ψ〉 is precisely what we previously called the wavefunction! The description of statesby wavefunctions is called the x-representation (or coordinate representation), and Schrodinger’swave mechanics is the form QM takes if the coordinates of a particle are all one cares about(e.g. if no spin, no antiparticle creation,...).It is worthwhile to emphasize that all aspects of wavemechanics can be derived from theDirac formalism. E.g. the concept of overlap is derived as follows:

〈ψ|φ〉 = 〈ψ|I|φ〉 = 〈ψ|(∫

dx|x〉〈x|)|φ〉

=∫dx〈ψ|x〉〈x|φ〉

=∫dx〈x|ψ〉∗〈x|φ〉

=∫dxψ∗(x)φ(x)

In summary, we have seen that the probability amplitude is the crucial object in QM that wemust compute to solve a problem fully.

1.7.3 Rules for amplitudes

• In the x-representation (i.e. usual wavefunctions) the amplitudes are just found by solv-ing the TISE with appropriate boundary conditions (and H!).

•Useful to state some very general rules for probability amplitudes that can be derived fromDirac formulation.

Let’s define an event in an experiment to be a situation in which all the initial andfinal conditions of the experiment are completely specified.

I.e., all positions, angular momenta, ..., of all participating particles are specified.

22

Rule 1 When an event can occur in several alternative ways, the amplitude is the sum ofthe amplitudes for each way considered separately (so get interference).

Rule 2 The amplitude for each separate way an event can occur can be written as theproduct of the amplitude for part of the event occurring that way with the amplitudeof the remaining part.

E.g. in the amplitude for at particle at position x going to position y:

Amp(x→ y) = Amp(x→ z) ·Amp(z → y)⇔ 〈y|x〉 = 〈y|z〉〈z|x〉.

Rule 3 If an experiment is performed which is capable in principle of determining whichof the alternative ways actually was taken (so in fact not all final conditions are thesame) then the total probability is the sum of probabilities for each alternative (NOTE! notamplitudes):

Ptotal = P1 + P2 + ...

(Interference is lost). Note: ’capable in principle’ doesn’t mean that it’s necessary fora human (or other sentient being) to check that all final conditions are the same (it’senough for the state of one atom to be different whether we’re aware or not!).

It is useful to manipulate amplitudes even when we don’t know (yet) exactly their value.E.g. two-slit interference:

〈1|s〉 = Amplitude for particle to go from source s to slit 1.〈x|1〉 = Amplitude for particle to go from slit 1 to point x on the screen.〈2|s〉 and 〈x|2〉 have analogous definitions.So the total amplitude from source s to screen coordinate x is

Amp1+2 = 〈x|1〉〈1|s〉︸ ︷︷ ︸multiply amps.

along route

add differentroutes︷︸︸︷

+ 〈x|2〉〈2|s〉

23

Prob(s→ x) = |total amp.|2 6=Prob(s→ 1→ x) + Prob(s→ 2→ x)

because of the existence of interference terms. Here, Prob(s→ 1→ x) = |〈x|1〉〈1|s〉|2 etc.

1.8 Lecture 8: Conserved quantities and the potential step

1.8.1 Conserved quantities

In classical physics there are many conserved quantities such as energy, momentum andangular momentum. What about QM?

• Consider expectation value of some operator Q which does not have any explicit time-dependence (e.g. x or px).

〈Q〉ψ =∫ +∞

−∞ψ∗(x, t)Qψ(x, t)dx

Because ψ is a function of t we will find, in general, that 〈Q〉ψ is a function of t.

d〈Q〉ψdt

=∫ +∞

−∞

dψ∗(x, t)

dtQψ(x, t) + ψ∗(x, t)Q

dψ(x, t)dt

dx

Using the TDSE Hψ(t) = i~∂ψ(t)∂t to replace the t-derivatives we get

d〈Q〉ψdt

=∫ +∞

−∞

(Hψ(x, t)

i~

)∗Qψ(x, t) + ψ∗(x, t)Q

Hψ(x, t)i~

dx

Then the RHS becomes:

RHS =1i~

∫ +∞

−∞

−ψ ∗ (x, t)HQψ(x, t) + ψ∗(x, t)QHψ

dx

=i

~

∫ +∞

−∞ψ∗(x, t)

(HQ− QH

)ψ(x, t)dx

=i

~〈HQ− QH〉ψ

where we have use the Hermiticity of the Hamiltonian in the first line.

• Q is a conserved quantity if d〈Q〉dt = 0 no matter what state ψ the particle is in.

This can only happen if (HQ− QH

)ψ = 0 (30)

for any normalizable function ψ.

• The object HQ− QH is called the commutator of H and Q, denoted

[H, Q] def= HQ− QH.

24

The statement that [H, Q] = 0 means equation (30). NB: when working out commutatorsof differential operators remember to have them act on some some arbitrary function toavoid mistakes.

ExampleThe simplest case is Q = 1:

d

dt〈ψ|ψ〉 =

i

~〈ψ|[H, 1]|ψ〉 = 0

since ∀A : [A, 1] = 0. Thus, the probability is conserved independent of time, 〈ψ|ψ〉 = 1. Infact, probability (and all other conserved quantities) are conserved locally⇒ it flows aroundsatisfying a local conservation equation.

1.8.2 Probability current

We define ρ = ψ∗ψ as the probability current.

⇒ ∂ρ

∂t=∂ψ∗

∂tψ + ψ∗

∂ψ

∂t(31)

But the TDSE from 3D QM says

i~∂ψ

∂t= − ~2

2m∇2ψ + V ψ

⇒ −i~∂ψ∗

∂t= − ~2

2m∇2ψ∗ + V ψ∗

V ∈ R. Thus, equation (31) becomes

∂ρ

∂t=i

~

(− ~2

2m∇2ψ∗ + V ψ∗

)ψ − ψ∗

(− ~2

2m∇2ψ + V ψ

)= − i~

2m(ψ∇2ψ∗ − ψ∗∇2ψ

)= − i~

2m∇ · (ψ∇ψ∗ − ψ∗∇ψ)

Defining:

j def=i~2m

(ψ∇ψ∗ − ψ∗∇ψ) (32)

we have the conservation equation:

∂ρ

∂t+∇ · j = 0 (33)

Example:Consider the plane wave ψ = ei(px−Et)~, then

jx = − i~2m

(e−ipx/~ ∂

∂xeipx/~ − c.c.

)= − i~

2m2ip~

=p

m= v

25

1.8.3 The potential step

Application of conservation of probability (and superposition of eigenstates).

Classically a particle incident from the left with KE = E with E < V will be reflected bybarrier. What happens in QM?Procedure• Find general solution to TDSE in each region.• Match solutions at x = 0 using the boundary conditions: (i) ψ(x, t) is continuous and (ii)∂ψ(x,t)∂x is continuous. Note: (ii) applies when the jump in V is not infinite.• [Normalize solution].

Region 1We already know that the free (V = 0) TDSE has solution

ψ1 = e−iEt/~ (aeikx + be−ikx)

where E = ~2k2

2m . For the first term (+ve traveling/incoming wave) p = ~k, for the secondterm (-ve traveling/reflected wave) p = −~k. Set a = 1 and b = r (we’ll see why soon).

Region 2

−~2

2m∂2φ2

∂x2+ V φ2 = E′φ2, ψ2 = φ2e

−iE′t/~

Suppose for the moment that E′ > V then

φ2 = ceik′x + de−ik

′x,~2k′2

2m= E − V ′

Now de−ik′x is the -ve traveling wave, but we want particles only incident from the left so

d = 0.

Now impose boundary conditions:(i) ∀t : ψ1(x = 0, t) = ψ2(x = 0, t): ⇒ E = E′ (so energy constant). Also, φ1(0) = φ2(0) so

1 + r = c (34)

(ii) Gives us φ′1(0) = φ′2(0) as the e−iEt/~ cancel. This implies:

ik(1− r) = ik′c (35)

Equations (34) and (35) jointly entail

r =1− k′/k1 + k′/k

(36)

26

c =2

1 + k′/k(37)

Note: we find c 6= 0 - a transmitted wave as expected (for E > V ), but also r 6= 0 - there is areflected wave too (classically, for E > V there is no reflection)!

Let’s now calculate the probability fluxes:

In region 1 we have ψ1 = e−iEt/~ (eikx + re−ikx)

so:

j(1)x =

i~2m

((eikx + re−ikx

) ∂

∂x

(e−ikx + reikx

)−(e−ikx + reikx

) ∂

∂x

(eikx + re−ikx

))=

~km︸︷︷︸

incident

− |r|2 ~km︸ ︷︷ ︸

reflected

In region 2 the wave is ψ1 = ceik′xe−iEt/~ whence:

j(2)x = |c|2 ~k′

m

- not just |c|2! What we care about is really the probability of reflection (R) and transmission(T ). These are given by the ratios:

R =reflected fluxincident flux

, T =transmitted flux

incident fluxSince we divide by the incident flux we do not need to normalize region 1 wave (easiest totake coefficients 1, r). Thus we find

R =|r|2~k/m

~k/m= |r|2 =

(k − k′

k + k′

)2

T =|c|2~k′/m

~k/m=k′

k|c|2 =

4kk′

(k + k′)2

Note that R+ T = 1 (as desired, as total probability = 1).

What happens if E < V ?

Consider region 2. Clearly the TISE d2φ2dx2 = 2m(V−E)

~2 φ2 has solution

φ2 = ce−Kx + deKx

where ~2K2

2m = V − E. The last term explodes as x → +∞, which is physically impossible⇒ d = 0.Again, matching solutions at the boundary

1 + r = c

ik(1− r) = −Kcso

27

r =1− iK/k1 + iK/k

, c =2

1 + iK/k

Now we find that R = |r|2 = 1 (all reflected) and

j(2)x = − i~

2m|c|2

(e−Kx

∂

∂xe−Kx − c.c.

)= 0

So none is transmitted. Note, however, that the wavefunction does penetrate into the clas-sically forbidden region (see figure)!

1.9 Lecture 9: The conservation equation and compatible observables

1.9.1 The Ehrenfest theorem

In the preceding subsection we derived the basic conservation equation

d

dt〈ψ|Q|ψ〉 =

i

~〈ψ|[H, Q]|ψ〉 (38)

So far we have studied the consequences for the simplest case Q = 1. Now, we’ll explorethe relation more generally (as usual, take H = p2

2m + V (x)):

i) The total energy is conserved because [H, H] = 0 so

∀|ψ〉 :d

dt〈ψ|H|ψ〉 = 0

ii) Now consider momentum p:

[H, p] = [ p2

2m +V (x), p] = [ p2

2m , p]+[V (x), p] = 12m

(p2p− pp2

)︸ ︷︷ ︸=0

+[V (x), p] = [V (x), p]

But [V (x), p] = [V (x),−i~ ∂∂x ]. To work this out, put an arbitrary function on the RHS:

[V (x), p]f(x) =V (x)

(−i~ ∂

∂x

)+ i~

∂

∂xV (x)

f(x)

= −i~V ∂f∂x

+ i~∂V

∂xf + i~

∂f

∂x

= i~∂V

∂xf

28

⇒ [V (x), p] = i~∂V∂x (not zero) and therefore [H, p] = i~∂V∂x and

d〈p〉dt

= −⟨∂V∂x

⟩(39)

Note the similarity of (39) to the classical equation of motion F = dpdt = −dVdx . (39)

shows that d〈p〉dt cannot be zero unless ∂V∂x = 0. But this is only true when V = const.,

and find linear momentum conservation equation only when H is independent ofposition (the system is independent of position).

iii) Now take Q = x.

[H, x] = [ p2

2m + V (x), x] = [ p2

2m , x] + [V (x), x]︸ ︷︷ ︸=0

It is easy to check that commutators satisfy

[AB, C] = A[B, C] + [A, C]B, ∀A, B, C

Applying this we find

[H, x] = 12m p[p, x] + [p, x]p

Now calculate [p, X] by acting on an arbitrary function f(x)

[p, x]f(x) =−i~∂x

∂x+ xi~

∂

∂x

f(x)

= −i~∂x

∂xf + x

∂f

∂x− x∂f

∂x

= −i~f(x)

Hence, we get the following fundamental property of x and p operators.

[p, x] = −i~ (40)

Hence,

[H, x] = 12m p(−i~) + (i~)p = −i~ p

m

Thus, (38) implies

d〈x〉dt

=〈p〉m

(41)

and position is not conserved unless in the special case with 〈p〉 = 0. Note: (41) is theQM analogue to the classical equation x = p

m .From (39) and (41) we see that classical equations of motion are obeyed on the averagein QM (Ehrenfest theorem).

29

1.9.2 More on commutators

We have seen that [H, Q] = 0 implies d〈Q〉dt = 0 in QM.

So far we have been working with expectation values (averages) of measurements. Does[H, Q] = 0 have special consequences for individual measurements as well? Yes...

•We know that eigenstates of H satisfy Qφn = Enφn. remember that this means that if

ψ(x, t) = e−iEnt/~φn(x)︸ ︷︷ ︸NOT a

superposition

then when we measure energy we will get the result En (with probability = 1).• Now suppose the same functions φn are also eigenstates of another operator Q i.e.

Qφn = qnφn

This means that if a particle is described by φn then it also has definite values of Q given by qn:

Prob(Q = qn) = 1

•What is the condition that this is possible? Act on Hφn = Enφn with Q:

QHφn = EnQφn = Enqnφn (42)

Alternatively act on Qφn = qnφn with H

HQφn = qnHφn = qnEnφn (43)

Subtract equation (42) from equation (43) to get(HQ− QH

)φn = 0

Since, by assumption, this is true ∀φn we get

[H, Q] = 0 (44)

Thus

If [H, Q] = 0 it is possible for a particle to be in a state of definite energy (astationary state) in which the value of Q is also definite.

•What happens if ψ(x, t) is not an energy eigenstate? E.g. superposition of two eigenstates

ψ(x, t) = a1φ1(x)e−iE1t/~ + a2φ2(x)e−iE2t/~

with 1 = |a1|2 + |a2|2 so Prob(E = Ei) = |ai|2 for i = 1, 2.• Let’s compute 〈Q〉:

30

〈Q〉 =∫ψ∗Qψdx

=∫ (

a∗1φ1(x)eiE1t/~ + a∗2φ2(x)eiE2t/~)(

a1q1φ1(x)e−iE1t/~ + a2q2φ2(x)e−iE2t/~)dx

= q1|a1|2 + q2|a2|2

That is just

〈Q〉 = q1 · (prob in state 1) + q2 · (prob in state 2)

So if we make a measurement of both H and Qwe find

(E1, q1) with Prob = |a1|2.

(E2, q2) with Prob = |a2|2.

The fundamental point is

If [H, Q] = 0 then we can measure both quantities and simultaneously findprecise values of both H and Q,

•What happens if [H, Q] 6= 0? In fact we already studied this when we looked at Q = p

in case of the infinite, square well. Recall [H, p] = i~∂V∂x 6= 0. In particular, for the well, atx = 0: ∂V∂x → −∞ and at x = a: ∂V∂x → ∞. In that case we saw that if a particle has definiteenergy, ψ(x, t) = e−iEnt/~φn(x), then

ψ is a superposition of states with different momenta, or equivalently: we cannotknow both E and p for sure since [H, p] 6= 0.

A general result in quantum mechanics is

Theorem: If A and B are physical (Hermitian) operators with [A, B] 6= 0 thenwe cannot in general simultaneously know values of A and B.

1.9.3 Complete sets of quantum numbers

Suppose we have two operators Q1 and Q2 satisfying [H, Q1] = 0 and [H, Q2] = 0, thend〈Q1〉dt = 0 and d〈Q2〉

dt = 0 and both observables are conserved.•However, if [Q1, Q2] 6= 0 then we will not be able to make definite measurements of both.• On the other hand, if [Q1, Q2] = 0 then it is always possible to find simultaneous eigenstatesof Q1, Q2 and H :

ψ(x, t) = e−iEnt/~φn(x)

which have

31

The quantum numbersof the state ψ =

Definite energy: En

Definite Q1 : q1

Definite Q2 : q2

• In the case [Q1, Q2] = 0, Q1 and Q2 are said to be compatible.• In the case [Q1, Q2] 6= 0, Q1 and Q2 are said to be incompatible, and it is possible for us tohave a state ψ(x, t) = e−iEt/~φ(x) with either

Definite energy: E

Definite Q1 : q1

or Definite energy: E

Definite Q2 : q2

but not both.• E.g. since [p, x] = i~ we can never have a state in which both p and x have definite values.• We say that E, ... form a complete set of quantum numbers iff the maximal number ofsimultaneous eigenvalues are specified.

1.10 Lecture 10: The uncertainty principle

1.10.1 The collapse postulate

We previously looked at states which are eigenfunctions of more than one observable.These are rather special states. Also we want to know what happens when we make mea-surements on some general superposition.Consider, e.g. the energy superposition

ψ(x, t) = a1φ1e−iE1t/~ + a2φ2e

−iE2t/~

Suppose a measurement at the time t = t0 gives the result E1,→ this means that we know forsure that the energy is E1. But this implies that now (i.e. for t ≥ t0) the system must be in theenergy eigenstate:

ψ(x, t > t0) = φ1e−iE1t/~

Thus,

1. The part of the original ψ which has a different energy to the one measured disappears.

2. The remaining part has its coefficient changed so that the new wavefunction is correctlynormalized.

This procedure is the so-called collapse of the wavefunction. It applies when any quantity (notjust energy) is measured. The collapse of the wavefunction (also known as the projectionpostulate of QM) appears to be a distinct type of time evolution (dynamical law) from the TDSE.

32

• The Copenhagen Interpretation of Bohr et al. accepts this.

• Others have tried to derive ’collapse’ from the TDSE for large objects (e.g. you) actingas measuring devices interacting with systems.

• Finally, there are radical proposals, such as Many Worlds, which abolish the collapsepostulate whilst trying to explain why we experience it.

In summary, the collapse postulate us still controversial (e.g. it leads to such fun as Schrodinger’scat), but it is important to stress: every experiment ever performed leads to results which areconsistent with the simple projection postulate and do not require anything fancier.

Let’s see what it predicts for a series of measurements:

i) Start with the ground state in the infinite potential well:

ψ(x, t) = φ1(x)e−iE1t/~

where En = ~2π2n2

2ma2 and φn =√

2a sin nπx

a .

ii) Measure the energy. Result: E1 and the wave remains ψa(x, t) = φ1(x)e−iE1t/~.

iii) Measure the momentum. To work out the possible results we must rewrite ψa asa sum of momentum eigenstates.

ψa = e−iE1t/~√

2a

12ieiπx/a︸ ︷︷ ︸p=+

π~a

− e−iπx/a︸ ︷︷ ︸p=−π~

a

Suppose we find p = +π~a , then the wavefunction collapses to

ψb =1√ae−iE1t/~e−iπx/a

where the prefactor is for normalization.

iv) Now we measure energy again. To work out possible results we must rewrite ψb as asum of energy eigenstates. Firstly, Euler’s formula gives:

1√aeiπx/a =

1√a

(cos

πx

a+ i sin

πx

a

)The second term on the RHS happens to be an energy eigenstate. The first term onthe RHS must be rewritten in terms of sin using Fourier series:

1√a

cosπx

a=∞∑n=1

an

√2a

sinnπx

a︸ ︷︷ ︸energy

eigenfunction φn

33

since ∫ a

0

sinmπx

asin

nπx

adx =

a

2δmn

we get

an =1√a

∫ a

0

√2a

sinnπx

acos

πx

adx =

0 if n = odd

2√

2nπ(n2−1) if n = even

Thus,

1√aeiπx/a =

i√2

(φ1(x)) +∑n even

2√

2nπ(n2 − 1)

(φn(x))

So when we measure the energy we get

E1 with probability P (E1) = 12 .

E2 with probability P (E2) = 329π2 .

...

En with probability P (En) = 8n2

(n2−1)2π2 .

Thus, even though we started with an energy eigenstate, by making a measurement ofmomentum (an operator incompatible with H) we have left the particle in a state of veryindefinite energy!

1.10.2 The uncertainty principle

Considerations such as these led Heisenberg to his famous uncertainty principle. The un-certainty principle is not a separate axiom of QM; it is simply a consequence of the rulesalready stated.When there is uncertainty in the outcome of a measurement it is useful to quantify. Defineuncertainty as:

(∆q)2 def= 〈q2〉 − 〈q〉2 (45)

∆q = 0 if q takes a single value.Heisenberg’s uncertainty principle considers x and p (remember [x, p] = i~) and states

∆x∆p ≥ ~2 . (46)

Before proving this let’s investigate some aspects of its meaning: it says that if we preparea particle in a state whereby its location x is known to within ∆x, then the uncertainty inits momentum is at least ~

2∆x .

Example: suppose we have a Gaussian wavefunction

φ(x) =1

a1/2π1/4e−

x2

2a2

34

Then we can compute ∆x and ∆p:

〈x〉 =∫ +∞

−∞x|φ(x)|2dx = 0

〈x2〉 =1

a√π

∫ +∞

−∞x2e−

x2

a2 dx =a2

2

⇒ ∆x =[〈x2〉 − 〈x〉2

]1/2=

a√2.

〈p〉 = − i~a√π

∫ +∞

−∞e−

x2

2a2∂

∂xe−

x2

2a2 dx = 0

〈p2〉 = − ~2

a√π

∫ +∞

−∞e−

x2

2a2∂2

∂x2e−

x2

2a2 dx

= − ~2

a√π

∫ +∞

−∞e−

x2

a2

(x2

a4− 1a2

)dx =

~2

2a2

⇒ ∆p =[〈p2〉 − 〈p〉2

]1/2=

~a√

2.

Thus,

∆x∆p =a√2

~a√

2=

~2

which exactly satisfies the Heisenberg uncertainty bound. In fact, a Gaussian wavefunction givesthe least possible value for ∆x∆p.

• Since ∆p 6= 0, it is interesting to ask what the momentum distribution is. We know thatthe x-space wavefunction is

φ(x) =1

a1/2π1/4e−

x2

2a2

In Dirac notation this is 〈x|φ〉 - the amplitude for a particle in state φ to be found at x.What we want is 〈p|φ〉 = φ(p) - the amplitude for a particle in state φ to be measured withmomentum p. This is just a Fourier transform of φ(x):

φ(p) =1√2π~︸ ︷︷ ︸

normalizations.t.

R +∞−∞ |φ|

2

∫ +∞

−∞eipx/~φ(x)dx (47)

or in Dirac notation

〈p|φ〉 =∫ +∞

−∞〈p|x〉〈x|φ〉dx

where

〈p|x〉 = 〈x|p〉∗ =(e−ipx/~√

2π~

)∗= eipx/~√

2π~ .

35

The integral for φ(p) can be done by completing the square in the exponent4:

φ(p) =1√2π~

1√a√π

∫ +∞

−∞e−x

2/2a2eipx/~

=1√2π~

1√a√π

∫ +∞

−∞exp

(x2

2a2 − p2a2

2~2

)dx

=√

a

~√π

exp(−p

2a2

2~2

)(where we have used the result

∫ +∞−∞ dxe−λx

2=√

πλ ). This is another Gaussian, as you

should recall from the theory of Fourier transforms.

1.10.3 Proof of general uncertainty principle

(Non-examinable) Suppose we have two Hermitian operators A and B with [A, B] 6= 0.Defining A := A− 〈A〉 and B := B − 〈B〉 and consider the following wavefunction

φλ = Aψ + iλBψ

(where λ is a real parameter we’re going to vary) and the normalization integral

I(λ) =∫ +∞

−∞dxφ∗λφλ ≥ 0

Since A and B are Hermitian, so are A and B, so

I(λ) =∫ +∞

−∞dx(Aψ + iλBψ

)∗ (Aψ + iλBψ

)=∫ +∞

−∞dx|Aψ|2 + λ2|Bψ|2 + iλ[(Aψ)∗(Bψ)− (Bψ)∗(Aψ)]

(expanding)

=∫ +∞

−∞dxψ∗

(A

2 + λ2B2 + iλ[A, B]

)ψ (Hermiticity)

= (∆A)2 + λ2(∆B)2 + iλ

∫ +∞

−∞dxψ∗[A, B]ψ (∆A,∆B definition)

= (∆A)2 + λ2(∆B)2 + iλ〈[A, B]〉 (constant pieces cancel)

Thus we learn that

(∆A)2 + λ2(∆B)2 + iλ〈[A, B]〉 ≥ 0 (48)

and a minimum must exist when d(LHS)dλ = 0:

2λ(∆B)2 + i〈[A, B]〉 = 0

4Recall that if we have the expression−xᵀAx± bᵀx, where A is symmetric, then we complete the square byperforming the mapping x 7→ x± 1

2A−1b. The expression becomes −xᵀAx + 1

4bᵀA−1b.

36

Substituting λ solution of this into equation (48) we get

(∆A)2 − 〈[A, B]〉2

4(∆B)2+〈[A, B]〉2

2(∆B)2≥ 0

⇒ (∆A)2(∆B)2 ≥ 14〈i[A, B]〉2 (49)

Useful to know! So in general any two operators with [ , ] 6= 0 will have an uncertaintyrelation (though the RHS will depend on the details). For the particular case of [p, x] = −i~we get

∆x∆p ≥ ~2 .

1.11 Lecture 11: The simple harmonic oscillator

1.11.1 Two approaches to the SHO

This is an important case because many systems to leading approximation are the S.H.O.(or many weakly coupled S.H.O.s).The potential energy is V (x) = 1

2kx2 so the TISE is

− ~2

2md2φ

dx2+

12kx2φ = Eφ

It is very convenient to rescale the x variable by x = αy:

− ~2

2mα2

d2φ

dy2+

12kα2y2φ = Eφ

and insist that the coefficients of the kinetic energy and potential energy terms are the same

~2

2mα2=

12kα2 ⇔ α2 =

~2

√mk

.

Then the TISE becomes

~√k

m

12

(−d

2φ

dy2+ y2φ

)= Eφ.

Now√

km = ω, the classical angular frequency, so let

ε = E/(~ω

2

)i.e. ε is the energy in units of ~ω

2 . Thus, our TISE becomes

−d2φ

dy2+ y2φ = εφ. (50)

It’s easy to check that

φ0 = e−y2/2

37

is a solution (clearly it is normalizable). Let’s do it:

φ′0 = −ye−y2/2

φ′′0 = y2e−y2/2 − e−y

2/2.

so TISE reads

−(y2e−y

2/2 − e−y2/2)

+ y2e−y2/2 = εe−y

2/2

⇒ Solution if ε = 1 - the energy eigenvalue. In fact, we’ll see soon that this is the groundstate:

E0 = ~ω2 > 0. (51)

There are two ways of getting the rest:

I Try φ = H(y)e−y2/2 then

φ′ = H ′e−y2/2 − yHe−y

2/2

φ′′ = H ′′e−y2/2 − 2yH ′e−y

2/2 −He−y2/2 + y2He−y

2/2

So the TISE, (50), becomes e−y2/2(−H ′′ + 2yH ′ +H) = εHe−y

2/2 or

H ′′ − 2yH ′ +H(ε− 1) = 0

This is called Hermite’s equation, and it can be solved by the Frobeinius series method.There are normalizable solutions for ε = 2n+ 1; Hn(y) is a Hermite polynomial.

II A much more instructive method which generalizes to many other problems. In the y-coordinatesour Hamiltonian operator is

H = − d2

dy2+ y2

Let’s try to factorize this. Define the operators:

a+def= − d

dy+ y, a−

def=d

dy+ y.

Acting on an arbitrary f(y):

a+a−f =(− ddy + y

)(ddy + y

)f

= −d2fdy2 + y dfdy − y

dfdy − f + y2f

=(−d

2fdy2 + y2f

)− f.

Similarly,

38

a−a+f =(−d

2fdy2 + y2f

)+ f.

So we learn

(a) [a+, a−] = a+a− − a−a+ = −2.(b) We can write the TISE, (50), as

(a+a− + 1)φ = εφ. (52)

(c) a−φ0 =(ddy + y

)e−y

2/2 = 0. Thus, (a+a− + 1)φ0 = 1 · φ0 and again we findε0 = 1.

(d) Now act on the TISE (52) with a+ and use the commutator [a+, a−]:

a+(a+a− + 1)φ = εa+φ

a+(a−a+ − 2 + 1)φ = εa+φ

a+a−a+ + (−2 + 1)a+φ = εa+φ

(a+a− + 1)a+φ = (ε+ 2)a+φ

so we have learnt thatIf φ has eigenvalue ε then a+φ has eigenvalue ε+ 2.

(e) Similarly, it is easy to show thatIf φ has eigenvalue ε then a−φ has eigenvalue ε− 2.

Since a−φ0 = 0 (we say that a− annihilates φ0) there is no state with lower energythan φ0. ⇒ φ0 is indeed the ground state with ε = 1.

(f) By repeated application of a+ on φ0 we generate the entire spectrum! a+ and a−are known as raising and lowering or ladder or creation and annihilation operators.

1.11.2 The spectrum of the SHO:

ε0 = 1 : E0 = ~ω2 φ0 = e−y

2/2

ε1 = 3 : E1 =(1 + 1

2

)~ω φ1 =

(− ddy + y

)φ0 = 2ye−y

2/2

ε2 = 5 : E2 =(2 + 1

2

)~ω φ2 =

(− ddy + y

)φ1 = 2(2y2 − 1)e−y

2/2

......

...

εn = 2n+ 1 : En =(n+ 1

2

)~ω φn =

(− ddy + y

)nφ0 = Hn(y)e−y

2/2

39

(a)

(b)

Figure 1: (a) The first three excited states of the simple harmonic oscillator. (b) The ground state with theclassically forbidden region clearly highlighted.

A complete solution of the problem!

• So far we haven’t normalized the wavefunctions. E.g. φ0(x) = Ce−x2/2α2

(note: we mustconvert back to x):

1 =∫ +∞

−∞dx|φ0|2 = C2

∫ +∞

−∞dxe−x

2/α2= C2

√πα.

so C = 1π1/2α1/4 , where we defined α = ~√

mk. Similarly we can normalize higher φn(x)s.

• Just as important our wavefunctions are also orthogonal:∫ +∞

−∞dxφn(x)φm(x) = 0 n 6= m (53)

This must be true: they are eigenfunctions of the Hermitian operator H with differenteigenvalues εn and εm, c.f. Sturm-Liouville theorem in the Mathematical Methods course.

40

• Note that the φn(x) go even, odd, even, ... and just like the infinite square well the nth

excited state has n nodes. See figure 1(a).• Let’s consider some features of the ground state, viz.

φ0(x) = 1α1/2π1/4 exp

(− x2

2α2

)E0 = ~ω

2 α2 = ~√mk

A classical particle of total energy ~ω2 would be confined to the region where

V (x) ≤ E0

12kx

2 ≤ ~ω2

⇒ x2 ≤ ~ωk = ~

k

√km = α2

i.e.

−α ≤ x ≤ α

If we go back to the TISE

− ~2

2md2φ

dx2+

12kx2φ = Eφ

at the point where E = V (x) (the limit of the classical motion) we see that d2φdx2 = 0 - a point of

inflection for φ. This is true for any state.Notice that there is a non-zero probability Po.c. of being outside the classically bounded region,which is most significant for low energy states. In particular Pn=0

o.c. = 0.157, Pn=1o.c. = 0.116,

Pn=2o.c. = 0.095 etc. The ground state is exemplified in figure 1(b).

1.12 Lecture 12: Scattering, tunneling, and finite potential wells

1.12.1 The potential step

Consider the following potential: V (x) = V0 if x ∈ [0, a] - otherwise V (x) = 0:

with a particle of energy E < V0 incident from the left. There are three distinct regions inwhich the solutions to the TISE are:

ψI = eikx︸︷︷︸incident

+ re−ikx︸ ︷︷ ︸reflected

, ~2k2

2m = E

ψII = AeKx +Be−Kx, ~2K2

2m = V0 − E

ψIII = te+ikx︸ ︷︷ ︸transmitted

, ~2k2

2m = E

41

Now we must impose the boundary conditions at x = 0 and x = a:

At x = 0: The continuity of ψ implies that 1 + r = A + B and the continuity of ∂ψ∂x implies

that ik(1− r) = K(A−B).

At x = a: The continuity of ψ implies that AeKa +Be−Ka = teika and the continuity of ∂ψ∂ximplies that K(AeKa −Be−Ka) = ikteika.

Classically, there is zero transmission through the barrier. We want to find the transmissioncoefficient t in the quantum mechanical case.Eliminating r from the x = 0 equations we get

2 = A(1− iK

k

)+B

(1 + iK

k

)(54)

Eliminating A and B respectively from the x = a equations we get the expressions

2AeKa =(1 + ik

K

)teika

2Be−Ka =(1− ik

K

)teika

(55)

Substituting equations (55) back into equation (54):

2 = teika

12e−Ka (1− iK

k

) (1 + ik

K

)+ 1

2eKa(1 + iK

k

) (1− ik

K

)= teika

12e−Ka (2 + i

[kK −

Kk

])+ 1

2eKa(2− i

[kK −

Kk

] )= teika

2 coshKa− i

(kK −

Kk

)sinhKa

Thus,

t =2e−ika

2 coshKa− i(kK −

Kk

)sinhKa

and the probability of transmission is

T = |t|2 =4

4 cosh2Ka+ (k2−K2)2

k2K2 sinh2Ka(56)

NB: the factors of ~km incident and transmitted flux cancel (as k is the same).

There is much physics in equation (56); let’s look at some cases:

• When Ka is large

T ≈ 4e−2Ka

1 + (k2−K2)2

4k2K2

> 0

This is typical quantum tunneling behaviour, in particular the exponential damping ofthe tunneling probability with the width of the barrier.

• When E = V0

K2 = 2m~2 (V0 − E) = 0

42

and this implies that

T =4

4 + k2a2< 1

Note that even though one classically would get perfect transmission, here T 6= 1,and thence, quantum mechanically, there is some reflection:

R = 1− T =k2a2

4 + k2a2.

• When E > V0 then K2 = 2m~2 (V0 − E) < 0 so write

K = ik = i[

2m~2 (E − V0)

]1/2Since cosh ix = cosx and sinh ix = i sinx this substitution gives

t =2e−ika

2 cos ka−(kk

+ kk

)i sin ka

Giving

T = |t|2 =4

4 cos2 ka+(kk

+ kk

)2

sin2 ka

which is less than 1 unless ka = nπ (a phenomenon known as resonant transmission).Classically, we of course have T = 1 for E > V0.

1.12.2 The finite square well:

Now lets us consider the potential where V (x) = 0 if x ∈ [−a, a], otherwise V (x) = V0.Again, writing down the respective wavefunctions for the three regions:

ψI = AeKx, ~2K2

2m = V0 − E

ψII = B cos kx+ C sin kx, ~2k2

2m = E

ψIII = De−Kx, ~2K2

2m = V0 − E

Applying the standard boundary conditions

43

Ae−Ka = B cos ka− C sin ka,

KAe−Ka = kB sin ka+ kC sin ka,

De−Ka = B cos ka+ C sin ka,

−KDe−Ka = −kB sin ka+ kC cos ka.

These equations are a pain to solve by hand - nowadays we’d use a symbolic math program(e.g. Maple, Mathematica, ...) but it is worthwhile to study some features. By writing theboundary conditions in matrix form

e−Ka − cos ka sin ka 0Ke−Ka −k sin ka −k cos ka 0

0 − cos ka − sin ka e−Ka

0 k sin ka −k cos ka −Ke−Ka

ABCD

= 0

the requirement for non-trivial solutions (det( ) = 0) gives e−2Ka(K cos ka−k sin ka)(K sin ka+k cos ka) = 0. This implies

K

k= tan ka (57)

in which case: even wavefunctions (⇐ C = 0, D = A,B = e−Ka

cos kaA). Or,

K

k= − cot ka (58)

in which case: odd wavefunctions (⇐ B = 0, D = −A,C = − e−Ka

sin kaA).We need to solve (57) and (58) to get the energies. But these transcendental equationscannot be solved algebraically, so use the graphical method. Let’s write

E = ε~2

2ma2V0 = λ

~2

2ma2

then equations (57) and (58) become√λε − 1 = tan

√ε and

√λε − 1 = − cot

√ε

Using this we can draw graphs of the method of solution (Figure 2) and graphs of thewavefunctions (Figure 3).Note that there is always one even bound state no matter how small V0 (this true of 1DTISE). Also, there is only a finite number of bound states for V0 < ∞ and the formula forenergies is not nearly as simple as for the infinite square potential well.

44

(a) (b)

Figure 2: (a) y = tan√ε plotted against y =

qλε− 1 (even case). (b) y = − cot

√ε plotted against y =

qλε− 1

(odd case).

(a) (b)

Figure 3: (a) The ground state of the finite square well. (b) The first excited state of the finite square well.

45

2 Part II: Hilary Term

2.1 Lecture 13: Three dimensional TDSE

2.1.1 The cubical box and degeneracy

In three dimensions the Hamiltonian becomes

H =1

2m(p2x + p2

y + p2z

)+ V (x, y, z)

and the TDSE is

Hψ(x, y, z) = i~∂ψ(x, y, z)

∂t.

How easy/hard this is to solve depends on V (x, y, z). However, the time-dependence isalways separable, ψ(x, t), and since H is not t-dependent we get

1φ(x)

Hφ(x) =i~T (t)

∂T (t)∂t

= E

So T (t) = e−iEt/~ and Hφ(x) = Eφ(x) similar to the 1D case.

The cubical box:

Let us consider the simple example of a particle inside a potential box of dimensions a×b×c,s.t. V =∞ outside the box (i.e. φ = 0) whilst V = 0 inside the box, where the TISE is

− ~2

2m

(∂2

∂x2+

∂2

∂y2+

∂2

∂z2

)φ = Eφ

To solve this 3D TISE we proceed by using the separation of variables method φ(x) =X(x)Y (y)Z(z) so

− ~2

2m

X ′′(x)X(x)

+Y ′′(y)Y (y)

+Z ′′(z)Z(z)

= E

so

− ~2

2mX ′′ = E1X, − ~2

2mY ′′ = E2Y, − ~2

2mZ ′′ = E3Z

with E = E1 +E2 +E3. Now as X(0) = X(a) = 0 the x-dependent part is exactly the sameas the 1D infinite square well of width a:

46

⇒ E1 =~2n2

1π2

2ma2, X =

√2a

sinn1πx

a.

Similarly for y and z:

E2 =~2n2

2π2

2mb2, Y =

√2b

sinn2πy

band E3 =

~2n23π

2

2mc2, Z =

√2c

sinn3πz

c.

Altogether we have

φ(x) =23/2

√abc

sinn1πx

asin

n2πy

bsin

n3πz

cwith E =

~2

2m

(n2

1

a2+n2

2

b2+n3

3

c2

). (59)

Note: the energy eigenfunctions are labelled by three quantum numbers n1, n2 and n3.Furthermore, the φn1,n2,n3s are correctly normalized:∫ a

0

dx

∫ b

0

dy

∫ c

0

dz|φ(x)|2 = 1

• Now look at the special case of an equal-sided cubical box (a = b = c), then we have:

En1,n2,n3 =~2

2ma2(n2

1 + n22 + n2

3)

and the ground state (∀i : ni = 1) has energy E111 = 3~2

2ma2 . The next state is n1 = 2, n2 =1, n3 = 1 where

E211 =~2

2ma2(22 + 12 + 12) =

6~2

2ma2

but notice that the quantum number values n1 = 1, n2 = 2, n3 = 1 and n1 = 1, n2 = 1, n3 =2 give rise to

E121 =6~2

2ma2, E112 =

6~2

2ma2

Thus, we have three states with the same energy. Such states are called degenerate states andhave 3-fold degeneracy.

• Note that although the energies are equal, the wavefunctions are not - in fact they areorthogonal! E.g.

∫ a

0

dxdydzψ∗112(x, t)ψ121(x, t) =∫ a

0

dxdydz

(√2a

)6

sinπx

asin

πy

asin

2πza

sinπx

asin

2πya

sinπz

a

=(

2a

)3 ∫ a

0

dx sin2 πx

a

∫ a

0

dy sinπy

asin

2πya︸ ︷︷ ︸

=0

∫ a

0

dz sin2πza

sinπz

a︸ ︷︷ ︸=0

= 0

47

2.1.2 The 3D SHO and degeneracy

For the 3D simple harmonic oscillator the potential is V = 12k1x

2 + 12k2y

2 + 12k3z

2 whencethe TISE is

px + py + pz2m

φ+12

(k1x2 + k2y

2 + k3z2)φ = Eφ.

This expression is separable: setting φ(x) = X(x)Y (y)Z(z) we obtain

− ~2

2mX ′′+ 1

2k1x2X = E1X, − ~2

2mY ′′+ 1

2k2y2Y = E2Y, − ~2

2mZ ′′+ 1

2k3z2Z = E3Z

with E = E1 + E2 + E3. Each of these is just the 1D SHO problem so

E1 = ~ω1

(n1 + 1

2

), ω1 =

√k1/m

E2 = ~ω2

(n2 + 1

2

), ω2 =

√k2/m

E3 = ~ω3

(n3 + 1

2

), ω3 =

√k1/m

Again something important happens if k1 = k2 = k3 = k:

E = ~ω(n1 + n2 + n3 + 3

2

), ω =

√k/m

Here, the ground state, n1 = n2 = n3 = 0 is unique with energy E = 32~ω, but the first

excited state, n1, n2, n3 ∈ 0, 0, 1, with energy E = 52~ω, has a 3-fold degeneracy. More-

over, the second excited state n1, n2, n3 ∈ 0, 0, 1 and 0, 0, 2, with energy E = 72~ω,

has a 6-fold degeneracy. In fact, the nth excited state has energy E =(n+ 3

2

)~ω with

degeneracy

dn =(n+ 1)(n+ 2)

2.

As in the cubical well case, all the wavefunctions for these degenerate states are orthogonal.

Why degeneracy? The SHO can be analyzed in a different way. If k = k1 = k2 = k3 then

V = 12k(x2 + y2 + z2) = 1

2kr2

where r is the radial coordinate of the spherical polars. So the TISE is

− ~2

2m∇2φ(r, θ, ϕ) + V (r)φ(r, θ, ϕ) = Eφ(r, θ, ϕ)

with V (r) spherically symmetric gives rise to the degeneracy of excited states (similarly thecubic symmetry of equal sided cubic box gives rise to it’s degeneracy).

48

2.2 Lecture 14: Spherical symmetry and angular momentum

2.2.1 L2 in classical and quantum mechanics

Last lecture we observed that spherically symmetric potentials V = V (r) (and not depen-dent upon θ, ϕ) are special.It is natural to think about angular momentum, L.

Classically: L := r × p so in CM we have

L2 = (r × p) · (r × p)= (r × p)i(r × p)i= εijkrjpkεimnrmpn

= (δjmδkn − δjnδkm)rjrmpkpn= rmrmpkpk − rnrmpmpn= r2p2 − (r · p)2

⇔ p2 =1r2

(r · p)2︸ ︷︷ ︸this just radial

part of p2

+1r2L2︸ ︷︷ ︸

tangential partdetermined by

ang. momentum

(60)

L2

in quantum mechanics: one must be careful as the components of r and p don’t commute:

[x, px] = [y, py] = [z, pz] = i~

[x, py] = [x, pz] = · · · = 0

Equivalently,

[ri, pj ] = i~δij i, j = x, y, z (61)

The representation of the p operator in terms of Cartesian coordinates is p = −i~∇ where∇ =

(∂∂x ,

∂∂y ,

∂∂z

). Classically, angular momentum is L = r × p so the corresponding

operator is

L = −i~r ×∇ (62)

It is easy to see that this is well-defined since in Lx = −i~(y∂z − z∂y), y commutes with ∂zand z commutes with ∂y . Similarly for Ly and Lz .Also, it is simple to check that L is Hermitian.Now, classically L2 = L ·Lwas important. What is L

2in QM?

49

L · L = (−i~)2(r ×∇) · (r ×∇)

= −~2(r ×∇)i(r ×∇)i= −~2εijkrj∂kεimnrm∂n

= −~2(δjmδkn − δjnδkm)rj∂krm∂n= −~2(rm∂nrm∂n − rn∂mrm∂n)

= −~2(rmδmn∂n + rmrm∂n∂n − rnδmm∂n − rnrm∂m∂n)

= −~2(rm∂m + rmrm∂n∂n − 3rn∂n − rn∂nrm∂m − rn∂n)= −~2r2∇2 − r · ∇ − (r · ∇)2

(63)

Rearranging equation (63) gives

−~2∇2 = −~2

r2(r · ∇)2 + r · ∇+

L2

r2(64)

It is most useful to write an explicit form in spherical polar coordinates:

x = r sin θ cosϕ, y = r sin θ sinϕ, z = r cos θ

We need r · ∇. In (x, y, z):

r · ∇ = x∂x + y∂y + z∂z

But

∂r =∂x

∂r

∣∣∣∣θ,ϕ

∂x +∂y

∂r

∣∣∣∣θ,ϕ

∂y +∂z

∂r

∣∣∣∣θ,ϕ

∂z

=x

r∂x +

y

r∂y +

z

r∂z

So r · ∇ = r∂r and thus we can write the radial part of equation (64) as

1r2(r · ∇)2 + r · ∇ =

1r2r∂rr∂r + r∂r =

1r2∂r(r2∂r)

Thus,

−~2∇2 = −~2

r2∂r(r2∂r) +

L2

r2(65)

But we know that∇2 in spherical polars is given by

∇2 =1r2∂r(r∂r) +

1r2 sin θ

∂θ(sin θ∂θ) +1

r2 sin2 θ∂2ϕ

Substituting this into equation (65)we see that the ∂r parts exactly match, and we concludethat

L2

= −~2

1

sin θ∂θ(sin θ∂θ) +

1sin2 θ

∂2ϕ

(66)

50

Note: this is indeed only tangential as it only contains ∂θ and ∂ϕ.Of course, we could also have calculated L

2from L

2= L

2

x + L2

y + L2

z and expressions forLx etc.

Lx = ypz − zpy, Ly = zpx − xpz, Lz = xpy − ypx.

For example, Lz = −i~(x∂y − y∂x) which can be simplified using

∂ϕ =∂x

∂ϕ

∣∣∣∣r,θ

∂x +∂y

∂ϕ

∣∣∣∣r,θ

∂y +∂z

∂ϕ

∣∣∣∣r,θ

∂z

= −y∂x + x∂y

Thus, we have the important relation

Lz = −i~(x∂y − y∂x) = −i~∂ϕ (67)

Proceeding in a similar fashion we would find the previous expression for L2.

2.2.2 The angular momentum quantum numbers

In the classical case Lx,Ly,Lz,L2 are all defined, and specifying any three of them (sayLx,Ly,L2 or Lx,Ly,Lz) completely determines angular momentum. This is not thecase in quantum mechanics. Indeed the observables Lx, Ly, Lz, L

2 do not all commute

with each other (recall: for a state to be an eigenstate of two different operators simultane-ously, those operators must commute). E.g.[

Lx, Ly

]= [(ypz − zpy), (zpx − xpz)]

= y[pz, z]px + (−1)2py[z, pz]x+ all others zero

= −i~ypx + i~pyx

= i~Lz.

In fact, as you should check for yourself, it is also the case that [Ly, Lz] = i~Lx and[Lz, Lx] = i~Ly , so no two operators of Lx, Ly, Lz commute. This can be written com-pactly as follows:

[Li, Lj ] = i~εijkLk (68)

So out of Lx, Ly, Lzwe can at most have one quantum number. It is conventional to pickLz and choose coordinates such that

Lz = −i~ ∂

∂ϕ

However, this is not yet a maximal commuting set of angular momentum operators. Con-sider

51

[Lz, L

2]

= [Lz, L2

x + L2

y + L2

z]

= [Lz, L2

x] + [Lz, L2

y] + [Lz, L2

z]

= Lx[Lz, Lx] + [Lz, Lx]Lx + Ly[Lz, Ly] + [Lz, Ly]Ly

= Lx(i~Ly) + (i~Ly)Lx + Ly(−i~Lx) + (−i~Lx)Ly= 0.

Thus, the maximal set of commuting operators is L2, Lzmax.

Note: we can easily show that [Lx, L2] = 0 and [Ly, L

2] = 0 (obvious by symmetry). So an

alternating set of commuting operators would be L2, Lxmax or L

2, Lymax.

2.3 Lecture 15: Quantum systems with V = V (r)

2.3.1 The angular momentum eigenvalue equations

The TISE with spherically symmetric potentials V = V (r) is

− ~2

2m∇2φ(r, θ, ϕ) + V (r)φ(r, θ, ϕ) = Eφ(r, θ, ϕ)

Recall from the last lecture that

−~2∇2 = −~2

r2∂r(r2∂r) +

L2

r2

So

− ~2

2m1r2∂r(r2∂rφ) +

L2

2mr2φ+ V (r)φ = Eφ

Since L2

depends only on θ, ϕ this equation is separable: φ(r, θ, ϕ) = R(r)Y (θ, ϕ). Substi-tuting this into the TISE and multiplying by 2mr2 we obtain

1YL

2Y︸ ︷︷ ︸

only θ,ϕdependent

= 2mr2(E − V (r)) +~2

R∂r(r2∂rR)︸ ︷︷ ︸

only rdependent

Thus, both sides must equal a constant ~2K2 and thence we get two equations dependingon r only and θ, ϕ only.

1. The radial equation

~2

R∂r(r2∂rR(r)) + 2mr2(E − V (r)) = ~2K2 (69)

the solution of which depends on V (r).

52

2. The angular equation

L2Y (θ, ϕ) = ~2K2Y (θ, ϕ) (70)

This does not depend on V (r). So the solution is universal for all spherically symmetricpotentials. It is an eigenvalue equation for total angular momentum squared.

• Let us study these angular eigenfunctions in (70) by substituting in equation (66) andmultiplying through by sin2 θ:

− sin θ∂θ(sin θ∂θY )− ∂2ϕY = K2 sin2 θY

This equation is yet again separable. Set Y (θ, ϕ) = P (θ)f(ϕ) and get

− sin θP

∂θ(sin θ∂θP )−K2 sin2 θ =1f∂2ϕf

Thus, both sides are equal to a constant, −m2 (NOT the mass squared), leading to twoequations

d2f

dϕ2= −m2f (71)

− sin θP

d

dθsin θ

dP

dθ−K2 sin2 θ = −m2 (72)

Solving equation (71) we get

f(ϕ) = e±imϕ

It is tempting to argue that f(ϕ) = f(ϕ + 2π) and therefore m ∈ Z (an integer), but thisis a false argument (even though the conclusion is correct!). In quantum mechanics theobservable is |φ(r, θ, ϕ)|2 so we only need

|f(ϕ)|2 = |f(ϕ+ 2π)|2

(i.e. that the probability density is single-valued) and this is true for any m! The correctargument is as follows (see e.g. page 322 of Shankar 1st edition):We want Lz = −i~∂ϕ to be an observable - thus it must be a Hermitian operator i.e.〈ψ1|Lz|ψ2〉 = 〈ψ2|Lz|ψ1〉∗.

⇒∫ 2π

0

ψ∗1(−i~∂ϕ)ψ2dϕ =[∫ 2π

0

ψ∗2(−i~∂ϕ)ψ1dϕ

]∗for all ψ1(ϕ) and ψ2(ϕ). Integrating the RHS by parts gives

RHS =∫ 2π

0

[(−i~∂ϕ)ψ2]ψ∗1dϕ+ i~∫ 2π

0

∂ϕ(ψ2ψ∗1)dϕ

The additional boundary term

i~∫ 2π

0

∂ϕ(ψ2ψ∗1)dϕ = i~ψ∗1(2π)ψ2(2π)− ψ∗1(0)ψ2(0).

53

must vanish if we are to have LHS = RHS (i.e. Hermitian Lz). But this is only true for allψ1, ψ2 if ψi(0) = ψi(2π) for i = 1, 2. This in turn implies that

eim2π = 1, m = 0,±1,±2, ...

after all! So it is the Hermitian nature of Lz that forces the eigenvalues of Lz to be

lz = m~, m = 0,±1,±2, ... (73)

• Now return to equation (72):

− sin θP

d

dθsin θ

dP

dθ−K2 sin2 θ = −m2

This is the Associated Legendre equation which you encountered in the mathematical methodscourse. Its solution is an Associated Legendre polynomial P`,m(cos θ).

•Why is K2 = `(`+ 1), ` ∈ Z?In quantum mechanics the wavefunction must be normalizable. This means that

1 =∫d(Vol)|R(r)P (θ)f(ϕ)|2

=∫ ∞

0

r2|R(r)|2dr∫ 2π

0

dϕ

∫ π

0

|P (θ)|2 sin θdθ

and each factor must be separately normalizable. It is convenient/conventional to choose∫ ∞0

r2|R(r)|2dr = 1, 2π∫ π

0

|P (θ)|2 sin θdθ = 1.

It can be shown that the solution to the Associated Legendre equation are only normalizablewhen

(i) K2 = `(`+ 1), where ` is an integer known as the orbital quantum number.

(ii) Furthermore, m must be an integer satisfying |m| ≤ ` i.e. m can only take thevalues−`,−`+ 1,−`+ 2, ...., 0, ..., `− 2, `− 1, `. We designate m the z-component of theangular momentum or the magnetic quantum number.

Notice that equation (71) is tantamount to (−i~)2∂2ϕf = m2~2f or simply L

2

zf = m2~2f .Hence, we finally obtain the eigenvalue equations for our maximal set of commuting an-gular momentum operators:

L2Y`,m(θ, ϕ) = ~2`(`+ 1)Y`,m(θ, ϕ) (74)

LzY`,m(θ, ϕ) = ~mY`,m(θ, ϕ) (75)

wherem and ` are quantum numbers s.t. |m| ≤ ` and Y`,m(θ, ϕ) = P`,m(θ)f(ϕ) = P`,m(θ)eimϕ

are known as the spherical harmonics.Note: (a) |Y`,m(θ, ϕ)|2 is independent of ϕ (superpositions may not be!). (b) |Y`,m(θ, ϕ)|2match the orbitals of A-level chemistry5.

5Summary: In solving the 3D TISE with a generic spherically symmetric potential V (r), we rewrite∇2 in terms

54

2.3.2 Appendix: The spherical harmonics

The spherical harmonics Y`,m(θ, ϕ) are eigenfunctions of angular momentum and satisfythe equations (74) and (75). They are normalized so that∫

Y ∗`,m(θ, ϕ)Y ∗`′,m′(θ, ϕ)dΩ = δ``′δmm′

The first few spherical harmonics are

Y0,0(θ, ϕ) =√

14π Y1,0(θ, ϕ) =

√3

4π cos θ

Y1,±1(θ, ϕ) =√

38π sin θe±iϕ Y2,0(θ, ϕ) =

√5

4π

12(3 cos2 θ − 1

)Y2,±1(θ, ϕ) =

√5

24π3 sin θ cos θe±iϕ Y2,±2(θ, ϕ) =√

596π3 sin2 θe±2iϕ

Figure 4: The above illustration shows |Y`m|2 = |Ym` |2 for the six spherical harmonics listed.

2.4 Lecture 16: Angular momentum: a better methodRecall that for the simple harmonic oscillator we had two ways of getting the eigenvaluesand eigenfunctions: (i) using differential equations and (ii) using the algebraic method with

of a ∂r(r2∂r) and L2

. This allows us to make use the separation of variables method φ(r, θ, ϕ) = R(r)Y (θ, ϕ)

such as to obtain a radial (r) and an angular (θ, ϕ) equation which must equal a constant ~2K2. The latter is the L2

eigenvalue relation which once again is amenable to the separation of variables technique Y (θ, ϕ) = P (θ)f(ϕ)whence we obtain two new differential equations equal to a second constant, −m2. The ϕ ODE encodes theeigenvalue relation for Lz and using the requirement of Hermiticity we discovered that m ∈ Z. Furthermore,the requirement that the wavefunction should be renormalizable entails that K2 = `(` + 1) where ` ∈ Z andthat |m| ≤ `. Thus, we have discovered that L2

, Lzmax have the common eigenfunctions Y`,m (the sphericalharmonics) and the eigenvalues ~2`(`+ 1) and ~m respectively.

55

raising and lowering operators a+, a−. Similarly for angular momentum there are twoways of getting the angular momentum quantization equations.

1. Represent Lx, Ly, Lz, L2

as differential operators and solve for the eigenfunctions Y`m(θ, ϕ)and the eigenvalues

L2Y`,m(θ, ϕ) = ~2`(`+ 1)Y`,m(θ, ϕ)

LzY`,m(θ, ϕ) = ~mY`,m(θ, ϕ)`,m ∈ Z where |m| ≤ `

This is what we did last time.

2. Alternatively, just like in the SHO case, we can introduce step or ladder operators andsolve for eigenvalues of L

2and Lz algebraically. There’s a big difference though: the

algebraic method shows us that there are more possible eigenvalues of L2

and Lz thanwere found by using differential operators (and functions of θ, ϕ)!

Let’s see how the latter can happen. Because [Lx, Ly] = i~Lz etc. the maximal set of com-muting operators is L

2, Lzmax and we’re interested in the following angular momentum

eigenvalue equations

L2ψ = ~2K2ψ, Lzψ = ~kψ. (76)

• To find out what K2 and k are consider the following angular momentum step operators:

L+def= Lx + iLy, L−

def= Lx − iLy (77)

and note that [L±, Lz

]= [Lx ± iLy, Lz]

= −i~Ly ± i(i~Lx)

= ∓~Lx − i~Ly= ∓~L±

•Now consider a ψk which satisfies Lzψk = ~kψk and define a new ψ′ := L+ψk (if it exists- it might be zero, ψ′ = 0. We’ll come back to this).What Lz value does ψ′ = L+ψk have, i.e. Lz(L+ψk) =? Rearranging the commutationrelation [Lz, L+] = +~L+ we get LzL+ = ~L+ + L+Lz . Thus,

Lz(L+ψk) = (~L+ + L+Lz)ψk

= ~ψ′ + Lz~kψk= (~ + ~k)ψ′.

So the state ψ′ must be proportional to one with Lz eigenvalue ~(k+ 1) (like the eigenstateψk+1!) i.e.

56

L+ψk = c+(k)ψk+1

where c+(k) is a normalization constant. Similarly, it is easy to show that L−ψk satisfies

Lz(L−ψk) = ~(k − 1)L−ψk

so L−ψk must be proportional to a state with L− eigenvalue ~(k − 1)

L−ψk = c−(k)ψk−1

So we conclude that L+ and L− step up and down tower of Lz eigenvalues by ~ unit.

• What do L+, L− do to L2

eigenvalues? Well, since L2

commutes with Lx and Ly , andsince L± = Lx ± iLy it is obvious that

[L2, L±] = 0.

This means that if ψk satisfies L2ψk = ~2K2ψk (so really, we should denote the state as

ψk,K2 ) then

L2L+ψk,K2 = L+L

2ψk,K2

= L+~2K2Lk,K2

= ~2K2(L+ψk,K2)

and similarly for L−ψk,K2 .

Thus L+, L− do NOT change the L2

eigenvalue.

• But this implies that the tower of Lz eigenstates with eigenvalues

. . . , ~(k − 2), ~(k − 1), ~k, ~(k + 1), ~(k + 2), . . .

generated by L+ or L− repeatedly acting on ψk must terminate at both ends. To see this, notethat

L2− L

2

z = L2

x + L2

y + L2

z − L2

z = L2

x + L2

y.

Now consider an arbitrary member of the tower (say ψk′,K2 ) and take the expectation value

of both sides of L2− L

2

z = L2

x + L2

y in this state:

LHS =∫ψ∗k′,K2(L

2− L

2

z)ψk′,K2dΩ

= 〈k′,K2|(L2− L

2

z)|k′,K2〉 (Dirac)

= (~2K2 − (k′~)2)〈k′,K2|k′,K2〉= ~2(K2 − k′2)

But

57

RHS = 〈k′,K2|(L2

x + L2

y)|k′,K2〉 ≥ 0

since it is a sum of expectation values of squares of Hermitian operators, whence

~2(K2 − k′2) ≥ 0 ⇒ k′2 ≤ K2.

I.e. the eigenvalues of Lz are bounded.Since k′2 is bounded by K2 there must exist both a state with highest Lz , ψkmax,K2 , and astate with lowest Lz , ψkmin,K2 .

• The only way this can consistent with the action of L+ and L− is if

I L+ψkmax,K2 = 0 i.e. c+(kmax) = 0 and

II L−ψkmin,K2 = 0 i.e. c−(kmin) = 0.

Figure 5: The tower of Lz eigenstates terminates at both ends. If |φ〉 is an Lz eigenstate with eigenvalue ~k,then the operator L+ will take us up the tower in increment steps, each time increasing the eigenvalue by oneunit of ~. Analogously, L− takes us down the tower in increment steps, each time reducing the eigenvalue by oneunit of ~.

• In fact there is a connection between the values kmax and K2 (or kmin and K2). To see this,we need to manipulate the operators L−L+ and L+L−. Firstly

58

L−L+ = (Lx − iLy)(Lx + iLy)

= L2

x + L2

y + i[Lx, Ly]

= L2

x + L2

y + i(i~Lz)

= L2− L

2

z − ~Lz.

(78)

Similarly

L+L− = L2− L

2

z + ~Lz. (79)

Now act on equation I with L− and use equation (78) to give

(L2− L

2

z − ~Lz)ψkmax,K2 = 0

⇒ ~2(K2 − k2max − kmax)ψkmax,K2 = 0

and thus we get

K2 = kmax(kmax + 1). (80)

• Imagine starting from the top of the tower ψkmax,K2 and acting q times with L− till we reachψkmin,K2 . But ψkmin,K2 satisfies L−ψkmin,K2 = 0 and acting with L+ on this equation (i.e.equation II) and using (79) gives

(L2− L

2

z + ~Lz)ψkmin,K2 = 0

⇒ ~2(K2 − k2min + kmin)ψkmin,K2 = 0

whence

K2 = kmin(kmin − 1). (81)

But comparing equations (80) and (81) we see kmax(kmax + 1) = kmin(kmin − 1) i.e.

−kmax = kmin. (82)

Moreover, as we go from kmax to kmin in q steps (using L−) we must have

kmax − kmin = 2kmax = q

⇒ kmax = q2 , q = 0, 1, 2, 3, ...

(83)

It is conventional to refer to q2 = j where j = 0, 1

2 , 1,32 , ... as the angular momentum state

(j = kmax).Finally, we had equation (80)

K2 = kmax(kmax + 1) = j(j + 1).

Thus, we’ve learnt

59

L2ψjm = ~2j(j + 1)ψjm

Lzψjm = ~mψjmj = 0, 1

2 , 1,32 , ...

m = −j,−j + 1,−j + 2, ..., j−, j.

2.5 Lecture 17: Spin and the radial equation

2.5.1 The meaning of the angular momentum operator results

Last lecture we showed that operator analysis of the angular momentum algebra allowed

L2ψjm = ~2j(j + 1)ψjm, Lzψjm = ~mψjm

(m = −j,−j+1, ..., j−1, j) with both (i) j ∈ Z|j ≥ 0 (integer ) and (ii) j ∈ Z+ + 12 |j ≥

12

(odd half integer).But didn’t we show earlier that Lz = −i~∂ϕ (and similar differential representation ofL

2) only had integer eigenvalues? Yes! And that result is quite correct, but the existence of

j = 12 ,

32 ,

52 , ... is also correct.

What’s going on? The extra half-integer eigenvalues arose because we have in fact solveda more general problem than that of Lx, Ly, Lz, L

2(although we didn’t know we were!).

Nowhere did we use explicit differential representation for Lz , L2

etc. We just used com-mutation relations [Li, Lj ] = i~εijk which reflect the law of the combination of rotations inthree dimensions and which must be satisfied whatever the nature of the wavefunctions theyrotate.

•What we have discovered is spin (j) which can take on both integral and odd-half-integralvalues. E.g. the photon has j = 1, the pion and the conjectured Higgs boson have j = 0,the electron and the proton have j = 1

2 whilst nuclei take on both integer and half-integerspin values.Importantly, spin wavefunctions are not functions of angular coordinates θ, ϕ; rather theymust be represented as column vectors

|ψ〉j=

127→(AB

), |ψ〉j=1 7→

ABC

, |ψ〉j=

327→

ABCD

etc.

s.t. a spin j particle must be represented by a (2j+1)-vector. Then the operators Lx, Ly, Lz, L2

act as matrices on these vectors and reshuffle their components.Note that spin essentially has forced us to adopt the Heisenberg formulation of QM (ma-trix mechanics) which is more general, though more abstract, than Schrodinger’s picture.Heisenberg thought of all operators as being matrices (in which case differential operatorslike px = −i~∂x require infinite dimensional matrices!). On the other hand, normal orbital

60

angular momentum is always integral and therefore can always be represented by Y`,m(θ, ϕ)and differential operators.

• It is helpful to have a notation which distinguishes orbital from spin (or intrinsic) angularmomentum. Let’s write

J i = Li + Si, i ∈ x, y, z (84)

where J is the total angular momentum (j ∈ 0, 12 , 1,

32 , ...), L is the orbital angular mo-

mentum (` ∈ 0, 1, 2, ...) and S is the spin angular momentum (s ∈ 0, 12 , 1,

32 , ...).

Since rotations in three dimensions always obey the same composition rules the operatorsSi obey the same commutation relations:

[Sx, Sy] = i~Sz and cyclic.

Thus, so does J i (as they must!):

[Jx, Jy] = i~Jz and cyclic.

So the eigenvalue equations are

Total =

J

2ψjjz = ~2j(j + 1)ψjjz

Jzψjjz = ~jzψjjz

Orbital =

L

2ψ`m = ~2`(`+ 1)ψ`m

Lzψ`m = ~mψ`m

Spin =

S

2ψssz = ~2s(s+ 1)ψssz

Szψssz = ~szψssz

j

jz

`

m

s

sz

= 0, 12 , 1, ...

= −j,−j + 1, ..., j

= 0, 12 , 1, ...

= −`,−`+ 1, ..., `

= 0, 12 , 1, ...

= −s,−s+ 1, ..., s

And as J i = Li + Si we have Jz = Lz + Sz so

jz = m+ sz (85)

Note: An equation which is important in atomic physics is

J2

=∑

i=x,y,z

(Li + Si

)2

= L2

+ S2

+ 2L · S

⇒ L · S = 12

(J

2− L

2− S

2)

More on spin later. For the moment we’ll ignore it.

2.5.2 The radial equation

• Recall that we were considering three dimensional problems of the form

H =p2

2m+ V (r)

which we were able to turn into the TISE

61

− ~2

2m1r2∂r(r2∂rψ

)+

L2

2mr2ψ + V (r)ψ = Eψ. (86)

We then wrote

ψ(r, θ, ϕ) = R(r)Y`,m(θ, ϕ) (87)

where

L2Y`,m(θ, ϕ) = ~2`(`+ 1)Y`,m(θ, ϕ). (88)

Since m = −`,−`+ 1, ..., `− 1, ` there are 2`+ 1 states of the same ` but varying m.Combining equations (86), (87) and (88) we are left with the radial wave equation

− ~2

2m1r2∂r(r2∂rR

)+

~2`(`+ 1)2mr2

R+ V (r)R = ER. (89)

One immediate consequence of this is that

Energy eigenvalues E cannot depend on the quantum number m.

when V = V (r) as the above equation is independent ofm.⇒ There are (2`+1) degeneratestates when the total orbital angular momentum = `.

This degeneracy is a consequence of rotational symmetry.

Infinitely deep spherical wellLet’s try to solve the radial equation for

V (r) =

0, r < a

∞, r ≥ a

This implies that ψ(r ≥ a) = 0 whence we have R(r ≥ a) = 0 as a boundary condition.For r < a R(r) satisfies

−R′′ − 2rR′ +

`(`+ 1)r2

R = εR

where ε = 2mE~2 . Now let r = ρ/

√ε then we’ll find

d2R

dρ2+

2ρ

dR

dρ+(

1− `(`+ 1)ρ2

)R = 0

This is the Spherical Bessel Equation. If we make the further substitution R = Q/√ρ we find

d2Q

dρ2+

1ρ

dQ

dρ+

(1−

(`+ 1

2

)2ρ2

)Q = 0 (90)

which is the ordinary Bessel Equation, but for 1/2-integer order.This has solutions J±(`+1/2)(ρ). And equivalently,

62

R±(ρ) =1√ρJ±(`+1/2)(ρ)

• Are both of these forms allowable?Recall from the mathematical methods course that Ju(x) ∼ xu as x→∞. Thus,

R+(r) ∼ r−1/2r(`+1/2) ∼ r`,

R−(r) ∼ r−1/2r−`−1/2 ∼ r−`−1.

Since R− blows up as r → ∞ it may seem as though it cannot be normalizable (andthus allowable), but the normalization integral has the form

∫∞0r2|R|2dr since d(Vol) =

r2 sin θdrdθdϕ. So the behaviour at r = 0 is∫0

r2|R+|2dr ∼∫

0

r2+2`dr ∼ finite∫0

r2|R−|2dr ∼∫

0

r−2`dr ∼

∞ if ` ≥ 1finite if ` = 0

Thus, the R−(r) solution with ` = 0, i.e. R−(r) ∼ 1√rJ−1/2(r) are allowed by normalizability

(on the other hand, for ` ≥ 0, all J−1/2+`).

However, R`=0− (r) ∼ r−1 as r → 0 and this is disallowed because it in fact does not solve

our equation anywhere.To see this, recall from electromagnetic theory that the potential 1

4πr of a point charge sat-isfies

∇2

(1

4πr

)= −δ(3)(r).

But we are solving exactly a radial equation of the form ∇2R− = (E − V (r))R−, so unlessV (r) contains a delta function charge at the origin (or similar construct leading to V ∼δ(3)(r)) R`=0

− is in fact not allowed.Thus, the allowed solutions of the spherical well are of the form

R(ρ) = 1√ρJ`+1/2(ρ) = ρ`

(−1ρ

d

dρ

)`( sin ρρ

)(91)

leading to general solutions of the form

ψ(r, θ, ϕ) =∞∑`=0

+∑m=−`

cm`1√ρJ`+1/2(ρ)Y`,m(θ, ϕ). (92)

But we must impose the boundary condition R(r = a) = 0. Let’s look at some specificcases:

` = 0: Using equation (91) together with ρ = r√ε:

R =1r√ε

sin(r√ε).

63

Setting R(a) = 0 gives sin(a√ε) which is true iff a

√ε = nπ for n = 1, 2, 3, ... Hence, as

for the 1D square well,

E =~2π2n2

2ma2.

` = 1:

R =1

(r√ε)2

sin(r√ε)− 1

r√ε

cos(r√ε)

The zeroes are now more complicated: a√ε = 4.49, 7.72, 10.90, ...

` = 2: The radial wave equation is found by setting ` = 2 in (91) remembering that(− 1ρddρ

)2

means − 1ρddρ

(− 1ρddρ

). The first five zeroes can be estimated from figure

6(a) and the first energy levels from figure 6(b).

What do we see?• For each ` value (orbital angular momentum) there is a tower of states (radial excitations)labelled by a principal quantum number which counts the number of nodes in R(r).• States with higher ` have higher energies for a given principal quantum number.• For each ` there are (2`+ 1) states of varying Lz , so each of the plotted levels is degenerate(apart from ` = 0).• Don’t forget that we have to normalize the radial R(r).

2.6 Lecture 18: The Hydrogen atomNow we start the study of a system of great physical importance. For the time being we’llstudy only the simplest leading approximation to the real hydrogen atom, i.e. we’ll ig-nore relativistic effects, electron spin, nuclear (proton) spin, the finite mass and size of thenucleus etc. Even so we’ll find a good and very useful description.

Using Coulomb’s law, the force exerted on the electron (charge −e) by the proton (charge+e) is

F = − 14πε0

e2

r2r.

where r is a unit vector from the proton towards the electron. Hence, since the differentialof the potential energy is dV = −Fxdx we find that

VH(r) = +∫ r

∞

14πε0

e2

r′2dr′ = − e2

4πε0r.

This means that we expect bound states to have negative energy (just a consequence oftaking the zero of energy to be at V (r →∞)).

The radial wave equation (89) is now

− 1r2∂r(r2∂rR

)+`(`+ 1)r2

R− 2me2

4πε0~2

R

r=

2mE~2

R. (93)

64

(a)

(b)

Figure 6: (a) The first three Spherical Bessel Functions j`(ρ) ≡ 1√ρJ`+1/2(ρ) which constitute the radial solu-

tionsR(ρ) to the infinitely deep spherical well. Explicit expressions for j`(ρ) can be found by computing equation(91). (b) The spectrum of the infinitely deep spherical well given in terms of E/E0 where E0 is the ground state.Also, remember that for each ` state there is a (2`+ 1)-degeneracy brought about by the quantum number m.

65

Introducing the Bohr radius

a0def=

4πε0~2

me2

and letting r = a0ρ we get

−R′′ − 2ρR′ +

(`(`+ 1)ρ2

− 2ρ

)R = εR, with ε =

2mEa20

~2(94)

Let’s start our analysis of this equation by finding the behaviour at large r (ρ 1). Thenthe equation reduces to

−R′′ = εR

which has the solution R ∼ exp(−√−ερ) for bound states with ε < 0 (notice exp(+

√−ερ)

is not normalizable).Now what happens for small ρ 1?

−R′′ − 2ρR′︸ ︷︷ ︸

− 1ρ2 ∂ρ(ρ2∂ρR)

+`(`+ 1)ρ2

R = 0

I.e. we can ignore the Coulomb and energy terms. Let’s try the power law solution R ∼ ρu.We find

−u(u+ 1)ρu−2 + `(`+ 1)ρu−2 = 0

whence u = ` or u = −(`+ 1).

•Normalizability disallows the u = −`−1 solution for ` ≥ 1 (as the integral would explodeat the origin). But as in the infinite spherical well case we could in principle have u = −1which is normalizable: ∫

0

r2(r−1)2dr ∼ finite.

However, just in the spherical well case a solution of the form R ∼ r−1 would imply thatV (r) ∼ δ(3)(r) near the origin. Unlike the infinite spherical well case there is now some-thing going on at the origin - the nucleus sits there! But fortunately, in reality (as experi-ments tell us) the interactions of the (finite sized) proton with the electron do not lead to aterm in the potential

V (r) = VCoulomb +

δ(3)(r)

which means that the u = −1 mode is also disallowed.Thus the allowed modes behave as R ∼ ρ` for ρ 1.

• Note that equation (7(b)) being written in terms of the dimensionless quantities (ρ, ε, `)must have solutions for ε which are pure numbers, so

66

E =~2

2ma−2

0 ε =~2

2mm2

~4

(e2

4πε0

)2

ε

=m

2~2

(e2

4πε0

)2

ε.

If we had a central charge Ze instead of e we’d get

E =m

2~2

(Ze2

4πε0

)2

ε

=mc2Z2

2

(e2

4πε0~c

)2

ε.

So we see that the energy of the states in the Coulomb potential is

1. Proportional to the mass of the orbiting particle.

2. Proportional to the nuclear charge squared.

3. Of order (in Hydrogen-like systems)

Z2

2mec

2α2 (useful to remember) (95)

where mec2 ≈ 0.5 MeV is the mass-energy of the electron and α ≈ 1/137 is the di-

mensionless fine structure constant. Thus,

EH ∼ 12mec

2α2 ∼ 12 · 0.5 · 106 ·

(1

140

)2 ≈ 12 eV.

Now let’s solve equation (7(b)) exactlyIt is reasonable to factor out the known large ρ and small ρ behaviour, so define F via

R = ρ`e−λρF (2λρ)

where λ =√−ε. Substitute this into (7(b)) gives that F (y) satisfies

yF ′′(y) + F ′(y) (2(`+ 1)− y)−(`+ 1− 1

λ

)F (y) = 0.

This is yet another differential equation which you can look up in books. It is the GeneralizedLaguerre Equation. The solutions are polynomials L(2`+1)

k (y) where λ = 1k+`+1 and k =

0, 1, 2, ...We have now found the spectrum as ε = −λ2 (by definition of λ) and E = ~2ε/2ma2

0

⇒ E =−~2

2ma20

1(k + `+ 1)2

, k = 0, 1, 2, ... (96)

(See the spectrum: Figure 7(a)).

Comments

67

(a)

(b)

Figure 7: (a) The spectrum of the hydrogen atom defined in terms of the quantum numbers k and `. (b)Examples of amplitudes, r2|Rn,`|2, of the Hydrogen radial wavefunctions.

68

• The energy levels obey the rule first proposed by Balmer, and also predicted by the Bohrmodel, i.e.

E = − Rn2, n = 1, 2, 3, ...

where R ≈ 13.6 eV. Note however that the angular momentum quantum numbers aredifferent from the Bohr model (which assumes mr2ω = n~) and that all levels except n = 1 arehighly degenerate.• The nomenclature is confusing. in atomic physics we use the quantum numbers n, ` (andm) to label the states (historical hangover), whereas in nuclear physics we use k, ` (and m)which makes more sense as k − 1 = No. of nodes of R(r).

In any case it is unfortunately standard to define

ndef= k + `+ 1, n ∈ Z+ (97)

as the principal quantum number, so at each n the allowed values of ` are

` = n− k − 1 = 0, 1, 2, ..., (n− 1).

Moreover, for each ` we have (2` + 1) degenerate m values, −`,−` + 1, ..., `, sothe total degeneracy at each n is

n−1∑`=0

(2`+ 1) = n2. (98)

• The fact that for a given n, different ` values are degenerate is a special feature of the exactCoulomb potential (there is in fact a hidden symmetry that leads to this degeneracy).• In the real Hydrogen atom this degeneracy of different `s is removed by all the (smaller)interactions and effects we have ignored in our idealized treatment (spin orbit coupling,relativistic corrections etc.) - these are the subject of the atomic physics course (applicationsof quantum mechanics).•As emphasized before, the degeneracy inm values at a given `: m = −`,−`+1, ..., `−1, `,is due to rotational symmetry.

2.7 Lecture 19: Hydrogen continued and spin

2.7.1 Normalization of the Hydrogen atom

This requires that

1 =∫d(Vol)ψ∗ψ

=∫

sin θdθdϕ∫ ∞

0

r2|Rn,`Y`,m|2dr

=∫|Y`,m|2dΩ︸ ︷︷ ︸

=1

∫ ∞0

r2|Rn,`|2dr.

69

This means that the probability of finding the electron between r and r + dr is

Pr(r, r + dr) = r2|Rn,`|2

- NOT simply |Rn,`|2!Let’s look at the implications for the ground state ψ = CY00e

−r/a0 . We get:

1 = C2

∫ ∞0

r2e−2r/a0dr = 2C2(a0

2

)3

⇒ C =2

a3/20

.

Now let’s calculate the expectation value of r for the ground state

〈r〉 =∫dV ψ∗rψ

=∫|Y00|2dΩ

4a3

0

∫ ∞0

r3e−2r/a0dr

=4a3

0

(a0

2

)4

=32a0

On the other hand, the single most probable value of r (mode) is at the maximum of the po-sition amplitude r2e−2r/a0 . I.e. differentiating with respect to r and setting the expressionequal to zero:

0 = 2re−2r/a0 − 2a0r2e−2r/a0 ⇔ r = a0

which is the Bohr radius.

Generalized features of the radial wavefunctions: See Figure 7(b). We have expressions of theform:

Rn,` = constant · e−r/na0 · r` · (1 +O(r))

The exponential term implies that as n gets larger, the probability density spreads out, butthe behaviour is always exponentially small at large distances. The final term is from thefact that F always is a polynomial in r, starting with 1.

2.7.2 Spin 1/2

We previously showed that it is possible to have states satisfying

J2| 12︸︷︷︸=j

,± 12 〉 = ~2j(j + 1)| 12 ,±

12 〉, Jz| 12 ,±

12 〉︸︷︷︸

=jz

= ±~ 12 |

12 ,±

12 〉

These do not turn up when we solve the Schrodinger equation, but require the matrix for-mulation of quantum mechanics. The spin 1/2 case is particularly important as electrons,protons, neutrons, quarks, ... all have spin 1/2. Since for j = 1

2 it is not possible to representangular momentum operators as differential operators, consider the matrices

70

Sx =~2

(0 11 0

)︸ ︷︷ ︸

=σx

, Sy =~2

(0 −ii 0

)︸ ︷︷ ︸

=σy

, Sz =~2

(1 00 −1

)︸ ︷︷ ︸

=σz

.

(σx, σy, σy are the Pauli spin matrices). The spin matrices are Hermitian

S†x = Sx, S

†y = Sy, S

†z = Sz,

so they have real eigenvalues. Now calculate the commutator[Sx, Sy

]=

~2

22

((0 11 0

)(0 −ii 0

)−(

0 −ii 0

)(0 11 0

))=

~2

4

((i 00 −i

)−(−i 00 i

))=

~2

42i(

1 00 −1

)= i~Sz

- exactly the angular momentum commutation relation. It is a simple exercise to check that[Sy, Sz] = i~Sx and [Sz, Sx] = i~Sy .

We can also calculate the matrix for S2:

S2

= S2

x + S2

y + S2

z

=~2

4

(0 11 0

)2

+(

0 −ii 0

)2

+(

1 00 −1

)2

=~2

4

(1 00 1

)+(

1 00 1

)+(

1 00 1

)=

~2

43(

1 00 1

)= ~2 1

2

(12 + 1

)I.

where I is the unit matrix. This is exactly what we want for S2

since then any two-

component vector(ab

)is an eigenvector of S

2with eigenvalue ~2 1

2

(12 + 1

). Thus,

(ab

)for any a, b (a, b nor both equal to zero) represents a state with total angular momentumj = 1/2.

Let’s look at the eigenstates and eigenvalues of the spin z operator: Sz| 〉 = λ| 〉. Solving

det(~

2 − λ 00 −~

2 − λ

)= 0.

we find that the eigenvalues are λ1 = +~2 and λ2 = −~

2 . The associated (normalized)

eigenvectors of these eigenvalues are | ↑ 〉 =(

10

)and | ↓ 〉 =

(01

).

In a similar fashion we can show that the eigenvalues of Sx are λ1 = +~2 and λ2 = −~

2 and

the eigenvectors are |→〉 = 1√2

(11

)and |←〉 = 1√

2

(1−1

).

71

Likewise, for Sy , the eigenvalues are λ1 = +~2 and λ2 = −~

2 and the eigenvectors are

|⊗〉 = 1√2

(1i

)and |〉 = 1√

2

(1−i

).

Important: the eigenvalues for Sx, Sy , Sz are the same but the eigenvectors are different,and this implies that the measurement of Sx, Sy , Sz have some interesting properties.E.g. suppose we start with the sx = +~/2 state | →〉 and then measure Sz . So we mustdecompose |→〉 into the Sz eigenvectors | ↑ 〉 and | ↓ 〉:

1√2

(11

)=

1√2

(10

)+

1√2

(01

)Thus upon measurement of Sz we could find either ±~/2 each with probability 1/2. Sup-pose we find sz = +~/2 then the after measurement state is | ↑ 〉. Now if we measure Sxwe must decompose the | ↑ 〉 into Sx eigenvectors:(

10

)=

1√2

(1√2

(11

)+

1√2

(1−1

))Thus a measurement of Sx can give sx = +~/2 with probability 1/2 or sx = −~/2 withprobability 1/2. Thus, even though we started with definite sx value +~/2 the measure-ment of Sz has generated the possibility that the system now has sx = −~/2.

2.8 Lecture 20: Dipole moments and the Stern-Gerlach experiment

2.8.1 Spin dipole moment

How do we know that the electron has spin 1/2? Well, classically a charged object withangular momentum has a magnetic dipole moment. In the most general case, an arbitrarycurrent distribution in space has the magnetic dipole moment

µ =12

∫r × Jd(Vol)

where d(Vol) is the volume element, r is the position vector from the origin to the volumeelement and J is the current density. The above formula can be applied to any assembly ofmoving charges by substituting

J(r, t) = ρ(r, t)v(r, t)

where ρ(r, t) is the charge density at position r at time t, and v(r, t) is the particles’ averagedrift velocity at position r at time t. E.g. the magnetic dipole moment produced by acharged particle q moving along a circular path is

µ =12qr × v =

q

2mqL

where r is the position of the charge q relative to the center of the circle, v is the instanta-neous velocity of the charge, mq is its mass and L is the angular momentum.

72

Quantum mechanics also leads to this conclusion (strictly speaking relativistic quantummechanics; equivalently quantum field theory is needed to give a truly consistent and pre-cise understanding the electron dipole moment). Here µ ∼ S - more particularly, theelectron (charge −e) has the magnetic dipole moment

µ = −gse

2meS.

where gs ≈ 2 is a correction coefficient due to quantum effects (which we need the Diracequation and QFT to calculate), whilst −e/2me is simply the classical coefficient of propor-tionality between µ and S.

Now if we have a magnetic field in the z direction, Bz , there are two possible values for theinteraction energy −B · µ:

Due to quantization of Sz

If sz = +~2

then −B · µ = +gse

2me

~2Bz

If sz = −~2

then −B · µ = − gse

2me

~2Bz

Hence, there are two discrete possible values of the magnetic dipole moment.

The combination

e~2me

def= µB

is known as the Bohr magneton. Hence an electron has a magnetic dipole moment of ap-proximately one Bohr magneton.

Of course, typically the potential energy will be a combination of a spin-independent po-tential V (r) and a spin-dependent term (arising, say, from a −B · µ interaction). Thus,typically, the form of the energy (Hamiltonian) for a spin 1/2 system (e.g. an electron in anatom) will be

H = H0I +gse

2meB · S

where H0 could be, e.g., the Coulomb potential and kinetic energy, I is the identity matrix(so the energy from the H0 term is independent of the spin state) and B · S = BxSx +BySy +BzSz .Thus, we can consider a modified form of the TDSE

HΨ = i~∂Ψ∂t.

H is now a 2 × 2 matrix as well as a differential operator, whilst Ψ is a two componentvector with each component being a function of (t, r, θ, ϕ), i.e.

Ψ =(ψ1(t, r, θ, ϕ)ψ2(t, r, θ, ϕ)

).

73

If theB-field is constant then the modified TDSE is simple to analyze. Write Ψ = Φ(t, r, θ, ϕ)u,where u is a constant two component vector. This gives

(H0φ)u+ φgse

2me(B · S)u = i~

∂φ

∂tu. (99)

This can only be satisfied if

(B · S)u = λu.

Moreover, if the rest of the problem (i.e. apart from B) is spherically symmetric, then wecan always choose the direction of the (x, y, z) axes such that the z-axis is in theB direction.With this coordinate choiceB = (0, 0,Bz) then

BzSzu = λu.

Since Sz = ~2

(1 00 −1

)this gives us

λ = ±~Bz

2for u =

(10

),

(01

).

Putting this back into equation (99) we get two equations(H0 +

gse

2me

~B2

)φ1(t, r, θ, ϕ) = i~

∂φ1

∂t(100)

(H0 −

gse

2me

~B2

)φ2(t, r, θ, ϕ) = i~

∂φ2

∂t(101)

where φ1 and φ2 are the top and bottom components of Φ = (φ1, φ2). Now we can separatevariables as usual for (100) and (101), e.g.

φ1(t, r, θ, ϕ) = e−iE1t/~ϕ1(r, θ, ϕ)

φ2(t, r, θ, ϕ) = e−iE2t/~ϕ2(r, θ, ϕ)

giving TISEs for ϕ1 and ϕ2. The interpretation is simply that the electron can either havespin aligned withB, or anti-aligned, and the energies are shifted by ±gsµBB/2.

2.8.2 The Stern-Gerlach experiment

Now consider a dipole in a non-uniform B-field. To get oriented with the physics consideran electric dipole (q,−q) in a non-uniform electric field E s.t. the displacement vector afrom −q to q points in the direction of the positive z-axis. Then the net force on the dipoleis

F = qE(z + a)− qE(z)

≈ q(E(z) + a

∂E

∂z

∣∣∣∣z

−E(z))

(Taylor)

= qa∂E

∂z= p

∂E

∂z

74

where p is the dipole moment.Similarly, suppose we send a beam of electrons through a non-uniform magnetic field gen-erated a pair of magnets, after which the electrons impinge on a photographic plate. Asbefore, the potential energy of the electron spin magnetic moment in a magnetic field ap-plied in the z-direction is given by±gsµBB/2. Hence, since F is equivalent to the negativeof the potential gradient:

F z = ∓µBgs2∂B

∂z. (102)

I.e. the electrons will experience a force either up depending on which way their µ (i.e. Sz)is pointing. This is the essence of the Stern-Gerlach experiment (see French+Taylor, section10.3, for more information as well as the HT problem set).

Figure 8: In the Stern-Gerlach experiment a beam of electrons is sent through an inhomogeneous magnetic fieldwhich divides the electrons into two separate paths depending on their magnetic moments (whether they arepositive or negative). Indeed, experiments corroborate that the electrons only display two discrete positions onthe observing screen contrary to our classical expectations of a continuous distribution of deflections. Assumingthat the magnet is a distance L across, it is a simple task to compute the total deflection of an electron inside themagnet. Under constant acceleration a we have z = 1

2at2. Also, a = Fz/m and t = L/v, where v is the initial

(horizontal) speed of the electron. Thus, z = 12Fzm

“Lv

”2= 1

4L2

EkinFz and an expression for Fz can be found in

equation (102).

Important points to note:

1. Classically one would expect to see a continuous distribution of e deflections on theobserving screen (as classically Sz , and thus µz , can take on any value - not quan-tized).

2. The experiment instead shows just two spots (for spin 1/2 systems) on the screenwhich implies that Sz is quantized (and similarly for any other axis) with two values.

3. In detail, when one calculates one finds that they are twice as far apart as one wouldexpect classically where

µ = − e

2meS.

75

This is one of many reason(s) why we are forced to introduce the correction factorgs ≈ 2 (the gyromagnetic ratio) so

µ = − gse

2meS = −gsµB

(±1

2

).

2.9 Lecture 21: Multi-particle systems

2.9.1 Two-particle systems

So far we have only studied the quantum mechanics of one particle in an external staticpotential. However, most of the world is not like this: usually we have systems of two ormore particles interacting with each other.The simplest case is two particles of masses m1 and m2 interacting via a potential whichdepends on their separation, e.g. the electron and proton in the Hydrogen atom. Then

H =p2

1

2m1+p2

2

2m2+ V (r1 − r2)

and ψ = ψ(r1, r2, t). The TDSE is now− ~2

2m1∇2

1 −~2

2m2∇2

2 + V (r1 − r2)ψ(r1, r2, t) = i~

∂

∂tψ(r1, r2, t).

Note that the time dependence is still separable:

ψ(r1, r2, t) = e−iEt/~Φ(r1, r2)

leading to − ~2

2m1∇2

1 −~2

2m2∇2

2 + V (r1 − r2)

Φ(r1, r2) = EΦ(r1, r2).

This can always be simplified by introducing a center of mass coordinate. Let

X = R =m1r1 +m2r2

m1 +m2, x = r = r1 − r2,

then6

∂

∂x1=∂X

∂x1

∂

∂X+

∂x

∂x1

∂

∂x=

m1

m1 +m2

∂

∂X+

∂

∂x∂

∂x2=∂X

∂x2

∂

∂X+

∂x

∂x2

∂

∂x=

m2

m1 +m2

∂

∂X− ∂

∂x

6A note on notation: expressions such as ∂X∂x1

∂∂X

actually encode matrix expressions of the form:0BB@∂X∂x1

. . . ∂Z∂x1

.... . .

...∂X∂z1

. . . ∂Z∂z1

1CCA0BB@

∂∂X

∂∂Y

∂∂Z

1CCA .

Here only the diagonal elements are non-zero (in fact, the matrix is readily seen to equate m1m1+m2

I3×3).

76

and therefore

−~2

2

1m1

∂2

∂x21

+1m2

∂2

∂x22

= −~2

2

m1 +m2

(m1 +m2)2

∂2

∂X2 +m1 +m2

m1m2

∂2

∂x2

.

So we get − ~2

2M∇2

R −~2

2µ∇2

r + V (r)

Φ = EΦ

where M = m1 + m2 is the total mass and µ = m1m2/(m1 + m2) is the reduced mass. Nowwe see that theX and x parts are separable:

Φ(r1, r2) = exp(− iP CM ·X

~

)φ(x)

where P CM is the centre of mass momentum and φ(x) is a function which only depends onx. So

E = − ~2

2M

(− iP CM

~

)2

︸ ︷︷ ︸K.E. of centre

of mass motion

+Eint

where Eint is the interacting energy which satisfies− ~2

2µ∇2

x + V (x)φ(x) = Eintφ(x).

• Note that this is of the form of a single particle TISE with external potential V (x), which weknow how to solve. The problem thus splits into a center of mass motion contribution plusan internal excitation contribution.• This also explains why we could solve the Hydrogen atom problem (which involves twoparticles) using just the single particle TISE.• The difference is that the internal excitation problem involves just the reduced mass, µ.For Hydrogen this means that we really should not use me in computing the spectrum, butrather

µH =memp

me +mp≈(

1− me

mp+ . . .

)which really is a small correction since me/mp ≈ 1/1800. However, in some systems thereduced mass effect is not small at all. E.g. in positronium (a system consisting of an electronand its anti-particle, the positron, bound together in a so-called excotic atom) we have

µ =m2e

2me=me

2.

77

2.9.2 The addition of classical angular momentum

So far, in two-particle systems, we have not looked at angular momentum. As an exampleof the usefulness of considering angular momentum for multi-particle systems let us con-sider the Helium atom in the limit where the mass of the nucleus is taken to infinity (i.e.we are only considering the motion of the two electrons).

The classical picture is like a star with two orbiting planets

where L(i) is the angular momentum of the mass mi for i = 1, 2 (of course, for Helium,m1 = m2 = me as the orbiting ”planets” are electrons).

• First suppose that the electrons feel only the central potential (i.e. we switch off theirrepulsion), then classically

L(1) is conserved, i.e. a constant vector.

L(2) is conserved, i.e. a constant vector.

The total angular momentum is L = L(1) + L(2) and this is conserved too (as it is just thevector sum of L(1) and L(2)).

•What happens (still classically) when we switch on the repulsion between the electrons?Well, electron number 1 is no longer in a central potential, whence its angular momentumL(1) no longer is conserved (it varies with time). Similarly, L(2) of the second electron failsto be conserved.

However, there is no external couple on the entire system and therefore L =L(1) +L(2) is still conserved.

Proof : suppose we have a generic coordinate system where r1 is the position vector ofelectron 1 and r2 is the position vector of electron 2 then

78

d

dt

(L(1) +L(2)

)=

d

dt(r1 × p1 + r2 × p2)

=d

dt(m1r1 × v1 +m2r2 × v2)

= (m1r1 × v1 +m2r2 × v2 +m1r1 × v1 +m2r2 × v2) (where r = v)= (r1 × F 1 + r2 × F 2) (since F = mv)= (r1 − r2)× F 1 (F 1 = −F 2)= 0 (as (r1 − r2)‖F 1)

QED.

2.10 Lecture 22: The addition of angular momentum in QM

2.10.1 The ang. mom. quantum numbers of 2-particle systems

When we switch on the interaction potential in the Helium atom, the TISE takes the form

HΦ(r1, r2) = EΦ(r1, r2)

where ri is the position vector of electron i with respect to the nucleus (i = 1, 2) and where

H = − ~2

2me∇2

1 −2e2

4πε0r1︸ ︷︷ ︸H1

− ~2

2me∇2

2 −2e2

4πε0r2︸ ︷︷ ︸H2

+e2

4πε0|r1 − r2|︸ ︷︷ ︸H int

.

This is nasty: because r1, r2 and |r1 − r2| appear in the potential, this equation is notseparable. In fact, it cannot be solved exactly.However, our classical discussion suggests that we might be able to say something aboutthe angular momentum of the system. We know, e.g., that

L(1)

z = −i~(x1

∂

∂y1− y1

∂

∂x1

)

commutes with H1 (just like in Hydrogen). Also, L(2)

z commutes with L(1)

x , L(1)

y , L(1)

z and

L(1)2

.

This is an example of the fact that operators associated purely with system no.1 (e.g. x1, p1 = −i~∇1, etc.) commute with those associated purely with systemno. 2 (e.g. x2, p2, L2,...).

• The problem is that L(1)

z does not commute with the H int = e2(4πε0|r1 − r2|2)−1 part ofH . Let’s show this explicitly:

79

[L

(1)

z , H int

]= − i~e

2

4πε0

(x1

∂

∂y1− y1

∂

∂x1

)((x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2

)−1/2

=i~e2

4πε0(y1x2 − x1y2)

1|r1 − r2|3

6= 0.

But we also expect (based on classical physics) that it should be L(1)

z + L(2)

z which is con-served. So let us compute

[L

(2)

z , H int

]= − i~e

2

4πε0

(x2

∂

∂y2− y2

∂

∂x2

)((x1 − x2)2 + (y1 − y2)2 + (z1 − z2)2

)−1/2

=i~e2

4πε0(y1x2 − x1y2)

1|r1 − r2|3

6= 0.

Clearly the sum Lz = L(1)

z + L(2)

z satisfies

[Lz, H int] = 0. (103)

Similarly Lx = L(1)

x + L(2)

x and Ly = L(1)

y + L(2)

y commute with H int, and thus H total =H1 + H2 + H int too, leading to the conclusion that

L = L(1)

+ L(2)

commutes with the Hamiltonian.

Since [L, H] = 0 eigenstates of H can also be eigenstates of L, whence

(Total) angular momentum quantum numbers are good quantum numbers.

• But what are they?Consider again the Helium atom with the repulsion turned off. The TISE is then

(H1 + H2)Φ(r1, r2) = EΦ(r1, r2).

This is separable: substitute Φ(r1, r2) = φ1(r1)φ2(r2) then we get:

1φ1

− ~2

2m∇2

1 + V (r1)φ1︸ ︷︷ ︸

=E1Hydrogen-like Z=2

+1φ2

− ~2

2m∇2

2 + V (r2)φ2︸ ︷︷ ︸

=E2Hydrogen-like Z=2

= E

so

Φ = φn1,`1,m1(r1)φn2,`2,m2(r2)

is an eigenstate.

80

Recall that electron number 1 has L(1)2

eigenvalue `1(`1 + 1)~ and likewise that electron

number 2 has L(2)2

eigenvalue `2(`2 + 1)~. What about the whole atom?Let’s look at the commutators of L:[

Lx, Ly

]= [L

(1)

x + L(2)

x , L(1)

y + L(2)

y ]

= [L(1)

x , L(1)

y ] + [L(2)

x , L(2)

y ]

= i~L(1)

z + i~L(2)

z

= i~Lzas expected (similarly for the others). Thus, we know that eigenstates Ψ must satisfy

L2Ψ = ~2L(L+ 1)Ψ L = 0, 1, 2, . . .

LzΨ = ~MΨ −L ≤M ≤ L

where L2

and Lz are the total angular momentum operators (note that we have been doingorbital angular momentum; however, this would also be true for spin).Thus, the states of the Helium atom have orbital angular momentum quantum numbersL,M just like the Hydrogen atom has orbital quantum numbers `,m. (This argumentworks for all atoms).

• It turns out that there is a connection between (`1,m1), (`2,m2) and (L,M):

LzΦ = ~MΦ

LzΦ = (L(1)

z + L(2)

z )Φ

= (L(1)

z + L(2)

z )φn1,`1,m1(r1)φn2,`2,m2(r2)= ~(m1 +m2)φn1,`1,m1(r1)φn2,`2,m2(r2)= ~(m1 +m2)Φ.

Thus, we obtain the addition rule for the z-component of angular momentum of two sys-tems:

M = m1 +m2. (104)

On the other hand, matters are not so simple for L2:

L2Φ = (L

(1)+ L

(2))2Φ

= (L(1)2

+ L(2)2

+ 2L(1)· L

(2))Φ

=(L

(1)2+ L

(2)2+ 2L

(1)

x L(2)

x + L(1)

y L(2)

y + L(1)

z L(2)

z )φn1,`1,m1(r1)φn2,`2,m2(r2)

Whilst φn1,`1,m1φn2,`2,m2 is an eigenstate of the first, second and fifth term, it is not an

eigenstate of the third and the fourth (i.e. of L(1)

i L(2)

i for i = x, y).

81

Because φn1,`1,m1φn2,`2,m2 is not an eigenstate of all of the L2

operator, when we measureL

2there is in general a superposition of different values we can get:

Φ is a linear superposition of eigenstates of L2

with different values of quantumnumber L.

In this course we won’t prove the following result:

If two angular momenta (with quantum numbers `1 and `2) are combined thenthe possible outcomes for L are:

L = `1 + `2, `1 + `2 − 1, . . . , |`1 − `2|. (105)

This result (which you need to know!) has a geometrical interpretation (much loved bychemists).

2.10.2 The triangle rule

Consider `1 and `2 as the lengths of two vectors which we shall add head to tail. If thesetwo vectors are parallel then the resulting vector has length L = `1 + `2 (extremal case).If `1 and `2 are not parallel then the resulting vector L will have a length which is smallerthan their sum (geometrically the three vectors `1, `2 and L will form a triangle). Thetriangle rule states that if `1, `2, L is a possible triangle (including the extremal case) thenL is a possible outcome when the total angular momentum quantum number is measured.Remember that all of these quantum numbers must be positive integers (or zero)!

Example 1: Suppose `1 = 1 and `2 = 2 then the possible L values are

3 = 1 + 22 = 1 + 2− 11 = 2− 1.

Note that the total number of states matches:

`1 = 1, m1 = 1, 0,−1 (3 states)`2 = 2, m1 = 2, 1, 0,−1,−2 (5 states)

whence the number of possible combinations (composite states) is 3 · 5 = 15. Comparing:

L = 3, M = 3, 2, 1, 0,−1,−2,−3 (7 states)L = 2, M = 2, 1, 0,−1,−2 (5 states)L = 1, M = 1, 0− 1 (3 states)

and indeed 7 + 5 + 3 = 15 states as desired.

Example 2: This also works for angular momentum including spin. E.g. for j1 = 12 and

j2 = 1 the possible J values are

82

32 = 1 + 1

212 = 1− 1

2 .

The number of possible states is

j1 = 12 , jz = + 1

2 ,−12 (2 states)

j2 = 1, jz = 1, 0,−1 (3 states)

i.e. 2 ·3 = 6 possible combinations. Again we find that the J, Jz calculation agrees with thisas

J = 32 , Jz = 3

2 ,12 ,−

12 ,−

32 (4 states)

J = 12 , Jz = 1

2 ,−12 (2 states)

and 4 + 2 = 6.

• Finally, return to Helium. Let’s now turn on the repulsion H int between the electrons. As

we’ve seen L(1)2

etc. do not commute with H int, but L2

does. Thus, if we measure L2

weget L(L+ 1) and if we measure Lz we get M .

But because of H int L(1)

and L(2)

are not individually meaningful (time independent).Only the counting of possible L,M values given by the angular momentum rules

M = m1 +m2

L = `1 + `2, `1 + `2 − 1, . . . , |`1 − `2|

is sensible.(Of course sometimes H int is relatively weak so classification via `1, `2,m1,m2 is approxi-mately valid. See the course on Intermediate Quantum Mechanics).

83