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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.
 Fall '13
 AkimasaMiyake

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