Part III Quantum Mechanics Preparatory

Workshop Solutions

Roland Bittleston Kai Roehrig

Michaelmas 2019

Please get in touch at r.bittleston@damtp.cam.ac.uk if you find typos/errorsor have further questions.

1 Principles

1.1

Using [x, p] = i we find[x, pn] = nipn−1

and hence[x, e−iap] = ae−iap .

Using this we find

eiapxe−iap = eiap(e−iapx+ [x, e−iap]

)= x+ a1 .

Assuming |x〉 is a position eigenvector, we find that the state e−iap|x〉 is alsoan eigenvector of x, with shifted eigenvalue x+ a

x(e−iap|x〉

)= (x+ a)

(e−iap|x〉

).

Soe−iap|x〉 = λ(x, a)|x+ a〉

for some complex valued function λ(x, a). Unitarity requires that |λ(x, a)|2 = 1,but this still leaves the phase of λ(x, a) unfixed. Fortunately we can absorbthis phase into our choice of position basis {|x〉}x∈R, i.e. we simply define

|x〉 = e−ixp|0〉 .

Acting on the previous line with 〈p| we find

〈p|x〉 = e−ixp〈p|0〉 . (1)

1

Now notice that the canonical commutation relation, [x, p] = i, is unchangedif we simultaneously exchange x and p, and replace i by −i. Applying thissimultaneous exchange and replacement to eq. (1), which we derived usingthe canonical commutation relation, we find

〈x|p〉 = eixp〈x|0〉 =⇒ 〈p|x〉 = e−ixp〈0|x〉

Together they imply 〈p|x〉 ∝ e−ixp, since eixp〈p|x〉 is independent of bothx and p and hence constant. We determine the constant of proportionalityusing unitarity, i.e. 1 =

∫dp |p〉〈p|

δ(x− y) = 〈x|y〉 =

∫dp 〈x|p〉〈p|y〉

so 〈x|p〉 = eixp/√

2π. (Up to an irrelevant constant phase.)

1.2

To show that eigenvalues are real consider

λi〈ψi|ψi〉 = 〈ψi|Oψi〉 = 〈O†ψi|ψi〉 = 〈Oψi|ψi〉 = λ∗i 〈ψi|ψi〉 .

Since 〈ψi|ψi〉 ≡ ‖ψi‖2 > 0 for ψi 6= 0 by our axioms, we learn that λi = λ∗i .To show that the eigenvectors |i〉 are mutually orthogonal consider

λi〈ψi|ψj〉 = 〈Oψi|ψj〉 = 〈ψi|O|ψj〉 = 〈ψi|Oψj〉 = λj〈ψi|ψj〉

which shows that(λi − λj)〈ψi|ψj〉 = 0 .

Hence we learn that eigenvectors of different eigenvalue are necessarily or-thogonal, 〈ψi|ψj〉 = 0 for λi 6= λj . If several eigenvectors share an eigenvalue,they span an eigenspace, and we can pick any basis for it, in particular alsoan orthonormal one. Hence, in general we have/can choose 〈ψi|ψj〉 = δij .

1.3

To find the components of a vector in a given basis, we take the inner productwith the basis elements:

ψ(x) := 〈x|ψ〉 , and ψ(p) := 〈p|ψ〉 ,

where a tilde indicates momentum basis. Next, since the bases are defined(in canonical quantization) via the action of x and p we get

(xψ)(x) := 〈x|x|ψ〉 = xψ(x) , and (pψ)(p) := 〈p|p|ψ〉 = pψ(p) .

2

From question 1.1 we know that the inner product between the basis vectorsis 〈x|p〉 = eixp/

√2π. Hence we can calculate

(pψ)(x) := 〈x|p|ψ〉 =

∫dp 〈x|p〉pψ(p) =

1√2π

∫dp eixppψ(p) = −i∂xψ(x) ,

and similarly

(xψ)(p) := 〈p|x|ψ〉 =

∫dx 〈p|x〉xψ(x) =

1√2π

∫dx e−ixpxψ(p) = i∂pψ(p) .

1.4

To evaluate ρ(x) we insert a completeness relation for the position basis

ρ(x) = 〈ψ|δ(3)(x− x)|ψ〉 =

∫d3y 〈ψ|δ(3)(x− x)|y〉〈y|ψ〉

=

∫d3y δ(3)(y − x)〈ψ|y〉〈y|ψ〉 = |ψ(x)|2 .

We recognize this as the probability density in position basis. To evaluate~j(x) we first notice

2m~j(x) = 〈ψ|{~p , δ(3)(x− x)}|ψ〉 = 2Re〈ψ|~p δ(3)(x− x)|ψ〉 ,

where { , } denotes the anticommutator, and then proceed similarly

m~j(x) = Re〈ψ|~p δ(3)(x− x)|ψ〉 = Re〈ψ|~p |x〉〈x|ψ〉

= Re[ψ(x)i~∇ψ∗(x)

]= Im

[ψ∗(x)~∇ψ(x)

]which we recognize as the probability current.

To write ∇ · j in this manner we can either work backwards throughsimilar steps as above, or remember that derivation in position space isrepresented by p, and its action is implemented by a commutator:

−2mi∇ · j = 〈ψ|[pi, {pi, δ(3)(~x− ~x)}]|ψ〉 = 〈ψ|[p2, δ(3)(~x− ~x)]|ψ〉 .

Note that in the second term above I am using summation notation.

1.5

We define (pψ)(x) = −i∂xψ(x) and then compute

〈ψ|p|ξ〉 = −i∫

dxψ∗(x)∂xξ(x) = i

∫dx (∂xψ

∗(x))ξ(x) = (〈ξ|p|ψ〉)∗

confirming that p is Hermitian. Note that we had to use negligibility ofboundary terms in the partial integration. This is the origin of the restrictionto square integrable functions in the Hilbert space. Look out for this subtletye.g. when you study AdS/CFT.

3

1.6

a) To find the classical EOMs for the Hamiltonian

H(x, p) =1

2m(~p− e ~A(x))2

we first note that

~x =∂H

∂~p=

~p

m− e

m~A(x)

and

~p = −∂H∂~x

=e

m(pi − eAi)~∇Ai = exi~∇Ai(x)

by the Hamilton-Jacobi equations (and using the first to simplify thesecond). Putting the two together we find

m~x = ~p− e ~A = e( ~E − (x · ∇) ~A+ xi~∇Ai) = e( ~E + ~x× ~B) ,

using the formula for the total time derivative ~A = ∂t ~A+ (x ·∇) ~A andthe vector-product relation ~a× (~b× ~c) = ~b a·c− ~c a·b.

b) From a) we see that the tangent vector is ~pA = ~p−e ~A(x). The Poissonbracket is

{(pA)i, (pA)j} = −e(∂xiAj + e∂xjAi) = −eFij

with Fij the space-space components of the electromagnetic field strengthFµν .

c) In promoting the functions p, x and A(x) to operators we have todecide which classical expression to start with. Different choices maydiffer in the ordering of the quantum operators. First note that we canquantize the electromagnetic potential by quantizing x, i.e. A(x) →A(x) in terms of its power series, with no ordering ambiguity. Nextobserve that classically we have

H =1

2m(p− eA(x))2 =

p2

2m− e

mp ·A(x) +

e2

2mA2(x) .

Quantum mechanically there will be a difference between these twoexpressions, because x and p no longer commute. Clearly this affectsthe middle term, where −e(p ·A(x)+A(x) ·p) = −2ep ·A(x) classically.The ambiguity arises because after quantizing p ·A(x) = A(x) · p is nolonger necessarily true. But is it? Now calculate

p ·A(x)−A(x) · p = [pi, Ai(x)] = −i∇ ·A

4

which is the Coulomb condition. In the second term above I haveadopted summation convention. Hence, in Coulomb gauge we cansimply write

H(x, p) =1

2m(p− eA(x))2

without any ordering ambiguity.

d) In position space i~pA = ~∇ − ie ~A(x), which goes by the name of co-variant derivative. You will encounter this in both QFT and GR andlearn about its beautiful origin in geometry and its place in physics.

1.7

We have PA(i) = |i〉〈i|, hence (PA(i))2 = |i〉〈i|i〉〈i| = |i〉〈i| = PA(i) asrequired. Also PA(i)† = (|i〉〈i|)† = |i〉〈i| = PA(i) using (|a〉〈b|)† = |b〉〈a|.For projecting on an arbitrary subset of the spectrum, say i = 1, · · · , N , wecan use the operator P =

∑Ni=1 |i〉〈i|. By orthogonality of the eigenvectors

(remember, A is an observable) this is indeed again a projection operator.

1.8

a) We calculate

d

dt〈ψ|ψ〉 =

d〈ψ|dt|ψ〉+ 〈ψ|d|ψ〉

dt= i〈ψ|H†|ψ〉 − i〈ψ|H|ψ〉 = 0

using that the Hamiltonian is Hermitian H† = H.

b) Using H† = H we find

∂tρ+∇·j = i〈ψ|[H− p2

2m, δ(3)(x−x)]|ψ〉 = i〈ψ|[V (x), δ(3)(x−x)]|ψ〉 = 0

because functions of x commute.

d) If the Hamiltonian is not Hermitian we get ddt〈ψ|ψ〉 = 〈ψ|Im H|ψ〉 and

hence loose total probability. This could arise in situations where Honly describes a subsystem and the imaginary part of the potentialarises from ignoring the remainder of the total system.

1.9

a) At time t = 0 the system is in a normalized eigenstate of A, whichwe shall denote by |ψ(0)〉 = |i〉. At a time t > 0 this has evolved into|ψ(t)〉 = U(t)|i〉. Hence the probability of measuring the system to bein the state |j〉 after a time T is

〈ψ(T )|j〉〈j|ψ(T )〉〈ψ(T )|ψ(T )〉

= |〈j|U(T )|i〉|2

5

b) If the states |i〉, |j〉 are energy eigenstates with different energies thenabove becomes |〈i|j〉|2 = δij . This is because U(T ) acts on an energyeigenstate by multiplying it by a phase.

1.10

We calculate, using Euclidean time tE = it,

〈x| exp

(− it

2mp2

)|0〉 =

∫dp 〈x|p〉〈p|0〉e−

it2m

p2 =

∫dp

2πexp

(− it

2mp2 + ipx

)=

∫dp

2πexp

(− tE

2mp2 + ipx

)=

√m

2πtEexp

(− m

2tEx2

)=

√m

2πitexp

(− m

2itx2).

In QFT you will repeat this calculation in a relativistic setting and haveto be more careful about regularizing this integral. This will reveal a deepphysical meaning behind this regularization procedure, related to causality.

1.11

Letting [p, f(x)] act on an arbitrary vector |ψ〉 and working in position basiswe find by appropriately inserting a 1 =

∫dx |x〉〈x|

〈x|[p, f(x)]|ψ〉 = 〈x|pf(x)|ψ〉 − 〈x|f(x)p|ψ〉= −i (∂x(f(x)ψ(x))− f(x)∂xψ(x))

= −if ′(x)ψ(x)

= −i〈x|f ′(x)|ψ〉 .

To calculate basis-independently we expand

[p, f(x)] =∑n

f (n)

n![p, xn] = −i

∑n

f (n)

n!nxn−1 = −if ′(x) .

We used that [p, xn] = −inxn−1, which we can show via induction from

[p, xn] = xn−1[p, x] + [p, xn−1]x

and the canonical commutation relation (which serves also as the inductionstarting point).

1.12

To show that σ2A(ψ) = 〈A2〉 − 〈A〉2 ≥ 0, we show that 〈A〉2ψ ≥ 〈A2〉ψ by

considering the norm of the state |ψλ〉 = A|ψ〉−λ|ψ〉 = (A−λ1)|ψ〉 for anyλ ∈ R. By the properties of the Hilbert space we have

0 ≤ ‖ψλ‖2 = 〈ψ|A2|ψ〉 − 2λ〈ψ|A|ψ〉+ λ2〈ψ|ψ〉

6

for all λ. Minimizing the right hand side (r.h.s.) with respect to (w.r.t.) λin order to get the strongest possible bound we learn

0 ≤ 〈ψ|A2|ψ〉 − 〈ψ|A|ψ〉2

〈ψ|ψ〉which is clearly what we were looking for. Note that again the properties ofthe Hilbert space saved us from worrying about the case 〈ψ|ψ〉 = ‖ψ‖2 = 0.

Finally, if we want to saturate the bound on σ2A, i.e. have σ2

A = 0 weneed to saturate the bound 0 ≤ ‖ψA‖2, which (in our Hilbert space) directlymeans that ψ is a eigenvector of A with eigenvalue λ.

1.13

Firstly we can apply the argument from above to the operator O = A+ λBto show that f(λ) = 〈O†0O0〉ψ = 〈(A0 + λ∗B0)(A0 + λB0)〉ψ ≥ 0. Then weminimize f(λ) w.r.t. to λ over all C to find

0 ≤ f(λmin) = 〈A20〉ψ −

〈A0B0〉ψ〈B0A0〉ψ〈B2

0〉ψ.

Note that we have to assume that ψ is not an eigenvector to B. (If it were,we could swap A↔ B.) Since 〈B2

0〉ψ > 0 we then find

σ2A(ψ)σ2

B(ψ) ≥ 〈A0B0〉ψ〈B0A0〉ψ =

(1

2〈{A0, B0}〉ψ

)2

+

(1

2i〈[A, B]〉ψ

)2

where we used |z|2 = (Re z)2 + (Im z)2. That last step is useful becauseit separates the classical and quantum contribution, in the following sense:going back to classical mechanics, the phase-space functions A,B which gaverise to the operators A and B now commute, so the second term (involvingthe commutator) vanishes. The term with the anti-commutator reduces tothe usual co-variance 〈A0B0〉ψ.

1.14

a) The Schrodinger equation reads i∂t|ψ〉 = H|ψ〉. Plugging in |ψ〉 =exp(−itH)|ψ0〉 we find

i∂t|ψ(t)〉 = i∂t exp(−itH)|ψ0〉 = H exp(−itH)|ψ0〉 = H|ψ(t)〉

which verifies that it’s a solution.

b) We can proceed by calculating explicitly

U(t)†U(t) = exp(−itH)† exp(−itH) = exp(+itH†) exp(−itH)

= exp(+itH) exp(−itH) = exp(−it(H − H)) = 1

if H = H†.

7

c) Again we directly find

U(t2)U(t1) = exp(−it2H) exp(−it1H) = exp(−i(t1+t2)H) = U(t1+t2)

Note that in all three computations we rely on the fact that [H, H] = 0, sothat the exponentials combine in the familiar fashion. Keep in mind thatthis is in general not the case!

1.15

Using the by now very familiar relations we find

exp(iap)|x〉 =

∫dp |p〉eiap〈p|x〉 =

∫dp |p〉eiap+ipx = |x+ a〉

and hence also, using Up(a)† = Up(a)−1 = Up(−a),

Up(a)†f(x)Up(a)|x〉 = Up(−a)f(x)|x+a〉 = f(x+a)Up(−a)|x+a〉 = f(x+a)|x〉.

1.16

The time dependence of observable expectation values should not dependon the picture we work in. We check that

d

dt〈O〉Ψ(t) =

1

‖Ψ‖2d

dt〈Ψ|O|Ψ〉 =

1

‖Ψ‖2(

(−iH|Ψ〉)†O|Ψ〉 − i〈Ψ|OH|Ψ〉)

= i〈[H, O]〉Ψ(t)

in the Schrodinger picture and

d

dt〈O(t)〉ψ =

⟨dO(t)

dt

⟩ψ

= i〈[H, O(t)]〉ψ

in the Heisenberg picture. Note that we assumed the operator to have noexplicit time dependence, but this could be included in the obvious way.

1.17

The Heisenberg EOM for the operators are given by

˙x = i[H, x] =i

2m[p2, x] =

i

2m(p[p, x] + [p, x]p)

!=

1

mp

and˙p = i[H, p] = [V (x), ip]

!= −V ′(x) ,

where the last term comes from demanding the “classical EOM” to hold forthe operators. Specializing to V (x) = x we find that we need [x, p] = i, andusing past results we see that this choice indeed solves both requirements.

8

1.18

The Hilbert space describing the spin degree of freedom of a spin-12 particle

is C2, which has dimension 2. Taking N identical particles the Hilbertspace becomes (C2)⊗N = C2N . Note that we eventually would have totake into account the statistics of the particles in question, i.e. take theanti-symmetric tensor product for fermions.

1.19

We note that〈ψ|H|ψ〉 =

ω

2‖aψ‖2 +

ω

2‖a†ψ‖2 ≥ 0

since ‖ · ‖2 ≥ 0 and the sum of two non-negative quantities is again non-negative, (and ω > 0) so the energy is indeed bounded from below.

1.20

Using the results from above we can swiftly write

a = x

√mω

2+ p

i√2mω

= x

√mω

2+

1√2mω

∂x

and

a† = x

√mω

2+ p

−i√2mω

= x

√mω

2− 1√

2mω∂x .

We calculate their anti-commutator in this basis via

a±f(x) =

√mω

2xf(x)∓ 1√

2mωf ′(x)

a∓a±f(x) =mω

2x2f(x)∓ x

2f ′(x)− f ′′(x)

2mω± 1

2f(x)± 1

2xf ′(x) ,

with a− := a and a+ := a†, hence

ω

2{a, a†}f(x) :=

ω

2(aa† + a†a)f(x) =

(− ∂2

x

2m+mω2

2

)f(x)

which we indeed recognize as the Hamiltonian for the harmonic oscillator.The commutator can be calculated using linearity and anti-symmetry of thecommutator, i.e. [a+ b, c] = [a, c] + [b, c] and [a, a] = 0. It is straightforwardto verify that

[a, a†] = 1 .

9

1.21

Looking back we see that in order to improve the bound to 〈ψ|H|ψ〉 > 0, wehave to show that no state simultaneously satisfies a|ψ〉 = 0 and a†|ψ〉 = 0.Of course we have to exclude the trivial case |ψ〉 = 0. (Note that |ψ〉 = 0is not considered a physical state, and in particular is is not to be confusedwith the vacuum state |0〉.) Let us show this by contradiction: Assume wehave a state which satisfies a|ψ〉 = 0 and a†|ψ〉 = 0. But then also

|ψ〉 = [a, a†]|ψ〉 = aa†|ψ〉 − a†a|ψ〉 = 0

which shows that the only vector for which this can be true is in fact 0.Hence we can improve our previous bound to

〈ψ|H|ψ〉 > 0 .

This positivity of the energy is called zero-point energy and is a hallmark ofquantum physics. The classical harmonic oscillator has minimal energy 0,corresponding to the particle being at rest. Note that we can trace the strictpositivity of the energy back to the non-vanishing commutator of x and p.This strict positivity is so ubiquitous in quantum physics that its absence(i.e. the existence of a state with zero energy) is usually an indicator forsupersymmetry.

To find a strict lower bound use the commutation relations to write

H =ω

2{a, a†} = ω

(a†a+

1

2

).

It is then clear that

〈H〉|ψ〉 = 〈ψ|H|ψ〉 = ω‖a|ψ〉‖2 +ω

2≥ ω

2,

for any normalized state |ψ〉. This lower bound is saturated when a|ψ〉 = 0.As part of the next question you are required to show that such a stateexists for the spectrum to be bounded below.

1.22

This result follows from a standard calculation.

1.23

a) We recall [a, a†] = 1 and calculate

〈n|m〉cncm

= 〈0|an(a†)m|0〉 = 〈0|an−1[a, (a†)m]|0〉

10

since a|0〉 = 0. Using induction we show that [a, (a†)m] = m(a†)m−1,and hence

〈0|an−1[a, (a†)m]|0〉 = m〈0|an−1(a†)m−1|0〉 = m〈0|an−2[a, (a†)m−1]|0〉= m(m− 1)〈0|an−2(a†)m−2|0〉 = · · · = m!〈0|an−m|0〉

where we had to assume that n ≥ m. (If n < m we can insteadcalculate 〈m|n〉 using the same steps.) Now, since a|0〉 = 0 we will geta non-zero result only if n = m. So

〈n|m〉 = n!cncmδnm .

To normalize the states we set cn = 1/√n!, whence 〈n|m〉 = δnm.

b) We can expand f in its Taylor series, so it suffices to show the desiredrelation for a monomial. We use the previous result and proceed byinduction

an(a†)m|0〉 = an−1[a, (a†)m]|0〉 = man−1(a†)m−1|0〉

= · · · =

{0 for n > mm!

(m−n)!(a†)m−n for n ≤ m

which is clearly the right behaviour for a derivative.

c) Using the realization of the commutation relation the computation isstreamlined

〈n|m〉 =1√n!m!

〈0|∂na†(a†)m|0〉 =

1√n!m!

m!δnm = δnm .

1.24

To show that |λ〉 = exp(λa†)|0〉 is a eigenvector of a, calculate

a|λ〉 = ∂a†eλa† |0〉 = λeλa

† |0〉 = λ|λ〉 .

Expanding the exponential in a power series and using the results from abovewe find

〈λ|κ〉 = 〈0|eλ∂a†eκa† |0〉 =∑n

λn

n!〈0|∂na†e

κa† |0〉 =∑n

(λκ)n

n!〈0|eκa† |0〉 = eλκ .

Alternatively we can use the orthogonality of the energy eigenstates

〈λ|κ〉 =

∞∑n,m=0

λnκm

n!m!〈0|an(a†)m|0〉 =

∞∑n,m=0

λnκm√n!m!

〈n|m〉

=

∞∑n,m=0

λnκm√n!m!

δnm = eλκ .

11

Likewise we can calculate

〈λ|H|λ〉 =

∞∑n,m=0

λnλm√n!m!

〈n|H|m〉 =

∞∑n=0

|λ|2n

n!

(n+

1

2

)=(|λ|2 +

1

2

)e|λ|

2

and hence find the average energy

〈H〉λ = |λ|2 +1

2.

2 Spin

2.1

The components of the angular momentum vector are Li = εijkxjpk. Incanonical quantization we have to arbitrarily pick an ordering for the oper-ators. The difference between the two orderings is however

εijkxj pk − εijkpkxj = εijk[xj , pk] = iεijkδjk = 0 ,

so there is no ordering ambiguity. The angular momentum algebra can beverified by straightforwardly plugging in the operator definitions on bothsides and using a Levi-Civita symbol identity.

2.2

We pick a basis for R3 where ~α = αez. Further we work in cylindricalcoordinates (ρ, φ, z), in which Lz = −i ∂∂φ . Then the action becomes

exp

(α∂

∂φ

)f(ρ, φ, z) = f(ρ, φ+ α, z) .

Hence we find that it rotates around the axis of ~α by an angle α = ‖~α‖.

2.3

Verify by explicit calculation.

2.4

Using σ2i = 1 we immediately find S2 = 3

41, and hence s = 12 . The Hilbert

space is C2. We can pick the basis elements 〈 ↑ | = (1, 0) and 〈 ↓ | = (0, 1)and write a generic element (c1, c2)T = c1|↑ 〉+ c2|↓ 〉.

12

2.5

The Hamiltonian is H = b2σ3. Clearly we have the eigenvectors |↑ 〉 = (1, 0)T

and |↓ 〉 = (0, 1)T with eigenvalues E± = ±b/2 respectively. Given an initialstate |ψ(0)〉 = (c1, c2)T we find the state at time t by applying the timeevolution operator

U(t) = exp(−itH) = exp(−itbσ3/2) =

(e−i

b2t 0

0 eib2t

).

Hence we find |ψ(t)〉 = (c1e−i b

2t, c2e

i b2t)T .

2.6

As a first glance suggests the range of ϕ is ϕ ∈ (0, 2π), with endpointsidentified. The range for ϑ is more tricky, because we have to keep in mindthat a state is only defined modulo a complex number. Note that if we sendϑ→ ϑ+4π the vector does not change at all. But, if we perform the smallershift ϑ → ϑ + 2π, the vector goes to an equivalent vector, describing thesame state which suggests the range ϑ ∈ (−π, π). However, if we performϑ → −ϑ and at the same time ϕ → ϕ + π, we get again to a physicallyequivalent state. We conclude that the correct range for ϑ is (0, π), withendpoints not identified. Note that at the endpoints of the ϑ range, thestate becomes independent of ϕ. Altogether this shows that the state spaceis actually a sphere, S2, called Bloch sphere in quantum information andRiemann sphere otherwise.

Under the time evolution above we see that ϑ(t) = ϑ0 and ϕ(t) = ϕ0 +bt,or ϑ = 0 and ϕ = b, so the state moves on a circle in a plane normal to thedirection of the magnetic field.

Hidden in the above question is some stunningly beautiful geometry. Tofind it note that the unit sphere in our Hilbert space is given by |c1|2+|c2|2 =1, which defines a 3-sphere S3 ⊂ C2. Identifying states related by a phasegives a 2-sphere, as argued above. Hence this process of identifying statesdefines a map

S3 → S2 .

The preimage of any point in S2 is a set of unit vectors in S3 all differingby a phase. Hence, this set of unit vectors is geometrically a circle, S1. Assuch the map above is sometimes expressed as

S1 ↪−→ S3 → S2 .

This notation is supposed to remind us that circles in S3 get mapped topoints in S2. This map is so important that it has its own name, the Hopffibration. If you look it up on YouTube you’ll find some great animationsgiving you an idea about how it works geometrically.

13

2.7

Using the above we find that ~S(t) = (cosϕ sinϑ, sinϕ sinϑ,− cosϑ)T , whichwe recognize as a unit vector in R3 in spherical coordinates. We noticethat Sz = − cosϑ = const., while ~S⊥(t) ∝ R(bt)(1, 0)T with the 2d-rotationmatrix R(α). Note the relation to the previous question.

The parametrization of a real 3d unit vector by a complex 2d vectorup to a phase is also used in the spinor-helicity formalism which you maylearn about in the context of modern techniques for scattering of masslessparticles.

2.8

The Heisenberg equation says that the components of ~S evolve as

˙Si = i[H, Si] = iBj [Sj , Si] = εijkBjSk = ( ~B × ~S)i .

In the coordinate system where ~B = bez this means Sz = const., while

d

dt

(SxSy

)= b

(0 −11 0

)(SxSy

)which we can integrate (solve) to give(

Sx(t)

Sy(t)

)=

(cos(bt) − sin(bt)sin(bt) cos(bt)

)(Sx(0)

Sy(0)

)which is consistent with our previous results.

2.9

The trick here is to choose a basis of C2 in which H = ~B · ~S takes a partic-ularly simple form. The best way to achieve this is to write

~B = b(cosϕ sinϑ, sinϕ sinϑ,− cosϑ)T

for some ϑ ∈ [0, π] and ϕ ∈ [0, 2π), where b = ‖B‖. Then the states

|+〉 = (sin(ϑ/2), cos(ϑ/2)eiϕ)T and |−〉 = (cos(ϑ/2),− sin(ϑ/2)eiϕ)T

obey H|±〉 = ±(b/2)|±〉. To see why note that H2 = (b/2)2, so the eigen-values of H can only be ±b/2. We also know from question 2.7 that〈H〉|±〉 = ±b/2. This is only possible if |±〉 are eigenstates of H with eigen-values ±b/2 respectively.

14

2.10

The main key point of this calculation is that (~a · ~σ)2 = ‖a‖21, which is aspecial case of

(~a · ~σ)(~b · ~σ) = aibjσiσj =aibj

2({σi, σj}+ [σi, σj ]) = ~a ·~b+ i(~a×~b) · ~σ .

Hence (~a · ~σ)2n = ‖a‖2n1, so that we find (with a = ~a/‖a‖)

exp(i~a · ~σ) = cos(~a · ~σ) + i sin(~a · ~σ) = 1 cos(‖a‖) + ia · ~σ sin(‖a‖) ,

since cos and sin arise from the even and odd terms in the exponentialrespectively.

3 Perturbation Theory

3.1

We want to show

exp(A+ εB) = eA + ε

∫ 1

0ds e(1−s)ABesA +O(ε2)

by counting powers on both sides. First set ε = 0 and notice that the O(ε0)piece is satisfied. Next we examine the piece at order O(ε1). Expanding theexponential in its Taylor series and collecting all terms containing a singleεB we find on the left hand side

∞∑n=0

1

(n+ 1)!

n∑i=0

AiBAn−i .

Note that we find one term for each constellation of A’s to the left/rightof B. Now we turn our attention to the right hand side. Expanding bothexponentials we find

∞∑m,n=0

AmBAn∫ 1

0ds

(1− s)msn

m!n!.

Notice that the double sum covers exactly the same terms as the doublesum on the l.h.s.. (We could manifest this by shifting the summation indicesappropriately.) Evaluating the integral

1

m!n!

∫ 1

0ds (1− s)msn =

1

(m+ n+ 1)!

shows moreover that each term appears with the correct coefficient, estab-lishing the equality of l.h.s. and r.h.s..

15

3.2

The result is obtained by performing the integral∫ t

0dt′ 〈j|e−i(t−t′)H0HIe

−it′H0 |i〉 =

∫ t

0dt′ e−i(t−t

′)Ej−it′Ei〈j|HI |i〉

and then taking the modulus squared.Note that the dependence on the concrete form of the perturbation re-

sides only and completely in the matrix element 〈j|HI |i〉. So from now onwe will just have to calculate this object.

3.3

At δ = 0 this is just the Hamiltonian of a spin-12 system with energies ±E

and eigenvectors |±〉. The transition amplitudes are hence 〈+|HI |−〉 = δand 〈−|HI |+〉 = δ and hence |〈±|HI |∓〉|2 = |δ|2.

3.4

The object of interest is the matrix element 〈j|x3|i〉 in the energy-eigenstatebasis. We recall that x = 1√

2mω(a+ a†) as well as

a|i〉 =√i|i− 1〉 and a†|i〉 =

√i+ 1|i+ 1〉 .

Hence the anharmonic perturbation will facilitate jumps across at most threeenergy levels (since 〈i|j〉 = δij). As an example we can calculate the proba-bility of such an maximal jump, which comes from only one term in x3

〈i+ 3|x3|i〉 =1

(2mω)3/2〈i+ 3|(a†)3|i〉 =

√(i+ 1)(i+ 2)(i+ 3)

(2mω)3/2.

Note that under this anharmonic perturbation it is not possible for statesto transition up or down exactly 2 energy levels at leading order in ε.

3.5

We now want to extend perturbation theory to second order in ε. To do thiswe need to extend the result of question 3.1 to second order in ε. This isgiven by

exp(A+ εB) = eA

+ ε

∫s,t∈[0,1]

ds dt δ(1− s− t) esABetA

+ ε2∫s,t,u∈[0,1]

ds dtdu δ(1− s− t− u) esABetABeuA

+O(ε3) .

(2)

16

The generalization to arbitrary order in ε is straightforward. To prove thiswe essentially repeat the argument used in 3.1, now making use of the inte-gral identity∫

s,t,u∈[0,1]ds dt du δ(1− s− t− u) sptqur =

p!q!r!

(p+ q + r + 2)!

for integers p, q, r.Using eq. (2) we can compute the second order contribution to the tran-

sition amplitude between two energy eigenstates |i〉, |j〉. Let’s start by ex-panding out the transition amplitude in powers of the perturbation

〈j| exp(−itH)|i〉 = Ai→j(t) = A(1)i→j(t) +A(2)

i→j(t) +O(ε3) .

We’ve already computed the first order term in question 3.2 to be

A(1)i→j(t) = (−i)e−itEij

sin(ωijt/2)

ωij/2〈j|HI |i〉 ,

where Eab = (Ea + Eb)/2 and ωab = Ea − Eb. The second order term isgiven by

i2A(2)i→j(t) =∫

t1,t2,t3∈[0,t]dt1 dt2 dt3 δ(t− t1 − t2 − t3) 〈j|e−it1H0HIe

−it2H0HIe−it3H0 |i〉 .

By applying the completeness relation for the energy eigenstates of the un-perturbed system we deduce

〈j|e−it1H0HIe−it2H0HIe

−it3H0 |i〉 =∑k

e−i(t1Ej+t3Ei)〈j|HI |k〉〈k|e−it2H0HI |i〉

=∑k

e−i(t1Ej+t2Ek+t3Ei)〈j|HI |k〉〈k|HI |i〉

We can now perform the integrals directly∫t1,t2,t3∈[0,t]

dt1 dt2 dt3 δ(t− t1 − t2 − t3) e−i(t1Ej+t2Ek+t1Ei) =

1

iωij

(e−itEjk

sin(ωjkt/2)

ωjk/2− e−itEki

sin(ωkit/2)

ωki/2

).

From this we conclude that

A(2)i→j(t) =

i

ωij

∑k

(e−itEjk

sin(ωjkt/2)

ωjk/2− e−itEik

sin(ωikt/2)

ωik/2

)〈j|HI |k〉〈k|HI |i〉 .

17

At this point we can find the next to leading order contribution to thetransition probability. We have

Pi→j(t) = |Ai→j(t)|2 = |A(1)i→j(t)|

2 + 2Re((A(1)i→j(t)

)∗A(2)i→j(t)

)+O(ε4) .

Note that for the contribution at third order in ε to be non-zero it must bethe case that both A(1)

i→j(t) and A(2)i→j(t) are non-vanishing for some i and j.

Of course, this requires that 〈j|HI |i〉 is non-zero, but there must also exista state in the unperturbed spectrum, labelled by k, such that 〈k|HI |i〉 and〈j|HI |k〉 are also non-zero. This requires that i− j, j − k, k − i ∈ {±1,±3},which is impossible. Hence there is no contribution at third order in ε.

3.6

This is a straightforward calculation similar in character to those involvedin transitioning between the Schrodinger and Heisenberg pictures.

4 The Feynman propagator

4.1

We want to calculate〈0|T (x(t)x(0))|0〉

for the harmonic oscillator. Start by writing

x(t) =a(t) + a†(t)√

2mω

where a(t) and a†(t) are the ladder operators in the Heisenberg picture.They are related to the Schrodinger picture via the time evolution operatorU(t) = exp(−itH) as

a(t) := U(t)†aU(t) = eitHae−itH .

We may calculate this either by resorting to the Heisenberg equation, whichis simply the time derivative of the above definition, or by explicit computa-tion. We’ll do both ways. The Heisenberg equation leads to the answer veryquickly. taking the derivative of the last equation we find the Heisenbergequation

a(t) = i[H, a(t)]

which, using [H, a(t)] = −ωa(t), we swiftly solve to find

a(t) = e−itωa(0) .

18

For the record, let’s do the explicit way as well. Using the definition of theexponential we write

eitHae−itH =∞∑

n,m=0

(it)n

n!

(−it)m

m!HnaHm

=∞∑n=0

(it)n

n!

n∑m=0

(n

m

)(−1)mHn−maHm .

Now we have to do a bit of gymnastics. First, let’s introduce the shorthandnotation for the commutator

[H,O] := adHO .

for any operator O. With this notation we may write e.g.

[H, [H, a]] = (adH)2a

and[H, [H, [H, a]]] = (adH)3a

and so on. Notice that under this so called adjoint action of the Hamiltonian,the operator a is an eigen-operator with eigenvalue −ω, in the sense thatadHa = −ωa. This leads to great simplifications soon, but first we want toshow that

(adB)nA =n∑

m=0

(n

m

)(−1)mBn−mABm

for any two operators A and B. Let’s show it via induction. Clearly it is truefor n = 0 and 1, pretty much by definition. Now let’s prove the inductivestep.

adB

(n∑

m=0

(n

m

)(−1)mBn−mABm

)

=

n∑m=0

(n

m

)(−1)mBn+1−mABm −

n∑m=0

(n

m

)(−1)mBn−mABm+1

= Bn+1A+

n∑m=1

((n

m

)+

(n

m− 1

))(−1)mBn+1−mABm + (−1)n+1ABn+1

=

n+1∑m=0

(n+ 1

m

)(−1)mBn+1−mABm = (adB)n+1A

where, going from the second to the third line, we collected terms of thesame power in B, and then used the identity(

n

m

)+

(n

m− 1

)=

(n+ 1

m

).

19

Now going back to the time evolution of a, we may hence write

eitHae−itH =∞∑n=0

(it)n

n!(adH)na =

∞∑n=0

(it)n

n!(−ω)na = e−iωta .

This confirms the result obtained using the Heisenberg equation.Substituting in the definition for the position operator we find further

that

x(t) =1√

2mω(e−iωta+ eiωta†) .

Going back to the Feynman propagator, we plug in this expression for x(t)and use the crucial property of the ladder operator, a|0〉 = 0, to find

〈0|x(t)x(0)|0〉 =1

2mω〈0|(e−iωta+ eiωta†)(a+ a†)|0〉

=1

2mωe−iωt〈0|aa†|0〉 =

1

2mωe−iωt

using 〈0|aa†|0〉 = 〈0|[a, a†]|0〉 = 1, and similarly

〈0|x(0)x(t)|0〉 =1

2mωeiωt .

Keeping in mind the time ordering

T (x(t)x(0)) := θ(t)x(t)x(0) + θ(−t)x(0)x(t)

with the Heavyside step function θ(t) we finally conclude

〈0|T (x(t)x(0))|0〉 =1

2mωe−iω|t| .

(Note that the Feynman propagator is not a real valued function, despitethe position operator x being Hermitian, due to the time ordering.)

4.2

First note that Feynman propagator depends only on the difference t − t′,which a consequence of time translation invariance. Abusing notation bywriting GF (t, t′) = GF (t− t′) this question amounts to showing that(

d2

dt2+ ω2

)GF (t) = − i

mδ(t) .

Let’s start by calculating

d2

dt2GF (t) =

d2

dt2

(1

2mωe−iω|t|

).

20

Differentiating once we have

d

dt

(1

2mωe−iω|t|

)= − i

2msign(t)e−iω|t| ,

where we have used the fact that

d

dt|t| = sign(t) .

Differentiating again we arrive at

d2

dt2

(1

2mωe−iω|t|

)=

d

dt

(− i

2msign(t)e−iω|t|

)= − i

mδ(t)e−iω|t| − ω

2m(sign(t))2e−iω|t| .

This can be simplified by noting that

δ(t)e−iω|t| = δ(t)

and similarly(sign(t))2 = 1

except perhaps when t = 0, but at this point the delta function is singularso we’ve got bigger problems. Hence we deduce that

d2

dt2GF (t) = − i

mδ(t)− ω

2me−iω|t| = − i

mδ(t)− ω2GF (t)

as required.

4.3

Now we want to use the Feynman propagator to see that

〈0|[x, p]|0〉 = i .

Start by rewriting the l.h.s. as follows

〈0|[x, p]|0〉 = limδt→0+

〈0|(x(0)p(−δt)− p(δt)x(0)

)|0〉

where on the r.h.s. we’re using the Heisenberg picture operators x(t), p(t).Note we are taking the limit δt→ 0 from above. This allows us to introducetime ordering symbols to rewrite the above as

〈0|[x, p]|0〉 = limδt→0+

(〈0|T (x(0)p(−δt))|0〉 − 〈0|T (x(0)p(δt))|0〉

).

21

Using the Heisenberg equation of motion p(t) = m ˙x(t) we have

〈0|T (x(0)p(t))|0〉 = md

dt〈0|T (x(0)x(t))|0〉 = m

d

dtGF (t) = − i

2sign(t)e−iω|t| .

From this we deduce that

〈0|[x, p]|0〉 = limδt→0+

(− i

2sign(−δt)e−iω|−δt| + i

2sign(δt)e−iω|−δt|

),

and since δt > 0 this can be simplified to give

〈0|[x, p]|0〉 = limδt→0+

ie−iω|δt| = i .

22