Midterm 5 Solutions

Statistics mean 556 median 6 stddev 259 problem

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Unformatted text preview: grees of freedom Stot and Stot, z , we have the energy eigenstates when w0 = 0 1, 1\ = + z, + z\ 1, 0\ = 1 2 1, -1\ = B + z, - z\ + -z, +z\F 0, 0\ = 1 2 B + z, - z\ - -z, +z\F - z, - z\ `2 ` where the first label j corresponds to the eigenvalue of Stot , and the second label m correspons to the eigenvalue of Stot, z . ü Statistics mean = 5.56 median = 6 stddev = 2.59 Problem 2 Given that Alice and Bob share the entangled state YH-L \ = 1 2 B + z, - z\ - -z, +z\F ü Part a ` When qA = q and qB = 0. Bob measures in the Sz eigenbasis, and Alice measures in the basis +n\ = cosHq ê 2L +z\ + sinHq ê 2L -z\ -n\ = sinHq ê 2L +z\ - cosHq ê 2L -z\ ñ +z\ = cosHq ê 2L +n\ + sinHq ê 2L -n\ -z\ = sinHq ê 2L +n\ - cosHq ê 2L -n\ Thus YH-L \ = 1 YH-L \ = 1 2 2 :@cosHq ê 2L +n\ + sinHq ê 2L -n\D -z\ - @sinHq ê 2L +n\ - cosHq ê 2L -n\D +z\> :cosHq ê 2LB +n, -z\ + -n, +z\F + sinHq ê 2LB -n, -z\ - Printed by Mathematica for Students +n, +z\F> 7 -n\ = sinHq ê 2L +z\ - cosHq ê 2L -z\ -z\ = sinHq ê 2L +n\ - cosHq ê 2L -n\ Thus 8 MidtermExam5Solutions.nb YH-L \ = 1 YH-L \ = 1 2 2 :@cosHq ê 2L +n\ + sinHq ê 2L -n\D -z\ - @sinHq ê 2L +n\ - cosHq ê 2L -n\D +z\> :cosHq ê 2LB +n, -z\ + -n, +z\F + sinHq ê 2LB -n, -z\ - +n, +z\F> ü Method 1: Using a global description (see example 5.1) — We have the probability that Alice will observe an outcome of + 2 p A, + — X+n, +z YH-L \ = 1 Asin2 Hq ê 2L 2 = p A, + 2 + X+n, -z YH-L \ 2 2 — = 2 + cos2 Hq ê 2LE 1 2 and p A, - — =1-p 2 A, + — = 1 2 = 1 2 2 Similarly p B, + — X+n, +z YH-L \ = 1 Asin2 Hq ê 2L 2 = p B, + 2 + X-n, +z YH-L \ 2 2 — = 2 + cos2 Hq ê 2LE 1 2 and p B, - — =1-p 2 B, + — 2 Given Alice’s probabilities, it shouldn’t be at all surprising that Bob’s probabilities are the same, since rotating the global coordinates by -q (so qA = 0 and qB = q) exchanges Alice’s and Bob’s roles, but such a rotation should not affect the probabilities they observe. It thus doesn’t matter if it is Alic...
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