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Annihilation Operator Hermitian?

  1. Apr 7, 2007 #1
    1. The problem statement, all variables and given/known data

    How do I show that the annihilation operator [tex] \hat{a} [/tex] is hermitian WITHOUT explicitly using the condition where an operator X is hermitian if its adjoint is also X ie. [tex] X=X^+ [/tex]

    2. Relevant equations

    none.

    3. The attempt at a solution

    I could show [tex] \hat{a} \hat{x} \neq \hat{x} \hat{a} [/tex] where [tex] \hat{x} [/tex] is the position operator, but that only shows non-hermiticity for that one operator...
    Is there a more elegant way to show non-hermiticity simply?
     
  2. jcsd
  3. Apr 7, 2007 #2

    Dick

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    Commuting with operators doesn't have much to do with hermiticity. In the last problem you posted you found an eigenstate of the 'a' operator. What were it's eigenvalues like? Hermitian operators have real eigenvalues.
     
  4. Apr 7, 2007 #3
    oh yes offcourse - the eigenvalue was real...stupid me. :biggrin:
     
  5. Apr 7, 2007 #4

    Dick

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    The point is that the operator has eigenvalues that AREN'T real. Hope you misspoke.
     
  6. Apr 7, 2007 #5
    yeah sorry I was meant to say: "...the eigenvalues are meant to be real..."

    And the question did give a complex eigenvalue...
     
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