Midterm 5 Solutions

Therefore p a 2 1 2 1 2 p a 2 trb

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e or Bob that measures in the ± n\ basis. ü Method 2: Using a reduced state We can also take p A, + — ` = TrB +n\ X+n rA F 2 where ` rA = TrB YH-L ] XYH-L = = ` rA = 1 TrB @ +z, 2 1 @ +z\ X+z 2 1` 1ཽ 2 -z\ X+z, -z - +z, -z\ X-z, +z - -z, +z\ X+z, -z + -z, +z\ X-z, +z D + -z\ X-z D ` (and by symmetry, we may argue that the same is true of rB ). Therefore p A, + — = 2 1 2 1 2 = p A, + — 2 = ` TrB +n\ X+n 1ཽF ` X+n 1ཽ +n\ 1 2 and p A, - — =1-p 2 A, + — 2 = 1 2 ` Using our symmetry argument above and the fact that Z+z 1ཽ +z^ = 1, we may therefore conclude that the probabilities that Bob — will measure the outcomes ± 2 are also Printed by Mathematica for Students p B, + — 2 =p B, - — 2 = 1 2 p A, + — = 1 2 = p A, + — = 2 TrB +n\ X+n 1ཽF 2 2 ` X+n 1ཽ +n\ 1 2 and p A, - — =1-p 2 A, + — = 2 MidtermExam5Solutions.nb 1 2 9 ` Using our symmetry argument above and the fact that Z+z 1ཽ +z^ = 1, we may therefore conclude that the probabilities that Bob — will measure the outcomes ± 2 are also p B, + — =p 2 B, - = — 2 1 2 ü Part b From Method 2 of (a), ` 1ཽ 2 ` ` r A = rB = which is independent of the q8A, B< , so regardless of which qB Bob measures along, his measurement outcomes will always be consistient with the probabilities given by this state, from which he can extract no information about qA . Therefore, Bob cannot determine the orientation of Alice’s device by changing his orientation if Alice and Bob do not communicate their outcomes. ü Part c If Alice and Bob share the classical mixture ` r= 1 @ 2 +z, +z\ X+z, +z + -z, -z\ X-z, -z D then we again have ` rA = ` rB = 1 @ 2 1` 1ཽ 2 +z\ X+z + -z\ X-z D = 1 2 ` 1ཽ which are the same marginals they would have if they had instead shared the entangled state YH-L \. Thus, if they cannot commu` nicate their outcomes, then they cannot determine whether they share the mixed state r or the entangled state YH-L \. If they can communicate their outcomes, then without loss of generality, assume Alice measures her degree of freedom in the z-direction — first and obtains the result + 2 . If...
View Full Document

Ask a homework question - tutors are online