MASSACHUSETTS INSTITUTE OF TECHNOLOGY
2.111J/18.435J/ESD.79
Quantum Computation
Problem 1. For the state ψ =
1
n /2
2
2n −1
∑ (−1)f (x )
x
1
x =0
g(x ) 2 , where g(x ) is a 1-1
function, find the partial trace ρ1 ≡ tr2 ( ψ ψ ) and calculate
⊗n
+ ρ1 +
⊗n
.
Solution:
1
ψ ψ = n
2
2n −1 2n −1
∑ ∑ (−1)f (x )+f (y )
x
1
x =0 y =0
y ⊗ g(x ) 2 g(y )
Now, the fact that g(x ) is a 1-1 function implies that for x ≠ y , we have g(x ) ≠ g(y ) ,
and therefore,
g(x ) g(y ) = δxy = tr( g(x ) g(y ) )
Using the above relation, we have
ρ1 ≡ tr2 ( ψ ψ )
1
= n
2
1
= n
2
1
= n
2
=
1
2n
2n −1 2n −1
∑ ∑ (−1)f (x )+ f (y )
x
1
y ⊗ tr ( g(x ) 2
g(y ) )
x
1
y δxy
x = 0 y = 0
2n −1 2n −1
∑ ∑ (−1)f (x )+ f (y )
x = 0 y = 0
2n −1
∑ (−1)f (x )+ f (x ) x 1
x
x =0
2n −1
∑
x 1
x
x =0
n
= I1 / 2 .
Now, the probability of measuring + ⊗n is as follows
tr(ρ1 + ⊗n
⊗n
+ ) =⊗n + ρ1 +
⊗n
1 ⊗n
+ I1 +
2
n
1
= n.
2
=
⊗n
Problem 2. Find H ⊗nRαH ⊗n and H ⊗nTαH ⊗n in simpler terms, where
2n −1
Rα =
∑ (−1)x iα
x x
x =0
and
2n −1
Tα =
∑
x ⊕α x .
x =0
Solution:
We know
H ⊗n =
1
2n −1
∑ (−1)a ib
2n / 2 a,b =0
b a
⇒
H
⊗n
TαH
⊗n
2n −1 2n −1 2n −1
1
= n ∑ ∑ ∑ (−1)a ib +c id c d x ⊕ α x b a
2 a,b =0 c,d =0 x =0
2n −1 2n −1 2n −1
1
= n ∑ ∑ ∑ (−1)a ib +c id c δd ,x ⊕α δxb a
2 a,b = 0 c,d = 0 x =0
1
= n
2
1
= n
2
=
2n −1 2n −1 2n −1
∑ ∑ ∑ (−
1)a ix +c i(x ⊕α) c
a = 0 c = 0 x =0
2n −1 2n −1
∑ ∑ (−1)
c iα
2n −1
c a
a =0 c =0
2n −1 2n −1
1
2n a∑
=0
∑ (−1)c iα
a
∑ (−1)a ix +c ix
x =0
c a 2n δac
c =0
2n −1
=
∑ (−1)a iα
a a
a =0
= Rα .
−1
Now, using H ⊗n = ( H ⊗n )
, we have
H ⊗n RαH ⊗n = Tα .
2n −1
Hint: ∑ (−1)a ix = 2n δ(a ) .
x =0
Problem 3. Find U p RpU p† and U pTpU p† in simpler terms, where
p−1
Rp =
∑ exp(2πxi / p) x
x =0
x
p−1
Tp =
∑ x + 1 mod p
x
x =0
p−1 p−1
Up =
1
∑ exp(2πixy / p) y x
p x∑
=0 y = 0
and p is a prime number.
Solution:
U pTαU p†
p−1
p−1 p−1
p−1
p−1 p−1
1
= ∑ ∑ ∑ exp ( 2πi(cd − ab)/ p ) c d x + 1 x b a , addition mod p
p a,b= 0 c,d = 0 x =0
1
= ∑ ∑ ∑ exp ( 2πi(cd − ab)/
p ) c δd ,x +1δxb a
p a,b= 0 c,d =0 x =0
p−1 p−1 p−1
1
= ∑ ∑ ∑ exp ( 2πi(cx + c − ax )/ p )
c a
p a = 0 c= 0 x =0
p−1 p−1
=
1
∑ exp ( 2πic / p ) c a
p a∑
= 0 c= 0
p−1
∑ exp ( 2πi(cx − ax )/ p )
x =0
p−1 p−1
1
= ∑ ∑ exp ( 2πic / p )
c a pδac
p a = 0 c= 0
p−1
=
∑ exp ( 2πia / p ) a
a
a =0
= Rp . Similarly, you can show that
U p†RpU p = Tp
and
U p RpU p† = Tp† .
p


Hint: ∑ exp ( 2πiax / p ) =  exp ( 2πia ) − 1

=0
x =0
 exp ( 2πia / p ) − 1
p−1
a ≡ 0 (mod p)
otherwise
, for a ∈
.
Problem 4. Show that U 2 ⊗ U 3 = PU 6P −1 where U p is defined in Problem 3, and P
is a permutation matrix (a matrix with only one nonzero element 1 in each row and
column).
Solution: It can be seen that
1 U 3
U 2 ⊗U 3 =
2 U 3

and
1

1


1
U3 
 = 1 
−U 3 
6 1


1


 1

1

1



1 1
U6 =

6 1

1


 1

where ω = exp(2πi / 6) .
1
1 1
1
ω2
ω 4 1 ω2
ω4
ω2 1 ω 4
1
1 ω3
ω3
ω2
ω4 ω3
ω5
ω4
ω2 ω 3
ω
1 

4 
ω 

ω2 

3
ω 

ω 

ω 5 

1
1
1
1
ω
ω2
ω3
ω4
ω2
ω4
1
ω2
ω3
1
ω3
1
ω4
ω2
1
ω4
ω5
ω4
ω3
ω2
1 

ω 5 

ω4 

3
ω 

ω 2 

ω 

There is no such P that satisfies U 2 ⊗U 3 = PU 6P −1 . You can however find P1 and P2
such that U 2 ⊗U 3 = PU
1 6P2 . For instance,
1

0

0

P1 = 
0

0


 0
0 0 0 0 0
1 0


0 0
0 1 0 0 0


0 0

0 0 0 1 0

 and P2 = 
0 0
0 0 1 0 0


0 1
0 0 0 0 1 



1 0 0 0 0 
 0 0

0 0 0 0

0 0 1 0

1 0 0 0 
.
0 1 0 0

0 0 0 0 

0 0 0 1 

−1
Also, you can verify that U 2 ⊗U 3† = PU
1 6P1 .