MASSACHUSETTS INSTITUTE OF TECHNOLOGY
MIT 2.111/8.411/6.898/18.435
Quantum Information Science I
October 28, 2010
Problem Set #7 Solution
(due in class, 04-Nov-10)
P1: (Continuous time Grover with noise) Consider a continuous time Grover algorithm on n qubits
employing two Hamiltonians, an oracle Hamiltonian, HO = EO |xihx|, and a driving Hamiltonian
HD = ED |ψi hψ|, where |xi is the target state and |ψi is state with all the qubits being |0i + |1i. Here,
we explore the impact of a simple noise model on this algorithm.
(a) Ideally, HD has the same strength as the oracle HO . However, we the strength of HD may
fluctuate.
√ Assuming ED = EO (1 + δ), calculate the probability of being in the target state at
time π 2n /(2EO ).
Answer:
H = HO + HD = EO |xihx| + ED |ψihψ|
|ψi =
⊗n
1 X
1
√ (|0i + |1i)
=√
|yi
2
2n y
where y are binary numbers from 0 to 2n − 1.
|ψi can be expanded as
1
|ψi = √ |xi +
2n
√
2n − 1 0
√
|x i
2n
|x0 i is orthogonal to |xi.
Therefore, in the basis |xi and |x0 i, H can be written as
#
"
#
"
√ n
√ n
2 −1
2 −1
1
1
1
+
(1
+
δ)
(1
+
δ)
1 0
n
n
n
n
2
√ n2
H = EO
+ ED √22n −1 2n2−1 = EO
n
0 0
(1 + δ) 22n−1 (1 + δ) 2 2−1
n
2n
2n
Writing in terms of Pauli operators,
√ n
2+δ
1+δ
δ
2 −1
H = EO (
I + ( n − )Z + (1 + δ)
X) = EO (a0 I + a1 Z + a2 X)
2
2
2
2n
√
After time t = π 2n /(2EO ), the evolution operator is
exp(−iHt) = exp(−ia0 EO tI −ia1 EO tZ −ia2 EO tX) = exp(−ia0 EO t)exp(−it0 (a01 EO Z +a02 EO X))
p
p
p
where a01 = a1 / a21 + a22 , a02 = a2 / a21 + a22 , t0 = a21 + a22 t.
Hence,
exp(−iHt) = exp(−ia0 EO t)(cos (EO t0 )I − i sin (EO t0 )(a01 Z + a02 X))
Therefore, the state after evolution is
|ψ(t)i = exp(−iHt)|ψi
√
√
a0 + a02 2n − 1
a0 − a01 2n − 1 0
√
√
= exp(−ia0 EO t) cos (EO t0 )|ψi − i sin (EO t0 )( 1
|xi + 2
|x i)
2n
2n
2
The probability of being in the target state is
p = |hx|ψ(t)i|2
√
a01 + a02 2n − 1 2
√
= | cos (EO t )hx|ψi − i sin (EO t )
|
2n
√
(a0 + a02 2n − 1)2
= cos2 (EO t0 )/2n + sin2 (EO t0 ) 1
2n
0
To first order in δ,
p
a21 + a22 =
0
1+δ/2
√
.
2n
0
Therefore EO t0 = (1 + δ/2)π/2.
0
2 2
cos (EO t ) = −πδ/4 , sin (EO t ) = 1 − π δ /32 ,
√
2
2n − 1 2
a01 + a02 2n − 1
√
=1−
δ
4
2n
1
π2
p(δ) = 1 − δ 2 (2n − 1)( + n ) = 1 − λδ 2
4 2 16
where λ = (2n − 1)( 14 +
π2
2n 16 ).
(b) Suppose we run the algorithm on an imperfect quantum computer in which each qubit experiences
1
a δ that has a random value given by the probability distribution √2πσ
exp(−δ 2 /(2σ 2 )).
Calculate how many times one needs to run the algorithm to know what the target state is with
probability 2/3, as a function of σ.
Answer:
The probability of getting the right answer each time is
Z∞
p(δ) √
−∞
1
exp(−δ 2 /(2σ 2 )) dx = 1 − λσ 2
2πσ
Therefore, after running the algorithm N times, the probability of NOT getting the target state
goes down exponentially as (λσ 2 )N . In order to succeed with probability 2/3, N ≥ log(λσ2 ) 31 .
P2: (Compositions of Hamiltonian operations) Consider a physical system with four energy levels
which are addressable, |0i, |1i, |2i, and |3i. You are provided with controls which turn one of two
Hamiltonians Hb , which couples {|0i, |2i}, and Ha , which couples {|2i, |3i} and {|0i, |1i}. Specifically,
0
α∗
Ha (α) = 
0
0

α
0
0
0
0
0
0
α∗

0
0
α
0
0
0
Hb (β) =  ∗
β
0

0
0
0
0
β
0
0
0

0
0
.
0
0
(1)
Note that α and β are both complex numbers. These may be visualized as transitions between some
subset of energy levels, eg in the hyperfine levels of an atomic system:
|0ñ
|3ñ
Ha
Hb
|1ñ
|2ñ
Your mission is to perform qubit rotations in the {|0i, |1i} qubit subspace, while leaving {|2i, |3i} alone.
3
(a) Suppose H1 and H2 are Hamiltonians such that tr|H1 | ≤ and tr|H2 | ≤ .
e−iH1 e−iH2 eiH1 eiH2 = e−iHc + O(3 ), where Hc = i[H1 , H2 ] = i(H1 H2 − H2 H1 ).
Answer:
Up to second order in Prove that
e−iH1 = I − iH1 − H12 /2 + O(3 ) , e−iH2 = I − iH2 − H22 /2 + O(3 )
eiH1 = I + iH1 − H12 /2 + O(3 ) , eiH2 = I + iH2 − H22 /2 + O(3 )
Retaining only up to second order in e−iH1 e−iH2 eiH1 eiH2 = (I − iH1 − H12 /2)(I − iH2 − H22 /2)(I + iH1 − H12 /2)(I + iH2 − H22 /2)
= I − H1 H2 + H2 H1 + O(3 )
= e−iHc + O(3 )
(b) Show our goal is possible in principle, by constructing a set of Hamiltonians from which H01 =
γ|0ih1| + γ ∗ |1ih0| can be generated. Specifically, compute Hc = i[Hb |β=1 , Hb |β=i ]/2 and Hd =
i[Ha |α=i , Hc ] and explain how to obtain H01 from this, by quantum simulation techniques.
Answer:

1
0
Hc = i[Hb |β=1 , Hb |β=i ]/2 = 
0
0

0
1
Hd = i[Ha |α=i , Hc ] = 
0
0
1
0
0
0
0
0
0
0
0
0
0
−1
0
0
−1
0

0
0
0
0

0
0
−1
0
Define He = i[Ha , Hc ]
0 −iα 0
0
iα∗ 0
He = i[Ha , Hc ] = 
0
0
0
0
0 −iα∗


0
0
iα
0
H01 can be obtained by
H01 = (Ha |α=γ + He |α=iγ )/2
Note that here we can take the sum of Ha |α=γ and He |α=iγ because they commute. eitH01 =
/2
eitHa |α=γ /2 eitHe |α=iγ
.
i
h
θ
(c) Let Rx (θ) = exp −i 2 (|0ih1| + |1ih0|) be a rotation about the x̂ axis of the {|0i, |1i} qubit subspace (it acts as identity on the {|2i, |3i} subspace). Give a sequence of individual Hamiltonian
evolutions, eg U = eit1 Ha |α=1 eit2 Hb |β=i · · · , turning on Ha and Hb sequentially (only one Hamiltonian on at a time), with specified values of α, β, and pulse durations, such that U = Rx (θ)
exactly.
Answer:
Hb |β=1

0
0
=
1
0
0
0
0
0
1
0
0
0


0
0 0
0
0 0
= X02 , Hb |β=−i = 
0
i 0
0 0
0
−i
0
0
0

0
0
= Y02
0
0
4
Composing X02 and Y02 we can get any operation on |0i and |2i. For example


eiθ/2 0
0
0
 0 1
0
0
iθ/4

Uz (−θ/2)01 ⊕ e−iθ/4 Uz (θ/2)23
eiY02 π/4 eiX02 θ/2 e−iY02 π/4 = eiZ02 θ/2 = 
 0 0 e−iθ/2 0 = e
0 0
0
1
Ha |α=−i = Y01 ⊕ Y23 , e
−iHa |α=−i π/4
1 1 −1
1 1 −1
=√
⊕√
2 1 1 01
2 1 1 23
eiHa |α=−i π/4 eiZ02 θ/2 e−iHa |α=−i π/4 = eiθ/4 Ux (θ/2)01 ⊕ e−iθ/4 Ux (−θ/2)23 = U1
But
e−iHa |α=1 θ/4 = Ux (θ/2)01 ⊕ Ux (θ/2)23 = U2
Combining U1 and U2 we get
U1 U2 = eiθ/4 Ux (θ)01 ⊕ e−iθ/4 I23
Therefore, the total pulse sequence for applying an X operation on |0i and |1i is
Rx (θ) = eiHa |α=−i π/4 eiHb |β=−i π/4 eiHb |β=1 θ/2 e−iHb |β=−i π/4 e−iHa |α=−i π/4 e−iHa |α=1 θ/4
i
h
(d) Do the same for Rz (θ) = exp −i θ2 (|0ih0| − |1ih1|) , such that you now have “pulse sequences”
for performing arbitrary operations on the {|0i, |1i} qubit.
Answer:
Rotation around Z axis can be obtained from rotaion around X axis as
Rz (θ) = eiHa |α=−i π/4 Rx (θ)eiHa |α=−i π/4
= eiθ/4 (eiY01 π/4 Ux (θ)01 e−iY01 π/4 ) ⊕ e−iθ/4 (eiY23 π/4 I23 e−iY23 π/4 )
= eiθ/4 Uz (θ)01 ⊕ e−iθ/4 I23