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Expressing the Klein Gordon Hamiltonian in terms of ladder operators

  1. Jun 3, 2009 #1
    Hi everyone

    I'm trying to express each term of the Hamiltonian

    [itex]H = \int d^{3}x \frac{1}{2}\left[\Pi^2 + (\nabla \Phi)^2 + m^2\Phi^2\right][/tex]

    in terms of the ladder operators [itex]a(p)[/itex] and [itex]a^{\dagger}(p)[/itex].

    This is what I get for the first term

    [tex]\int d^{3}x \frac{E_{p}}{2}\left[a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)-a(p)a(-p)-a^{\dagger}(p)a^{\dagger}(-p)\right][/tex]

    whereas the book I'm reading from says

    [tex]\int d^{3}x \frac{E_p}{2}\left[-a(p)a(-p)e^{-2iE_{p}t} + a(p)a^{\dagger}(p) + a^{\dagger}(p)a(p)-a^{\dagger}(p)a^{\dagger}(-p)e^{-2iE_{p}t}\right][/tex]

    Is this because the time dependence must be explicitly accounted for? It so happens that the explicit time dependence goes away through the other two terms...but is my own computation correct?

    Thanks.

    (PS -- This is not homework.)
     
    Last edited: Jun 3, 2009
  2. jcsd
  3. Jun 3, 2009 #2
    Ok I think I get it. I have to convert from the Schrodinger to the Heisenberg picture. Is that correct?
     
  4. Jun 3, 2009 #3

    Avodyne

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    Science Advisor

    You can do it in either picture. The H-picture is more commonly used in field theory, and then you get those time-dep phases.
     
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