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Creation operator

  1. Feb 6, 2009 #1

    KFC

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    For creation operator of hamonic oscillator, we have

    [tex]a^\dagger |n> = \sqrt{n+1}|n+1>[/tex]

    if I consider the creation operator operate on the bar vector, should I also get the same thing? namely

    [tex]<n|a^\dagger = \sqrt{n+1}<n+1|[/tex]
     
  2. jcsd
  3. Feb 6, 2009 #2

    Avodyne

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    No. Take the hermitian conjugate of your first equation; what do you get?
     
  4. Feb 6, 2009 #3

    malawi_glenn

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    You have to consider the dual correspondence:

    [tex] X|\alpha > [/tex] corresponds to [tex] < \alpha | X^{\dagger} [/tex]

    Also you can consider the number operator: [tex] N = a^{\dagger}a [/tex], with [tex] N|n> = n |n> [/tex]

    Sandwhich it between: [tex] <n|N|n> [/tex], try it!
     
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