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Operator equation

  1. Sep 12, 2006 #1
    message deleted
    Last edited: Sep 13, 2006
  2. jcsd
  3. Sep 12, 2006 #2


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    Proof by induction (on m) should work.

    If m= 1, then
    [tex]a^{m}(a\dagger)^{m}= aa\dagger= n+1[/tex]
    which is the correct formula.

    Now, assume that, for some k,
    and look at
    [tex]a^{k+1}(a\dagger)^{k+1}= a(a^{k}(a\dagger)^k) a\dagger[/tex]
    [tex]= a((n+1)(n+2)...(n+k))a\dagger[/tex]
    and apply the commutativity relation to that.
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