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MASSACHUSETTS INSTITUTE OF TECHNOLOGY
MIT 2.111/8.411/6.898/18.435
Quantum Information Science I
September 16, 2010
Problem Set #2 Solution
1. Density matrices. A density matrix (also sometimes known as a density operator) is a representation
of statistical mixtures of quantum states. This exercise introduces some examples of density matrices,
and explores some of their properties.
(a) Let |ψi = a|0i + b|1i be a qubit state. Give the matrix ρ = |ψihψ|, which you may compute using
linear algebra using the vector representations of |ψi and hψ|. What are the eigenvectors and
eigenvalues of ρ?
Answer:
ρ = |ψihψ| =
2
a
|a| ab∗
a∗ b∗ =
b
ba∗ |b|2
The eigenvalues of ρ are 1 and 0 and the corresponding eigenvectors are
∗ a
b
,
b
−a∗
(b) Let ρ0 = |0ih0| and ρ1 = |1ih1|. Give the matrix σ =
eigenvalues of σ?
Answer:
1
1
1
σ = |0ih0| + |1ih1| =
2
2
2
The eigenvalues of σ are both
1
2
ρ0 +ρ1
2 .
What are the eigenvectors and
1 0
0 1
and the eigenvectors are
|0i, |1i
or any two orthogonal vectors in this Hilbert space.
(c) Compute tr(ρ2 ) and tr(σ 2 ). In general, tr(M 2 ) ≤ 1, with equality if and only if M is a pure state.
Answer:
trρ2 = tr(|ψihψ|)2 = tr(|ψihψ|) = 1,
1
1
1
tr(σ 2 ) = tr(|0ih0| + |1ih1|)2 = tr(|0ih0| + |1ih1|) =
4
4
2
2. Exponential of the Pauli matrices. Let ~v be any real, three-dimensional unit vector and θ a real
number. Prove that
exp(iθ~v · ~σ ) = cos(θ)I + i sin(θ)~v · ~σ ,
where ~v · ~σ ≡
P3
i=1
vi σi , and σi are the Pauli matrices, σ1 = X, σ2 = Y , σ3 = Z.
(1)
2
Id: hw2.tex,v 1.4 2009/02/09 04:31:40 ike Exp
Answer:
exp(iθ~v · ~σ ) =
0
0
∞
∞
∞
X
X
X
0
1
(−1)k
i(−1)k
2k0
(iθ~v · ~σ )k =
(θ~
v
·
~
σ
)
+
(θ~v · ~σ )2k +1
0
0
k!
(2k )!
(2k + 1)!
0
0
k=0
k =0
k =0
The powers of Pauli matrices have a simple form. Note that,
3
3
X
X
(~v · ~σ )2 = (
vi σi )2 =
vi vj σ i σ j
i=1
i,j=1
The commutation relation between Pauli matrices satisfies that
σi σj + σj σi = 0 and σi2 = I
Hence,
(~v · ~σ )2 =
3
X
vi2 I = I
i=1
Therefore,
0
0
0
0
(θ~v · ~σ )2k = θ2k I (θ~v · ~σ )2k +1 = θ2k +1~v · ~σ
and
0
0
∞
∞
X
X
0
(−1)k
(−1)k 2k0
θ I +i
θ2k +1~v · ~σ = cos θI + i sin θ~v · ~σ
exp(iθ~v · ~σ ) =
0 )!
0 + 1)!
(2k
(2k
0
0
k =0
k =0
3. Hadamard operator on n qubits. The Hadamard operator on one qubit may be written as
i
1 h
H = √ (|0i + |1i)h0| + (|0i − |1i)h1| .
2
(2)
Show explicitly that the Hadamard transform on n qubits, H ⊗n , may be written as
1 X
H ⊗n = √
(−1)x·y |xihy|.
2n x,y
Write out an explicit matrix representation for H ⊗2 .
Answer:
Hadamard transform on 1 qubit can be written as
i
X
1 h
H = √ (|0i + |1i)h0| + (|0i − |1i)h1| =
(−1)x1 ·y1 |x1 ihy1 |
2
x1 ,y1
(3)
3
Id: hw2.tex,v 1.4 2009/02/09 04:31:40 ike Exp
Here x1 · y1 is multiplication mod 2.
Hadamard transform on n qubits can be written as
X
H ⊗n = ⊗ni=1 (
(−1)xi ·yi |xi ihyi |)
xi ,yi
Define |xi = ⊗|xi i, |yi = ⊗|yi i, x · y =
Pn
i=1
xi · yi
Therefore,
H ⊗n = ⊗ni=1 (
X
(−1)xi ·yi |xi ihyi |) =
xi ,yi
X
X
(⊗ni=1 (−1)xi ·yi |xi ihyi |) =
(−1)x·y |xihy|
x,y
x,y
In particular,
1
H ⊗2 = √
2
1
1 1
⊗√
1 −1
2
1 1
1 −1

1
1 1
= 
2 1
1
1
−1
1
−1
1
1
−1
−1

1
−1
−1
1
4. Single qubit rotations. Define the rotation operator
θ
θ
Rn̂ (θ) ≡ exp(−iθ n̂ · ~σ /2) = cos
I − i sin
(nx X + ny Y + nz Z) ,
2
2
(4)
where n̂ is a real three-dimensional unit vector.
1. Prove that an arbitrary single qubit unitary operator can be written in the form U =
exp(iα)Rn̂ (θ), for some real numbers α and θ.
Answer:
a b
A single qubit operator U =
can be expanded in terms of {I, X, Y, Z} as U = a0 I + a1 X +
c d
a2 Y + a3 Z, where
a0 =
a+d
b+c
c−b
a−d
a1 =
a2 =
a3 =
2
2
2i
2
U is unitary if U † U = I.
U † U = (a∗0 I + a∗1 X + a∗2 Y + a∗3 Z)(a0 I + a1 X + a2 Y + a3 Z)
= (|a0 |2 + |a1 |2 + |a2 |2 + |a3 |2 )I + (a∗0 a1 + a∗1 a0 + ia∗2 a3 − ia∗3 a2 )X +
(a∗0 a2 − ia∗1 a3 + a∗2 a0 + ia∗3 a1 )Y + (a∗0 a3 + ia∗1 a2 − ia∗2 a1 + a∗3 a0 )Z
This requires that
|a0 |2 + |a1 |2 + |a2 |2 + |a3 |2
a∗0 a1 + a∗1 a0 + ia∗2 a3 − ia∗3 a2
a∗0 a2 − ia∗1 a3 + a∗2 a0 + ia∗3 a1
a∗0 a3 + ia∗1 a2 − ia∗2 a1 + a∗3 a0
=
=
=
=
1
0
0
0
4
Id: hw2.tex,v 1.4 2009/02/09 04:31:40 ike Exp
Define cos
Define
θ
2
= |a0 |, then |a1 |2 + |a2 |2 + |a3 |2 = | sin
θ
2
|.
θ
|
2
θ
ny = |a2 |/| sin
|
2
θ
nz = |a3 |/| sin
|
2
nx = |a1 |/| sin
Obviously, n2x + n2y + n2z = 1 and n̂ is a three dimensional unit vector.
Define exp(iα) = a0 / cos θ2 . Denote the phase of a1 , a2 , a3 as α1 , α2 , α3 respectively.
As
0 = a∗0 a1 + a∗1 a0 + ia∗2 a3 − ia∗3 a2
θ
θ
θ
θ
sin
nx + exp(i(α − α1 )) cos
sin
nx +
= exp(i(α1 − α)) cos
2
2
2
2
θ
θ
ny nz − i exp(i(α2 − α3 )) sin2
ny nz
i exp(i(α3 − α2 )) sin2
2
2
θ
θ
θ
= 2 cos(α − α1 ) cos
sin
nx + 2i sin(α3 − α2 ) sin2
ny nz
2
2
2
Therefore, cos(α − α1 ) = 0, sin(α2 − α3 ) = 0. Hence α1 = α − π/2, α2 = α3 .
Similarly, we find, α2 = α3 = α − π/2, α1 = α2 = α3 .
Therefore,
θ
a0 = exp(iα) cos
2
θ
a1 = −i exp(iα) sin
nx
2
θ
ny
a2 = −i exp(iα) sin
2
θ
a3 = −i exp(iα) sin
nz
2
That is,
θ
θ
U = exp(iα) cos
I − i sin
(nx X + ny Y + nz Z) = exp(iα)Rn̂ (θ)
2
2
2. Find values for α, θ, and n̂ giving the Hadamard gate H.
Answer:
1
H = √ (X + Z)
2
1
1
α = π/2, θ = π, nx = √ , ny = 0, nz = √
2
2
5
Id: hw2.tex,v 1.4 2009/02/09 04:31:40 ike Exp
3. Find values for α, θ, and n̂ giving the phase gate
1 0
S=
.
0 i
Answer:
S=
1+i
1−i
I+
Z
2
2
α = π/4, θ = π/2, nx = 0, ny = 0, nz = 1
(5)