Homework 1 Solutions

B1. For B1 this measurement of √ 3 almost tells us the whole density matrix. It
sets ρ 1 13 and as a consequence there is only one value of ξ, ξ = 1/6, that satsifies the
Eq. 37
where
ρ 1 00 = 1/4 (99)
ρ 1 11 = 1/2 (100)
ρ 1 33 = 1/4
√
(101)
ρ 1 01 = ρ 1 10 = 1/8 (102)
ρ 1 03 = ρ 1∗
30 = |ρ 1 13|e −iφ 13
(103)
√
ρ 1 13 = ρ 1 31 = 1/8 (104)
(105)
|ρ 1 03| ≤ 1/4 (106)
ρ 1 could still be a pure state so the max T r[ρ 1 ρ 1 ]=1 and max T r[|ψ〉 〈ψ| |ψ〉 〈ψ|]=1.
For the minimum purity, we set ρ 1 13=0 and find T r[ρ 1 ρ 1 ] = 1−2(1/16) = 7/8. For the
minimum overlap, we set ρ 1 13=-1/4 and find T r[|psi〉 〈psi| ρ 1 ] = 1 − 4(1/16) = 0.75.
B2.
For B2 we have
ρ 2 00 = 7/12 − 2ξ (107)
ρ 2 11 = 3ξ (108)
ρ 2 33 = 5/12 − ξ (109)
ρ 2 01 = ρ 2 10 = 0 (110)
ρ 2 03 = ρ 2∗
30 = |ρ 1 03|e −iφ 03
(111)
ρ 2 13 = −ρ 2∗
31 = ±i|ρ 1 13| (112)
(113)
where
|ρ 2 03| ≤
|ρ 2 13| ≤
√
7/12 − 2ξ
√
3ξ
√
5/12 − ξ (114)
√
5/12 − ξ (115)
7/24 ≥ ξ ≥ 0 (116)
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