1
Solutions of selected exercises
from ‘Quantum Physics’
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
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=hc
λ =
6.63× 10−34× 3 × 108
10−10× 1.6 × 10−19 = 1.24× 10
4eV = 12.4 keV
In the neutron case, we use p = h/λ, that is Eneut= p 2 2mn = h 2 2mnλ2 = 8.2× 10−2eV = 82 meV
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
me
= 151 eV
2. The frequency of a wave with wavelength k = 1 nm is ω = 5× 1012rad.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× 10−24kg and its wavelength
λ = h
mv = 2.5× 10
−12m
This wavelength is smaller than the molecule size by a factor∼ 1/300. 4. The distance between the mass M1and the molecule center-of-mass is
r1=
M2
M1+ M2
r0
and the molecule moment of inertia is
I = M1r21+ M2r22=
M1M2
M1+ M2
r2 0= µr20
The rotational kinetic energy is given as a function of the angular momentum J = Iω by Erot= 1
2Iω
2=J2
2I
and choosing J = ~ leads to εrot= ~2/(2I) εrot= ~2 2µr2 0 = ~ 2 2µb2a2 0 = m b2µR∞
using R∞= e2/(2a0) and a0= ~2/(me2).
5. The elastic constant K is
K = 2cR∞ b2a2 0 =4cmR 2 ∞ ~2b2 that is ~ωv= 2r c b2 rm µ R∞
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
dimension is given by M−1L3T−2. The quantityp~c/G has the dimension of a mass, which gives for
the Planck energy
EP = r ~c G c 2= 1.9 × 109J = 1.2× 1019GeV and for the Planck length
lP = ~ c r G ~c = 1.6× 10 −35m
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∼ 10−18m.
1.6.4 Neutron diffraction by a crystal
1. The incident wave arriving at point ~risuffers 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− x N 1− x one can, for example, evaluate the sum
Σx= N −1 X n=0 e−iqxna= 1− e −iqxaN 1− e−iqxa = e
−iqxa(N −1)/2sin qxaN/2
sin qxa/2
from which one deduces F (aqx, bqy) as given in the statement of the problem.
3. Suppose that qxdiffers very little from 2πnx/a, where nx is an integer: qxa = 2πnx+ ε. Then
sinqxaN 2 = sin πnxN + ε N 2 =± sinεN 2 sinqxa 2 = sin πnx+1 2ε =± sinε2
7 The peak width is then ε∼ 1/N and its height is obtained taking the limit ε → 0
lim
ε→0
sin2εN/2 sin2ε/2 = N
2
which gives an intensity within the peak∼ N2× 1/N = N. The same calculation can be repeated in the
y direction.
4. The condition for elastic scattering is
~k2= ~k′2= (~k + ~q)2= k2+ 2~q
· ~k + q2 that is q2+ 2~q
· ~k = 0. Suppose that nx= 0 (or qx= 0) and thus k′x= kx
kx′ = kx ky′ = ky−2πny
b The condition for elastic scattering is|k′
y| = |ky| which implies ky = πny b k ′ y =− 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 kxthrough kx′ = kx+ qx. One would then obtain diffraction maxima for any angle of incidence
5. The sum over the cells is
N −1 X n=0 M−1 X m=0 e−i(2aqxn+2bqym)= F (2aq x, 2bqy)
while the scattering amplitude due to the first cell is f1 1 + e−i(aqx+bqy)+ f 2 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
f = f11 + (−1)nx+ny + f2[(−1)nx+ (−1)ny]
The final result is
• nx and ny even: f = 2(f1+ f2)
• nx et ny odd: f = 2(f1− f2)
• nx even (odd) et ny odd (even): f = 0
6. When the lattice nodes are occupied randomly by the two kinds of atom, the lattice spacing is (a, b) and not (2a, 2b) as in the preceding question. One has in fact f1= f2 and half of the diffraction peaks
are lost.
1. Let b1(b2) the probability amplitude for finding the photon in the upper (lower) arm of the
interferom-eter. The probability amplitudes a1 (a2) that the photon triggers the detector D1 (D2) are obtained by
examining the transmission by the beam splitter S2: for example, a1is 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 = rb1eiδ+ tb2
a2 = tb1eiδ+ 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 t2eiδ+ r2
Using the values of t and r given in the statement of the problem and choosing the (arbitrary) normal-ization|a0| = 1 (the photon has unit probability for arriving at S1)
p1 = |a1|2= 1 2(1 + cos δ) p2 = |a2|2= 1 4 e
2iαeiδ+ e2iβ
2
= 1
2[1 + cos(2α− 2β + δ)]
By letting δ vary, one can manage that all the photons are detected by D1 (or by D2).
2. We must have p1+ p2= 1 whatever δ, because the photon must trigger one of the detectors, which
implies that cos 2(α− β) = −1, that is
α− β = π
2 mod nπ
1.6.7 Neutron interferometer and gravity
1. and 2. Let us compute the probability amplitudes a1 and a2for triggering D1 and D2
a1 = a0r2t eiχ+ 1
a2 = a0r t2eiχ+ r2
and the probabilities by taking the modulus squared
p1 = A(1 + cos χ) A = 2|r2t|2 p2 = B + A′cos χ A′= 2|r2
|Re t2(r∗)2eiχ The sum p1+ p2must 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
2 mod nπ
3. At elevation z the neutron energy is K = K0+ mgz if its energy is taken, by convention, to be K0for
z = 0. Its momentum is p =√2mK =p2m(K0+ mgz)≃p2mK0 1 +mgz 2K0
9 The approximation is justified, because, for z = 1 m
mgz≃ 10−7eV
≪ K0∼ 0.1 eV
The variation ∆k of the wave vector is
∆k = kmgz 2K0 ∆k k = mgz 2K0
On a path with length L, the phase shift accumulated between the two arms, one at elevation z and the other at elevation z = 0 is ∆φ = ∆kL = mgzLk 2K0 =mgkS 2K0 = m 2g S ~2k
because zL is the rhombus area and2K0= ~2k2/m. The numerical values is ∆φ = .59 rad.
4. It is enough to replace z by z cos θ in the preceding results. The phase shift becomes χ = ∆φ = m
2gS
~2k cos θ
and one will thus observe oscillations in the neutron detection rate by varying θ.
1.6.8 Coherent and incoherent scattering from a crystal
1. One observes that α2
i = αi. If i = j,hα2ii = hαii = p1, while if i6= j
hαiαji = hαiihαji = p21
the two results being summarized in
hαiαji = δijp1+ (1− δij)p21= p21+ p1p2δij
2. The scattering probabiilty by the crystal is h|ftot|2i = X i,j D αif1+ (1− αi)f2αjf1+ (1− αj)f2 E ei~q·(~ri−~rj) = X i,j (p2 1+ p1p2δij)f12+ 2p1p2(1− δij)f1f2+ (p22+ p1p2δij)f22ei~q·(~ri−~rj) = X i,j (p1f1+ p2f2)2ei~q·(~ri−~rj)+N p1p2(f1− f2)2
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 dtA(0)e Bt= A(0)eBtB = A(t)B The solution of dA dt = BA(t) is A(t) = eBtA(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
det 1 + A11δt A12δt A21δt 1 + A22δt = 1 + (A11+ A22)δt + (A11A22− A12A21)(δt)2 Let g(t) = det[exp(At)] g′(t) = lim δt→0 1 δt
det eA(t+δt)− det eAt
= 1 δt det e Aδt − 1 det eAt= 1 δt[δtTr A] det e At= Tr Ag(t)
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= deteA
2.4.10 Positive matrices
1. Let us decompose A into a Hermitian part and an antiHermitian part
A = B + C B = B† C =
−C†
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∗1C12x2+ x∗2C21x1= 2i Im(x∗1C12x2) = 0 =⇒ C12= 0
Since A is Hermitiian, it can be diagonalized. Let ϕ be an eigenvector of A, Aϕ = aϕ. The positivity condition implies (ϕ, Aϕ) = a||ϕ||2≥ 0 and thus a ≥ 0.
2. In the case of a real and antisymmetric matrix CT =
−C
(x, Cx) = x1C12x2+ x2C21x1= x1(C12+ C21)x2= 0
One can thus have a positive matrix of the form
A = B + C BT = B CT =−C 6= 0
2.4.11 Operator identities
1. Let us compute df /dt df dt = e tAABe−tA− etABAe−tA= etA[A, B]e−tA
The second derivative is computed in the same way, and the general case is obtained by recursion. 2. Let us compute dg/dt
dg dt = e
tA(A + B)etB
and use the result of the preceding question
etAB = etABe−tAetA = (B + t[A, B])etA
Indeed, because of the commutation relations of [A, B] with A and B, the series expansion stops after the second term. We get the differential equation
dg dt =
A + B + t[A, B]g(t)
Taking the commutation relations into account, this equation has the solution g(t) = e(A+B)t+12[A,B]t
2
Note that this solution holds only because the commutator [[A, B], A + B] vanishes. Setting t = 1 g(1) = eAeB= eA+B+[A,B]/2= eA+Be[A,B]/2
2.4.12 A beam splitter
1.The condition that there are no losses reads
13 The norm of the vector (AD, AG) is conserved, which implies that the matrix R′ is unitary. The
deter-minant of R′ then obeys
| det R′
| = 1, which can be written det R′= exp(iθ)
2. One defines R through
R = ie−iθ/2R′ det R =−e−iθdet 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|e−iφ |t|eiφ −|r|e−iχ 1 0 =|r|e iχ |t|eiφ and then ψ = (0, 1) Rψ = |r|eiχ |t|e−iφ |t|eiφ −|r|e−iχ 0 1 = |t|e−iφ −|r|e−iχ
One deduces the phase shifts for the reflected (R) and transmitted (T ) waves δD
R = χ δTD= φ δD= χ− φ
δRG = −(χ − π) δTG=−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
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 δ = e no− n ′ e cos α = 0.9248 mm
2. The index difference is no− ne= 0.17102 and the thickness of the intermediary plate
D =2× 0.9248 no− ne = 10.815 mm 3. Let β = 1/ cos α D =2e(no− βn ′ e) n0− ne
One infers from this the relative error δD/D δD D = δe e + δn0− βδn′e n0− βn′e − δno− δne no− ne
In order to simplify the error calculation, we neglect the difference between neand n′e
δD D = δe e + (β− 1)(neδno− noδne) (no− ne)(no− βne which leads to δD D ≃ 0.7 |δn| and to |δD| ≃ 7 × 10−5mm ≪ 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)
1. The action of U on σxand σy is U†σ xU = 0 eiψ e−iψ 0 U†σ yU = 0 −ieiψ ie−iψ 0 σz is clearly unchanged.
2. The possible solutions of (3.45) are
cos(α− αx) = cos φ α− αx= φ or α− αx=−φ cos(α− αy) = sin φ α− αy =π 2 − φ or α − αy= φ− π 2 The difference (α− αx)− (α − αy) =−(αx− αy)
must be independent of φ because αx and αy are given data independent of α. There are then two
possible solutions • Solution 1 α− αx= φ α− αy = φ−π 2 that is αx= αy−π 2 and σx= 0 e−iαx eiαx 0 σy = 0 −ie−iαx ie−iαx 0
From question 1, this new form corresponds to a rotation of the axes by an angle αxabout Oz.
• Solution 2
α− αx=−φ α− αy=
π 2 − φ Choosing as a reference solution αx= 0 et αy =−π/2
σx= 0 1 1 0 σy= 0 i −i 0
The change of sign of σy corresponds to an inversion of the Oy axis: one goes from a right 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· ˆp)2= I (~σ· ˆp)3= (~σ· ˆp) . . . the series expansion of the exponential is
exp −iθ2~σ· ˆp = I− iθ2~σ· ˆp + 2!1 −iθ2 2 I + 1 3! −iθ2 3 ~σ· ˆp· · · = I cosθ 2− i(~σ · ˆp) sin θ 2 Taking ˆp = (− sin φ, cos φ, 0) we get
exp
−iθ2~σ· ˆp
= I cosθ
2 + iσxsin φ sin θ
2 − iσycos φ sin θ 2 =
cosθ2 e−iφ sinθ 2 eiφsinθ 2 cos θ 2
17 2. We must have U = a1I + ia2σz+ ib2σx+ ib1σy = I cosθ 2− i sin θ 2 nz nx− iny nx+ iny −nz
and we deduce from this a1= cos θ 2 a2=−nzsin θ 2 b2=−nxsin θ 2 b1=−nysin θ 2 These equations have solutions because
a2
1+ a22+ b21+ b22= 1
3. The product of two exponentials of Pauli matrices is
e−iα(~σ·ˆa)e−iβ(~σ·ˆb)= cos α cos β− i sin α cos β(~σ · ˆa) − i sin β cos α(~σ · ˆb) − sin α sin β[ˆa · ˆb + i~σ · (ˆa × ˆb)] On the other hand
e−i[α(~σ·ˆa)+β(~σ·ˆb)]= I cos||αˆa + βˆb|| − isin||αˆa + βˆb||
||αˆa + βˆb|| [α(~σ· ˆa) + β(~σ · ˆb)] In order to ensure the equality of the two factors, we must
• get rid of the sines;
• have cos α cos β = cospα2+ β2
One can choose, for example
α = 3π β = 4π pα2+ β2= 5π
with
e−3iπσx =
−I e−4iπσy = I e−i(3πσx+4πσy)=
−I
3.3.9 Neutron scattering from spin 1/2 nuclei
1. When the nucleus spin does not flip, it is not possible to tell from which nucleus the neutron was scattered, and we must add the amplitudes
f = fa X i ei~q·~ri I = fa2 X i,j ei~q·(~ri−~rj)
2. If the scattering is accompanied with a spin flip, it leaves the nucleus in a state which is 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
I =X
i
fb2=N fb2
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
f =X i (αifa+ (1− αi)fc) ei~q·~ri+ X i αifbei~q·~ri
Were the configuration{αi} fixed, the intensity would be Iαi = X i,j (αifa+ (1− αi)fc)2ei~q·(~ri−~rj)+ X i α2 ifb2
Observing that αi= α2i the average values arehαii = hα2ii = 1/2 and hαiαji = 1/4 if i 6= j, whence
I =X i,j αiαjfa2+ 2αi(1− αj)fafc+ (1− αi)(1− αj)fc2 ei~q·(~ri−~rj)+ X i hαiifb2
We then get the result given in the statement of the problem I = 14(fa+ fc)2 X i,j ei~q·(~ri−~rj)+N 4 [(fa− fc) 2+ 2f2 b]
4. From rotational invariance, we have, for example
fa : neutron ↓ + nucleus ↑ → neutron ↓ + nucleus ↑
and a similar result for the two other amplitudes. One finds again the result of the preceding 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
i ˙c+ = Ac++ Bc−
i ˙c− = 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 accounth+|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/2cos
θ 2− µe iΩt/2sinθ 2 3. If c+(0) = 0 λ cosθ 2− µ sin θ 2 = 0 and a possible solution is
λ = sinθ
2 µ = cos
θ 2 which gives the following for c+(t)
c+(t) =− sin θ 2cos θ 2 e−iΩt/2
− eiΩt/2= i sin θ sinΩt
2 The probability p+(t) is p+(t) =|c+(t)|2= sin2θ sin2 Ωt 2 = B2 A2+ B2 sin 2Ωt 2 4. If c+(0) = 1, a possible solution is λ = cosθ 2 µ =− sin θ 2 19
and one obtains for c+(t) c+(t) = cos2θ 2e −iΩt/2+ sin2θ 2e iΩt/2 = cosΩt 2 − i cos θ sin Ωt 2 The probability p+(t) is p+(t) =|c+(t)|2= cos2 Ωt 2 + cos 2θ sin2Ωt 2 = 1− sin 2θ cos2Ωt 2
4.4.5 Unstable states
1. Let us use a series expansion of exp(−iHt/~) for small values of t c(t) = 1−~ hi Hit −2~12hH 2 it2+ O(t3) so that |c(t)|2 = 1 −hH 2i − hHi2 ~2 t 2+ O(t3) = 1−(∆H) 2 ~2 t 2+ O(t3)
2. From (4.27) we derive, with the substitution A→ P ∆P∆H ≤ 12~ dP dt p(t) =hPi(t) while ∆P = (hP2 i − hPi2)1/2= ( hPi − hPi2)1/2=p p(1− p) We thus obtain the differential inequality
dp
pp(1 − p) ≥ 2 ∆H
~ dt which integrates into
cos−1[1− 2p(t)] ≥ 2∆H~ t whence p(t)≥ cos2 t∆H ~ 0≤ t ≤ π~ 2∆H
3. Inserting a complete set of eigenstates|ni of H in the expression for c(t), we obtain c(t) = X
n
hϕ(0)|e−iHt/~|nihn|ϕ(0)i
= X
n
|hϕ(0)|ni|2e−iEnt/~
Let the spectral function w(E) be the inverse Fourier transform of c(t) w(E) = Z dt 2πe iEt/~c(t) = Z dt 2πe iEt/~X n |hϕ(0)|ni|2e−iEnt/~ = X n |hϕ(0)|ni|2δ(E − En)
21 The expectation values of H and H2 are
hHi = X n En|hϕ(0)|ni|2= Z dE E w(E) hH2i = Z dE E2w(E)
4. Use the following identity, obtained from a suitable contour in the complex x-plane Z +∞ −∞ dx e itx x2+ α2 = π αe −α|t|
4.4.6 The solar neutrino enigma
1. The Hamiltonian reads, in the reference frame where the neutrino is at rest 1 c2H = me+ mµ 2 I + me−mµ 2 m m −me−mµ 2
Comparison with exercise 4.4.4 leads to the correspondence
A→me− m2 µ B→ m tan θ = B A →
2m me− mµ
2. Since the states|ν1i and |ν2i are eigenvectors of H, the time evolution of |ϕ(t)i is
|ϕ(t)i = cosθ2e−iE1t/~|ν
1i − sin
θ 2e
−iE2t/~|ν
2i
Taking into account
hνe|ν1i = cos
θ
2 hνe|ν2i = − sin θ 2 one finds for the amplitude ce(t)
ce(t) =hνe|ϕ(t)i = e−iE1t/~ cos2θ 2 + sin 2θ 2e −i(E2−E1)t/~
and for the probability|ce(t)|2
|ce(t)|2 = cos4θ 2 + sin 4θ 2+ 2 cos 2θ 2sin 2θ 2cos ∆Et ~ = 1−12 sin2θ 1− cos∆Et~
= 1− sin2θ sin2∆Et
2~ 3. When p≫ mc one obtains the following approximate expression for E
E = (m2c4+ p2c2)1/2= cp 1 +m 2c2 p2 1/2 ≃ cp +m 2c3 2p and ∆E = E2− E1=(m 2 2− m21)c3 2p = ∆m2c3 2p
Substituting the distance L travelled by the neutrino to ct, the oscillation takes the form sin2 ∆m
2c2L
2p~
If half of an oscillation is observed between the Sun and the Earth ∆m2c2L 2p~ = π ∆m 2c4=2π~c L cp≃ 7 × 10 −11eV2
and thus ∆mc2∼ 10−5eV.
4.4.8 The neutral K meson system
1. Let us compute the productC−1MC for a generic matrix M
C = C−1= σx= 0 1 1 0 σx a b c d σx= d c b a
The commutation relation implies a = d and b = c. 2. The eigenvalues and eigenvectors of M are
λ+ = A + B |K1i = √1 2 |K0 i + |K0i λ− = A− B |K1i = 1 √ 2 |K0 i − |K0i
The time evolution of the states|K1i et |K2i is given by
|K1(t)i = e−i(A+B)t|K1i = exp −iE~1t−Γ21t |K1i |K2(t)i = e−i(A−B)t|K1i = exp −iE~2t−Γ22t |K1i
Let us start at time t = 0 from |ϕ(t = 0)i =√1
2 (c(0) + c(0))|K1i + 1 √
2 (c(0)− c(0)) |K2i One gets at time t
|ϕ(t)i = 12(c(0) + c(0))|K0 i + |K0ie−i(E1/~−iΓ1/2)t + 1 2(c(0)− c(0)) |K0 i − |K0ie−i(E2/~−iΓ2/2)t
which gives the coeficients c(t) and c(t) c(t) = hK0
|ϕ(t)i = 12(c(0) + c(0)) e−i(E1/~−iΓ1/2)t+1
2(c(0)− c(0)) e −i(E2/~−iΓ2/2)t c(t) = hK0|ϕ(t)i = 1 2(c(0) + c(0)) e −i(E1/~−iΓ1/2)t −12(c(0)− c(0)) e−i(E2/~−iΓ2/2)t
3. In the case which is considered in the statement of the problem c(0) = 1, c(0) = 0 and c(t) = 1
2 h
e−i(E1/~−iΓ1/2)t− e−i(E2/~−Γ2/2)ti
The probability to observe a K0 meson at time t is
p(t) =|c(t)|2=1 4 e−Γ1t+ e−Γ2t − 2e−(Γ1+Γ2)t/2cos∆Et ~
Chapter 5
Exercises from Chapter 5
5.5.3 Butadiene
1.The matrix form of H is
H = E0 −A 0 0 −A E0 −A 0 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|ϕ2i + · · · + cN −1ϕNi + c0|ϕ0i − Ac2|ϕ1i + · · · cN|ϕN −1i + cN +1|ϕNi = E0|χi − A N X n=1 cn−1+ cn+1 |ϕni
3. Taking into acount the postulated form for the coefficients cn, we get
cn−1+ cn+1 = c
2i h
ei(n−1)δ+ ei(n+1)δ− e−i(n−1)δ− e−i(n+1)δi
= c
2i2 cos δe
inδ
− e−inδ = 2c ncos δ
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|χsi = E0|χsi − 2A cos
πs N + 1|χi whence the values of the energy
Es= E0− 2A cos
πs N + 1 Let us compute the normalization of|χsi
X = N X n=1 sin2 πsn N + 1= 1 2 N X n=1 1− cos 2πsn N + 1 23
The sum over n is readily computed N X n=1 cos 2πsn N + 1 = Re N X n=1 exp 2iπsn N + 1 = Re 1− e 2iπs 1− e2iπs/(N +1 − 1 = −1
and X = (N + 1)/2. The normalized vectors|χsi are then
|χsi = r 2 N + 1 N X n=1 sin(nδs)|ϕni
5. In the case of butadiene N = 4 cosπ
5 = 0.809 cos 2π
5 = 0.309 which gives for the two lowest levels
Es=1= E0− 1.62A Es=2= E0− 0.62A
The energy of the four π electrons is then
E = 4E0− 2(1.62A + 0.62A) = 4(E0− A) − 0.48A
The delocalization energy is−0.48A.
7. The coefficients of the normalized eigenvectors|χ1i are
(0.372, 0.601, 0.601, 0.372) while those of|χ2i are
(0.601, 0.372,−0.372, −0.601) The order of the 1-2 bond is
1 + 2hhϕ1|χ1ihχ1|ϕ2i + hϕ1|χ2ihχ2|ϕ2i
i = 1.89 which is very close to that of a double bond, while for the 2-3 bond
1 + 2hhϕ2|χ1ihχ1|ϕ3i + hϕ2|χ2ihχ2|ϕ3i
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 H
2+ 1. The potential V (x) is V (x) =−e2 1 |x + r/2|+ 1 |x − r/2|Its value is−∞ when x = ±r/2 and its maximal value is −4e2/r for x = 0.
2. l≃ a0, characteristic size of the hydrogen atom.
3. Eigenvalues and eigenvectors
E+ = E0− A |χ+i = 1 √ 2 |ϕ1i + |ϕ2i E− = E0+ A |χ−i = 1 √ 2 |ϕ1i − |ϕ2i
25 4. A is a tunnel effect transmission coefficient. The width of the potential barrier decreases as r decreases: the transmission coefficient increases when r decreases.
5. e2/r is the (repulsive) potential energy between the two protons
E±′ (r) = E±(r) +
e2
r = E0∓ A(r) + e2
r 6. The approximate expression for E′
+(r) is E′+(r) = E0+ e2 1 r − c e −b/r = E0+ ∆E(r)
Let us look for the minimum of ∆E(r) d∆E(r) dr = e 2 −r12 + c be −r/b= ⇒ r12 0 = c be −r0/b One obtains ∆E(r0) = e2 1 r0 − c e −r0/b = e2 1 r0 − b r2 0 whence b = 6 5r0= 12 5 a0 c = 3 5a0 e5/6 ≃1.38 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 = ~ 2ωσz| ˜ϕ(t)i + e −iωσzt/2Heiωσzt/2 | ˜ϕ(t)i = ~ 2ωσz| ˜ϕ(t)i + ˜H(t)| ˜ϕ(t)i 2. Let us study ˜σ±(t) d˜σ±(t) dt =− iω 2 e −iωσzt/2[σ z, σ±]eiωσzt/2=∓iω˜σ±(t)
because [σz, σ±] =±2σ±. Solving the differential equation gives
˜
σ±(t) = e∓iωtσ±
Let us rewrite H1 in terms of σ+ and σ−, using
σx= σ++ σ− σy =−iσ++ iσ−
H1=−
~
2ω1(σ++ σ−) e
iωte−iφ+ e−iωteiφ whence
˜
H1(t) = −
~
2ω1σ+ e
−iφ+ e−2iωteiφ − ~2ω1σ− e2iωte−iφ+ eiφ
and in the rotating wave approximation, where one neglects terms in exp(±2iωt) ˜ H1 = − ~ω1 2 σ+e −iφ+ σ −eiφ = −~2ω1(σxcos φ + σysin φ)
3. From equation (3.67), exp[−iθ(~σ · ˆp)/2] is the rotation operator of a spin 1/2 about an axis ˆp. By identification, we find ˆp = (cos φ, sin φ, 0) and θ = −ω1t. The vector ˆn being normalized (ˆn2 = 1), we
have
exp(−i ˜Ht/~) = I cosΩt
2 − i(~σ · ˆn) sin Ωt 2 with ~σ· ˆn = −ω1 Ω σx+ δ Ωσz so that the matrix form of exp(−i ˜Ht/~) is
e−i ˜Ht/~= cosΩt 2 + δ Ωsin Ωt 2 iω1 Ω sin Ωt 2 iω1 Ω sin Ωt 2 cos Ωt 2 − δ Ωsin Ωt 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 giveshϕ|ρ|ϕi ≥ 0, which implies that the 2 × 2 sub-matrix A
A =
ρii ρij
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≤ ρiiρjj− |ρij|2≤
1 4 If ρii= 0, this implies ρij = 0.
2. The condition for a maximal test with 100% success implies that there exists a vector |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, i6= 1
because the test |ii, |ii 6= |1i has zero probability of success. From the preceding question, all the nondiagonal elements vanish, ρij = 0, i6= 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 σxand σy on the vectors|ε1ε2i
σ1xσ2x| + +i = | − −i σ1yσ2y| + +i = −| − −i
σ1xσ2x| + −i = | − +i σ1yσ2y| + −i = | − +i
σ1xσ2x| − +i = | + −i σ1yσ2y| − +i = | + −i
σ1xσ2x| − −i = | + +i σ1yσ2y| − −i = −| + +i
and σ1zσ2z|ε1ε2i = ε1ε2|ε1ε2i, whence the action of ~σ1· ~σ2
~σ1· ~σ2| + +i = | + +i
~σ1· ~σ2| + −i = 2| − +i − | + −i
~σ1· ~σ2| − +i = 2| + −i − | − +i
~σ1· ~σ2| − −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 =
1 √
2(| + −i + | − +i) = |IIi ~σ1· ~σ2|IV i = −3
1 √
2(| + −i − | − +i) = −3|IV i
4. The projectorsP1 et P−3 as well as ~σ1· ~σ2 are diagonalized in the basis {|Ii, |IIi, |IIIi, |IV i}
P1= 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 P −3= 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 ~σ1· ~σ2= 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 −3
IfP1= λI + µ~σ1· ~σ2, we must have
λ + µ = 1 and λ− 3µ = 0 that is, λ = 3/4 and µ = 1/4. One deduces
P1= 1
4(3I + ~σ1· ~σ2) P−3= 1
4(I− ~σ1· ~σ2) 5. We immediately getP12| + +i = | + +i and P12| − −i = | − −i, while
P12| + −i =
1
2(| + −i + 2| − +i − | + −i) = | − +i and in generalP12|ε1ε2i = |ε2ε1i.
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
H = H0− (~µe−+ ~µe+)· ~B = H0−
qe~
2m(σ1z− σ2z) Let us examine the action of the operator (σ1z− σ2z) on the basis vectors of H0
(σ1z− σ2z)| + +i = 0 (σ1z− σ2z)| − −i = 0
29 and
hII|H1|IV i = hIV |H1|IIi = 2
−q2me~
= 2Ax The matrix form of H is in the basis{|Ii, |IIi, |IIIi, |IV i}
H = E0+ A 0 0 0 0 E0+ A 0 2Ax 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
H′ = E0I + A 1 2x 2x −3 = E0I + AM
The eigenvalue equation of M is
λ2+ 2λ− (3 + 4x2) = 0
which gives the values of the energy
E±= E0− A ± 2A
p 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 ≥12N A−12N A = 0
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)|Φ0i = |Φ0i
2. The operatorPn,n+1exchanges 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|Ψni, this constant is N − 1. The operator I − Pn,n+1 applied to|Ψni 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 Pn−1,n|Ψni = |Ψn−1i Pn−1,n|Ψn−1i = |Ψni we obtain (I− Pn−1,n)|Ψni = |Ψni − |Ψn−1i (I− Pn,n+1)|Ψni = |Ψni − |Ψn+1i
This gives the action of H on|Ψni
H|Ψni = −A
|Ψn−1i + |Ψn+1i − 2|Ψni
3. Let us work out the action of H on|ksi
H|ksi = N −1 X n=0 eiksnlH|Ψ ni = −A N −1 X n=0 eiksnl |Ψn−1i + |Ψn+1i − 2|Ψni
On the other hand X n eiksnl |Ψn−1i = X n eiks(n−1)leiksl |Ψn−1i = eiksl X n eiksnl |Ψni = eiksl|ksi We thus have H|ksi = 2A(1 − cos ksl)|ksi
The eigenfrequencies are
ωk= 2A(1− cos ksl) |ks| → 0 ωk ≃ (Al2)k2s
It is also interesting to apply the method of§ 5.1.2, observing that H can be cast into the form H =−A UP+ UP−1− 2
where UP makes a circular permutation n→ n + 1 and to look for eigenvectors |Ψsi of UP. Writing
|Ψsi = X n cs n|Ψni UP|Ψsi = eiδs|Ψsi with δs= 2πs N = ksl We must have cn+1= eiδscn
6.5.7 Calculation of E(ˆa, ˆb)
1. Let us compute, for example, the amplitude a+−(θ)
a+−(θ) =h+ ⊗ [−, ˆb]|Φi = − sin2θh+ + | + cosθ2h+ − |√1 2 | + −i − | − +i=√1 2cos θ 2 2. The rotation operator by an angle θ about an axis ˆn of the ensemble of two spins is, from (3.67)
Unˆ(θ) = exp
−θ2 ~σ(a)+ ~σ(b)·ˆn
The rotational invariance of|Φi implies
Unˆ(θ)|Φi = |Φi =⇒
h
~σ(a)+ ~σ(b) · ˆni|Φi = 0 that is
~σ(a)· ˆn|Φi = − ~σ(b)· ˆn|Φi
As a consequence
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 + cos2θ
|xyi − sin2θ
|yxi + sin θ cos θ|yyi To obtain|θ⊥θi, it is enough to exchange x ↔ y
|θ⊥θi = − sin θ cos θ|xxi + cos2θ|yxi − sin2θ|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 |xi = √1 2 −|Ri + |Li |yi =√i 2 |Ri + |Li For photon 2 we must change|yi into −|yi
|xi = √1 2 −|Ri + |Li |yi = −√i 2 |Ri + |Li and one finds for photons propagating in opposite directions
|Φi = √i
2 |RRi − |LLi
Remark: if the two photons propagate in the same direction, as is the case for pairs obtained from parametric conversion, we have
|Φi = −√i
2 |RLi − |LRi
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 ˆnα along Oz while ˆnβ 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 a++(φ) = 1 √ 2
cos φhxx| + sin φhxy||xyi − |yxi=√1 2 sin φ and thus
p++(φ) = 1 2sin
2φ
Using symmetry properties in the exchange +↔ − p++(φ) = p−−(φ) =
1 2sin
2φ
Since the sum of all probabilities must add up to one p+−(φ) = p−+(φ) =
1 2cos
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 |Ψi = √1 2 |θθi + |θ⊥θ⊥i = 1 √ 2 |RRi + |LLi
Applying Σzagain 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) = −2Dl (x + y) =−θ(x + y) and the phase shift
φ(x, y)− φ(0, 0) = 2πλ(δ(x, y)− δ(0, 0)) = −kθ(y + x) The amplitude for detecting the photon 1 at point y is proportional to
1 2 exp 2iπ λ (δ(x, y)− δ(0, 0) + exp −2iπλ (δ(x, y)− δ(0, 0) = cos kθ(y + x)
3. If the pair is emitted at point x, the probability amplitude for detection in coincidence is obtained 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
a(y, z) = 1 d
Z d/2
−d/2
dx cos kθ(y + x) cos kθ(z + x)
= 1
2d Z d/2
−d/2
dxhcos[kθ((y + z + 2x)] + cos kθ(y− z)i
= 1
2d 1
kθsin kθd cos[kθ(y + z)] + d cos kθ(y− z)
5. If d ≫ 1/(kθ), the second term of a(y, z) dominates over the first one and the probability p(y, z) 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
I(z) = Z d/2 −d/2 dy cos2[kθ(y− z)] = cst 1 + O 1 kθd
33 6. In the limit d≪ 1/(kθ), the probability becomes
p(y, z) ∝ 1 4d2
h
d cos[kθ(y + z)] + d cos[kθ(y− z)]i2 = cos2(kθy) cos2(kθz)
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
a(c, c′) =√1 2 reiα r + t teiβ If r = it and|t| = |r| = 1/√2 p(c, c′) =|a(c, c′)|2= 1 8 − eiα+ eiβ 2 = 1 2 sin 2α− β 2 In a similar way, one finds
a(c, d′) = 1 √ 2rt e iα+ eiβ p(c, d′) = |a(c, d′) |2= 1 2 cos 2α− β 2 The detection of a single photon does not lead to any interference
p(c) = p(c, c′) + p(c, d′) =1 2 A symmetry argument gives at once the other probabilities
p(d, d′) =1 2 sin 2α− β 2 p(c ′, d) = 1 2 cos 2α− β 2 whence the quantity E(α, β)
E(α, β) = sin2α− β
2 − cos
2α− β
2 =− cos(α − β)
One will thus get a maximal violation of Bell’s inequalities with a choice of angles (α, β) identical to 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.
2. One may write four different probability amplitudes for the joint detection of photons in D1 et D2
(S = short, L = long)
A = a1Sa2S B = a1Sa2L
A′ = eiδa1LaL2 B′= eiδa1La2S
Amplitude ai
S corresponds to a photon path (i) going through the shorter arm of the interferometer,
ai
L 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 D10.7 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=|a1Sa2S+ eiδa1La2L|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 D2be 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
|Ψi = H|0i ⊗ H|1i =12 |0i + |1i ⊗ |0i − |1i =12
1
X
x=0
|xi⊗ |0i − |1i
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. (6.1)
The state Uf|Ψi is then a tensor product (and not an entangled state)
Uf|Ψi =
1 2
X1
x=0
35 The net result for the input register is
|xi Uf
−−→ (−1)f (x)
|xi The state of the qubit of the input register then is
|ϕi =√1 2 (−1)f (0) |0i + (−1)f (1) |1i .
Before measuring the input register, we apply a Hadamard gate (see Fig. 6.19): H|ϕi = 1 2(−1) f (0) |0i + |1i + (−1)f (1) |0i − |1i = 1 2(−1) f (0)+ ( −1)f (1)|0i +12(−1)f (0) − (−1)f (1)|1i.
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.
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 − AnB || = n||An−1 || whence ||An|| ≥ n2||An−1|| We deduce the following bounds for||An
|| n
2 ||A
n−1
|| ≤ ||An|| ≤ ||A|| ||An−1|| since ||An|| ≤ ||A|| ||An−1|| and thus||A|| ≥ n/2. It is impossible for A to be bounded.
2. The problem is that if A is not bounded, the vector B|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 0 dx ϕ∗(x) i ∂ ∂xxϕ(x) 6= Z 1 0 i∂ϕ(x) ∂x ∗ xϕ(x) 37
the difference being|ϕ(1)|2.
3. The function
ϕ(x) = ei(2πn+α)x C = eiα
obeys
ACϕ(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
Chapter 8
Exercises from Chapter 8
8.5.2 Rotations and SU(2)
1. Let us start from the most general 2× 2 matrix U =
a b c d
and compute U†U which must be identical to I
U†U = |a|2+|b|2 ac∗+ bd∗ ca∗+ db∗ |c|2+|d|2 = 1 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
τ = 1 2 3 X i=1 θiσi
where the angles θi are infinitesimal since τ itself is infinitesimal.
3. Since θ/N is infinitesimal. we may write Uˆn θ N = I− iθ 2N(~σ· ˆn) and using lim N →∞ 1− x N N = e−x we deduce Unˆ(θ) = exp −iθ2(~σ· ˆn)
4. The determinant ofV is equal to minus the length squared of the vector ~V : det V = −~V2 and since
det(UVU−1) = det
V
we obtain ~W2= det
W = ~V2, which shows that the transformation preserves the lengths of vectors. It is
thus, either a rotation, or a rotation combined with a parity operation. 5. SinceW is Hermitian and has zero trace, because
Tr (UVU−1) = Tr V we may writeW = ~σ · ~W = ~σ· ~V (θ). d dθ~σ· ~V (θ) = − i 2[~σ· ˆn, ~σ · ~V (θ)] = ~σ · (ˆn × ~V (θ))
where we have used (3.52), which shows that ~V (θ) is deduced from ~V through a rotation by an angle θ about ˆn
d~V
dθ = ˆn× ~V (θ)
To any rotationRnˆ(θ) correspond two matrices Unˆ : Unˆ(θ) and Uˆn(θ + 2π) =−Unˆ(θ).
8.5.4 The Lie algebra of a continuous group
1Since 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 f a(θ, θ)
fa(θ, θ) = θa+ θa+ λabcθbθc+ λabcθbθc+ fabcθbθc+ O(θ3, θ2θ, θ θ 2
, θ3) The condition fa(θ, θ = 0) = θa implies λabc= 0 and similarly λabc = 0.
3. One the one hand, we have, neglecting terms of order (θ3, θ2θ, θ θ2, θ3)
U (θ)U (θ) = I− iθaTa− iθaTa−
1
2(θbθc+ θbθc)Tbc− θaθbTaTb and, on the other hand
U (f (θ, θ)) = I− iTa(θa+ θa+ fabcθbθc)−1
2(θb+ θb)(θc+ θc)Tbc
Relabeling the summation indices and taking into account the symmetry property Tbc= Tcb, because
Tbc=− ∂ 2U ∂θa∂θb θa=θb=0
the comparison between the two expressions gives
θbθcTbTc= ifabcTaθbθc+ θbθcTbc
We deduce from this
TbTc = Tbc+ ifabcTa
TcTb = Tbc+ ifacbTa
and, subtracting the second equation from the first one
[Tb, Tc] = i[fabc− facb]Ta
The structure constant Cabc is
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
[P2, X] =−2i~P and [[P2, X], X] =−2i[P, X] = −2~2 whence P2 2m+ V (X), X = [H, X] =−i~mP [[H, X], X] =−~ 2 m 2. On the other hand, the commutator is also expressed as
[[H, X], X] = HX2− 2XHX + X2H and usinghϕn|H|ϕmi = Enδnm hϕ0|HX2|ϕ0i = X n,m hϕ0|H|ϕnihϕn|X|ϕmihϕm|X|ϕ0i = E0 X n |Xn0|2 hϕ0|XHX|ϕ0i = X n,m hϕ0|X|ϕnihϕn|H|ϕmihϕm|X|ϕ0i = X n En|Xn0|2
whence the result
X
n
2m|Xn0|2
Chapter 9
Exercises from Chapter 9
9.7.2 Wave packet spreading
1.
[P2, X] = P [P, X] + [P, X]P =−2i~P One can also use
[f (P ), X] =−i~f′(P ) 2. From the Ehrenfest theorem (4.26), choosing A = X2
d dthX 2 i(t) = ~ hi [H, X2]i =~iDhP 2 2m+ V (X), X 2iE = i 2~mh[P 2, X2] i Furthermore [P2, X2] =−2i~(XP + P X) from which we derive the final result
d dthX
2
i(t) =m1hXP + P Xi Going to the x representation
hP Xi = Z dx ϕ∗(x) −i∂x∂ (xϕ(x) = i Z dxxϕ(x)∂ϕ ∗(x) ∂x
where the second expression is obtained from an integration by parts. Combining with hXP i = Z dx ϕ∗(x)x −i∂ϕ(x)∂x we finally get d dthX 2 i(t) = i~m Z ∞ −∞ dx x ϕ∂ϕ ∗ ∂x − ϕ ∗∂ϕ ∂x
These results are valid for a particle in a non zero potential, and not only for a free particle.
3. On the contrary, the results that follow are only valid for a free particle, V (X) = 0. The Hamiltonian is then reduced to the kinetic Hamiltonian H = K = P2/(2m). We compute the second derivative of
hX2 i(t) d2 dt2hX 2 i(t) =~i dtdh[K, X2] i = −~12 d dth[K, [K, X 2]] i 43
where we have used Ehrenfest’s theorem twice. Taking into account [P2, XP + P X] =−4i~P2 we find d2 dt2hX 2 i(t) = m2 hP2 i The third derivative ofhX2
i(t) and higher order derivatives vanish dn dtnhX 2 i(t) = 0 n≥ 3 because [K, [K, X2]] ∝ P2 and [K, P2] = 0. hX2
i(t) is thus a second order polynomial in t
hX2 i(t) = hX2 i(t = 0) + tdtdhX2 i(t) t=0+ 1 2t 2d 2 dt2hX 2 i(t) t=0 In order to compute the dispersion, we use for a free particle
∆x2(t) = hX2 i(t) − [hXi(t)]2 and d dthXi(t) = i ~[K, X] = DP m E which gives hXi(t) = hXi(t = 0) + tDPmE 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−α2x2 = √π α ∆x = 1 √ 2 α 1. Setting k′= k − k Z dk|A(k)|2= 1 πσ2 1/2Z dk′ exp −k ′2 σ2 ! = 1
so that ∆k = σ/√2. Let us compute the wave function at time t = 0
ϕ(x, 0) = 1 πσ2 1/4 eikx Z dk′ √ 2π exp ik ′x − k ′2 2σ2 ! = σ 1/2 π1/4 exp ikx−1 2σ 2x2
The modulus squared of the wave function is
|ϕ(x, t)|2=√σπe−σ2x2 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 πσ2 1/4Z dk √ 2π exp ikx− i ~k2 2mt− (k− k)2 2σ2 !
The exponent is rewritten, within a factor of i and with k′ = k
− k kx−~k 2 2m t− (k− k)2 2σ2 = kx + k ′x −~k 2 2mt− ~kk′ m t− ~k′2 2m t− k′2 σ2 ϕ(x, t) reads ϕ(x, t) = 1 πσ2 1/4 exp ikx− i~k 2 2mt ! Z dk′ √ 2π exp ik′ x− ~k 2mt exp " −k ′2 2 1 σ2 + i~t 2m #
If we can neglect the term i~t/2m in the last exponential, we simply obtain
ϕ(x, t) = 1 πσ2 1/4 exp ikx− i~k 2 2mt ! exp " −σ 2 2 x−~k m t 2# = exp i~k 2 2mt ! ϕ(x− vgt, 0) = eiω(k)tϕ(x− vgt, 0)
3. In the general case, we set
1 σ′2 = 1 σ2 + i~t m and we find, after doing the integration
ϕ(x, t) = 1 πσ2 1/4 σ′exp −12σ′2(x − vgt)2 expi(kx − ω(k)t) Taking the modulus squared
|ϕ(x, t)|2= 1 πσ2 1/2 |σ′|2exp−Re σ′2(x− vgt)2
The peak of|ϕ(x, t)|2is centered at x = v
gt and has a width
∆x2(t) = 1 2 Re σ′2 that is, ∆x2(t) = 1 2σ2 1 + ~ 2σ4t2 m2
The width of the wave packet increases with time, because of the term ~2σ4t2/m2 in the bracket.
4. ∆x2(t) doubles for t = m ~σ2 = 2m∆x2(t = 0) ~ = 3.2× 10 −11s
9.7.7 A delta-function potential
1. The dimension of δ(x) isL−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
which is indeed the dimension of an energy.
2. We integrate the Schr¨odinger written in the form d2 dx2 + 2mE ~2 ϕ(x) = g δ(x) ϕ(x) 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 x6= 0 so that ϕ′(0+) − ϕ′(0−) = −2κA = −|g|ϕ(0) = −|g|A whence 2κ =|g| and E E =−~ 2κ2 2m =− ~2g2 8m
There is no odd solution because ϕ(0) = 0 for an odd solution. Let us retrieve the result by taking the limit a→ 0, V0a→ ~2g/(2m) of the square well potential whose energy levels are given by (9.82)
κ = k tanka 2 which leads to k≃ p2m|V~ 0| tanka 2 ≃ tan p2m|V0|a 2~ → 0 and κ≃p2m|V0| ~ p2m|V0| a 2~ = m|V0|a ~2 = m ~2 ~2|g| 2m = 1 2|g|
3. Since the diatomic molecule potential is even, one looks for even and odd solutions. For the 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−κltanh 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| 2 1 + e
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 A sinh κl = e−κl
−κe−κl
− Aκ cosh κl = −|g|e−κl
which leads to
κ = |g| 2 1− e
−2κl
This equation has a unique solution if the derivative of |g|[1 − exp(−2κl)]/2 > κ at κ = 0, that is, if |g|l > 1. There is no odd solution if |g|l < 1.
4. Let us consider two deep and narrow potential wells of width a separated by a distance l, with a≪ l. The two wells can then be approximated by delta-functions, which leads us back to the potential of the preceding question, and we assume κl ≫ 1. Then there exist two bound states, one with an even wave-function corresponding to
κ+= |g|
2 (1 + e
−2κl)
and the other one with an odd wave-function corresponding to κ−= |g|
2 (1− e
−2κl)
The energy difference between the two states is then E−− E+=− ~2 2m(κ 2 −− κ2+) = ~2 2mg 2e−2κl
the average enrgy E0 being
E0=
1
2(E++ E−)≃ − ~2g2
8m We can thus write
E+ ≃ − ~2g2 8m 1 + 2e −2κl E− ≃ − ~2g2 8m 1− 2e −2κl These are the eigenvalues of a two-level Hamiltonian
H =−~ 2g2 8m 1 −2e−2κl −2e−2κl 1
From (9.100) and (9.105), the tunneling transmission coefficient of a particle with energy≃ 0 through 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
We derive
A = 1 + ig
2k B =−
ig 2k whence the matrix elements of the transmission matrix
M11= 1 + ig
2k M12=
ig 2k
We check that the results do agree with the limit a→ 0, V0a→ ~2g/(2m) of equations (9.96) and (9.97).
6. The condition
ϕq(x) = eiqlϕq(x− l)
gives
F eikx+ Ge−ikx= eiqlAeik(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
Ah1− ei(q−k)li+ Bh1
− ei(q+k)li = 0
Ahg + ik1− ei(q−k)li+ Bhg
− ik1− ei(q+k)li = 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
∆ = det 1− eiα 1− eiβ g + ik 1− eiα g− ik 1 − eiβ = 0 Computing ∆ gives
∆ = g eiβ− eiα − 2ik1
− 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 2k sin kl
9.7.13 Study of the Stern-Gerlach experiment
1. Since the yOz plane is symmetry plane of the problem, Bzmust be an even function of x and we must
have ∂Bz ∂x x=0= 0 Translation invariance along Oy gives
∂Bz ∂y x=0= 0
The two nonzero components of the magnetic field in the vicinity of x = 0 are Bx=−bx Bz= B0+ bz
This field obeys the two Maxwell equations (1.8) and (1.9) in vacuum ~∇ × ~B = 0 and ~
∇ · ~B = ∂Bx ∂x +
∂Bz
49 The potential energy is
−~µ · ~B =−µxBx− µzBz= bµxx− bµzz
whence the force ~F with components
Fx=∂(−~µ · ~B)
∂x = bµx Fz=
∂(−~µ · ~B)
∂z =−bµz
The B0z component of the magnetic field leads to Larmor precession of the spin about the Oz axis(ˆ § 3.25)
in which µzstays constant. By contrast, due to this precession, the average value of µxvanishes: hµxi = 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 ±ˆz. The splitting between the trajectories of an up spin and a down spin at the exit of the magnet gap is then δ = 21 2 F mt 2= F m L v 2 =µb m L v 2 Let us also evaluate the product ∆z∆pz
∆z∆pz∼ 10−4 1.8× 10−25 (10) = 1.8 × 10−28MKSA∼ 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
i~∂ϕ± ∂t = −~ 2 2m∇ 2 ∓ µbz ϕ±= Hϕ±
Using Ehrenfest’s theorem (4.26), we obtain d dth ~R±i(t) = i ~[H, ~R±] = 1 mh ~P±i d dthPx,y,±i(t) = i ~[H, Px,y,±] = 0 d dthPz,±i(t) = i ~[H, Pz,±] =±µb This last result is deduced from
∓[µbZ, Pz] =∓i~µb
We finally get
hZ±i = ±µb
2mt
2
and the center of the wave packet does follow the classical trajectory.
4. Let us make a reflection with respect to the xOy plane. In this reflection, ~µ does not change, 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 −ˆx. However, the image of the trajectory in the mirror always leave in the +ˆx 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
which integrates into U (t, t0) = exp −~iAP Z t t0 g(t′)dt′
Between times ti and tf we thus have
U (tf, ti)≃ exp −~iAP Z +∞ −∞ g(t′)dt′ = exp −~i gAP
2. The action of U (tf, ti) on the vector |n ⊗ ϕi is
U (tf, ti)|n ⊗ ϕi = e−iganP/~|n ⊗ ϕi
Furthermore, exp(−iganP/~) is a translation operator by gan and from (9.13)
e−iganP/~ϕ(x) = ϕ(x
− gan)
3. Because of the linearity property of quantum mechanics, the final state vector is |χfi =
X
n
cn|n ⊗ ϕni
The reduced state operator of S is from (6.66) ρ(1)=X n,m cnc∗m|mihn|hϕm|ϕni = X n |cn|2|nihn|
becausehϕm|ϕni = δnm. The system S is thus a statistical mixture of states|ni with a weight |cn|2, and
Chapter 10
Exercises from chapter 10
We remind the reader that we use in chapter 10 a system of units where ~ = 1
10.7.5 Orbital angular momentum
1. From the expression of the orbital angular momentum operator as a function of the position ~R and 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))
e−iαLxf (~r) = f R−1
ˆ
x [α](~r) = f (Rˆx[−α](~r))
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′ = z cos α− y sin α dz =−ydα 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 tan θdα This allows us too identify Lx
[−idαLxf ](~r) = dα sin φ∂ ∂θ + cos φ tan θ ∂ ∂φ f (~r) that is Lx= i sin φ∂ ∂θ + cos φ tan θ ∂ ∂φ
To compute Ly we can use the same method, or make use of the commutation relation
Ly =−i[Lz, Lx] =−
∂ ∂φ, Lx
