Klein-Gordon operator on a time-ordered product

[Dixanadu]

Homework Statement

Hey guys,

So here's the problem I'm faced with. I have to show that

(\Box + m^{2})<|T(\phi(x)\phi^{\dagger}(y))|>=-i\delta^{(4)}(x-y),

by acting with the quabla (\Box) operator on the following:

T(\phi(x)\phi^{\dagger}(y))=\theta(x_{0}-y_{0})\phi(x)\phi^{\dagger}(y)+\theta(y_{0}-x_{0})\phi^{\dagger}(y)\phi(x)

Homework Equations

\partial_{0}\theta(x_{0}-y_{0})=\delta(x_{0}-y_{0})

The Attempt at a Solution

So I've split the quabla into its time and spatial derivatives: \Box = \partial_{0}^{2}-\nabla^{2} and I'm applying the time derivative first, using the product rule:

<br /> \partial_{0}T(\phi(x)\phi^{\dagger}(y))<br /> =\delta(x_{0}-y_{0})\phi(x)\phi^{\dagger}(y) +\theta(x_{0}-y_{0})\dot{\phi}(x)\phi^{\dagger}(y)\\<br /> +\theta(x_{0}-y_{0})\phi(x)\dot{\phi}^{\dagger}(y) -\delta(x_{0}-y_{0})\phi^{\dagger}(y)\phi(x)<br /> +\theta(y_{0}-x_{0})\dot{\phi}^{\dagger}(y)\phi(x) +\theta(y_{0}-x_{0})\phi^{\dagger}(y)\dot{\phi}(x)<br />

However, the two delta terms vanish as the commutator of two fields is 0. so I'm left with

<br /> \partial_{0}T(\phi(x)\phi^{\dagger}(y))<br /> =\theta(x_{0}-y_{0})\dot{\phi}(x)\phi^{\dagger}(y) +\theta(x_{0}-y_{0})\phi(x)\dot{\phi}^{\dagger}(y)\\<br /> +\theta(y_{0}-x_{0})\dot{\phi}^{\dagger}(y)\phi(x) +\theta(y_{0}-x_{0})\phi^{\dagger}(y)\dot{\phi}(x)<br />

At this point I'm meant to be using the equal-time commutation relation: [\phi(x),\dot{\phi}^{\dagger}(y)] = i\delta^{(3)}(x-y) but all my signs are positive...so what do I do?

Thanks guys...

 

[Greg Bernhardt]

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?