-
1Solutions of Exercises
from A Short Introduction to Quantum
Information and Quantum Computation
Michel Le Bellac
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2
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Chapter 2
Exercises from Chapter 2
2.6.1 Determination of the polarization of a light wave
1. We can choose x = 0, y = . The equation of the ellipse
x = cos cost y = sin cos(t )reads in Cartesian coordinates
x2
cos2 2xy cos
sin cos +
y2
sin2 = sin2
The direction of the axes is obtained by looking for the eigenvectors of the matrix
A =
1
cos2 cos sin cos
cos sin cos
1
sin2
which make angles and + /2 with the x-axis, where is given by
tan = cos tan 2
The vector product of the position ~r with the velocity ~v, ~r ~v, is easily seen to be
~r ~v = 12z sin 2 sin
so that the sense of rotation is given by the sign of the product sin 2 sin .
2. The intensity at the entrance of the polarizer is
I0 = k(E20 cos
2 + E20 sin2 )= kE20
where k is a proportionality factor. At the exit of the polarizer it is
I = kE20 cos2 = I0 cos
2
The reduction of the intensity allows us to determine | cos |.3. The projection of the electric field on the polarizer axis is
E02
[cos cost+ sin cos(t )]
and the intensity is given by the time average
I = kE20
cos2 cos2 t+ sin2 cos2(t ) + 2 sin cos cost cos(t )
=
1
2kE20 (1 + sin 2 cos ) =
1
2I0(1 + sin 2 cos )
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4 CHAPTER 2. EXERCISES FROM CHAPTER 2
From the measurement of I we deduce cos , which allows us to deduce up to a sign. The remainingambiguities are lifted if one remarks that the ellipse is invariant under the transformations
+ and
+ .
2.6.2 The (, ) polarizer
1. The components E x and E y are given by
E x = Ex cos2 + Ey sin cos ei = ||2Ex + Ey,E y = Ex sin cos ei + Ey sin2 = Ex + ||2Ey.
2. This operation amounts to projection on |. In fact, if we choose to write the vectors |x and |y ascolumn vectors
|x =(1
0
), |y =
(0
1
)then the projector P
P = || =(|x + |y)(x|+ y|)
is represented by the matrix
P =( ||2
||2)
3. Since P = ||, we clearly have P| = | and P| = 0, because| = + = 0
2.6.3 Circular polarization and the rotation operator
1. In complex notation the fields Ex and Ey are written as
Ex = 12E0, Ey = 1
2E0 e
ipi/2 =i2E0,
where the (+) sign corresponds to right-handed circular polarization and the () to left-handed. Theproportionality factor E0 common to Ex and Ey defines the intensity of the light wave and plays no rolein describing the polarization, which is characterized by the normalized vectors
|R = 12(|x+ i|y), |L = 1
2(|x i|y)
2. Let us compute |R. We have
|R = 12
(cos |x+ sin |y i sin |x+ i cos |y)
=12
(ei |x+ iei |y) = ei |R
and similarly |L = exp(i)|L. The vectors |R and L differ from |R and L by a phase factor only,and they do not represent different physical states.
3. The projectors on the vectors |R and |L are given by
PD = 12
(1 ii 1
)PG = 1
2
(1 ii 1
)
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5and is
= PD PG =(
0 ii 0
)
This operator has the states |R and |L as its eigenvectors, and their respective eigenvalues are +1 and1:
|R = |R, |L = |LThus exp(i)|R = exp(i)|R and exp(i)|L = exp(i)|L4. From the form of in the {|x, |y} basis we get at once 2 = I, and thus
ei = I i + (i)2
2!I +
(i)33!
+
The series is easily summed with the result
exp(iz) =(
cos sin sin cos
)
If we apply the operator exp(i) to the vectors of the {|x, |y} basis, we get the rotated vectors |and |, so that this operator represents a rotation by an angle about the z axis.
2.6.4 An optimal strategy for Eve?
1. If Alice uses the |x basis, the probability that Eve guesses correctly is px = cos2 . If she uses thethe | /4 basis, this probability is
ppi/4 = || /4|2 =1
2(cos+ sin)2
The probability that Eve guesses correctly is
p() =1
2(px + ppi/4) =
1
4
[2 cos2 + (cos+ sin)2
]=
1
4[2 + cos 2+ sin 2]
The maximum of p() is given by = 0 = /8, which is evident from symmetry considerations: themaximum must be given by the bisector of the Ox and /4 axes. The maximum value is
pmax =1
2
(1 +
12
) 0.854
2. If Alice sends a | (|) photon, Eve obtains the correct result with probability cos2 (sin2 ), andthe probability that Bob receives the correct polarization is cos4 (sin4 ). The probability of success forEve is
ps =1
2
(1 + cos4 + sin4
)and the probability of error
pe = 1 ps = sin2 cos2 =1
4sin2 2
Evess error are maximal for = /4.
Heisenberg inequalities
1. The commutator of A and B is of the form iC, where C is a Hermitian operator because
[A,B] = [B, A] = [B,A] = [A,B].
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6 CHAPTER 2. EXERCISES FROM CHAPTER 2
We can then write[A,B] = iC, C = C. (2.1)
2. Let us define the Hermitian operators of zero expectation value (a priori specific to the state |):
A0 = A AI, B0 = B BI.
Their commutator is also iC, [A0, B0] = iC, because A and B are numbers. The squared norm ofthe vector
(A0 + iB0)|,where is chosen to be real, must be positive:
||(A0 + iB0)|||2 = ||A0|||2 + i|A0B0| i|B0A0|+ 2||B0|||2= A20 C + 2B20 0.
The second-degree polynomial in must be positive for any , which implies
C2 4A20B20 0.
This demonstrates the Heisenberg inequality
(A) (B) 12
C . (2.2)3. In a finite dimensional space, the trace of a commutator vanishes because Tr (AB) = Tr (BA), so thatthe equality
[X,P ] = i~I
cannot be realized in a finite dimensional space.
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Chapter 3
Exercises from Chapter 3
3.5.1 Rotation operator for spin 1/2
1. We use x|0 = |1, x|1 = |0, y |0 = i|1, y|1 = i|1, z |0 = |0, z |1 = |1 to obtain
x| = ei/2 sin 2|0+ ei/2 cos
2|1
y| = iei/2 sin 2|0+ iei/2 cos
2|1
z | = ei/2 cos 2|0 ei/2 sin
2|1
so that
|x| = sin 2cos
2
(ei + ei
)= sin cos
|y | = sin 2cos
2
(iei + iei) = sin sin|z | = cos2
2 sin2
2= cos
2. From (3.8) we derive the identity
(~ ~a)(~ ~b) = ~a ~b I + i~ (~a~b)
so that(~ p)2 = I (~ p)3 = (~ p)
and the series expansion of the exponential reads
exp
(i
2~ p
)= I +
(i2
)(~ p) + 1
2!
(i2
)2I +
1
3!
(i2
)3(~ p) +
= I cos
2 i(~ p) sin
2
The action of the operator exp (i ~ p/2) on the vector |0 is
exp
(i
2~ p
)|0 = cos
2|0+ ei sin
2|1
which is the same as (3.4) up to a physically irrelevant phase factor exp(i/2). Thus exp (i ~ p/2) isthe operator which rotates the vector |0, the eigenvector of z with eigenvalue 1, on |, the eigenvectors
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8 CHAPTER 3. EXERCISES FROM CHAPTER 3
of ~ n with the same eigenvalue. The same result holds for the eigenvalue 1, corresponding to |1 andthe rotated vector U [Rp()]|1.3. Let us specialize the above results to the case = /2, which corresponds to rotations about thex-axis
U [Rx()] = exp(i
2x
)=
(cos(/2) i sin(/2)i sin(/2) cos(/2)
)
A rotation about the x-axis transforms |0 into the vector
| = cos 2|0 i sin
2|1
Taking = 1t, this corresponds exactly to (3.31) with the initial conditions a = 1, b = 0.
3.5.2 Rabi oscillations away from resonance
1. Substituting in the differential equation the exponential form of (t), we get the second order equationfor
22 2 1
221 = 0
whose solutions are
=1
2
[
2 + 21
]=
1
2[ ]
2. The solution of the differential equation for is a linear combination of exp(i+t) and exp(it)
(t) = a exp(i+t) + b exp(it).
Let us choose the initial conditions (0) = 1, (0) = 0. Since (0) d/dt(0), these initial conditionsare equivalent to
a+ b = 1 and a+ b = 0,and so
a =
, b =+
The final result can be written as
(t) =e it/2
[cos
t
2 i sin t
2
],
(t) =i1
eit/2 sint
2,
which reduces to (3.31) when = 0. The factor exp(it/2) arises because is the Larmor frequency inthe rotating reference frame. The second equation shows that if we start from the state |0 at t = 0, theprobability of finding the spin in the state |1 at time t is
p01(t) =212
sin2(t
2
)
We see that the maximum probability of making a transition from the state |0 to the state |1 fort/2 = /2 is given by a resonance curve of width :
pmax =
212
=21
21 + 2=
2121 + ( 0)2
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Chapter 4
Exercises from Chapter 4
Basis independence of the tensor product
The tensor product |iA jB is given by
|iA jB =m,n
RimSjn|mA nB
Let us define |A B by using the {|iA, |jB} bases
|A B =i,j
cidj |iA jB
=
i,j,m,n
cidjRimSin|mA nB
We can now use the transformation law of the components in a change of basis
ci =k
R1ki ck dj =l
S1lj dl
to show that|A B =
m,n
cmdn|mA nB = |A B
Thus the tensor product is independent of the choice of basis.
4.6.2 Properties of the state operator
1. Since the pi are real, is clearly Hermitian. Furthermore Tr =
i pi = 1, and finally is positive as
|| =i
pi||i|2 0
Let us first compute Tr (M |ii|)
Tr (M |ii|) =j
j|M |ii|j = i|M |i
whence
Tr
(i
piM |ii|)=i
pii|M |i
The expectation value of M in the state |i appears in the sum over i with the weight pi, as expected.
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10 CHAPTER 4. EXERCISES FROM CHAPTER 4
2. In the |i basis, has a diagonal form with matrix elements ii = pi, so that 2 = can only holdif one of the probabilities is one, as the equation p2i = pi has solutions pi = 1 and pi = 0. Furthermore,Tr 2 =
i p
2i and
i p
2i
i pi, where the equality holds if and only if one of the pi is equal to one.
Let us assume, for example, that p1 = 1, pi = 0, i 6= 1. Then = |11|, which corresponds to the purestate |1. One can also remark that 2 = implies that is a projector P , and the rank of this projectoris one, because TrP is the dimension of the subspace on which P projects.
4.6.3 The state operator for a qubit and the Bloch vector
The condition for a Hermitian 2 2 matrix is 01 = 10, so that
=
(a cc 1 a
)
is indeed the most general 22 Hermitian matrix with trace one. The eigenvalues + and of satisfy
+ + = 1, + = det = a(1 a) |c|2,
and we must have + 0 and 0. The condition det 0 implies that + and have the samesign, and the condition + + = 1 implies that + reaches its maximum for + = 1/4, so thatfinally
0 a(1 a) |c|2 14
The necessary and sufficient condition for to describe a pure state is
det = a(1 a) |c|2 = 0.
The coefficients a and c for the state matrix describing the normalized state vector | = |+ + |with ||2 + ||2 = 1 are
a = ||2 c =
so that a(1 a) = |c|2 in this case.2. Since any 2 2 Hermitian matrix can be written as a linear combination of the unit matrix I and thei with reals coefficients, we can write the state matrix as
=I
2+i
bii =1
2
(I +~b ~
)=
1
2
(1 + bz bx ibybx + iby 1 bz
)
where we have used Tri = 0. The vector ~b, called the Bloch vector, must satisfy |~b|2 1 owing to theresults of question 1, and a pure state corrresponds to |~b|2 = 1. Let us calculate the expectation value of~ using Tr ij = 2ij . We find
i = Tr ( i) = biso that ~b is the expectation value ~.3. With ~B parallel to Oz, the Hamiltonian reads
H = 12z
The evolution equation
i~d|(t)dt
= H |translates into the following for the state matrix
i~d(t)
dt= [H, ]
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11
so thatd
dt=
1
i~[H, ] = 1
2B(bxy byx)
which is equivalent todbxdt
= Bby dbydt
= Bbxdbxdt
= 0
This can be put in vector form
d~b
dt= ~B ~b
This equation shows that the Bloch vector rotates about the Oz axis with an angular frequency = B.
4.6.4 The SWAP operator
1. Let us write explicitly the action of x et y on the vectors |12
1x2x|++ = | 1y2y|++ = | 1x2x|+ = | + 1y2y|+ = | +1x2x| + = |+ 1y2y| + = |+1x2x| = |++ 1y2y| = |+ +
Furthermore, 1z2z |12 = 12|12, whence the action of ~1 ~2 on the basis vectors
~1 ~2|++ = |++~1 ~2|+ = 2| + |+~1 ~2| + = 2|+ | +~1 ~2| = |
Then one obtains immediately1
2(I + ~A ~B)|iAjB = |jAiB
4.6.5 The Schmidt purification theorem
Let us choose as a basis of HA a set {|mA} which diagonalizes the reduced state operator A:
A = TrB|ABAB | =NSm=1
pm|mAmA|
If the number NS of nonzero coefficients pm is smaller than the dimension NA of HA, we complete theset {|mA} by a set of (NA NS) orthonormal vectors, chosen to be orthogonal to the space spanned bythe vectors |mA. We use (4.12) to compute A from |AB
A =m,n
nB |mB|mAnA|
On comparing the two expressions of A we see that
nB|mB = pmmn,
and with our choice of basis {|mA} it turns out that the vectors {|mB} are, after all, orthogonal. Toobtain an orthonormal basis, we only need to rescale the vectors |nB
|nB = p1/2n |nB,
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12 CHAPTER 4. EXERCISES FROM CHAPTER 4
where we may assume that pn > 0 because, as explained above, it is always possible to complete thebasis of HB by a set of (NB NS) orthonormal vectors. We finally obtain Schmidts decomposition of|AB on an orthonormal basis of HA HB:
|AB =n
p1/2n |nA nB.
Any pure state |AB may be written in the preceding form, but the bases {|nA} and {|nB} will ofcourse depend on the state under consideration. If some of the pn are equal, then the decompositionis not unique, as is the case for the spectral decomposition of a Hermitian operator with degenerateeigenvalues. The reduced state operator B is readily computed from (4.12) using the orthogonalitycondition mA|nA = mn:
B = TrA|ABAB | =n
pn|nBnB|
4.6.6 A model for phase damping
The state matrix at time t is
(t) =
( |(t)|2 (t)(t)(t)(t) |(t)|2
)
where stands for an average over all the realizations of the random function. Clearly |(t)|2 and|(t)|2 are time-independent, so that the populations are time-independent. However, the coherencesdepend on time. Let us compute the average of (t)(t)
(t)(t) = 00exp
(i
t0
(t)dt)
= 00 exp (i 0t) exp
(12
t0
C(t t)dtdt)
where we have used a standard property of Gaussian random functions. We thus obtain
01(t) = 01(t = 0) exp (i 0t) exp(12
t0
C(t t)dtdt)
If we assume that t , then t0
dtdte|tt|/ 2t
0
dtet/ = 2t
and01(t) = 01(t = 0)e
i0t eCt
4.6.7 Amplitude damping channel
1 The evolution of | during t is
| U | = |0A 0E+ 1 p |1A 0E+ p |1A 1E
In order to obtain the state matrix of system A, we take the trace over the environment
TrE(U ||U ) = (||2 + p||2)|0A0A|+ 1 p |0A1A|
+ 1 p |1A0A|+ (1 p)||2|1A1A|
or, in matrix from
(1) = (t) =
(1 (1 p)||2 1 p
1 p (1 p)||2)
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13
After n iterations we get
(n) = (nt) =
(1 (1 p)n||2 (1 p)n/2 (1 p)n/2 (1 p)n||2
)
Using in the limit t 0 the relation
limt0
(1 t)t/t = et
we get the expression given in the statement of the problem. We clearly have T1 = 1/ and T2 = 2/, sothat T2 = 2T1.
2. If we detect no photons, we know that we have prepared the atom in the (unnormalized) state
|0A+ 1 p |1A
The failure to detect a photon has changed the state of the atom!
4.6.8 Invariance of the Bell states under rotation
We have
|xAxB = (cos |A sin |A)(cos |B sin |B)|yAxB = (sin |B+ cos |B)(sin |B+ cos |B)
and an explicit calculation immediately gives
| = 12(|xAxB+ |yAyB) = 1
2(|AB+ |AB)
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14 CHAPTER 4. EXERCISES FROM CHAPTER 4
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Chapter 5
Exercises from chapter 5
5.10.1 Justification of the figures of Fig. 5.4
1. The upper circuit of Fig. 5.4 reads in matrix form
M =
(C 00 C
)(I 00 x
)(B 00 B
)(I 00 x
)(A 00 A
)
=
(CBA 00 CxBxA
)
where the matrices have been written in block diagonal form with 2 2 matrices. Then we must findthree matrices, A, B and C such that
CBA = I CxBxA = U
Action of the cNOT gate
cNOT
[12(|00+ |10)
]=
12(|00+ |11)
This is an entangled state (in fact it is one of the four Bell states).
2. Let us define the unitary matrix U by
U =
(
)
with||2 + ||2 = ||2 + ||2 = 1 + = + = 0
and start from the most general two-qubit state
| = a|00+ b|01+ c|10+ d|11Assume we measure the control bit and find it in the |0 state. Then the state vector of the target bit is
|0 = a|0+ b|1If we find the control bit in state |1, then we apply U to the state | = c|0+ d|1
U | = |1 = (c+ d)|0+ (c+ d)|1On the other hand, if we apply the cU gate to |, the result is
cU| = a|00+ b|01+ (c+ d)|10+ (c+ d)|11= |0 0+ |1 1
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16 CHAPTER 5. EXERCISES FROM CHAPTER 5
3. It is clear that the vectors |000 and |001 are not modified by the lower circuit on the left of Fig. 5.4,so that we may start from the vector
| = a|010+ b|011+ c|100+ d|101+ e|110f |111We apply on | the first gate on the left, the c2U3 gate (with obvious notations)
c2U3| = |1 = (a+ b)|010+ (a+ b)|011+ c|100+ d|101+ (e+ f)|110+ (e+ f)|111
where the matrix U is defined in the preceding question. The transformation law is thus
a a+ b b a+ b c cd d e e+ f f e+ f
Let us give as an intermediate result of the calculation
|4 = (c1NOT2)(c2U3)(c1NOT2)(c2U3)|We find
|4 = a|010+ b|011+ (c+ d)|100+ (c+ d)|101+ (e+ f)|110+ (e+ f)|111
Finally
|5 = c1U3|4 = a|010+ b|011+ c|100+ d|101+ (U211e+ U
212f)|110+ (U221e+ U222f)|111
where U2ij is a matrix element of of the matrix U2, for example
U211 = 2 +
This gives preciseley the action of the Toffoli gate TU2 . A non trivial action is obtained only if bothcontrol bits 1 and 2 are in the |1 state
TU2(|110+ |111) = (U211e+ U212f)|110+ (U221e+ U222f)|111
5.10.2 The Deutsch-Jozsa algorithm
1. Before entering the box Uf , the two upper qubits are in the state
H2|00 = 12(|0+ |1) 1
2(|0+ |1)
=1
2(|00+ |01+ |10+ |11) = 1
2
3x=0
|x
2. From the results of Sec. 5.5
Uf | =(1
2
3x=0
(1)f(x)|x) 1
2(|0 |1)
so that, calling | the state of the two upper qubits(i) f(x) = cst
| = 12(|00+ |01+ |10+ |11)
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17
(i) f(x) = xmod 2
| = 12(|00 |01+ |10 |11)
3. Since H2 = I, in case (i) we getH2| = |00
while in case (ii) we may write
| = 12(|0 (|0 |1) + (|0 |1) |1)
and
H2| = H2 12[(|0+ |1) (|0 |1)] = |01
The first qubit is in the state |0 and the second in the state |1. Note that the result of the finalmeasurement of the upper qubits is unambiguous only if | is a non entangled state, so that we musthave
(1)f(0)+f(3) = (1)f(1)+f(2)
5.10.3 Grover algorithm and constructive interference
Let us first apply the oracle O on |
O| = 1N
N1x=0
(1)f(x) |x = 1N
x
ax|x
Then we apply G = 2|| I using |x = 1/N
GO| =(
2
N
y
ay
)| 1
N
x
ax|x
=1N
x
[2
N
y
ay ax]|x = 1
N
x
a(1)x |x
This gives us the relation, with a(0)x = 1
a(1)x =2
N
(y
(1)f(y)a0y) (1)f(x)a(0)x
which leads to the recursion relation
a(n+1)x =2
N
(y
(1)f(y)any) (1)f(x)a(n)x
If, for example, N = 16, then
(i) For xi 6= x0, a(1)i = 34(ii) For xi = x0, a
(1)i =
114
The probability (ii) for finding x0 is greater than the probability (i) for finding xi by a factor 121/9 13.4.A good check of the calculation is that the final sate vector is normalized to one: 15(3/4)2+(11/4)2 = 1!
5.10.4 Example of finding yj
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18 CHAPTER 5. EXERCISES FROM CHAPTER 5
The probability for finding yj is given by (5.45) with, in our specific case, K = 5, n = 4 and r = 3
p(yj) =1
2nK
sin2(jKr/2n)
sin2(jr/2n)=
1
80
sin2(15j/16)
sin2(3j/16)
The possible values of j are j = 0, j = 1, j = 2 and j = 3. To the first value corresponds yj = 0 andj = 0. To the second one corresponds yj = 5 with |j | = .33 and to the third one yj = 11 with |j | = .33,while there is no yj with |j | < 1/2 for the last one. We obtain for the probabilities
p(0) =5
16p(1) = p(2) = .225
so thatp(0) + p(1) + p(2) = 0.76 > 0.4
Assume, for example, that the measurement of the final qubits gives yj = 11. Then we deduce that j = 2and r = 3.
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Chapter 6
Exercises from chapter 6
6.5.1 Off-resonance Rabi oscillations
From exercise 3.5.1, we kow that exp(i(~ p)/2) is the rotation operator by of a spin 1/2 about anaxis p. The vector n being normalized (n2 = 1), we have
exp(iHt/~) = I cos t2 i(~ n) sin t
2
with
~ n = 1
x +
z
so that the matrix form of exp(iHt/~) is
eiHt/~ =
cos
t
2+
sin
t
2i1
sint
2
i1
sint
2cos
t
2
sin
t
2
6.5.2 Commutation relations between the a and a
1. The commutator of a and a is, from the definition (6.26)
[a, a] =Mz2~
[z +
ipzMz
, z ipzMz
]
=Mz2~
2iMz
[z, pz] = I
2. To compute the commutator [a, a], we use the identity
[AB,C] = A[B,C] + [A,C]B
and we find[aa, a] = a[a, a] + [a, a]a = a
6.5.2 Quantum computing with trapped ions
1. We write the interaction Hamiltonian in terms of + and
Hint = 12~1[+ + ]
[ei(tkz) + ei(tkz)
]
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20 CHAPTER 6. EXERCISES FROM CHAPTER 6
and go to the interaction picture using (6.5)
eiH0t/~ eiH0t/~ = ei0t
In the rotating wave approximation, we can neglect terms which behave as exp[i(0 + )t] and we areleft with
Hint ~21
[+ e
i(t) eikz + ei(t) e ikz
]
2. z =~/(2Mz) is the spread of the wave function in the harmonic well. Thus, = kz is the
ratio of this spread to the wavelength of the laser light. We may write
kz = k
~
2Mz(a+ a) = (a+ a)
The matrix element of Hint between the states |1,m+m and |0,m is
1,m+m|Hint|m = 12~1e
i(t) m+m|ei(a+a)|m
The Rabi frequency for oscillations between the two levels is
mm+m
1 = 1|m+m|ei(a+a)|m|
|1, 1
+
+
0
|1, 2
|1, 0
|0, 2
|0, 1|0, 0
Figure 6.1: The level scheme. The transitions which are used are (0, 0) (0, 1) and (0, 1) (1.2):bluesideband, + = 0 + z and (0, 1) (1, 1): red sideband, = 0 z.
3. Writing
ei(a+a) I i(a+ a)
and keeping terms to first order in we get
Hint =i
2~1
[+ a e
i(z)t ei aei(z)t ei
+ + aei(+z)t ei a ei(+z)t ei
]
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21
The first line of Hint corresponds to a resonance at = 0 = z, that is, = 0 + z, a blue sideband, and the second line to a resonance at = 0 z, that is, a red sideband. The +a term of theblue sideband induces transitions from |0,m+ 1 to |1,m, and the a term from |1,m to |0,m+ 1.Now
m|a|m+ 1 = m+ 1|a|m = m+ 1so that we get H+int as written in the statement of the problem with
ab =a
m+ 1ab =
am+ 1
The Rabi frequency is then 1m+ 1. The same reasoning may be applied to the red sideband.
4. The rotation operators R(, ) are given by
R(, = 0) = I cos
2 ix sin
2
R(, =
2
)= I cos
2 iy sin
2
so thatR(, 0) = ix R
(,
2
)= iy
We have, for example,
R(,
2
)R(, 0)R
(,
2
)= (iy)
(I cos
2 ix sin
2
)(iy)
= (I cos
2 ix sin
2
)= R(, 0)
Let us call A the transition |0, 0 |1, 1 and B the transition |0, 1 |1, 2. The Rabi frequenciesare linked by B =
2A. Thus, if the rotation angle is A for transition A, it will be B =
2 A for
transition B. For transition A, we choose = /2 and =
R(
2,
2
)R(, 0)R
( 2,
2
)R(, 0) = I
For transition B we shall have = and = 2
R(,
2
)R(
2, 0)R
(,
2
)R(
2, 0) = I
The state |1, 0 is not affected because the transition |0, 0 |1, 0 does not resonate on the blue sidebandfrequency. Thus we have
|00 |0, 0 |0, 1 |0, 1 |1, 0 +|1, 0 |1, 1 |1, 1
5. R( , /2
)= iy so that
R( ,
2
)|0, 1 = |1, 0 R
( ,
2
)|1, 0 = |0, 1
Let us start from the general two ion state, where both ions are in the vibrational ground state
| = (a|00+ b|01+ c|10+ d|11) |0= a|00, 0+ b|01, 0+ c|10, 0+ d|11, 0
The action of R(2)(, /2) on ion 2 gives| = R(2)(, /2)| = a|00, 0+ b|00, 1+ c|10, 0+ d|10, 1
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22 CHAPTER 6. EXERCISES FROM CHAPTER 6
Then we apply R+(1) on ion 1
| = R+(1) | = a|00, 0 b|00, 1+ c|10, 0 d|10, 1
and finally R(2)(, /2) on ion 2
| = R(2)(, /2)| = a|00, 0 b|01, 0+ c|10, 0 d|11, 0= (a|00 b|01+ c|10 d|11) |0
This is the result of applying a cZ gate, within trivial phase factors.
6.5.4 Vibrational modes of two ions in a trap
Setting z1 = z0 + u, z2 = z0 + v and expanding to second order in powers of u and v we get
V 12M2z
(2z20 + 2z0(u v) + u2 + v2
)+e2
z0
(1 u v
2z0+
(u v)24z20
)
with e2 = q2/(40). The equilibrium condition is given by the condition that the terms linear in u andv vanish
M2zz0 e2
2z20= 0
so that
z0 =
(1
2
)1/3l l =
(e2
M2z
)1/3The normal modes are obtained by examining the terms quadratic in u and v, which lead to a potentialenergy
U(u, v) =1
2M2z
(u2 + v2
)+
e2
4z30(u v)2
The equations of motion are
Mu = M2zue2
2z30(u v) = M2z(2u v)
Mv = M2zv e2
2z30(v u) = M2z(2v u)
The center of mass mode (u+ v)/2 oscillates at frequency z
(u + v) = 2z(u + v)
while the breathing mode (u v) oscillates with frequency 3z(u v) = 32z(u v)
6.5.5 Meissner effect and flux quantization
1. We start from the expression (6.43) of the electromagnetic current
~em =~q
m
(~(~r) q
~
~A(~r))(~r)
Let us take the curl of the preceding equation, assuming (~r) to be constant
~ ~em = q2
m~B
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23
From the Maxwell equation ~ ~B = 0~em we also have
2 ~B = 0~ ~emand comparing the two equations we obtain
2 ~B == q2
m~B =
1
2L~B 2L =
m
0q2=
me20q2e
Taking a one-dimensional geometry, where the region z > 0 is superconducting, we see that the magneticfield must decrease as
B(z) = B(z = 0)ez/
From ~ ~B = 0~em, we see that the electromagnetic current must also vanish in the bulk of a super-conductor.
2. Let us take a ring geometry and draw a contour C well inside the ring. Then we have
0 =
C
~em d~l = ~qm
C
~ d~l q2
m
C
~A d~l
Since exp(i) is single valued, we must have + 2n after a full turn, and~q
m(2n) =
q2
m
~B d~S n = ,1, 0, 1, 2,
6.5.6 Josephson current
Let us start from (6.45) and writei = ie
ii i = 1, 2
The first of the equations (6.45) becomes
i~
2
d1dt
~1 d1dt
=1
2qV 1 +K
12 e
i
with = 2 1. Taking the real and imaginary parts of this equation and the corresponding equationfor i = 2, we obtain
d1dt
=2K
~(12)
1/2 sin ,
d2dt
= 2K~
(12)1/2 sin ,
d1dt
= K~
(21
)1/2cos qcV
2~,
d2dt
=K
~
(12
)1/2cos +
qcV
2~
and subtracting the last but one equation from the last one
d
dt=
qcV
~
6.5.7 Charge qubits
From the relation 2pi0
d
2n||m =
2pi0
d
2ei(nm) = nm
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24 CHAPTER 6. EXERCISES FROM CHAPTER 6
we derive 2pi0
d
2|| = I
Furthermore
N | =n
nein|n = i
(n
ein|n)
so that
N = i
We can also use the commutation relation
[N,] = iI
to obtaineiN ei = N i[, N ] = N I
and to deriveN(ei|n) = ei(N I)|n = (n 1) (ei|n)
We may then choose the phases of the states |n such that
ei|n = |n 1 ei|n = |n+ 1
and thus
cos|n = 12(|n 1+ |n+ 1)
2. In the vicinity of ng = 1/2, the Hamiltonian becomes
H = 14EcI + Ec
(ng 1
2
)|00| Ec
(ng 1
2
)|11| 1
2EJ(|01|+ |10|)
In the {|0, |1} basis, this can be written, omitting the (irrelevant) constant term
H Ec(ng 1
2
)z 1
2EJx
If ng is far enough from 1/2, the eigenvectors of H are approximately the vectors |0 and 1, due to thecondition Ec EJ . When ng comes close to 1/2, tunneling becomes important, and at ng = 1/2, theeigenvectors are those | of x with eigenvalues EJ
| = 12(|0 |1) x| = |
One observes the standard phenomenon of level repulsion around ng = 1/2. It is usual to exchange thex and z bases, so that the control parameter appears as the coefficient of x
H 12EJz + Ec
(ng 1
2
)x
-
Chapter 7
Exercises from Chapter 7
7.5.1 Superdense coding
1. From the identities
x|0 = |1 x|1 = |0 z |0 = |0 z |1 = |1
we immediately get
A00| = |A01| = 1
2
(|0A 0B |1A 1B)A10| = 1
2
(|1A 0B+ |0A 1B)A11| = 1
2
(|1A 0B |0A 1B)
Let us first examine the action of the cNOT-gate on the four states Aij |
cNOT[A00|
]=
12
(|0A+ |1A) |0BcNOT
[A01|
]=
12
(|0A |1A) |0BcNOT
[A10|
]=
12
(|0A+ |1A) |1BcNOT
[A11|
]= 1
2
(|0A |1A) |1BThe measurement of qubit B has the result |0B for i = 0 (A00 and A01) and |1B for i = 1 (A10 andA11), so that this measurement gives the value of i. Furthermore
H12
(|0A+ |1A) = |0AH
12
(|0A |1A) = |1Aand the measurement of qubit A gives the value of j.
7.5.2 Shannnon entropy versus von Neumann entropy
25
-
26 CHAPTER 7. EXERCISES FROM CHAPTER 7
The state matrix is given by
=
(p+ (1 p) cos2 /2 (1 p) sin /2 cos /2(1 p) sin /2 cos /2 (1 p) sin2 /2
)
and its eigenvalues are
=1
2
(1
1 4p(1 p) sin2 /2
)=
1
2(1 x)
This allows us to write the von Neumann entropy as (it is convenient to use ln rather than log)
HvN = 1 + x2
ln1 + x
2+
1 x2
ln1 x2
Let us compute the x-derivative of HvN
ddx
HvN(x) =1
2ln
1 + x
1 x = tanh1(x)
Thus HvN(x) is a concave function of x which has a maximum at x = 0: HvN(x = 0) = ln 2. For thisvalue of x, we have HvN = HSh. Let us write p = (1 + p)/2, so that
HSh =1 + p
2ln
1 + p
2+
1 p2
ln1 p2
and
x =
1 (1 p)2 sin2 /2
Now we have the inequalityp2 x2 = (1 p2) cos2 /2 0
Thus |x| p and HSh HvN.
7.5.3 Information gain of Eve
1. Alice uses the bases {|0, |1} and {|+, |}, while Eve uses the basis {|0, |1}. The conditionalprobabilities p(r|i) are
p(0|0) = 1 p(1|0) = 0 p(0|1) = 0 p(1|1) = 1and
p(0|+) = 1/2 p(1|+) = 1/2 p(0|) = 1/2 p(1|) = 1/2We obtain p(r) from
p(r) =i
p(r|i)p(i) = 12r
Let us now turn to the conditional probabilities p(i||r) (we use a double vertical bar to underline thedifference with p(r|i))
p(i||r) = p(r|i)p(i)p(r)
=1
2p(r|i)
We find
p(0||0) = 12
p(1||0) = 0 p(+||0) = 14
p(||0) = 14
p(1||0) = 0 p(1||1) = 12
p(+||1) = 14
p(||1) = 14
Before Eves measurement, the (Shannon) entropy is H() = 2, after Eves measurement, the entropyH(|) is, from (7.10)
H(|) = r
p(r)i
p(i||r) log p(i||r)
= 12log
1
2 2
(1
4log
1
4
)=
3
2
-
27
The information gain of Eve is
I( : ) = H()H(|) = 12
2. Eve uses a {|+ /8, | /8} basis, so that p(r|i) is given byp(0|1) = p(0|+) = .854 p(1|0) = p(1|) = 0.146p(1|1) = p(1|+) = .146 p(1|1) = p(1|) = 0.854
We then compute p(i||r)p(0||0) = p(+||0) = 0.427 p(1||0) = p(||0) = .073
andH(|) =
r
p(r)i
p(i||r) log p(i||r) = 1.600
so that the information gain is nowI( : ) = 0.400
7.5.4 Symmetry of the fidelity
Let us take = ||, a pure state, and write as
=
p||
Observing that 1/2 = ||, we obtain1/21/2 = ||||
from which it follows thatF(, ) = ||
Let us now take = || and use the diagonal form of
=i
pi|ii|
Since = || is a rank one operator, the same is true for (1/21/2) whose eigenvalue equation is,in a space of dimension N
N Tr(1/21/2)N1 = 0whence
= Tr (1/21/2)
is the only non zero eigenvalue. Let us decompose | on the {|i} basis
| =i
ci|i
and write the matrix 1/21/2 in this basis
(1/21/2)ij =pipj cic
j
From this we deduceTr (1/21/2) =
i
pi|ci|2 = || =
and (Tr1/21/2
)2= = || = F(, )
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28 CHAPTER 7. EXERCISES FROM CHAPTER 7
7.5.5 Quantum error correcting code
From X | = |, we obtain, for example
XAXB|A = XAXB(| +++ |+) = |AXAXC |A = XAXC(| +++ |+) = |A
Let us also check
cNOTBcNOTC(HA HB HC)|C = cNOTBcNOTC(|001+ |110)= |001+ |101) = (|0+ |1) |01