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Homework Help: Useful Raising/Lowering Operator Equation, Quick derivation help

  1. Oct 4, 2007 #1

    Show that,

    [tex]a (a^{ \dagger})^{n} = n (a^{\dagger})^{n-1}+(a^{\dagger})^{n} a[/tex]


    [tex] a = \sqrt{\frac{m \omega}{2 \hbar}}(x+\frac{\imath p}{m \omega})[/tex]

    [tex] a^{\dagger} = \sqrt{\frac{m \omega}{2 \hbar}}(x-\frac{\imath p}{m \omega})[/tex]

    [tex] [a,a^{\dagger}]= a a^{\dagger}-a^{\dagger}a=1[/tex]


    This is just one step in a long derivation from another problem. The author (Goswami, pg 147) uses it without proof, but I would like to modify it – and hence I need to understand where he got it from.

    I can see how you might pull [tex] a^{\dagger}[/tex]’s out from the first and third term, and try to use the commutation relation, but the n’s don’t seem to work out right.

    Any thoughts?
    Last edited: Oct 4, 2007
  2. jcsd
  3. Oct 4, 2007 #2


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    The formula you are trying to prove is the same as [a,(a^+)^n]=n*(a^+)^(n-1), right? For any three operators A,B and C we know [A,BC]=[A,B]C+B[A,C]. This looks like the product rule for differentiation with [A, acting like a derivative. You can extend this to any number of operators, [A,BCDEF...]. So what does this say about [A,B^n] when [A,B]=1? BTW I think [a,a^+]=1, not the order you wrote.
  4. Oct 4, 2007 #3
    Ah, I didn't think to look up some commutator identities. And, yes, we proved the differentiation/commutation relationship last homework assignment. Right, I'll try it again. Thanks.
  5. Oct 4, 2007 #4
    So then,

    [tex][a^{2}, (a^{\dagger})^{n}] = a [a, (a^{\dagger})^{n}]+[a, (a^{\dagger})^{n}]a= a n (a^{\dagger})^{n-1} + n (a^{\dagger})^{n-1} a[/tex]






    [tex]a^{2}|n>=\frac{1}{\sqrt{n!}}a^{2}( a^{\dagger})^{n}|0> [/tex]
    [tex]= \frac{1}{\sqrt{n!}}( a n (a^{\dagger})^{n-1} + n (a^{\dagger})^{n-1} a
    )| 0>[/tex]
    [tex]= \frac{1}{\sqrt{n!}} a n (a^{\dagger})^{n-1}| 0>[/tex]
    [tex]=\frac{1}{\sqrt{n!}} n (n-1) (a^{\dagger})^{n-2}|0>[/tex]
    [tex]= \sqrt {n (n-1)} |n-2>[/tex]

    I know that I may have not provided enough of a primer in the problem to have anyone check this, but just puttin’ it out there!
    Last edited: Oct 4, 2007
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