Homework 1 Solutions

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We have two inequalities √ ( 1/8 − η)/ √ 2 ≤ √ η ≤ 3/8 − ξ √ 2ξ √ 2ξ (88) √ 5/8 − ξ (89) We will comeback to these. step 3:Finally we calculate the min and max of T r[|ψ〉〈ψ| ρ] First, we note that if the measurements are consistent we reach the max when ρ = |ψ〉〈ψ| To calculate the min and max generally we do the following: First, we write out |ψ〉〈ψ| |ψ〉〈ψ| = 1 2 |1〉〈1|+1 1 (|0〉〈0|+|3〉〈3|+|0〉〈3|+|0〉〈3|)+ 4 2 √ (|0〉〈1|+|0〉〈1|+|1〉〈3|+|1〉〈3|) 2 (90) We can then write for any ρ T r[|ψ〉〈ψ| ρ] = 1 2 ρ 11 + 1 4 (ρ 00 + ρ 33 + 2Re(ρ 30 )) + 1 2 √ 2 (2Re(ρ 10) + 2Re(ρ 13 )) (91) B1. For B1 the measurements are consistent with ρ 1 = |ψ〉〈ψ| Therefore maxT r[|ψ〉〈ψ| ρ 1 ] = 1 (92) To calculate the min we examine Eq. 91. To minimize this function we set Re(ρ 30 ) and Re(ρ 10 ) to negative values with the largest magnitude allowed by the constraints. minT r[|ψ〉〈ψ| ρ] = 1 √ 2 (3ξ)+1 4 (1−3ξ−2 (7/12 − 2ξ)(5/12 − ξ))+ 1 2 √ 2 ( √ 1 √ −2 (3ξ)(5/12 − ξ)) 2 (93) Now we need to pick an ξ that minimizes this. I usually plot it in matlab, mathematica, maple, or scilab. We can see from the espression that the function is basically linear in ξ so we suspect that the answer will be either 1/6 or 1/8. For the ρ = |ψ〉〈ψ|, η = 1/6, so I would guess the minimum would occur for η = 1/8. This is true as shown below The min T r[|ψ〉〈ψ| ρ] = 0.204 B2. 10
0.25 B1 0.245 0.24 0.235 min Tr[|ψ>
Page 1 and 2: Homework 1 Solutions January 29, 20
Page 3 and 4: Substituting Eq. 12 into Eq. 11, we
Page 5 and 6: = x 0 (〈ψ ′ |ψ〉 + 〈ψ |ψ
Page 7 and 8: step 1: use N measurement and T r[
Page 9: For B2 we have ρ 2 00 = 7/12 − 2
Page 13 and 14: maxminT r[|ψ〉〈ψ| ρ] = 1 2 (
Page 15 and 16: We now know the offdiagonal element
Page 17 and 18: 3.1 Solution First calculate the Jo

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