Le Bellac M. a Short Introduction to Quantum Information and Quantum Computation.. Solutions of...

1Solutions of Exercises

from A Short Introduction to Quantum

Information and Quantum Computation

Michel Le Bellac

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 )

3

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

)

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].


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.

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

7


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

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.

9


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, ]

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,


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)

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)

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

15


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)

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


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.

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)

]

19


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

]

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


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

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


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


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


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(, )


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

Le Bellac M. a Short Introduction to Quantum Information and Quantum Computation.. Solutions of...

Documents

Transcript of Le Bellac M. a Short Introduction to Quantum Information and Quantum Computation.. Solutions of...