Solutions of selected exercises from ‘Quantum Physics’ Michel Le Bellac

2

Contents 1 Exercices from Chapter 1

5

2 Exercices from Chapter 2

11

3 Exercices from Chapter 3

15

4 Exercices from Chapter 4

19

5 Exercises from Chapter 5

23

6 Exercises from Chapter 6

27

7 Exercises from Chapter 7

37

8 Exercises from Chapter 8

39

9 Exercises from Chapter 9

43

10 Exercises from chapter 10

51

11 Exercises from Chapter 11

59

12 exercises from Chapter 12

71

13 Exercises from Chapter 13

83

14 Exercises from Chapter 14

87

15 Exercises from Chapter 15

95

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 numerical computations. In the case of photons, the energy is in eV Ephot =

6.63 × 10−34 × 3 × 108 hc = 1.24 × 104 eV = 12.4 keV = λ 10−10 × 1.6 × 10−19

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

p2 h2 = = 8.2 × 10−2 eV = 82 meV 2mn 2mn λ2

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

mn = 151 eV me

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 with energy of a few ten meV, rather than to that of a photon with an energy of 10 keV in order to detect the creation of one phonon. 3. The mass of a fullerene molecule is M = 1.2 × 10−24 kg and its wavelength λ=

h = 2.5 × 10−12 m mv

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 r0 M1 + M2

and the molecule moment of inertia is I = M1 r12 + M2 r22 =

M1 M2 2 r = µr02 M1 + M2 0

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

1 2 J2 Iω = 2 2I

5

6

CHAPTER 1. EXERCICES FROM CHAPTER 1

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

~2 m ~2 = = 2 R∞ 2 2 2 2µr0 2µb a0 b µ

using R∞ = e2 /(2a0 ) and a0 = ~2 /(me2 ). 5. The elastic constant K is K=

2 2cR∞ 4cmR∞ = 2 b 2 a0 ~2 b 2

that is ~ωv = 2

r

c b2

r

m R∞ µ

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, which are indeed numbers close to one. 6. The dimension of G is easily obtained by observing that Gm2 /r is an energy. One finds that this p −1 3 −2 dimension is given by M L T . The quantity ~c/G has the dimension of a mass, which gives for the Planck energy r ~c 2 c = 1.9 × 109 J = 1.2 × 1019 GeV EP = G and for the Planck length r ~ G lP = = 1.6 × 10−35 m c ~c 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 compared to 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 at point ~r = 0, and the scattered wave suffers a phase shift δsc = −~k ′ · ~ri with respect to the wave scattered by 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 X n=0

xn =

1 − xN 1−x

one can, for example, evaluate the sum Σx =

N −1 X n=0

e−iqx na =

1 − e−iqx aN sin qx aN/2 = e−iqx a(N −1)/2 1 − e−iqx a sin qx a/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: qx a = 2πnx + ε. Then qx aN εN N sin = ± sin = sin πnx N + ε 2 2 2 1 ε qx a = sin πnx + ε = ± sin sin 2 2 2

7 The peak width is then ε ∼ 1/N and its height is obtained taking the limit ε → 0 sin2 εN/2 = N2 ε→0 sin2 ε/2 lim

which gives an intensity within the peak ∼ N 2 × 1/N = N . The same calculation can be repeated in the y direction. 4. The condition for elastic scattering is ~k 2 = ~k ′ 2 = (~k + ~q)2 = k 2 + 2~q · ~k + q 2 that is q 2 + 2~ q · ~k = 0. Suppose that nx = 0 (or qx = 0) and thus kx′ = kx kx′ = kx

ky′ = ky −

2πny b

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

πny b

ky′ = −

πny b

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

πny bk

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 kx′ would not be linked to kx through kx′ = kx + qx . One would then obtain diffraction maxima for any angle of incidence 5. The sum over the cells is N −1 M−1 X X

e−i(2aqx n+2bqy m) = F (2aqx , 2bqy )

n=0 m=0

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 =

πnx a

qy =

πny b

and it follows that The final result is

f = f1 1 + (−1)nx +ny + f2 [(−1)nx + (−1)ny ]

• 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 peaks are lost.

1.6.6 Mach-Zehnder interferometer

8

CHAPTER 1. EXERCICES FROM CHAPTER 1

1. Let b1 (b2 ) the probability amplitude for finding the photon in the upper (lower) arm of the interferometer. The probability amplitudes a1 (a2 ) that the photon triggers the detector D1 (D2 ) are obtained by examining the transmission by the beam splitter S2 : for example, a1 is obtained by adding the amplitude rb1 where the photon originating from the upper arm is reflected by the beam splitter and the amplitude tb2 where the photon originating from the lower arm is transmitted by the beam splitter a1 a2

= =

rb1 eiδ + tb2 tb1 eiδ + rb2

We have made explicit the possibility of a variable phase shift δ in the upper arm. One calculates now b1 et b2 as functions of a0 b1 = ta0 b2 = ra0 and plugs the result in the preceding equations a1

= rta0 1 + eiδ

a2

= a0 t2 eiδ + r2

Using the values of t and r given in the statement of the problem and choosing the (arbitrary) normalization |a0 | = 1 (the photon has unit probability for arriving at S1 ) p1

=

p2

= =

1 (1 + cos δ) 2 2 1 |a2 |2 = e2iα eiδ + e2iβ 4 1 [1 + cos(2α − 2β + δ)] 2

|a1 |2 =

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, which implies that cos 2(α − β) = −1, that is α−β =

π mod nπ 2

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 = a0 r2 t eiχ + 1 a2 = a0 r t2 eiχ + r2

and the probabilities by taking the modulus squared p1 p2

=

A(1 + cos χ)

=

′

B + A cos χ

A = 2|r2 t|2 A′ = 2|r2 |Re t2 (r∗ )2 eiχ

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

cos 2(α − β) cos χ − sin 2(α − β) sin χ = − cos χ and thus α−β =

π mod nπ 2

3. At elevation z the neutron energy is K = K0 + mgz if its energy is taken, by convention, to be K0 for z = 0. Its momentum is p p √ mgz p = 2mK = 2m(K0 + mgz) ≃ 2mK0 1 + 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 = k

mgz 2K0

∆k mgz = k 2K0

On a path with length L, the phase shift accumulated between the two arms, one at elevation z and the other at elevation z = 0 is mgzLk mgkS m2 gS ∆φ = ∆kL = = = 2 2K0 2K0 ~ k because zL is the rhombus area and2K0 = ~2 k 2 /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 =

m2 gS cos θ ~2 k

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, hα2i i = hαi i = p1 , while if i 6= j hαi αj i = hαi ihαj i = p21 the two results being summarized in hαi αj i = δij p1 + (1 − δij )p21 = p21 + p1 p2 δij 2. The scattering probabiilty by the crystal is X D E αi f1 + (1 − αi )f2 αj f1 + (1 − αj )f2 ei~q·(~ri −~rj ) h|ftot |2 i = i,j

=

X i,j

=

X i,j

(p21 + p1 p2 δij )f12 + 2p1 p2 (1 − δij )f1 f2 + (p22 + p1 p2 δij )f22 ei~q·(~ri −~rj )

(p1 f1 + p2 f2 )2 ei~q·(~ri −~rj ) + N p1 p2 (f1 − f2 )2

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

10

CHAPTER 1. EXERCICES FROM CHAPTER 1

Chapter 2

Exercices from Chapter 2 2.4.3 Determinant and trace 1. Let A(t) = A(0) exp(Bt). Let us compute the derivative d A(0)eBt = A(0)eBt B = A(t)B dt The solution of

dA = BA(t) dt

is A(t) = eBt A(0) 2. One remarks that for infinitesimal δt det eAδt ≃ det(I + Aδt) = 1 + δtTr A + O(δt)2 For example, for a 2 × 2 matrix 1 + A11 δt A12 δt det = 1 + (A11 + A22 )δt + (A11 A22 − A12 A21 )(δt)2 A21 δt 1 + A22 δt Let g(t) = det[exp(At)] g ′ (t)

1 det eA(t+δt) − det eAt δt→0 δt 1 1 = det eAδt − 1 det eAt = [δtTr A] det eAt = Tr Ag(t) δt δt

=

lim

and one obtains for g(t) the differential equation

g ′ (t) = [Tr A]g(t) =⇒ g(t) = etTr A using the boundary conditiom g(0) = 1. Setting t = 1 we find g(1) = eTr A = 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†

11

C = −C †

12

CHAPTER 2. EXERCICES FROM CHAPTER 2

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 a more explicit proof. Let, for example, x = (x1 , x2 , 0, . . . , 0) Then (x, Cx) = x∗1 C12 x2 + x∗2 C21 x1 = 2i Im(x∗1 C12 x2 ) = 0 =⇒ C12 = 0 Since A is Hermitiian, it can be diagonalized. Let ϕ be an eigenvector of A, Aϕ = aϕ. The positivity condition implies (ϕ, Aϕ) = a||ϕ||2 ≥ 0 and thus a ≥ 0. 2. In the case of a real and antisymmetric matrix C T = −C

(x, Cx) = x1 C12 x2 + x2 C21 x1 = x1 (C12 + C21 )x2 = 0 One can thus have a positive matrix of the form BT = B

A=B+C

C T = −C 6= 0

2.4.11 Operator identities 1. Let us compute df /dt df = etA ABe−tA − etA BAe−tA = etA [A, B]e−tA dt The second derivative is computed in the same way, and the general case is obtained by recursion. 2. Let us compute dg/dt dg = etA (A + B)etB dt and use the result of the preceding question etA B = etA Be−tA etA = (B + t[A, B])etA Indeed, because of the commutation relations of [A, B] with A and B, the series expansion stops after the second term. We get the differential equation dg = A + B + t[A, B] g(t) dt

Taking the commutation relations into account, this equation has the solution 1

2

g(t) = e(A+B)t+ 2 [A,B]t

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 determinant of R′ then obeys | det R′ | = 1, which can be written det R′ = exp(iθ) 2. One defines R through R = ie−iθ/2 R′

det R = −e−iθ det R′ = −1

This redefinition corresponds to a global change of phase of the state vectors phase. One checks that with the form of R given in the statement of the problem R† R = I

det R = −|r|2 − |t|2 = −1

Let us first choose ψ = (1, 0) Rψ =

|r|eiχ |t|eiφ

|t|e−iφ −|r|e−iχ

iχ 1 |r|e = 0 |t|eiφ

and then ψ = (0, 1) Rψ =

|r|eiχ |t|eiφ

|t|e−iφ −|r|e−iχ

0 |t|e−iφ = 1 −|r|e−iχ

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

= =

χ −(χ − π)

δTD = φ δTG = −φ

δD = χ − φ δG = π − (χ − φ)

so that δD + δG = π

√ If the beam splitter is symmetric, one must have t = t∗ et r = −r∗ as well as |t| = |r| = 1/ 2 and δD = δG = π/2.

14

CHAPTER 2. EXERCICES FROM CHAPTER 2

Chapter 3

Exercices from 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 tan α = 1.09 mm the beams are well separated. The difference in optical paths is n′e = 0.9248 mm δ = e no − cos α 2. The index difference is no − ne = 0.17102 and the thickness of the intermediary plate D=

2 × 0.9248 = 10.815 mm no − ne

3. Let β = 1/ cos α D= One infers from this the relative error δD/D

2e(no − βn′e ) n0 − ne

δD δno − δne δe δn0 − βδn′e − = + ′ D e n0 − βne no − ne In order to simplify the error calculation, we neglect the difference between ne and n′e δe (β − 1)(ne δno − no δne ) δD = + D e (no − ne )(no − βne which leads to

and to

δD ≃ 0.7 |δn| D

|δD| ≃ 7 × 10−5 mm ≪ λ 5. The beam polarization is elliptic in the region where the two beams overlap, and linear in the region where they don’t.

3.3.4 Other solutions of (3.45)

15

16

CHAPTER 3. EXERCICES FROM CHAPTER 3

1. The action of U on σx and σy is 0 † U σx U = e−iψ

eiψ 0

†

U σy U =

−ieiψ 0

0 ie−iψ

σz is clearly unchanged. 2. The possible solutions of (3.45) are cos(α − αx )

= cos φ

cos(α − αy )

= sin φ

α − αx = φ or α − αx = −φ π π α − αy = − φ or α − αy = φ − 2 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 two possible solutions • Solution 1

α − αx = φ

α − αy = φ −

that is αx = αy − and σx =

0 eiαx

e−iαx 0

π 2

π 2

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 2

α − αx = −φ

α − αy =

π −φ 2

Choosing as a reference solution αx = 0 et αy = −π/2 0 1 0 i σx = σy = 1 0 −i 0 The change of sign of σy corresponds to an inversion of the Oy axis: one goes from a right handed referential to a left handed one. The other solutions are obtained from the reference solution by a rotation about the Oz axis.

3.3.6 Exponentials of Pauli matrices 1. From (3.50) (~σ · pˆ)2 = I

(~σ · pˆ)3 = (~σ · pˆ) . . .

the series expansion of the exponential is 2 3 1 θ 1 θ θ θ I+ ~σ · pˆ · · · = I − i ~σ · pˆ + −i −i exp −i ~σ · pˆ 2 2 2! 2 3! 2 θ θ = I cos − i(~σ · pˆ) sin 2 2 Taking pˆ = (− sin φ, cos φ, 0) we get θ exp −i ~σ · pˆ = 2 =

θ θ θ I cos + iσx sin φ sin − iσy cos φ sin 2 2 2 cos 2θ e−iφ sin θ2 cos θ2 eiφ sin 2θ

17 2. We must have U

=

a1 I + ia2 σz + ib2 σx + ib1 σy θ θ nz I cos − i sin n 2 2 x + iny

=

nx − iny −nz

and we deduce from this a1 = cos

θ 2

a2 = −nz sin

θ 2

b2 = −nx sin

θ 2

b1 = −ny sin

θ 2

These equations have solutions because a21 + a22 + b21 + b22 = 1 3. The product of two exponentials of Pauli matrices is ˆ e−iα(~σ ·ˆa) e−iβ(~σ·b) = cos α cos β − i sin α cos β(~σ · a ˆ) − i sin β cos α(~σ · ˆb) − sin α sin β[ˆ a · ˆb + i~σ · (ˆ a × ˆb)]

On the other hand sin ||αˆ a + βˆb|| ˆ [α(~σ · a ˆ) + β(~σ · ˆb)] e−i[α(~σ ·ˆa)+β(~σ·b)] = I cos ||αˆ a + βˆb|| − i ||αˆ a + βˆb|| In order to ensure the equality of the two factors, we must • get rid of the sines; • have cos α cos β = cos

p α2 + β 2

One can choose, for example

α = 3π

β = 4π

with e−4iπσy = I

e−3iπσx = −I

p α2 + β 2 = 5π 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 was scattered, and we must add the amplitudes X X ei~q·(~ri −~rj ) I = fa2 ei~q·~ri f = fa i,j

i

2. If the scattering is accompanied with a spin flip, it leaves the nucleus in a state which is different from its initial state. If all the nuclei had initially a down spin, the nucleus which scattered the neutron could be in principle identified (even though this identification would impossible in practice). Neutron scattering from a given nucleus rather than from another one corresponds to different nucleus final states, and we must add probabilities X fb2 = N fb2 I= i

3. Let {αi } define the spin configuration in the crystal. If a neutron is scattered by the crystal in the configuration {αi }, the scattering amplitude is X X αi fb ei~q·~ri (αi fa + (1 − αi )fc ) ei~q·~ri + f= i

i

18

CHAPTER 3. EXERCICES FROM CHAPTER 3

Were the configuration {αi } fixed, the intensity would be X X 2 α2i fb2 (αi fa + (1 − αi )fc ) ei~q·(~ri −~rj ) + Iαi = i,j

i

Observing that αi = α2i the average values are hαi i = hα2i i = 1/2 and hαi αj i = 1/4 if i 6= j, whence X X

hαi ifb2 αi αj fa2 + 2αi (1 − αj )fa fc + (1 − αi )(1 − αj )fc2 ei~q·(~ri −~rj ) + I= i

i,j

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

X 1 N ei~q·(~ri −~rj ) + (fa + fc )2 [(fa − fc )2 + 2fb2 ] 4 4 i,j

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 question if the neutrons are all polarized with a down spin. The result for unpolarized neutrons is obtained by taking the average of the result with spin up and spin down, and one again finds the result of question 3.

Chapter 4

Exercices from 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)i is decomposed on the eigenvectors |χ± i as |ϕ(t = 0)i = λ|χ+ i + µ|χ− i then the time evolution is |ϕ(t)i = λe−iΩt/2 |χ+ i + µeiΩt/2 |χ− i Taking into account h+|χ+ i = cos θ/2 and h+|χ− i = − sin θ/2, one finds for c+ (t) c+ (t) = h+|ϕ(t)i = λe−iΩt/2 cos

θ θ − µeiΩt/2 sin 2 2

3. If c+ (0) = 0 λ cos

θ θ − µ sin = 0 2 2

and a possible solution is λ = sin

θ 2

µ = cos

θ 2

which gives the following for c+ (t) c+ (t) = − sin

Ωt θ θ −iΩt/2 e − eiΩt/2 = i sin θ sin cos 2 2 2

The probability p+ (t) is p+ (t) = |c+ (t)|2 = sin2 θ sin2

Ωt Ωt B2 sin2 = 2 2 A + B2 2

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

θ 2

µ = − sin

19

θ 2

20

CHAPTER 4. EXERCICES FROM CHAPTER 4

and one obtains for c+ (t) θ θ −iΩt/2 e + sin2 eiΩt/2 2 2 Ωt Ωt cos − i cos θ sin 2 2

cos2

c+ (t) = = The probability p+ (t) is p+ (t) = |c+ (t)|2 = cos2

Ωt Ωt Ωt + cos2 θ sin2 = 1 − sin2 θ cos2 2 2 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 1 hHit − 2 hH 2 it2 + O(t3 ) ~ 2~

so that |c(t)|2

hH 2 i − hHi2 2 t + O(t3 ) ~2 (∆H)2 2 t + O(t3 ) = 1− ~2

= 1−

2. From (4.27) we derive, with the substitution A → P 1 dP ∆P∆H ≤ ~ 2 dt

p(t) = hPi(t)

while

∆P = (hP 2 i − hPi2 )1/2 = (hPi − hPi2 )1/2 =

We thus obtain the differential inequality

p p(1 − p)

dp ∆H p dt ≥2 ~ p(1 − p)

which integrates into

cos−1 [1 − 2p(t)] ≥ 2 whence p(t) ≥ cos2

t∆H ~

∆H t ~

0≤t≤

π~ 2∆H

3. Inserting a complete set of eigenstates |ni of H in the expression for c(t), we obtain X c(t) = hϕ(0)|e−iHt/~ |nihn|ϕ(0)i n

X

=

n

|hϕ(0)|ni|2 e−iEn t/~

Let the spectral function w(E) be the inverse Fourier transform of c(t) Z dt iEt/~ w(E) = e c(t) 2π Z dt iEt/~ X e |hϕ(0)|ni|2 e−iEn t/~ = 2π n X 2 = |hϕ(0)|ni| δ(E − En ) n

21 The expectation values of H and H 2 are hHi

=

hH 2 i

=

X n

Z

2

En |hϕ(0)|ni| =

Z

dE E w(E)

dE E 2 w(E)

4. Use the following identity, obtained from a suitable contour in the complex x-plane Z +∞ π eitx = e−α|t| dx 2 2 x + α α −∞

4.4.6 The solar neutrino enigma 1. The Hamiltonian reads, in the reference frame where the neutrino is at rest me −mµ 1 me + mµ m 2 H= I+ m −m c2 2 m − e2 µ Comparison with exercise 4.4.4 leads to the correspondence A→

me − mµ 2

B→m

tan θ =

B 2m → A me − mµ

2. Since the states |ν1 i and |ν2 i are eigenvectors of H, the time evolution of |ϕ(t)i is θ θ |ϕ(t)i = cos e−iE1 t/~ |ν1 i − sin e−iE2 t/~ |ν2 i 2 2 Taking into account hνe |ν1 i = cos

θ 2

hνe |ν2 i = − sin

θ 2

one finds for the amplitude ce (t) ce (t) = hνe |ϕ(t)i = e

−iE1 t/~

2 θ −i(E2 −E1 )t/~ 2 θ + sin e cos 2 2

and for the probability |ce (t)|2 |ce (t)|2

= =

θ θ θ ∆Et θ + sin4 + 2 cos2 sin2 cos 2 2 2 2 ~ 1 ∆Et ∆Et 1 − sin2 θ 1 − cos = 1 − sin2 θ sin2 2 ~ 2~

cos4

3. When p ≫ mc one obtains the following approximate expression for E 1/2 m 2 c3 m 2 c2 ≃ cp + E = (m2 c4 + p2 c2 )1/2 = cp 1 + 2 p 2p and ∆E = E2 − E1 =

∆m2 c3 (m22 − m21 )c3 = 2p 2p

Substituting the distance L travelled by the neutrino to ct, the oscillation takes the form ∆m2 c2 L sin2 2p~

22

CHAPTER 4. EXERCICES FROM CHAPTER 4

If half of an oscillation is observed between the Sun and the Earth ∆m2 c2 L =π 2p~

∆m2 c4 =

2π~c cp ≃ 7 × 10−11 eV2 L

and thus ∆mc2 ∼ 10−5 eV.

4.4.8 The neutral K meson system 1. Let us compute the product C −1 M C for a generic matrix M a b d 0 1 σx = C = C −1 = σx = σx c d b 1 0

c a

The commutation relation implies a = d and b = c. 2. The eigenvalues and eigenvectors of M are λ+

= A+B

λ−

= A−B

1 0 |K1 i = √ |K 0 i + |K i 2 1 0 |K1 i = √ |K i − |K 0 i 2

The time evolution of the states |K1 i et |K2 i is given by E1 t Γ1 t |K1 i − e−i(A+B)t |K1 i = exp −i ~ 2 E2 t Γ2 t e−i(A−B)t |K1 i = exp −i |K1 i − ~ 2

|K1 (t)i = |K2 (t)i = Let us start at time t = 0 from

1 1 |ϕ(t = 0)i = √ (c(0) + c(0)) |K1 i + √ (c(0) − c(0)) |K2 i 2 2 One gets at time t |ϕ(t)i

= +

1 (c(0) + c(0)) |K 0 i + |K 0 i e−i(E1 /~−iΓ1 /2)t 2 1 (c(0) − c(0)) |K 0 i − |K 0 i e−i(E2 /~−iΓ2 /2)t 2

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

1 (c(0) + c(0)) e−i(E1 /~−iΓ1 /2)t + 2 1 hK 0 |ϕ(t)i = (c(0) + c(0)) e−i(E1 /~−iΓ1 /2)t − 2

hK 0 |ϕ(t)i =

1 (c(0) − c(0)) e−i(E2 /~−iΓ2 /2)t 2 1 (c(0) − c(0)) e−i(E2 /~−iΓ2 /2)t 2

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

i 1 h −i(E1 /~−iΓ1 /2)t e − e−i(E2 /~−Γ2 /2)t 2

The probability to observe a K 0 meson at time t is 1 −Γ1 t ∆Et p(t) = |c(t)|2 = e + e−Γ2 t − 2e−(Γ1 +Γ2 )t/2 cos 4 ~ 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

E0 −A 0 0 −A E0 −A 0 H= 0 −A E0 −A 0 0 −A E0

2. Let us write the action of H on the vector |χi H|χi = E0 |χi − A c1 |ϕ2 i + · · · + cN −1 ϕN i + c0 |ϕ0 i − A c2 |ϕ1 i + · · · cN |ϕN −1 i + cN +1 |ϕN i =

E0 |χi − A

N X

n=1

cn−1 + cn+1 |ϕn i

3. Taking into acount the postulated form for the coefficients cn , we get i c h i(n−1)δ e + ei(n+1)δ − e−i(n−1)δ − e−i(n+1)δ cn−1 + cn+1 = 2i c = 2 cos δ einδ − e−inδ = 2cn cos δ 2i The condition c0 = 0 is satisfied by construction. The condition cN +1 = 0 implies δs =

πs N +1

s = 0, 1, · · · , N

4. From the results of question 3 we have H|χs i = E0 |χs i − 2A cos whence the values of the energy Es = E0 − 2A cos

πs |χi N +1

πs N +1

Let us compute the normalization of |χs i N 2πsn 1X πsn 1 − cos = X= sin N +1 2 n=1 N +1 n=1 N X

2

23

24

CHAPTER 5. EXERCISES FROM CHAPTER 5

The sum over n is readily computed N X 2iπsn 1 − e2iπs 2πsn = Re exp = Re cos − 1 = −1 N +1 N +1 1 − e2iπs/(N +1 n=1 n=1 N X

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

r

N X 2 sin(nδs )|ϕn i N + 1 n=1

5. In the case of butadiene N = 4 cos

π = 0.809 5

cos

2π = 0.309 5

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 i are (0.372, 0.601, 0.601, 0.372) while those of |χ2 i are

(0.601, 0.372, −0.372, −0.601)

The order of the 1-2 bond is h i 1 + 2 hϕ1 |χ1 ihχ1 |ϕ2 i + hϕ1 |χ2 ihχ2 |ϕ2 i = 1.89

which is very close to that of a double bond, while for the 2-3 bond h i 1 + 2 hϕ2 |χ1 ihχ1 |ϕ3 i + hϕ2 |χ2 ihχ2 |ϕ3 i = 1.45

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

5.5.5 The molecular ion H2+ 1. The potential V (x) is V (x) = −e2

1 1 + |x + r/2| |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

E−

=

E0 + A

1 |χ+ i = √ |ϕ1 i + |ϕ2 i 2 1 |χ− i = √ |ϕ1 i − |ϕ2 i 2

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 e2 = E0 ∓ A(r) + r r

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

= E0 + e

2

1 − c e−b/r r

= E0 + ∆E(r)

Let us look for the minimum of ∆E(r) d∆E(r) 1 1 c −r/b c 2 =⇒ 2 = e−r0 /b =e − 2 + e dr r b r0 b One obtains ∆E(r0 ) = e whence b=

2

1 − c e−r0 /b r0

12 6 r0 = a0 5 5

c=

=e

2

1 b − 2 r0 r0

1.38 3 5/6 e ≃ 5a0 a0

One must have b > r0 for the H2+ ion to be a bound state.

5.5.6 The rotating wave approximation in NMR The evolution equation for the state vector |ϕ(t)i ˜ in the rotating reference frame is i~

d|ϕ(t)i ˜ dt

= =

~ ωσz |ϕ(t)i ˜ + e−iωσz t/2 He iωσz t/2 |ϕ(t)i ˜ 2 ~ ˜ ωσz |ϕ(t)i ˜ + H(t)| ϕ(t)i ˜ 2

2. Let us study σ ˜± (t) iω d˜ σ± (t) = − e−iωσz t/2 [σz , σ± ]e iωσz t/2 = ∓iω˜ σ± (t) dt 2 because [σz , σ± ] = ±2σ± . Solving the differential equation gives σ ˜± (t) = e∓iωt σ± Let us rewrite H1 in terms of σ+ and σ− , using σx = σ+ + σ−

whence

σy = −iσ+ + iσ−

~ H1 = − ω1 (σ+ + σ− ) eiωt e−iφ + e−iωt eiφ 2 ˜ 1 (t) = H −

~ − ω1 σ+ e−iφ + e−2iωt eiφ 2 ~ ω1 σ− e2iωt e−iφ + eiφ 2

26

CHAPTER 5. EXERCISES FROM CHAPTER 5

and in the rotating wave approximation, where one neglects terms in exp(±2iωt) ˜1 H

~ω1 σ+ e−iφ + σ− eiφ 2 ~ = − ω1 (σx cos φ + σy sin φ) 2

= −

3. From equation (3.67), exp[−iθ(~σ · pˆ)/2] is the rotation operator of a spin 1/2 about an axis pˆ. By identification, we find pˆ = (cos φ, sin φ, 0) and θ = −ω1 t. The vector n ˆ being normalized (ˆ n2 = 1), we have Ωt Ωt ˜ exp(−iHt/~) = I cos − i(~σ · n ˆ ) sin 2 2 with ω1 δ ~σ · n ˆ = − σx + σz Ω Ω ˜ so that the matrix form of exp(−iHt/~) is ˜

e−iHt/~ =

δ Ωt cos Ωt 2 + Ω sin 2 iω1 Ωt Ω sin 2

cos

iω1 Ωt Ω sin 2 Ωt δ Ωt 2 − Ω sin 2

Chapter 6

Exercises from Chapter 6 6.5.3 Properties of state operators 1. Let us consider a vector |ϕi of the form |ϕi = (0, · · · , ai , 0, · · · , 0, aj , 0, · · · , 0) Positivity of ρ gives hϕ|ρ|ϕi ≥ 0, which implies that the 2 × 2 sub-matrix A ρii ρij A= ρji 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 ≤ ρii ρjj − |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 |ϕi such that Tr ρPϕ = 1, with Pϕ = |ϕihϕ|, and thus hϕ|ρ|ϕi = 1. We choose an orthonormal basis, where, for example, |ϕi is the first basis vector, |ϕi ≡ |1i. In that case, the diagonal elements of ρ obey ρ11 = 1, ρii = 0, i 6= 1 because the test |ii, |ii = 6 |1i has zero probability of success. From the preceding question, all the nondiagonal elements vanish, ρij = 0, i 6= j and ρ = |ϕihϕ| = |1ih1|.

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 |ε1 ε2 i σ1x σ2x | + +i = σ1x σ2x | + −i =

σ1x σ2x | − +i = σ1x σ2x | − −i =

| − −i | − +i

σ1y σ2y | + +i = −| − −i σ1y σ2y | + −i = | − +i

| + −i | + +i

σ1y σ2y | − +i = | + −i σ1y σ2y | − −i = −| + +i

27

28

CHAPTER 6. EXERCISES FROM CHAPTER 6

and σ1z σ2z |ε1 ε2 i = ε1 ε2 |ε1 ε2 i, whence the action of ~σ1 · ~σ2 ~σ1 · ~σ2 | + +i = ~σ1 · ~σ2 | + −i =

| + +i 2| − +i − | + −i

~σ1 · ~σ2 | − −i =

| − −i

~σ1 · ~σ2 | − +i =

2| + −i − | − +i

3. From the results of the preceding question we get ~σ1 · ~σ2 |Ii = |Ii

~σ1 · ~σ2 |IIIi = |IIIi

as well as ~σ1 · ~σ2 |IIi = ~σ1 · ~σ2 |IV i = 4. The projectors 1 0 P1 = 0 0

1 √ (| + −i + | − +i) = |IIi 2 1 −3 √ (| + −i − | − +i) = −3|IV i 2

P1 et P−3 as well as ~σ1 · ~σ2 0 0 0 1 0 0 P−3 = 0 1 0 0 0 0

If P1 = λI + µ~σ1 · ~σ2 , we must have

are diagonalized in the basis {|Ii, |IIi, |IIIi, |IV i} 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 ~σ1 · ~σ2 = 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 −3

λ + µ = 1 and λ − 3µ = 0 that is, λ = 3/4 and µ = 1/4. One deduces P1 =

1 (3I + ~σ1 · ~σ2 ) 4

P−3 =

1 (I − ~σ1 · ~σ2 ) 4

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

1 (| + −i + 2| − +i − | + −i) = | − +i 2

and in general P12 |ε1 ε2 i = |ε2 ε1 i. 6. We know the eigenvalues and eigenvectors of ~σ1 · ~σ2 , and thus those of H • |Ii, |IIi, |IIIi are eigenvectors H with eigenvalue E0 + A • |IV i 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 ~ = H0 − H = H0 − (~ µe− + µ ~ e+ ) · B

qe ~ (σ1z − σ2z ) 2m

Let us examine the action of the operator (σ1z − σ2z ) on the basis vectors of H0 (σ1z − σ2z )| + +i = 0

(σ1z − σ2z )| − −i = 0

(σ1z − σ2z )|IIi = 2|IV i

29 and

qe ~ = 2Ax hII|H1 |IV i = hIV |H1 |IIi = 2 − 2m

The matrix form of H is in the basis {|Ii, |IIi, |IIIi, |IV i} E0 + A 0 0 0 0 E + A 0 2Ax 0 H= 0 0 E0 + A 0 0 2Ax 0 E0 − 3A

Two eigenvectors are obvious: |Ii and |IIIi with eigenvalues E0 + A. The other two are obtained through diagonalization of the 2 × 2 matrix 1 2x H ′ = E0 I + A = E0 I + AM 2x −3 The eigenvalue equation of M is which gives the values of the energy

λ2 + 2λ − (3 + 4x2 ) = 0 p 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 those of (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 1 NA − NA = 0 2 2

If the state vector is such that (~σn · ~σn+1 ) = 1, we obtain the ground state E0 = 0. This state vector is Φ0 = | + + + · · · + +i because (~σn · ~σn+1 )|Φ0 i = |Φ0 i 2. The operator Pn,n+1 exchanges spins n et n+ 1: in the case of two spins, we have seen in the preceding question that P12 | + +i = | + +i P12 | + −i = | − +i

P12 | − −i = | − −i P12 | − +i = | + −i

and the number of up spins minus the number of down spins must stay unchanged. The eigenvectors of H are thus such that the number of up spins minus the number of down spins is a constant. In particular, for the state |Ψn i, this constant is N − 1. The operator I − Pn,n+1 applied to |Ψn i gives zero on any pair of 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 | + + + + − + + +i = | + + + − + + + +i that is we obtain

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

Pn−1,n |Ψn−1 i = |Ψn i = |Ψn i − |Ψn−1 i = |Ψn i − |Ψn+1 i

30

CHAPTER 6. EXERCISES FROM CHAPTER 6

This gives the action of H on |Ψn i H|Ψn i = −A |Ψn−1 i + |Ψn+1 i − 2|Ψn i 3. Let us work out the action of H on |ks i H|ks i =

N −1 X n=0

eiks nl H|Ψn i = −A

N −1 X n=0

eiks nl |Ψn−1 i + |Ψn+1 i − 2|Ψn i

On the other hand X X X eiks nl |Ψn−1 i = eiks (n−1)l eiks l |Ψn−1 i = eiks l eiks nl |Ψn i = eiks l |ks i n

n

n

We thus have H|ks i = 2A(1 − cos ks l)|ks i The eigenfrequencies are |ks | → 0 ωk ≃ (Al2 )ks2

ωk = 2A(1 − cos ks l)

It is also interesting to apply the method of § 5.1.2, observing that H can be cast into the form H = −A UP + UP−1 − 2

where UP makes a circular permutation n → n + 1 and to look for eigenvectors |Ψs i of UP . Writing X |Ψs i = csn |Ψn i UP |Ψs i = eiδs |Ψs i n

with δs =

2πs = ks l N

We must have cn+1 = eiδs cn

6.5.7 Calculation of E(ˆ a, ˆb) 1. Let us compute, for example, the amplitude a+− (θ) 1 1 θ θ θ a+− (θ) = h+ ⊗ [−, ˆb]|Φi = − sin h+ + | + cos h+ − | √ | + −i − | − +i = √ cos 2 2 2 2 2 2. The rotation operator by an angle θ about an axis n ˆ of the ensemble of two spins is, from (3.67) θ n Unˆ (θ) = exp − ~σ (a) + ~σ (b) ·ˆ 2

The rotational invariance of |Φi implies

Unˆ (θ)|Φi = |Φi =⇒ that is

h

i ~σ (a) + ~σ (b) · n ˆ |Φi = 0

~σ (a) · n ˆ |Φi = − ~σ (b) · n ˆ |Φi

As a consequence hΦ| ~σ (a) · a ˆ ⊗ ~σ (b) · ˆb |Φi = −hΦ| ~σ · a ˆ ~σ · ˆb |Φi = −hΦ|ˆ a · ˆb + i~σ · (ˆ a × ˆb)|Φi = −ˆ a · ˆb = − 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⊥ i |θθ⊥ i = cos θ|xi + sin θ|yi − sin θ|xi + cos θ|yi

= − sin θ cos θ|xxi + cos2 θ|xyi − sin2 θ|yxi + sin θ cos θ|yyi

To obtain |θ⊥ θi, it is enough to exchange x ↔ y |θ⊥ θi = − sin θ cos θ|xxi + cos2 θ|yxi − sin2 θ|xyi + sin θ cos θ|yyi and, subtracting the the second equation from the first we get |θθ⊥ i − |θ⊥ θi = |xyi − |yxi 2. For photon 1 1 |xi = √ −|Ri + |Li 2

For photon 2 we must change |yi into −|yi

1 |xi = √ −|Ri + |Li 2

i |yi = √ |Ri + |Li 2 i |yi = − √ |Ri + |Li 2

and one finds for photons propagating in opposite directions

i |Φi = √ |RRi − |LLi 2

Remark: if the two photons propagate in the same direction, as is the case for pairs obtained from parametric conversion, we have i |Φi = − √ |RLi − |LRi 2 In the two cases, one checks that the Oz component of angular momentum vanishes: (Σ1z + Σ2z )|Φi = 0, where Σz is given by (3.26). 3. Rotational invariance allows us to choose n ˆ α along Oz while n ˆ β makes an angle (β − α) with the Ox axis. Let us define φ = β − α and the amplitude a++ (φ) for finding |Φi in the state |x ⊗ φi = cos φ|xxi + sin φ|xyi is

and thus

1 1 a++ (φ) = √ cos φhxx| + sin φhxy| |xyi − |yxi = √ sin φ 2 2 p++ (φ) =

1 sin2 φ 2

Using symmetry properties in the exchange + ↔ − p++ (φ) = p−− (φ) =

1 sin2 φ 2

Since the sum of all probabilities must add up to one p+− (φ) = p−+ (φ) =

1 cos2 φ 2

32

CHAPTER 6. EXERCISES FROM CHAPTER 6

This gives E(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 used for spin 1/2. 4. We find

1 1 |Ψi = √ |θθi + |θ⊥ θ⊥ i = √ |RRi + |LLi 2 2 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 is uncertain by ±d/2. 2. One recovers a standard interference calculation in optics. The difference of optical path for the photon 1 going through the upper slit is, with θ = l/(2D) δ(x, y) − δ(0, 0) = −

l (x + y) = −θ(x + y) 2D

and the phase shift

2π (δ(x, y) − δ(0, 0)) = −kθ(y + x) λ The amplitude for detecting the photon 1 at point y is proportional to 1 2iπ 2iπ exp (δ(x, y) − δ(0, 0) + exp − (δ(x, y) − δ(0, 0) = cos kθ(y + x) 2 λ λ φ(x, y) − φ(0, 0) =

3. If the pair is emitted at point x, the probability amplitude for detection in coincidence is obtained by multiplying 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 the plate CD, one must add the amplitudes corresponding to all possible positions of this emission Z 1 d/2 a(y, z) = dx cos kθ(y + x) cos kθ(z + x) d −d/2 Z d/2 h i 1 dx cos[kθ((y + z + 2x)] + cos kθ(y − z) = 2d −d/2 1 1 sin kθd cos[kθ(y + z)] + d cos kθ(y − z) = 2d kθ 5. If d ≫ 1/(kθ), the second term of a(y, z) dominates over the first one and the probability p(y, z) is obtained 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 is observed, there is no interference: integrating the probability p(y, z) over y, the result is a constant with respect to the variable z Z d/2 1 I(z) = dy cos2 [kθ(y − z)] = cst 1 + O kθd −d/2

33 6. In the limit d ≪ 1/(kθ), the probability becomes i2 1 h d cos[kθ(y + z)] + d cos[kθ(y − z)] 4d2 = cos2 (kθy) cos2 (kθz)

p(y, z) ∝

which corresponds to a product of two independent interference patterns. The position at which the pair is emitted is very precisely fixed, so that the angular aperture for each photon is very large. There is no longer any constrain from momentum conservation. Entanglement has no influence: if photon 1, for example, passes through the upper slit, there is no guarantee that that photon 2 passes through the lower slit. The spread in transverse momentum is too large to allow one photon to tag the trajectory of the other one. The condition d ≪ λ/θ is the condition for a good fringe visibility in a standard Young’s slit experiment: this condition is complementary to that of trajectory tagging. Conversely, if the emission position 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. The simultaneous 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 two photons are transmitted. The condition d ≫ 1/(kθ) must clearly be verified. The probability amplitude is obtained by summing over the two paths, because the paths are not distinguishable 1 iα re r + t teiβ a(c, c′ ) = √ 2

√ If r = it and |t| = |r| = 1/ 2

p(c, c′ ) = |a(c, c′ )|2 = In a similar way, one finds a(c, d′ ) p(c, d′ )

2 α−β 1 1 − eiα + eiβ = sin2 8 2 2

1 √ rt eiα + eiβ 2 1 α−β = |a(c, d′ )|2 = cos2 2 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 α−β sin2 2 2

p(c′ , d) =

1 α−β cos2 2 2

whence the quantity E(α, β) E(α, β) = sin2

α−β α−β − cos2 = − cos(α − β) 2 2

One will thus get a maximal violation of Bell’s inequalities with a choice of angles (α, β) identical to that of 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 difference between the two arms, one can certainly not observe interference in an individual interferometer. But there exists a deeper reason which will be explained at the end of the next question.

34

CHAPTER 6. EXERCISES FROM CHAPTER 6

2. One may write four different probability amplitudes for the joint detection of photons in D1 et D2 (S = short, L = long) A = ′

A

=

a1S a2S e

iδ

a1L a2L

B = a1S a2L B ′ = eiδ a1L a2S

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 to distinguish between (SL) and (LS); even if the emission time of the pair is unknown, the photon taking the upper arm arrives earlier than that taking the shorter arm. For example, in the (CL) case, photon 1 arrives at D1 0.7 ns before the second photon at D2 , which is much larger than the resolution time 0.1 ns of the detectors. On the contrary, the experimental set up does not allow one to distinguish, even in principle, between paths (SS) et (LL). We must then add the amplitudes for these paths to obtain the probability of detection in coincidence p(D1 , D2 ) = |A + A′ |2 = |a1S a2S + eiδ a1L a2L |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 the photon 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 would simply 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 important observation is that, if one suppresses the beam splitters of the left hand interferometer, one still has an information on the travel time: if the left hand photon arrives before the right hand one, we know that the 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 any possibility of interference, even if the coherence length is smaller than ∆l. In fact, it is not even necessary that 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 as the 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 |Ψi defined in Fig. 6.19 1 1 X 1 |xi ⊗ |0i − |1i |Ψi = H|0i ⊗ H|1i = |0i + |1i ⊗ |0i − |1i = 2 2 x=0

We apply Uf to this state with the following result: 1. If f (x) = 0, then |0i − |1i → |0i − |1i ; 2. If f (x) = 1, then |0i − |1i = |1i − |0i → − |0i − |1i , or, to summarize,

|0i − |1i → (−1)f (x) |0i − |1i .

The state Uf |Ψi is then a tensor product (and not an entangled state) Uf |Ψi =

1 1 X (−1)f (x) |xi ⊗ |0i − |1i 2 x=0

(6.1)

35 The net result for the input register is Uf

|xi −−→ (−1)f (x) |xi The state of the qubit of the input register then is 1 |ϕi = √ (−1)f (0) |0i + (−1)f (1) |1i . 2

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

= =

1 (−1)f (0) |0i + |1i + (−1)f (1) |0i − |1i 2 1 1 (−1)f (0) + (−1)f (1) |0i + (−1)f (0) − (−1)f (1) |1i. 2 2

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

36

CHAPTER 6. EXERCISES FROM CHAPTER 6

Chapter 7

Exercises from 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|| be its norm. We get, using the inequality which is valid for two operators C et D ||C|| ||D|| ≥ ||CD|| the relation 2||An || ||B|| ≥ ||BAn − An B|| = n||An−1 || whence ||An || ≥

n ||An−1 || 2

We deduce the following bounds for ||An || n ||An−1 || ≤ ||An || ≤ ||A|| ||An−1 || since ||An || ≤ ||A|| ||An−1 || 2 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|ϕi does not belong to the domain of A and the product AB|ϕi is not defined: we cannot apply Hermitian conjugation and write hϕ|AB|ϕi = hAϕ|Bϕi If the Hilbert space is identified, for example, with L(2) [0, 1] and if B = X, which is bounded on that space, while A = AC defined in § 7.2.2, (Xϕ)(x) = xϕ(x) = ψ(x) does not belong to the domain of AC which 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 ∗ Z 1 Z 1 ∂ϕ(x) ∂ ∗ i xϕ(x) xϕ(x) = 6 dx ϕ (x) i ∂x ∂x 0 0

37

38

CHAPTER 7. EXERCISES FROM CHAPTER 7

the difference being |ϕ(1)|2 . 3. The function ϕ(x) = ei(2πn+α)x

C = eiα

obeys AC ϕ(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 of AC . The von Neumann theorem does not apply because AB|ϕi is not defined whatever |ϕi ∈ H, while the domain of definition of AB should be dense in H.

Chapter 8

Exercises from Chapter 8 8.5.2 Rotations and SU(2) 1. Let us start from the most general 2 × 2 matrix a U= c and compute U † U which must be identical to I |a|2 + |b|2 U †U = ca∗ + db∗

b d

ac∗ + bd∗ |c|2 + |d|2

=

1 0

0 1

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

From c = −b∗ d/a∗ and det U = ad − bc = 1 one deduces that d = a∗ . 2. To order τ U † U = (I + iτ † )(I − iτ ) ≃ I − i(τ − τ † )

and the condition U † U = I implies τ = τ † . Moreover, the condition det U = 1 implies Tr τ = 0. The decomposition (3.54) together with the condition Tr τ = 0 allows us to write 3

τ=

1X θi σi 2 i=1

where the angles θi are infinitesimal since τ itself is infinitesimal. 3. Since θ/N is infinitesimal. we may write θ iθ Unˆ =I− (~σ · n ˆ) N 2N and using lim

N →∞

we deduce

1−

x N = e−x N

θ ˆ) Unˆ (θ) = exp −i (~σ · n 2

~ : det V = −V ~ 2 and since 4. The determinant of V is equal to minus the length squared of the vector V det(U VU −1 ) = det V

39

40

CHAPTER 8. EXERCISES FROM CHAPTER 8

~ 2 = det W = V ~ 2 , which shows that the transformation preserves the lengths of vectors. It is we obtain W thus, either a rotation, or a rotation combined with a parity operation. 5. Since W is Hermitian and has zero trace, because Tr (U VU −1 ) = Tr V ~ = ~σ · V~ (θ). we may write W = ~σ · W d ~ (θ) = − i [~σ · n ~ (θ)] = ~σ · (ˆ ~σ · V ˆ , ~σ · V n × V~ (θ)) dθ 2 ~ (θ) is deduced from V ~ through a rotation by an angle θ where we have used (3.52), which shows that V about n ˆ ~ dV ~ (θ) =n ˆ×V dθ To any rotation Rnˆ (θ) correspond two matrices Unˆ : Unˆ (θ) and Unˆ (θ + 2π) = −Unˆ (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) 2

3

fa (θ, θ) = θ a + θa + λabc θb θc + λabc θ b θc + fabc θ b θc + O(θ3 , θ2 θ, θ θ , θ ) The condition fa (θ, θ = 0) = θa implies λabc = 0 and similarly λabc = 0. 2

3

3. One the one hand, we have, neglecting terms of order (θ3 , θ2 θ, θ θ , θ ) 1 U (θ)U (θ) = I − iθa Ta − iθa Ta − (θ b θc + θb θc )Tbc − θa θb Ta Tb 2 and, on the other hand 1 U (f (θ, θ)) = I − iTa (θa + θa + fabc θ b θc ) − (θb + θb )(θc + θc )Tbc 2 Relabeling the summation indices and taking into account the symmetry property Tbc = Tcb , because Tbc = −

∂ 2 U ∂θa ∂θb θa =θb =0

the comparison between the two expressions gives

θb θc Tb Tc = ifabc Ta θb θc + θb θc Tbc We deduce from this Tb Tc Tc Tb

= =

Tbc + ifabc Ta Tbc + ifacb Ta

and, subtracting the second equation from the first one [Tb , Tc ] = i[fabc − facb ]Ta The structure constant Cabc is Cabc = [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

i~ P2 + V (X), X = [H, X] = − P 2m m

[[H, X], X] = −

~2 m

2. On the other hand, the commutator is also expressed as [[H, X], X] = HX 2 − 2XHX + X 2 H and using hϕn |H|ϕm i = En δnm hϕ0 |HX 2 |ϕ0 i = hϕ0 |XHX|ϕ0 i = whence the result

X

n,m

X

n,m

hϕ0 |H|ϕn ihϕn |X|ϕm ihϕm |X|ϕ0 i = E0 hϕ0 |X|ϕn ihϕn |H|ϕm ihϕm |X|ϕ0 i =

X 2m|Xn0 |2 n

~2

(En − E0 ) = 1

X

X n

n

|Xn0 |2

En |Xn0 |2

42

CHAPTER 8. EXERCISES FROM CHAPTER 8

Chapter 9

Exercises from 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 = X 2 d hX 2 i(t) dt

= =

iE i i Dh P 2 h[H, X 2 ]i = + V (X), X 2 ~ ~ 2m i 2 2 h[P , X ]i 2~m

Furthermore [P 2 , X 2 ] = −2i~(XP + P X) from which we derive the final result 1 d hX 2 i(t) = hXP + P Xi dt m Going to the x representation hP Xi =

Z

Z ∂ ∂ϕ∗ (x) dx ϕ∗ (x) −i (xϕ(x) = i dxxϕ(x) ∂x ∂x

where the second expression is obtained from an integration by parts. Combining with Z ∂ϕ(x) hXP i = dx ϕ∗ (x)x −i ∂x we finally get d i~ hX 2 i(t) = dt m

∞

∂ϕ ∂ϕ∗ − ϕ∗ dx x ϕ ∂x ∂x −∞

Z

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 Hamiltonian is then reduced to the kinetic Hamiltonian H = K = P 2 /(2m). We compute the second derivative of hX 2 i(t) i d 1 d d2 hX 2 i(t) = h[K, X 2 ]i = − 2 h[K, [K, X 2 ]]i 2 dt ~ dt ~ dt

43

44

CHAPTER 9. EXERCISES FROM CHAPTER 9

where we have used Ehrenfest’s theorem twice. Taking into account [P 2 , XP + P X] = −4i~P 2 we find

2 d2 hX 2 i(t) = hP 2 i dt2 m

The third derivative of hX 2 i(t) and higher order derivatives vanish dn hX 2 i(t) = 0 dtn

n≥3

because [K, [K, X 2 ]] ∝ P 2 and [K, P 2 ] = 0. hX 2 i(t) is thus a second order polynomial in t hX 2 i(t) = hX 2 i(t = 0) + t

1 d2 d hX 2 i(t) + t2 2 hX 2 i(t) dt 2 dt t=0 t=0

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

∆x2 (t) = hX 2 i(t) − [hXi(t)]2 and

DP E i d hXi(t) = [K, X] = dt ~ m

which gives

hXi(t) = hXi(t = 0) + t

DP E 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 Z

+∞

dx e

−α2 x2

−∞

1. Setting k ′ = k − k

Z

2

dk|A(k)| =

√ π = α

1 πσ 2

1/2 Z

1 ∆x = √ 2α

2

k′ dk exp − 2 σ ′

!

=1

√ so that ∆k = σ/ 2. Let us compute the wave function at time t = 0 ϕ(x, 0)

=

=

σ 1/2 π 1/4

1/4

2

dk ′ k′ √ exp ik ′ x − 2 e 2σ 2π 1 exp ikx − σ 2 x2 2

1 πσ 2

ikx

Z

The modulus squared of the wave function is 2 2 σ |ϕ(x, t)|2 = √ e−σ x π √ which is indeed normalized to one with ∆x = 1/( 2 σ), and thus

∆x ∆k =

1 2

!

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

1 πσ 2

1/4 Z

2

~k 2 dk (k − k) √ exp ikx − i t− 2m 2σ 2 2π

!

The exponent is rewritten, within a factor of i and with k ′ = k − k 2

2

2

2

~k ~k 2 (k − k) ~kk ′ ~k ′ k′ ′ kx + k x − kx − = t− t − t − t − 2m 2σ 2 2m m 2m σ2 ϕ(x, t) reads ϕ(x, t) =

1 πσ 2

1/4

2

~k exp ikx − i t 2m

!Z

" # ′2 ~k 1 i~t k dk ′ √ exp ik ′ x − + t exp − 2m 2 σ2 2m 2π

If we can neglect the term i~t/2m in the last exponential, we simply obtain ! " 1/4 2 # 2 ~k ~k σ2 1 x− exp ikx − i t exp − t ϕ(x, t) = πσ 2 2m 2 m ! 2 ~k t ϕ(x − vg t, 0) = eiω(k)t ϕ(x − vg t, 0) = exp i 2m 3. In the general case, we set

1 i~t 1 2 = σ2 + m ′ σ

and we find, after doing the integration ϕ(x, t) =

1 πσ 2

1/4

Taking the modulus squared 2

|ϕ(x, t)| =

1 ′2 2 σ exp − σ (x − vg t) exp i(kx − ω(k)t) 2 ′

1 πσ 2

1/2

2 |σ ′ |2 exp −Re σ ′ (x − vg t)2

The peak of |ϕ(x, t)|2 is centered at x = vg t and has a width ∆x2 (t) = that is, 1 ∆x (t) = 2σ 2 2

1 2 Re σ ′ 2

~ 2 σ 4 t2 1+ m2

The width of the wave packet increases with time, because of the term ~2 σ 4 t2 /m2 in the bracket. 4. ∆x2 (t) doubles for t=

m 2m∆x2 (t = 0) = = 3.2 × 10−11 s ~σ 2 ~

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 dimension of V (x) is M2 L4 T −2 L−1 L−1 = M2 L2 T −2

46

CHAPTER 9. EXERCISES FROM CHAPTER 9

which is indeed the dimension of an energy. 2. We integrate the Schr¨odinger written in the form 2 d 2mE + 2 ϕ(x) = g δ(x) ϕ(x) dx2 ~ between x = −ε and x = +ε Z

+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, we must have ϕ(x) = Ae−κ|x| for x = 6 0 so that ϕ′ (0+ ) − ϕ′ (0− ) = −2κA = −|g|ϕ(0) = −|g|A whence 2κ = |g| and E

~2 g 2 ~2 κ2 =− 2m 8m There is no odd solution because ϕ(0) = 0 for an odd solution. Let us retrieve the result by taking the limit a → 0, V0 a → ~2 g/(2m) of the square well potential whose energy levels are given by (9.82) E=−

κ = k tan which leads to

and

p 2m|V0 | k≃ ~ κ≃

p

2m|V0 | ~

ka 2

ka tan ≃ tan 2

p 2m|V0 |a →0 2~

p 2m|V0 | a m ~2 |g| m|V0 |a 1 = = = |g| 2~ ~2 ~2 2m 2

3. Since the diatomic molecule potential is even, one looks for even and odd solutions. For the even solutions we have x < −l : ϕ(x) = eκx

− l < x < l : ϕ(x) = A cosh κx

x > l : ϕ(x) = e−κx

The continuity of ϕ at x = l gives A cosh κl = e−κl and the continuity of its derivative at the same point −κe−κl − Aκ sinh κl = −|g|e−κl We remark that A sinh κl = A cosh κl tanh κl = e−κl tanh κl whence the equation for the energy eigenvalue (1 + tanh κl) =

|g| κ

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

|g| 1 + e−2κl 2

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 sinh κx

x > l : ϕ(x) = e−κx

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

−κl

−κe

A sinh κl

=

e−κl

− Aκ cosh κl

=

−|g|e−κl

which leads to

|g| 1 − e−2κl 2 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. κ=

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 of the preceding question, and we assume κl ≫ 1. Then there exist two bound states, one with an even wave-function corresponding to |g| (1 + e−2κl ) κ+ = 2 and the other one with an odd wave-function corresponding to κ− =

|g| (1 − e−2κl ) 2

The energy difference between the two states is then E− − E+ = − the average enrgy E0 being E0 =

~2 2 ~2 2 −2κl (κ− − κ2+ ) = g e 2m 2m

~2 g 2 1 (E+ + E− ) ≃ − 2 8m

We can thus write E+ E−

~2 g 2 1 + 2e−2κl 8m ~2 g 2 1 − 2e−2κl ≃ − 8m ≃ −

These are the eigenvalues of a two-level Hamiltonian ~2 g 2 1 H =− −2e−2κl 8m

−2e−2κl 1

From (9.100) and (9.105), the tunneling transmission coefficient of a particle with energy ≃ 0 through a barrier 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 condition at x = 0 gives A+B =1 while the condition on the derivative is ik − ik(A − B) = g

48

CHAPTER 9. EXERCISES FROM CHAPTER 9

We derive

ig ig B=− 2k 2k whence the matrix elements of the transmission matrix A=1+

M11 = 1 +

ig 2k

M12 =

ig 2k

We check that the results do agree with the limit a → 0, V0 a → ~2 g/(2m) of equations (9.96) and (9.97). 6. The condition ϕq (x) = eiql ϕq (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 i i h h = A 1 − ei(q−k)l + B 1 − ei(q+k)l i i h h = A g + ik 1 − ei(q−k)l + B g − ik 1 − ei(q+k)l

0 0

We thus obtain a system of equations with two unknowns A et B whose discriminant ∆ must be zero for a non trivial solution. Setting α = (q − k)l et β = (q + k)l 1 − eiα 1 − eiβ ∆ = det =0 g + ik 1 − eiα g − ik 1 − eiβ 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 sin kl 2k

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 must have ∂Bz =0 ∂x x=0 Translation invariance along Oy gives ∂Bz =0 ∂y x=0

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

Bz = B0 + bz

~ ×B ~ = 0 and This field obeys the two Maxwell equations (1.8) and (1.9) in vacuum ∇ ~ ·B ~ = ∂Bx + ∂Bz = −b + b = 0 ∇ ∂x ∂z

49 The potential energy is ~ = −µx Bx − µz Bz = bµx x − bµz z −~ µ·B

whence the force F~ with components Fx =

~ ∂(−~ µ · B) = bµx ∂x

Fz =

~ ∂(−~µ · B) = −bµz ∂z

The B0 zˆ 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 µx vanishes: hµx i = 0, and the force along Ox averages to zero, if the transit time ≫ 1/ω, as the spin makes a large number of rotations about Oz. 2. The force on the magnetic moment is vertical and constant; its value is F = ±µb for a spin lying along ±ˆ z . The splitting between the trajectories of an up spin and a down spin at the exit of the magnet gap is then 2 2 1 F 2 F L µb L δ=2 t = = 2 m m v m v 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 Schr¨odinger equations for ϕ± are ~2 2 ∂ϕ± = − ∇ ∓ µbz ϕ± = Hϕ± i~ ∂t 2m Using Ehrenfest’s theorem (4.26), we obtain d ~ hR± i(t) = dt d hPx,y,± i(t) = dt d hPz,± i(t) = dt

i ~ ± ] = 1 hP~± i [H, R ~ m i [H, Px,y,± ] = 0 ~ i [H, Pz,± ] = ±µb ~

This last result is deduced from ∓[µbZ, Pz ] = ∓i~µb We finally get µb 2 t 2m and the center of the wave packet does follow the classical trajectory. hZ± i = ±

4. Let us make a reflection with respect to the xOy plane. In this reflection, ~µ does not change, because ~ the orientation of a current loop lying in the xOy plane stays unchanged. In the same operation, B ~ changes its direction, but not its gradient, and thus ∇B lies along −ˆ x. However, the image of the trajectory in the mirror always leave in the +ˆ x direction, and then the image of the trajectory in the mirror 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 U (t, t0 ) = [g(t)AP ] U (t, t0 ) dt

50

CHAPTER 9. EXERCISES FROM CHAPTER 9

which integrates into

i U (t, t0 ) = exp − AP ~

Z

t

′

g(t )dt

t0

′

Between times ti and tf we thus have Z +∞ i i ′ ′ U (tf , ti ) ≃ exp − AP g(t )dt = exp − gAP ~ ~ −∞ 2. The action of U (tf , ti ) on the vector |n ⊗ ϕi is U (tf , ti )|n ⊗ ϕi = e−igan P/~ |n ⊗ ϕi Furthermore, exp(−igan P/~) is a translation operator by gan and from (9.13) e−igan P/~ ϕ (x) = ϕ(x − gan ) 3. Because of the linearity property of quantum mechanics, the final state vector is X |χf i = cn |n ⊗ ϕn i n

The reduced state operator of S is from (6.66) X X ρ(1) = cn c∗m |mihn|hϕm |ϕn i = |cn |2 |nihn| n,m

n

because hϕm |ϕn i = δnm . The system S is thus a statistical mixture of states |ni with a weight |cn |2 , and the probability to observe S in the state |ni 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 ~ and 1. From the expression of the orbital angular momentum operator as a function of the position R ~ momentum 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)) r ) = f (Rxˆ [−α](~r)) e−iαLx f (~r) = f Rx−1 ˆ [α](~

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

=

z sin α + y cos α

dy = zdα

′

=

z cos α − y sin α

dz = −ydα

z

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

=

r cos θ sin φ dθ + r sin θ cos φ dφ

dz

=

−r sin θ dθ

which leads to dθ = sin φ dα

dφ =

cos φ dα tan θ

This allows us too identify Lx cos φ ∂ ∂ f (~r) + [−idαLx f ](~r) = dα sin φ ∂θ tan θ ∂φ that is

∂ cos φ ∂ Lx = i sin φ + ∂θ tan θ ∂φ

To compute Ly we can use the same method, or make use of the commutation relation ∂ , Lx Ly = −i[Lz , Lx ] = − ∂φ

51

52

CHAPTER 10. EXERCISES FROM CHAPTER 10

Taking into account

∂ , f (φ) ∂φ ∂ ∂ , f (φ) ∂φ ∂φ we obtain

=

f ′ (φ)

=

f ′ (φ)

∂ ∂φ

sin φ ∂ ∂ + Ly = i − cos φ ∂θ tan θ ∂φ

3. The operator Lz = −i∂/∂φ is defined on the periodic functions f (φ) = f (φ + 2π) and it is selfadjoint on this domain (see § 7.2.2). By contrast, the function φf (φ) is not periodic and does not belong to the 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 the integration 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 Lz f (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 hlm|θ, φi = Ylm ∗ (θ, φ) but we also have hlm|θ, φi

= hlm|e−iφLz e−iθLy |θ = 0, φ = 0i X = hlm|e−iφLz e−iθLy |l′ m′ ihl′ m′ |θ = 0, φ = 0i l′ ,m′

and introducing a complete set of states X

l′ ,m′

|l′ m′ ihl′ m′ | = I

From the result of question 1 hl′ m′ |θ = 0, φ = 0i ∝ δm′ 0 and from (10.32) (l)

hlm|e−iφLz e−iθLy |l′ , 0i = δl′ l Dm0 (θ, φ) In fact, we can easily obtain the proportionality coefficient because ′

′

hl m |θ = 0, φ = 0i =

δm′ 0 Yl0 (θ

= 0, φ = 0) =

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

10.7.8 The spherical well 1. Setting E = −B and (~ = 1) k=

p 2m(V0 − B)

κ=

√ 2mB

r

2l + 1 4π

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 − k 2 . The eigenvalue equation becomes √ U − k2 cot kR = − 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 mass mp /2 and writing ~ explicitly π2 mp V0 R2 = 2 ~ 4 One finds the numerical value V0 ≃ 100 MeV ≫ B 3. The radial wave equation is

A B 1 d2 + 2 − u(r) = Eu(r) − 2m dr2 r 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′ = hL, θ = 0|T |j = 1, m = −1i

a = hR, θ = 0|T |j = 1, m = 1i

If the transition is of the electric dipole kind, we have seen in § 10.5.2 that a = a′ (with our phase conventions). We find, using the rotational invariance of the transition matrix, [U (R), T ] = 0 am=1 (θ) R

= = =

hR, θ|T |j = 1, m = 1i = hR, θ = 0|T U † [Ryˆ(θ)]|j = 1, m = 1i hR, θ = 0|T |j = 1, m = 1ihj = 1, m = 1|U † [Ryˆ(θ)]|j = 1, m = 1i 1 (1) ad11 (θ) = a(1 + cos θ) 2

Similarly we find (1)

am=1 (θ) = ad1,−1 (θ) = L

1 a(1 − cos θ) 2

If the photon is emitted in the direction n ˆ = (θ, φ), we must use the rotation operator U [R(θ, φ)] and we get am=1 (θ, φ) R

=

am=1 (θ, φ) L

=

1 a(1 + cos θ)eiφ 2 1 a(1 − cos θ)eiφ 2

2. If the atom absorbs the photon, the two nonzero amplitudes are, from angular momentum conservation b = hj = 1, m = 1|T ′ |Ri

b′ = hj = 1, m = −1|T ′ |Li

54

CHAPTER 10. EXERCISES FROM CHAPTER 10

If the transition is of the electric dipole kind, b′ = b from the results of § 10.5.2. Introducing a complete set of intermediate states X cP →P ′ (θ) = cP →(jm) c(jm)→P ′ (θ) = hP ′ , θ|S|P i = hP ′ , θ|T |1mih1m|T ′|P i m

hR, θ|S|Ri = hR, θ|S|Li = hL, θ|S|Ri = hL, θ|S|Li =

1 ab(1 + cos θ) 2 1 hR, θ|T |j = 1, m = −1ihj = 1, m = −1|T ′ |Li = ab(1 − cos θ) 2 1 ′ hL, θ|T |j = 1, m = 1ihj = 1, m = 1|T |Ri = ab(1 − cos θ) 2 1 ′ hL, θ|T |j = 1, m = −1ihj = 1, m = −1|T |Li = ab(1 + cos θ) 2

hR, θ|T |j = 1, m = 1ihj = 1, m = 1|T ′ |Ri =

In the two possible initial cases, |Ri and |Li, the angular distribution is W (θ) =

1 2 2 |a| |b| (1 + cos2 θ) 2

If the initial photon is polarized along Ox we find hx, θ|S|xi = hy, θ|S|xi =

ab cos θ 0

In a classical model of photon scattering by an electric charge, a photon with linear polarization along Ox sets the charge into motion along this axis, and the charge radiates an electromagnetic wave polarized along Ox with an angular distribution ∝ cos2 θ. If the photon is emitted in the direction n ˆ = (θ, φ), we find hR; θ, φ|S|Ri

=

hR; θ, φ|S|Li

=

hL; θ, φ|S|Ri

=

hL; θ, φ|S|Li =

1 ab(1 + cos θ) eiφ 2 1 ab(1 − cos θ) e−iφ 2 1 ab(1 − cos θ) eiφ 2 1 ab(1 + cos θ) e−iφ 2

If the initial photon is polarized along Ox, we obtain the following amplitudes hx; θ, φ|S|xi

hy; θ, φ|S|xi

=

ab cos θ cos φ

=

ab sin φ

These results can be understood by observing that the cosine of the angle between the initial and final polarizations is cos θ cos φ for a final polarization along Ox and sin φ 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 angular momentum vanishes hθ = 0, m′ |T |mi ∝ δmm′ The amplitude b is obtained from a through a reflection with respect to a plane xOz: |a| = |b| if parity is conserved.

55 2. If the proton is emitted along a direction making an angle θ with the Oz axis in the xOz plane, we can compute the decay amplitude using the rotation operator U [Ryˆ(θ)] by an angle θ about Oy hθ, m′ |T |mi = hθ = 0, m′ |U † [Ryˆ(θ)]T |mi X = hθ = 0, m′ |T |m′′ ihm′′ |eiθJy |mi m′′

(1/2)

= am′ ,m′ (θ = 0)dm′ m (θ)

With the definitions of question 1 a 21 , 21 = a

a− 21 ,− 12 = b

and taking (10.38) into account (1/2)

a++ (θ) = ad 1 , 1 (θ) = a cos 2 2

θ 2

(1/2)

a−+ (θ) = ad 1 ,− 1 (θ) = −b sin 2

2

θ 2

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

θ θ + |b|2 sin2 2 2 1 1 (|a|2 + |b|2 ) + (|a|2 − |b|2 ) cos θ 2 2

= |a|2 cos2 =

whence w0 =

1 (|a|2 + |b|2 ) 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 angular distribution of the decay is thus a proof that parity conservation is violated. ~ which is a pseudovector, must necessarily be 4. ~p × p~Λ0 is the only available pseudovector, and hSi, oriented along this direction : hSx i = hSy i = 0. 5. The spin Hamiltonian in the magnetic field is ~ = −γ S ~·B ~ H = −~µ · B ~ with an angular velocity ω = γB in the xOz plane. At time t = τ , The proton spin precesses around B it will have rotated by an angle λ = ωτ = γBτ The decay angle must be measured from the direction n ˆ in the xOz plane n ˆ=x ˆ sin λ + zˆ cos λ The proton is emitted in the direction pˆ = x ˆ sin θ cos φ + yˆ sin θ sin φ + zˆ cos θ and cos Θ = pˆ · n ˆ = cos θ cos λ + sin θ sin λ cos φ The measurement of the angular distribution allows us to infer the direction of n ˆ (or the angle λ) and to deduce γ from λ = γBτ .

10.7.15 Production and decay of the ρ+ meson

56

CHAPTER 10. EXERCISES FROM CHAPTER 10

1. Calculation of am (θ, φ) : R(θ, φ) is the rotation (10.30) which brings the Oz axis along the direction of emission for the π + meson = hθ, φ|T |mi = hθ = 0, φ = 0|U † [R(θ, φ)]T |mi X = hθ = 0, φ = 0|T |m′ ihm′ |U † [R(θ, φ)]|mi

am (θ, φ)

m′

where we have used the rotational invariance of the transition matrix [U, T ] = 0. From angular momentum conservation hθ = 0, φ = 0|T |m′i ∝ δm′ 0 = aδm′ 0

because if π + meson is emitted in the Oz direction, its angular momentum along this direction vanishes. Furthermore h i∗ (1) hm′ = 0|U † [R(θ, φ)]|mi = hm|U [R(θ, φ)]|m′ = 0i∗ = Dm0 (θ, φ) (1)

= eimφ dm0 (θ)

One obtains the different decay amplitudes using (10.39) a1 (θ, φ)

=

a (1) a eiφ d10 (θ) = − √ eiφ sin θ 2

a0 (θ, φ)

=

a d00 (θ) = a cos θ

a−1 (θ, φ)

=

(1)

a (1) a e−iφ d−10 (θ) = √ e−iφ sin θ 2

whence the angular distributions W1 = W−1 =

|a|2 sin2 θ 2

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 angular distribution is isotropic, which must be the case as there is no privileged direction. In what follows, W will be normalized as W1 + W0 + W−1 = |a|2 = 1 2. If the initial state vector is given by X |λi =

m=−1,0,1

cm |1mi

X

m=−1,0,1

|cm |2 = 1

the decay amplitude aλ will be aλ (θ, φ) = hθ, φ|T |λi =

X m

hθ, φ|T |1mih1m|λi =

X

cm am (θ, φ)

m

and the angular distribution Wλ (θ, φ) =

X

cm c∗m′ am (θ, φ)a∗m′ (θ, φ)

m,m′

The explicit expression of Wλ is given in the next section. 3. For every component of the statistical mixture pλ , the angular distribution is X ∗ ∗ (λ) Wλ = c(λ) m c m′ am (θ, φ)am′ (θ, φ) m,m′

and the angular distribution corresponding to the state operator ρ is X X X ∗ (λ) ∗ Wρ (θ) = pλ Wλ = pλ c(λ) ρmm′ am (θ, φ)a∗m′ (θ, φ) m c m′ am (θ, φ)am′ (θ, φ) = λ

λ,m,m′

m,m′

57 The result is simplified by observing that ρmm′ = ρ∗m′ m m 6= m′

ρmm′ am a∗m′ + ρm′ m am′ a∗m = 2Re (ρmm′ am a∗m′ )

For example Re (ρ10 a1 a∗0 ) = −Re

ρ10 iφ √ e sin θ cos θ 2

The final result is Wρ (θ, φ)

1 sin2 θ(ρ11 + ρ−1,−1 ) 2 1 √ sin 2θ Re e −iφ ρ−10 − e iφ ρ10 − sin2 θ Re e 2iφ ρ1,−1 2

= ρ00 cos2 θ + +

To obtain the angular distribution of question 2, it is enough to make the sustitutions ρ11 → |c1 |2 , ρ10 → c1 c∗0 etc.

~ which is a pseudovector, lies necessarily along n 4. The only available pseudovector is n ˆ , and hJi, ˆ . Using the 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, the reaction kinematics is unchanged, and since the target is unpolarized and parity is conserved, the reaction is identical to its image in the xOz plane. We must then have [ρ, Z] = 0 ou Z −1 ρZ = ρ Furthermore, we use Π|1mi = η|1mi where η is the ρ meson intrinsic parity (η = −1 because the ρ meson is, as the photon, a vector meson). We then derive ′ h1m|ΠeiπJz ρ e−iπJz Π|1m′ i = |η|2 e−iπ(m −m) ρmm′ that is

′

ρmm′ = (−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 angular momentum vanishes, and angular momentum conservation holds for the amplitudes a and b, but not for c 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 the following way on the photon states which are odd under parity, ηγ = −1 (see (10.104)) Y|Ri = −|Li

Y|Li = −|Ri

while for the Σ0 and Λ0 particles (see (10.102)) Y|jmi

Y|jm′ i

= ηΣ (−1)1/2−m |j, −mi ′

= ηΛ (−1)1/2−m |j, −m′ i

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

ηΣ ηΛ ηγ (−1)1−m−m = ηΣ ηΛ = η

58

CHAPTER 10. EXERCISES FROM CHAPTER 10

We can also use directly (10.119) ησ ηλ ηγ (−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 −ˆ p with a spin projection m′ on pˆ for a right handed polarized photon is ′

am R (θ)

= = =

hR, m′ ; θ|T |m = 1/2i

hR, m′ ; θ = 0|U † [Ryˆ(θ)]|m = 1/2i = (1/2)

a d1

1 2 2

(θ) = a cos

θ 2

X m′′

hR, m′ ; θ = 0|T |m′′ ihm′′ |U † [Ryˆ (θ)]|1/2i

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

(1/2)

am L (θ) = b d 1 ,− 1 (θ) = −a sin 2

m′ =−1/2

Only the amplitudes aR

m′ =1/2

(θ) and aL

2

(θ) are nonzero.

θ 2

Chapter 11

Exercises from 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† )q ap and let us compute its commutator with N [N, (a† )q ap ] = [N, (a† )q ]ap + (a† )q [N, ap ] = (q − p)(a† )q ap which vanishes only if p = q. As any function of a and a† is a sum of such monomials, using if necessary the 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 of terms of the form (a† )p ap . Any monomial of the form (a† )p ap can be written as a function of a† a using the 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 hn′ |A|ni = hn′ |f (N )|ni = f (n)δnn′ 2. Let |ϕi be a vector orthogonal to all vectors |ni, hϕ|ni = 0 ∀n P ′ |ni = |ni

P ′ |ϕi = 0

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

hϕ|[P ′ , a]|ϕi = 0

Let us examine hϕ|[P ′ , a]|ni and hϕ|[P ′ , a† ]|ni hϕ|P ′ a|ni

hϕ|aP ′ |ni hn|a† P ′ |ni

= hϕ|P ′ a† |ni = 0 √ = hϕ|a|ni = n hϕ|n − 1i = 0 √ = hn|a† |ϕi = n + 1 hϕ|n + 1i = 0

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

11.5.3 Coherent states

1 When 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

60

CHAPTER 11. EXERCISES FROM CHAPTER 11

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

d hai(t) = h[a, H]i = ~ωhai(t) dt

because [a, H] = ~ωa. If the initial condition is hai(t = 0) = hϕ(0)|a|ϕ(0)i = z0 the solution of the preceding differential equation is hai(t) = z0 e−iωt 3. In addition, we want that hϕ(0)|a† a|ϕ(0)i = |z0 |2 , which implies, with b(z0 ) = a − z0 hϕ(0)|b† (z0 )b(z0 )|ϕ(0)i

= =

hϕ(0)|a† a|ϕ(0)i − z0 hϕ(0)|a† |ϕ(0)i − z0∗ hϕ(0)|a|ϕ(0)i + |z0 |2 hϕ(0)|(a† a − |z0 |2 )|ϕ(0)i = ||b(z0 )ϕ(0)||2 = 0

which is possible only if b(z0 )|ϕ(0)i = 0 : |ϕ(0)i is an eigenvector of b(z0 ) with eigenvalue 0 b(z0 )|ϕ(0)i = 0, that is, a|ϕ(0)i = z0 |ϕ(0)i 4. Writing D(z) = exp A with A = −z ∗ a + za†

A† = −za† + z ∗ a = −A

then D† (z) = exp A† = exp(−A) and D† (z)D(z) = I. Using (2.55) we may rewrite D(z) D(z) = e−|z|

2

/2 za†

e

e−z

∗

a

From the definition (11.31) of a coherent state |zi we have D(z)|0i = e−|z|

2

/2 za†

e

∗

e−z a |0i = e−|z|

2

/2 za†

e

|0i = |zi

because exp(−z ∗ a)|0i = 0 taking a|0i = 0 into account. 5. We use (2.55) 1

eA eB = eA+B e 2 [A,B] with

and

c A = √ (z − z ∗ )(a + a† ) 2 1 c = −c′ = √ 2

c′ B = √ (z + z ∗ )(a − a† ) 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 A et B. We use a system of units such that ~ = mω = 1 ˆ or, equivalently, we use the operators Qand Pˆ of (11.4) instead of Q et P i 1 [A, B] = − (z − z ∗ )(z + z ∗ )[Q, P ] = (z 2 − z ∗2 ) 2 2 We must then choose

1 f (z, z ∗ ) = exp − z 2 − z ∗2 4

61 Given all these conditions, the wave function of the coherent state ϕz (q) in the q-representation is i 1 1 hϕz (q) = hq|D(z)|0i = exp − z 2 − z ∗2 exp √ [z − z ∗ ]q hQ| exp − √ [z + z ∗ ]P |0i 4 2 2 1 2 1 1 = exp − z − z ∗2 exp √ [z − z ∗ ] q ϕ0 q − √ [z + z ∗ ] 2 2 2

where ϕ0 (q) is the wave function (11.23) of the harmonic oscillator ground state. From (11.37), the expectation values of Q and P are 1 hP i = √ (z − z ∗ ) i 2

1 hQi = √ (z + z ∗ ) 2

and we can rewrite the preceding result as 1 1 i ϕz (q) = 1/4 exp − hQihP i exp (ihP iq) exp − (q − hQi)2 2 2 π Reestablishing the dimensionful factors mω 1/4 1 mω i hP iq exp − ϕz (q) = exp − hQihP i exp i (q − hQi)2 π~ 2~ ~ 2 ~ The global phase factor i exp − hQihP i 2~ is physically irrelevant and may be omitted. 6. We write, with z = ρ exp(iθ) and Anm = hn|A|mi X hz|A|zi = hz|nihn|A|mihm|zi n,m

X Anm z m z ∗n √ n!m! n,m 2 X Anm √ ei(m−n)θ ρn+m e−ρ n!m! n,m e−|z|

= =

2

where we made use of (11.31) hm|zi = e−|z|

2

/2

zm √ m!

Writing the expansion of hz|A|zi under the general form 2

hz|A|zi = e−ρ we derive cpq and we can make the identification

1 = q!

Z

∞ X ∞ X

cpq eipθ ρq

q=0 p=−∞

dRe zIm z −ipθ e hz|A|zi π

Anm =

√ n!m! cm−n,m+n

with (m − n) and (m + n) integers, (m + n) ≥ 0. Thus, we can get from hz|A|zi all the matrix elements of Amn .

11.5.4 Coupling to a classical force

62

CHAPTER 11. EXERCISES FROM CHAPTER 11

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) U0 UI

= =

i~

dUI dUI dU0 UI + U0 i~ = H0 U0 UI + U0 i~ dt dt dt

and derive (11.126) i~

dUI = U0−1 W (t)U0 UI = WI (t)UI dt

(11.1)

2. Let us show that aI (t) = e iH0 t/~ a e −iH0 t/~ = a e−iωt It is easy to derive a differential equation for aI (t) daI (t) dt

= =

i i [H0 , aI ] = e iH0 t/~ [H0 , a] e −iH0 t/~ ~ ~ −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) for UI (t) becomes dUI = − a e −iωt + a† e iωt f (t)UI (t) = WI (t)UI (t) i~ dt with the boundary condition UI (t = 0) = I. If the commutator of two operators A1 and A2 is a multiple of I, from the identity (2.55) of Exercise 2.4.11 1

eA2 eA1 = eA2 +A1 e 2 [A2 ,A1 ] Repeating this operation for n operators A1 , . . . , An (we may use a recursion reasoning) 1

eAn eAn−1 · · · eA1 = eAn +···+A1 e 2

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 i i ′ ′ ′ 2 ′ UI (t + ∆t, t ) = I − WI (t )∆t + O(∆t) = exp − WI (t )∆t + O(∆t)2 ~ ~ and it follows that

n Y i exp − WI (tj )∆t UI (t) ≃ ~ j=1

Since [WI (tj ), WI (ti )] is a multiple of I, we have n 2 X X (∆t) i WI (tj ) exp − WI (tj ), WI (ti ) UI (t) ≃ exp − ∆t 2 ~ 2~ t >t j=1 j

i

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 X j=1

t

Z

WI (tj ) →

0

dt′ WI (t′ ) = −

Z

0

−~az ∗ (t) − ~a† z(t)

=

t

′ ′ dt′ a e −iωt + a† e iωt f (t′ )

In a similar way (∆t)2

X

tj >ti

[WI (tj ), WI (ti )] →

Z

t

Z

dt′

0

t

′

′′

dt′′ e −iω(t −t

0

)

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) = X

=

X exp i az (t) + a z(t) exp − 2 2~ Z t Z t ′ ′′ dt′ dt′′ e −iω(t −t ) f (t′ )f (t′′ )ε(t′ − t′′ ) ∗

0

†

0

5. We can write this result in a form which is more directly useful thanks to the identity † 1 ∗ † ∗ ∗ exp i az (t) + a z(t) = exp ia z(t) exp [iaz (t)] exp − z(t)z (t) 2 and we find, using once more (2.55) UI (t)

=

Y

=

†

e ia Z t 0

=

Z

t

0

2

∗

z(t)

e iaz (t) e−Y /~ Z t ′ ′′ dt′ dt′′ e −iω(t −t ) f (t′ )f (t′′ )θ(t′ − t′′ ) 0

dt′

Z

t′

′

′′

dt′′ e −iω(t −t

)

f (t′ )f (t′′ )

0

where θ(x) is the Heaviside (or step) function. It is worth checking by explicit differentiation that this result does obey the differential equation we started from. This expression fully determines the time evolution of the forced harmonic oscillator. A particular case is that where the initial state is a coherent state: 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 |ni of H0 . We assume that the force acts only during a finite time interval [t1 , t2 ], and we choose to observe the oscillator at a time t > t2 : to summarize 0 < t1 < t2 < t. We can then integrate from −∞ to +∞: z(t) is independent of t and is nothing else than the Fourier transform of f (t)/~ 1 f˜(ω) = ~

Z

∞

′

dt′ e iωt f (t′ )

−∞

The Y factor is computed using the Fourier representation of the θ-function θ(t) = lim

η→0+

Z

+∞

−∞

dE e itE 2iπ E − iη

(11.2)

64

CHAPTER 11. EXERCISES FROM CHAPTER 11

Introducing this expression in that of Y Z ′ ′′ 1 1 dE 1 Y = e i(E−ω)t e −i(E−ω)t f (t′ )f (t′′ ) dt′ dt′′ ~2 ~2 2iπ E − iη Z dE 1 = f˜(E − ω)f˜∗ (E − ω) 2iπ E − iη Z dE ˜ 1 = P |f (E − ω)|2 + |f˜(ω)|2 2iπE 2 1 = iφ + |f˜(ω)|2 2 where P denotes a Cauchy principal value, and we have made use of 1 1 = P + iπδ(E) E − iη E 7. We finally obtain an operator UI (t) independent of t for t > t2 1 UI (t) = exp ia† f˜(ω exp iaf˜∗ (ω exp(−iφ) exp − |f˜(ω)|2 2

This allows us to compute the n → m transition amplitude Amn (t) = =

hm|U (t)|ni = hm|U0 (t)UI (t)|ni e −iEm t/~ hm|UI (t)|ni

The result is particularly simple if the oscillator is in its ground state at time t = 0 since UI (t)|0i is then a coherent state 2 † ˜ ˜ UI (t)|0i = e −iφ e −|f (ω)| /2 e ia f (ω) |0i = e −iφ |if˜(ω)i The probability to observe a state |mi is given by a Poisson law (11.34) m |f˜(ω)|2 exp −|f˜(ω)|2 p(m) = 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 result on the left by exp(iHt/~) and on the right by exp(−iHt/~) to obtain the result for any t. We note ˙ r ). We find for the time derivative of Φ ΦH (~r, t = 0) = Φ(~r), Φ˙ H (~r, t = 0) = Φ(~ s X√ ˙ r ) = −i ~ ωk ak − a†k ϕk (~r) Φ(~ 2µ k

˙ is and the commutation relation between Φ and µΦ i X r ωk h ˙ r ′ )] = − i~ aj + a†j , ak − a†k ϕj (~r)ϕk (~r ′ ) [Φ(~r), µΦ(~ 2 ωj j,k r i~ X ωk 2 δjk ϕj (~r)ϕk (~r ′ ) = 2 ωj j,k

= i~δ(~r − ~r ′ )I

2. Orthogonality of the eigenfunctions: Z Z L L 1 L ′ dx cos[(k − k ′ )x] − cos[(k + k ′ )x] = δk,k′ dx sin kx sin k x = 2 0 2 0

65 Completeness relation: let f (x) =

X

ck sin kx

k

j≥1

πj j = 1, 2, · · · L

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

Z

=

L

dx f (x)

0

X

=

k′

X

′

′

sin kx sin k x = sin kx

Z

L

0

k

X

ck′ sin k ′ x sin kx

k,k′

LX L L ′ ck ′ = ck sin kx′ = f (x) sin k x 2 2 2 k

It follows that

2X (sin kx sin kx′ ) = δ(x − x′ ) L k

The functions

r

2 sin kx e±iωt L 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 of the quantized field ΦH (x, t) s ~ X 1 ak e−iωt + a†k eiωt sin kx ΦH (x, t) = √ µL ωk ϕk (x, t) =

k≥1

3. The functions ϕ~k (~r, t) =

r

8 sin xkx sin yky sin zkz e±iω~k t V

V = Lx Ly Lz

obey the vanishing boundary conditions as well as the wave equation 2 ∂ 2 2 − c ∇ ϕ~k (~r, t) = 0 ∂t2 ~ H (~r, t) by analogy with (11.79) if ω~k2 = c2~k 2 . We can then write the expansion of the quantized field A ~AH (~r, t) =

r

2 i 4~ X X 1 h ˆ −iωk t + a† ~e ∗ (k)e ˆ iωk t sin xkx sin yky sin zkz a~ks~es (k)e √ s ~ ks ε0 V ωk s=1 ~ k

If we consider a single space dimension, the normalization is

11.5.11 Non Abelian gauge transformations

p ~/ε0 L.

1. The expression for ~ ′ is Let us first remark that We wish to have ~ = ~ ′ whence the condition

~ − qA ~ ′ ΩΦ ~ ′ = Φ† Ω−1 −i∇ ~ ~ ∇(ΩΦ) = (∇Ω)Φ + Ω∇Φ ~ − iΩ−1 (∇Ω) ~ ~ ′Ω Φ − q Ω−1 A ~ ′ = Φ† −i∇ ~ − qA ~ = −i∇ ~ − iΩ−1 (∇Ω) ~ ~ ′Ω −i∇ − q Ω−1 A

66

CHAPTER 11. EXERCISES FROM CHAPTER 11

that is

−1 ~ ~ ′ = ΩAΩ ~ −1 − i (∇Ω)Ω A q

~ is a number, not a matrix In the Abelian case, the field A ~ −1 = A ~ ΩAΩ and we recover

~′ = A ~ − ∇Λ ~ A

2. Let us choose an infinitesimal gauge transformation X 1 Ω = I − iq Λa (~r) σa = I − iqT 2 a Then

~ −1 ≃ A ~ − iq[T, A] ~ ΩAΩ

and −1 ~ (∇Ω)Ω ≃ −iq

X a

~ a ∇Λ

1 σa 2

The commutation relations (3.52) give " # X 1 X 1 X 1 ~ ~ ~ [T, A] = Λb =i εabc Λb Ac Ac σb , σc σa 2 2 2 c b

b,c

whence, by identifying the coefficient of σa /2 ~a = A ~ ′a − A ~ a = −∇Λ ~ a+q δA

X

~c εabc Λb A

b,c

3. We established the form of the covariant derivative by requiring that ~ ′ = ~, and, from our construction ~ −1 = D ~′ ΩDΩ We can write the time-independent Schr¨odinger equation (~ = m = 1)

in the form

1 ~ 2 −iD Φ = EΦ 2

Φ = Ω−1 Φ′

1 −1 ~ 2 −1 ′ Ω Ω −iD Ω Φ = EΩ−1 Φ′ 2 Multiplying the two sides of the equation by Ω we get 1 ~ ′ 2 ′ −iD Φ = EΦ′ 2

11.5.12 The Casimir effect 1. The only available physical parameters are L, ~ (we are dealing with a quantum problem) and c (the speed of electromagnetic waves). With these three parameters, we can form a unique combination with the dimension of a pressure, that is, ~c/L4 . 2. The stationary modes of the electric field have the form i h ˆ e±iωK t ~ r , t) = ei(xkx +yky ) sin πnz ~es (K) E(~ L

67 ~ When n 6= 0, the three dimensional wave vector K ~ is of the where n is an integer ≥ 0 and ωK = c|K|. form ~ = kx , ky , ± πn K L ˆ The vanishing of the transverse component and there are two independent orthogonal directions for ~es (K). ~ of the electric field E for z = 0 and z = L is then guaranteed by the factor sin(πnz/L). When n = 0, ˆ must be parallel to Oz due to the condition that the transverse component of E ~ must vanish. In ~es (K) ~ ˆ 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 X ′ ~ ωn (~k) E0 (L) = 2 2 n,~ k

4.Taking the continuum limit, (cf. Exercice 9.7.11), when the dimensions Lx , Ly → ∞ Z X S d2 k → (2π)2 ~ k

However, since L stays finite, the sum over n remains discrete, and Z ~S X′ d2 k ωn (~k) E0 (L) = (2π)2 n,~ k

This integral is divergent for large frequencies (it is called an ultraviolet divergence). We introduce a cutoff χ, which physically represents the fact that a real metal does not remain a perfect conductor at large frequencies ! ∞ Z ~k) ~S X ′ ω ( n 2 E0 (L) = d k ωn (~k)χ (2π)2 n=0 ωc Setting

c2 π 2 n 2 + c2 k 2 = ωn2 + c2 k 2 L2 and making the change of variables 2π d2 k = 2πkdk = 2 ωdω c we obtain ∞ Z ~S X ′ ∞ ω 2 E0 (L) = dω ω χ 2πc2 n=0 ωn ωc ω2 =

ω ≥ ωn

ωn =

πcn L

5. E0 (L) depends on L only through ωn = πnc/L and πnc ~S 2 dωn ~S 2 ω ω dE0 (L) = ω χ ω χ =− dL dL 2πc2 n ωc L2 2πc2 n ωc We derive the internal pressure Pint

∞ ∞ 1 dE0 (L) π 2 ~c X′ π 2 ~c X ′ 3 ω =− = − =− n χ g(n) S dL 2L4 n=0 ωc 2L4 n=0

The calculation of the external pressure is done by taking the the limit L → ∞, since outside the condenser, the field is not confined between plates. We may then replace the discrete sum over n with an integral Z π 2 ~c ∞ g(n) Pext = − 2L4 0

68

CHAPTER 11. EXERCISES FROM CHAPTER 11

The value of the total pressure is Ptot = Pint − Pext

∞ X

π 2 ~c =− 2L4

′

n=0

g(n) −

Z

∞

!

g(n) 0

From the Euler-Mac Laurin formula ∞ X

′

n=0

g(n) −

Z

∞

g(n) =

0

1 +O 5!

πc Lωc2

which leads to

π 2 ~c 240 L4 An integration allows us to derive the zero point energy Ptot = −

E0 (L) = −

π 2 ~c 720 L3

11.5.13 Quantum computing with trapped ions 1. We write the interaction Hamiltonian in terms of σ+ and σ− i h 1 Hint = − ~ω1 [σ+ + σ− ] ei(ωt−kz−φ) + e−i(ωt−kz−φ) 2

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

eiH0 t/~ σ± e−iH0 t/~ = e∓iω0 t σ± In the rotating wave approximation, we can neglect terms which behave as exp[±i(ω0 + ω)t] and we are left with i h ˜ int ≃ − ~ ω1 σ+ e i(δt−φ) e−ik˜z + σ− e−i(δt−φ) e ik˜z H 2 p 2. ∆z = ~/(2M ωz ) is the spread of the wave function in the harmonic well. Thus, η = k∆z is the ratio of this spread to the wavelength of the laser light. We may write r ~ (a + a† ) = η(a + a† ) k˜ z=k 2M ωz ˜ int between the states |1, m + m′ i and |0, mi is The matrix element of H ˜ int |mi = − 1 ~ω1 ei(δt−φ) hm + m′ |e−iη(a+a† ) |mi h1, m + m′ |H 2 The Rabi frequency for oscillations between the two levels is ′

†

ω1m→m+m = ω1 |hm + m′ |e−iη(a+a ) |mi| 3. Writing †

e±iη(a+a

)

≃ I ± iη(a + a† )

and keeping terms to first order in η we get ˜ int H

= +

h i η~ω1 σ+ a ei(δ−ωz )t e−iφ − σ− a† e−i(δ−ωz )t eiφ 2 i

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

69 ˜ int corresponds to a resonance at δ = ω − ω0 = ωz , that is, ω = ω0 + ωz , a blue sideband, The first line of H and the second line to a resonance at ω = ω0 − ωz , that is, a red sideband. The σ+ a term of the blue sideband induces transitions from |0, m + 1i to |1, mi, and the σ− a† term from |1, mi to |0, m + 1i. Now √ hm|a|m + 1i = hm + 1|a† |mi = m + 1 ˜ + as written in the statement of the problem with so that we get H int a ab = √ 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, 2i |1, 1i |1, 0i

ω+ ω0

ω+ ω−

ω

|0, 2i |0, 1i |0, 0i

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

so that

R(θ, φ = 0) = π R θ, φ = = 2 R(π, 0) = −iσx

We have, for example, π π R π, R(β, 0)R π, = 2 2

=

θ θ − iσx sin 2 2 θ θ I cos − iσy sin 2 2

I cos

π = −iσy R π, 2 β β + iσx sin )(−iσy ) 2 2 β β −(I cos + iσx sin ) = −R(−β, 0) 2 2 (−iσy )(I cos

Let us call A the transition |0, 0i ↔ |1, 1i and B the transition |0, 1i ↔ |1, 2i. The Rabi frequencies √ √ are linked by ωB = 2 ωA . Thus, if the rotation angle is θA for transition A, it will be θB = 2 θA for

70

CHAPTER 11. EXERCISES FROM CHAPTER 11

√ transition B. For transition A, we choose α = π/ 2 and β = π π π π π R √ , R(π, 0)R √ , R(π, 0) = −I 2 2 2 2 √ For transition B we shall have α = π and β = π 2 π π √ √ R(π 2, 0)R π, R(π 2, 0) = −I R π, 2 2

The state |1, 0i is not affected because the transition |0, 0i ↔ |1, 0i does not resonate on the blue sideband frequency. Thus we have |00i ↔ −|0, 0i

|0, 1i ↔ −|0, 1i

5. R ± π, π/2 = ∓iσy so that

π R ± π, |0, 1i = ∓|1, 0i 2

|1, 0i ↔ +|1, 0i

|1, 1i ↔ −|1, 1i

π R ± π, |1, 0i = ±|0, 1i 2

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

=

(a|00i + b|01i + c|10i + d|11i) ⊗ |0i

a|00, 0i + b|01, 0i + c|10, 0i + d|11, 0i

The action of R−(2) (−π, π/2) on ion 2 gives |Ψ′ i = R−(2) (−π, π/2)|Ψi = a|00, 0i + b|00, 1i + c|10, 0i + d|10, 1i +(1)

Then we apply Rαβ

on ion 1 +(1)

|Ψ′′ i = Rαβ |Ψ′ i = −a|00, 0i − b|00, 1i + c|10, 0i − d|10, 1i and finally R−(2) (π, π/2) on ion 2 |Ψ′′′ i = R−(2) (π, π/2)|Ψ′′ i =

=

−a|00, 0i − b|01, 0i + c|10, 0i − d|11, 0i (−a|00i − b|01i + c|10i − d|11i) ⊗ |0i

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 2 1 ~c e = × 200 MeV ≃ 1.5 MeV E0 = ~c R 137 The temperature at the center of the Sun is of order of 1.5 × 107 K, corresponding to a kinetic energy E of 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=

Z

RN

dr

R

e2 −E r

1/2

We make the change of variables u2 = The integration limits are

e2 −E r

dr = −

2e2 udu (u2 + E)2

e2 e2 −E ≃ R R " r r # 2 1 πe2 1 e R √ I = 2e2 √ tan−1 ≃ − RE 2 e2 2 E 2 E u = 0 et u2 =

which leads to

The probability for tunneling is ln pT (E) = − Defining EB =

√ 2µ πe2 √ ~ E

2µπ 2 e4 ~2

we cast the probability for tunneling into the form pT (E) ≃ exp −

r

EB E

!

The numerical values are, with µ = 6mp /5 EB = 2π 2 α2 µc2 = 1.18 MeV

71

72

CHAPTER 12. EXERCISES FROM CHAPTER 12

3. The factor ∼ 4π/k 2 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. Here we must in addition take tunneling into account: the potential barrier must be crossed, otherwise the reaction does not occur. An order of magnitude of the total cross section is obtained if we multiply the geometrical factor by the tunneling probability σ(E) ∼

4π p (E) k2 T

3. The angular integration gives a factor 4πv 2 and the average value of vσ is given by 3/2 Z ∞ µv 2 µ 3 hvσi = 4π dv v σ(v) exp − 2πkB T 2kB T 0 We make the change of variables v → E 2π~2 2π~2 σ(E) = pT (E) = exp − µE µE

1 E = µv 2 2 which leads to hvσi =

16π 2 ~2 µ3

We must study the integral

J=

3/2 Z

µ 2πkB T

Z

∞

∞

dE e−E/(kB T ) e−

r

EB E

!

√

EB /E

0

dE e−E/(kB T ) e−

√

EB /E

0

Let us define the function f (E) by (β = 1/(kB T )) r EB f (E) = βE − E

√ EB f (E) = β − 2E 3/2 ′

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

and we find f (E0 ) = −

p 1 kB T EB 2

1 2

2EB kB T

2/3

1/3

≃ −5.8

The width of the peak exp[−f (E)] is obtained from the second derivative f ′′ (E0 ) = and the width of the peak is

3 3p −5/2 EB E0 = 4 4

−5/3 1 −1/3 EB (kB T )−5/3 2

1/6

∆E ∼ EB (kB T )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 ~ R 2

r1 ≃ r −

1 ~ R · rˆ 2

The amplitude for finding the scattered wave at the detector after scattering by nucleus 1 is proportional to ! ~ · rˆ R 1 ikr 1 ikr1 ~′ ~ 1+ e−ik ·R/2 ≃ e e r1 r 2r

73 with ~k ′ = kˆ r . We must multiply this result by the amplitude for finding the incident plane wave at ~r1 ~ which is exp(i~k · R/2) and by a1 . The scattering amplitude by nucleus 1 is finally ! ! ~ · rˆ ~ · rˆ R R i i ~ ~′ ~ a1 ikr a1 ikr ~ 1+ 1+ exp exp − ~q · R e (k − k ) · R = e r r 2 r 2r 2 By adding the amplitudes from nuclei 1 and 2, we obtain the scattering amplitude a as the coefficient of (exp ikr)/r i ~ a = a1 + a2 − (a1 − a2 )~q · R 2 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=

~2 q 2 = 2 meV 2mn

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

= =

1 (as + 3at )I + 4 1 (as + 3at )I + 2

1 1 1 (at − as )(~σn · ~σ1 ) + (as + 3at )I + (at − as )(~σn · ~σ2 ) 4 4 4 1 ~ (at − as )(~σn · Σ) 2

3. The reduced mass is (mp ≃ mn = m) µH2 =

2m 2m2 = m + 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 2π~2 a g= µ For the same value of g the ratio of the scattering lengths is then aH2 µH2 4 = = ap µp 3 4. The total cross section is obtained using a ˆ2 a ˆ2 =

1 1 ~ + 1 (at − as )2 (~σn · Σ) ~ 2 (as + 3at )2 I + (as + at )(at − as )(~σn · Σ) 4 2 4

and, for neutron scattering on polarized molecules σtot = 4πTr a ˆ2 ~ vanishes, so that In the parahydrogen case, the average value Σ para σtot = π(as + 3at )2

In the orthohydrogen case, one remarks that Tr (A ⊗ B)2 = Tr (A2 ⊗ B 2 ) = Tr A2 Tr B 2 Since Tr σni σnj Σi Σj ∝ δij

74

CHAPTER 12. EXERCISES FROM CHAPTER 12

we obtain ~ 2 Tr (~σn · Σ)

2 2 2 2 = Tr σnx Σx + σny Σ2y + σnz Σ2z ~ 2 = 12 = 2Tr Σ2x + Σ2y + Σ2z = 2Tr Σ

The unpolarized cross section contains a factor 1/6 owing to the average over the initial spins (1/2 from the neutron and 1/3 for the orthohydrogen) and 1 4π 1 2 2 ortho (as + 3at ) × 6 + (at − as ) × 12 σtot = 6 4 4 para σtot + 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) = N e−κr

The continuity condition for the wave function at r = R gives A sin k ′ R = N e−κR and that of its logarithmic derivative cot κR = −

k′ sin k ′ R = p κ2 + k ′ 2

κ k′

that is

κ cos k ′ R = − p κ2 + k ′ 2

p κ2 + k ′ 2 A = Ne k′ The normalization is obtained from the two integrals −κR

2

J< = A

Z

R

dr sin2 k ′ r =

0

J> = N 2

Z

∞

i N 2 e−2κR h ′2 2 κ + R(k + κ ) 2k ′

dr e−2κr =

R

and the sum J< + J> has the value J< + J> =

N 2 −2κR e 2κ

N 2 e−2κR ′ 2 (k + κ2 )(1 + κR) 2k ′ 2 κ

whence N 2

2

N2 =

2κk ′ e2κR (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 Schr¨odinger equation is a linear combination of these two solutions, and because 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 i h d2 ∗ ∗2 − 2mV (r) g ∗ (k, r) g (k, r) + k dr2 i h d2 g(−k ∗ , r) + k ∗ 2 − 2mV (r) g(−k ∗ , r) 2 dr

= 0 = 0

Writing k = k1 + ik2 , we find the following asymptotic behaviors ∗ g ∗ (k, r) ∝ eikr = e−ik1 r e−k2 r g(−k ∗ , r)

∝

e−ik

∗

r

= e−ik1 r e−k2 r

The two functions obey the same differential equations and have the same behavior at infinity, so that they must be identical. We thus derive g ∗ (k ∗ ) = g(−k), whence g ∗ (k ∗ ) g(−k) 1 = = g ∗ (−k ∗ ) g(k) S(k)

S ∗ (k ∗ ) = and S(−k) =

1 g(−k) = = S ∗ (k ∗ ) g(k) S(k)

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

′

′

Ae−ik r + Ceik r e−ikr

g(k, r) = g(k, r) =

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

k′ with k ′ =

′

Ae−ik R + Ceik R ′ ′ Ae−ik R − Ceik R

=

e−ikR

=

ke−ikR

p M (V0 + E). From this we deduce

k g(k) = A + C = e−ikR cos k ′ R + i ′ sin k ′ R k

One observes that g(k) is indeed an entire function of k. The only delicate point could come from the point k ′ = 0 owing to the square root in the definition of k ′ , but there is in fact no problem as cos k ′ R 2 and (1/k ′ ) sin k ′ R are analytic functions of k ′ . 5. Let us assume that S(k) possesses a pole on the positive imaginary axis at k = iκ, κ > 0. We then have 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/2 then give the bound state energies. Let us now assume a pole of S(k) at k = h−ib. Owing to the properties

76

CHAPTER 12. EXERCISES FROM CHAPTER 12

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 the Schr¨odinger equation, they are orthogonal Z ∞ Z ∞ dr g(h + ib, r)g(−h + ib, r) = dr |g(h + ib, r)|2 = 0 0

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) ≃ for k ∼ h (k − h + ib)(k + h + ib) k − h + ib

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

e2iδ + 1 S +1 h−k =i = e2iδ − 1 S −1 b

that is δ(k) = tan−1

−b k−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 = Let us set

4πb2 4π 4π 2 = sin δ = k2 k 2 [(k − h)2 + b2 ] k 2 (1 + cot2 δ)

2mE0 2mE h2 = 2 ~ ~2 2 2 2 2 k −h m(E 2 − E02 ) k −h √ ≃ = k−h= k+h 2h ~ 2mE0 k2 =

We find σ(E) = provided we set

2π~2 ~2 Γ2 /4 M E (E − E0 )2 + ~2 Γ2 /4 ~ 2 Γ2 2E0 ~2 b2 = 4 m

7. Let us start from the radial Schr¨odinger equation u′′ − 2mV u + k 2 u = 0 and let us differentiate this equation with respect to k ∂u′′ ∂u ∂u − 2mV + k2 = −2uk ∂k ∂k ∂k We then multiply the first equation by ∂u/∂k, the second one by u and subtract the second equation from the first ∂u′′ ∂u −u = 2ku2 u′′ ∂k ∂k that is ∂ ∂u′ ′ ∂u u = 2ku2 −u ∂r ∂k ∂k

Integrating over r we get

∂u′ −u = 2k u ∂k ∂k ′ ∂u

Z

0

r

u2 (r′ )dr′

77 When r → 0 we have Furthermore

∂k ′ ∂u ′ ′ ′ cos k r =0 = A (k) sin k r + A(k)r ∂k r=0 ∂k r=0 ∂u = g(k, r)g ′ (−k) + O e−κr ∂k

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′ −u → 2 i k g(iκ)[g ′ (−iκ)] ∂k ∂k In the vicinity of the pole k = iκ, g(−k) is proportional to (k − iκ) u′

g(−k) ≃ D(k − iκ) and then g ′ (−k) = −D. Setting g(iκ) = F , we obtain Z ∞ −2ikDF = 2k u2 (k, r) dr 0

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 as N exp(−κr), where N is the constant defined at question 1, N = F G. One the one hand we have Z ∞ dr u2 (k, r) = −iDF 0

and on the other hand Z Z ∞ 1= dr u2 (k, r) = G2

∞

0

0

dr u2 (k, r) = −iF G2 = −i(F G)2

D D = −iN 2 F F

In the vicinity of the pole at k = iκ S(k) ≃

−iN 2 F = D(k − iκ) 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. We may then write its Taylor expansion as 1 1 k cot δ(k) = − + r0 k 2 + O(k 4 ) a 2 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 1 − r0 κ2 = −κ a 2

78

CHAPTER 12. EXERCISES FROM CHAPTER 12

that is

1 1− κa

2 r0 = κ

Let us calculate the residue at the pole k cot δ − ik

= =

that is S(k) =

∂ k cot δ − ik + ik0 − ik0 ∂k k=k0 −i(k − k0 ) + (k − k0 )iκr0 = −i(k − k0 )(1 − κr0 )

k0 cot δ0 + (k − k0 )

2κ −2κ =⇒ N 2 = −i(k − k0 )(1 − κr0 ) 1 − κr0

12.5.5 Neutron optics

√ 1.The distance between the nucleus and the observation point is r = s2 + z 2 . The total probability amplitude at the observation point is obtained by adding coherently the amplitudes from each nucleus Z ∞ ikr Z ∞ ikr e e 2πsds = −2πaρδ rdr ϕd = −aρδ r r z 0 as rdr = sds. The upper bound of the integral is an oscillating function. However, this oscillation has no physical meaning and we obtain the total amplitude in the form aρ δ ϕ(z) = 1 − 2iπ e ikz k 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−

aρλ2 2πaρ = 1 − k2 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 − that is

aρλ2 1 2 θc = 1 − n = 2 2π

1 2 θ =n 2 c

θc = λ

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

ρa 1/2 π

1 1 fˆ = − (as + 3at )I − (at − as )(~σn · ~σp ) = −as Ps − at Pt 4 4 As | + +i is a triplet state

fc = h+ + |fˆ| + +i = −at

If |χs i et χt i are the singlet and triplet states corresponding to m = 0, the relation inverse to (10.125) et (10.126) is | + −i = | − +i =

1 √ (|χt i + |χs i) 2 1 √ (|χt i − |χs i) 2

79 and fa

=

fb

=

1 h+ − |fˆ| + −i = − (at + as ) 2 1 h+ − |fˆ| − +i = − (at − as ) 2

The weights 3/4 and 1/4 are determined by the degeneracy of the triplet (3) and singlet (1) states. The coherent scattering length 1 3 aeff = at + as 4 4 has the numerical value −1.9 fm, and the refraction index is larger than one: there cannot be total reflection. 4. Taking the square of ~ I~ = J~ + ~σ 2 that is

~2 I~2 = J~2 + ~σ 2 + ~J~ · ~σ 4

we obtain 1 ~ J · ~σ ~ 1 ~ J · ~σ ~

= j if I = j +

1 2

= −(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 [(j + 1)a+ + ja− ] 2j + 1

b=

1 (a+ − a− ) 2j + 1

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

=

σinc

=

4π 2 [(j + 1)a+ + ja− ] (2j + 1)2 j(j + 1) (a+ − a− )2 4π (2j + 1)2

12.5.6 The cross section for neutrino absorption 1. The matrix element hϕf |ϕi i is given by Z 1 1 hϕf |ϕi i = 2 d3 r ei(~p1 +~p2 +~p3 )·~r/~ = δp~1 +~p2 +~p3 ,0 V V where ~pi represent the momenta of the final particles. The symbol δ represents a Kronecker δ since we have used plane waves in a box. Taking the square of the preceding equation, we obtain |hϕf |ϕi i|2 =

1 δp~ +~p +~p ,0 V2 1 2 3

80

CHAPTER 12. EXERCISES FROM CHAPTER 12

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, the electron kinetic energy k is p2 p2 ≫K∼ k∼ 2me 2mp The final density of states is then D(E) = V 2

Z

d3 p d3 q δ(K + E + cq − E0 ) 3 (2π~) (2π~)3

Let us integrate over the angles V

d3 p (2π~)3

→

4πV 2 4πV pEdE p dp = (2π~)3 (2π~)3 c2

V

d3 q (2π~)3

→

4πV 2 q dq (2π~)3

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

V2 pE (E − E0 )2 (2π 2 )2 ~6 c2 c3

This gives for the decay rate par unit energy E dΓ G2 h|Mf i |2 i = 2π F 4 7 5 pE(E0 − E)2 dE 4π ~ c and integrating over E while neglecting the electron mass we obtain the expression for the lifetime 1 G2F E05 =Γ∼ τ 60π 3 ~(~c)6 From the expression

G2F 60π 3 ~Γ = (~c)6 E05

we deduce that G2F /(~c)6 has the dimension of minus the fourth power of an energy and is expressed, for example, in MeV−4 . For the numerical evaluation, we convert the inverse lifetime in MeV ~ = ~Γ = 0.73 × 10−24 MeV τ which leads to

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

in qualitative agreement with the exact value. 2. The transition matrix element is 1 hϕf |ϕi i = (V)2

Z

d3 rei(~p1 +~p2 −~p

′

p ′ 2 )·~ r /~ 1 −~

=

1 δp~ +~p ,~p ′ +~p ′ V 1 2 1 2

where p~ and p~ ′ represent the initial and final particles momenta. The differential cross section is 1 2π dσ = h|Mf i |2 ihϕf |ϕi i|2 D(E) dΩ F ~

81 The final density of states for an electron of energy E emitted in the direction Ω is D(E) =

V E2 (2π~)3 c3

while the flux is given by F = c/V, as the neutrinos propagate with very nearly the speed of light and their density is 1/V. One finds for the differential cross section dσ G2 h|Mf i |2 i 2 E = F2 dΩ 4π (~c)4 and integrating over angles we obtain the total cross section for neutrino absorption σtot =

G2F h|Mf i |2 i 2 E π (~c)4

We can get an order of magnitude, in a system of units where ~ = c = 1, 1 fm = 1/(200 Mev), σ ∼ G2F E 2 , GF = 10−5 GeV−2 . For E = 10 MeV, σ ∼ 10−20 MeV−2 = 4 × 10−16 fm2 = 4 × 10−46 m2 −3

Choosing the density to be that of iron, n ≃ 1029 atoms/m ℓ=

, we find for the mean free path ℓ

1 ∼ 1015 m 30nσ

where we have taken into account the fact that an iron nucleus contains 30 neutrons. The mean free path is about 1/10th of a light year, and detecting a neutrino is really an achievement! 3. The maximum inelastic cross section is σin,max = We must have

π π ≃ 2 k2 E

1 E< ≃ 300GeV ∼ √ GF

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

82

CHAPTER 12. EXERCISES FROM CHAPTER 12

Chapter 13

Exercises from 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 any exchange of two spins: for example, the state j = 3/2, jz = 3/2 is 3 E 1 1 1 E 3 = , , j = , jz = 2 2 2 2 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 quarks must 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 of different color: it is the color wave function which is antisymmetrical.

13.4.2 Parity of the π-meson 1. The π − -meson-deut´eron system is analogous to an hydrogen atom, apart from the reduced mass, which is different mD mπ 2 µ= = 129 MeV/c mD + mπ As a consequence, the energy levels are given by En = −

R∞ 3.42 µ R∞ = −253 2 = − 2 keV 2 me n n n

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 3 S1 , 3 P1 , 1 P1 et 3 D1 , but only the state 3 P1 is antisymmetric in space+spin in the exchange of the two final neutrons. The parity of the final 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 m2e e4 1 4 ~2 n2

83

84

CHAPTER 13. EXERCISES FROM CHAPTER 13

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 conservation along 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 of the 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 global state vector must not change sign in this operation. With Y = exp(−iπJy ), we have from (10.102) Y |jmi = (−1)j−m |j, −mi The initial state changes sign if j = 1, because m = 0, and does not change sign if j = 0. As a consequence, 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 initial state vector changes sign, because the operator Y which implements this reflection is Y = Π exp(−iπJz ). In the photon case, from (10.104) and taking into account the odd parity ηγ = −1 of the photon Y|Ri = −|Li

Y|Li = −|Ri

This shows that |Φ+ i does not change sign, while |Φ− i changes sign. Thus it is the two photon entangled state |Φ− i 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 to be reflected is r = i sin θ, where t and r are defined in Exercise 1.6.6. The factor i takes into account the π/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 [G, [G, a]] + · · · 2!

where we have made use of (2.54). A straightforward calculation gives the commutators [G, a] = [ab† + a† b, a] = b so that e iGθ a e−iGθ = a + iθb −

[G, [G, a]] = [ab† + a† b, b] = a 1 2 θ a + · · · = a cos θ + ib sin θ 2!

3. If the initial state is a two photon state |Ψ0 i = |1a , 1b i = a† b† |Ωi

85 where |Ωi is the vacuum state, then the beam splitter transforms this state into |Ψi =

= = =

where we have used (a† )2 |0i =

U (θ)|Ψ0 i = U (θ)a† b† |0a , 0b i = U (θ)a† b† U † (θ)|Ωi (a† cos θ + ib† sin θ)(ia† sin θ + b† cos θ)|Ωi i √ sin 2θ (a† )2 + (b† )2 + cos 2θ a† b† |Ωi 2 i √ |2a , 0b i + |0a , 2b i + cos 2θ|1a , 1b i 2

√ 2|2i [see (11.18)]. For a symmetric beam splitter with θ = π/4 we obtain i |Ψi = √ |2a , 0b i + |0a , 2b i 2

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 |0i) or occupied (state |1i), 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 they can be represented by the 2 × 2 matrices 0 0 0 1 a= a† = 1 0 0 0 which leads to the anticommutation relation {a, a† } = I. When we consider two modes, the antisymmetry is ensured owing to the anticommutation relations. For example, |1a , 1b i = a† b† |Ωi = −b† a† |Ωi = −|1b , 1a i In order to compute e iGθ a e−iGθ we use the identity [AB, C] = A{B, C} − {A, C}B from which we find [G, b] = −a

[G, a] = b so that e iGθ a e−iGθ

= =

1 2 θ a + ··· 2! a cosh θ + ib sinh θ

a + iθb +

Repeating the calculation of question 3, we obtain |Ψi = (a† cosh θ − ib† sinh θ)(b† cosh θ + ia† sinh θ)|Ωi = a† b† |Ωi = |1a , 1b i

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

86

CHAPTER 13. EXERCISES FROM CHAPTER 13

Chapter 14

Exercises from 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)i

=

(H0 + λW )|ϕ(λ)i

=

(E0 + λE1 + λ2 E2 )|ϕ(λ)i

with |ϕ(λ)i = |ϕ0 i + λ|ϕ1 i + λ2 |ϕ2 i

and the auxiliary condition hϕ0 |ϕ(λ)i = 1, whence

hϕ0 |ϕ1 i = −λhϕ0 |ϕ2 i We deduce from these two equations, to order λ2 (H0 − E0 )|ϕ2 i = (E1 − W )|ϕ1 i + E2 |ϕ0 i Multiplying on the left by the bra hϕ0 | and taking into account that hϕ0 |ϕ1 i is of order λ E2 = hϕ0 |W |ϕ1 i

(14.1)

Furthermore, from the identification of the terms of order λ for |ϕ(λ)i gives (H0 − E0 )|ϕ1 i = (E1 − W )|ϕ0 i Let us write the identity operator under the form (|ϕ0 i ≡ |ni) X I = |nihn| + (H0 − E0 )−1 |kihk| (H0 − E0 ) k6=n

and let us use the expression (14.1) of E2 E2 = hn|W |nihn|ϕ1 i +

X

k6=n

hn|W |

1 |kihk|(H0 − E0 )|ϕ1 i H0 − E0

Neglecting the first term, which is justified because hn|ϕ1 i = O(λ) and using (14.2) E2 =

X |hn|W |ki|2 E0 − Ek

k6=n

~ = 0 creates at point R ~ an electric field 2. A dipole moment d~ = qe~r1 located at R i h ~ − 3(d~ · R) ˆ R ˆ ~ =− 1 d E 4πε0 R3

87

(14.2)

88

CHAPTER 14. EXERCISES FROM CHAPTER 14

and the potential energy of the two atom system is ~ W = −qe~r2 · E 3. hϕ01 ϕ02 |W |ϕ01 ϕ02 i = 0 because the average values of X1 , Y1 et Z1 vanish owing to parity conservation hϕ01 |X1 |ϕ01 i = 0 and the same property holds for the average values of X2 , Y2 and Z2 . 3. Using the completeness relation X α

we obtain E2 ≃ −

|ϕα ihϕα | = I

1 hϕ01 ϕ02 |W 2 |ϕ01 ϕ02 i 2R∞

The only terms of W 2 whose average value does not vanish are X12 X22 , Y12 Y22 et 4Z12 Z22 . Using rotational invariance we get 1 ~ 2 |ϕ0 i = 1 hR2 i = a20 hϕ01 |X12 |ϕ01 i = hϕ0 |R 3 3 The characteristic time for a fluctuation of the dipole moment of one of the atoms is τ ≃ ~/R∞ . In order that a static calculation such as that developed above be valid, one must be able to neglect the propagation 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 mµ /me ≃ 1 + mµ /mA 1 + mµ /(Amp )

that is m′µ = α(A)me

α(A) =

mµ /me 1 + mµ /(Amp )

Applications • Aluminium: aZ=13 = 19.8 fm µ • Lead: aZ=82 = 3.1 fm µ

R = 3.6 fm

R = 7.1 fm

2. Role of the electrons from internal shells. The wave function of the innermost electrons is ϕZ 0 (r) = p

1

Z

3 π(aZ e)

e−r/ae

aZ e =

a0 Z

The electric charge contained within an orbit of radius aZ µ is Q ≃ 2qe

aZ µ aZ e

!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 aluminium atom, its value is 3 ∆EµAl = (13)2 α(27) R∞ = 354.9 keV 4 and in the case of lead 3 ∆EµPb = (82)2 α(208) R∞ = 14.2 MeV 4

89 The approximation of a point nucleus is evidently not valid for lead, because the orbit radius is about half of the nuclear radius! This approximation cannot even be used as a zeroth order approximation. It would 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 2 Ze2 1 r 2 −3+ W (r) = 2 R R r The W potential being nonzero only for r < R, the contribution of a state whose wave function vanishes at r = 0 (p wave, d wave etc.) is negligible. For an ns wave we use Z Z 3 2 2 d r W (r)|ϕns (~r)| ≃ |ϕns (0)| d3 r W (r) Taking

into account, we find

Z

d3 r W (r) =

2π Ze2 R2 5

2π Ze2 R2 |ϕns (0)|2 5 which give the following numerical value for the 2p → 1s transition 2 2 R 4 mµ R 2Ze2 2 = = 12.6 keV R Z δE1s = ∞ Z Z 5aZ a 5 m a e µ µ µ δEns =

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 ordinary and muonic atoms: in both cases, it is proportional to α2 . By contrast, the ratio of the hyperfine structure to 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 into account a factor 1/a3 ∼ (mµ /me )3 and a factor me /mµ coming from the ratio of the electron magnetic moment 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 n r r unl (r) = c0 exp − a0 na0 l=n−1

Let us set x = r/a0 and let us study the function f (x) = xn exp(−x/n), or, in an equivalent way its logarithm g(x). The function f (x) displays a sharp maximum at x0 which is determined by studying g ′ (x), g ′ (x0 ) = 0 1 n x0 = n2 g ′ (x) = − x n Let us also compute the second derivative g ′′ (x) = − and thus

n x2

g ′′ (x0 ) = −

1 n3

(x − x0 )2 f (x) ≃ f (x0 ) exp − 2n3

90

CHAPTER 14. EXERCISES FROM CHAPTER 14

The dispersion around the maximum at x0 is ∆x = n3/2 . When l = n − 1, the radial wave function is localized around a value a0 n2 with a dispersion a0 n3/2 . 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 l 1 2 1 2 ≃ exp − lδ sin θ = cos δ ≃ 1 − δ 2 2 l

l

√ √ The wave function is thus concentrated within an angular opening δθ ≃ 1/ l ≃ 1/ n, which gives the following dispersion along Oz 1 ∆z ≃ √ a0 n2 = a0 n3/2 n The horizontal dispersion (question 1) and the vertical one (question 2) being both proportional to a0 n3/2 , the wave function is indeed concentrated in a torus of radius a0 n3/2 drawn around a circle of radius a0 .

14.6.6 Vacuum Rabi oscillations 2. The energy of the state |ϕgn i is Eng = n~ω and that of the state |ϕen i, Ene = ~ω0 + (n − 1)~ω. The energy difference between the two levels is then ∆En = Eng − Ene = ~δ and the two levels are almost degenerate if δ ≪ ω0 . 3. The coupling of the electromagnetic field with the dipole is r ~ω W = −id a(b + b† ) − a† (b + b† ) ε0 V i~ = − ΩR a(b + b† ) − a† (b + b† ) 2 W applied to a state |g ⊗ ni give two types of contribution (n) (i) Weg (n)

(ii) W eg

= =

√ i~ i~ ΩR he ⊗ (n − 1)|ab† |g ⊗ ni = − ΩR n 2 2 √ i~ i~ † † he ⊗ (n + 1)|W |g ⊗ ni = − ΩR he ⊗ (n + 1)|a b |g ⊗ ni = − ΩR n + 1 2 2 he ⊗ (n − 1)|W |g ⊗ ni = −

The second term changes the energy by 2~ω. Let us compare the time evolution of the operators ab† and a† b† in the Heisenberg picture, with H0 = Hat + Hfield ab† → eiH0 t/~ a|eihg| e−iH0 t/~ = ab† ei(ω0 −ω)t = ab† e−iδt where we have used (11.67), while

a† b† → eiH0 t/~ a† |eihg| e−iH0 t/~ = 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 only the term corresponding to the combination ab† . Owing to Hermitian conjugation, we must also keep the combination a† b. We finally obtain the Jaynes-Cummings Hamiltonian H = ~ω0 |eihe| + ~ωN −

i~ ΩR (ab† − a† b) 2

In the subspace H(n) the matrix form of this Hamiltonian is 1 1 δ√ H = n~ωI − ~δI + ~ −iΩR n 2 2

√ iΩR n −δ

91 4. From (2.35), the eigenvalues and eigenvectors in the basis {|ϕgn i, |ϕen i} are q 1 En− = − ~ δ 2 + nΩ2R |χn− i = sin θn |ϕgn i + i cos θn |ϕen i 2 q 1 ~ δ 2 + nΩ2R En+ = |χn+ i = cos θn |ϕgn i − i sin θn |ϕen i 2

When δ = 0

1 En± = ± ~δ 2

1 g e |χ± n i = √ (|ϕn i ∓ i|ϕn i) 2

5. We decompose the state |ei ≡ |e ⊗ 0i on the states |χ± 1i i − |ei = √ (|χ+ 1 i − |χ1 i) 2 which gives the time evolution for δ = 0, with E1± = ±~ΩR /2 i iΩR t/2 − |χ1 i |ei → |ψ(t)i = √ e−iΩR t/2 |χ+ 1i−e 2 The probabilty for finding |ei after a time t spent in the cavity is then pe (t) = |he|ψ(t)i|2 =

2 1 −iΩR t/2 ΩR t + eiΩR t/2 = cos2 e 4 2

6. Off resonance, pe (t) is given by the Rabi formula (4.36) pe (t) = 1 −

Ω2 t Ω2R sin2 2 Ω 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 i et |ϕgn i with a frequency q Ωn+1 = δ 2 + (n + 1)Ω2R

If the cavity contains a coherent state with an average photon number hni, we shall observe a superposition of oscillations whose frequencies are given by the results of the preceding question, the probability of each frequency being given by (11.34) hnin −hni pn = e n!

14.6.7 Reactive forces 1. The eigenvalues and eigenvectors are given by (2.35). We find q 1 |χ1n (z)i : E1n = − ~ δ 2 + nΩ21 (z) 2 q 1 |χ2n (z)i : E2n = ~ δ 2 + nΩ21 (z) 2 The force on the atom in the state |χ1n i, for example, is F1n = − and F2n = −F1n .

∂E1n 1 1 ∂Ω21 p = ~n ∂z 4 ∂z δ 2 + nΩ21 (z)

92

CHAPTER 14. EXERCISES FROM CHAPTER 14

2. The transition amplitudes are given by an11 an21

= =

an12 an22

= =

hχ1,n−1 |(b + b† )|χ1n i = − sin θn−1 cos θn hχ2,n−1 |(b + b† )|χ1n i = cos θn−1 cos θn

hχ1,n−1 |(b + b† )|χ2n i = − sin θn−1 sin θn hχ2,n−1 |(b + b† )|χ2n i = − 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 field EH (z, t) in the coherent state |zi, with |z|2 = hni r ~ω hz|EH (z, t)|zi = 2hni cos ωt sin kz ε0 V which leads to

s

~ωhni 1 = E0 ε0 V 2

and for the atom-field coupling p ~Ω1 (z) hni = dE0 sin kz = ~ω1 (z)

where ω1 (z) is the usual Rabi frequency (cf. for example (14.74)). 3. We find at once (θhni = θ) pst 1 =

sin4 θ cos4 θ + sin4 θ

pst 2 =

cos4 θ cos4 θ + sin4 θ

From question 1, the force on the atom in state |χ1 i is F1 =

1 ∂ω12 (z) 1 ~ 4 ∂z Ω1hni (z)

and F2 = −F1 . The force on an atom then is st F = F1 (pst 1 − p2 ) =

with

sin4 θ − cos4 θ 1 ∂ω12 (z) 1 p ~ 4 ∂z δ 2 + ω12 (z) cos4 θ + sin4 θ

2δΩ1hni sin4 θ − cos4 θ =− 2 4 4 δ + Ω21hni (z) cos θ + sin θ

Assembling all the factors

δ 1 ∂ω 2 (z) F =− ~ 1 2 2 ∂z 2δ + ω12 (z)

or, in a vector form

δ 1 ~ 2 r) 2 F~ = − ~∇ω 1 (~ 2 2δ + ω12 (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) ≃

r δ a a pr + δ 1− =1+ =1− =− pr pr r r a

2. Electric dipole transitions are suppressed at very low energy because of the centrifugal barrier: a P wave 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 the a~† term kλ

h1 photon|a~† |0 photonsi = 1 kλ

′

Thus we are left with W , as given in the statement of the problem. 3. The term

2π |hf |W |ii|2 δ(~ω − (Ei − Ef )) ~ comes from the Fermi Golden rule (9.170), F is the flux factor and Vω 2 dω (2π)3 c3 is the final photon space phase. 4. The angular momentum is J~ = (~/2)(~σp + ~σn ) and J~|χs i = 0 as |χs i is a state with zero angular momentum. In the same way hχs |~σp |χs i = 0 because ~σp is a vector operator whose matrix elements are zero from the Wigner-Eckart theorem, if it is sandwiched between two zero angular momentum states. If ψi (~r) is the spatial wave function of a 3 S1 state, the potential is the same as in the deuteron case, and from the orthogonality of wave functions corresponding to two different values of the energy, we have Z Z ∗ (~r)ψi (~r) = 0 d3 r ψf∗ (~r)ψi (~r) = d3 r ψD If, on the contrary, the initial wave function corresponds to a 1 S0 state, the integral does not vanish Z Z ∞ Z e−κr as as ND ∗ dΩ r2 dr 1− d3 r ψD (~r)ψi (~r) = − √ r r r 4π 0 √ √ Z ∞ as ND 4π = −ND as 4π dr e−κr 1 − (1 − κas ) = r κ2 0

5. We use the completeness relation (11.80) X eiλ (~k)ejλ (~k) = δij − kˆi kˆj λ

and X m

|hχm σp |χs i|2 t |~

=

X m

=

X m

m hχs~σp |χm σp |χs i t ihχt |~ m hχs~σp |χm σp |χs i + hχs |~σp |χs i · hχs |~σp |χs i t i · hχt |~

where we have used hχs |~σp |χs i = 0. Then we use the completeness relation in the four dimensional Hilbert space of the two spins X m |χm t ihχt | + |χs ihχs | = I m

so that

′ |h|Wspin |2 i =

ˆ 2 = 1. because ~σp2 = 3 and (~σp · k) 6. Let us summarize the various factors

1 ˆ 2 |χs i = 1 hχs |~σp2 − (~σp · k) 4 2

94

CHAPTER 14. EXERCISES FROM CHAPTER 14 1. A factor

~ 2ε0 V

originates in the expression of the quantized magnetic field. 2. A factor

1 qp2 ~2 (gp − gn )2 4 4M 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 factor

2 4πND 8π (1 − κas )2 = 3 (1 − κas )2 κ4 κ 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

Exercises from Chapter 15 15.5.1 POVM as a projective measurement in a direct sum Multiplying the normalized two component vectors of (15.31) by ˜ = M

p 2/3, we form the 3 × 2 matrix

p p p 2/3 −p 1/6 −p1/6 0 1/2 − 1/2

Let us call |v1 i and |v2 i the three-dimensional vectors whose components are the first rows of the matrix ˜ . They obey ||v1 ||2 = ||v2 ||2 = 1 and hv1 |v2 i = 0. We complete the matrix M ˜ to a 3 × 3 matrix M by M adding a third row made of the components of a vector |v3 i p p p |v3 i = ( 1/3, 1/3, 1/3)

which is normalized and orthogonal to |v1 i and |v2 i. By construction, M is and orthogonal matrix, so that the vectors |uα i whose components are given by the columns of M are normalized and mutually orthogonal. Now, let us consider a projective measurement in H(3) P1 = |u1 ihu1 |

P2 = |u2 ihu2 |

P3 = |u3 ihu3 |

Assume that the state operator ρ has non vanishing matrix elements only in H(2) . Then the probabilty of result α is p(α) = Tr (ρ|uα ihuα | = huα |ρ|uα i = h˜ α|ρ|˜ αi

15.5.3 A POVM with two arbitrary qubit states 1. The two projectors are sin2 α Pa⊥ = − sin α cos α

− sin α cos α cos2 α

Pb⊥ =

cos2 α − sin α cos α

To build a POVM with a third vector |ci we must have A + |λ|2 B −2 sin α cos α + Bλµ∗ =I −2 sin α cos α + Bλ∗ µ A + |µ|2 B

− sin α cos α sin2 α

√ √ in order that (15.23) be satisfied, whence |λ| = |µ| = 1/ 2. Choosing λ = µ = 1/ 2 we obtain A

=

B

=

1 1 = 1 + sin 2α 1+S 2S 2 sin 2α = 1 + sin 2α 1+S

95

96

CHAPTER 15. EXERCISES FROM CHAPTER 15

The POVM is Qa⊥

=

Qb⊥

=

Qc

=

1 |a⊥ iha⊥ | 1+S 1 |b⊥ ihb⊥ | 1+S S |cihc| 1+S

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

1 1 |aiha| + |bihb| 2 2

If Bob measures the result a⊥ , he knows with certainty that the spin was in the state |bi, because he would have got zero had the spin been in state |ai. The probability for finding a⊥ is 1 1 Tr (|a⊥ iha⊥ |ρ) = (1 − S) 1+S 2

p(a⊥ ) = Tr (Qa⊥ ρ) =

If sin α = 1/2, then p(a⊥ ) = p(b⊥ ) = 1/4 and in 50% of the √ cases Bob will make the right guess. In quantum 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 decides randomly, with a 50% probality of success. Thus she will get the correct result in (58+21)%=79% of the cases.

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 {|ni} basis. Furthermore hn|a† aρ + ρa† a|ni =

2nρnn

†

hn|aρa |ni =

(n + 1)ρn+1,n+1

whence the time evolution of ρnn dρnn = −nΓρnn + (n + 1)Γρn+1,n+1 dt If we choose n = 0, we find dρ/dt = Γρ11 , which means that the population of the ground state increases at a rate proportional to that of the first excited state times Γ. The corresponding physical process is the spontaneous emission of a photon, so that Γ is the rate for spontaneous emission. The evolution equation for ρn+1,n is obtained from hn + 1|[H0 , ρ]|ni = †

hn + 1|aρa |ni =

hn + 1|{a† a, ρ}|ni = so that

(2n + 1)ρn+1,n+1

p 1 dρn+1,n = −iω0 ρn+1,n + Γ (n + 1)(n + 2) ρn+2,n+1 − Γ(2n + 1)ρn+1,n dt 2

2. Using (2.54) we obtain

∗

eλ that is

~ω0 ρn+1,n p (n + 1)(n + 2) ρn+2,n+1

a

∗

a† e−λ

a

∗

a† e−λ

= a† + λ∗ [a, a† ] = a† + λ∗ a

∗

= e−λ

a

a† + λ∗

97 Taking the derivative of C(λ, λ∗ ; t) with respect to λ we obtain † ∗ ∂ = Tr ρ e λa e−λ a ∂λ

† ∗ Tr ρ a† e λa e−λ a † ∗ Tr ρ e λa e−λ a (a† + λ∗ ) † ∗ Tr (a† + λ∗ )ρ e λa e−λ a

=

=

where we have used the invariance of the trace under circular permutations to derive the last line. This equation can be written schematically as ∂ ∂ ∗ − λ → a† ρ while clearly → ρ a† ∂λ ∂λ Similarly we get for ∂/∂λ∗ ∂ λa† −λ∗ a Tr ρ e e ∂λ∗ which can be rewritten as

† ∗ = −Tr ρ e λa a e−λ a † ∗ = −Tr ρ (a − λ)e λa a e−λ a

∂ ∂ λ− → ρa while − → aρ ∂λ∗ ∂λ∗

3. Let us examine the different terms in the RHS of (15.79). From the results of the preceding question ∂ ∂ ∂2 ∂ a† aρ → − ∗ =− + λ∗ ∗ − λ∗ ∂λ ∂λ ∂λ∂λ∗ ∂λ ∂2 ∂ ∂ ∂ =− +λ ρa† a → λ− ∗ ∂λ ∂λ ∂λ∂λ∗ ∂λ so that [a† a, ρ] → λ∗

∂ ∂ −λ ∗ ∂λ ∂λ

Similarly aρa† → and {a† a, ρ} = −

∂ − λ∗ ∂λ

∂2 ∂λ∂λ∗

∂2 ∂ ∂ ∂ ∂ ∂ =− − ∗ + λ− + λ∗ ∗ + λ ∗ ∗ ∂λ ∂λ ∂λ ∂λ∂λ ∂λ ∂λ

Assembling all these results, we finally get the partial differential equation ∂ ∂ Γ Γ ∗ ∂ C(λ, λ∗ ; t) = 0 + − iω0 λ + + iω0 λ ∂t 2 ∂λ 2 ∂λ∗ or

∂ + ∂t

Γ − iω0 2

∂ + ∂ ln λ

Γ + iω0 2

∂ C(λ, λ∗ ; t) = 0 ∂ ln λ∗

To implement the method of characteristics we write dt d ln λ d ln λ = = 1 Γ/2 − iω0 Γ/2 + iω0 whence λ = λ0 exp[(Γ/2 − iω0 )t]

λ∗ = λ∗0 exp[(Γ/2 + iω0 )t]

98

CHAPTER 15. EXERCISES FROM CHAPTER 15

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 differential 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 |zihz| e λa e−λ a = hz|e λa e−λ a |zi = exp(λz ∗ − λ∗ z) We then have at time t

which can be written as

h i C(λ, λ∗ ; t) = exp z ∗ λe−(Γ/2−iω0 )t , λ∗ ze−(Γ/2+iω0 )t C(λ, λ∗ ; t) = exp [λz ∗ (t) − λ∗ z(t)]

with z(t) = ze−(Γ/2+iω0 )t Thus C(λ, λ∗ ; t) corresponds to the coherent state |z(t)i = |z e−iω0 t e−Γt/2 i 5. When |Φi is a superposition of coherent states |Φi = c1 |z1 i + c2 |z2 i then ∗

∗

C(λ, λ∗ ; t = 0) = |c1 |2 e(λz1 −λ

z1 )

∗

∗

+ |c2 |2 e(λz2 −λ

z2 )

∗

∗

+ c1 c∗2 hz2 |z1 ie(λz2 −λ

z1 )

∗

∗

+ c∗1 c2 hz1 |z2 ie(λz1 −λ

z2 )

The last two terms originate in the fact that |Φi is a coherent superposition, while these two terms would be absent in an incoherent superposition of the two coherent states. At time t we have C(λ, λ∗ ; t)

∗

∗

∗

∗

= |c1 |2 e[λz1 (t)−λ z1 (t)] + |c2 |2 e[λz2 (t)−λ z2 (t)] ∗ ∗ ∗ ∗ + c1 c∗2 hz2 |z1 ie[λz2 (t)−λ z1 (t)] + c∗1 c2 hz1 |z2 ie[λz1 (t)−λ z2 (t)]

Note that the scalar products in the last two terms are hz2 |z1 i and hz1 |z2 i, and not hz2 (t)|z1 (t)i and hz1 (t)|z2 (t)i. In order to recover the same form as at t = 0 form, we must write, for example hz2 |z1 i =

hz2 |z1 i hz2 (t)|z1 (t)i = η(t)hz2 (t)|z1 (t)i hz2 (t)|z1 (t)i

We can then describe the final state as a linear superposition of two coherent states, but the coherence is reduced by a factor Γ 1 |η(t)| = exp − |z1 − z2 |2 1 − e−Γt ≃ exp − |z1 − z2 |2 2 2

The coherence is then damped with a rate which is |z1 − z2 |2 larger that the damping rate Γ of the individual 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)i = c1 |0i + c2 |zi The global state vector at t = 0 is |Ψ(t = 0)i = c1 |0 ⊗ |0F i + c2 |z ⊗ 0F i where |0F i is the state vector (vacuum state) of the radiation field, since at T = 0 there are no available photons (or phonons). The first component of |Ψi stays unchanged under the time evolution, because spontaneous emission cannot exist. By contrast, a photon will be emitted on average after a time ∼ Γ|z|2 due the second component of |Ψi. Indeed, the time evolution of ρnn in question 1 tells us that the decay amplitude of an excited state |ni is nΓ, and the average number hni is equal to |z|2 , |z|2 = hni in the coherent state |zi. As soon as one photon is emitted, the two components of |Φi become entangled to orthogonal states of the environment, and the reduced state matrix of the oscillator loses all phase coherence. The decoherence time is thus the average time for the emission of a single photon, and τdec ≃ 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) Z +∞ y y 1 e−ipy/~ hx + |ρ(t)|x − idy (W1) w(x, p; t) = 2π~ −∞ 2 2 and

1 w(x, p; t) = 2π~

Z

+∞

−∞

z z e−ixz/~ hp + |ρ(t)|x − idz (W2 ) 2 2

We use the completeness relation and (9.22) to write Z +∞ Z +∞ y 1 1 y |x − i = dq |qihq|x − i = √ dq e−iq(x−y/2) |qi 2 2π~ −∞ 2 2π~ −∞ and an analogous formula for hx + y/2|. Plugging the two formulae in (W1) and integrating over y leads to Z +∞ 2 dq e−2i(p−q)x h2p − q|ρ(t)|qi w(x, p; t) = 2π~ −∞ 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 πσ 2

1/4

x2 exp − 2 2σ

A straightforward calculation shows that w(x, p) =

2 2 2 p σ x 1 exp − 2 exp − 2 π~ σ ~

Let us now consider the superposition (15.143) 1 ϕ(x) ≃ √ 2

1 πσ 2

1/4 h (x − a)2 i h (x + a)2 i exp − + exp − 2σ 2 2σ 2

with a ≫ σ. We find 2 2 2 h (x − a)2 i h (x + a)2 i x 2ap p σ 1 exp − + exp − + 2 exp − cos exp − 2 w(x, p) = √ ~ 2σ 2 2σ 2 σ2 ~ 2 π~

100

CHAPTER 15. EXERCISES FROM CHAPTER 15

The first two terms would correspond to an incoherent superposition of two Gaussian wave packets, but the 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 can be dealt with using exactly the same techniques. Using the (W1) form of w we obtain at once w2

=

1 2π~

=

i ∂ 2π ∂p

Z

+∞

y y e−ipy/~ hx + |{P, ρ}|x − i ydy 2 2

−∞ Z +∞ −∞

y y e−ipy/~ hx + |{P, ρ}|x − idy 2 2

We then use the W2 form w2

=

1 ∂ 2π~ ∂p

Z

+∞

−∞

y y e−ipy/~ hx + |{P, ρ}|p − idy 2 2

Z +∞ y y i ∂ p e−ixy/~ hp + |{P, ρ}|x − idy = 2π ∂p 2 2 −∞ ∂ = i~ [pw(x, p; t)] ∂p 4. One has only to observe that w(x, p; t) must vanish when x → ±∞.

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