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Number variance

  1. Oct 22, 2007 #1
    I need to find the number variance [tex]\langle (\Delta N ) ^2 \rangle [/tex] for the state [tex]| \beta \rangle = e^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle [/tex]

    we know:
    [tex]\langle (\Delta N ) ^2 \rangle [/tex]
    [tex]\langle a^{\dagger} a a^{\dagger} a \rangle [/tex]
    [tex]\langle a^{\dagger} (a^{\dagger}a +1) a \rangle [/tex]
    [tex]\langle a^{\dagger} a^{\dagger}a a+a^{\dagger} a \rangle [/tex]
    [tex]\langle \beta| a^{\dagger} (a^{\dagger}a +1) a |\beta \rangle [/tex]

    I know the relation (since this was derived):
    [tex][a^{\dagger},e^{\alpha a}]=-\alpha e^{\alpha a} [/tex]
    [tex][a,e^{\alpha a^{\dagger}}]=\alpha e^{\alpha a^{\dagger}} [/tex]

    I could expand:
    [tex]\langle 1|e^{\alpha a^{\dagger}-\alpha^{*} a} a^{\dagger} (a^{\dagger}a +1) a e^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle [/tex]

    But i'm not sure how to apply the relation. any ideas would be appreciated.
     
  2. jcsd
  3. Oct 22, 2007 #2

    Physics Monkey

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    Hi indigojoker,

    The first thing I would do is translate those commutator statements into something a little more concrete. It would be nice to know how a (or a^+) acts on your state. You might hope for a simple relationship since a annihilates the ground state and has a simple commutator with the operator that creates your state from the ground state.
     
  4. Oct 23, 2007 #3
    hmm, I'm not sure what you mean by something more concrete. Do you mean figuring out what:

    [tex]a| \beta \rangle = ae^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle [/tex]
    and
    [tex]a^{\dagger}| \beta \rangle = a^{\dagger} e^{\alpha a^{\dagger}-\alpha^{*} a}|1 \rangle [/tex]

    is when simplified using:
    [tex][a^{\dagger},e^{\alpha a}]=a^{\dagger}e^{\alpha a}-e^{\alpha a} a^{\dagger}=-\alpha e^{\alpha a} [/tex]
    [tex][a,e^{\alpha a^{\dagger}}]=a e^{\alpha a^{\dagger}}-e^{\alpha a^{\dagger}}a=\alpha e^{\alpha a^{\dagger}} [/tex]
     
  5. Oct 24, 2007 #4

    Physics Monkey

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    Yes, that is what I had in mind. Here is a hint: since a annihilates the ground state, one can write [a,B]|ground> = (aB-Ba)|ground> = aB|ground> where B is anything!
     
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