MASSACHUSETTS INSTITUTE OF TECHNOLOGY
2.111J/18.435J/ESD.79
Quantum Computation
Problem 1. For any unit vector j = ( jx , jy , jz ) we can define the following operator
σ j = jx σX + jy σY + jz σZ
which corresponds to a π-radian rotation about j-axis.
(a) Show that σ 2j = I .
(b) σ j has two eigenvalues: +1 and –1.
(c) Find the eigenvectors +
+1 and –1.
j
and −
j
respectively corresponding to eigenvalues
Solution:
(a) Using the anti-commutator relation for Pauli matrices,
{σi , σ j } ≡ σiσ j + σ j σi = 0 for i ≠ j ∈ {X ,Y , Z }
and the fact that σX2 = σY2 = σZ2 = I we have
σ 2j = jx2σX2 + jy2σY2 + jz2σZ2
+ jx jy {σX , σY } + jz jy {σY , σZ } + jx jz {σZ , σX }
= ( jx2 + jy2 + jz2 )I
=I.
(b) If β is an eigenstate of σ j with eigenvalue β , then we have
σj β = β β
⇒ σ 2j β = βσ j β
⇒ I β = β2 β
⇒ β2 = 1
⇒ β = ±1
(c) In the basis { 0 , 1 } ,
0 1
1 0 
 0 −i 
 + jy 

 + jz 
σ j = jx 

 0 −1 
i 0 
1
0






jx − ijy 
 jz

= 

−j
j
+
ij
x
y
z


a 
Therefore, assuming + j =   , a ∈ \ + , and a 2 + b 2 = 1 , we have
b 
 
σj + j = + j
⇒
ajz + b( jx − ijy ) = a
1 − jz
b =a
jx − ijy
⇒
⇒
2

 1 + (1 − jz ) a 2 = 1
jx2 + jy2 

⇒
1 − jz2 + 1 − 2jz + jz2 2
a =1
1 − jz2
⇒
a =
⇒
+j =
Similarly, one can obtain
−j =
(1 + jz )/ 2
1

(1 + jz ) 

 (1 − jz )/(jx − ijy ) 
2


1

(1 + jz ) 

 −(1 + jz )/(jx − ijy ) 
2


You can verify that j + − j =0. You can also verify that + j and − j are respectively
the unit vectors in the positive and negative directions of the j-axis on the Bloch sphere.
This fact was predictable because these are the only two vectors that keep their
orientations unchanged under the rotation about j-axis.
Problem 2. Find an approximation to
e iθσY / 2e iθσX / 2e −iθσY / 2e −iθσX / 2
up to the second order of θ for θ 1 .
Solution:
Defining
f (θ) = e i θσY / 2e iθσX / 2
2 (I + i θσY / 2 − θ 2σY
/ 8)(I + i θσX / 2 − θ 2σX2 / 8)
= (I + i θσY / 2 − θ 2I / 8)(I + i θσX / 2 − θ 2I / 8)
I − θ 2I / 4 + i θ(σX + σY )/ 2 − θ 2σY σX / 4
= I − θ 2I / 4 + i θ(σX + σY )/ 2 + i θ 2σZ / 4
we have
e i θσY / 2e iθσX / 2e −iθσY / 2e −iθσX / 2 = f (θ)f (−θ)
I − θ 2I / 2 + i θ 2σZ / 2 + θ 2 (σX + σY )2 / 4
But using Problem 1.a for jx = jy = 1/ 2 , and jz = 0 , we have
(σX + σY )2 / 2 = I .
Therefore,
e i θσY / 2e iθσX / 2e −iθσY / 2e −iθσX / 2 I + i θ 2σZ / 2
where all signs stand for the approximation up to the second order of θ .
Problem 3. Rewrite
{+
A
1
( 0 A ⊗ 0 B + 1 A ⊗ 1 B ) ≡ ψ in the following basis
2
⊗ + B , + A ⊗ − B , − A ⊗ + B , − A ⊗ − B }.
What are Pr(++) , Pr(+−) , Pr(−+) , and Pr(−−) ?
Solution:
We need to find the inner products of ψ and the basis vectors. Using the fact that
+ 0 = + 1 = − 0 = − − 1 = 1/ 2
we have
B
B
B
B
1
[ + 0AB
2 A
= 1/ 2
1
− ⊗A + ψ =
[ + 0AB
2 A
=0
1
+ ⊗A − ψ =
[ −0AB
2 A
=0
1
− ⊗A − ψ =
[ −0AB
2 A
+ ⊗A + ψ =
+ 0
+
A
+ 1A
B
+ 1B]
−0B +
A
+ 1A
B
−1B]
+ 0B +
A
−1A
B
+ 1B]
−0
A
−1A
B
−1B]
B
B
+
= 1/ 2
Therefore,
ψ =
Consequently,
and
1
( + A ⊗ + B + − A ⊗ − B ).
2
Pr(++) = Pr(−−) = 1/ 2
Pr(+−) = Pr(−+) = 0 .
Problem 4. For any two qubits ψ
following property:
U ψ
A
and φ , find a unitary operator U with the
⊗ φ
= φ
B
A
⊗ ψ
B.
Write down U in the basis { 0 A ⊗ 0 B , 0 A ⊗ 1 B , 1 A ⊗ 0 B , 1 A ⊗ 1 B } .
Solution:
To find the matrix representation of an operator U, it suffices to find its action on the
basis vectors. In general, if the set of basis vectors is { ψi ,i ∈ I } for an index set I, we
have U ij = ψi U ψj , where U ij is the element on the ith row and the jth column of the
matrix representation of U. Therefore, using the following relations obtained from the
main property of U,
U
U
U
U
0A⊗
0A⊗
1A ⊗
1A ⊗
=
=
=
=
0B
1B
0B
1B
0A⊗
1A ⊗
0A⊗
1A ⊗
0B
0B
1B
1B
one can obtain
1

0

U =
0

 0

For instance,
but,
U 23 = B 1 ⊗
= B 1 ⊗
=1
0 0 0

0 1 0

1 0 0 

0 0 1 

A
A
0U 1A ⊗ 0B
0 0A ⊗ 1B
U 34 = B 0 ⊗ A 1 U 1 A ⊗ 1 B
= B 0 ⊗ A11A ⊗ 1B
= A 1 1A×B 0 1B
= 0.
Problem 5. Write out
ψ
ABC
=
1
( 0 A ⊗ 1 C + 1
A ⊗ 0 C ) ⊗ + B
2
in the following basis:
{ 000
ABC
, 001 ABC , 010 ABC , 011 ABC , 100
ABC , 101
ABC , 110
ABC , 111
ABC } .
Solution:
ψ
ABC
1
1
( 0 A ⊗ 1 C + 1
A ⊗ 0 C ) ⊗
( 0 + 1 B)
2
2
B
1
= ( 001 ABC + 100
ABC + 011 ABC + 110
ABC )
2
=
Problem 6.
Solution:
A
+(0
A
⊗ 1B + 1A ⊗ 0
B
)/ 2 = + B / 2 .