Midterm 3 Solutions

# Thus n sy q 1 s rhq j l y 0 n h qln n split

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Unformatted text preview: sis 0 1 0 001 ¬༼ be careful of this term ` ` Â` Sy = - 2 JS+ - S- N Ø8 j, mz \< basis `2 Sy Ø8 j, mz \< basis `3 Sy Ø8 j, mz \< basis 0 -Â 0 Â 0 -Â 0Â0 — 2 1 0 -1 020 -1 0 1 0 -2 Â 0 —3 2 Â 0 -2 Â 22 0 2Â 0 —2 2 ` = —2 S y ` `n So Sy does not square to the identity, but Sy as a function of n does have a period of 2 (up to normalization and excepting n = 0, which is the identity). Thus `n Sy Âq ` ` 1 -S RHq j L = ‰ — y = ⁄¶= 0 n! H-Â qLn n ¬༼ Split into identity term and the rest into n-even/odd — ` 2m Sy ` 1 = 1ཽ + ⁄¶ = 1 H2 mL! H-Â qL2 m m — 1 + ⁄¶ = 0 H2 m + 1L! H-Â qL2 m + 1 m `2 Sy — H-1Lm - Â B⁄¶ = 0 H2 m + 1L! q2 m + 1 F m = cos q - 1 ` RHq j L Ø8 j, mz \< basis ` Sy — = sin q 100 010 001 + 1 2 1 0 -1 Hcos q - 1L 0 2 0 -1 0 1 - Â 2 0 -Â 0 sin q Â 0 -Â 0Â0 Therefore 1 2 ` RHq j L Ø8 j, mz \< basis H1 + cos qL 1 2 — ` JS y ê—N ` 2 = JS y ê—N ` H-1Lm = 1ཽ + B⁄¶ = 1 H2 mL! q2 m F m ` 2m+1 Sy sin q 1 2 1 Printed M sin q by H1athematicaqfL Students - cos or 2 cos q - 1 2 sin q 8 MidtermExam3Solutions.nb Therefore 1 2 ` RHq j L Ø8 j, mz \< basis H1 + cos qL 1 2 1 2 sin q H1 - cos qL 1 2 sin q 1 2 cos q 1 2 sin q H1 - cos qL - 1 2 1 2 sin q H1 + cos qL and we have ` 1, 1n \ = RHq j L 1, 1z \ = 1 2 H1 + cos qL 1, 1z \ + 1 2 sin q 1, 0z \ + 1 2 H1 - cos qL 1, -1z \ ü Part b We see from (a) that the fraction of particles that survive the third measurement will be maximized at 1 16 when sin4 q = 1 p ﬂ q= 2 ü Part c The fraction of particles that survive the last measurement if the SGn device were removed is zero as 1, ± 1z \ are orthogonal. ü Statistics Mean = 5.22 Stddev = 3.00 Printed by Mathematica for Students MidtermExam3Solutions.nb Total Mean = 16.11 Stddev = 7.51 Printed by Mathematica for Students 9...
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## This note was uploaded on 02/01/2014 for the course PHYC 491/496 taught by Professor Akimasamiyake during the Fall '13 term at New Mexico.

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