# Klein-Gordon operator on a time-ordered product

1. Nov 4, 2014

1. The problem statement, all variables and given/known data
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:

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

2. Relevant equations
$\partial_{0}\theta(x_{0}-y_{0})=\delta(x_{0}-y_{0})$

3. 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:

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

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

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

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...

2. Nov 9, 2014