Homework 1 Solutions

Now using our measurements we can differentiate between B1 and B2.
For B1 we have
This puts additional constraints on ξ by Eq. 37.
We look at the squared equation
ρ 1 00 = 7/12 − 2ξ (53)
ρ 1 11 = 3ξ (54)
ρ 1 33 = 5/12 − ξ (55)
√
ρ 1 01 = ρ 1 10 = 1/8 (56)
√ √ √
1/8 ≤ 3ξ 7/12 − 2ξ (57)
1/8 ≤ 7/4ξ − 6ξ 2 (58)
First, we find values of ξ where it is equal. We find ξ = 1/8 and ξ = 1/6.
This means that 1/6 ≥ ξ ≥ 1/8
We have no information about the other elements. Therfore at this point our
density matrix is
ρ 1 00 = 7/12 − 2ξ (59)
ρ 1 11 = 3ξ (60)
ρ 1 33 = 5/12 − ξ (61)
√
ρ 1 01 = ρ 1 10 = 1/8 (62)
ρ 1 03 = ρ 1∗
30 = |ρ 1 03|e −iφ 03
(63)
ρ 1 13 = ρ 1∗
31 = |ρ 1 13|e −iφ 13
(64)
(65)
where
|ρ 1 03| ≤
|ρ 1 13| ≤
√
7/12 − 2ξ
√
3ξ
√
5/12 − ξ (66)
√
5/12 − ξ (67)
1/6 ≥ ξ ≥ 1/8 (68)
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