Midterm 7 Solutions

# Bn ha l f a a continuing this recursion relation we

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Unformatted text preview: Ha L F a ` `† n - 1 `† `† n = Ha L + BN , Ha L F a ` `† n ` `† n - 2 `† `† `† n `† n - 1 BN , Ha L F = Ha L + :Ha L + BN , Ha L F a > a Continuing this recursion relation, we find that ` `† n `† n BN , Ha L F = n Ha L ` Therefore, the eigenstates of N are of the form n\ = `† Ha Ln n! 0\ where ` N 0^ = 0 as ` N `† Ha Ln ` N `† Ha Ln n! 0\ = Bn `† Ha Ln n! 0\ + `† Ha Ln n! ` N 0\F =0 n! 0\ = n `† Ha Ln n! 0\ Printed by Mathematica for Students MidtermExam7Solutions.nb ü Part c The condition that we have equal probability of measuring E = yH0L\ = 1 2 1 2 — w or E = 3 2 — w implies (from (b)) B 0\ + ‰Â f 1\F where we determine the relative phase ‰Â f using mw— 2 X px \ = as ` X px \ = YyH0L px yH0L] ` ¬༼ px = Â m—w 2 `† ` Ja - aN from (a) = Â 2 mw— 2 `† ` ` BX0 + ‰-Â f X1 F Ja - aNB 0\ + ‰Â f 1\F ¬༼ a 0] = 0 and inner product on 2\ vanishes = Â 2 mw— 2 BX0 + ‰-Â f X1 FB 1\ - ‰Â f 0\F ‰Â f - ‰- Â f N 2Â = mw— 2 J X px \ = mw— 2 sin f Thus, the constraint on the average momentum implies sin f = 1 ‰Â f = Â Therefore yH0L\ = 1 2 B 0\ + Â 1\F ü Part d We have YHx, tL = Xx yHtL\ = Xx ‰ - ` ÂH t — y H0L\ where ‰ - ‰ - ` ÂH t — y H0L\ = 1 y H0L\ = 1 ` ÂH t — 2 2 ‰ - B‰ ` ÂH t - — B 0\ + Â 1\F...
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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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