# Which operator corresponds to the Green function in QFT?

• A
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.

Demystifier
Gold Member
If ##\phi(x)## is a non-free interacting field, then it satisfies a nonlinear equation, e.g. ##(\Box+m^2)\phi(x)+\lambda\phi^3(x)=0##. Hence there is no linear operator that would satisfy your requirement.

• jordi
A. Neumaier
If ##\phi(x)## is a non-free interacting field, then it satisfies a nonlinear equation, e.g. ##(\Box+m^2)\phi(x)+\lambda\phi^3(x)=0##. Hence there is no linear operator that would satisfy your requirement.
In QFT, this equation is ill-defined.

• jordi
If ##\phi(x)## is a non-free interacting field, then it satisfies a nonlinear equation, e.g. ##(\Box+m^2)\phi(x)+\lambda\phi^3(x)=0##. Hence there is no linear operator that would satisfy your requirement.
It does not need to be linear, the operator could be non-linear.

Could one say that these functions are then correlations but not Green functions (except in the free case)?

• Demystifier
king vitamin
Gold Member
Could one say that these functions are then correlations but not Green functions (except in the free case)?
Pretty much. Sometimes people still call them "Green's functions" even though they satisfy a nonlinear equation like the one Demystifier gave, but this is an abuse of language since Green's functions are technically only defined for linear operators.

• Demystifier and jordi
A. Neumaier
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?
All Green's functions are inverses of the same operator, but with different bounday condiions.

• jordi

A. Neumaier
The boundary conditions depend on the complex integration path in the inversion formula. See here.

• jordi
Not sure if it is what you are asking for, but in "on gauge invariance and vacuum polarization" (1951) Schwinger defines and calculates an operator G with $$\left\langle x\right|G\left|x'\right\rangle =G(x,x')$$ where the right hand side is the Green's function of the corresponding equation (Dirac or Klein-Gordon).

• vanhees71 and jordi
Does it correspond to the free or to the "full-perturbative" equation?

You can find the exact solutions to G in the case of the free particle and a particle under the effect of some "classical" electromagnetic field configurations. I think that's the best that can be done non-perturbatively.

• vanhees71