Quantum Computers Final Exam Solution
1. Define dµ = p(θ, φ)dθdφ = 4π1 sin θdθdφ.
(a) For |ψ(θ, φ)i and |ϕ(θ0, φ0)i, hF i =
ZZ
|hψ(θ, φ)|ϕ(θ0, φ0)i|2dµdµ0
=
ZZ
cosθ
2h0| + e−iφsinθ 2h1|
!
cosθ0
2|0i + e−iφ0sinθ0 2|1i
!
2
dµdµ0
= 1
16π2
Z 2π 0
Z π 0
Z 2π 0
Z π 0
cos2 θ
2cos2θ0
2 + 2 cos(φ0− φ) sinθ 2cosθ
2sinθ0 2 cosθ0
2 + sin2 θ
2sin2 θ0 2
!
sin θ sin θ0dθdφdθ0dφ0
= 1 4
Z π
0
Z π
0
cos2 θ
2cos2θ0
2 + sin2 θ
2sin2 θ0 2
!
sin θ sin θ0dθdθ0
= 1 4
Z π
0
cos2 θ
2sin θdθ
!2
+
Z π 0
sin2 θ
2sin θdθ
!2
= 1 2.
Or, using density operators ρ(θ, φ) and ρ(θ0, φ0), hF i =
ZZ
tr (ρ(θ, φ)ρ(θ0, φ0)) dµdµ0
=
ZZ
tr
1
2(I + sin θ cos φσ1+ sin θ sin φσ2 + cos θσ3) 1
2(I + sin θ0cos φ0σ1+ sin θ0sin φ0σ2+ cos θ0σ3)
dµdµ0
= 1
16
Z π 0
Z π 0
tr ((I + cos θσ3)(I + cos θ0σ3)) sin θ sin θ0dθdθ0
= 1 8
Z π 0
Z π 0
(1 + cos θ cos θ0) sin θ sin θ0dθdθ0
= 1 8
Z π
0
sin θdθ
2
= 1 2.
In the third equality terms involving cos φ, sin φ, cos φ0, or sin φ0 vanish after integration. Also note that tr (σ3) = 0 and σ3σ3 = I.
Notice that the chosen |ψi and |ϕi is independent, hence hρψρϕi = hρψi hρϕi, that is, we can consider the two states separately. Since the probability of ρ(θ, φ) is p(θ, φ), we have
hρψi = 1 4π
ZZ 1
2(I + sin θ cos φσ1+ sin θ sin φσ2+ cos θσ3) sin θdθdφ
= 1
4
Z π 0
(I + cos θσ3) sin θdθ
= 1
2I.
So
hF i = tr
1 2I1
2I
= 1 2. (b)
hF i =
ZZ
tr (ρψρ) 1
4πsin θdθdφ
= 1
4π
Z 2π 0
Z π 0
|hψ|0i|4+ |hψ|1i|4sin θdθdφ
= 1
4π
Z 2π 0
Z π 0
cos4 θ
2 + sin4 θ 2
!
sin θdθdφ
= 1 2
Z π 0
cos4 θ
2+ sin4 θ 2
!
sin θdθ
= 1 2
Z π 0
1 2 +1
2cos2θ
sin θdθ
= 2 3.
(c) The difference represents the information gained by making a mea- surement in terms of fidelity.
2. The state after rearrangement is
|ΦiAB = 1 2√
2|00i +
√3 2√
2|01i +
√3 2√
2|10i + 1 2√
2|11i, (a)
ρA= ρB = 1
2|0ih0|+
√3
4 |0ih1|+
√3
4 |1ih0|+1
2|1ih1| =
1 2
√3
√ 4 3 4
1 2
!
.
(b) The eigenvalues are λ+ = 12 +
√ 3
4 and λ− = 12 −
√ 3
4 , with corre- sponding eigenstates | ↑xi ≡ √1
2(|0i+|1i) and | ↓xi ≡ √1
2(|0i−|1i).
(c)
|ΦiAB =qλ+| ↑xiA| ↑xiB+qλ−| ↓xiA| ↓xiB.
3. (a) ρ0A= a (1 − p)b (1 − p)c d
!
.
(b) a 0 0 d
!
.
(c) ~p0 = ((1 − p)p1, (1 − p)p2, p3).
(d) (0, 0, p3).
4. (a)
U|ϕ1iA|βiB= |ϕ1iA|β1iB U|ϕ2iA|βiB= |ϕ2iA|β2iB
, since U is unitary, we have
(Ahϕ1|Bhβ|) (|ϕ2iA|βiB) = (Ahϕ1|Bhβ1|) (|ϕ2iA|β2iB)
⇒ Ahϕ1|ϕ2iA =Ahϕ1|ϕ2iABhβ1|β2iB
⇒ 1 =Bhβ1|β2iB
because Ahϕ1|ϕ2iA6= 0. That is, |β1iB = |β2iB.
(b) With Ahϕ1|ϕ2iA = 0, there is no restriction on Bhβ1|β2iB. This reflects the fact that orthogonal states can be distinguished with perfect accuracy.
5. (a)
perror = p1tr (ρ1E2) + p2tr (ρ2E1)
= tr (p1ρ1(I − E1) + p2ρ2E1)
= tr (p1ρ1+ (p2ρ2− p1ρ1)E1)
= tr (p1ρ1) + tr ((p2ρ2− p1ρ1)E1)
= p1+ tr X
i
λi|iihi|E1
!
= p1+X
i
λihi|E1|ii
(b)
E1 = X
i:λi<0
|iihi|.
(c) p2− p1 = 1 − 2p1 =Piλi, so p1 = 1
2 − 1 2
X
i
λi.
(perror)optimal = p1+ X
i:λi<0
λi
= 1 2 − 1
2
X
i
λi+ X
i:λi<0
λi
= 1 2 − 1
2
X
i:λi>0
λi+1 2
X
i:λi<0
λi
= 1 2 − 1
2kp2ρ2− p1ρ1ktr.
6. (a) Since ρ is a density operator, all its eigenvalues λ0i > 0. Then p2ρ2− p1ρ1 = (p2− p1)ρ = (p2− p1)X
i
λ0i|iihi| =X
i
λi|iihi|.
So λi = (p2− p1)λ0i for all i.
When p2 ≥ p1, the λi’s are all non-negative, so (perror)optimal = p1. When p2 < p1, the λi’s are all negative, so
(perror)optimal = p1+X
i
λi = p1+ (p2− p1) = p2.
(b)
kp2ρ2− p1ρ1ktr = tr
h
(p2ρ2− p1ρ1)†(p2ρ2− p1ρ1)i
1 2
= tr[(p2ρ2− p1ρ1)(p2ρ2 − p1ρ1)]12
= tr
h
(p22ρ22− p1ρ1p2ρ2− p2ρ2p1ρ1+ p21ρ21)i
1 2
= tr
h
(p22ρ22+ p21ρ21)i
1 2
= tr
h
(p22ρ22+ p1ρ1p2ρ2+ p2ρ2p1ρ1+ p21ρ21)i
1 2
= tr[(p2ρ2+ p1ρ1)(p2ρ2+ p1ρ1)]12
= tr (p2ρ2+ p1ρ1)
= p2+ p1
= 1 So we have
(perror)optimal = 1 2− 1
2kp2ρ2− p1ρ1ktr= 1 2− 1
2 = 0.
(c) When ρ1 = ρ2, there is no way to distinguish between the two, hence the best we can do is guess the one with the higher proba- bility, so
perror = min(p1, p2).
When ρ1 and ρ2 are orthogonal, they can be distinguished with no error, hence
perror = 0.
