Time ordered product & Green's function

by johnahn
Tags: function, green, ordered, product, time
johnahn is offline
Nov7-12, 05:45 PM
P: 8
The problem is showing

(□+m^2)<0| T(∅(x)∅(y)) |0> = -δ^4 (x-y)

I know that it is relavent to Green's function, but the problem is that it should be alternatively solved without any information of Green's function, and using equal time commutation relations.

Does Anyone know that?
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Einj is offline
Nov8-12, 01:53 AM
P: 230
The point is to use the Klein Gordon operator on the T-product. Note that it is defined as [itex]T(\phi(x)\phi(y))=\theta(x^0-y^0)\phi(x)\phi(y)+\theta(y^0-x^0)\phi(y)\phi(x)[/itex] which means that when you apply the spatial derivation it passes through the thetas and acts directly on the fields. The time derivative needs a little more attention. You can perform an explicit calculation:

\partial_0 T(\phi(x)\phi(y)) &= \theta(x^0-y^0)\dot{\phi}(x)\phi(y)+\delta(x^0-y^0)\phi(x)\phi(y)+\theta(y^0-x^0)\phi(y)\dot{\phi}(x)-\delta(x^0-y^0)\phi(y)\phi(x)\\

where I have used the commutation relation at equal time and the fact that the delta is the derivative of the theta.
To have the KG operator you have to perform another time derivative and thus obtain, with the same procedure:

$$\partial_0^2 T(\phi(x)\phi(y))=T(\ddot{\phi}(x)\phi(y))+\delta(x^0-y^0)[\dot{\phi}(x),\phi(y)]=T(\ddot{\phi}(x)\phi(y))-\delta^4(x-y)$$

where I have used again the commutation realtion at equal time (because of the delta).

Now, every other term in KG operator pass through the thetas and so you have:

$$(\Box+m^2)T(\phi(x)\phi(y))=T((\Box+m^2)\phi(x) \phi(y))-\delta^4(x-y)=-\delta^4(x-y)$$

because [itex]\phi[/itex] is solution of the KG equation.

I hope this was the answer you were looking for.

ehms is offline
Nov12-13, 09:34 AM
P: 1
Thanks a lot Einj !

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