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Formal integration of Heissenberg equation?

  1. Feb 8, 2013 #1

    I am reading a paper and can't get the same answer they do, they have;

    H= -i\hbar \sum_{\lambda} \int dK (a_{(k,\lambda)} g_{(1;K,\lambda)} \sigma_{1}^{+} e^{-i(\omega_{k}- \omega)t} - H.C )

    Then states they do a formal integration of the Heissenberg equation for [tex] a_{k,\lambda}(t) [/tex] using the above Hamiltonian, they get;

    a_{k,\lambda}(t) = a_{k,\lambda}(t_{0}) + \int_{t_{0}}^{t} dt^{'} g^{*} \sigma_{1}^{-}(t^{'}) e^{-i(\omega_{k}- \omega)t^{'}}

    Could anyone help me as to how they get this answer from the Heissenberg equation...you be greatly appreciated!!

  2. jcsd
  3. Feb 9, 2013 #2
    Does [itex] a_{(k,\lambda)}[/itex] and [itex]a^\dagger_{(k,\lambda)}[/itex] follow some kind of orthogonality relation?
    In fact, you get the correct result if [itex]a_{(k,\lambda)}\cdot a_{(q,\lambda')}=0[/itex] and [itex]a_{(k,\lambda)}\cdot a^\dagger_{(q,\lambda')}=\delta(k-q)\delta_{\lambda\lambda'}[/itex].

    However, I'm just guessing.
  4. Feb 9, 2013 #3


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    It would be much easier to help you, if you'd give the reference to the paper!
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