Advanced Quantum Mechanics

• 14 December 2020 - colloquium

• 2 February 2021 - egzamination

Note: Thursday 12 November and Friday 8 January are Wednesdays

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Epigraph:Ty podarisz mnie smiertnuju droza nie blednuju droz s ladostrastijai mienia na wsiegda uwiedioszk ostrowiam sowierszennowo sczastija.Niko laj Gumilow, Beatricze

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Here will appear material which is not included in Chapters of QFT (posted separately)Problem 0.1 Consider the Hamiltonian of the form2 H = H0 + Vint, where

H0 = Eb|b〉〈b| +∞∑

l=−∞

l∆|l〉〈l| ,

Vint = g∞∑

l=−∞

(|b〉〈l| + |l〉〈b|) .

It is understood that 〈l′|l〉 = δl′l and 〈b|l〉 = δbl. Assuming that at t = 0 the system wasprepared in the isolated state |b〉, |ψ(0)〉 = |b〉, find in the first order of the perturbativeexpansion the probability of finding it at a further instant t in any of the states |l〉 usingthe Poisson summation formula instead of the usual reasoning which leads to the FermiGolden Rule. Compare the result of this first order calculation with the exact result.

Solution to Problem 0.Writing the solution of the Schrodinger equation

i~d

dt|ψ(t)〉 = (H0 + Vint)|ψ(t)〉 ,

as a series |ψ(t)〉 = |ψ(t)〉(0) + |ψ(t)〉(1) + . . . in powers of the coupling g, one has to solvethe equations

i~d

dt|ψ(t)〉(0) = H0|ψ(t)〉(0) ,

i~d

dt|ψ(t)〉(1) = H0|ψ(t)〉(1) + Vint|ψ(t)〉(0) ,

with the boundary conditions |ψ(0)〉(0) = |b〉, |ψ(0)〉(1) = 0. The solution of the firstequation |ψ(t)〉(0) = |b〉 exp(−(i/~)Ebt) is obvious. Inserting this in the second one andclosing it from the left with 〈l| one gets the linear inhomogeneous equation

i~d

dt〈l|ψ(t)〉(1) = El 〈l|ψ(t)〉(1) + g e−iEbt/~ ,

which can be solved in the standard way. Thus, in the first order the evolution of thestate of the system is given by3

|ψ(t)〉 = |b〉 e−iEbt/~ + g∞∑

l=−∞

|l〉 1 − ei(El−Eb)t/~

El −Eb

e−iElt/~ .

1This is taken from the paper “Fermi’s golden rule: its derivation and breakdown by an ideal model”by Zhang J.M. and Liu Y., arXiv:1604.06916 [quant-ph].

2The Hamiltonian is somewhat unphysical in that it has now lowest energy eigenstate - the spectrumis unbounded from below - nevertheless can be used for illustrative purposes.

3In the first order there is no correction proportional to |b〉 because 〈b|Vint|b〉 = 0.

3

The probability of finding the system in any of the states |l〉 can be then obtained as

P =

∞∑

l=−∞

|〈l|ψ(t)〉|2 = g2∞∑

l=−∞

4 sin2((El −Eb)t/2~)

(El −Eb)2= g2

t2

~2

∞∑

l=−∞

(

sin xlxl

)2

,

where xl = (El−Eb)t/2~ = (l∆−Eb)t/2~. Instead of appealing to the standard argumentsand using the ensuing Fermi Golden rule, one can take advantage of the simplicity of theH0 spectrum and evaluate the probability given by this formula exactly. Introducing theparameter T ≡ t(∆/2~) and writing xl = (l − α)T with4 α = Eb/∆ the probability canbe written in the form

P =4g2

∆2Wα(T ) ,

with

Wα(T ) = T 2

∞∑

l=−∞

(

sin xlxl

)2

.

To perform the sum one can now use the Poisson summation formula wchich states that

l∑

l=−∞

f(a+ lT ) =1

T

∞∑

l=−∞

f

(

2π

Tl

)

exp

(

2πi

Tla

)

,

where f(q) is the Fourier transform of the function f(x):

f(q) =

∫ ∞

−∞

dx e−iqx f(x) .

In the case at hands, one has to find the Fourier transform of the function (x−1 sin x)2. Thiscan be done by noticing that the Fourier transform g(q) of a product g(x) = f1(x)f2(x)of two functions is the convolution of the fourier transforms f1(q) and f2(q) of thesefunctions:

g(q) =

∫ ∞

−∞

dx e−iqx f1(x)f2(x) =

∫ ∞

−∞

dx e−iqx

∫

dq12π

eiq1x f1(q1)

∫

dq22π

eiq2x f2(q2)

=

∫

dq12π

∫

dq22π

f1(q1)f2(q2)

∫ ∞

−∞

dx e−ix(q−q1−q2) .

Since the innermost integral equals 2πδ(q − q1 − q2),

g(q) =

∫

dp

2πf1(p)f2(q − p) .

4Because the sum over l extends indefinitely in both directions, one can always restrict Eb/∆ in thedefinition of α to the range [0, 1).

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It is clear that the Fourier transform of f1(x) = x−1 sin x is5

f1(q) =

π if |q| ≤ 10 if |q| > 1

,

and it requires little imagination6 to see that the convolution of the two transforms f(q)required in our problem is

f(q) =π

2·

2 − |q| if |q| ≤ 20 if |q| > 2

.

It follows that in the Poisson summation formula only a finite number of terms is nonzeroand the sought probability is given by

P =2π

~

g2

∆t

∞∑

l=−∞

θ(

1 − π

T|l|)(

1 − π

T|l|)

exp(−2πiαl).

This result is exact in the first order of the perturbative expansion. It is now clear thatif 0 < T ≤ π that is, if 0 < t < tH = 2π~/∆, only the l = 0 term contributes to thesum and the probability P grows linearly with time with the slope (2π/~)(g2/∆) which iscorrectly reproduced by the Fermi golden rule because in the considered case the densityof states is just constant: ρ(E) = 1/∆ and g2 = |〈l|Vint|b〉|2 (the matrix element does notdepend on l). If, however, t is larger, more and more terms in the Poisson summationformula contribute. The probability still grows then piecewise linearly with time but thecoefficient of this linear growth changes each time when T = t(∆/2~) passes a multipleof π. In fact it can be checked that when nπ ≤ T ≤ (n + 1)π,

Wα(T ) = πsin[(2n+ 1)πα]

sin(πα)(T − nπ) + π2 sin2(nπα)

sin2(πα),

that is the function Wα(T ) is continuous (the probability P does not “jump”) and piece-wise linear. It should be noted that only in the first time interval, 0 < t < tH is thebahaviour of the probability P (t) independent of α = Eb/∆ (modulo unity): already inthe next interval, tH < t < 2tH, the slope of the linear dependence of P (t) on time dependson this parameter and can even be negative (i.e. the probability P (t) may temporarilydecrease with time)! It is important to stress that if the coupling g is sufficiently small,the first order approximation to the time evolution of the state can be valid beyond theHeisenberg time tH, so for some systems the Fermi Golden Rule can fail while the firstorder approximation is still valid. Of course, if the spectrum of states |l〉 tends, to be

5It is easier to check that f1(q) yields f1(x).6f(q−p) equals π on the segment [−1+ q, 1+ q]. Therefore one computes the area under the constant

function equal π2/2π = π/2 on the segment which is the overlap of the two segments: this one and[−1, 1]; it is clear that that the overlap is zero if 1 + q < −1, starts to grow linearly up to q = 0 wherethe length of the overlap is 2 and then decreases linearly to zero up to q = 2; when q > 2 the overlap isagain zero.

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continuous, ∆ → 0, the Heisenberg time tH ∝ 1/∆ becomes infinite and the first orderapproximation to the evolution necessarily breaks down before the Fermi rule fails.

The time evolution of the probability P (t) in the considered example can be computedexactly as 1 − Pb→b(t), where Pb→b(t) = |〈b|UI(t, 0)|b〉|2 is the probability of finding thesystem in the state |b〉 at the instant t > 0. The interaction operator in the interactionpicture is given by (ωbl = (Eb −El)/~)

V Iint(t) = eiH0t/~ Vint e

−iH0t/~ = g

∞∑

l=−∞

(

|b〉〈l|eiωblt + |l〉〈b|e−iωblt)

,

and the matrix element between the states |b〉 of the interaction picture evolution operatorUI(t, 0) can be computed directly using the formula

UI(t, 0) = 1 +1

i~

∫ t

0

dt1 VIint(t1) +

(

1

i~

)2∫ t

0

dt2

∫ t2

0

dt1 VIint(t2)V

Iint(t1) + . . .

Since 〈b|Vint|b〉 = 〈l1|Vint|l2〉 = 0, only terms with even number of interactions contribute.Insterting the complete sets of intermediate states one gets

〈b|V Iint(t2s) . . . V

Iint(t1)|b〉 = g2s

∑

ls

. . .∑

l1

expiωbls(t2s − t2s−1) + . . .+ iωbl1(t2 − t1) .

Again the summations can be performed with the help of the Poisson summation formula.They are all of the form

∞∑

l=−∞

e−il(τ∆/~) =

∞∑

l=−∞

f(lT ) ,

with T = τ∆/~ and τ = t2s − t2s−1 or τ = t2s−2 − t2s−3, etc. The Fourier transform off(x) = exp(−ix) is just 2πδ(1 + q), so

∞∑

l=−∞

e−il(τ∆/~) =~

τ∆

∞∑

l=−∞

2πδ

(

1 +2π~

τ∆l

)

= tH

∞∑

l=−∞

δ(τ + ltH) .

The 2s-th term of the expansion of the matrix element of the operator UI(0, t) takestherefore the form

(

−g2tH~2

)s ∫ t

0

dt2s . . .

∫ t2

0

dt1 eiEb(t2s−t2s−1)/~ . . . eiEb(t2−t1)/~

×∑

ls

. . .∑

l1

δ(t2s − t2s−1 + lstH) . . . δ(t2 − t1 + l1tH).

If t < tH, then none of the differences t2s − t2s−1, . . . , t2 − t1 cannot exceed tH and to thesums only the terms with li’s= 0 contribute, all exponential factors are equal unity andthis reduces to

(

−2πg2

~∆

)s ∫ t

0

dt2s . . .

∫ t2

0

dt1 δ(t2s − t2s−1) . . . δ(t2 − t1).

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With the prescription that7

∫ t2

0

dt1δ(t2 − t1) =1

2,

it is straightforward to convince oneself (work out the case s = 3 for instance) that themultiple integral yields the factor (1/s!)(t/2)s. In this case, therefore,

〈b|UI(t, 0)|b〉 = exp

(

−πg2

~∆t

)

.

The probability Pb→b(t) = exp(−(2π/~)(g2/∆)t) decreases exponentially with the decre-ment coefficient given by the Fermi Golden Rule, in agreement with the statistical rea-soning applied to the ensemble of such systems all prepared at t = 0 in the state |b〉. Thisof course gives

P (t) = 1 − Pb→b(t) =2π

~

g2

∆t + . . .

in agreement with the first order result. If, however, t > tH, for instance tH < t < 2tH,the dependence of Pb→b(t) on time is no longer exponential. Indeed, in this case onecontribution comes from all l′is = 0 and is the same, as found above, but in additionone li can be equal −1. Working out two or more terms one notices that the additionalcontribution to the term of order g2s takes the form

1

2s−1(s− 1)!(t− tH)s e2πiα ,

and, therefore, for tH < t < 2tH the computed element of the (interaction picture) evolu-tion operator is

〈b|UI(t, 0)|b〉 = e−γt/2 − γ (t− tH) e−γ(t−tH)/2 e2πiα .

This can be continued but it is already clear that if t > tH the probability of findng thesystem in the state |b〉 (i.e. in the statistical approach, the change of the population ofthis state) is no longer purely exponential and can even temporarily increase, dependingon the value of α = Eb/∆ (modulo 1).

7The possibility of finding the matrix element of the evolution operator relies here on doing the sum-mations before taking the integrals. Taking the integrals first would lead to combinations of individuallydivergent sums, the same which would be encountered in trying to extend the metod of finding |ψ(t)〉(s)used initially beyond the first nontrivial order. It would be an interesting problem to regularize thesedivergent summations (e.g. by restricting them to |l| < L) and to see whether it is possible removing thecutoff to recover the same results as the ones obtained by performing the summations first and relyingon adopting the somewhat controversial prescription given here for integrals with the delta function.

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Problem I.1A vector V rotated by an angle φ around the axis n (where |n| = 1) can be written as

V′ = V cosφ+ n (n·V)(1 − cosφ) + n×V sinφ

≈ V + φ×V − 1

2φ2V +

1

2φ (φ·V) + . . .

where φ ≡ φn. Justify this formula. Find the vector φ corresponding to the compositionof two successive infinitesimal rotations characterized by φ1 and φ2 of a vector V. Usingthe result find the structure constants of the rotation group. Show also that the matrix

Rij(φ) = δij cos φ+ (1 − cosφ)ninj + ǫikjnk sinφ ,

such that V ′i = Rij(φ)V j , is just the matrix exp(

−iφkJ kvec

)

, where (J kvec)

ij = iǫikj arethe rotation group generators in the defining (vector) representation.

Solution to Problem I.1Rotating V around n by an angle φ means decomposing V into V|| = n (V · n) andV⊥ = V −V|| (the components parallel and perpendicular to n, respectively) and thenadding them back, after having rotated V⊥ around n by φ ≡ |φ|:

V′ = n (V·n) + Vrotated⊥ .

In turn, Vrotated⊥ ≡ [V − n (V·n)]rotated is given by

Vrotated⊥ = V⊥ cosφ+ (n×V⊥) sinφ = [V − n (V·n)] cosφ+ (n×V) sinφ ,

(upon taking into account the equality |n×V⊥| = |V⊥|). Thus,

V′ = V cosφ+ n (V·n)(1 − cosφ) + (n×V) sinφ ,

or, in the infinitesimal form,

V′ ≈ V + φ×V − 1

2φ2V +

1

2φ (V·φ) + . . .

Two successive rotations: around n1 by φ1 and around n2 by φ2 give

V′′ ≈ V′ + φ2 ×V′ − 1

2φ2

2V′ +

1

2φ2(V

′ ·φ2) + . . .

≈ V + (φ2 + φ1) ×V + φ2 × (φ1 ×V)

−1

2φ2

2V +1

2φ2(V·φ2) −

1

2φ2

1V +1

2φ1(V·φ1) + . . .

This should be compared with

V′′ ≈ V + φ(φ2,φ1) ×V − 1

2φ2(φ2,φ1)V +

1

2φ(φ2,φ1)[V·φ(φ2,φ1)] + . . .

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It is obvious, that to the first order φ(φ2,φ1) = φ2 + φ1, so equating the two forms ofV′′ with

φ(φ2,φ1) = φ2 + φ1 + δφ ,

(where δφ is at least of the second order in φi) we obtain the relation

φ2 × (φ1 ×V) + . . . = δφ×V − (φ2 ·φ1)V +1

2φ2(V·φ1) +

1

2φ1(V·φ2) + . . .

(the ellipses stand for terms of higher order in φi). It is then a simple exercise in thevector algebra to check that

δφ =1

2(φ2 × φ1) .

The symmetry group structure constants f ijk are given by f ijk = cijk − cikj, where thecoefficients cijk appearing in the general formula

φi(φ2,φ1) ≈ φi2 + φi

1 + cijkφj2φ

k1 .

can be read off from the formula for φ(φ2,φ1) obtained above. For the rotation groupconsidered here one finds cijk = 1

2ǫijk and, therefore, f ijk = ǫijk as expected.

To find the matrix exp(

−iφkJ kvec

)

we write explicitly the exponent:

A ≡ −i3∑

k=1

φkJ kvec =

0 −φz φy

φz 0 −φx

−φy φx 0

,

and its square

A2 =

−(φy2 + φz2) φxφy φxφz

φxφy −(φx2 + φz2) φyφz

φxφz φyφz −(φx2 + φy2)

.

It is then easy to check that A3 = −φ2A (where φ2 = φ2). This implies that A2n =(−1)n+1φ2n−2A2 and A2n+1 = (−1)nφ2nA and, therefore,

eA = I − A2

φ2

∞∑

n=1

(−1)n

(2n)!φ2n +

A

φ

∞∑

n=0

(−1)n

(2n+ 1)!φ2n+1 = I +

A2

φ2(1 − cos φ) +

A

φsin φ .

Since Aij = φkǫikj, A2 = −φ2 I+φ⊗φ, and φi/φ = ni one indeed gets the matrix Rij(φ)which can be written as

Rij(φ) = δij + (ninj − δij)(1 − cosφ) + nkǫikj sinφ .

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Problem I.2A left-invariant measure dµ(g) on a group G has the property

∫

dµ(g) f(g) =

∫

dµ(g) f(g′g)

(g denotes an element of G and f(g) is a function defined on the group G). In a concreteparametrization g = g(θ) of the group elements by some parameters θa, a = 1, . . . , n(where n is the dimension of the Lie algebra ofG) the measure is given by dµ(g) = dnθρ(θ).Find the left-invariant measure (i.e. the density ρ) on the rotation group SO(3) in theparametrization given by the components of the vector φ = (φx, φy, φz) defined in ProblemI.1. To this end justify and exploit the formula

ρ(θ) = ρ(0) det−1

(

∂h(θ, θ)

∂θ

)

θ=0

,

in which h(θ, θ) is the group composition function appropriate for the chosen para-metrization g = g(θ) of the group elements. Compute the rotation group volume assumingthat the measure is normalized so that ρ(0) = 1.

Solution to Problem I.2In a concrete parametrization by the parameters θa the condition of left-invariance

∫

dµ(g)f(g′g) =∫

dµ(g)f(g) reads

∫

dnθ ρ(θ)f(θ(θ′, θ)) =

∫

dnθ ρ(θ)f(θ) ≡∫

dnθ ρ(θ)f(θ) .

The second form on the right hand side is obtained just by renaming the integrationvariables θ → θ. After changing in the rightmost integral the integration variables to θby substituting θa = ha(θ′, θ), where ha is the group composition function (in the chosenparametrization), we get the condition

ρ(θ) = ρ(h(θ′, θ)) det

(

∂ha(θ′, θ)

∂θb

)

.

For the left-invariant measure the right-hand side must be of course independent of θ′.Assuming that a density ρ fulfilling this condition exists, one can determine it just byconsidering infinitesimal values of the parameters θa → 0. Evaluating the above formulafor θa = 0 and using the equality θa = ha(θ′, 0) = θa′ we get

ρ(θ′) = ρ(0) det−1

(

∂ha(θ′, θ)

∂θb

)

θ=0

.

This, together with the condition ρ(0) = 1, fixes ρ uniquely.

We now apply this general result to the rotation group SO(3). Since we have toevaluate the Jacobian (i.e. the derivatives of the composition function θa = ha(θ′, θ))

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at θ = 0, it is sufficient to find the components of the vector φ (playing the role of θ)resulting from the composition of a rotation by of φ′ (playing the role of θ′) with aninfinitesimal rotation by φ. To find the vector φ we can multiply the matrix8 Rij(φ) fromProblem I.1 from the right by a matrix of an infinitesimal rotation

R(φ) =

c+ aφx′φx′ aφx′φy′ − b φz′ aφx′φz′ + b φy′

aφy′φx′ + b φz′ c+ aφy′φy′ aφy′φz′ − b φx′

aφz′φx′ − b φy′ aφz′φy′ + b φx′ c+ aφz′φz′

1 −φz φy

φz 1 −φx

−φy φx 1

.

We have introduced here the abbreviations

c = cos |φ′| , b =sin |φ′||φ′| , a =

1 − cos |φ′||φ′|2 .

Comparing the traces of both sides one gets that

1 + 2 cos |φ| = 1 + 2 cos |φ′| − 2sin |φ′||φ′| φ′ ·φ

≈ 1 + 2 cos |φ′ + φ| .

Hence |φ| = |φ′ + φ|, to the first order in φ. Furthermore, to establish the direction ofthe vector (φx, φy, φz) we notice that a rotation does not change its own axis, i.e.

R(φ) · φ = φ ,

To simplify the determination of the direction of φ we will use the condition

[

R(φ) −RT (φ)]

· φ = 0 .

While it is possible to extract from the above formulae also the individual componentsφx, φy and φz of the vector φ, the actual computation would be rather cumbersome. Toproceed we can take advantage of the fact that the measure density ρ(φ′) should dependonly on |φ′| and take the vector φ′ in the direction of the z-axis, so that φz′ = θ′ andφx′ = φy′ = 0. The matrix R(φ) simplifies then to

R(φ) =

cos θ′ − φz sin θ′ −φz cos θ′ − sin θ′ φy cos θ′ + φx sin θ′

sin θ′ + φz cos θ′ −φz sin θ′ + cos θ′ φy sin θ′ − φx cos θ′

−φy φx 1

.

Acting with R(φ) − RT (φ) on (φx, φy, φz) we get three equations:

−2φy (φz cos θ′ + sin θ′) + φz [φy(1 + cos θ′) + φx sin θ′] = 0 ,

2φx (sin θ′ + φz cos θ′) + φz [φy sin θ′ − φx(1 + cos θ′)] = 0 ,

8We take here the active view on rotations.

11

−φx [φy(1 + cos θ′) + φx sin θ′] + φy [−φy sin θ′ + φx(1 + cos θ′)] = 0 .

In other words, the ratios of the components of the vector φ are given by

φx

φy=

−φy sin θ′ + φx(1 + cos θ′)

φy(1 + cos θ′) + φx sin θ′,

φx

φz=

−φy sin θ′ + φx(1 + cos θ′)

2(sin θ′ + φz cos θ′),

φy

φz=φy(1 + cos θ′) + φx sin θ′

2(φz cos θ′ + sin θ′).

(Obviously, these three equations are mutually consistent). Using the fact that φi areinfinitesimal and expanding the last two equalities to first order in φi we get

φx ≈ 1

2

(

−φy + φx 1 + cos θ′

sin θ′

)

φz ,

φy ≈ 1

2

(

φx + φy 1 + cos θ′

sin θ′

)

φz .

With our choice of φ′, the length |φ| is to the first order in φi equal to |φz| = |θ′ + φz|and, therefore,

φx ≈ 1

2

(

−φy + φx 1 + cos θ′

sin θ′

)

θ′ ,

φy ≈ 1

2

(

φx + φy 1 + cos θ′

sin θ′

)

θ′ ,

φz ≈ θ′ + φz ,

(It is easy to see that the length of φ to this order is still |θ′ + φz|, as it should be).The rest is straightforward: the Jacobian J = det(∂φi/∂φj) reads

J =

∣

∣

∣

∣

∣

∣

det

12θ′ 1+cos θ′

sin θ′−1

2θ′ 0

12θ′ 1

2θ′ 1+cos θ′

sin θ′0

0 0 1

∣

∣

∣

∣

∣

∣

=1

2(θ′)2

1 + cos θ′

sin2 θ′=

(θ′)2

2(1 − cos θ′).

Since we have worked to first order in φ, this is just the result for φi = 0 which is needed.Hence, as in the general formula, we identify θ′ with |φ| and drop the tildas. In this way,in the parametrization of the rotations by the vector φ = (φx, φy, φz) we find

dφxdφydφz ρ(φ) = dφxdφydφz 2(1 − cos |φ|)|φ|2 ≡ d|φ|dΩφ 2(1 − cos |φ|) .

The volume of the rotation group SO(3) then is (the length |φ| varies from 0 to π):

Vol(SO(3)) =

∫ 2π

0

dϕ

∫ π

0

dϑ sinϑ

∫ π

0

d|φ| 2(1 − cos |φ|) = 8π2 .

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For functions defined genuinely on SU(2) instead of SO(3), i.e. functions taking dif-ferent values on those SU(2) group elements which get identified in SO(3), the integrationover |φ| ranges from 0 to 2π and the SU(2) group volume is 16π2.

13

Problems to the lecture Advanced Quantum Mechanics

Problem 0Prove the following expansion

e−AB eA = B − [A, B] +1

2![A, [A, B]] − 1

3![A, [A, [A, B]]] + . . .

Prove also the Baker-Hausdorff operator identity

eA+B = eA eBe−1

2[A, B] ,

holding for operators A and B commuting with [A, B]. Finally, prove the general for-mula,9

et(A+B) = etA T exp

(∫ t

0

dτ e−τAB eτA)

,

valid for any two operators A and B, in which T denotes the “time” ordered product.Hints: To prove the expansion solve iteratively the differential equation satisfied by theoperator function C(λ) = e−λAB eλA. Similarly, to prove the Baker-Hausdorff formulaconsider the function F (λ) = e−λBe−λAeλ(A+B) and simplify the differential equation sat-isfied by it using the fact that owing to the assumption, in the expansion of eBA e−B inpowers of the operator B only two first terms are nonvanishing.

Problem 1Let |Ψ(t)〉S be an eigenvector with the eigenvalue a(t) of the Schrodinger picture operatorAS. Show that |Ψ〉H is the eigenvector of AH(t) with the same eigenvalue a(t). Prove alsothat if [AH(t0), B

H(t0)] = CH(t0), then the same holds for any t.

Problem 2Find the Heisenberg picture operators xH(t) and pH(t) of a particle of mass M movingin one dimension ifa) it is a free particle (H = p2/2M),b) H = p2/2M − xF (t), where F (t) is an external, time dependent force,c) H = p2/2M +Mω2x2/2.In all these cases compute the commutators

[xH(t), xH(t′)] , [pH(t), pH(t′)] , [xH(t), pH(t′)] .

Using the Heisenberg picture operators compute in cases a) and c) the dispersion ofthe particle’s position at the instant t expressing it through matrix elements of somecombinations of the position and momentum operators at t = 0.

9The Baker-Hausdorff formula is its special case with t = 1 and [A, [A, B]] = [B, [A, B]] = 0.

14

Problem 3Find the Heisenberg picture operators xH(t) and pH(t) of the one-dimensional harmonicoscillator whose dynamics is set by the time dependent Hamiltonian

H(t) =p2

2M+

1

2Mω2x2 − xF (t) ,

using the solution10 of the corresponding classical motion with the initial conditions x(0) =x0 and p(0) = p0. To this end, recalling that x0 and p0 are also canonical variables relatedto the standard ones, x(t) and p(t), by the canonical transformation (whose generatingfunction is just the properly understood action), promote them to operators x0 and p0on which the standard commutation rules [x0, p0] = i~, etc. are imposed and representthem in terms of the creation and annihilation operators. Since the classical Hamiltonianwritten in terms of the canonical variables x0 and p0 vanishes (this is precisely what isensured by solving the Hamilton-Jacobi equation, but one does not need to do it explicitlyhere), the operators x(t) and p(t) obtained from the classical solution in which x0 and p0are substituted by the operators x0 and p0 (expressed, in turn, through the creation andannihilation operators) are just the Heisenberg picture operators. The Heisenberg pictureoperators aH(t) and a†H(t) can be then read off from the form of xH(t) and pH(t).A reassuring remark: the description of the problem is long but the steps to do areentirely trivial. After you do it, you will have, perhaps, a better understanding of what“quantization” means.

Problem 4A particle of mass m and electric charge q (in units of e > 0) moves in the con-stant magnetic field B = ezB. Find the Heisenberg picture operators xH(t), yH(t) andzH(t) and compute the commutators [xH(t), xH(t′)], [yH(t), yH(t′)], [xH(t), yH(t′)] and[xH(t), zH(t′)]. Do these commutators depend on the choice of the potential A (the choiceof the gauge)? Consider also the operators pxH(t), pyH(t), pzH(t) and their commutators.Do they depend on the gauge?Hint: If it is too difficult to work without specifying a gauge, set e.g. A = eyξBx −ex(1 − ξ)By with an arbitrary parameter ξ in order to follow the gauge (in)dependenceat least within a restricted class of gauges. To construct the Heisenberg picture operatorsxH(t), yH(t), pxH(t), pyH(t), take the inspiration from Problem 3.

Problem 5A particle of mass M and electric charge q (in units of e > 0) moves in the electric andmagnetic fields represented by the potentials ϕ(t, r) and A(t, r). Find the equation ofmotion satisfied by the Heisenberg picture operator rH(t), that is compute d2rH(t)/dt2.Establish how this derivative differs from the classical formula (written here in the Gausssystem of units)

Md2r(t)

dt2= qe

[

E(t, r) +v

c×B(t, r)

]

.

10The solution can be found e.g. in my notes to Classical Mechanics, somewhere on this web page.

15

Problem 6Express the difference EΩ − EΩ0

of ground state energies of H = H0 + λVint and of H0

through the derivative with respect to λ of the operator11

Sε0 ≡ U−ε

I (+∞, 0)UεI (0, −∞) = [U−ε

I (0, + ∞, )]†UεI (0, −∞) ,

that is, prove the so-called Sucher formula

EΩ − EΩ0=

1

2i~ ε λ

∂

∂λln〈Ω0|Sε

0|Ω0〉 ,

Problem 6′

By considering the differential equation satisfied by it, find the complete evolution op-erator Uε(t, 0), including its phase, corresponding to the Gell-Mann - Low modificationVint −→ eεtVint of the Hamiltonian

H = H0 + Vint = ~ω a†a+ ∆ω + λa† + λ∗ a ,

of the linearly perturbed harmonic oscilator.Hint: In order to ensure the proper transformation to the Heisenberg picture of the basicoperators a and a†, the sought operator must have the form

Uε(t, 0) = eiϕ(t) e−iH0t/~ eh(t) a†−h∗(t) a , h(t) = − i

~

∫ t

0

dτ λ e(ε+iω)τ ,

so only the phase ϕ(t) has to be determined.

Problem 7 (TRK)The Hamiltonian of a set of N identical nonrelativistic spinless particles of mass M hasthe general form

H =N∑

a=1

p2a

2M+ V (r1, . . . , rN) .

Defining the operators

Or(a) = O†r(a) =

N∑

a=1

a·ra , Op(a) = O†p(a) =

N∑

a=1

a·pa ,

in which a can be any (real for Hermiticity) vector, prove that if |s〉 is a normalizable(and normalized to unity) eigenvector of H , the following sum rules

∑

l

|〈l |Op(a)|s〉|2 =M2

~2

∑

l

(El − Es)2 |〈l |Or(a)|s〉|2 ,

11In U−εI (+∞, 0) the original time independent interaction is replaced by λVint e

−εt and the limitε→ 0+ is taken.

16

∑

l

(El −Es) |〈l |Or(n)|s〉|2 =N~

2

2M,

∑

l

(El −Es)∣

∣〈l |eiOr(q)|s〉∣

∣

2= N

~2q2

2M,

hold. The second one, in which it is assumed that n2 = 1, is called the Thomas-Reiche-Kuhn sum rule. The summations over l, where |l〉 are eigenvectors of H , mean alsointegrations over the continuous part of the Hamiltonian spectrum.Hint: Prove first the identity Op(a) = i(M/~)[H, Or(a)]. To prove the TRK rule computein two ways the |s〉 state expectation value of the double commutator [[H, Or(n)], Or(n)]and to prove the last one work out the operator e−iOr(q)HeiOr(q) − H using the expan-sion proved in Problem 0 and take the expectation value of both sides in a normalizedeigenvectors |s〉 od H .

Problem 8Justify the identity12

a†a =∞∑

n=0

|n〉n〈n| ,

in which |n〉 are the normalized eigenvectors of the operator a†a, where a and a† are thestandard annihilation and creation operators.

Problem 9The harmonic oscillator of mass M and frequency ω is acted upon by an external forceF (t) (vanishing as t → ∞). The oscillator state was at t = 0 prepared in the coherentstate |z〉 (the eigenvector of the annihilation operator with the eigenvalue z). Show thatthe oscillator remains in a coherent state at any instant t and find z(t).

Problem 10The harmonic oscillator of mass M and frequency ω on which acts an external force F (t)vanishing as t → ∓∞ (i.e. the Hamiltonian as in Problem 3; one can consider concreteforces like F (t) = F0/(1 + t2/τ 2) or F (t) = F0 exp(−t2/τ 2)) was at t = −∞ in the groundstate of H0. Compute the mean oscillator energy E at t = ∞ and the energy dispersion

squared E2−E2

using the formalism of the in and out operators. Obtain the same resultusing the S-matrix.

Problem 11The work W done on the harmonic oscillator of frequency ω and mass M , whose classicalmotion as t→ −∞ is given by x(t) = A cos(ωt+ δ), by a force F (t) vanishing as t→ ∓∞can be computed as the difference of the oscillator energies at t = ∞ and and t = −∞. IfF (t) = F0 exp(−t2/τ 2) this work is equal (see Kotkin & Serbo Problem 5.12 or my notes

12This is taken from the BMW, but there the problem is formulated with a misprint...

17

to classical mechanics, Problem 2.8)

W =πF 2

0

2Mω2ω2τ 2 e−

1

2ω2τ2 −

√π F0Aωτ e

− 1

4ω2τ2 sin δ .

Find the analogous result in quantum mechanics of the harmonic oscillator, that is com-pute the mean value W of the work done by the force F (t) on the oscillator, whose(Schrodinger picture) state-vector |ψ(t)〉 was at t→ −∞ such that

〈ψ(t)| x |ψ(t)〉 = A cos(ωt+ δ) .

Hint: Express the matrix elemet 〈ψ(t)| x |ψ(t)〉 in the Heisenberg picture and use theformalism of the in and out operators.

Problem 12The center of the one-dimensional harmonic oscillator force gets suddenly displaced bythe distance d (the time τ in which the displacement takes place is much smaller than1/ω), that is the Hamiltonian undergoes the change

H0 ≡p2

2M+

1

2Mω2x2 −→ Hd ≡

p2

2M+

1

2Mω2(x− d)2.

Compute the probability that after the displacement the oscillator which initially was inthe n-th state of H0 will after the change be found in the m-th state of Hd. Do this usingthe relation between the corresponding creation and annihilation operators a, a† ad, a

†d

(in my notes this is done in a more complicated way, using the wave functions etc.)

Problem 13The frequency ω(t) of a one-dimensional harmonic oscillator of mass M varies with timeaccording to the formula

ω2(t) = ω20 + ∆ω2

0 arctg(t/τ) , ω20 >

π

2∆ω2

0 ,

so that H(−∞) = H(−) and H(∞) = H(+). Compute in the lowest order of the pertur-bative expansion the probability of finding the oscillator in the k-th excited state of H(+)

if at t = −∞ it was in the n-th eigenstate of H(−).

Problem 14An atom, initially, i.e. in the far past, in the state |i〉, is placed beetween the plates of acapacitor. The electric field E(t) is switched on and off according to the formula:

E(t) = E0 exp(−t2/τ 2) .

Using the first order of the time-dependent perturbative expansion express the probabilityof finding the atom in the far future in the state |f〉 through the matrix element of theelectric dipole operator. Using the obtained formula, compute explicitly the probabilities

18

of finding the Hydrogen atom in the far future in the |2P 〉 states if it was initially in the|2S〉 state.

Problem 15Compute approximately the probability that an atom, initially in an |i〉 state, will getexcited to a state |f〉 (belonging to the discrete or continuous part of the spectrum ofthe free atom Hamiltonian) by the variable electric field produced by a heavy chargedparticle (treated classically and without taking into account its small deflection due tothe interaction - this is possible in view of the neutrality of the atom as a whole) passingnear the atom with the impact parameter b≫ aB (b is counted with respect to the atom’snucleus) with a constant velocity v. The approximation should consist of truncating theformula suggested below to the lowest l.Hint: Use the formula

1

|r−R| =

∞∑

l=0

|r|l|R|l+1

Pl(cosϑ) ,

where ϑ is the angle between the vectors r and R and Pl(z)’s are the Legendre polynomials.The formula is written assuming that |r| < |R|.

Problem 16Combine the result of the preceding Problem with the Thomas-Reiche-Kuhn sum rule(Problem TRK) to estimate the mean energy loss per unit length of the trajectory of aheavy charged (classical) particle which passes through a medium in which there are Nidentical atoms (each having Z electrons) per unit volume.

Problem 17The Hilbert space of a system is two-dimensional. The Hamiltonian is H = H0 + Vint.The two normalized eigenstates |1〉 and |2〉 of H0 (corresponding to its eigenvalues E1 andE2) can be taken for the basis of the Hilbert space. At t = 0 the system was prepared inthe state |1〉. Compute the probability of finding the system in the states |1〉 and |2〉 atany instant t. Compare the exact result with the one obtained in the lowest order of thetime-dependent perturbative expansion.

Problem 18As in the preceding Problem the Hilbert space of a system is two-dimensional. Theunperturbed Hamiltonian H0 has two degenerate (normalized to unity) eigenvectors |1〉and |2〉, both corresponding to the energy E0. The perturbation has the form Vint =f(t)O, where O is some hermitian operator and f(t) is a c-number function. What isthe probability of finding the system at the instant t > 0 in the the state |2〉, if it wasprepared at t = 0 in the state |1〉? Establish the conditions in which the transitionprobability obtained using the first order of the time-dependent perturbative expansionis a good approximation to the one computed exactly.

Problem A.

19

Consider the evolution of the magnetic moment represented by the operator

µ = µ1

2σ ,

of a spin 12

particle13 in a variable magnetic field

B(t) = B0 ez +B1 (ex cos Ωt + ey sin Ωt) .

Find the exact evolution of the state which at t = 0 is the lower energy eigenstate of thesystem’s Hamiltonian at that instant. Find the state-vector ψ(t) of the spin (magneticmoment) after the complete rotation of the direction of the magnetic field (Ωt = 2π) andtaking the adiabatic limit Ω → 0 t→ ∞ with Ωt = 2π, identify the Berry’s phase.Hint: Find first the evolution of the state-vector ψ′(t) related to ψ(t) by the time-dependent unitary transformation S(t) = exp( i

2σzΩt): ψ′(t) = S(t)ψ(t).

Problem BApplying the Fermi’s Golden Rule calculate the probability of the Hydrogen atom ioniza-tion by the spatially constant and uniform electric field E(t) = 2E0 sinωt (produced e.g.in a capacitor). Use the plane waves as the final states of the electron (to avoid technicalcomplications).

Problem 19Consider the electron bound in the Coulomb potential −Ze2/r (a Hydrogen-like atom)interacting with the electromagnetic plane wave represented by the vector potential

A(t, r) = A0 ǫ(k, λ) e−i(ωt−k·r) + c.c. ,

of frequency ω such that ~ω is higher than the atom’s ionization energy (ǫ(k, λ) is theunit polarization vector). Approximating the electron asymptotic states by plane waves,compute the differential cross section of the atom ionization process in which the electronis found with the momentum in the element dΩk of the solid angle.Hint: The cross section is given by the transition probability per unit time divided bythe flux of photons. If the calculation is done within the semiclassical radiation theory,the “flux of photons” should be identified with the energy flux (averaged over the period)carried by the electromagnetic wave divided by the energy ~ω od a single quantum.

Problem 20Solve the preceding problem using the quantum theory of radiation, i.e. compute the crosssection of the process in which the atom gets ionized as a result of absorbing one photon(the initial state consists of the atom in the ground state and one photon). Average thecross section over polarizations of the initial photon.

13If the particle has a nonzero electric charge Qe (e > 0 is the elementary charge) one usually writesµ = Q(e~/2mc)g = QgµB, where µB is the Bohr magneton and g some numerical factor (g = 2 pluspennies, if the particle is the electron); if the particle is neutral (e.g. neutron), one can always writeµ = gµB admitting either sign of g.

20

Problem 21Consider the quantum three dimensional isotropic harmonic oscillator of mass M andfrequency ω carrying the electric charge q. Compute the width (or the lifetime) of theoscillator |nx, ny, nz〉 states using the electric dipole approximation. When is the dipoleapproximation reliable?

Problem 22Estimate the probability per unit time of the spontaneous electric dipole transition (withthe emission of one photon) between the Hydrogen atom 2S1/2 and 2P1/2 states. Thedifference of energies of these two energy levels is ∆E = 4.4 × 10−6 eV. This spliting,called the Lamb shift, is due to higher order corrections, calculable only in full QuantumElectrodynamics (the details of this calculation are, however, entirely irrelevant for thepresent Problem).

Problem 23In the dipole approximation derive the general formula giving the probability per unit timethat the Hydrogen atom excited to the atomic level characterized by the quantum numbers(n, l) makes the spontaneous transition (with the emission of one photon) to another levelcharacterized by the numbers (n′, l′). Average over the transition probability over the ml

and sum over the m′l quantum numbers.

Problem 24A neutron at rest is placed in the constant uniform magnetic field B0. This results inspliting its spin up and spin down states. The energy difference of these split states is∆E = 2|µ||B0|, where |µ| is the neutron magnetic moment equal14 (e~/2Mnc)gn withthe appropriate dimensionless factor gn (find its value somewhere!). Assuming that theneutron is initially in the higher energy state and using

Vint = −µ·B(0) ,

as the interaction, compute the probability per unit time wfi of the spontaneous transi-tion to the lower energy level. Give the answer in the form wfi = (. . .)× [|B0|/Gauss]a orwfi = (. . .) × [|B0|/Tesla]a (if you are a legalist) with the appropriate power a.

Remark: Setting to zero the space argument of the B field operator results from approx-imating the wave function of the neutron at rest by a Gaussian packet strongly peaked atr = 0.

Problem 25Compute the probability per unit time of the spontaneous transition with the emission ofone photon between the triplet and singlet Hydrogen atom 1S states. The triplet state hasenergy higher than the singlet one by ∆E = 1420 MHz (that is - notice the old-fashioned

14The quantity e~/2Mpc is called the nuclear magneton (in analogy to the Bohr magneton e~/2Mecof the electron); notice that “magnetons” defined in this way already include the factors of 2, that is, ifproton were a pointlike particle the factor gp would be (close to) unity.

21

conversion factor! - ∆E = 1420 MHz·h ≡ 1420 · 2π~). This so-called hyperfine splittingis due to the interaction between the electron and nucleus (proton) spins.Remark: Write the wave functions of the singlet atomic state in the form

Ψsinglet =1√2

[

(

10

)

e

⊗(

01

)

p

−(

01

)

e

⊗(

10

)

p

]

2

a3/2B

e−r/aBY00 ,

and analogously the functions of the triplet state and take for the interaction

Vint =e

2Mecge

~

2(σe ⊗ Ip)·B(r) ,

(Ip is a unit 2 × 2 matrix acting on the proton spinors; the proton magnetic moment

coupling to the magnetic field operator B can be neglected because the proton magneticmoment is much smaller than that of the electron.

Problem 26Taking the electron spin into account (i.e. using the Pauli equation instead of theSchrodinger one) compute the probability per unit time of the (magnetic dipole) transi-tion between the 2S1/2 and 1S1/2 states in the Hydrogen atom with the emission of onephoton.Remarks: Approximate the operator e−ik·r by −1

2(k·r)2 - in the relevant matrix element

the first two terms of the expansion give zeros. Sum the transition matrix element squaredover the final state photon polarizations, over the final state electron spin directions andaverage it over the initial state electron spin projections.

Problem 27Find the contribution of the interaction term15

V(2)int =

e2

2Mc2A2(r) ,

to the probability per unit time of the spontaneous transition between the 2S and 1Slevels of the Hydrogen atom with the emission of two photons.Remark: Approximate the matrix element of the operator e−i(k1+k1)·r between the 2Sand 1S atomic states by expanding the exponent up to the second order (the first twoterms of this expansion give vanishing contributions). If the complete calculation is toocomplicated, try at least to make the estimate by finding the power of αEM = 1/137 andof other dimensional factors (like ~, c, Me) to which this rate is proportional.

Problem 28Compute the ratio of the intensities of the two first lines of the Balmer series of spectrallines emitted by the Hydrogen atom. The first, called Hα, line of this series is due to the

15The complete calculation of the rate of the 2S → 1S transition should take into account the contri-

bution (which is also of the order e4) of the term V(1)int = (e/Mc)A(r) · p).

22

transitions |3S〉 → |2P 〉, |3P 〉 → |2S〉, and |3D〉 → |2P 〉, while the second, Hβ, line isdue to the transitions |4S〉 → |2P 〉, |4P 〉 → |2S〉, and |4D〉 → |2P 〉.

Problem 29Taking the electron spin into account (i.e. using the Pauli equation instead of theSchrodinger one) compute (in order e4) the differential cross section of the low frequency(~ωk ≪ Mec

2) photon Compton scattering on free electron at rest. Average the crosssection over the two possible spin projections of the initial electron and sum over the spinprojections of the final electron. Compare the result with the Klein-Nishijina formula.

Problem 30Compute (in order e4) the differential and total cross sections (averaged over the initialphoton polarization and summed over the polarizations of the final photon) of the elasticlow frequency (~ωk ≪ Mc2) photon scattering on a spinless charged particle of mass Mbound in the spherical harmonic oscillator potential V (r) = 1

2Mω2

0r2.

Problem 31Compute the differential cross sections of the low frequency (~ωk ≪ Mnc

2) photon scat-tering on a free neutron in a definite spin state and no neutron spin flip (the final neutronhas the same spin direction as the initial one) and with the spin flip. Average these crosssections over the polarizations of the incoming photon and sum over the polarizations ofthe final one. Compute also the total cross sections (i.e. integrate over the directions ofthe final photon). The recoil of the neutron can be neglected - use only the spin part ofthe neutron wave function.

Problem 33Check by direct calculation that if ψ(t, r) satisfies the Schrodinger equation with thepotential V (r), then

ψ′(t, r) = exp

(

i

~V i(−Mxi + tP i)

)

ψ(t, r)

= exp

(

−iMV2

2~t

)

exp

(

− i

~MV ixi

)

ψ(t, r + Vt) ,

satisfies the Schrodinger equation with V ′(r+Vt). How are related the probability densityand the probability currents constructed out of ψ′ and ψ? Repeat the check in the abstractlanguage of state-vectors, that is show that if i~d|ψ(t)〉/dt = [p2/2M + V (r)]|ψ(t)〉, then

|ψ′(t)〉 = e−iMV2

2~t e−

i~MV·r e

i~tV·p|ψ(t)〉 ,

satisfies i~d|ψ′(t)〉/dt = H ′|ψ′(t)〉 with H ′ = p2/2M + V (r + tV).

Problem 34The nucleus of the Hydrogen atom (in the 1S state) gets a sudden kick and starts movingwith the velocity v. Assuming that the time τ of the action of the kicking force is veryshort compared to all relevant characteristic times (including aB/|v|) derive a general

23

formula valid for an N -electron atom of finding the atom in a concrete stationary state.Applying it to the Hydrogen-like one-electron atom compute explictly the probability thatthe it will not remain in the initial 1S state.Hint: Transform the initial wave function of the electron to the frame in which thenucleus is at rest after having received the kick.

Problem 35Show that if the rotation represented by the 3 × 3 matrix R is generated by the 2 × 2SU(2) matrix M through the formula

M · (σ ·r) ·M † = σ ·(R·r) ,

and the rotation matrix R arises in the same way from M , then the SU(2) matrix M ·Mgenerates in the rotation matrix R · R.

Problem 36Let16 O(θ,n) with n2 = 1 be the (active) rotation around the direction n by the angle θ.Show that (k2 = 1)

O(θ,n)·O(ψ,k)·O−1(θ,n) = O(ψ,Ovec(θ,n)·k) ,

where Ovec means the rotation realized on vectors (in this formula O(θ,n) stand for anabstract rotation which can be realized in any vector space, in particular in a Hilbertspace).

Problem CUsing the commutation rules of the rotation group generators Jk with k = x, y, z showthat

e−iφJz

e−iθJy

e+iφJz

= e−iθ(Jy cos φ−Jx sinφ) .

Problem 37Using the result of Problem 36 show that the (active) rotation O(αβγ) parametrized bythree Euler angles and composed of three successive rotations: first by the angle α aroundthe axis n1 ≡ ez, then by the angle β around the axis n2 ≡ −ex sinα+ey cosα and finallyby γ around the axis n3 ≡ ex cosα sin β + ey sinα sin β + ez cos β (these are the famousthree moves of the paw - who attended my Classical Mechanics course, know what I mean)is equivalent to the composition of three other successive rotations: first by γ around ez,then by β around ey and finally by α again around ez:

O(γ,n3)·O(β,n2)·O(α,n1) = O(α, ez)·O(β, ey)·O(γ, ez) .

16Since it is desirable to denote differently active rotations (which are linear mapping of the vectorspace into itself and, hence, their matrices are written in a fixed basis) and passive rotations (whosematrices interpreted as the matrices of the change of bases are matrices of the identity mapping butwritten in two different bases - see my famous Algebra notes), we choose to denote the active ones by O(from polish - let proud Poland getting up from knees contribute also to physics - “obrot”).

24

Show also formally (what is obvious) that in the reference frame rotated by the anglesα, β and γ the components of the rotated vector are the same as the compotents of theoriginal vector in the original reference frame.

Problem DLet the two-parameter group G of transformations of the real axis R be defined by theformula

x′ = (1 + ξ1)x+ ξ2 ,

Find the left- and right- invariant measures on G. Are they identical?

Problem DProve that if the dimension n of the group G is odd the formula dµ(g) ≡ dnξ ρ(ξ1, . . . , ξn)with

ρ(ξ1, . . . , ξn) ∝ ǫi1i2...in tr

(

O−1 · ∂O∂ξi1

· O−1 · ∂O∂ξi2

· . . . ·O−1 · ∂O∂ξin

)

,

where O(ξ1, . . . , ξn) is a matrix representation of the group element parametrized by theparameters ξi, defines on G a left-invariant measure (if n even the measure defined in thisway vanishes as a result of the antisymmetry of ǫi1i2...in and the cyclicity of the trace).Use this result to find explicitly the density ρ(α, β, γ) of the left-invariant measure on theSO(3) and SU(2) groups parametrized by the three Euler angles α, β and γ.

Problem 39Making the “minimal substitution”

∂

∂xµ→ ∂

∂xµ+ i

qe

~cAµ , Aµ = (ϕ, A) ,

with A = 0 and ϕ = Ze/r, in the Klein - Gordon wave equation (∂µ∂µ+M2c2/~2)φ(t, r) =

0 and postulating its stationary solutions in the form φ(t,x) = exp(−iEt/~)χ(r) find thespectrum Enl of the energy levels of the Hydrogen atom (in this case q = −1).Hint: Instead of solving directly the time independent equation resulting from the KGequation, write it in the form parallel to the time independent Schrodinger equation withV (r) = −Ze2/r in which the operator L2/r2 in −∇

2 has been replaced by its eigenvaluel(l + 1)/r2 (setting ~ = c = 1):

(

− ∂2

∂r2− 2

r

∂

∂r+l(l + 1)

r2− 2MZe2

r− 2ME

)

ψnl = 0 .

Make the appropriate substitutions of the parameters in the Schrodinger equation con-verting it into the KG one. Write the parameter k in the coefficient k(k + 1), playing inthe KG equation the same role as does l(l + 1) in the Schrodinger equation, in the formk = l − δl with the appropriate factor δl. Remember that the quantum number n in the

25

Schrodinger result equals nr + l + 1, where nr = 0, 1, . . . is the radial quantum number(counting the nunber of nodes of the radial wave function).

Problem 40Compute the commutators (L = r× p)

[L, HDir] , [L2, HDir] and [~

2Σ, HDir] ,

of the Dirac Hamiltonian

HDir = −i~c αi∂i + β mc2 ≡ cα·p + β mc2 .

Check that HDir commutes with the total angular momentum operator J = r× p + ~

2Σ

and also with the helicity operator h ≡ J · p = ~

2Σ·p.

Problem 41Check that the two two-component spinors

ϕljm =

√

l+ 1

2+m

2l+1Yl,m− 1

2√

l+ 1

2−m

2l+1Yl,m+ 1

2

, with l = l− = j − 1

2,

ϕljm =

√

l+ 1

2−m

2l+1Yl,m− 1

2

−√

l+ 1

2+m

2l+1Yl,m+ 1

2

, with l = l+ = j +1

2,

which are eigenfunctions of L2, S2, J2 = (L + S)2 and Jz, are related by

ϕl±

jm = (er ·σ)ϕl∓

jm , where er ≡r

r.

Problem 42Find the exact energy levels of the Hydrogen-like atom (electron bound in the potentialϕ = Ze/r) predicted by the Dirac equation. To this end act first on the Dirac equation(~ = c = 1)

[γµ(i∂µ − qeAµ) −M ]ψ = 0 ,

with the matrix-differential operator [γµ(i∂µ − qeAµ) + M ]. Then set A = 0 and A0 =ϕ = Ze/r and substitute ψ(t, r) = χ(r) exp(−iEt). To decouple the equation satisfiedby the “large” χ spinor components from the one satisfied by its “small” components usethe Weyl (chiral) representation of the gamma matrices. The resulting equation involvesa nondiagonal differential-(2 × 2)matrix operator. To diagonalize it, evaluate for fixed jits matrix elements between the two two-components functions of Problem 41 and findthe eigenvalues of the resulting c-number matrix. For each of these eigenvalues writethe c-number differential equation in the form parallel to the radial Schrodinger eqution

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with V (r) = −Ze2/r and making the steps as in Problem 39, find the Hydrogen-atomrelativistic spectrum.

Problem 43Check directly that the Hamiltonian obtained from the Dirac one using the Foldy-Wouthuysenprocedure

H ′′′Dir = Mc2 +

1

2M

(

p− qe

cA)2

+ qeϕ− qe

Mc

~

2σ ·B

− p4

8M3c2− qe~2

8M2c2divE− i

qe~

8M2c2σ ·rotE− qe

4M2c2σ ·(E×p) ,

is Hermitian. In particular, check that the two last terms are only together Hermitian.

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Problem 44Consider the Dirac equation with the additional term (~ and c are set to 1)

(

i 6∂ − eq 6A− qeκ

4MσµνFµν −M

)

ψ = 0 .

where σµν ≡ i2[γµ, γν ], Fµν = ∂µAν − ∂νAµ and κ is some parameter. Check that the

probability density and the probability current still satisfy the continuity equation. Whatis the magnetic moment of the particle whose wave function satisfies this equation?

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