A short introduction to quantum information and quantum computation: solutions of exercises

28 Pages·2006·0.186 MB·English

by  Le Bellac M.

1
Solutions of Exercises
from ‘A Short Introduction to Quantum
Information and Quantum Computation’
Michel Le Bellac
2
Chapter 2
Exercises from Chapter 2
2.6.1 Determination of the polarization of a light wave
1. We can choose δ =0, δ =δ. The equation of the ellipse
x y
x=cosθ cosωt y =sinθ cos(ωt δ)
−
reads in Cartesian coordinates
x2 cosδ y2
2xy + =sin2δ
cos2θ − sinθ cosθ sin2θ
The direction of the axes is obtained by looking for the eigenvectors of the matrix
1 cosδ
A= cos2θ −sinθ cosθ
 cosδ 1 
−sinθ cosθ sin2θ
 
 
which make angles α and α+π/2 with the x-axis, where α is given by
tanα=cosδtan2θ
The vector product of the position~r with the velocity~v,~r ~v, is easily seen to be
×
1
~r ~v = ωzˆsin2θsinδ
× 2
so that the sense of rotation is given by the sign of the product sin2θsinδ.
2. The intensity at the entrance of the polarizer is
I =k E2cos2θ+E2sin2θ =kE2
0 0 0 0
where k is a proportionality factor. At th(cid:0)e exit of the polarize(cid:1)r it is
I =kE2cos2θ =I cos2θ
0 0
The reduction of the intensity allows us to determine cosθ .
| |
3. The projection of the electric field on the polarizer axis is
E
0
cosθ cosωt+sinθ cos(ωt δ)
√2 −
(cid:2) (cid:3)
and the intensity is given by the time average
I′ = kE2 cos2θ cos2ωt+sin2θ cos2(ωt δ)+2sinθcosθ cosωtcos(ωt δ)
0 − −
1 D 1 E
= kE2(1+sin2θcosδ)= I (1+sin2θcosδ)
2 0 2 0
3
4 CHAPTER 2. EXERCISES FROM CHAPTER 2
From the measurement of I′ we deduce cosδ, which allows us to deduce δ up to a sign. The remaining
ambiguities are lifted if one remarks that the ellipse is invariant under the transformations
θ θ+π δ δ
→ →
and
θ θ δ δ+π
→− →
.
2.6.2 The (λ,µ) polarizer
1. The components ′ and ′ are given by
Ex Ey
′ = cos2θ+ sinθcosθe−iη = λ2 +λµ∗ ,
Ex Ex Ey | | Ex Ey
′ = sinθcosθeiη+ sin2θ =λ∗µ + µ2 .
Ey Ex Ey Ex | | Ey
2. This operation amounts to projection on Φ . In fact, if we choose to write the vectors x and y as
| i | i | i
column vectors
1 0
x = , y =
| i 0 | i 1
(cid:18) (cid:19) (cid:18) (cid:19)
then the projector
Φ
P = Φ Φ = λx +µy λ∗ x +µ∗ y
Φ
P | ih | | i | i h | h |
is represented by the matrix (cid:0) (cid:1)(cid:0) (cid:1)
λ2 λµ∗
PΦ = λ|∗|µ µ2
(cid:18) | | (cid:19)
3. Since = Φ Φ, we clearly have Φ = Φ and Φ =0, because
Φ Φ Φ ⊥
P | ih | P | i | i P | i
ΦΦ = λ∗µ∗+µ∗λ∗ =0
⊥
h | i −
2.6.3 Circular polarization and the rotation operator
1. In complex notation the fields and are written as
x y
E E
1 1 i
= E , = E e±iπ/2 = ± E ,
x 0 y 0 0
E √2 E √2 √2
where the (+) sign corresponds to right-handed circular polarization and the ( ) to left-handed. The
−
proportionality factor E common to and defines the intensity of the light wave and plays no role
0 x y
E E
in describing the polarization, which is characterizedby the normalized vectors
1 1
R = (x +iy ), L = (x iy )
| i √2 | i | i | i √2 | i− | i
2. Let us compute R′ . We have
| i
1
R′ = cosθ x +sinθ y isinθ x +icosθ y
| i √2 | i | i− | i | i
1 (cid:0) (cid:1)
= e−iθ x +ie−iθ y =e−iθ R
√2 | i | i | i
(cid:0) (cid:1)
and similarly L′ = exp(iθ)L . The vectors R′ and L′ differ from R and L by a phase factor only,
| i | i | i i | i i
and they do not represent different physical states.
3. The projectors on the vectors R and L are given by
| i | i
1 1 i 1 1 i
PD = 2 i −1 PG = 2 i 1
(cid:18) (cid:19) (cid:18) − (cid:19)
5
and Σ is
0 i
Σ=PD−PG = i −0
(cid:18) (cid:19)
This operator has the states R and L as its eigenvectors, and their respective eigenvalues are +1 and
| i | i
1:
−
ΣR = R , ΣL = L
| i | i | i −| i
Thus exp( iθΣ)R =exp( iθ)R and exp( iθΣ)L =exp(iθ)L
− | i − | i − | i | i
4. From the form of Σ in the x , y basis we get at once Σ2 =I, and thus
{| i | i}
( iθ)2 ( iθ)3
e−iθΣ =I iθΣ+ − I + − Σ+
− 2! 3! ···
The series is easily summed with the result
cosθ sinθ
exp(−iθΣz)= sinθ −cosθ
(cid:18) (cid:19)
If we apply the operator exp( iθΣ) to the vectors of the x , y basis, we get the rotated vectors θ
− {| i | i} | i
and θ , so that this operator represents a rotation by an angle θ about the z axis.
⊥
| i
2.6.4 An optimal strategy for Eve?
1. If Alice uses the x basis, the probability that Eve guesses correctly is p = cos2φ. If she uses the
| i x
the π/4 basis, this probability is
|± i
1
p = φ π/4 2 = (cosφ+sinφ)2
π/4 |h |± i| 2
The probability that Eve guesses correctly is
1 1
p(φ) = (p +p )= 2cos2φ+(cosφ+sinφ)2
2 x π/4 4
1 (cid:2) (cid:3)
= [2+cos2φ+sin2φ]
4
The maximum of p(φ) is given by φ = φ = π/8, which is evident from symmetry considerations: the
0
maximum must be given by the bisector of the Ox and π/4 axes. The maximum value is
1 1
p = 1+ 0.854
max 2 √2 ≃
(cid:18) (cid:19)
2. If Alice sends a θ (θ ) photon, Eve obtains the correct result with probability cos2θ (sin2θ), and
⊥
the probabilitythat|Biob| reiceivesthe correctpolarizationis cos4θ (sin4θ). The probabilityof successfor
Eve is
1
p = 1+cos4θ+sin4θ
s 2
and the probability of error (cid:0) (cid:1)
1
p =1 p =sin2θcos2θ = sin22θ
e − s 4
Eves’s error are maximal for θ =π/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].
−
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 ϕ ):
| i
A =A A I, B =B B I.
0 ϕ 0 ϕ
−h i −h i
Their commutator is also iC, [A ,B ]=iC, because A and B are numbers. The squared norm of
0 0 ϕ ϕ
h i h i
the vector
(A +iλB )ϕ ,
0 0
| i
where λ is chosen to be real, must be positive:
(A +iλB )ϕ 2 = A ϕ 2+iλ ϕA B ϕ iλ ϕB A ϕ +λ2 B ϕ 2
0 0 0 0 0 0 0 0
|| | i|| || | i|| h | | i− h | | i || | i||
= A2 λ C +λ2 B2 0.
h 0iϕ− h iϕ h 0iϕ ≥
The second-degree polynomial in λ must be positive for any λ, which implies
C 2 4 A2 B2 0.
h iϕ− h 0iϕh 0iϕ ≤
This demonstrates the Heisenberg inequality
1
(∆ A)(∆ B) C . (2.2)
ϕ ϕ ϕ
≥ 2 h i
(cid:12) (cid:12)
(cid:12) (cid:12)
3. Ina finite dimensionalspace,the traceofacommutatorvanishes becauseTr(AB)=Tr(BA), sothat
the equality
[X,P]=i~I
cannot be realized in a finite dimensional space.
Chapter 3
Exercises from Chapter 3
3.5.1 Rotation operator for spin 1/2
1. We use σ 0 = 1 , σ 1 = 0 , σ 0 =i1 , σ 1 = i1 , σ 0 = 0 , σ 1 = 1 to obtain
x x y y z z
| i | i | i | i | i | i | i −| i | i | i | i −| i
θ θ
σ ϕ = eiφ/2 sin 0 +e−iφ/2 cos 1
x
| i 2| i 2| i
θ θ
σ ϕ = ieiφ/2 sin 0 +ie−iφ/2 cos 1
y
| i − 2| i 2| i
θ θ
σ ϕ = e−iφ/2 cos 0 eiφ/2 sin 1
z
| i 2| i− 2| i
so that
θ θ
ϕσ ϕ = sin cos eiφ+e−iφ =sinθcosφ
x
h | | i 2 2
θ θ (cid:0) (cid:1)
ϕσ ϕ = sin cos ieiφ+ie−iφ =sinθsinφ
y
h | | i 2 2 −
θ (cid:0) θ (cid:1)
ϕσ ϕ = cos2 sin2 =cosθ
z
h | | i 2 − 2
2. From (3.8) we derive the identity
(~σ ~a)(~σ ~b)=~a ~bI+i~σ (~a ~b)
· · · · ×
so that
(~σ pˆ)2 =I (~σ pˆ)3 =(~σ pˆ)
· · · ···
and the series expansion of the exponential reads
2 3
θ iθ 1 iθ 1 iθ
exp i ~σ pˆ = I + − (~σ pˆ)+ − I + − (~σ pˆ)+
− 2 · 2 · 2! 2 3! 2 · ···
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
θ θ
= I cos i(~σ pˆ) sin
2 − · 2
The action of the operator exp( iθ~σ pˆ/2) on the vector 0 is
− · | i
θ θ θ
exp i ~σ pˆ 0 =cos 0 +eiφ sin 1
− 2 · | i 2| i 2| i
(cid:18) (cid:19)
which is the same as (3.4) up to a physically irrelevant phase factor exp(iφ/2). Thus exp( iθ~σ pˆ/2) is
− ·
theoperatorwhichrotatesthevector 0 ,theeigenvectorofσ witheigenvalue1,on ϕ ,theeigenvectors
z
| i | i
7
8 CHAPTER 3. EXERCISES FROM CHAPTER 3
of~σ nˆ with the same eigenvalue. The same result holds for the eigenvalue 1, correspondingto 1 and
· − | i
the rotated vector U[ (θ)]1 .
pˆ
R | i
3. Let us specialize the above results to the case φ = π/2, which corresponds to rotations about the
−
x-axis
θ cos(θ/2) isin(θ/2)
U[Rx(θ)]=exp −i2σx = isin(θ/2) −cos(θ/2)
(cid:18) (cid:19) (cid:18) − (cid:19)
A rotation about the x-axis transforms 0 into the vector
| i
θ θ
ϕ =cos 0 isin 1
| i 2| i− 2| i
Taking θ = ω t, this corresponds exactly to (3.31) with the initial conditions a=1,b=0.
1
−
3.5.2 Rabi oscillations away from resonance
1. Substitutinginthedifferentialequationtheexponentialformofλˆ(t),wegetthesecondorderequation
for Ω
±
1
2Ω2 2δΩ ω2 =0
±− ±− 2 1
whose solutions are
1 1
Ω = δ δ2+ω2 = [δ Ω]
± 2 ± 1 2 ±
(cid:20) q (cid:21)
2. The solution of the differential equation for λˆ is a linear combination of exp(iΩ t) and exp(iΩ t)
+ −
λˆ(t)=a exp(iΩ t)+b exp(iΩ t).
+ −
Let us choose the initial conditions λˆ(0) = 1, µˆ(0) = 0. Since µˆ(0) dλˆ/dt(0), these initial conditions
∝
are equivalent to
a+b=1 and aΩ bΩ =0,
+ −
−
and so
Ω Ω
− +
a= , b=
− Ω Ω
The final result can be written as
eiδt/2 Ωt Ωt
λˆ(t) = Ωcos iδsin ,
Ω 2 − 2
(cid:20) (cid:21)
iω Ωt
µˆ(t) = 1e−iδt/2sin ,
Ω 2
which reduces to (3.31) when δ =0. The factor exp( iδt/2) arises because δ is the Larmor frequency in
±
the rotating reference frame. The second equation shows that if we start from the state 0 at t=0, the
| i
probability of finding the spin in the state 1 at time t is
| i
ω2 Ωt
p (t)= 1 sin2
0→1 Ω2 2
(cid:18) (cid:19)
We see that the maximum probability of making a transition from the state 0 to the state 1 for
| i | i
Ωt/2=π/2 is given by a resonance curve of width δ:
ω2 ω2 ω2
pmax = 1 = 1 = 1
− Ω2 ω2+δ2 ω2+(ω ω )2
1 1 − 0
Chapter 4
Exercises from Chapter 4
Basis independence of the tensor product
The tensor product i j is given by
A B
| ⊗ i
i j = R S m n
A B im jn A B
| ⊗ i | ⊗ i
m,n
X
Let us define ϕ χ ′ by using the i , j bases
A B A B
| ⊗ i {| i | i}
ϕ χ ′ = cˆdˆ i j
A B i j A B
| ⊗ i | ⊗ i
i,j
X
= cˆdˆR S m n
i j im in A B
| ⊗ i
i,j,m,n
X
We can now use the transformation law of the components in a change of basis
cˆ = R−1c dˆ = S−1d
i ki k j lj l
k l
X X
to show that
ϕ χ ′ = c d m n = ϕ χ
A B m n A B A B
| ⊗ i | ⊗ i | ⊗ i
m,n
X
Thus the tensor product is independent of the choice of basis.
4.6.2 Properties of the state operator
1. Since the p arereal,ρ is clearlyHermitian. FurthermoreTrρ= p =1,andfinallyρis positiveas
i i i
P
ϕρϕ = p ϕi 2 0
h | | i i|h | i| ≥
i
X
Let us first compute Tr(M i i)
| ih |
Tr(M i i)= j M i ij = iM i
| ih | h | | ih | i h | | i
j
X
whence
Tr p M i i = p iM i
i | ih |! ih | | i
i i
X X
The expectation value of M in the state i appears in the sum over i with the weight p , as expected.
| i i
9
10 CHAPTER 4. EXERCISES FROM CHAPTER 4
2. In the i basis, ρ has a diagonal form with matrix elements ρ = p , so that ρ2 = ρ can only hold
| i ii i
if one of the probabilities is one, as the equation p2 = p has solutions p = 1 and p = 0. Furthermore,
i i i i
Trρ2 = p2 and p2 p , where the equality holds if and only if one of the p is equal to one.
i i i i ≤ i i i
Let us assume, for example, that p = 1, p = 0,i = 1. Then ρ = 1 1, which corresponds to the pure
P P P 1 i 6 | ih |
state 1 . One can alsoremarkthat ρ2 =ρ implies thatρ is a projector , andthe rankof this projector
| i P
is one, because Tr is the dimension of the subspace on which projects.
P P
4.6.3 The state operator for a qubit and the Bloch vector
The condition for a Hermitian 2 2 matrix is ρ =ρ∗ , so that
× 01 10
a c
ρ=
c∗ 1 a
(cid:18) − (cid:19)
isindeedthemostgeneral2 2Hermitianmatrixwithtraceone. Theeigenvaluesλ andλ ofρsatisfy
+ −
×
λ +λ =1, λ λ =detρ=a(1 a) c2,
+ − + −
− −| |
and we must have λ 0 and λ 0. The condition detρ 0 implies that λ and λ have the same
+ − + −
≥ ≥ ≥
sign, and the condition λ +λ = 1 implies that λ λ reaches its maximum for λ λ = 1/4, so that
+ − + − + −
finally
1
0 a(1 a) c2
≤ − −| | ≤ 4
The necessary and sufficient condition for ρ to describe a pure state is
detρ=a(1 a) c2 =0.
− −| |
The coefficients a and c for the state matrix describing the normalized state vector ψ = λ+ +µ
| i | i |−i
with λ2+ µ2 =1 are
| | | |
a= λ2 c=λµ∗
| |
so that a(1 a)= c2 in this case.
− | |
2. Since any 2 2 Hermitianmatrix canbe written as a linear combinationofthe unit matrix I andthe
×
σ with reals coefficients, we can write the state matrix as
i
I 1 1 1+b b ib
ρ= 2 + biσi = 2 I+~b·~σ = 2 bx+ibzy 1x−bzy
Xi (cid:16) (cid:17) (cid:18) − (cid:19)
where we have used Trσ = 0. The vector~b, called the Bloch vector, must satisfy ~b2 1 owing to the
i
| | ≤
results of question1, and a pure state corrrespondsto ~b2 =1. Let us calculate the expectation value of
| |
~σ using Trσ σ =2δ . We find
i j ij
σ =Tr(ρσ )=b
i i i
h i
so that~b is the expectation value ~σ .
h i
3. With B~ parallel to Oz, the Hamiltonian reads
1
H = γσ
z
−2
The evolution equation
dϕ(t)
i~ | i =H ϕ
dt | i
translates into the following for the state matrix
dρ(t)
i~ =[H,ρ]
dt