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Calculating a matrix element & first order shift

  1. Nov 12, 2014 #1
    So I saw this a moment ago:
    Sem Título.png
    How can the second and third terms yield (n+1)+(n-1)=2n and not (n+1)+n=2n+1?
    PS: I solved the problem by using [a,a(+)]=1.

    Sorry, this is very simple but I cannot figure out what I did wrong.
     
  2. jcsd
  3. Nov 12, 2014 #2

    dextercioby

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    It's not clear to me what A|n> is equal to. This is very important, since (dropping the finesse of domain issues linked to unbounded operators on Hilbert space)

    [tex] \langle n, A^{\dagger} A n\rangle = \langle A^{\dagger\dagger} n, A n\rangle = \langle A n, A n\rangle [/tex]

    [tex] \langle n, A A^{\dagger} n\rangle = \langle A^{\dagger} n, A^{\dagger} n\rangle = \sqrt{n+1}^2 = n+1 [/tex]
     
    Last edited: Nov 12, 2014
  4. Nov 12, 2014 #3
    Thank you, I understood your answer: simple and pratical.
    Here, A|n> is equal to sqrt(n)|n-1> and A(+)|n> is equal to sqrt(n+1)|n+1>.

    PS: Sorry for not using the math font but I am too lazy right now to insert it in latex.
     
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