Expressing the Klein Gordon Hamiltonian in terms of ladder operators

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SUMMARY

The discussion focuses on expressing the Klein-Gordon Hamiltonian in terms of ladder operators \(a(p)\) and \(a^{\dagger}(p)\). The user presents their formulation of the Hamiltonian, highlighting discrepancies with a reference book's version, particularly regarding the treatment of time dependence in the terms. The conclusion reached is that transitioning from the Schrödinger picture to the Heisenberg picture is necessary to account for explicit time dependence in the Hamiltonian's expression.

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maverick280857
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Hi everyone

I'm trying to express each term of the Hamiltonian

H = \int d^{3}x \frac{1}{2}\left[\Pi^2 + (\nabla \Phi)^2 + m^2\Phi^2\right][/tex]<br /> <br /> in terms of the ladder operators a(p) and a^{\dagger}(p).<br /> <br /> This is what I get for the first term<br /> <br /> \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]<br /> <br /> whereas the book I&#039;m reading from says<br /> <br /> \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]<br /> <br /> 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?<br /> <br /> Thanks.<br /> <br /> (PS -- This is not homework.)
 
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Ok I think I get it. I have to convert from the Schrödinger to the Heisenberg picture. Is that correct?
 
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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