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A Commutator of annihilation operator

  1. Jun 5, 2014 #1
    Hi, everybody:

    I encountered a problem when I am reading a book.
    It's about the atom-photon interaction.
    Let the Hamiltonian for the free photons be
    [itex]H_0=\hbar \omega(a^{\dagger}a+\frac{1}{2})[/itex].
    so the commutator of the annihilation operator and the Hamiltonian is
    [itex][a,H_0]=\hbar\omega a [/itex]
    and I have no problem with that.

    In the book I am reading is a commutation as
    [itex][a,H_{I}]=\frac{\partial}{\partial a^{\dagger}}H_I[/itex]
    This is the thing I do not understand.
    [itex]H_{I}[/itex] is the Hamiltonian that describing the interaction between the atom and the photon,
    and it is a little tedious to type it here. However, If I change [itex]H_I[/itex] into [itex]H_0[/itex] in the second commutator, it is correct.

    My question is, what's the condition for the second commutator stands?
    Is it correct for all the Hamiltonian, or just the special one such as [itex]H_I[/itex]?
    If it is just for some special Hamiltonian, What is the key properties of such Hamiltonian to satisfy that commutator?

    Thanks
    Robert
     
    Last edited: Jun 5, 2014
  2. jcsd
  3. Jun 5, 2014 #2

    strangerep

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    Science Advisor

    Let ##f(z)## be any complex-analytic function of ##z##. Then it can be proven that $$[a , f(a^\dagger) ] ~=~ \frac{\partial f(a^\dagger)}{\partial a^\dagger} ~.$$(Actually, there might also be some factors of ##i \hbar## depending on your convention for the commutation relations.)

    It's fairly easy to prove this by induction if ##f## is a polynomial. For more general analytic functions, one must work a bit harder.

    This can be extended to functions like ##f(a, a^\dagger)## provided you're careful with the operator ordering.
     
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