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Question to a calculation rule -> Dirac formalism

  1. Nov 12, 2012 #1
    Hi everyone

    1. The problem statement, all variables and given/known data

    I have a question: Am I allowed to do the following, where the a's are operators and the alphas are quantum states.

    [tex] \langle \alpha \mid a^\dagger a^\dagger a a \mid \alpha \rangle = \langle \alpha \mid a^\dagger a^\dagger \mid \alpha \rangle \langle \alpha \mid a a \mid \alpha \rangle[/tex]


    Well, I think I am allowed to, but I'm not sure how to verify this.

    Thanks for your help
     
  2. jcsd
  3. Nov 12, 2012 #2

    vela

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    No, you can't. For that to be true, you'd have to have ##\lvert \alpha \rangle \langle \alpha \rvert = 1##, which is almost certainly not the case.
     
  4. Nov 12, 2012 #3

    and when the alpha is a coherent state?
     
  5. Nov 12, 2012 #4

    dextercioby

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    Not even then, it would require the annihilation operator to be a self-adjoint operator (moreover a projector), which is not the case.
     
  6. Nov 12, 2012 #5
    Ok thank you.

    my problem is the following: I want to simplyfy the following expression

    [tex] \langle \alpha \mid a^\dagger a^\dagger a a + a^\dagger a \mid \alpha \rangle [/tex]

    I'm very uncertain about calculation rules when it comes to operators and states

    is the expression above the same as?:
    [tex] \langle \alpha \mid (a^\dagger a^\dagger a a \mid \alpha \rangle + a^\dagger a \mid \alpha \rangle) [/tex]

    or doesn't that help at all?

    I think the aim here is to get an expressin which only has the alpha (as an eigen value) in it. The task is about calculating the expectation value
     
    Last edited: Nov 12, 2012
  7. Nov 12, 2012 #6
    [tex] \langle \alpha \mid a^\dagger a^\dagger a a + a^\dagger a \mid \alpha \rangle = \langle \alpha \mid a^\dagger a^\dagger a a \mid \alpha \rangle + \langle \alpha \mid a^\dagger a \mid \alpha \rangle
    = \alpha \alpha^* \langle \alpha \mid a^\dagger a \mid \alpha \rangle +|\alpha|^2
    = |\alpha|^4 +|\alpha|^2[/tex]

    sorry for making a new reply, I tried to edit, but didn't come out in latex format.
    To get to this solution I used that

    [tex] a\mid \alpha \rangle = \alpha \mid \alpha \rangle [/tex]

    and the equivalent for the bra vector
     
  8. Nov 12, 2012 #7

    dextercioby

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    It looks OK to me.
     
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