作者phs (世故人情情难还...)
看板Physics
标题Re: [问题] 量子简谐振子
时间Thu Nov 5 18:56:26 2009
※ 引述《musicbox810 (结束是一种开始)》之铭言：
: 我导到一半卡住了
: <n│a+ a+ a a│n>要怎麽得出<N>^2 N : number operator
: 我的想法
: a│n> = √n│n-1>
: aa│n> = √n√(n-1) => 右边的bra哪去了? 应该是 aa│n> = √n√(n-1)|n-2>
: 所以<n│a+ a+ a a│n> = n(n-1)
: 并不是<N>^2 = n^2
: 请问我错在哪里
: 感谢解答
(解)
From the commutation relation [a,a+] =1,
we have a+ a = a a+ -1 . Thus
<n│a+ a+ a a│n>
= <n|a+ (a a+ -1)a |n>
= <n|a+ a a+ a|n> - <n|a+ a|n>
= √n√n <n-1|a a+|n-1> - √n√n <n-1|n-1>
= n* √n√n - n
= n(n-1)
,where we have used properties of arising and lowering operators:
a|n> = √n|n-1> , <n|a+ = <n-1|√n ;
a+|n> = √n+1|n+1> , <n|a = <n+1|√n+1
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◆ From: 140.112.102.3
※ 编辑: phs 来自: 140.112.102.3 (11/05 19:10)
1F：推 allencce:推荐这篇文章~ 11/05 21:44
2F：推 Glamdring:想问为什麽 <n│(a+)(a+)aa│n> = <N>^2 11/06 00:24
※ 编辑: phs 来自: 123.193.215.124 (11/06 01:55)
3F：→ musicbox810:谢谢你 11/08 21:37