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University of Illinois at Urbana-Champaign last updated 14 April 2007
Problem Set # Question 1 Given the following density matrices:
1 2 3 4
ρ ρ ρ ρ
= ^ ^ = ^ ^ = ^ ^ = ^
Within the 10 combinations F ( ρ j , ρ k )= 1 for j = k. So we just need to consider 6
terms, ( 1 , 2 )^3
F ρ ρ = , ( 1 , 3 )^1
F ρ ρ = , ( 1 , 4 )^1
F ρ ρ = , (^) ( 2 , (^3) ) (^1) ( 2 3 ) 0. 4
F ρ ρ = + = ,
( 2 ,^4 ) (^1) ( 2 3 ) 0. 8
F ρ ρ = + = and ( 3 , 4 )^1
F ρ ρ =.
Question 2 (a)
Given the linear entropy (^) ( ) (^4) ( 1 ( )^2 ) L 3 S x = − Tr ρ x ,
SL ( 1 ) = 0 , S L ( 2 )= 1/ 2, S L ( 3 ) = 2 / 3 and S L ( 4 )= 1.
(b)
Given the von Neumann entropy S = − tr ρ log 2 ρ, let EV ( ρ k ) be the list of
eigenvalues of the density matrix ρ (^) k ,
1 1
2 2 2 2
3 3 2 2
4 3 2
(^1) , 3 , 0, 0 1 log 1 3 log 3 0.81, 4 4 4 4 4 4 (^1) , 3 , 0, 0 1 log 1 1 log 1 1, 4 4 2 2 2 2 (^1) , 1 , 1 , 1 log 4 2, 4 4 4 4
EV S
EV S
EV S
EV S
ρ ρ
ρ ρ
ρ ρ
ρ ρ
= ^ ^ ⇒ = − ^ ^ − ^ =
= ^ ^ ⇒ = − ^ ^ − ^ =
= ^ ⇒ = =
which are more or less correlated with the linear entropy.
University of Illinois at Urbana-Champaign last updated 14 April 2007
Question 3 (a) We first consider how to systematically get the reduced density matrix: (let’s ignore the matrix elements)
00 00 01 00 10 00 11 00 00 01 01 01 10 01 11 01 . 00 10 01 10 10 10 11 10 00 11 01 11 10 11 11 11
ρ
= ^
To take the partial trace, using the rule: tr p q = q p , we then have
0 0 0 1 0 0 0 0 0 0 1 0 . 0 1 0 1 1 0 0 0 1 0 1 1
ρ
→ ^
Therefore, a systematic way to obtain the reduced density matrix is to sum over the diagonal elements of the submatrices (not true for matrices other than the 4x4 one).
So we have
1 2
3 4
ρ ρ
ρ ρ
= ^ ^ → = ^ →
= ^ ^ → = ^ →
They are all the same. The entropy E ɶ^ = − Tr ρ red (^) log 2 ρ red = 1 (which of course cannot
be considered as entanglement).
(b)
The eigenvalues for the R (^) ( ρ (^2) ) are 3 / 4 and 1/ 4 , so C (^) ( ρ 2 )= 1/ 2, x = (^) ( 2 + 3 / 4) ,
E ( ρ 2 )= 0.35. The eigenvlaues for R ( ρ 3 ) are 1/2 and 1/2, so E ( ρ 3 )= 0.
University of Illinois at Urbana-Champaign last updated 14 April 2007
3 1 2 (^2 1 ) 2 1 2 2 12 2 12 1 2 1 2 (^23 ) 2 1
z x z x
F I F F
I I I
I
n n
σ σ σ σ
= ^ + +
Therefore 2 2 1
p =
and n 3 = cos ( 8 π^ ) 0 + sin ( π 8 ) 1 = 22 + 21 0 + 22 − 211.
*Recall that any 2x2 matrix M can be decomposed by the Pauli matrices:
M = m I 0 + mx σ x + m y σ y + mz σ z , where mk = ( 1/ 2) tr M ( σ k ) for k = x y z , , and
m 0 = ( 1/ 2) trM. (The vector m =( m x , my , mz )
(^) is usually mapped to the Bloch sphere as
mz = cosθ, mx = sin θ cosφ and my = sin θ sinφ.) Therefore we write 1 1 = ( I − σ z )/ 2,
and − − = ( 1 − σ x )/ 2.
(b) Given 1/ 2 1/ 2 1 2 3 3
u a u b u p n c
+^ +
the constraint 1 2 2 1 2 1 j j 2 1 2 1 u u = ⇒ a = b = − =
and 2 2 1 2 1
c = −
. So we
write a ɶ = a / a , b ɶ^ = b / b and c ɶ = c / c , and u ɶ 1 ≡ u 1 (^) / a and so on
1 2 1/ 4^1 2 ,^2 2 1/ 4^2 ,^3 23 2.
u ≡ + a u ≡ − + b u ≡ n + c −
University of Illinois at Urbana-Champaign last updated 14 April 2007
1 2
1 3 1/ 4^ *^ *
2 3 1/ 4^ *^ *
u u a b a b
u u a c a c
u u b c b c
⇒ ^ − + = ⇒ =
⇒ ^ − + = ⇒ = −
− ^
ɶ ɶ ɶ ɶ^ ɶ ɶ
ɶ ɶ ɶ^ ɶ ɶ ɶ
where − n 3 = ( 1/ 2 )( 22 + 21 − 22 − 21 ). The solution is a ɶ^ = 1 , b ɶ^ = 1 and c ɶ^ = − 1 and
hence
(^1) & 2 1. 2 1 2 1
a = b = c = − −
(c) By the suggested substitutions, we have
1/ 2 (^1 1 2 1 ) 1/ 2 (^2 1 2 1 )
(^3 3 1 2 1 ) (^4 1 )
u a
u b
u p n c u
Note that u 4 is orthogonal to all other states. The POVM is implemented by
constructing the measurement operator as M = n 1 (^) n 1 (^) + n 2 (^) n 2 (^) + n 3 (^) n 3 (^) + n 4 (^) n 4
and initializing the second qubit as 0 2. In this way, the process of measurement is
represented by
1 1 2 2 3 3 1 2 3
n n n n n n Tr F Tr F Tr F
ρ ρ ρ ρ ρ ρ ρ
