- #1
jordi
- 197
- 14
The Feynman propagator:
$$D_{F}(x,y) = <0|T\{\phi_{0}(x) \phi_{0}(y)\}|0> $$
is the Green's function of the operator (except maybe for a constant):
$$ (\Box + m^2)$$
In other words:
$$ (\Box + m^2) D_{F}(x,y) = - i \hbar \delta^{4}(x-y)$$
My question is:
Which is the operator that corresponds to:
$$<\Omega |T\{\phi(x) \phi(y)\}|\Omega> $$
being the Green's function (understood as "operator times Green's function equals to delta") in QFT?
I assume that the answer to my question has something to do with the Schwinger-Dyson equations, but I cannot find it out.
$$D_{F}(x,y) = <0|T\{\phi_{0}(x) \phi_{0}(y)\}|0> $$
is the Green's function of the operator (except maybe for a constant):
$$ (\Box + m^2)$$
In other words:
$$ (\Box + m^2) D_{F}(x,y) = - i \hbar \delta^{4}(x-y)$$
My question is:
Which is the operator that corresponds to:
$$<\Omega |T\{\phi(x) \phi(y)\}|\Omega> $$
being the Green's function (understood as "operator times Green's function equals to delta") in QFT?
I assume that the answer to my question has something to do with the Schwinger-Dyson equations, but I cannot find it out.