Expressing the Klein Gordon Hamiltonian in terms of ladder operators

[maverick280857]

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]<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'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.)[/itex]

 

[maverick280857]

Ok I think I get it. I have to convert from the Schrödinger to the Heisenberg picture. Is that correct?

 

[Avodyne]

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.