Homework 1 Solutions

maxminT r[|ψ〉 〈ψ| ρ] = 1 2 (2ξ) + 1 1
(3/8 − ξ) +
4 2 √ (2η) (95)
2
Notice how all the terms vanish since |ψ〉 〈ψ| has no overlap with |2〉. Recall that
0 ≤ η ≤ 1/( √ 8) and 3/8 ≥ ξ ≥ 0 We could simply bound the min and max with
these bounds but we have more information.
Using Eq. √88, we need the quadratic problem to have real solutions. After some
algebra this 9/16 − 8η must be real. This means η ≤ 9/(8 · 16) can solve for ξ and
√
find ξ pm = 3/4± 9/16−8η
4
Maximum We choose the maximum ξ for each η and plot the trace. We see that
η = 0 and ξ = 3/8 is the best leading to a trace of 3/8 = 0.375
Minimum We √ set a minimum η=0. Then we need to refer to the other inequality
Eq. 89, 1/4 ≤ 5/4ξ − 2ξ 2 . The smallest value of ξ allowed is ξ = 1/16(5 − √ 17).
We then calculate the minimum Tr is 0.138
3. Bob buys a new detector. It can measure 〈a † a † + aa〉. The values obtained are
B1= √ 3 (ACTUAL HOMEWORK 6/ √ 2 WAS WRITTEN in ERROR THIS VALUE
MEANS THE MEASUREMENT WAS BAD), B2=0, and B3=0. What does this
measure in terms of position and momentum
Update the density matrices.
Calculate the min and max of T r[ρρ] for each case. (One way to measure the
purity. A pure state yields 1 and the most mixed state, 1/3 I, yields 1/3).
Calculate the min and max of T r[|ψ〉 〈ψ| ρ B1 ].
2.3 Solution
Another measurement givesus more insight into the density matrix
We further note
a † a † + aa = ( ˆX 2 /x 0 − ˆP 2 x /P x0 )/2 (96)
and
a † a † + aa = √ 2(|2〉 〈0| + |0〉 〈2|) √ 6(|3〉 〈1| + |1〉 〈3|) (97)
T r[(a † a † + aa)ρ] = 2 √ 2Re(ρ 02 ) + 2 √ 6(|3〉 〈1| + |1〉 〈3|) (98)
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