Quantum Physics by Bellac

1

Solutions of selected exercises

from ‘Quantum Physics’

Michel Le Bellac

2

Contents

1 Exercices from Chapter 1 5




5 Exercises from Chapter 5 23





10 Exercises from chapter 10 51


12 exercises from Chapter 12 71




3

4 CONTENTS

Chapter 1

Exercices from Chapter 1

1.6.1 Orders of magnitude

1. One must use particles whose wavelength λ be 1 A or less. We shall use λ = 1 A in numericalcomputations. In the case of photons, the energy is in eV

Ephot =hc

λ=

6.63 × 10−34 × 3 × 108

10−10 × 1.6 × 10−19= 1.24 × 104 eV = 12.4 keV

In the neutron case, we use p = h/λ, that is

Eneut =p2

2mn=

h2

2mnλ2= 8.2 × 10−2 eV = 82 meV

This energy is of order of that of thermal neutrons, 25 meV. In the case of electrons, it is sufficient tomultiply the preceding result by the mass ratio mn/me

Eel = Eneutmn

me= 151 eV

2. The frequency of a wave with wavelength k = 1 nm is ω = 5 × 1012 rad.s−1 and the phonon energy~ω = 3.3 meV. It is much easier to compare experimentally such an energy to that of a neutron withenergy of a few ten meV, rather than to that of a photon with an energy of 10 keV in order to detect thecreation of one phonon.

3. The mass of a fullerene molecule is M = 1.2 × 10−24 kg and its wavelength

λ =h

mv= 2.5 × 10−12 m

This wavelength is smaller than the molecule size by a factor ∼ 1/300.

4. The distance between the mass M1 and the molecule center-of-mass is

r1 =M2

M1 +M2r0

and the molecule moment of inertia is

I = M1r21 +M2r

22 =

M1M2

M1 +M2r20 = µr20

The rotational kinetic energy is given as a function of the angular momentum J = Iω by

Erot =1

2Iω2 =

J2

2I

5

6 CHAPTER 1. EXERCICES FROM CHAPTER 1

and choosing J = ~ leads to εrot = ~2/(2I)

εrot =~

2

2µr20=

~2

2µb2a20

=m

b2µR∞

using R∞ = e2/(2a0) and a0 = ~2/(me2).

5. The elastic constant K is

K =2cR∞

b2a20

=4cmR2

∞

~2b2

that is

~ωv = 2

√

c

b2

√

m

µR∞

In the case of the HCl molecule, µ = 0.97mp and m/µ = 5.6 × 10−4. Then b = 2.4 and c = 1.75, whichare indeed numbers close to one.

6. The dimension of G is easily obtained by observing that Gm2/r is an energy. One finds that thisdimension is given by M−1L3T −2. The quantity

√

~c/G has the dimension of a mass, which gives forthe Planck energy

EP =

√

~c

Gc2 = 1.9 × 109 J = 1.2 × 1019 GeV

and for the Planck length

lP =~

c

√

G

~c= 1.6 × 10−35 m

The Planck energy is huge compared to the highest energies available in elementary particle physics(roughly 2 TeV=2000 GeV as of today), and, as a consequence, Planck’s length is quite tiny comparedto the distances which are explored today in elementary particle physics which are ∼ 10−18 m.

1.6.4 Neutron diffraction by a crystal

1. The incident wave arriving at point ~ri suffers a phase shift δinc = ~k ·~ri with respect to that arriving atpoint ~r = 0, and the scattered wave suffers a phase shift δsc = −~k ′ ·~ri with respect to the wave scatteredby the nucleus at the point ~r = 0.

2. The scalar product ~q · ~ri is given by

~q · ~ri = naqx +mbqy

Using the formula for summation of a geometric series

N−1∑

n=0

xn =1 − xN

1 − x

one can, for example, evaluate the sum

Σx =N−1∑

n=0

e−iqxna =1 − e−iqxaN

1 − e−iqxa= e−iqxa(N−1)/2 sin qxaN/2

sin qxa/2

from which one deduces F (aqx, bqy) as given in the statement of the problem.

3. Suppose that qx differs very little from 2πnx/a, where nx is an integer: qxa = 2πnx + ε. Then

sinqxaN

2= sin

[

πnxN + εN

2

]

= ± sinεN

2

sinqxa

2= sin

[

πnx +1

2ε

]

= ± sinε

2

7

The peak width is then ε ∼ 1/N and its height is obtained taking the limit ε→ 0

limε→0

sin2 εN/2

sin2 ε/2= N2

which gives an intensity within the peak ∼ N2 × 1/N = N . The same calculation can be repeated in they direction.

4. The condition for elastic scattering is

~k2 = ~k ′2 = (~k + ~q)2 = k2 + 2~q · ~k + q2

that is q2 + 2~q · ~k = 0. Suppose that nx = 0 (or qx = 0) and thus k′x = kx

k′x = kx k′y = ky −2πnyb

The condition for elastic scattering is |k′y| = |ky| which implies

ky =πnyb

k′y = −πnyb

Since ky = k sin θB, where θB is the angle of incidence, one must have

sin θB =πnybk

One finds solutions only if ny is small enough or k large enough.

With only the first column of atoms, there would not be any constraint on kx, because k′x would not belinked to kx through k′x = kx + qx. One would then obtain diffraction maxima for any angle of incidence

5. The sum over the cells is

N−1∑

n=0

M−1∑

m=0

e−i(2aqxn+2bqym) = F (2aqx, 2bqy)

while the scattering amplitude due to the first cell is

f1

(

1 + e−i(aqx+bqy))

+ f2

(

e−iaqx + e−ibqy

)

Because of the argument of the F function, the condition for a diffraction peak is

qx =πnxa

qy =πnyb

and it follows thatf = f1

[

1 + (−1)nx+ny]

+ f2 [(−1)nx + (−1)ny ]

The final result is

• nx and ny even: f = 2(f1 + f2)

• nx et ny odd: f = 2(f1 − f2)

• nx even (odd) et ny odd (even): f = 0

6. When the lattice nodes are occupied randomly by the two kinds of atom, the lattice spacing is (a, b)and not (2a, 2b) as in the preceding question. One has in fact f1 = f2 and half of the diffraction peaksare lost.

1.6.6 Mach-Zehnder interferometer


1. Let b1 (b2) the probability amplitude for finding the photon in the upper (lower) arm of the interferom-eter. The probability amplitudes a1 (a2) that the photon triggers the detector D1 (D2) are obtained byexamining the transmission by the beam splitter S2: for example, a1 is obtained by adding the amplituderb1 where the photon originating from the upper arm is reflected by the beam splitter and the amplitudetb2 where the photon originating from the lower arm is transmitted by the beam splitter

a1 = rb1eiδ + tb2

a2 = tb1eiδ + rb2

We have made explicit the possibility of a variable phase shift δ in the upper arm. One calculates nowb1 et b2 as functions of a0

b1 = ta0 b2 = ra0

and plugs the result in the preceding equations

a1 = rta0

(

1 + eiδ)

a2 = a0

(

t2eiδ + r2)

Using the values of t and r given in the statement of the problem and choosing the (arbitrary) normal-ization |a0| = 1 (the photon has unit probability for arriving at S1)

p1 = |a1|2 =1

2(1 + cos δ)

p2 = |a2|2 =1

4

∣

∣

∣e2iαeiδ + e2iβ

∣

∣

∣

2

=1

2[1 + cos(2α− 2β + δ)]

By letting δ vary, one can manage that all the photons are detected by D1 (or by D2).

2. We must have p1 + p2 = 1 whatever δ, because the photon must trigger one of the detectors, whichimplies that cos 2(α− β) = −1, that is

α− β =π

2mod nπ

1.6.7 Neutron interferometer and gravity

1. and 2. Let us compute the probability amplitudes a1 and a2 for triggering D1 and D2

a1 = a0r2t(

eiχ + 1)

a2 = a0r(

t2eiχ + r2)

and the probabilities by taking the modulus squared

p1 = A(1 + cosχ) A = 2|r2t|2p2 = B +A′ cosχ A′ = 2|r2|Re

(

t2(r∗)2eiχ)

The sum p1 + p2 must be independent of χ, so that A+A′ = 0, or

cos 2(α− β) cosχ− sin 2(α− β) sinχ = − cosχ

and thusα− β =

π

2mod nπ

3. At elevation z the neutron energy is K = K0 +mgz if its energy is taken, by convention, to be K0 forz = 0. Its momentum is

p =√

2mK =√

2m(K0 +mgz) ≃√

2mK0

(

1 +mgz

2K0

)

9

The approximation is justified, because, for z = 1 m

mgz ≃ 10−7 eV ≪ K0 ∼ 0.1 eV

The variation ∆k of the wave vector is

∆k = kmgz

2K0

∆k

k=mgz

2K0

On a path with length L, the phase shift accumulated between the two arms, one at elevation z and theother at elevation z = 0 is

∆φ = ∆kL =mgzLk

2K0=mgkS2K0

=m2gS~2k

because zL is the rhombus area and2K0 = ~2k2/m. The numerical values is ∆φ = .59 rad.

4. It is enough to replace z by z cos θ in the preceding results. The phase shift becomes

χ = ∆φ =m2gS~2k

cos θ

and one will thus observe oscillations in the neutron detection rate by varying θ.

1.6.8 Coherent and incoherent scattering from a crystal

1. One observes that α2i = αi. If i = j, 〈α2

i 〉 = 〈αi〉 = p1, while if i 6= j

〈αiαj〉 = 〈αi〉〈αj〉 = p21

the two results being summarized in

〈αiαj〉 = δijp1 + (1 − δij)p21 = p2

1 + p1p2δij

2. The scattering probabiilty by the crystal is

〈|ftot|2〉 =∑

i,j

⟨

[

αif1 + (1 − αi)f2][

αjf1 + (1 − αj)f2]

⟩

ei~q·(~ri−~rj)

=∑

i,j

[

(p21 + p1p2δij)f

21 + 2p1p2(1 − δij)f1f2 + (p2

2 + p1p2δij)f22

]

ei~q·(~ri−~rj)

=∑

i,j

(p1f1 + p2f2)2ei~q·(~ri−~rj) + Np1p2(f1 − f2)

2

The first term gives rise to diffraction peaks, but the second one gives rise to a continuous background.


Chapter 2


2.4.3 Determinant and trace

1. Let A(t) = A(0) exp(Bt). Let us compute the derivative

d

dtA(0)eBt = A(0)eBtB = A(t)B

The solution ofdA

dt= BA(t)

isA(t) = eBtA(0)

2. One remarks that for infinitesimal δt

det eAδt ≃ det(I +Aδt) = 1 + δtTrA+O(δt)2

For example, for a 2 × 2 matrix

det

(

1 +A11δt A12δtA21δt 1 +A22δt

)

= 1 + (A11 +A22)δt+ (A11A22 −A12A21)(δt)2

Let g(t) = det[exp(At)]

g′(t) = limδt→0

1

δt

(

det eA(t+δt) − det eAt)

=1

δt

(

det eAδt − 1)

det eAt =1

δt[δtTrA] det eAt = TrAg(t)

and one obtains for g(t) the differential equation

g′(t) = [TrA]g(t) =⇒ g(t) = etTrA

using the boundary conditiom g(0) = 1. Setting t = 1 we find

g(1) = eTrA = det[

eA]

2.4.10 Positive matrices

1. Let us decompose A into a Hermitian part and an antiHermitian part

A = B + C B = B† C = −C†

11


One notes that (x,Cx) is pure imaginary

(x,Cx) = (C†x, x) = (x,C†x)∗ = −(x,Cx)∗

while (x,Bx) is real. If we want (x,Ax) to be real and ≥ 0, it is necessary that C = 0. Let us give amore explicit proof. Let, for example,

x = (x1, x2, 0, . . . , 0)

Then(x,Cx) = x∗1C12x2 + x∗2C21x1 = 2i Im(x∗1C12x2) = 0 =⇒ C12 = 0

Since A is Hermitiian, it can be diagonalized. Let ϕ be an eigenvector of A, Aϕ = aϕ. The positivitycondition implies (ϕ,Aϕ) = a||ϕ||2 ≥ 0 and thus a ≥ 0.

2. In the case of a real and antisymmetric matrix CT = −C

(x,Cx) = x1C12x2 + x2C21x1 = x1(C12 + C21)x2 = 0

One can thus have a positive matrix of the form

A = B + C BT = B CT = −C 6= 0

2.4.11 Operator identities

1. Let us compute df/dt

df

dt= etAABe−tA − etABAe−tA = etA[A,B]e−tA

The second derivative is computed in the same way, and the general case is obtained by recursion.

2. Let us compute dg/dtdg

dt= etA(A+B)etB

and use the result of the preceding question

etAB = etABe−tAetA = (B + t[A,B])etA

Indeed, because of the commutation relations of [A,B] with A and B, the series expansion stops afterthe second term. We get the differential equation

dg

dt=(

A+B + t[A,B])

g(t)

Taking the commutation relations into account, this equation has the solution

g(t) = e(A+B)t+ 12[A,B]t2

Note that this solution holds only because the commutator [[A,B], A +B] vanishes. Setting t = 1

g(1) = eA eB = eA+B+[A,B]/2 = eA+B e[A,B]/2

2.4.12 A beam splitter

1.The condition that there are no losses reads

|BD|2 + |BG|2 = |AD|2 + |AG|2

13

The norm of the vector (AD, AG) is conserved, which implies that the matrix R′ is unitary. The deter-minant of R′ then obeys | detR′| = 1, which can be written detR′ = exp(iθ)

2. One defines R through

R = ie−iθ/2R′ detR = −e−iθ detR′ = −1

This redefinition corresponds to a global change of phase of the state vectors phase. One checks thatwith the form of R given in the statement of the problem

R†R = I detR = −|r|2 − |t|2 = −1

Let us first choose ψ = (1, 0)

Rψ =

(

|r|eiχ |t|e−iφ

|t|eiφ −|r|e−iχ

)(

1

0

)

=

(|r|eiχ

|t|eiφ

)

and then ψ = (0, 1)

Rψ =

(

|r|eiχ |t|e−iφ

|t|eiφ −|r|e−iχ

)(

0

1

)

=

( |t|e−iφ

−|r|e−iχ

)

One deduces the phase shifts for the reflected (R) and transmitted (T ) waves

δDR = χ δDT = φ δD = χ− φ

δGR = −(χ− π) δGT = −φ δG = π − (χ− φ)

so thatδD + δG = π

If the beam splitter is symmetric, one must have t = t∗ et r = −r∗ as well as |t| = |r| = 1/√

2 andδD = δG = π/2.


Chapter 3


3.3.1 Decomposition and recombination of polarizations

1. Let e be the thickness of the plate. Since the separation of the centers of the beams is

y = e tanα = 1.09 mm

the beams are well separated. The difference in optical paths is

δ = e

(

no −n′e

cosα

)

= 0.9248 mm

2. The index difference is no − ne = 0.17102 and the thickness of the intermediary plate

D =2 × 0.9248

no − ne= 10.815 mm

3. Let β = 1/ cosα

D =2e(no − βn′

e)

n0 − ne

One infers from this the relative error δD/D

δD

D=δe

e+δn0 − βδn′

e

n0 − βn′e

− δno − δneno − ne

In order to simplify the error calculation, we neglect the difference between ne and n′e

δD

D=δe

e+

(β − 1)(neδno − noδne)

(no − ne)(no − βne

which leads to∣

∣

∣

δD

D

∣

∣

∣≃ 0.7 |δn|

and to

|δD| ≃ 7 × 10−5 mm ≪ λ

5. The beam polarization is elliptic in the region where the two beams overlap, and linear in the regionwhere they don’t.

3.3.4 Other solutions of (3.45)

15


1. The action of U on σx and σy is

U † σx U =

(

0 eiψ

e−iψ 0

)

U † σy U =

(

0 −ieiψ

ie−iψ 0

)

σz is clearly unchanged.

2. The possible solutions of (3.45) are

cos(α− αx) = cosφ α− αx = φ or α− αx = −φcos(α− αy) = sinφ α− αy =

π

2− φ or α− αy = φ− π

2

The difference(α− αx) − (α− αy) = −(αx − αy)

must be independent of φ because αx and αy are given data independent of α. There are then twopossible solutions

• Solution 1α− αx = φ α− αy = φ− π

2

that isαx = αy −

π

2

and

σx =

(

0 e−iαx

eiαx 0

)

σy =

(

0 −ie−iαx

ie−iαx 0

)

From question 1, this new form corresponds to a rotation of the axes by an angle αx about Oz.

• Solution 2α− αx = −φ α− αy =

π

2− φ

Choosing as a reference solution αx = 0 et αy = −π/2

σx =

(

0 11 0

)

σy =

(

0 i−i 0

)

The change of sign of σy corresponds to an inversion of the Oy axis: one goes from a right handedreferential to a left handed one. The other solutions are obtained from the reference solution by arotation about the Oz axis.

3.3.6 Exponentials of Pauli matrices

1. From (3.50)(~σ · p)2 = I (~σ · p)3 = (~σ · p) . . .

the series expansion of the exponential is

exp

(

−iθ

2~σ · p

)

= I − iθ

2~σ · p+

1

2!

(

−iθ

2

)2

I +1

3!

(

−iθ

2

)3

~σ · p · · ·

= I cosθ

2− i(~σ · p) sin

θ

2

Taking p = (− sinφ, cosφ, 0) we get

exp

(

−iθ

2~σ · p

)

= I cosθ

2+ iσx sinφ sin

θ

2− iσy cosφ sin

θ

2

=

(

cos θ2 e−iφ sin θ2

eiφ sin θ2 cos θ2

)

17

2. We must have

U = a1I + ia2σz + ib2σx + ib1σy

= I cosθ

2− i sin

θ

2

(

nz nx − inynx + iny −nz

)

and we deduce from this

a1 = cosθ

2a2 = −nz sin

θ

2b2 = −nx sin

θ

2b1 = −ny sin

θ

2

These equations have solutions because

a21 + a2

2 + b21 + b22 = 1

3. The product of two exponentials of Pauli matrices is

e−iα(~σ·a)e−iβ(~σ·b) = cosα cosβ − i sinα cosβ(~σ · a) − i sinβ cosα(~σ · b) − sinα sinβ[a · b+ i~σ · (a× b)]

On the other hand

e−i[α(~σ·a)+β(~σ·b)] = I cos ||αa+ βb|| − isin ||αa+ βb||||αa+ βb||

[α(~σ · a) + β(~σ · b)]

In order to ensure the equality of the two factors, we must

• get rid of the sines;

• have cosα cosβ = cos√

α2 + β2

One can choose, for example

α = 3π β = 4π√

α2 + β2 = 5π

withe−3iπσx = −I e−4iπσy = I e−i(3πσx+4πσy) = −I

3.3.9 Neutron scattering from spin 1/2 nuclei

1. When the nucleus spin does not flip, it is not possible to tell from which nucleus the neutron wasscattered, and we must add the amplitudes

f = fa∑

i

ei~q·~ri I = f2a

∑

i,j

ei~q·(~ri−~rj)

2. If the scattering is accompanied with a spin flip, it leaves the nucleus in a state which is differentfrom its initial state. If all the nuclei had initially a down spin, the nucleus which scattered the neutroncould be in principle identified (even though this identification would impossible in practice). Neutronscattering from a given nucleus rather than from another one corresponds to different nucleus final states,and we must add probabilities

I =∑

i

f2b = Nf2

b

3. Let αi define the spin configuration in the crystal. If a neutron is scattered by the crystal in theconfiguration αi, the scattering amplitude is

f =∑

i

(αifa + (1 − αi)fc) ei~q·~ri +∑

i

αifbei~q·~ri


Were the configuration αi fixed, the intensity would be

Iαi=∑

i,j

(αifa + (1 − αi)fc)2ei~q·(~ri−~rj) +

∑

i

α2i f

2b

Observing that αi = α2i the average values are 〈αi〉 = 〈α2

i 〉 = 1/2 and 〈αiαj〉 = 1/4 if i 6= j, whence

I =∑

i,j

⟨ (

αiαjf2a + 2αi(1 − αj)fafc + (1 − αi)(1 − αj)f

2c

) ⟩

ei~q·(~ri−~rj) +∑

i

〈αi〉f2b

We then get the result given in the statement of the problem

I =1

4(fa + fc)

2∑

i,j

ei~q·(~ri−~rj) +N4

[(fa − fc)2 + 2f2

b ]

4. From rotational invariance, we have, for example

fa : neutron ↓ + nucleus ↑ → neutron ↓ + nucleus ↑

and a similar result for the two other amplitudes. One finds again the result of the preceding questionif the neutrons are all polarized with a down spin. The result for unpolarized neutrons is obtained bytaking the average of the result with spin up and spin down, and one again finds the result of question 3.

Chapter 4


4.4.4 Time evolution of a two-level system

1. The system of differential equations obeyed by c±(t) is

ic+ = Ac+ +Bc−

ic− = Bc+ −Ac−

2. If |ϕ(t = 0)〉 is decomposed on the eigenvectors |χ±〉 as

|ϕ(t = 0)〉 = λ|χ+〉 + µ|χ−〉

then the time evolution is

|ϕ(t)〉 = λe−iΩt/2|χ+〉 + µeiΩt/2|χ−〉

Taking into account 〈+|χ+〉 = cos θ/2 and 〈+|χ−〉 = − sin θ/2, one finds for c+(t)

c+(t) = 〈+|ϕ(t)〉 = λe−iΩt/2 cosθ

2− µeiΩt/2 sin

θ

2

3. If c+(0) = 0

λ cosθ

2− µ sin

θ

2= 0

and a possible solution is

λ = sinθ

2µ = cos

θ

2

which gives the following for c+(t)

c+(t) = − sinθ

2cos

θ

2

(

e−iΩt/2 − eiΩt/2)

= i sin θ sinΩt

2

The probability p+(t) is

p+(t) = |c+(t)|2 = sin2 θ sin2 Ωt

2=

B2

A2 +B2sin2 Ωt

2

4. If c+(0) = 1, a possible solution is

λ = cosθ

2µ = − sin

θ

2

19


and one obtains for c+(t)

c+(t) = cos2θ

2e−iΩt/2 + sin2 θ

2eiΩt/2

= cosΩt

2− i cos θ sin

Ωt

2

The probability p+(t) is

p+(t) = |c+(t)|2 = cos2Ωt

2+ cos2 θ sin2 Ωt

2= 1 − sin2 θ cos2

Ωt

2

4.4.5 Unstable states

1. Let us use a series expansion of exp(−iHt/~) for small values of t

c(t) = 1 − i

~〈H〉t− 1

2~2〈H2〉t2 +O(t3)

so that

|c(t)|2 = 1 − 〈H2〉 − 〈H〉2~2

t2 +O(t3)

= 1 − (∆H)2

~2t2 +O(t3)

2. From (4.27) we derive, with the substitution A→ P

∆P∆H ≤ 1

2~

∣

∣

∣

dPdt

∣

∣ p(t) = 〈P〉(t)

while∆P = (〈P2〉 − 〈P〉2)1/2 = (〈P〉 − 〈P〉2)1/2 =

√

p(1 − p)

We thus obtain the differential inequality

dp√

p(1 − p)≥ 2

∆H

~dt

which integrates into

cos−1[1 − 2p(t)] ≥ 2∆H

~t

whence

p(t) ≥ cos2(

t∆H

~

)

0 ≤ t ≤ π~

2∆H

3. Inserting a complete set of eigenstates |n〉 of H in the expression for c(t), we obtain

c(t) =∑

n

〈ϕ(0)|e−iHt/~|n〉〈n|ϕ(0)〉

=∑

n

|〈ϕ(0)|n〉|2e−iEnt/~

Let the spectral function w(E) be the inverse Fourier transform of c(t)

w(E) =

∫

dt

2πeiEt/~c(t)

=

∫

dt

2πeiEt/~

∑

n

|〈ϕ(0)|n〉|2e−iEnt/~

=∑

n

|〈ϕ(0)|n〉|2δ(E − En)

21

The expectation values of H and H2 are

〈H〉 =∑

n

En|〈ϕ(0)|n〉|2 =

∫

dE E w(E)

〈H2〉 =

∫

dE E2 w(E)

4. Use the following identity, obtained from a suitable contour in the complex x-plane

∫ +∞

−∞

dxeitx

x2 + α2=π

αe−α|t|

4.4.6 The solar neutrino enigma

1. The Hamiltonian reads, in the reference frame where the neutrino is at rest

1

c2H =

me +mµ

2I +

( me−mµ

2 m

m −me−mµ

2

)

Comparison with exercise 4.4.4 leads to the correspondence

A→ me −mµ

2B → m tan θ =

B

A→ 2m

me −mµ

2. Since the states |ν1〉 and |ν2〉 are eigenvectors of H , the time evolution of |ϕ(t)〉 is

|ϕ(t)〉 = cosθ

2e−iE1t/~|ν1〉 − sin

θ

2e−iE2t/~|ν2〉

Taking into account

〈νe|ν1〉 = cosθ

2〈νe|ν2〉 = − sin

θ

2

one finds for the amplitude ce(t)

ce(t) = 〈νe|ϕ(t)〉 = e−iE1t/~

(

cos2θ

2+ sin2 θ

2e−i(E2−E1)t/~

)

and for the probability |ce(t)|2

|ce(t)|2 = cos4θ

2+ sin4 θ

2+ 2 cos2

θ

2sin2 θ

2cos

∆Et

~

= 1 − 1

2sin2 θ

(

1 − cos∆Et

~

)

= 1 − sin2 θ sin2 ∆Et

2~

3. When p≫ mc one obtains the following approximate expression for E

E = (m2c4 + p2c2)1/2 = cp

(

1 +m2c2

p2

)1/2

≃ cp+m2c3

2p

and

∆E = E2 − E1 =(m2

2 −m21)c

3

2p=

∆m2c3

2p

Substituting the distance L travelled by the neutrino to ct, the oscillation takes the form

sin2

(

∆m2c2L

2p~

)


If half of an oscillation is observed between the Sun and the Earth

∆m2c2L

2p~= π ∆m2c4 =

2π~c

Lcp ≃ 7 × 10−11 eV2

and thus ∆mc2 ∼ 10−5 eV.

4.4.8 The neutral K meson system

1. Let us compute the product C−1MC for a generic matrix M

C = C−1 = σx =

(

0 11 0

)

σx

(

a bc d

)

σx =

(

d cb a

)

The commutation relation implies a = d and b = c.

2. The eigenvalues and eigenvectors of M are

λ+ = A+B |K1〉 =1√2

(

|K0〉 + |K0〉)

λ− = A−B |K1〉 =1√2

(

|K0〉 − |K0〉)

The time evolution of the states |K1〉 et |K2〉 is given by

|K1(t)〉 = e−i(A+B)t|K1〉 = exp

(

−iE1t

~− Γ1t

2

)

|K1〉

|K2(t)〉 = e−i(A−B)t|K1〉 = exp

(

−iE2t

~− Γ2t

2

)

|K1〉

Let us start at time t = 0 from

|ϕ(t = 0)〉 =1√2

(c(0) + c(0)) |K1〉 +1√2

(c(0) − c(0)) |K2〉

One gets at time t

|ϕ(t)〉 =1

2(c(0) + c(0))

(

|K0〉 + |K0〉)

e−i(E1/~−iΓ1/2)t

+1

2(c(0) − c(0))

(

|K0〉 − |K0〉)

e−i(E2/~−iΓ2/2)t

which gives the coeficients c(t) and c(t)

c(t) = 〈K0|ϕ(t)〉 =1

2(c(0) + c(0)) e−i(E1/~−iΓ1/2)t +

1

2(c(0) − c(0)) e−i(E2/~−iΓ2/2)t

c(t) = 〈K0|ϕ(t)〉 =1

2(c(0) + c(0)) e−i(E1/~−iΓ1/2)t − 1

2(c(0) − c(0)) e−i(E2/~−iΓ2/2)t

3. In the case which is considered in the statement of the problem c(0) = 1, c(0) = 0 and

c(t) =1

2

[

e−i(E1/~−iΓ1/2)t − e−i(E2/~−Γ2/2)t]

The probability to observe a K0 meson at time t is

p(t) = |c(t)|2 =1

4

[

e−Γ1t + e−Γ2t − 2e−(Γ1+Γ2)t/2 cos∆Et

~

]

with ∆E = E1 − E2. One obtains the result given in the statement of the problem if Γ1 ≫ Γ2.

Chapter 5

Exercises from Chapter 5

5.5.3 Butadiene

1.The matrix form of H is

H =

E0 −A 0 0−A E0 −A 00 −A E0 −A0 0 −A E0

2. Let us write the action of H on the vector |χ〉

H |χ〉 = E0|χ〉 −A(

c1|ϕ2〉 + · · · + cN−1ϕN 〉)

+ c0|ϕ0〉

− A(

c2|ϕ1〉 + · · · cN |ϕN−1〉)

+ cN+1|ϕN 〉

= E0|χ〉 −A

N∑

n=1

(

cn−1 + cn+1

)

|ϕn〉

3. Taking into acount the postulated form for the coefficients cn, we get

cn−1 + cn+1 =c

2i

[

ei(n−1)δ + ei(n+1)δ − e−i(n−1)δ − e−i(n+1)δ]

=c

2i2 cos δ

[

einδ − e−inδ]

= 2cn cos δ

The condition c0 = 0 is satisfied by construction. The condition cN+1 = 0 implies

δs =πs

N + 1s = 0, 1, · · · , N

4. From the results of question 3 we have

H |χs〉 = E0|χs〉 − 2A cosπs

N + 1|χ〉

whence the values of the energy

Es = E0 − 2A cosπs

N + 1

Let us compute the normalization of |χs〉

X =

N∑

n=1

sin2 πsn

N + 1=

1

2

N∑

n=1

(

1 − cos2πsn

N + 1

)

23

24 CHAPTER 5. EXERCISES FROM CHAPTER 5

The sum over n is readily computed

N∑

n=1

cos2πsn

N + 1= Re

N∑

n=1

exp

(

2iπsn

N + 1

)

= Re1 − e2iπs

1 − e2iπs/(N+1− 1 = −1

and X = (N + 1)/2. The normalized vectors |χs〉 are then

|χs〉 =

√

2

N + 1

N∑

n=1

sin(nδs)|ϕn〉

5. In the case of butadiene N = 4

cosπ

5= 0.809 cos

2π

5= 0.309

which gives for the two lowest levels

Es=1 = E0 − 1.62A Es=2 = E0 − 0.62A

The energy of the four π electrons is then

E = 4E0 − 2(1.62A+ 0.62A) = 4(E0 −A) − 0.48A

The delocalization energy is −0.48A.

7. The coefficients of the normalized eigenvectors |χ1〉 are

(0.372, 0.601, 0.601, 0.372)

while those of |χ2〉 are(0.601, 0.372, −0.372, −0.601)

The order of the 1-2 bond is

1 + 2[

〈ϕ1|χ1〉〈χ1|ϕ2〉 + 〈ϕ1|χ2〉〈χ2|ϕ2〉]

= 1.89

which is very close to that of a double bond, while for the 2-3 bond

1 + 2[

〈ϕ2|χ1〉〈χ1|ϕ3〉 + 〈ϕ2|χ2〉〈χ2|ϕ3〉]

= 1.45

This bond is weaker than the preceding one, which explains why it is closer to a single bond, and thuslonger.

5.5.5 The molecular ion H+2

1. The potential V (x) is

V (x) = −e2(

1

|x+ r/2| +1

|x− r/2|

)

Its value is −∞ when x = ±r/2 and its maximal value is −4e2/r for x = 0.

2. l ≃ a0, characteristic size of the hydrogen atom.

3. Eigenvalues and eigenvectors

E+ = E0 −A |χ+〉 =1√2

(

|ϕ1〉 + |ϕ2〉)

E− = E0 +A |χ−〉 =1√2

(

|ϕ1〉 − |ϕ2〉)

25

4. A is a tunnel effect transmission coefficient. The width of the potential barrier decreases as r decreases:the transmission coefficient increases when r decreases.

5. e2/r is the (repulsive) potential energy between the two protons

E′±(r) = E±(r) +

e2

r= E0 ∓A(r) +

e2

r

6. The approximate expression for E′+(r) is

E′+(r) = E0 + e2

(

1

r− c e−b/r

)

= E0 + ∆E(r)

Let us look for the minimum of ∆E(r)

d∆E(r)

dr= e2

(

− 1

r2+c

be−r/b

)

=⇒ 1

r20=c

be−r0/b

One obtains

∆E(r0) = e2(

1

r0− c e−r0/b

)

= e2(

1

r0− b

r20

)

whence

b =6

5r0 =

12

5a0 c =

3

5a0e5/6 ≃ 1.38

a0

One must have b > r0 for the H+2 ion to be a bound state.

5.5.6 The rotating wave approximation in NMR

The evolution equation for the state vector |ϕ(t)〉 in the rotating reference frame is

i~d|ϕ(t)〉

dt=

~

2ωσz |ϕ(t)〉 + e−iωσzt/2He iωσzt/2|ϕ(t)〉

=~

2ωσz |ϕ(t)〉 + H(t)|ϕ(t)〉

2. Let us study σ±(t)

dσ±(t)

dt= − iω

2e−iωσzt/2 [σz , σ±]e iωσzt/2 = ∓iωσ±(t)

because [σz , σ±] = ±2σ±. Solving the differential equation gives

σ±(t) = e∓iωtσ±

Let us rewrite H1 in terms of σ+ and σ−, using

σx = σ+ + σ− σy = −iσ+ + iσ−

H1 = −~

2ω1(σ+ + σ−)

(

eiωte−iφ + e−iωteiφ)

whence

H1(t) = −~

2ω1σ+

(

e−iφ + e−2iωteiφ)

− ~

2ω1σ−

(

e2iωte−iφ + eiφ)


and in the rotating wave approximation, where one neglects terms in exp(±2iωt)

H1 = −~ω1

2

(

σ+e−iφ + σ−eiφ)

= −~

2ω1(σx cosφ+ σy sinφ)

3. From equation (3.67), exp[−iθ(~σ · p)/2] is the rotation operator of a spin 1/2 about an axis p. Byidentification, we find p = (cosφ, sinφ, 0) and θ = −ω1t. The vector n being normalized (n2 = 1), wehave

exp(−iHt/~) = I cosΩt

2− i(~σ · n) sin

Ωt

2

with

~σ · n = −ω1

Ωσx +

δ

Ωσz

so that the matrix form of exp(−iHt/~) is

e−iHt/~ =

(

cos Ωt2 + δ

Ω sin Ωt2

iω1

Ω sin Ωt2

iω1

Ω sin Ωt2 cos Ωt

2 − δΩ sin Ωt

2

)

Chapter 6


6.5.3 Properties of state operators

1. Let us consider a vector |ϕ〉 of the form

|ϕ〉 = (0, · · · , ai, 0, · · · , 0, aj , 0, · · · , 0)

Positivity of ρ gives 〈ϕ|ρ|ϕ〉 ≥ 0, which implies that the 2 × 2 sub-matrix A

A =

(

ρii ρijρji ρjj

)

must be positive. On the other hand, (ρii + ρjj), which is the sum of the eigenvalues of A, obeys(ρii + ρjj) ≤ 1. One deduces that the product of the eigenvalues of A must be less than 1/4

0 ≤ ρiiρjj − |ρij |2 ≤ 1

4

If ρii = 0, this implies ρij = 0.

2. The condition for a maximal test with 100% success implies that there exists a vector |ϕ〉 such thatTr ρPϕ = 1, with Pϕ = |ϕ〉〈ϕ|, and thus 〈ϕ|ρ|ϕ〉 = 1. We choose an orthonormal basis, where, forexample, |ϕ〉 is the first basis vector, |ϕ〉 ≡ |1〉. In that case, the diagonal elements of ρ obey

ρ11 = 1, ρii = 0, i 6= 1

because the test |i〉, |i〉 6= |1〉 has zero probability of success. From the preceding question, all thenondiagonal elements vanish, ρij = 0, i 6= j and ρ = |ϕ〉〈ϕ| = |1〉〈1|.

6.5.4 Fine structure and Zeeman effect in positronium

1. The reduced mass is half that of the electron, and the energy levels are deduced from (1.36)

En = −R∞

2n2

2. Let us work out explicitly the action of σx and σy on the vectors |ε1ε2〉

σ1xσ2x| + +〉 = | − −〉 σ1yσ2y| + +〉 = −| − −〉σ1xσ2x| + −〉 = | − +〉 σ1yσ2y| + −〉 = | − +〉σ1xσ2x| − +〉 = | + −〉 σ1yσ2y| − +〉 = | + −〉σ1xσ2x| − −〉 = | + +〉 σ1yσ2y| − −〉 = −| + +〉

27


and σ1zσ2z |ε1ε2〉 = ε1ε2|ε1ε2〉, whence the action of ~σ1 · ~σ2

~σ1 · ~σ2| + +〉 = | + +〉~σ1 · ~σ2| + −〉 = 2| − +〉 − | + −〉~σ1 · ~σ2| − +〉 = 2| + −〉 − | − +〉~σ1 · ~σ2| − −〉 = | − −〉

3. From the results of the preceding question we get

~σ1 · ~σ2|I〉 = |I〉 ~σ1 · ~σ2|III〉 = |III〉

as well as

~σ1 · ~σ2|II〉 =1√2(| + −〉 + | − +〉) = |II〉

~σ1 · ~σ2|IV 〉 = −31√2(| + −〉 − | − +〉) = −3|IV 〉

4. The projectors P1 et P−3 as well as ~σ1 · ~σ2 are diagonalized in the basis |I〉, |II〉, |III〉, |IV 〉

P1 =

1 0 0 00 1 0 00 0 1 00 0 0 0

P−3 =

0 0 0 00 0 0 00 0 0 00 0 0 1

~σ1 · ~σ2 =

1 0 0 00 1 0 00 0 1 00 0 0 −3

If P1 = λI + µ~σ1 · ~σ2, we must have

λ+ µ = 1 and λ− 3µ = 0

that is, λ = 3/4 and µ = 1/4. One deduces

P1 =1

4(3I + ~σ1 · ~σ2) P−3 =

1

4(I − ~σ1 · ~σ2)

5. We immediately get P12| + +〉 = | + +〉 and P12| − −〉 = | − −〉, while

P12| + −〉 =1

2(| + −〉 + 2| − +〉 − | + −〉) = | − +〉

and in general P12|ε1ε2〉 = |ε2ε1〉.6. We know the eigenvalues and eigenvectors of ~σ1 · ~σ2, and thus those of H

• |I〉, |II〉, |III〉 are eigenvectors H with eigenvalue E0 +A

• |IV 〉 is eigenvector of H with eigenvalue E0 − 3A

7. The gyromagnetic of the positron has a sign opposite to that of the electron: γe+ = −γe− = −qe/m.The total Hamiltonian reads

H = H0 − (~µe− + ~µe+) · ~B = H0 −qe~

2m(σ1z − σ2z)

Let us examine the action of the operator (σ1z − σ2z) on the basis vectors of H0

(σ1z − σ2z)| + +〉 = 0 (σ1z − σ2z)| − −〉 = 0

(σ1z − σ2z)|II〉 = 2|IV 〉

29

and

〈II|H1|IV 〉 = 〈IV |H1|II〉 = 2

(

−qe~2m

)

= 2Ax

The matrix form of H is in the basis |I〉, |II〉, |III〉, |IV 〉

H =

E0 +A 0 0 00 E0 +A 0 2Ax0 0 E0 +A 00 2Ax 0 E0 − 3A

Two eigenvectors are obvious: |I〉 and |III〉 with eigenvalues E0+A. The other two are obtained throughdiagonalization of the 2 × 2 matrix

H ′ = E0I +A

(

1 2x2x −3

)

= E0I +AM

The eigenvalue equation of M isλ2 + 2λ− (3 + 4x2) = 0

which gives the values of the energy

E± = E0 −A± 2A√

1 + x2

When x = 0 we recover the values E0 +A and E0−3A, while for |x| → ∞, the eigenvectors tend to thoseof (σ1z − σ2z) with eigenvalues ±2Ax.

6.5.4 Spin waves and magnons

1. Since the eigenvalues of (~σn · ~σn+1) lie between −3 and +1, we must have

E ≥ 1

2NA− 1

2NA = 0

If the state vector is such that (~σn · ~σn+1) = 1, we obtain the ground state E0 = 0. This state vector is

Φ0 = | + + + · · · + +〉

because(~σn · ~σn+1)|Φ0〉 = |Φ0〉

2. The operator Pn,n+1 exchanges spins n et n+1: in the case of two spins, we have seen in the precedingquestion that

P12| + +〉 = | + +〉 P12| − −〉 = | − −〉P12| + −〉 = | − +〉 P12| − +〉 = | + −〉

and the number of up spins minus the number of down spins must stay unchanged. The eigenvectors ofH are thus such that the number of up spins minus the number of down spins is a constant. In particular,for the state |Ψn〉, this constant is N − 1. The operator I −Pn,n+1 applied to |Ψn〉 gives zero on any pairof up spins, and only the pairs (n− 1, n) and (n, n+ 1) are going to give a nonzero result. Since Pn−1,n,for example, exchanges spins n− 1 and n

Pn−1,n| + + + + − + + +〉 = | + + + − + + + +〉

that isPn−1,n|Ψn〉 = |Ψn−1〉 Pn−1,n|Ψn−1〉 = |Ψn〉

we obtain

(I − Pn−1,n)|Ψn〉 = |Ψn〉 − |Ψn−1〉(I − Pn,n+1)|Ψn〉 = |Ψn〉 − |Ψn+1〉


This gives the action of H on |Ψn〉

H |Ψn〉 = −A(

|Ψn−1〉 + |Ψn+1〉 − 2|Ψn〉)

3. Let us work out the action of H on |ks〉

H |ks〉 =

N−1∑

n=0

eiksnlH |Ψn〉 = −AN−1∑

n=0

eiksnl(

|Ψn−1〉 + |Ψn+1〉 − 2|Ψn〉)

On the other hand∑

n

eiksnl|Ψn−1〉 =∑

n

eiks(n−1)l eiksl|Ψn−1〉 = eiksl∑

n

eiksnl|Ψn〉 = eiksl|ks〉

We thus haveH |ks〉 = 2A(1 − cos ksl)|ks〉

The eigenfrequencies are

ωk = 2A(1 − cos ksl) |ks| → 0 ωk ≃ (Al2)k2s

It is also interesting to apply the method of § 5.1.2, observing that H can be cast into the form

H = −A(

UP + U−1P − 2

)

where UP makes a circular permutation n→ n+ 1 and to look for eigenvectors |Ψs〉 of UP . Writing

|Ψs〉 =∑

n

csn|Ψn〉 UP |Ψs〉 = eiδs |Ψs〉

with

δs =2πs

N= ksl

We must havecn+1 = eiδscn

6.5.7 Calculation of E(a, b)

1. Let us compute, for example, the amplitude a+−(θ)

a+−(θ) = 〈+ ⊗ [−, b]|Φ〉 =(

− sinθ

2〈+ + | + cos

θ

2〈+ − |

) 1√2

(

| + −〉 − | − +〉)

=1√2

cosθ

2

2. The rotation operator by an angle θ about an axis n of the ensemble of two spins is, from (3.67)

Un(θ) = exp

[

−θ2

(

~σ(a) + ~σ(b))

·n]

The rotational invariance of |Φ〉 implies

Un(θ)|Φ〉 = |Φ〉 =⇒[

(

~σ(a) + ~σ(b))

· n]

|Φ〉 = 0

that is(

~σ(a) · n)

|Φ〉 = −(

~σ(b) · n)

|Φ〉As a consequence

〈Φ|(

~σ(a) · a)

⊗(

~σ(b) · b)

|Φ〉 = −〈Φ|(

~σ · a)(

~σ · b)

|Φ〉 = −〈Φ|a · b+ i~σ · (a× b)|Φ〉 = −a · b = − cos θ

31

because the expectation value of ~σ vanishes in a rotational invariant state.

6.5.8 Bell inequalities for photons

1. Let us work out the explicit form of the vector |θθ⊥〉

|θθ⊥〉 =(

cos θ|x〉 + sin θ|y〉)(

− sin θ|x〉 + cos θ|y〉)

= − sin θ cos θ|xx〉 + cos2 θ|xy〉 − sin2 θ|yx〉 + sin θ cos θ|yy〉

To obtain |θ⊥θ〉, it is enough to exchange x↔ y

|θ⊥θ〉 = − sin θ cos θ|xx〉 + cos2 θ|yx〉 − sin2 θ|xy〉 + sin θ cos θ|yy〉

and, subtracting the the second equation from the first we get

|θθ⊥〉 − |θ⊥θ〉 = |xy〉 − |yx〉

2. For photon 1

|x〉 =1√2

(

−|R〉 + |L〉)

|y〉 =i√2

(

|R〉 + |L〉)

For photon 2 we must change |y〉 into −|y〉

|x〉 =1√2

(

−|R〉 + |L〉)

|y〉 = − i√2

(

|R〉 + |L〉)

and one finds for photons propagating in opposite directions

|Φ〉 =i√2

(

|RR〉 − |LL〉)

Remark: if the two photons propagate in the same direction, as is the case for pairs obtained fromparametric conversion, we have

|Φ〉 = − i√2

(

|RL〉 − |LR〉)

In the two cases, one checks that the Oz component of angular momentum vanishes: (Σ1z +Σ2z)|Φ〉 = 0,where Σz is given by (3.26).

3. Rotational invariance allows us to choose nα along Oz while nβ makes an angle (β − α) with theOx axis. Let us define φ = β − α and the amplitude a++(φ) for finding |Φ〉 in the state

|x⊗ φ〉 = cosφ|xx〉 + sinφ|xy〉

is

a++(φ) =1√2

(

cosφ〈xx| + sinφ〈xy|)(

|xy〉 − |yx〉)

=1√2

sinφ

and thus

p++(φ) =1

2sin2 φ

Using symmetry properties in the exchange + ↔ −

p++(φ) = p−−(φ) =1

2sin2 φ

Since the sum of all probabilities must add up to one

p+−(φ) = p−+(φ) =1

2cos2 φ


This givesE(α, β) = (p++ + p−−) − (p+− − p−+) = − cos 2φ = cos 2(β − α)

In order to get maximal violation of Bell’s inequalities, one must use angles that are half of those usedfor spin 1/2.

4. We find

|Ψ〉 =1√2

(

|θθ〉 + |θ⊥θ⊥〉)

=1√2

(

|RR〉 + |LL〉)

Applying Σz again shows that the Oz component of the angular momentum of this state vanishes.

6.5.9 Two photon interference

1. The dispersion of the Ox component of the wave vector is ∆kx ≃ 1/d, as the vertical position isuncertain by ±d/2.

2. One recovers a standard interference calculation in optics. The difference of optical path for thephoton 1 going through the upper slit is, with θ = l/(2D)

δ(x, y) − δ(0, 0) = − l

2D(x + y) = −θ(x+ y)

and the phase shift

φ(x, y) − φ(0, 0) =2π

λ(δ(x, y) − δ(0, 0)) = −kθ(y + x)

The amplitude for detecting the photon 1 at point y is proportional to

1

2

[

exp

(

2iπ

λ(δ(x, y) − δ(0, 0)

)

+ exp

(

−2iπ

λ(δ(x, y) − δ(0, 0)

)]

= cos kθ(y + x)

3. If the pair is emitted at point x, the probability amplitude for detection in coincidence is obtained bymultiplying the probability amplitudes for detecting photon 1 at y and photon 2 at z

a(x|y, z) = cos[kθ(y + x)] cos[kθ(x+ z)]

4. Since it is not possible to know the vertical position where the photon pair has been emitted in theplate CD, one must add the amplitudes corresponding to all possible positions of this emission

a(y, z) =1

d

∫ d/2

−d/2

dx cos kθ(y + x) cos kθ(z + x)

=1

2d

∫ d/2

−d/2

dx[

cos[kθ((y + z + 2x)] + cos kθ(y − z)]

=1

2d

[

1

kθsin kθd cos[kθ(y + z)] + d cos kθ(y − z)

]

5. If d ≫ 1/(kθ), the second term of a(y, z) dominates over the first one and the probability p(y, z) isobtained by taking the modulus squared of a

p(y, z) ∝ cos2[kθ(y − z)]

Observation of photon 1 at y determines the interference pattern of 2: if one observes photon 2 in

coincidence with photon 1, one will observe an interference pattern. However, if only one photon isobserved, there is no interference: integrating the probability p(y, z) over y, the result is a constant withrespect to the variable z

I(z) =

∫ d/2

−d/2

dy cos2[kθ(y − z)] = cst

[

1 +O

(

1

kθd

)]

33

6. In the limit d≪ 1/(kθ), the probability becomes

p(y, z) ∝ 1

4d2

[

d cos[kθ(y + z)] + d cos[kθ(y − z)]]2

= cos2(kθy) cos2(kθz)

which corresponds to a product of two independent interference patterns. The position at which the pairis emitted is very precisely fixed, so that the angular aperture for each photon is very large. There isno longer any constrain from momentum conservation. Entanglement has no influence: if photon 1, forexample, passes through the upper slit, there is no guarantee that that photon 2 passes through the lowerslit. The spread in transverse momentum is too large to allow one photon to tag the trajectory of theother one. The condition d≪ λ/θ is the condition for a good fringe visibility in a standard Young’s slitexperiment: this condition is complementary to that of trajectory tagging. Conversely, if the emissionposition is very uncertain, momentum conservation allows trajectory tagging.

8. Let r be the amplitude for reflection by the plates S and S′, and t the transmission amplitude. Thesimultaneous detection of the two photons by c et c′ implies that, either the two photons are reflected(the left hand photon taking the upper path and the right hand photon the lower path), or that the twophotons are transmitted. The condition d≫ 1/(kθ) must clearly be verified. The probability amplitudeis obtained by summing over the two paths, because the paths are not distinguishable

a(c, c′) =1√2

[(

reiα)

r + t(

teiβ)]

If r = it and |t| = |r| = 1/√

2

p(c, c′) = |a(c, c′)|2 =1

8

∣

∣− eiα + eiβ∣

∣

2=

1

2sin2 α− β

2

In a similar way, one finds

a(c, d′) =1√2rt(

eiα + eiβ)

p(c, d′) = |a(c, d′)|2 =1

2cos2

α− β

2

The detection of a single photon does not lead to any interference

p(c) = p(c, c′) + p(c, d′) =1

2

A symmetry argument gives at once the other probabilities

p(d, d′) =1

2sin2 α− β

2p(c′, d) =

1

2cos2

α− β

2

whence the quantity E(α, β)

E(α, β) = sin2 α− β

2− cos2

α− β

2= − cos(α− β)

One will thus get a maximal violation of Bell’s inequalities with a choice of angles (α, β) identical to thatof spin 1/2.

6.5.10 Interference of emission times

1. Since the coherence length of the converted photons is small compared to the optical path differencebetween the two arms, one can certainly not observe interference in an individual interferometer. Butthere exists a deeper reason which will be explained at the end of the next question.


2. One may write four different probability amplitudes for the joint detection of photons in D1 et D2

(S = short, L = long)

A = a1Sa

2S B = a1

Sa2L

A′ = eiδ a1La

2L B′ = eiδ a1

La2S

Amplitude aiS corresponds to a photon path (i) going through the shorter arm of the interferometer,aiL to the path which follows the longer arm. Measuring the arrival times of the photons allows one todistinguish between (SL) and (LS); even if the emission time of the pair is unknown, the photon takingthe upper arm arrives earlier than that taking the shorter arm. For example, in the (CL) case, photon 1arrives at D1 0.7 ns before the second photon at D2, which is much larger than the resolution time 0.1 nsof the detectors. On the contrary, the experimental set up does not allow one to distinguish, even inprinciple, between paths (SS) et (LL). We must then add the amplitudes for these paths to obtain theprobability of detection in coincidence

p(D1, D2) = |A+A′|2 = |a1Sa

2S + eiδ a1

La2L|2

which clearly shows a dependence with respect to to δ.

In the experiment described in the statement of the problem, the coincidence window is less than thephoton travel time, which allows one to distinguish paths (LS) and (SL) from paths (LL) et (SS).However, this condition is not essential for observing interference; if it were not realized, one wouldsimply add a background noise

p(D1, D2) = |B|2 + |B′|2 + |A+A′|2

corresponding to the first two terms which do not interfere in the preceding equation. Another importantobservation is that, if one suppresses the beam splitters of the left hand interferometer, one still has aninformation on the travel time: if the left hand photon arrives before the right hand one, we know thatthe photon took the the longer arm. There is thus no dependence with respect to δ and no interference.The information available on the path followed by the photon in the right hand interferometer erases anypossibility of interference, even if the coherence length is smaller than ∆l. In fact, it is not even necessarythat detector D2 be present! It is enough that the information on the arrival time be available in principle,and, as we often emphasized, it is not necessary that the arrival times are effectively observed! As long asthe information on the arrival times is available in principle, and it is available because of entanglement,in no case can we have interference in one individual interferometer.

6.5.11 The Deutsch algorithm

We find for the vector |Ψ〉 defined in Fig. 6.19

|Ψ〉 =(

H |0〉)

⊗(

H |1〉)

=1

2

(

|0〉 + |1〉)

⊗(

|0〉 − |1〉)

=1

2

(

1∑

x=0

|x〉)

⊗(

|0〉 − |1〉)

We apply Uf to this state with the following result:

1. If f(x) = 0, then(

|0〉 − |1〉)

→(

|0〉 − |1〉)

;

2. If f(x) = 1, then(

|0〉 − |1〉)

=(

|1〉 − |0〉)

→ −(

|0〉 − |1〉)

,

or, to summarize,(

|0〉 − |1〉)

→ (−1)f(x)(

|0〉 − |1〉)

. (6.1)

The state Uf |Ψ〉 is then a tensor product (and not an entangled state)

Uf |Ψ〉 =1

2

(

1∑

x=0

(−1)f(x) |x〉)

⊗(

|0〉 − |1〉)

35

The net result for the input register is

|x〉 Uf−−→ (−1)f(x)|x〉

The state of the qubit of the input register then is

|ϕ〉 =1√2

(

(−1)f(0)|0〉 + (−1)f(1)|1〉)

.

Before measuring the input register, we apply a Hadamard gate (see Fig. 6.19):

H |ϕ〉 =1

2

[

(−1)f(0)(

|0〉 + |1〉)

+ (−1)f(1)(

|0〉 − |1〉)]

=1

2

[

(−1)f(0) + (−1)f(1)]

|0〉 +1

2

[

(−1)f(0) − (−1)f(1)]

|1〉.

If measurement of the qubit gives |0〉, then f(0) = f(1), i.e., the function is a ‘constant’ one. If itgives |1〉, then f(0) 6= f(1) and the function is a ‘balanced’ one. The important point is that quantumparallelism has allowed us to bypass the explicit calculation of the function f(x); the measurement of asingle qubit contains the two possible results.


Chapter 7


7.4.3 Canonical commutation relations

1. If we assume that B is a bounded operator, we can define B′ = B/||B||, ||B′|| = 1, and A′ = A||B||without modification of the commutation relations: [A′, B′] = iI. It is thus legitimate to assume that||B|| = 1. Let us use a recursion by assuming that

[B,An] = inAn−1

We then have[B,An+1] = [B,AAn] = A[B,An] + [B,A]An = i(n+ 1)An

which shows the validity of our starting hypothesis. Let us assume that A is bounded, and let ||A|| beits norm. We get, using the inequality which is valid for two operators C et D

||C|| ||D|| ≥ ||CD||

the relation2||An|| ||B|| ≥ ||BAn −AnB|| = n||An−1||

whence||An|| ≥ n

2||An−1||

We deduce the following bounds for ||An||n

2||An−1|| ≤ ||An|| ≤ ||A|| ||An−1|| since ||An|| ≤ ||A|| ||An−1||

and thus ||A|| ≥ n/2. It is impossible for A to be bounded.

2. The problem is that if A is not bounded, the vector B|ϕ〉 does not belong to the domain of A and theproduct AB|ϕ〉 is not defined: we cannot apply Hermitian conjugation and write

〈ϕ|AB|ϕ〉 = 〈Aϕ|Bϕ〉

If the Hilbert space is identified, for example, with L(2)[0, 1] and if B = X , which is bounded on thatspace, while A = AC defined in § 7.2.2, (Xϕ)(x) = xϕ(x) = ψ(x) does not belong to the domain of ACwhich is such that ϕ(1) = Cϕ(0), |C| = 1: the boundary conditions of ψ are

ψ(1) = ϕ(1) = Cϕ(0) while ψ(0) = 0

and ψ(1) 6= Cψ(0). The difficulty is immediately seen in the x representation

∫ 1

0

dxϕ∗(x)

(

i∂

∂xxϕ(x)

)

6=∫ 1

0

(

i∂ϕ(x)

∂x

)∗

xϕ(x)

37


the difference being |ϕ(1)|2.3. The function

ϕ(x) = ei(2πn+α)x C = eiα

obeysACϕ(x) = (2πn+ α)ϕ(x) ϕ(1) = Cϕ(0)

It is thus a normalizable eigenvector with eigenvalue (2πn + α) of AC , which belongs to the domain ofAC . The von Neumann theorem does not apply because AB|ϕ〉 is not defined whatever |ϕ〉 ∈ H, whilethe domain of definition of AB should be dense in H.

Chapter 8


8.5.2 Rotations and SU(2)

1. Let us start from the most general 2 × 2 matrix

U =

(

a bc d

)

and compute U †U which must be identical to I

U †U =

(

|a|2 + |b|2 ac∗ + bd∗

ca∗ + db∗ |c|2 + |d|2)

=

(

1 00 1

)

This gives|a|2 + |b|2 = |c|2 + |d|2 = 1

From c = −b∗d/a∗ and detU = ad− bc = 1 one deduces that d = a∗.

2. To order τU †U = (I + iτ†)(I − iτ) ≃ I − i(τ − τ†)

and the condition U †U = I implies τ = τ†. Moreover, the condition detU = 1 implies Tr τ = 0. Thedecomposition (3.54) together with the condition Tr τ = 0 allows us to write

τ =1

2

3∑

i=1

θiσi

where the angles θi are infinitesimal since τ itself is infinitesimal.

3. Since θ/N is infinitesimal. we may write

Un

(

θ

N

)

= I − iθ

2N(~σ · n)

and using

limN→∞

(

1 − x

N

)N

= e−x

we deduce

Un(θ) = exp

[

−iθ

2(~σ · n)

]

4. The determinant of V is equal to minus the length squared of the vector ~V : detV = −~V 2 and since

det(UVU−1) = detV

39


we obtain ~W 2 = detW = ~V 2, which shows that the transformation preserves the lengths of vectors. It isthus, either a rotation, or a rotation combined with a parity operation.

5. Since W is Hermitian and has zero trace, because

Tr (UVU−1) = TrV

we may write W = ~σ · ~W = ~σ · ~V (θ).

d

dθ~σ · ~V (θ) = − i

2[~σ · n, ~σ · ~V (θ)] = ~σ · (n× ~V (θ))

where we have used (3.52), which shows that ~V (θ) is deduced from ~V through a rotation by an angle θabout n

d~V

dθ= n× ~V (θ)

To any rotation Rn(θ) correspond two matrices Un : Un(θ) and Un(θ + 2π) = −Un(θ).

8.5.4 The Lie algebra of a continuous group

1 Since g(θ = 0) = I, the composition law reads

g(θ)I = g(θ) = g(f(θ, 0) =⇒ fa(θ, 0) = θa

We write an expansion to order θ2 of fa(θ, θ)

fa(θ, θ) = θa + θa + λabcθbθc + λabcθbθc + fabcθbθc +O(θ3, θ2θ, θ θ2, θ

3)

The condition fa(θ, θ = 0) = θa implies λabc = 0 and similarly λabc = 0.

3. One the one hand, we have, neglecting terms of order (θ3, θ2θ, θ θ2, θ

3)

U(θ)U(θ) = I − iθaTa − iθaTa −1

2(θbθc + θbθc)Tbc − θaθbTaTb

and, on the other hand

U(f(θ, θ)) = I − iTa(θa + θa + fabcθbθc) −1

2(θb + θb)(θc + θc)Tbc

Relabeling the summation indices and taking into account the symmetry property Tbc = Tcb, because

Tbc = − ∂2U

∂θa∂θb

∣

∣

∣

θa=θb=0

the comparison between the two expressions gives

θbθcTbTc = ifabcTaθbθc + θbθcTbc

We deduce from this

TbTc = Tbc + ifabcTa

TcTb = Tbc + ifacbTa

and, subtracting the second equation from the first one

[Tb, Tc] = i[fabc − facb]Ta

The structure constant Cabc isCabc = [fabc − facb] = −Cacb

41

8.5.5 The Thomas-Reiche-Kuhn sum rule

1. From the general relation (see(8.41))

[X, f(P )] = i~∂f

∂P

we deduce[P 2, X ] = −2i~P and [[P 2, X ], X ] = −2i[P,X ] = −2~

2

whence[

P 2

2m+ V (X), X

]

= [H,X ] = − i~

mP [[H,X ], X ] = −~

2

m

2. On the other hand, the commutator is also expressed as

[[H,X ], X ] = HX2 − 2XHX +X2H

and using 〈ϕn|H |ϕm〉 = Enδnm

〈ϕ0|HX2|ϕ0〉 =∑

n,m

〈ϕ0|H |ϕn〉〈ϕn|X |ϕm〉〈ϕm|X |ϕ0〉 = E0

∑

n

|Xn0|2

〈ϕ0|XHX |ϕ0〉 =∑

n,m

〈ϕ0|X |ϕn〉〈ϕn|H |ϕm〉〈ϕm|X |ϕ0〉 =∑

n

En|Xn0|2

whence the result∑

n

2m|Xn0|2~2

(En − E0) = 1


Chapter 9


9.7.2 Wave packet spreading

1.

[P 2, X ] = P [P,X ] + [P,X ]P = −2i~P

One can also use

[f(P ), X ] = −i~f ′(P )

2. From the Ehrenfest theorem (4.26), choosing A = X2

d

dt〈X2〉(t) =

i

~〈[H,X2]〉 =

i

~

⟨[ P 2

2m+ V (X), X2

]⟩

=i

2~m〈[P 2, X2]〉

Furthermore

[P 2, X2] = −2i~(XP + PX)

from which we derive the final result

d

dt〈X2〉(t) =

1

m〈XP + PX〉

Going to the x representation

〈PX〉 =

∫

dxϕ∗(x)

[

−i∂

∂x(xϕ(x)

]

= i

∫

dxxϕ(x)∂ϕ∗(x)

∂x

where the second expression is obtained from an integration by parts. Combining with

〈XP 〉 =

∫

dxϕ∗(x)x

[

−i∂ϕ(x)

∂x

]

we finally getd

dt〈X2〉(t) =

i~

m

∫ ∞

−∞

dxx

[

ϕ∂ϕ∗

∂x− ϕ∗ ∂ϕ

∂x

]

These results are valid for a particle in a non zero potential, and not only for a free particle.

3. On the contrary, the results that follow are only valid for a free particle, V (X) = 0. The Hamiltonianis then reduced to the kinetic Hamiltonian H = K = P 2/(2m). We compute the second derivative of〈X2〉(t)

d2

dt2〈X2〉(t) =

i

~

d

dt〈[K,X2]〉 = − 1

~2

d

dt〈[K, [K,X2]]〉

43


where we have used Ehrenfest’s theorem twice. Taking into account

[P 2, XP + PX ] = −4i~P 2

we findd2

dt2〈X2〉(t) =

2

m〈P 2〉

The third derivative of 〈X2〉(t) and higher order derivatives vanish

dn

dtn〈X2〉(t) = 0 n ≥ 3

because [K, [K,X2]] ∝ P 2 and [K,P 2] = 0. 〈X2〉(t) is thus a second order polynomial in t

〈X2〉(t) = 〈X2〉(t = 0) + td

dt〈X2〉(t)

∣

∣

∣

t=0+

1

2t2

d2

dt2〈X2〉(t)

∣

∣

∣

t=0

In order to compute the dispersion, we use for a free particle

∆x2(t) = 〈X2〉(t) − [〈X〉(t)]2

andd

dt〈X〉(t) =

i

~[K,X ] =

⟨P

m

⟩

which gives

〈X〉(t) = 〈X〉(t = 0) + t⟨P

m

⟩

Indeed we have just seen that derivatives of order ≥ 2 vanish.

9.7.3 A Gaussian wave packet

Let us recall two results on Gaussian integrals

∫ +∞

−∞

dx e−α2x2

=

√π

α∆x =

1√2α

1. Setting k′ = k − k∫

dk|A(k)|2 =

(

1

πσ2

)1/2 ∫

dk′ exp

(

−k′2

σ2

)

= 1

so that ∆k = σ/√

2. Let us compute the wave function at time t = 0

ϕ(x, 0) =

(

1

πσ2

)1/4

eikx

∫

dk′√2π

exp

(

ik′x− k′2

2σ2

)

=σ1/2

π1/4exp

[

ikx− 1

2σ2 x2

]

The modulus squared of the wave function is

|ϕ(x, t)|2 =σ√π

e−σ2x2

which is indeed normalized to one with ∆x = 1/(√

2σ), and thus

∆x∆k =1

2

45

2. Let us start from the expression of ϕ(x, t)

ϕ(x, t) =

(

1

πσ2

)1/4 ∫dk√2π

exp

(

ikx− i~k2

2mt− (k − k)

2

2σ2

)

The exponent is rewritten, within a factor of i and with k′ = k − k

kx− ~k2

2mt− (k − k)

2

2σ2= kx+ k′x− ~k

2

2mt− ~kk′

mt− ~k′

2

2mt− k′

2

σ2

ϕ(x, t) reads

ϕ(x, t) =

(

1

πσ2

)1/4

exp

(

ikx− i~k

2

2mt

)

∫

dk′√2π

exp

[

ik′(

x− ~k

2mt

)]

exp

[

−k′2

2

(

1

σ2+

i~t

2m

)

]

If we can neglect the term i~t/2m in the last exponential, we simply obtain

ϕ(x, t) =

(

1

πσ2

)1/4

exp

(

ikx− i~k

2

2mt

)

exp

[

−σ2

2

(

x− ~k

mt

)2]

= exp

(

i~k

2

2mt

)

ϕ(x− vgt, 0) = eiω(k)tϕ(x− vgt, 0)

3. In the general case, we set1

σ′2=

1

σ2+

i~t

m

and we find, after doing the integration

ϕ(x, t) =

(

1

πσ2

)1/4

σ′ exp

(

−1

2σ′2(x− vgt)

2

)

exp[

i(kx− ω(k)t)]

Taking the modulus squared

|ϕ(x, t)|2 =

(

1

πσ2

)1/2

|σ′|2 exp(

−Reσ′2(x− vgt)2)

The peak of |ϕ(x, t)|2 is centered at x = vgt and has a width

∆x2(t) =1

2 Reσ′2

that is,

∆x2(t) =1

2σ2

(

1 +~

2σ4t2

m2

)

The width of the wave packet increases with time, because of the term ~2σ4t2/m2 in the bracket.

4. ∆x2(t) doubles for

t =m

~σ2=

2m∆x2(t = 0)

~= 3.2 × 10−11 s

9.7.7 A delta-function potential

1. The dimension of δ(x) is L−1 and if the dimension of g is the inverse of a length, then the dimensionof V (x) is

M2L4T −2L−1L−1 = M2L2T −2


which is indeed the dimension of an energy.

2. We integrate the Schrodinger written in the form

[

d2

dx2+

2mE

~2

]

ϕ(x) = g δ(x)ϕ(x)

between x = −ε and x = +ε

∫ +ε

−ε

ϕ′′(x)dx = ϕ′(−ε) − ϕ′(−ε) = gϕ(0)

The function ϕ(x) is continuous at x = 0 but its derivative ϕ′(x) is not. In the case of a bound state, wemust have

ϕ(x) = Ae−κ|x| for x 6= 0

so thatϕ′(0+) − ϕ′(0−) = −2κA = −|g|ϕ(0) = −|g|A

whence 2κ = |g| and E

E = −~2κ2

2m= −~

2g2

8m

There is no odd solution because ϕ(0) = 0 for an odd solution. Let us retrieve the result by taking thelimit a→ 0, V0a→ ~

2g/(2m) of the square well potential whose energy levels are given by (9.82)

κ = k tanka

2

which leads to

k ≃√

2m|V0|~

tanka

2≃ tan

√

2m|V0|a2~

→ 0

and

κ ≃√

2m|V0|~

√

2m|V0| a2~

=m|V0|a

~2=m

~2

~2|g|2m

=1

2|g|

3. Since the diatomic molecule potential is even, one looks for even and odd solutions. For the evensolutions we have

x < −l : ϕ(x) = eκx − l < x < l : ϕ(x) = A coshκx x > l : ϕ(x) = e−κx

The continuity of ϕ at x = l givesA coshκl = e−κl

and the continuity of its derivative at the same point

−κe−κl −Aκ sinhκl = −|g|e−κl

We remark thatA sinhκl = A coshκl tanhκl = e−κl tanhκl

whence the equation for the energy eigenvalue

(1 + tanhκl) =|g|κ

The solution is unique, and is given by the intersection of the curves (1 + tanhκl) and g/κ drawn asfunctions of κ. One can rewrite the value of κ as

κ =|g|2

(

1 + e−2κl)

47

Let us now look for the odd solutions, which have the following form

x < −l : ϕ(x) = −eκx − l < x < l : ϕ(x) = A sinhκx x > l : ϕ(x) = e−κx

The condition of continuity for ϕ(x) and the condition on its derivative at x = l are now

A sinhκl = e−κl

−κe−κl −Aκ coshκl = −|g|e−κl

which leads to

κ =|g|2

(

1 − e−2κl)

This equation has a unique solution if the derivative of |g|[1 − exp(−2κl)]/2 > κ at κ = 0, that is, if|g|l > 1. There is no odd solution if |g|l < 1.

4. Let us consider two deep and narrow potential wells of width a separated by a distance l, with a≪ l.The two wells can then be approximated by delta-functions, which leads us back to the potential ofthe preceding question, and we assume κl ≫ 1. Then there exist two bound states, one with an evenwave-function corresponding to

κ+ =|g|2

(1 + e−2κl)

and the other one with an odd wave-function corresponding to

κ− =|g|2

(1 − e−2κl)

The energy difference between the two states is then

E− − E+ = − ~2

2m(κ2

− − κ2+) =

~2

2mg2e−2κl

the average enrgy E0 being

E0 =1

2(E+ + E−) ≃ −~

2g2

8m

We can thus write

E+ ≃ −~2g2

8m

(

1 + 2e−2κl)

E− ≃ −~2g2

8m

(

1 − 2e−2κl)

These are the eigenvalues of a two-level Hamiltonian

H = −~2g2

8m

(

1 −2e−2κl

−2e−2κl 1

)

From (9.100) and (9.105), the tunneling transmission coefficient of a particle with energy ≃ 0 through abarrier of height V0 and of width 2l is

T ≃ e−4κl

and the nondiagonal elements of the Hamiltonian are ∝√T .

5. As in § 9.4.1, we write the wave functions by selecting the case F = 1, G = 0. The continuity conditionat x = 0 gives

A+B = 1

while the condition on the derivative is

ik − ik(A−B) = g


We derive

A = 1 +ig

2kB = − ig

2k

whence the matrix elements of the transmission matrix

M11 = 1 +ig

2kM12 =

ig

2k

We check that the results do agree with the limit a→ 0, V0a→ ~2g/(2m) of equations (9.96) and (9.97).

6. The condition

ϕq(x) = eiqlϕq(x− l)

gives

F eikx +Ge−ikx = eiql(

Aeik(x−l) +Beik(x+l))

that is

F = A ei(q−k) G = B ei(q+l)

The continuity conditions on ϕq(x) and on its derivative read

A[

1 − ei(q−k)l]

+B[

1 − ei(q+k)l]

= 0

A[

g + ik(

1 − ei(q−k)l)]

+B[

g − ik(

1 − ei(q+k)l)]

= 0

We thus obtain a system of equations with two unknowns A et B whose discriminant ∆ must be zero fora non trivial solution. Setting α = (q − k)l et β = (q + k)l

∆ = det

(

1 − eiα 1 − eiβ

g + ik(

1 − eiα)

g − ik(

1 − eiβ)

)

= 0

Computing ∆ gives

∆ = g(

eiβ − eiα)

− 2ik(

1 − eiα − eiβ + ei(α+β))

= 0

and multiplying by exp(−iql)

2ig sin kl− 4ik(cos ql − cos kl) = 0

One thus recovers (9.126)

cos ql = cos kl +g

2ksinkl

9.7.13 Study of the Stern-Gerlach experiment

1. Since the yOz plane is symmetry plane of the problem, Bz must be an even function of x and we musthave

∂Bz∂x

∣

∣

∣

x=0= 0

Translation invariance along Oy gives∂Bz∂y

∣

∣

∣

x=0= 0

The two nonzero components of the magnetic field in the vicinity of x = 0 are

Bx = −bx Bz = B0 + bz

This field obeys the two Maxwell equations (1.8) and (1.9) in vacuum ~∇× ~B = 0 and

~∇ · ~B =∂Bx∂x

+∂Bz∂z

= −b+ b = 0

49

The potential energy is

−~µ · ~B = −µxBx − µzBz = bµxx− bµzz

whence the force ~F with components

Fx =∂(−~µ · ~B)

∂x= bµx Fz =

∂(−~µ · ~B)

∂z= −bµz

The B0z component of the magnetic field leads to Larmor precession of the spin about the Oz axis(§ 3.25)in which µz stays constant. By contrast, due to this precession, the average value of µxvanishes: 〈µx〉 = 0,and the force along Ox averages to zero, if the transit time ≫ 1/ω, as the spin makes a large number ofrotations about Oz.

2. The force on the magnetic moment is vertical and constant; its value is F = ±µb for a spin lying along±z. The splitting between the trajectories of an up spin and a down spin at the exit of the magnet gapis then

δ = 21

2

F

mt2 =

F

m

(

L

v

)2

=µb

m

(

L

v

)2

Let us also evaluate the product ∆z∆pz

∆z∆pz ∼(

10−4) (

1.8 × 10−25)

(10) = 1.8 × 10−28 MKSA ∼ 106~

The description by classical trajectories is legitimate.

3. The potential energy of an up spin (down spin) is −µbz, and the Schrodinger equations for ϕ± are

i~∂ϕ±

∂t=

(

− ~2

2m∇2 ∓ µbz

)

ϕ± = Hϕ±

Using Ehrenfest’s theorem (4.26), we obtain

d

dt〈~R±〉(t) =

i

~[H, ~R±] =

1

m〈~P±〉

d

dt〈Px,y,±〉(t) =

i

~[H,Px,y,±] = 0

d

dt〈Pz,±〉(t) =

i

~[H,Pz,±] = ±µb

This last result is deduced from

∓[µbZ, Pz] = ∓i~µb

We finally get

〈Z±〉 = ± µb

2mt2

and the center of the wave packet does follow the classical trajectory.

4. Let us make a reflection with respect to the xOy plane. In this reflection, ~µ does not change, becausethe orientation of a current loop lying in the xOy plane stays unchanged. In the same operation, ~Bchanges its direction, but not its gradient, and thus ~∇B lies along −x. However, the image of thetrajectory in the mirror always leave in the +x direction, and then the image of the trajectory in themirror does not represent a physically allowed motion, unless there is no deviation of the trajectory.

The von Neumann measurement model

1. From (4.17), the evolution operator U(t, t0) obeys

i~d

dtU(t, t0) = [g(t)AP ]U(t, t0)


which integrates into

U(t, t0) = exp

(

− i

~AP

∫ t

t0

g(t′)dt′)

Between times ti and tf we thus have

U(tf , ti) ≃ exp

(

− i

~AP

∫ +∞

−∞

g(t′)dt′)

= exp

(

− i

~gAP

)

2. The action of U(tf , ti) on the vector |n⊗ ϕ〉 is

U(tf , ti)|n⊗ ϕ〉 = e−iganP/~|n⊗ ϕ〉

Furthermore, exp(−iganP/~) is a translation operator by gan and from (9.13)

(

e−iganP/~ϕ)

(x) = ϕ(x− gan)

3. Because of the linearity property of quantum mechanics, the final state vector is

|χf 〉 =∑

n

cn|n⊗ ϕn〉

The reduced state operator of S is from (6.66)

ρ(1) =∑

n,m

cnc∗m|m〉〈n|〈ϕm|ϕn〉 =

∑

n

|cn|2|n〉〈n|

because 〈ϕm|ϕn〉 = δnm. The system S is thus a statistical mixture of states |n〉 with a weight |cn|2, andthe probability to observe S in the state |n〉 is |cn|2.

Chapter 10

Exercises from chapter 10

We remind the reader that we use in chapter 10 a system of units where ~ = 1

10.7.5 Orbital angular momentum

1. From the expression of the orbital angular momentum operator as a function of the position ~R andmomentum ~P operators, we have for the commutator [Lx, Ly]

[Lx, Ly] = [Y Pz − ZPy, ZPx −XPz]

= [Y Pz , ZPx] + [ZPy, XPz]

= i[−Y Px +XPy] = iLz

2. Let us start from equation (cf. (10.40))

(

e−iαLxf)

(~r) = f(

R−1x [α](~r)

)

= f (Rx[−α](~r))

where Rx[α] is a rotation by angle α about Ox. We choose α to be infinitesimal, α → α + dα; in arotation of angle −α about Ox

y′ = z sinα+ y cosα dy = zdα

z′ = z cosα− y sinα dz = −ydα

In this rotation, θ → θ + dθ and φ→ φ+ dφ, which are determined through

dy = r cos θ sinφdθ + r sin θ cosφdφ

dz = −r sin θ dθ

which leads to

dθ = sinφdα dφ =cosφ

tan θdα

This allows us too identify Lx

[−idαLxf ](~r) = dα

(

sinφ∂

∂θ+

cosφ

tan θ

∂

∂φ

)

f(~r)

that is

Lx = i

(

sinφ∂

∂θ+

cosφ

tan θ

∂

∂φ

)

To compute Ly we can use the same method, or make use of the commutation relation

Ly = −i[Lz, Lx] = −[

∂

∂φ, Lx

]

51


Taking into account

[

∂

∂φ, f(φ)

]

= f ′(φ)

[

∂

∂φ, f(φ)

∂

∂φ

]

= f ′(φ)∂

∂φ

we obtain

Ly = i

(

− cosφ∂

∂θ+

sinφ

tan θ

∂

∂φ

)

3. The operator Lz = −i∂/∂φ is defined on the periodic functions f(φ) = f(φ+ 2π) and it is selfadjointon this domain (see § 7.2.2). By contrast, the function φf(φ) is not periodic and does not belong tothe domain of Lz.Therefore, we cannot define φLz and write a commutation relation between φ and Lz.One cannot either use the method of Exercise 9.7.1 because the integration bounds contribute to theintegration by parts.

10.7.6 Relation between the rotation matrices and spherical harmonics

1. The function f(0, 0, z) is invariant under rotations about Oz. The action of exp(−iαLz) on f(0, 0, z)is equivalent to the identity operation, and thus Lzf(0, 0, z) = 0. In classical mechanics

lz = xpy − ypx

and lz = 0 if x = y = 0, that is, if the particle trajectory follows the z axis.

By definition

〈lm|θ, φ〉 = Y ml∗(θ, φ)

but we also have

〈lm|θ, φ〉 = 〈lm|e−iφLz e−iθLy |θ = 0, φ = 0〉=

∑

l′,m′

〈lm|e−iφLz e−iθLy |l′m′〉〈l′m′|θ = 0, φ = 0〉

and introducing a complete set of states

∑

l′,m′

|l′m′〉〈l′m′| = I

From the result of question 1

〈l′m′|θ = 0, φ = 0〉 ∝ δm′0

and from (10.32)

〈lm|e−iφLz e−iθLy |l′, 0〉 = δl′lD(l)m0(θ, φ)

In fact, we can easily obtain the proportionality coefficient because

〈l′m′|θ = 0, φ = 0〉 = δm′0Y0l (θ = 0, φ = 0) =

√

2l + 1

4π

using (10.61) and the property Pl(1) = 1.

10.7.8 The spherical well

1. Setting E = −B and (~ = 1)

k =√

2m(V0 −B) κ =√

2mB

53

we write the radial wave function in the s-wave

r < R : u(r) = A sin kr r > R : u(r) = Be−κr

The continuity of the logarithmic derivative leads to the relation

k cot kR = −κ

As in § 9.3.3, we define U = 2mV0 and κ2 = U − k2. The eigenvalue equation becomes

cot kR = −√U − k2

k

and its solutions are given by Figure 9.12 using the odd solutions of the one-dimensional square well(dotted lines). Solutions exist only if kR > π/2.

2. Assuming the deuteron binding energy B ≪ V0, taking as the reduced mass half of the proton massmp/2 and writing ~ explicitly

mpV0R2

~2=π2

4

One finds the numerical value V0 ≃ 100 MeV ≫ B

3. The radial wave equation is

[

− 1

2m

d2

dr2+A

r2− B

r

]

u(r) = Eu(r)

which is analogous to that (10.86) for the hydrogen atom with l(l + 1)/(2m) → A and B → e2.

10.7.13 Light scattering

1. If the photon is emitted along the Oz axis with a right handed (R) or left handed (L) polarization,angular momentum conservation allows only two nonzero amplitudes

a = 〈R, θ = 0|T |j = 1,m = 1〉 a′ = 〈L, θ = 0|T |j = 1,m = −1〉

If the transition is of the electric dipole kind, we have seen in § 10.5.2 that a = a′ (with our phaseconventions). We find, using the rotational invariance of the transition matrix, [U(R), T ] = 0

am=1R (θ) = 〈R, θ|T |j = 1,m = 1〉 = 〈R, θ = 0|T U †[Ry(θ)]|j = 1,m = 1〉

= 〈R, θ = 0|T |j = 1,m = 1〉〈j = 1,m = 1|U †[Ry(θ)]|j = 1,m = 1〉

= ad(1)11 (θ) =

1

2a(1 + cos θ)

Similarly we find

am=1L (θ) = ad

(1)1,−1(θ) =

1

2a(1 − cos θ)

If the photon is emitted in the direction n = (θ, φ), we must use the rotation operator U [R(θ, φ)] and weget

am=1R (θ, φ) =

1

2a(1 + cos θ)eiφ

am=1L (θ, φ) =

1

2a(1 − cos θ)eiφ

2. If the atom absorbs the photon, the two nonzero amplitudes are, from angular momentum conservation

b = 〈j = 1,m = 1|T ′|R〉 b′ = 〈j = 1,m = −1|T ′|L〉


If the transition is of the electric dipole kind, b′ = b from the results of § 10.5.2. Introducing a completeset of intermediate states

cP→P ′(θ) = cP→(jm)c(jm)→P ′(θ) = 〈P ′, θ|S|P 〉 =∑

m

〈P ′, θ|T |1m〉〈1m|T ′|P 〉

〈R, θ|S|R〉 = 〈R, θ|T |j = 1,m = 1〉〈j = 1,m = 1|T ′|R〉 =1

2ab(1 + cos θ)

〈R, θ|S|L〉 = 〈R, θ|T |j = 1,m = −1〉〈j = 1,m = −1|T ′|L〉 =1

2ab(1 − cos θ)

〈L, θ|S|R〉 = 〈L, θ|T |j = 1,m = 1〉〈j = 1,m = 1|T ′|R〉 =1

2ab(1 − cos θ)

〈L, θ|S|L〉 = 〈L, θ|T |j = 1,m = −1〉〈j = 1,m = −1|T ′|L〉 =1

2ab(1 + cos θ)

In the two possible initial cases, |R〉 and |L〉, the angular distribution is

W (θ) =1

2|a|2|b|2(1 + cos2 θ)

If the initial photon is polarized along Ox we find

〈x, θ|S|x〉 = ab cos θ

〈y, θ|S|x〉 = 0

In a classical model of photon scattering by an electric charge, a photon with linear polarization alongOx sets the charge into motion along this axis, and the charge radiates an electromagnetic wave polarizedalong Ox with an angular distribution ∝ cos2 θ.

If the photon is emitted in the direction n = (θ, φ), we find

〈R; θ, φ|S|R〉 =1

2ab(1 + cos θ) eiφ

〈R; θ, φ|S|L〉 =1

2ab(1 − cos θ) e−iφ

〈L; θ, φ|S|R〉 =1

2ab(1 − cos θ) eiφ

〈L; θ, φ|S|L〉 =1

2ab(1 + cos θ) e−iφ

If the initial photon is polarized along Ox, we obtain the following amplitudes

〈x; θ, φ|S|x〉 = ab cos θ cosφ

〈y; θ, φ|S|x〉 = ab sinφ

These results can be understood by observing that the cosine of the angle between the initial and finalpolarizations is cos θ cosφ for a final polarization along Ox and sinφ for a final polarization along Oy.

10.7.14 Measurement of the Λ0 magnetic moment

1. Angular momentum conservation along Oz implies m′ = m, since the z component of the angularmomentum vanishes

〈θ = 0,m′|T |m〉 ∝ δmm′

The amplitude b is obtained from a through a reflection with respect to a plane xOz: |a| = |b| if parityis conserved.

55

2. If the proton is emitted along a direction making an angle θ with the Oz axis in the xOz plane, wecan compute the decay amplitude using the rotation operator U [Ry(θ)] by an angle θ about Oy

〈θ,m′|T |m〉 = 〈θ = 0,m′|U †[Ry(θ)]T |m〉=

∑

m′′

〈θ = 0,m′|T |m′′〉〈m′′|eiθJy |m〉

= am′,m′(θ = 0)d(1/2)m′m (θ)

With the definitions of question 1

a 12, 12

= a a− 12,− 1

2= b

and taking (10.38) into account

a++(θ) = ad(1/2)12, 12

(θ) = a cosθ

2a−+(θ) = ad

(1/2)12,− 1

2

(θ) = −b sinθ

2

3. If the Λ0 particle is produced in a state m = 1/2, the angular distribution w(θ) is as follows (becausethe final states m′ = 1/2 et m′ = −1/2 are distinguishable)

w(θ) = |a|2 cos2θ

2+ |b|2 sin2 θ

2

=1

2(|a|2 + |b|2) +

1

2(|a|2 − |b|2) cos θ

whence

w0 =1

2(|a|2 + |b|2) α =

|a|2 − |b|2|a|2 + |b|2

Were parity conserved, we would have |a|2 = |b|2 and α = 0. Observation of a cos θ term in the angulardistribution of the decay is thus a proof that parity conservation is violated.

4. ~p × ~pΛ0 is the only available pseudovector, and 〈~S〉, which is a pseudovector, must necessarily beoriented along this direction : 〈Sx〉 = 〈Sy〉 = 0.

5. The spin Hamiltonian in the magnetic field is

H = −~µ · ~B = −γ~S · ~B

The proton spin precesses around ~B with an angular velocity ω = γB in the xOz plane. At time t = τ ,it will have rotated by an angle

λ = ωτ = γBτ

The decay angle must be measured from the direction n in the xOz plane

n = x sinλ+ z cosλ

The proton is emitted in the direction

p = x sin θ cosφ+ y sin θ sinφ+ z cos θ

and

cosΘ = p · n = cos θ cosλ+ sin θ sinλ cosφ

The measurement of the angular distribution allows us to infer the direction of n (or the angle λ) and todeduce γ from λ = γBτ .

10.7.15 Production and decay of the ρ+ meson


1. Calculation of am(θ, φ) : R(θ, φ) is the rotation (10.30) which brings the Oz axis along the directionof emission for the π+ meson

am(θ, φ) = 〈θ, φ|T |m〉 = 〈θ = 0, φ = 0|U †[R(θ, φ)]T |m〉=

∑

m′

〈θ = 0, φ = 0|T |m′〉〈m′|U †[R(θ, φ)]|m〉

where we have used the rotational invariance of the transition matrix [U, T ] = 0. From angular momentumconservation

〈θ = 0, φ = 0|T |m′〉 ∝ δm′ 0 = aδm′ 0

because if π+ meson is emitted in the Oz direction, its angular momentum along this direction vanishes.Furthermore

〈m′ = 0|U †[R(θ, φ)]|m〉 = 〈m|U [R(θ, φ)]|m′ = 0〉∗ =[

D(1)m0(θ, φ)

]∗

= eimφ d(1)m0(θ)

One obtains the different decay amplitudes using (10.39)

a1(θ, φ) = a eiφd(1)10 (θ) = − a√

2eiφ sin θ

a0(θ, φ) = a d(1)00 (θ) = a cos θ

a−1(θ, φ) = a e−iφd(1)−10(θ) =

a√2

e−iφ sin θ

whence the angular distributions

W1 = W−1 =|a|22

sin2 θ W0 = |a|2 cos2 θ

One observes that W1 +W0 +W−1 = |a|2: as a consequence, if the ρ meson is not polarized, the angulardistribution is isotropic, which must be the case as there is no privileged direction. In what follows, Wwill be normalized as

W1 +W0 +W−1 = |a|2 = 1

2. If the initial state vector is given by

|λ〉 =∑

m=−1,0,1

cm|1m〉∑

m=−1,0,1

|cm|2 = 1

the decay amplitude aλ will be

aλ(θ, φ) = 〈θ, φ|T |λ〉 =∑

m

〈θ, φ|T |1m〉〈1m|λ〉 =∑

m

cmam(θ, φ)

and the angular distribution

Wλ(θ, φ) =∑

m,m′

cmc∗m′am(θ, φ)a∗m′ (θ, φ)

The explicit expression of Wλ is given in the next section.

3. For every component of the statistical mixture pλ, the angular distribution is

Wλ =∑

m,m′

c(λ)m c∗

(λ)m′ am(θ, φ)a∗m′ (θ, φ)

and the angular distribution corresponding to the state operator ρ is

Wρ(θ) =∑

λ

pλWλ =∑

λ,m,m′

pλc(λ)m c∗

(λ)m′ am(θ, φ)a∗m′ (θ, φ) =

∑

m,m′

ρmm′am(θ, φ)a∗m′ (θ, φ)

57

The result is simplified by observing that ρmm′ = ρ∗m′m

m 6= m′ ρmm′ama∗m′ + ρm′mam′a∗m = 2Re (ρmm′ama

∗m′)

For example

Re (ρ10a1a∗0) = −Re

(

ρ10√2

eiφ sin θ cos θ

)

The final result is

Wρ(θ, φ) = ρ00 cos2 θ +1

2sin2 θ(ρ11 + ρ−1,−1)

+1√2

sin 2θRe(

e−iφ ρ−10 − e iφ ρ10

)

− sin2 θRe(

e 2iφ ρ1,−1

)

To obtain the angular distribution of question 2, it is enough to make the sustitutions ρ11 → |c1|2,ρ10 → c1c

∗0 etc.

4. The only available pseudovector is n, and 〈 ~J〉, which is a pseudovector, lies necessarily along n. Usingthe expression (10.24) of Jx and Jy for j = 1 one finds

Tr ρJx = 2(Re ρ10 + Re ρ0,−1) = 0

Tr ρJy = 2(Im ρ10 + Im ρ0,−1) = 0

that is, ρ10 + ρ0,−1 = 0. In the operation Z, which is a reflection with respect to the xOy plane, thereaction kinematics is unchanged, and since the target is unpolarized and parity is conserved, the reactionis identical to its image in the xOz plane. We must then have

[ρ,Z] = 0 ou Z−1ρZ = ρ

Furthermore, we useΠ|1m〉 = η|1m〉

where η is the ρ meson intrinsic parity (η = −1 because the ρ meson is, as the photon, a vector meson).We then derive

〈1m|ΠeiπJz ρ e−iπJzΠ|1m′〉 = |η|2e−iπ(m′−m)ρmm′

that isρmm′ = (−1)m−m′

ρmm′

10.7.17 Σ0 decay

1. If the photon is emitted along the Oz axis, the component along this axis of the orbital angularmomentum vanishes, and angular momentum conservation holds for the amplitudes a and b, but not forc and d, which must then vanish.

2. The amplitudes a et b are deduced from each other through a reflection with respect to the xOz plane,and if parity is conserved in the decay, then |a| = |b|. The reflection operator Y (10.100) acts in thefollowing way on the photon states which are odd under parity, ηγ = −1 (see (10.104))

Y|R〉 = −|L〉 Y|L〉 = −|R〉

while for the Σ0 and Λ0 particles (see (10.102))

Y|jm〉 = ηΣ(−1)1/2−m|j,−m〉Y|jm′〉 = ηΛ(−1)1/2−m

′ |j,−m′〉

and since m′ = −m, we obtain an overall factor

ηΣηΛηγ(−1)1−m−m′

= ηΣηΛ = η


We can also use directly (10.119)ησηληγ(−1)jΣ−jΛ−jγ = η

The transition is of the magnetic dipole kind, since the parities of the initial and final states are the same.

3. If ~p is the photon momentum, the emission amplitude for the Λ0 particle in the direction −p with aspin projection m′ on p for a right handed polarized photon is

am′

R (θ) = 〈R,m′; θ|T |m = 1/2〉= 〈R,m′; θ = 0|U †[Ry(θ)]|m = 1/2〉 =

∑

m′′

〈R,m′; θ = 0|T |m′′〉〈m′′|U †[Ry(θ)]|1/2〉

= a d(1/2)12

12

(θ) = a cosθ

2

because it is only the value m′′ = 1/2 which gives a nonzero contribution. We then have m′ = −1/2. Ananalogous calculation gives for a left handed polarized photon

am′

L (θ) = b d(1/2)12,− 1

2

(θ) = −a sinθ

2

Only the amplitudes am′=−1/2R (θ) and a

m′=1/2L (θ) are nonzero.

Chapter 11


11.5.2 Mathematical properties

1. We use a recursion method assuming that [N, ap] = −pap

[N, ap+1] = [N, aap] = [N, a]ap + a[N, ap] = −(p+ 1)ap+1

Let us consider a monomial1 in a and a†, P = (a†)qap and let us compute its commutator with N

[N, (a†)qap] = [N, (a†)q]ap + (a†)q[N, ap] = (q − p)(a†)qap

which vanishes only if p = q. As any function of a and a† is a sum of such monomials, using if necessarythe commutation relation [a, a†] = I to put the creation and annihilation operators in a suitable order,we see that the only possibility to have a non vanishing commutator is that this function be a sum ofterms of the form (a†)pap. Any monomial of the form (a†)pap can be written as a function of a†a usingthe commutation relations (11.8). If an operator A commutes with N , it is necessarily a function of N :A = f(N). There does not exist an operator independent of N and commuting with N , so that

〈n′|A|n〉 = 〈n′|f(N)|n〉 = f(n)δnn′

2. Let |ϕ〉 be a vector orthogonal to all vectors |n〉, 〈ϕ|n〉 = 0 ∀n

P ′|n〉 = |n〉 P ′|ϕ〉 = 0

Let us show that [P ′, a] = 0. It is clear that

〈n|[P ′, a]|n〉 = 0 〈ϕ|[P ′, a]|ϕ〉 = 0

Let us examine 〈ϕ|[P ′, a]|n〉 and 〈ϕ|[P ′, a†]|n〉

〈ϕ|P ′a|n〉 = 〈ϕ|P ′a†|n〉 = 0

〈ϕ|aP ′|n〉 = 〈ϕ|a|n〉 =√n 〈ϕ|n− 1〉 = 0

〈n|a†P ′|n〉 = 〈n|a†|ϕ〉 =√n+ 1 〈ϕ|n+ 1〉 = 0

and thus [P ′, a] = 0 and [P ′, a†] = 0. The projector P ′ then commutes with a and a†, and from vonNeumann’s theorem, it is a multiple of the identity operator, that is, either P ′ = I, or P ′ = 0. Thesecond possibility being excluded, we are left with P ′ = I.

11.5.3 Coherent states1When we write a combination of a and a

† with all the a to the right and all a† to the left, we say that this combination

has been written in normal from.

59


2. Using Ehrenfest’s theorem (4.26), we get

i~d

dt〈a〉(t) = 〈[a,H ]〉 = ~ω〈a〉(t)

because [a,H ] = ~ωa. If the initial condition is

〈a〉(t = 0) = 〈ϕ(0)|a|ϕ(0)〉 = z0

the solution of the preceding differential equation is

〈a〉(t) = z0e−iωt

3. In addition, we want that 〈ϕ(0)|a†a|ϕ(0)〉 = |z0|2, which implies, with b(z0) = a− z0

〈ϕ(0)|b†(z0)b(z0)|ϕ(0)〉 = 〈ϕ(0)|a†a|ϕ(0)〉 − z0〈ϕ(0)|a†|ϕ(0)〉 − z∗0〈ϕ(0)|a|ϕ(0)〉 + |z0|2= 〈ϕ(0)|(a†a− |z0|2)|ϕ(0)〉 = ||b(z0)ϕ(0)||2 = 0

which is possible only if b(z0)|ϕ(0)〉 = 0 : |ϕ(0)〉 is an eigenvector of b(z0) with eigenvalue 0

b(z0)|ϕ(0)〉 = 0, that is, a|ϕ(0)〉 = z0|ϕ(0)〉

4. Writing D(z) = expA with

A = −z∗a+ za† A† = −za† + z∗a = −A

thenD†(z) = expA† = exp(−A)

and D†(z)D(z) = I. Using (2.55) we may rewrite D(z)

D(z) = e−|z|2/2 eza†

e−z∗a

From the definition (11.31) of a coherent state |z〉 we have

D(z)|0〉 = e−|z|2/2 eza†

e−z∗a|0〉 = e−|z|2/2 eza

† |0〉 = |z〉

because exp(−z∗a)|0〉 = 0 taking a|0〉 = 0 into account.

5. We use (2.55)

eA eB = eA+Be12[A,B]

with

A =c√2

(z − z∗)(a+ a†) B =c′√2

(z + z∗)(a− a†)

and

c = −c′ =1√2

A+B = −z∗a+ za†

The identity (2.55) is valid because [A,B] is a multiple of the identity operator, which commutes with Aet B. We use a system of units such that

~ = mω = 1

or, equivalently, we use the operators Qand P of (11.4) instead of Q et P

[A,B] = − i

2(z − z∗)(z + z∗)[Q,P ] =

1

2(z2 − z∗2)

We must then choose

f(z, z∗) = exp

(

−1

4

[

z2 − z∗2]

)

61

Given all these conditions, the wave function of the coherent state ϕz(q) in the q-representation is

〈ϕz(q) = 〈q|D(z)|0〉 = exp

(

−1

4

[

z2 − z∗2]

)

exp

(

1√2[z − z∗]q

)

〈Q| exp

(

− i√2[z + z∗]P

)

|0〉

= exp

(

−1

2

[

z2 − z∗2]

)

exp

(

1√2

[z − z∗] q

)

ϕ0

(

q − 1√2[z + z∗]

)

where ϕ0(q) is the wave function (11.23) of the harmonic oscillator ground state. From (11.37), theexpectation values of Q and P are

〈Q〉 =1√2(z + z∗) 〈P 〉 =

1

i√

2(z − z∗)

and we can rewrite the preceding result as

ϕz(q) =1

π1/4exp

(

− i

2〈Q〉〈P 〉

)

exp (i〈P 〉q) exp

(

−1

2(q − 〈Q〉)2

)

Reestablishing the dimensionful factors

ϕz(q) =(mω

π~

)1/4

exp

(

− i

2~〈Q〉〈P 〉

)

exp

(

i〈P 〉q

~

)

exp

(

−1

2

mω

~(q − 〈Q〉)2

)

The global phase factor

exp

(

− i

2~〈Q〉〈P 〉

)

is physically irrelevant and may be omitted.

6. We write, with z = ρ exp(iθ) and Anm = 〈n|A|m〉

〈z|A|z〉 =∑

n,m

〈z|n〉〈n|A|m〉〈m|z〉

= e−|z|2∑

n,m

Anmzmz∗n√

n!m!

= e−ρ2∑

n,m

Anm√n!m!

ei(m−n)θ ρn+m

where we made use of (11.31)

〈m|z〉 = e−|z|2/2 zm√m!

Writing the expansion of 〈z|A|z〉 under the general form

〈z|A|z〉 = e−ρ2

∞∑

q=0

∞∑

p=−∞

cpq eipθ ρq

we derive

cpq =1

q!

∫

dRe zIm z

πe−ipθ 〈z|A|z〉

and we can make the identificationAnm =

√n!m! cm−n,m+n

with (m− n) and (m+ n) integers, (m+ n) ≥ 0. Thus, we can get from 〈z|A|z〉 all the matrix elementsof Amn.

11.5.4 Coupling to a classical force


1. Let us express Q as a function of the operators a and a† (cf. (11.6)), and write

H(t) = H0 −(

a+ a†)

f(t)

Differentiating with respect to t the equation

U(t) = U0(t)UI(t)

we obtain

i~dU

dt= H(t)U =

[

H0 +W (t)]

U0UI

= i~dU0

dtUI + U0i~

dUIdt

= H0U0UI + U0i~dUIdt

and derive (11.126)

i~dUIdt

= U−10 W (t)U0UI = WI(t)UI (11.1)

2. Let us show thataI(t) = e iH0t/~ a e−iH0t/~ = a e−iωt

It is easy to derive a differential equation for aI(t)

daI(t)

dt=

i

~[H0, aI ] =

i

~e iH0t/~ [H0, a] e

−iH0t/~

= −iωaI(t)

where we made use of (11.11). We then derive, with the initial condition aI(t = 0) = a

aI(t) = a e−iωt a†I(t) = a† e iωt

the formula for a†I being obtained from Hermitian conjugation. The differential equation (11.126) forUI(t) becomes

i~dUIdt

= −(

a e−iωt + a† e iωt)

f(t)UI(t) = WI(t)UI(t)

with the boundary condition UI(t = 0) = I. If the commutator of two operators A1 and A2 is a multipleof I, from the identity (2.55) of Exercise 2.4.11

eA2eA1 = eA2+A1 e12[A2,A1]

Repeating this operation for n operators A1, . . . , An (we may use a recursion reasoning)

eAneAn−1 · · · eA1 = eAn+···+A1 e12

P

j>i[Aj,Ai]

3. Let us divide the interval [0, t] into n infinitesimal intervals ∆t; the time evolution in the interval[t′, t′ + ∆t] is simple

UI(t′ + ∆t, t′) = I − i

~WI(t

′)∆t+O(∆t)2 = exp

(

− i

~WI(t

′)∆t

)

+O(∆t)2

and it follows that

UI(t) ≃n∏

j=1

[

exp

(

− i

~WI(tj)∆t

)]

Since [WI(tj),WI(ti)] is a multiple of I, we have

UI(t) ≃ exp

− i

~∆t

n∑

j=1

WI(tj)

exp

− (∆t)2

2~2

∑

tj>ti

[

WI(tj),WI(ti)]

63

4. Taking into acount

[

WI(t′),WI(t

′′)]

= f(t′)f(t′′)(

exp[−iω(t′ − t′′)] − exp[iω(t′ − t′′)])

we obtain UI(t) by taking the limit ∆t→ 0

∆t

n∑

j=1

WI(tj) →∫ t

0

dt′WI(t′) = −

∫ t

0

dt′(

a e−iωt′ + a†e iωt′)

f(t′)

= −~az∗(t) − ~a†z(t)

In a similar way

(∆t)2∑

tj>ti

[WI(tj),WI(ti)] →∫ t

0

dt′∫ t

0

dt′′ e−iω(t′−t′′) f(t′)f(t′′)ε(t′ − t′′)

where ε(x) is the sign function: ε(x) = 1 if x > 0, ε(x) = −1 if x < 0. The final result for UI(t) is then

UI(t) = exp[

i(

az∗(t) + a†z(t))]

exp

[

− X

2~2

]

X =

∫ t

0

dt′∫ t

0

dt′′ e−iω(t′−t′′) f(t′)f(t′′)ε(t′ − t′′)

5. We can write this result in a form which is more directly useful thanks to the identity

exp[

i(

az∗(t) + a†z(t))]

= exp[

ia†z(t)]

exp [iaz∗(t)] exp

[

−1

2z(t)z∗(t)

]

and we find, using once more (2.55)

UI(t) = e ia†z(t) e iaz∗(t)e−Y/~2

Y =

∫ t

0

dt′∫ t

0

dt′′ e−iω(t′−t′′) f(t′)f(t′′)θ(t′ − t′′)

=

∫ t

0

dt′∫ t′

0

dt′′ e−iω(t′−t′′) f(t′)f(t′′)

where θ(x) is the Heaviside (or step) function. It is worth checking by explicit differentiation that thisresult does obey the differential equation we started from. This expression fully determines the timeevolution of the forced harmonic oscillator. A particular case is that where the initial state is a coherentstate: then the final state is also a coherent state.

6. Let us examine the case where the initial state at t = 0 is an eigenvector |n〉 of H0. We assume thatthe force acts only during a finite time interval [t1, t2], and we choose to observe the oscillator at a timet > t2: to summarize 0 < t1 < t2 < t. We can then integrate from −∞ to +∞: z(t) is independent of tand is nothing else than the Fourier transform of f(t)/~

f(ω) =1

~

∫ ∞

−∞

dt′ e iωt′f(t′) (11.2)

The Y factor is computed using the Fourier representation of the θ-function

θ(t) = limη→0+

∫ +∞

−∞

dE

2iπ

e itE

E − iη


Introducing this expression in that of Y

1

~2Y =

1

~2

∫

dt′ dt′′dE

2iπ

1

E − iηe i(E−ω)t′ e−i(E−ω)t′′f(t′)f(t′′)

=

∫

dE

2iπ

1

E − iηf(E − ω)f∗(E − ω)

= P

∫

dE

2iπE|f(E − ω)|2 +

1

2|f(ω)|2

= iφ+1

2|f(ω)|2

where P denotes a Cauchy principal value, and we have made use of

1

E − iη= P

1

E+ iπδ(E)

7. We finally obtain an operator UI(t) independent of t for t > t2

UI(t) = exp(

ia†f(ω)

exp(

iaf∗(ω)

exp(−iφ) exp

(

−1

2|f(ω)|2

)

This allows us to compute the n→ m transition amplitude

Amn(t) = 〈m|U(t)|n〉 = 〈m|U0(t)UI(t)|n〉= e−iEmt/~ 〈m|UI(t)|n〉

The result is particularly simple if the oscillator is in its ground state at time t = 0 since UI(t)|0〉 is thena coherent state

UI(t)|0〉 = e−iφ e−|f(ω)|2/2 e ia†f(ω)|0〉 = e−iφ |if(ω)〉The probability to observe a state |m〉 is given by a Poisson law (11.34)

p(m) =

(

|f(ω)|2)m

exp(

−|f(ω)|2)

m!

11.5.9 Quantization in a cavity

1. It is enough to demonstrate the commutation relation at t = 0, because we can multiply the resulton the left by exp(iHt/~) and on the right by exp(−iHt/~) to obtain the result for any t. We noteΦH(~r, t = 0) = Φ(~r), ΦH(~r, t = 0) = Φ(~r). We find for the time derivative of Φ

Φ(~r) = −i

√

~

2µ

∑

k

√ωk

(

ak − a†k

)

ϕk(~r)

and the commutation relation between Φ and µΦ is

[Φ(~r), µΦ(~r ′)] = − i~

2

∑

j,k

√

ωkωj

[

aj + a†j , ak − a†k

]

ϕj(~r)ϕk(~r′)

=i~

2

∑

j,k

√

ωkωj

2 δjk ϕj(~r)ϕk(~r′)

= i~δ(~r − ~r ′)I

2. Orthogonality of the eigenfunctions:

∫ L

0

dx sin kx sin k′x =1

2

∫ L

0

dx(

cos[(k − k′)x] − cos[(k + k′)x])

=L

2δk,k′

65

Completeness relation: let

f(x) =∑

j≥1

ck sinkx kπj

Lj = 1, 2, · · ·

be the Fourier expansion of a function such that f(0) = f(L) = 0. Let us compute the integral

I =

∫ L

0

dx f(x)∑

k

sin kx sink′x = sinkx′∫ L

0

∑

k,k′

ck′ sin k′x sin kx

=∑

k′

sin k′x

(

L

2

)

ck′ =L

2

∑

k

ck sinkx′ =L

2f(x)

It follows that2

L

∑

k

(sin kx sinkx′) = δ(x− x′)

The functions

ϕk(x, t) =

√

2

Lsin kx e±iωt

obey the vanishing boundary conditions, the orthogonality and completeness relations on the interval[0, 1] as well as the wave equation (11.59) with ωk = c|k|. Thus, they can be used for the expansion ofthe quantized field ΦH(x, t)

ΦH(x, t) =

√

~

µL

∑

k≥1

1√ωk

(

ak e−iωt + a†k eiωt)

sin kx

3. The functions

ϕ~k(~r, t) =

√

8

V sinxkx sin yky sin zkze±iω~k

t V = LxLyLz

obey the vanishing boundary conditions as well as the wave equation

(

∂2

∂t2− c2∇2

)

ϕ~k(~r, t) = 0

if ω2~k

= c2~k2. We can then write the expansion of the quantized field ~AH(~r, t) by analogy with (11.79)

~AH(~r, t) =

√

4~

ε0V∑

~k

2∑

s=1

1√ωk

[

a~ks~es(k)e−iωkt + a†~ks~e

∗s (k)eiωkt

]

sinxkx sin yky sin zkz

If we consider a single space dimension, the normalization is√

~/ε0L.

11.5.11 Non Abelian gauge transformations

1. The expression for ~ ′ is

~ ′ = Φ†Ω−1(

−i~∇− q ~A′)

ΩΦ

Let us first remark that∇(ΩΦ) = (~∇Ω)Φ + Ω~∇Φ

We wish to have ~ = ~ ′

~ ′ = Φ†(

−i~∇− iΩ−1(~∇Ω) − qΩ−1 ~A′Ω)

Φ

whence the condition−i~∇− q ~A = −i~∇− iΩ−1(~∇Ω) − qΩ−1 ~A′Ω


that is~A′ = Ω~AΩ−1 − i

q(~∇Ω)Ω−1

In the Abelian case, the field ~A is a number, not a matrix

Ω ~AΩ−1 = ~A

and we recover~A′ = ~A− ~∇Λ

2. Let us choose an infinitesimal gauge transformation

Ω = I − iq∑

a

Λa(~r)

(

1

2σa

)

= I − iqT

ThenΩ~AΩ−1 ≃ ~A− iq[T, ~A]

and

(~∇Ω)Ω−1 ≃ −iq∑

a

~∇Λa

(

1

2σa

)

The commutation relations (3.52) give

[T, ~A] =

[

∑

b

Λb

(

1

2σb

)

,∑

c

~Ac

(

1

2σc

)

]

= i∑

b,c

εabcΛb ~Ac

(

1

2σa

)

whence, by identifying the coefficient of σa/2

δ ~Aa = ~A′a − ~Aa = −~∇Λa + q

∑

b,c

εabcΛb ~Ac

3. We established the form of the covariant derivative by requiring that ~ ′ = ~, and, from our construction

Ω~DΩ−1 = ~D′

We can write the time-independent Schrodinger equation (~ = m = 1)

1

2

(

−i~D)2

Φ = EΦ Φ = Ω−1Φ′

in the form1

2Ω−1Ω

(

−i~D)2

Ω−1Φ′ = EΩ−1Φ′

Multiplying the two sides of the equation by Ω we get

1

2

(

−i ~D′)2

Φ′ = EΦ′

11.5.12 The Casimir effect

1. The only available physical parameters are L, ~ (we are dealing with a quantum problem) and c (thespeed of electromagnetic waves). With these three parameters, we can form a unique combination withthe dimension of a pressure, that is, ~c/L4.

2. The stationary modes of the electric field have the form

~E(~r, t) =[

ei(xkx+yky) sinπnz

L

]

~es(K) e±iωKt

67

where n is an integer ≥ 0 and ωK = c| ~K|. When n 6= 0, the three dimensional wave vector ~K is of theform

~K =(

kx, ky,±πn

L

)

and there are two independent orthogonal directions for ~es(K). The vanishing of the transverse component

of the electric field ~E for z = 0 and z = L is then guaranteed by the factor sin(πnz/L). When n = 0,

~es(K) must be parallel to Oz due to the condition that the transverse component of ~E must vanish. In

addition, since ~es(K) is orthogonal to ~k, there exist a unique direction for polarization.

3. To any vector ~k correspond two polarization states (except if n = 0) and the zero point energy is

E0(L) =~

2

2∑

n,~k

′ωn(~k)

4.Taking the continuum limit, (cf. Exercice 9.7.11), when the dimensions Lx, Ly → ∞∑

~k

→ S(2π)2

∫

d2k

However, since L stays finite, the sum over n remains discrete, and

E0(L) =~S

(2π)2

∑

n,~k

′∫

d2k ωn(~k)

This integral is divergent for large frequencies (it is called an ultraviolet divergence). We introduce acutoff χ, which physically represents the fact that a real metal does not remain a perfect conductor atlarge frequencies

E0(L) =~S

(2π)2

∞∑

n=0

′

∫

d2k ωn(~k)χ

(

ωn(~k)

ωc

)

Setting

ω2 =c2π2n2

L2+ c2k2 = ω2

n + c2k2 ω ≥ ωn

and making the change of variables

d2k = 2πkdk =2π

c2ωdω

we obtain

E0(L) =~S

2πc2

∞∑

n=0

′

∫ ∞

ωn

dω ω2χ

(

ω

ωc

)

ωn =πcn

L

5. E0(L) depends on L only through ωn = πnc/L and

dE0(L)

dL= −dωn

dL

~S2πc2

ω2nχ

(

ω

ωc

)

=πnc

L2

~S2πc2

ω2nχ

(

ω

ωc

)

We derive the internal pressure

Pint = − 1

SdE0(L)

dL= −π

2~c

2L4

∞∑

n=0

′n3χ

(

ω

ωc

)

= −π2~c

2L4

∞∑

n=0

′g(n)

The calculation of the external pressure is done by taking the the limit L → ∞, since outside thecondenser, the field is not confined between plates. We may then replace the discrete sum over n withan integral

Pext = −π2~c

2L4

∫ ∞

0

g(n)


The value of the total pressure is

Ptot = Pint − Pext = −π2~c

2L4

(

∞∑

n=0

′g(n) −∫ ∞

0

g(n)

)

From the Euler-Mac Laurin formula

∞∑

n=0

′g(n) −∫ ∞

0

g(n) =1

5!+O

(

πc

Lω2c

)

which leads to

Ptot = − π2

240

~c

L4

An integration allows us to derive the zero point energy

E0(L) = − π2

720

~c

L3

11.5.13 Quantum computing with trapped ions

1. We write the interaction Hamiltonian in terms of σ+ and σ−

Hint = −1

2~ω1[σ+ + σ−]

[

ei(ωt−kz−φ) + e−i(ωt−kz−φ)]

and go to the interaction picture using (see exercise 5.5.6)

eiH0t/~ σ± e−iH0t/~ = e∓iω0t σ±

In the rotating wave approximation, we can neglect terms which behave as exp[±i(ω0 + ω)t] and we areleft with

Hint ≃ −~

2ω1

[

σ+ e i(δt−φ) e−ikz + σ− e−i(δt−φ) e ikz]

2. ∆z =√

~/(2Mωz) is the spread of the wave function in the harmonic well. Thus, η = k∆z is theratio of this spread to the wavelength of the laser light. We may write

kz = k

√

~

2Mωz(a+ a†) = η(a+ a†)

The matrix element of Hint between the states |1,m+m′〉 and |0,m〉 is

〈1,m+m′|Hint|m〉 = −1

2~ω1e

i(δt−φ) 〈m+m′|e−iη(a+a†)|m〉

The Rabi frequency for oscillations between the two levels is

ωm→m+m′

1 = ω1|〈m+m′|e−iη(a+a†)|m〉|

3. Writing

e±iη(a+a†) ≃ I ± iη(a+ a†)

and keeping terms to first order in η we get

Hint =i

2η~ω1

[

σ+ a ei(δ−ωz)t e−iφ − σ− a†e−i(δ−ωz)t eiφ

+ σ+ a†ei(δ+ωz)t e−iφ − σ− a e−i(δ+ωz)t eiφ

]

69

The first line of Hint corresponds to a resonance at δ = ω−ω0 = ωz, that is, ω = ω0+ωz, a blue sideband,and the second line to a resonance at ω = ω0 − ωz, that is, a red sideband. The σ+a term of the bluesideband induces transitions from |0,m+ 1〉 to |1,m〉, and the σ−a

† term from |1,m〉 to |0,m+ 1〉. Now

〈m|a|m+ 1〉 = 〈m+ 1|a†|m〉 =√m+ 1

so that we get H+int as written in the statement of the problem with

ab =a√m+ 1

a†b =a†√m+ 1

The Rabi frequency is then ω1

√m+ 1. The same reasoning may be applied to the red sideband.

|1, 1〉

ω+

ω+

ω−

ω0

|1, 2〉

|1, 0〉

|0, 2〉

ω|0, 1〉|0, 0〉

Figure 11.1: The transitions which are used are (0, 0) ↔ (0, 1) and (0, 1) ↔ (1, 2): blue sideband,ω+ = ω + ω0 and (0, 1) ↔ (1, 1): red sideband, ω− = ω−ω0.

4. The rotation operators R(θ, φ) are given by

R(θ, φ = 0) = I cosθ

2− iσx sin

θ

2

R(

θ, φ =π

2

)

= I cosθ

2− iσy sin

θ

2

so thatR(π, 0) = −iσx R

(

π,π

2

)

= −iσy

We have, for example,

R(

π,π

2

)

R(β, 0)R(

π,π

2

)

= (−iσy)(I cosβ

2+ iσx sin

β

2)(−iσy)

= −(I cosβ

2+ iσx sin

β

2) = −R(−β, 0)

Let us call A the transition |0, 0〉 ↔ |1, 1〉 and B the transition |0, 1〉 ↔ |1, 2〉. The Rabi frequenciesare linked by ωB =

√2ωA. Thus, if the rotation angle is θA for transition A, it will be θB =

√2 θA for


transition B. For transition A, we choose α = π/√

2 and β = π

R( π√

2,π

2

)

R(π, 0)R( π√

2,π

2

)

R(π, 0) = −I

For transition B we shall have α = π and β = π√

2

R(

π,π

2

)

R(π√

2, 0)R(

π,π

2

)

R(π√

2, 0) = −I

The state |1, 0〉 is not affected because the transition |0, 0〉 ↔ |1, 0〉 does not resonate on the blue sidebandfrequency. Thus we have

|00〉 ↔ −|0, 0〉 |0, 1〉 ↔ −|0, 1〉 |1, 0〉 ↔ +|1, 0〉 |1, 1〉 ↔ −|1, 1〉

5. R(

± π, π/2)

= ∓iσy so that

R(

± π,π

2

)

|0, 1〉 = ∓|1, 0〉 R(

± π,π

2

)

|1, 0〉 = ±|0, 1〉

Let us start from the general two ion state, where both ions are in the vibrational ground state

|Ψ〉 = (a|00〉 + b|01〉+ c|10〉 + d|11〉) ⊗ |0〉= a|00, 0〉+ b|01, 0〉+ c|10, 0〉+ d|11, 0〉

The action of R−(2)(−π, π/2) on ion 2 gives

|Ψ′〉 = R−(2)(−π, π/2)|Ψ〉 = a|00, 0〉+ b|00, 1〉+ c|10, 0〉+ d|10, 1〉

Then we apply R+(1)αβ on ion 1

|Ψ′′〉 = R+(1)αβ |Ψ′〉 = −a|00, 0〉 − b|00, 1〉+ c|10, 0〉 − d|10, 1〉

and finally R−(2)(π, π/2) on ion 2

|Ψ′′′〉 = R−(2)(π, π/2)|Ψ′′〉 = −a|00, 0〉 − b|01, 0〉+ c|10, 0〉 − d|11, 0〉= (−a|00〉 − b|01〉 + c|10〉 − d|11〉) ⊗ |0〉

This is the result of applying a cZ gate, within trivial phase factors.

Chapter 12

exercises from Chapter 12

12.5.1 The Gamow peak

1. Let us compute the Coulomb energy for R = 1 fm

E0 =

(

e2

~c

)(

~c

R

)

=1

137× 200 MeV ≃ 1.5 MeV

The temperature at the center of the Sun is of order of 1.5 × 107 K, corresponding to a kinetic energy Eof about 1.5 keV. We thus have E ≪ e2/R.

2. We define the distance RN as the distance where the Coulomb potential is equal to the kinetic energy:e2/RN = E. The integral to be evaluated for computing the tunnel effect is

I =

∫ RN

R

dr

(

e2

r− E

)1/2

We make the change of variables

u2 =e2

r− E dr = − 2e2udu

(u2 + E)2

The integration limits are

u = 0 et u2 =e2

R− E ≃ e2

R

which leads to

I = 2e2

[

1

2√E

tan−1

√

e2

RE− 1

2

√

R

e2

]

≃ πe2

2√E

The probability for tunneling is

ln pT (E) = −√

2µ πe2

~√E

Defining

EB =2µπ2e4

~2

we cast the probability for tunneling into the form

pT (E) ≃ exp

(

−√

EBE

)

The numerical values are, with µ = 6mp/5

EB = 2π2α2µc2 = 1.18 MeV

71


3. The factor ∼ 4π/k2 is a geometrical factor which must appear in the total cross section: cf. (12.52).It gives the order of magnitude of the total cross section in the absence of any other information. Herewe must in addition take tunneling into account: the potential barrier must be crossed, otherwise thereaction does not occur. An order of magnitude of the total cross section is obtained if we multiply thegeometrical factor by the tunneling probability

σ(E) ∼ 4π

k2pT (E)

3. The angular integration gives a factor 4πv2 and the average value of vσ is given by

〈vσ〉 = 4π

(

µ

2πkBT

)3/2 ∫ ∞

0

dv v3σ(v) exp

(

− µv2

2kBT

)

We make the change of variables v → E

E =1

2µv2 σ(E) =

2π~2

µEpT (E) =

2π~2

µEexp

(

−√

EBE

)

which leads to

〈vσ〉 =16π2

~2

µ3

(

µ

2πkBT

)3/2 ∫ ∞

0

dE e−E/(kBT )e−√EB/E

We must study the integral

J =

∫ ∞

0

dE e−E/(kBT ) e−√EB/E

Let us define the function f(E) by (β = 1/(kBT ))

f(E) = βE −√

EBE

f ′(E) = β −√EB

2E3/2

The maximum of exp[−f(E)] is reached for E = E0, where E0 obeys f ′(E0) = 0

E0 =

[

1

2kBT

√

EB

]2/3

and we find

f(E0) = −1

2

(

2EBkBT

)1/3

≃ −5.8

The width of the peak exp[−f(E)] is obtained from the second derivative

f ′′(E0) =3

4

√

EB E−5/20 =

3

4

(

1

2

)−5/3

E−1/3B (kBT )−5/3

and the width of the peak is

∆E ∼ E1/6B (kBT )5/6 = 4.5 keV

12.5.2 Low energy neutron scattering by a hydrogen molecule

1. Let r1 be the distance between nucleus 1 and the detector

~r1 = ~r − 1

2~R r1 ≃ r − 1

2~R · r

The amplitude for finding the scattered wave at the detector after scattering by nucleus 1 is proportionalto

1

r1eikr1 ≃ 1

reikr

(

1 +~R · r2r

)

e−i~k ′·~R/2

73

with ~k ′ = kr. We must multiply this result by the amplitude for finding the incident plane wave at ~r1which is exp(i~k · ~R/2) and by a1. The scattering amplitude by nucleus 1 is finally

a1

reikr

(

1 +~R · rr

)

exp

(

i

2(~k − ~k ′) · ~R

)

=a1

reikr

(

1 +~R · r2r

)

exp

(

− i

2~q · ~R

)

By adding the amplitudes from nuclei 1 and 2, we obtain the scattering amplitude a as the coefficientof (exp ikr)/r

a = a1 + a2 −i

2(a1 − a2)~q · ~R

The terms proportional to R/r are negligible as r → ∞.

2. The distance between the two protons is on the order of 1 A. If q = 1010 m−1, the neutron energy is

E =~

2q2

2mn= 2 meV

which corresponds to a temperature of 1 K. Neglecting the term proportional to (~q · ~R), we have

a =1

4(as + 3at)I +

1

4(at − as)(~σn · ~σ1) +

1

4(as + 3at)I +

1

4(at − as)(~σn · ~σ2)

=1

2(as + 3at)I +

1

2(at − as)(~σn · ~Σ)

3. The reduced mass is (mp ≃ mn = m)

µH2=

2m2

m+ 2m=

2m

3

while the reduced mass of the neutron-proton system is µp = m/2. The effective potential is of the form(12.41) with a constant g given by

g =2π~

2

µa

For the same value of g the ratio of the scattering lengths is then

aH2

ap=µH2

µp=

4

3

4. The total cross section is obtained using a2

a2 =1

4(as + 3at)

2I +1

2(as + at)(at − as)(~σn · ~Σ) +

1

4(at − as)

2(~σn · ~Σ)2

and, for neutron scattering on polarized molecules

σtot = 4πTr a2

In the parahydrogen case, the average value ~Σ vanishes, so that

σparatot = π(as + 3at)

2

In the orthohydrogen case, one remarks that

Tr (A⊗ B)2 = Tr (A2 ⊗B2) = TrA2 TrB2

SinceTrσniσnjΣiΣj ∝ δij


we obtain

Tr (~σn · ~Σ)2 = Tr[

σ2nxΣ

2x + σ2

nyΣ2y + σ2

nzΣ2z

]

= 2Tr[

Σ2x + Σ2

y + Σ2z

]

= 2Tr ~Σ2 = 12

The unpolarized cross section contains a factor 1/6 owing to the average over the initial spins (1/2 fromthe neutron and 1/3 for the orthohydrogen) and

σorthotot =

4π

6

[

1

4(as + 3at)

2 × 6 +1

4(at − as)

2 × 12

]

= σparatot + 2π(at − as)

2

12.5.3 Analytical properties of the neutron-proton scattering amplitude

1. The radial wave function is

r < R : u(r) = A sin k′R r > R : u(r) = Ne−κr

The continuity condition for the wave function at r = R gives

A sin k′R = Ne−κR

and that of its logarithmic derivative

cotκR = − κ

k′sin k′R =

k′√

κ2 + k′2cos k′R = − κ

√

κ2 + k′2

that is

A = Ne−κR

√

κ2 + k′2

k′

The normalization is obtained from the two integrals

J< = A2

∫ R

0

dr sin2 k′r =N2e−2κR

2k′

[

κ+R(k′2+ κ2)

]

J> = N2

∫ ∞

R

dr e−2κr =N2

2κe−2κR

and the sum J< + J> has the value

J< + J> =N2e−2κR

2k′2κ(k′

2+ κ2)(1 + κR)

whence N2

N2 =2κk′

2e2κR

(κ2 + k′2)(1 + κR)

2. The function g(−k, r) is linearly independent of g(k, r) because it behaves at infinity as exp(−ikr).The most general solution of the Schrodinger equation is a linear combination of these two solutions, andbecause u(k, r) must in addition vanish at r = 0, we have

u(k, r) = g(−k, r)g(k) − g(k,−r)g(−k) g(k) = g(k, r = 0)

The r → ∞ behavior of u(k, r) is then

r → ∞ : u(k, r) ∝ eikr g(k) − e−ikr g(−k)

75

Comparing with (12.22) for l = 0 then shows that

S(k) = e2iδ(k) =g(k)

g(−k)

3. Let us continue g(k, r) to complex values of k: g(k, r) and g(−k∗, r) obey the differential equations

d2

dr2g∗(k, r) +

[

k∗2 − 2mV (r)]

g∗(k, r) = 0

d2

dr2g(−k∗, r) +

[

k∗2 − 2mV (r)]

g(−k∗, r) = 0

Writing k = k1 + ik2, we find the following asymptotic behaviors

g∗(k, r) ∝(

eikr)∗

= e−ik1r e−k2r

g(−k∗, r) ∝ e−ik∗r = e−ik1r e−k2r

The two functions obey the same differential equations and have the same behavior at infinity, so thatthey must be identical. We thus derive g∗(k∗) = g(−k), whence

S∗(k∗) =g∗(k∗)

g∗(−k∗) =g(−k)g(k)

=1

S(k)

and

S(−k) =g(−k)g(k)

=1

S(k)= S∗(k∗)

4. The behavior of the function g(k, r) is given for r < R and r > R by

r < R g(k, r) = Ae−ik′r + Ceik′r

r > R g(k, r) = e−ikr

The continuity conditions for g(k, r) and of its derivative leads to the following

Ae−ik′R + Ceik′R = e−ikR

k′(

Ae−ik′R − Ceik′R)

= ke−ikR

with k′ =√

M(V0 + E). From this we deduce

g(k) = A+ C = e−ikR

(

cos k′R+ ik

k′sin k′R

)

One observes that g(k) is indeed an entire function of k. The only delicate point could come from thepoint k′ = 0 owing to the square root in the definition of k′, but there is in fact no problem as cos k′Rand (1/k′) sin k′R are analytic functions of k′

2.

5. Let us assume that S(k) possesses a pole on the positive imaginary axis at k = iκ, κ > 0. We thenhave g(−k) = g(−iκ) = 0 and the asymptotic behavior of u(k, r) is

r → ∞ : u(k, r) ∼ eikr g(k) + e−ikr g(−k)∼ e−κr g(k) + eκr g(−k)

u(k, r) explodes if r → ∞, unless g(−k) = 0, and then u(k, r) is a bound state wave function

u(k, r) = g(k)g(−k, r)

which automatically vanishes at r = 0. The poles of S(k) on the imaginary axis such that 0 < Im k < µ/2then give the bound state energies. Let us now assume a pole of S(k) at k = h−ib. Owing to the properties


demonstrated at question 3, S(k) also has a pole at k = −h− ib. If b < 0, g(h+ ib, r) and g(−h+ ib, r)are square integrable functions (they behave as exp(−|b|r) at infinity) and since they are solutions of theSchrodinger equation, they are orthogonal

∫ ∞

0

dr g(h+ ib, r)g(−h+ ib, r) =

∫ ∞

0

dr |g(h+ ib, r)|2 = 0

and we have a contradiction if b < 0. If h 6= 0, the only possibility is to have poles such that Im k < 0.

6. The following choice for S(k)

S(k) =(k − h− ib)(k + h− ib)

(k − h+ ib)(k + h+ ib)≃ k − h− ib

k − h+ ibfor k ∼ h

obeys the properties stated in question 3. The relation between cot δ and S is

cot δ = ie2iδ + 1

e2iδ − 1= i

S + 1

S − 1=h− k

b

that is

δ(k) = tan−1 −bk − h

The phase shift δ increase from δ(k) = tan−1 b/h for k = 0 to π for k → ∞, going through π/2 for k = h.The total cross section is

σtot =4π

k2sin2 δ =

4π

k2(1 + cot2 δ)=

4πb2

k2[(k − h)2 + b2]

Let us set

k2 =2mE

~2h2 =

2mE0

~2

k − h =k2 − h2

k + h≃ k2 − h2

2h=m(E2 − E2

0)

~√

2mE0

We find

σ(E) =2π~

2

ME

~2Γ2/4

(E − E0)2 + ~2Γ2/4

provided we set~

2Γ2

4=

2E0~2b2

m

7. Let us start from the radial Schrodinger equation

u′′ − 2mV u+ k2u = 0

and let us differentiate this equation with respect to k

∂u′′

∂k− 2mV

∂u

∂k+ k2 ∂u

∂k= −2uk

We then multiply the first equation by ∂u/∂k, the second one by u and subtract the second equationfrom the first

u′′∂u

∂k− u

∂u′′

∂k= 2ku2

that is∂

∂r

(

u′∂u

∂k− u

∂u′

∂k

)

= 2ku2

Integrating over r we get

u′∂u

∂k− u

∂u′

∂k= 2k

∫ r

0

u2(r′)dr′

77

When r → 0 we have∂u

∂k

∣

∣

∣

r=0=

[

A′(k) sin k′r +A(k)r∂k′

∂kcos k′r

]

r=0

= 0

Furthermore∂u

∂k= g(k, r)g′(−k) +O

(

e−κr)

and for a bound state

u(k, r) = g(−k, r)g(k) =⇒ if r → ∞ u′(k, r) = iku(k, r)

As a consequence, for r → ∞u′∂u

∂k− u

∂u′

∂k→ 2 ik g(iκ)[g′(−iκ)]

In the vicinity of the pole k = iκ, g(−k) is proportional to (k − iκ)

g(−k) ≃ D(k − iκ)

and then g′(−k) = −D. Setting g(iκ) = F , we obtain

−2ikDF = 2k

∫ ∞

0

u2(k, r) dr

The behavior r → ∞ of u(k, r) is

r → ∞ : u(k, r) ≃ g(iκ)e−κr = F e−κr

Let u(k, r) = Gu(k, r) be the normalized bound state wave function, which behaves at infinity asN exp(−κr), where N is the constant defined at question 1, N = FG. One the one hand we have

∫ ∞

0

dr u2(k, r) = −iDF

and on the other hand

1 =

∫ ∞

0

dr u2(k, r) = G2

∫ ∞

0

dr u2(k, r) = −iFG2 = −i(FG)2D

F= −iN2D

F

In the vicinity of the pole at k = iκ

S(k) ≃ F

D(k − iκ)=

−iN2

k − iκ

8. Let us express k cot δ as a function of g(k)

k cot δ(k) =ik[g(k) + g(−k)]g(k) − g(−k)]

This function is analytical for k ∼ 0, it tends to a constant for k → 0 and it is an even function of k. Wemay then write its Taylor expansion as

k cot δ(k) = −1

a+

1

2r0k

2 +O(k4)

Taking

S(k) =k cot δ(k) + ik

k cot δ(k) − ik

into acount, the existence of a pole of S(k) at k0 = iκ implies

k0 cot δ(k0) = ik0 = −κ − 1

a− 1

2r0κ

2 = −κ


that is

r0 =2

κ

(

1 − 1

κa

)

Let us calculate the residue at the pole

k cot δ − ik = k0 cot δ0 + (k − k0)∂

∂kk cot δ

∣

∣

∣

k=k0− ik + ik0 − ik0

= −i(k − k0) + (k − k0)iκr0 = −i(k − k0)(1 − κr0)

that is

S(k) =−2κ

−i(k − k0)(1 − κr0)=⇒ N2 =

2κ

1 − κr0

12.5.5 Neutron optics

1.The distance between the nucleus and the observation point is r =√s2 + z2. The total probability

amplitude at the observation point is obtained by adding coherently the amplitudes from each nucleus

ϕd = −aρδ∫ ∞

0

eikr

r2πsds = −2πaρδ

∫ ∞

z

eikr

rrdr

as rdr = sds. The upper bound of the integral is an oscillating function. However, this oscillation hasno physical meaning and we obtain the total amplitude in the form

ϕ(z) =

(

1 − 2iπaρ δ

k

)

e ikz

2. If the neutrons cross a medium with index n and thickness δ with k′ = nk

ϕ(z) = eik(z−δ) eik′δ = eikz ei(k′−k)δ)

= eikz ei(n−1)kδ ≃ eikz [1 + i(n− 1)kδ]

and by identification with the result of question 1 we get

n = 1 − 2πaρ

k2= 1 − aρλ2

2π

As the index of refraction is very close to one, the critical reflection angle is close to π/2

sin(π/2 − θc) = cos θc ≃ 1 − 1

2θ2c = n

that is1

2θ2c = 1 − n =

aρλ2

2πθc = λ

(ρa

π

)1/2

3. The neutron-proton scattering matrix in spin space is

f = −1

4(as + 3at)I −

1

4(at − as)(~σn · ~σp) = −asPs − atPt

As | + +〉 is a triplet state

fc = 〈+ + |f | + +〉 = −atIf |χs〉 et χt〉 are the singlet and triplet states corresponding to m = 0, the relation inverse to (10.125) et(10.126) is

| + −〉 =1√2(|χt〉 + |χs〉)

| − +〉 =1√2(|χt〉 − |χs〉)

79

and

fa = 〈+ − |f | + −〉 = −1

2(at + as)

fb = 〈+ − |f | − +〉 = −1

2(at − as)

The weights 3/4 and 1/4 are determined by the degeneracy of the triplet (3) and singlet (1) states. Thecoherent scattering length

aeff =3

4at +

1

4as

has the numerical value −1.9 fm, and the refraction index is larger than one: there cannot be totalreflection.

4. Taking the square of

~I = ~J +~

2~σ

that is

~I2 = ~J2 +~

2

4~σ2 + ~ ~J · ~σ

we obtain

1

~

~J · ~σ = j if I = j +1

21

~

~J · ~σ = −(j + 1) if I = j − 1

2

The scattering lengths are given by

a+ = a+ bj a− = a− b(j + 1)

or conversely

a =1

2j + 1[(j + 1)a+ + ja−] b =

1

2j + 1(a+ − a−)

5. The a amplitude corresponds to a scattering without spin flip, and thus to coherent scattering, andthe b amplitude to a scattering with spin flip, and thus incoherent. Another way of finding the result isto observe that the probability for scattering in the state (j + 1/2) is (j + 1)/(2j + 1) and j/(2j + 1) inthe state (j − 1/2). Using the results of 1.6.8 we find

σcoh =4π

(2j + 1)2[(j + 1)a+ + ja−]

2

σinc = 4πj(j + 1)

(2j + 1)2(a+ − a−)2

12.5.6 The cross section for neutrino absorption

1. The matrix element 〈ϕf |ϕi〉 is given by

〈ϕf |ϕi〉 =1

V2

∫

d3r ei(~p1+~p2+~p3)·~r/~ =1

V δ~p1+~p2+~p3,0

where ~pi represent the momenta of the final particles. The symbol δ represents a Kronecker δ since wehave used plane waves in a box. Taking the square of the preceding equation, we obtain

|〈ϕf |ϕi〉|2 =1

V2δ~p1+~p2+~p3,0


Let us show that the recoil kinetic energy K of the proton is negligible. The relation

~P + ~q + ~p = 0

shows that the three momenta are a priori of the same order of magnitude, ∼ p. On the other hand, theelectron kinetic energy k is

k ∼ p2

2me≫ K ∼ p2

2mp

The final density of states is then

D(E) = V2

∫

d3p

(2π~)3d3q

(2π~)3δ(K + E + cq − E0)

Let us integrate over the angles

V d3p

(2π~)3→ 4πV

(2π~)3p2dp =

4πV(2π~)3

pEdE

c2

V d3q

(2π~)3→ 4πV

(2π~)3q2dq

Energy conservation gives q = (E − E0)/c and on integrating the level density over q

D(E0) =V2

(2π2)2~6

pE

c2(E − E0)

2

c3

This gives for the decay rate par unit energy E

dΓ

dE= 2π

G2F 〈|Mfi|2〉4π4~7c5

pE(E0 − E)2

and integrating over E while neglecting the electron mass we obtain the expression for the lifetime

1

τ= Γ ∼ G2

FE50

60π3~(~c)6

From the expressionG2F

(~c)6=

60π3~Γ

E50

we deduce that G2F /(~c)

6 has the dimension of minus the fourth power of an energy and is expressed, forexample, in MeV−4. For the numerical evaluation, we convert the inverse lifetime in MeV

~

τ= ~Γ = 0.73 × 10−24 MeV

which leads toGF

(~c)3= 2.3 × 10−11 MeV−2 = 2.3 × 10−5 GeV−2

in qualitative agreement with the exact value.

2. The transition matrix element is

〈ϕf |ϕi〉 =1

(V)2

∫

d3rei(~p1+~p2−~p′1−~p

′2)·~r/~ =

1

V δ~p1+~p2,~p ′1+~p ′

2

where ~p and ~p ′ represent the initial and final particles momenta. The differential cross section is

dσ

dΩ=

1

F2π

~〈|Mfi|2〉〈ϕf |ϕi〉|2D(E)

81

The final density of states for an electron of energy E emitted in the direction Ω is

D(E) =V

(2π~)3E2

c3

while the flux is given by F = c/V , as the neutrinos propagate with very nearly the speed of light andtheir density is 1/V . One finds for the differential cross section

dσ

dΩ=G2F

4π2

〈|Mfi|2〉(~c)4

E2

and integrating over angles we obtain the total cross section for neutrino absorption

σtot =G2F

π

〈|Mfi|2〉(~c)4

E2

We can get an order of magnitude, in a system of units where ~ = c = 1, 1 fm = 1/(200 Mev), σ ∼ G2FE

2,GF = 10−5 GeV−2. For E = 10 MeV,

σ ∼ 10−20 MeV−2 = 4 × 10−16 fm2 = 4 × 10−46 m2

Choosing the density to be that of iron, n ≃ 1029 atoms/m−3

, we find for the mean free path ℓ

ℓ =1

30nσ∼ 1015 m

where we have taken into account the fact that an iron nucleus contains 30 neutrons. The mean free pathis about 1/10th of a light year, and detecting a neutrino is really an achievement!

3. The maximum inelastic cross section is

σin,max =π

k2≃ π

E2

We must have

E <∼1√GF

≃ 300GeV

Fermi’s theory is undoubtely limited to energies less than 300 GeV.


Chapter 13


13.4.1 The Ω− particle and color

If we build a spin 3/2 particle from three spins 1/2, the state vector is symmetrical with respect to anyexchange of two spins: for example, the state j = 3/2, jz = 3/2 is

∣

∣

∣j =3

2, jz =

3

2

⟩

=∣

∣

∣

1

2,1

2,1

2

⟩

If the spatial wave function has no zeroes, it is necessarily symmetrical. Indeed, if it were antisymmetrical,for example with respect to the exchange of particles 1 et 2

ϕ(~r1, ~r2, ~r3) = −ϕ(~r2, ~r1, ~r3)

it would vanish at the point ~r1 = ~r2. The state vector space⊗spin of a system of three identical quarksmust be antisymmetrical, and this is impossible if the three quarks are identical. As a matter of fact,quarks possess an additional quantum number, color, and the three quarks of the Ω− particle are ofdifferent color: it is the color wave function which is antisymmetrical.

13.4.2 Parity of the π-meson

1. The π−-meson-deuteron system is analogous to an hydrogen atom, apart from the reduced mass,which is different

µ =mDmπ

mD +mπ= 129 MeV/c

2

As a consequence, the energy levels are given by

En = − µ

me

R∞

n2= −253

R∞

n2= −3.42

n2keV

The transitions are in the x-ray domain, whose energy lie between ∼ 0.1 and ∼ 100 keV.

2. As the orbital angular momentum vanishes, the total angular momentum is that of the deuteron,j = 1. The possible final states with angular momentum j = 1 are 3S1,

3P1,1P1 et 3D1, but only the

state 3P1 is antisymmetric in space+spin in the exchange of the two final neutrons. The parity of thefinal state is −1 (angular momentum Lfin = −1) and that of the initial sate is

ηπ(−1)Linit = ηπ

Parity conservation implies ηπ = −1.

13.4.4 Positronium decay

1. Since the reduced mass is me/2, the energy levels are of the form

En = −1

4

m2ee

4

~2

1

n2

83


2. Adding two spins 1/2 gives either j = 1 (triplet state ), or j = 0 (singlet state).

3. As the projection along z of the orbital angular momentum vanishes, angular momentum conservationalong this direction reads

m = m1 +m2

where m is the projection along Oz of the positronium spin, m1 and m2 the projections of the spins ofthe two photons. One can a priori contemplate the following situations

• two right handed photons: m1 = 1 et m2 = −1, m1 +m2 = 0

• two left handed photons: m1 = −1 et m2 = 1, m1 +m2 = 0

• one left handed photon and one right handed photon: m1 +m2 = ±2

This last possibility is excluded as m = −1, 0 ou +1.

4. In a rotation by π about Oy, the two photons are exchanged. Since the photons are bosons, the globalstate vector must not change sign in this operation. With Y = exp(−iπJy), we have from (10.102)

Y |jm〉 = (−1)j−m|j,−m〉

The initial state changes sign if j = 1, because m = 0, and does not change sign if j = 0. As aconsequence, only the singlet state j = 0 can decay into two photons. Generally speaking, a spin one

particle cannot decay into two photons.

5. The parity of the positronium ground state is

Π = ηe+ηe−(−1)l = ηe+ηe−

where l = 0 is the orbital angular momentum. In a reflection with respect to a xOz plane, the initialstate vector changes sign, because the operator Y which implements this reflection is Y = Πexp(−iπJz).In the photon case, from (10.104) and taking into account the odd parity ηγ = −1 of the photon

Y|R〉 = −|L〉 Y|L〉 = −|R〉

This shows that |Φ+〉 does not change sign, while |Φ−〉 changes sign. Thus it is the two photon entangledstate |Φ−〉 which is produced in the decay.

Quantum statistics and beam splitters

1. The amplitude for mode a to be transmitted by the beam splitter is t = cos θ, and the amplitude tobe reflected is r = i sin θ, where t and r are defined in Exercise 1.6.6. The factor i takes into account theπ/2 phase shift between the transmitted and reflected beams.

2. Let us compute, for example,

e iGθ a e−iGθ = a+ iθ[G, a] − 1

2![G, [G, a]] + · · ·

where we have made use of (2.54). A straightforward calculation gives the commutators

[G, a] = [ab† + a†b, a] = b [G, [G, a]] = [ab† + a†b, b] = a

so that

e iGθ a e−iGθ = a+ iθb− 1

2!θ2a+ · · · = a cos θ + ib sin θ

3. If the initial state is a two photon state

|Ψ0〉 = |1a, 1b〉 = a†b†|Ω〉

85

where |Ω〉 is the vacuum state, then the beam splitter transforms this state into

|Ψ〉 = U(θ)|Ψ0〉 = U(θ)a†b†|0a, 0b〉 = U(θ)a†b†U †(θ)|Ω〉= (a† cos θ + ib† sin θ)(ia† sin θ + b† cos θ)|Ω〉

=

(

i√2

sin 2θ[

(a†)2 + (b†)2]

+ cos 2θ a†b†)

|Ω〉

=i√2

(

|2a, 0b〉 + |0a, 2b〉)

+ cos 2θ|1a, 1b〉

where we have used (a†)2|0〉 =√

2|2〉 [see (11.18)]. For a symmetric beam splitter with θ = π/4 we obtain

|Ψ〉 =i√2

(

|2a, 0b〉 + |0a, 2b〉)

The two photons stick together at the output of the beam splitter.

4. Let us first consider a single fermionic mode. The mode can be either unoccupied (state |0〉) oroccupied (state |1〉), as, from the Pauli principle, there cannot be more than one fermion in the mode.The operators a and a† act in the two-dimensional Hilbert space spanned by these two vectors and theycan be represented by the 2 × 2 matrices

a =

(

0 01 0

)

a† =

(

0 10 0

)

which leads to the anticommutation relation a, a† = I. When we consider two modes, the antisymmetryis ensured owing to the anticommutation relations. For example,

|1a, 1b〉 = a†b†|Ω〉 = −b†a†|Ω〉 = −|1b, 1a〉

In order to compute

e iGθ a e−iGθ

we use the identity[AB,C] = AB,C − A,CB

from which we find[G, a] = b [G, b] = −a

so that

e iGθ a e−iGθ = a+ iθb+1

2!θ2a+ · · ·

= a cosh θ + ib sinh θ

Repeating the calculation of question 3, we obtain

|Ψ〉 = (a† cosh θ − ib† sinh θ)(b† cosh θ + ia† sinh θ)|Ω〉= a†b†|Ω〉 = |1a, 1b〉

The two fermions must choose different outputs of the beam splitter: otherwise they would be in thesame state!


Chapter 14


14.6.1 Second order perturbation theory and van der Waals forces

1. Let us start from the eigenvalue equation to order λ2

H(λ)|ϕ(λ)〉 = (H0 + λW )|ϕ(λ)〉= (E0 + λE1 + λ2E2)|ϕ(λ)〉

with|ϕ(λ)〉 = |ϕ0〉 + λ|ϕ1〉 + λ2|ϕ2〉

and the auxiliary condition 〈ϕ0|ϕ(λ)〉 = 1, whence

〈ϕ0|ϕ1〉 = −λ〈ϕ0|ϕ2〉

We deduce from these two equations, to order λ2

(H0 − E0)|ϕ2〉 = (E1 −W )|ϕ1〉 + E2|ϕ0〉

Multiplying on the left by the bra 〈ϕ0| and taking into account that 〈ϕ0|ϕ1〉 is of order λ

E2 = 〈ϕ0|W |ϕ1〉 (14.1)

Furthermore, from the identification of the terms of order λ for |ϕ(λ)〉 gives

(H0 − E0)|ϕ1〉 = (E1 −W )|ϕ0〉 (14.2)

Let us write the identity operator under the form (|ϕ0〉 ≡ |n〉)

I = |n〉〈n| + (H0 − E0)−1(

∑

k 6=n

|k〉〈k|)

(H0 − E0)

and let us use the expression (14.1) of E2

E2 = 〈n|W |n〉〈n|ϕ1〉 +∑

k 6=n

〈n|W | 1

H0 − E0|k〉〈k|(H0 − E0)|ϕ1〉

Neglecting the first term, which is justified because 〈n|ϕ1〉 = O(λ) and using (14.2)

E2 =∑

k 6=n

|〈n|W |k〉|2E0 − Ek

2. A dipole moment ~d = qe~r1 located at ~R = 0 creates at point ~R an electric field

~E = − 1

4πε0R3

[

~d− 3(~d · R)R]

87


and the potential energy of the two atom system is

W = −qe~r2 · ~E

3. 〈ϕ01ϕ02|W |ϕ01ϕ02〉 = 0 because the average values of X1, Y1 et Z1 vanish owing to parity conservation

〈ϕ01|X1|ϕ01〉 = 0

and the same property holds for the average values of X2, Y2 and Z2.

3. Using the completeness relation∑

α

|ϕα〉〈ϕα| = I

we obtain

E2 ≃ − 1

2R∞〈ϕ01ϕ02|W 2|ϕ01ϕ02〉

The only terms of W 2 whose average value does not vanish are X21X

22 , Y 2

1 Y22 et 4Z2

1Z22 . Using rotational

invariance we get

〈ϕ01|X21 |ϕ01〉 =

1

3〈ϕ0|~R2|ϕ0〉 =

1

3〈R2〉 = a2

0

The characteristic time for a fluctuation of the dipole moment of one of the atoms is τ ≃ ~/R∞. Inorder that a static calculation such as that developed above be valid, one must be able to neglect thepropagation time of light between the two atoms, and we must have R≪ cτ = ~c/R∞.

14.6.3 Muonic atoms

1. The reduced mass of the problem is

m′µ =

mµ

1 +mµ/mA≃ mµ/me

1 +mµ/(Amp)

that is

m′µ = α(A)me α(A) =

mµ/me

1 +mµ/(Amp)

Applications

• Aluminium: aZ=13µ = 19.8 fm R = 3.6 fm

• Lead: aZ=82µ = 3.1 fm R = 7.1 fm

2. Role of the electrons from internal shells. The wave function of the innermost electrons is

ϕZ0 (r) =1

√

π(aZe )3e−r/a

Ze aZe =

a0

Z

The electric charge contained within an orbit of radius aZµ is

Q ≃ 2qe

(

aZµaZe

)3

≃ 2qe

(

me

mµ

)3

∼ qe × 10−6

The binding energy 2p→ 1s of the hydrogen atom is ∆EH ≃ (3/4)R∞. In the case of a muonic aluminiumatom, its value is

∆EAlµ = (13)2α(27)

3

4R∞ = 354.9 keV

and in the case of lead

∆EPbµ = (82)2α(208)

3

4R∞ = 14.2 MeV

89

The approximation of a point nucleus is evidently not valid for lead, because the orbit radius is abouthalf of the nuclear radius! This approximation cannot even be used as a zeroth order approximation. Itwould be wiser to use an harmonic oscillator approximation, using the potential of the following question.

3. The potential to be used in the perturbative calculation for r < R is

W (r) =Ze2

2

[

1

R

( r

R

)2

− 3 +2

r

]

The W potential being nonzero only for r < R, the contribution of a state whose wave function vanishesat r = 0 (p wave, d wave etc.) is negligible. For an ns wave we use

∫

d3rW (r)|ϕns(~r)|2 ≃ |ϕns(0)|2∫

d3rW (r)

Taking∫

d3rW (r) =2π

5Ze2R2

into account, we find

δEns =2π

5Ze2R2|ϕns(0)|2

which give the following numerical value for the 2p→ 1s transition

δE1s =2Ze2

5aZµ

(

R

aZµ

)2

=4

5R∞Z

2

(

mµ

me

)(

R

aZµ

)2

= 12.6 keV

Taking vacuum polarization into account, we have

∆E2p→1s = 354.9− 12.6 + 2.2 = 344.5 keV

in very good agreement with the experimental result.

4. The ratio of the fine structure characteristic energy to the ground state energy is the same for ordinaryand muonic atoms: in both cases, it is proportional to α2. By contrast, the ratio of the hyperfine structureto the ground state energy is larger by a factor ∼ mµ/me for muonic atoms. Indeed, if we look at (14.32)taking into account the fact that the ground state energy is proportional to 1/a, one must take intoaccount a factor 1/a3 ∼ (mµ/me)

3 and a factor me/mµ coming from the ratio of the electron magneticmoment to the muon magnetic moment.

14.6.4 Rydberg atoms

1. In the case l = n− 1 the expansion of unl(r) is reduced to a single term

unl(r)∣

∣

∣

l=n−1= c0

(

r

a0

)n

exp

(

− r

na0

)

Let us set x = r/a0 and let us study the function f(x) = xn exp(−x/n), or, in an equivalent way itslogarithm g(x). The function f(x) displays a sharp maximum at x0 which is determined by studyingg′(x), g′(x0) = 0

g′(x) =n

x− 1

nx0 = n2

Let us also compute the second derivative

g′′(x) = − n

x2g′′(x0) = − 1

n3

and thus

f(x) ≃ f(x0) exp

(

− (x− x0)2

2n3

)


The dispersion around the maximum at x0 is ∆x = n3/2. When l = n − 1, the radial wave function islocalized around a value a0n

2 with a dispersion a0n3/2. When l 6= n − 1, the exponential in unl(r) is

multiplied by a polynomial in r, and not by a monomial, which makes the curve wider.

2. In the vicinity of θ = π/2 and setting δ = π/2 − θ we obtain

sinl θ = cosl δ ≃(

1 − 1

2δ2)l

≃ exp

(

−1

2lδ2)

The wave function is thus concentrated within an angular opening δθ ≃ 1/√l ≃ 1/

√n, which gives the

following dispersion along Oz

∆z ≃ 1√na0n

2 = a0n3/2

The horizontal dispersion (question 1) and the vertical one (question 2) being both proportional toa0n

3/2, the wave function is indeed concentrated in a torus of radius a0n3/2 drawn around a circle of

radius a0.

14.6.6 Vacuum Rabi oscillations

2. The energy of the state |ϕgn〉 is Egn = n~ω and that of the state |ϕen〉, Een = ~ω0 + (n − 1)~ω. Theenergy difference between the two levels is then

∆En = Egn − Een = ~δ

and the two levels are almost degenerate if δ ≪ ω0.

3. The coupling of the electromagnetic field with the dipole is

W = −id

√

~ω

ε0V[

a(b+ b†) − a†(b+ b†)]

= − i~

2ΩR

[

a(b+ b†) − a†(b+ b†)]

W applied to a state |g ⊗ n〉 give two types of contribution

(i) W (n)eg = 〈e⊗ (n− 1)|W |g ⊗ n〉 = − i~

2ΩR 〈e⊗ (n− 1)|ab†|g ⊗ n〉 = − i~

2ΩR

√n

(ii) W(n)

eg = 〈e⊗ (n+ 1)|W |g ⊗ n〉 = − i~

2ΩR 〈e⊗ (n+ 1)|a†b†|g ⊗ n〉 = − i~

2ΩR

√n+ 1

The second term changes the energy by 2~ω. Let us compare the time evolution of the operators ab† anda†b† in the Heisenberg picture, with H0 = Hat +Hfield

ab† → eiH0t/~ a|e〉〈g| e−iH0t/~ =(

ab†)

ei(ω0−ω)t =(

ab†)

e−iδt

where we have used (11.67), while

a†b† → eiH0t/~ a†|e〉〈g| e−iH0t/~ =(

a†b†)

ei(ω0+ω)t

and this term is negligible in the rotating wave approximation (cf. (Sec. 5.3.2)). We are left with onlythe term corresponding to the combination ab†. Owing to Hermitian conjugation, we must also keep thecombination a†b. We finally obtain the Jaynes-Cummings Hamiltonian

H = ~ω0|e〉〈e| + ~ωN − i~

2ΩR (ab† − a†b)

In the subspace H(n) the matrix form of this Hamiltonian is

H = n~ωI − 1

2~δI +

1

2~

(

δ iΩR√n

−iΩR√n −δ

)

91

4. From (2.35), the eigenvalues and eigenvectors in the basis |ϕgn〉, |ϕen〉 are

E−n = −1

2~

√

δ2 + nΩ2R |χn−〉 = sin θn|ϕgn〉 + i cos θn|ϕen〉

E+n =

1

2~

√

δ2 + nΩ2R |χn+〉 = cos θn|ϕgn〉 − i sin θn|ϕen〉

When δ = 0

E±n = ±1

2~δ |χ±

n 〉 =1√2

(|ϕgn〉 ∓ i|ϕen〉)

5. We decompose the state |e〉 ≡ |e⊗ 0〉 on the states |χ±1 〉

|e〉 =i√2(|χ+

1 〉 − |χ−1 〉)

which gives the time evolution for δ = 0, with E±1 = ±~ΩR/2

|e〉 → |ψ(t)〉 =i√2

(

e−iΩRt/2|χ+1 〉 − eiΩRt/2|χ−

1 〉)

The probabilty for finding |e〉 after a time t spent in the cavity is then

pe(t) = |〈e|ψ(t)〉|2 =1

4

∣

∣

∣e−iΩRt/2 + eiΩRt/2∣

∣

∣

2

= cos2ΩRt

2

6. Off resonance, pe(t) is given by the Rabi formula (4.36)

pe(t) = 1 − Ω2R

Ω2sin2 Ω2t

2

Figure (4.5b) illustrates the decrease in the amplitudes of the oscillations off resonance.

7. If the cavity contains n photons, we shall observe oscillations between the states |ϕen〉 et |ϕgn〉 with afrequency

Ωn+1 =√

δ2 + (n+ 1)Ω2R

If the cavity contains a coherent state with an average photon number 〈n〉, we shall observe a superpositionof oscillations whose frequencies are given by the results of the preceding question, the probability of eachfrequency being given by (11.34)

pn =〈n〉nn!

e−〈n〉

14.6.7 Reactive forces

1. The eigenvalues and eigenvectors are given by (2.35). We find

|χ1n(z)〉 : E1n = −1

2~

√

δ2 + nΩ21(z)

|χ2n(z)〉 : E2n =1

2~

√

δ2 + nΩ21(z)

The force on the atom in the state |χ1n〉, for example, is

F1n = −∂E1n

∂z=

1

4~n

∂Ω21

∂z

1√

δ2 + nΩ21(z)

and F2n = −F1n.


2. The transition amplitudes are given by

an11 = 〈χ1,n−1|(b+ b†)|χ1n〉 = − sin θn−1 cos θn

an21 = 〈χ2,n−1|(b+ b†)|χ1n〉 = cos θn−1 cos θn

an12 = 〈χ1,n−1|(b+ b†)|χ2n〉 = − sin θn−1 sin θn

an22 = 〈χ2,n−1|(b+ b†)|χ2n〉 = − cos θn−1 sin θn

By choosing in a suitable way the phase φ in (11.93), we obtain for the expectation value of the fieldEH(z, t) in the coherent state |z〉, with |z|2 = 〈n〉

〈z|EH(z, t)|z〉 = 2〈n〉√

~ω

ε0Vcosωt sinkz

which leads to√

~ω〈n〉ε0V

=1

2E0

and for the atom-field coupling

~Ω1(z)√

〈n〉 = dE0 sin kz = ~ω1(z)

where ω1(z) is the usual Rabi frequency (cf. for example (14.74)).

3. We find at once (θ〈n〉 = θ)

pst1 =

sin4 θ

cos4 θ + sin4 θpst2 =

cos4 θ

cos4 θ + sin4 θ

From question 1, the force on the atom in state |χ1〉 is

F1 =1

4~∂ω2

1(z)

∂z

1

Ω1〈n〉(z)

and F2 = −F1. The force on an atom then is

F = F1(pst1 − pst

2 ) =1

4~∂ω2

1(z)

∂z

1√

δ2 + ω21(z)

sin4 θ − cos4 θ

cos4 θ + sin4 θ

withsin4 θ − cos4 θ

cos4 θ + sin4 θ= − 2δΩ1〈n〉

δ2 + Ω21〈n〉(z)

Assembling all the factors

F = −1

2~∂ω2

1(z)

∂z

δ

2δ2 + ω21(z)

or, in a vector form

~F = −1

2~~∇ω2

1(~r)δ

2δ2 + ω21(z)

in agreement with (14.98) if Γ ≪ ω1, that is, if the laser intensity is large enough.

14.6.8 Radiative capture of neutrons by hydrogen

1. When r → 0

ψ(r) ≃ pr + δ

pr= 1 +

δ

pr= 1 − a

r= −a

r

(

1 − r

a

)

2. Electric dipole transitions are suppressed at very low energy because of the centrifugal barrier: a Pwave is suppressed near the origin. Starting from the expression (11.84) of the quantized magnetic field,

93

we have to retain transitions between zero photon states and one photon states, which are driven by thea†~kλ

term

〈1 photon|a†~kλ|0 photons〉 = 1

Thus we are left with W ′, as given in the statement of the problem.

3. The term2π

~|〈f |W |i〉|2δ(~ω − (Ei − Ef ))

comes from the Fermi Golden rule (9.170), F is the flux factor and

Vω2dω

(2π)3c3

is the final photon space phase.

4. The angular momentum is ~J = (~/2)(~σp + ~σn) and ~J |χs〉 = 0 as |χs〉 is a state with zero angularmomentum. In the same way

〈χs|~σp|χs〉 = 0

because ~σp is a vector operator whose matrix elements are zero from the Wigner-Eckart theorem, if it issandwiched between two zero angular momentum states. If ψi(~r) is the spatial wave function of a 3S1

state, the potential is the same as in the deuteron case, and from the orthogonality of wave functionscorresponding to two different values of the energy, we have

∫

d3r ψ∗f (~r)ψi(~r) =

∫

d3r ψ∗D(~r)ψi(~r) = 0

If, on the contrary, the initial wave function corresponds to a 1S0 state, the integral does not vanish

∫

d3r ψ∗D(~r)ψi(~r) = − ND√

4π

∫

dΩ

∫ ∞

0

r2dre−κr

r

asr

(

1 − asr

)

= −NDas√

4π

∫ ∞

0

dr e−κr(

1 − asr

)

=ND

√4π

κ2(1 − κas)

5. We use the completeness relation (11.80)

∑

λ

eiλ(~k)ejλ(

~k) = δij − kikj

and

∑

m

|〈χmt |~σp|χs〉|2 =∑

m

〈χs~σp|χmt 〉〈χmt |~σp|χs〉

=∑

m

〈χs~σp|χmt 〉 · 〈χmt |~σp|χs〉 + 〈χs|~σp|χs〉 · 〈χs|~σp|χs〉

where we have used 〈χs|~σp|χs〉 = 0. Then we use the completeness relation in the four dimensionalHilbert space of the two spins

∑

m

|χmt 〉〈χmt | + |χs〉〈χs| = I

so that

|〈|W ′spin|2〉 =

1

4〈χs|~σ2

p − (~σp · k)2|χs〉 =1

2

because ~σ2p = 3 and (~σp · k)2 = 1.

6. Let us summarize the various factors


1. A factor~

2ε0Voriginates in the expression of the quantized magnetic field.

2. A factor1

4

q2p~2

4M2(gp − gn)

2

originates in the coupling of the magnetic moments with the quantized magnetic field.

3. A factor 1/2 comes from the spin summation.

4. A factor4πN2

D

κ4(1 − κas)

2 =8π

κ3(1 − κas)

2

originates in the overlap integral of the spatial wave functions.

5. As dσ/dΩ is a isotropic, a factor 4π arises from the dΩ integration.

6. ~ω = B from energy conservation.

One finds the following for the numerical value of the theoretical estimate

σ = 7.72 × 10−4 (MeV)2 = 30.9 (fm)2

Chapter 15


15.5.1 POVM as a projective measurement in a direct sum

Multiplying the normalized two component vectors of (15.31) by√

2/3, we form the 3 × 2 matrix

M =

( √

2/3 −√

1/6 −√

1/6

0√

1/2 −√

1/2

)

Let us call |v1〉 and |v2〉 the three-dimensional vectors whose components are the first rows of the matrixM . They obey ||v1||2 = ||v2||2 = 1 and 〈v1|v2〉 = 0. We complete the matrix M to a 3 × 3 matrix M byadding a third row made of the components of a vector |v3〉

|v3〉 = (√

1/3,√

1/3,√

1/3)

which is normalized and orthogonal to |v1〉 and |v2〉. By construction, M is and orthogonal matrix, sothat the vectors |uα〉 whose components are given by the columns of M are normalized and mutuallyorthogonal. Now, let us consider a projective measurement in H(3)

P1 = |u1〉〈u1| P2 = |u2〉〈u2| P3 = |u3〉〈u3|

Assume that the state operator ρ has non vanishing matrix elements only in H(2). Then the probabiltyof result α is

p(α) = Tr (ρ|uα〉〈uα| = 〈uα|ρ|uα〉 = 〈α|ρ|α〉

15.5.3 A POVM with two arbitrary qubit states

1. The two projectors are

Pa⊥ =

(

sin2 α − sinα cosα− sinα cosα cos2 α

)

Pb⊥ =

(

cos2 α − sinα cosα− sinα cosα sin2 α

)

To build a POVM with a third vector |c〉 we must have

(

A+ |λ|2B −2 sinα cosα+Bλµ∗

−2 sinα cosα+Bλ∗µ A+ |µ|2B

)

= I

in order that (15.23) be satisfied, whence |λ| = |µ| = 1/√

2. Choosing λ = µ = 1/√

2 we obtain

A =1

1 + sin 2α=

1

1 + S

B =2 sin 2α

1 + sin 2α=

2S

1 + S

95


The POVM is

Qa⊥ =1

1 + S|a⊥〉〈a⊥|

Qb⊥ =1

1 + S|b⊥〉〈b⊥|

Qc =S

1 + S|c〉〈c|

2. The state operator of the qubits sent by Alice is

ρ =1

2|a〉〈a| + 1

2|b〉〈b|

If Bob measures the result a⊥, he knows with certainty that the spin was in the state |b〉, because hewould have got zero had the spin been in state |a〉. The probability for finding a⊥ is

p(a⊥) = Tr (Qa⊥ρ) =1

1 + STr (|a⊥〉〈a⊥|ρ) =

1

2(1 − S)

If sinα = 1/2, then p(a⊥) = p(b⊥) = 1/4 and in 50% of the cases Bob will make the right guess. Inquantum cryptography, Eve uses α = π/8, so that S = 1/

√2, (1 − S)/2 ≃ 0.145. Thus, if she uses a

POVM, Eve can be sure of the state sent by Alice in 58% of the cases. In the remaining 42%, she decidesrandomly, with a 50% probality of success. Thus she will get the correct result in (58+21)%=79% of thecases.

15.5.7 Superposition of coherent states

1. The term [H0, ρ] does not contribute to the evolution of ρnn because H0 is diagonal in the |n〉 basis.Furthermore

〈n|a†aρ+ ρa†a|n〉 = 2nρnn

〈n|aρa†|n〉 = (n+ 1)ρn+1,n+1

whence the time evolution of ρnn

dρnndt

= −nΓρnn + (n+ 1)Γρn+1,n+1

If we choose n = 0, we find dρ/dt = Γρ11, which means that the population of the ground state increasesat a rate proportional to that of the first excited state times Γ. The corresponding physical process is thespontaneous emission of a photon, so that Γ is the rate for spontaneous emission. The evolution equationfor ρn+1,n is obtained from

〈n+ 1|[H0, ρ]|n〉 = ~ω0ρn+1,n

〈n+ 1|aρa†|n〉 =√

(n+ 1)(n+ 2) ρn+2,n+1

〈n+ 1|a†a, ρ|n〉 = (2n+ 1)ρn+1,n+1

so thatdρn+1,n

dt= −iω0ρn+1,n + Γ

√

(n+ 1)(n+ 2) ρn+2,n+1 −1

2Γ(2n+ 1)ρn+1,n

2. Using (2.54) we obtain

eλ∗a a† e−λ

∗a = a† + λ∗[a, a†] = a† + λ∗

that is

a† e−λ∗a = e−λ

∗a(

a† + λ∗)

97

Taking the derivative of C(λ, λ∗; t) with respect to λ we obtain

∂

∂λTr(

ρ eλa†

e−λ∗a)

= Tr(

ρ a† eλa†

e−λ∗a)

= Tr(

ρ eλa†

e−λ∗a(a† + λ∗)

)

= Tr(

(a† + λ∗)ρ eλa†

e−λ∗a)

where we have used the invariance of the trace under circular permutations to derive the last line. Thisequation can be written schematically as

(

∂

∂λ− λ∗

)

→ a† ρ while clearly∂

∂λ→ ρ a†

Similarly we get for ∂/∂λ∗

∂

∂λ∗Tr(

ρ eλa†

e−λ∗a)

= −Tr(

ρ eλa†

a e−λ∗a)

= −Tr(

ρ (a− λ)e λa†

a e−λ∗a)

which can be rewritten as(

λ− ∂

∂λ∗

)

→ ρa while − ∂

∂λ∗→ a ρ

3. Let us examine the different terms in the RHS of (15.79). From the results of the preceding question

a†aρ →(

∂

∂λ− λ∗

)(

− ∂

∂λ∗

)

= − ∂2

∂λ∂λ∗+ λ∗

∂

∂λ∗

ρa†a →(

λ− ∂

∂λ∗

)(

∂

∂λ

)

= − ∂2

∂λ∂λ∗+ λ

∂

∂λ

so that

[a†a, ρ] → λ∗∂

∂λ∗− λ

∂

∂λ

Similarly

aρa† → ∂2

∂λ∂λ∗

and

a†a, ρ = −(

∂

∂λ− λ∗

)(

− ∂

∂λ∗

)

+

(

λ− ∂

∂λ∗

)(

∂

∂λ

)

= − ∂2

∂λ∂λ∗+ λ∗

∂

∂λ∗+ λ

∂

∂λ

Assembling all these results, we finally get the partial differential equation

[

∂

∂t+

(

Γ

2− iω0

)

λ∂

∂λ+

(

Γ

2+ iω0

)

λ∗∂

∂λ∗

]

C(λ, λ∗; t) = 0

or[

∂

∂t+

(

Γ

2− iω0

)

∂

∂ lnλ+

(

Γ

2+ iω0

)

∂

∂ lnλ∗

]

C(λ, λ∗; t) = 0

To implement the method of characteristics we write

dt

1=

d lnλ

Γ/2 − iω0=

d lnλ

Γ/2 + iω0

whenceλ = λ0 exp[(Γ/2 − iω0)t] λ∗ = λ∗0 exp[(Γ/2 + iω0)t]


or solving for λ0, λ∗0

λ0 = λ exp[−(Γ/2 − iω0)t] λ0 = λ∗ exp[−(Γ/2 + iω0)t]

λ exp[−(Γ/2− iω0)t] and λ exp[−(Γ/2 + iω0)t] are constants along the characteristics. The partial differ-ential equation for C(λ, λ∗; t) tells us that this function is constant along the characteristics

C(λ, λ∗; t) = C0(λ, λ∗; t = 0) = C0(λ, λ

∗)

4. The state operator at time t = 0 is

C0(λ, λ∗) = Tr

(

|z〉〈z| eλa† e−λ∗a)

= 〈z|eλa† e−λ∗a|z〉 = exp(λz∗ − λ∗z)

We then have at time t

C(λ, λ∗; t) = exp[

z∗λe−(Γ/2−iω0)t, λ∗ze−(Γ/2+iω0)t]

which can be written asC(λ, λ∗; t) = exp [λz∗(t) − λ∗z(t)]

withz(t) = ze−(Γ/2+iω0)t

Thus C(λ, λ∗; t) corresponds to the coherent state

|z(t)〉 = |z e−iω0t e−Γt/2〉

5. When |Φ〉 is a superposition of coherent states

|Φ〉 = c1|z1〉 + c2|z2〉

then

C(λ, λ∗; t = 0) = |c1|2e(λz∗1−λ∗z1) + |c2|2e(λz∗2−λ

∗z2) + c1c∗2〈z2|z1〉e(λz∗2−λ

∗z1) + c∗1c2〈z1|z2〉e(λz∗1−λ∗z2)

The last two terms originate in the fact that |Φ〉 is a coherent superposition, while these two terms wouldbe absent in an incoherent superposition of the two coherent states. At time t we have

C(λ, λ∗; t) = |c1|2e[λz∗1 (t)−λ∗z1(t)] + |c2|2e[λz∗2(t)−λ∗z2(t)]

+ c1c∗2〈z2|z1〉e[λz∗2 (t)−λ∗z1(t)] + c∗1c2〈z1|z2〉e[λz∗1 (t)−λ∗z2(t)]

Note that the scalar products in the last two terms are 〈z2|z1〉 and 〈z1|z2〉, and not 〈z2(t)|z1(t)〉 and〈z1(t)|z2(t)〉. In order to recover the same form as at t = 0 form, we must write, for example

〈z2|z1〉 =〈z2|z1〉

〈z2(t)|z1(t)〉〈z2(t)|z1(t)〉 = η(t)〈z2(t)|z1(t)〉

We can then describe the final state as a linear superposition of two coherent states, but the coherenceis reduced by a factor

|η(t)| = exp

[

−1

2|z1 − z2|2

(

1 − e−Γt)

]

≃ exp

[

−Γ

2|z1 − z2|2

]

The coherence is then damped with a rate which is |z1 − z2|2 larger that the damping rate Γ of theindividual coherent states. The decoherence time is then

τdec =2

Γ|z1 − z2|2

99

5. At time t = 0, we have for the oscillator the superposition

|Φ(t = 0)〉 = c1|0〉 + c2|z〉

The global state vector at t = 0 is

|Ψ(t = 0)〉 = c1|0 ⊗ |0F 〉 + c2|z ⊗ 0F 〉

where |0F 〉 is the state vector (vacuum state) of the radiation field, since at T = 0 there are no availablephotons (or phonons). The first component of |Ψ〉 stays unchanged under the time evolution, becausespontaneous emission cannot exist. By contrast, a photon will be emitted on average after a time ∼ Γ|z|2due the second component of |Ψ〉. Indeed, the time evolution of ρnn in question 1 tells us that thedecay amplitude of an excited state |n〉 is nΓ, and the average number 〈n〉 is equal to |z|2, |z|2 = 〈n〉 inthe coherent state |z〉. As soon as one photon is emitted, the two components of |Φ〉 become entangledto orthogonal states of the environment, and the reduced state matrix of the oscillator loses all phasecoherence. The decoherence time is thus the average time for the emission of a single photon, andτdec ≃ 1/|z|2Γ.

15.5.11 The Fokker-Planck-Kramers equation for a Brownian particle

1. Let us show the equivalence of the two formulae for w(x, p; t)

w(x, p; t) =1

2π~

∫ +∞

−∞

e−ipy/~ 〈x+y

2|ρ(t)|x − y

2〉dy (W1)

and

w(x, p; t) =1

2π~

∫ +∞

−∞

e−ixz/~ 〈p+z

2|ρ(t)|x − z

2〉dz (W2)

We use the completeness relation and (9.22) to write

|x− y

2〉 =

1

2π~

∫ +∞

−∞

dq |q〉〈q|x − y

2〉 =

1√2π~

∫ +∞

−∞

dq e−iq(x−y/2)|q〉

and an analogous formula for 〈x+ y/2|. Plugging the two formulae in (W1) and integrating over y leadsto

w(x, p; t) =2

2π~

∫ +∞

−∞

dq e−2i(p−q)x〈2p− q|ρ(t)|q〉

A change of variable p− q = z/2 allows us to recover (W2).

2. Let us consider a Gaussian wave packet (Exercise 9.7.3)

ϕ(x) =

(

1

πσ2

)1/4

exp

(

− x2

2σ2

)

A straightforward calculation shows that

w(x, p) =1

π~exp

(

−x2

σ2

)

exp

(

−p2σ2

~2

)

Let us now consider the superposition (15.143)

ϕ(x) ≃ 1√2

(

1

πσ2

)1/4(

exp[

− (x− a)2

2σ2

]

+ exp[

− (x+ a)2

2σ2

]

)

with a≫ σ. We find

w(x, p) =1√2 π~

exp

(

−p2σ2

~2

)[

exp[

− (x− a)2

2σ2

]

+ exp[

− (x+ a)2

2σ2

]

+ 2 exp

(

−x2

σ2

)

cos

(

2ap

~

)]


The first two terms would correspond to an incoherent superposition of two Gaussian wave packets, butthe last one reflects the coherence of the two wave packets.

3. We limit ourselves to the second term [X, P, ρ] in the RHS of (15.142), as the other two terms canbe dealt with using exactly the same techniques. Using the (W1) form of w we obtain at once

w2 =1

2π~

∫ +∞

−∞

e−ipy/~ 〈x+y

2|P, ρ|x− y

2〉 ydy

=i

2π

∂

∂p

∫ +∞

−∞

e−ipy/~ 〈x+y

2|P, ρ|x− y

2〉dy

We then use the W2 form

w2 =1

2π~

∂

∂p

∫ +∞

−∞

e−ipy/~ 〈x+y

2|P, ρ|p− y

2〉dy

=i

2π

∂

∂pp

∫ +∞

−∞

e−ixy/~ 〈p+y

2|P, ρ|x− y

2〉dy

= i~∂

∂p[pw(x, p; t)]

4. One has only to observe that w(x, p; t) must vanish when x→ ±∞.

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