# Two-point correlation function in path integral formulation

1. Sep 6, 2014

### besprnt

Suppose that I have already calculated the two-point correlation function for a Lagrangian with no interations using the path integral formulation.
$$\langle \Omega | T[\phi(x)\phi(y)] | \Omega \rangle = \frac{ \int \mathcal{D}\phi \phi(x)\phi(y) \exp[iS_0] }{ \int \mathcal{D}\phi \exp[iS_0] }.$$
If I now add an interaction, such that the new action may be written as $S = S_0 + S_I,$ the new two-point correlation function is obviously
$$\langle \Omega | T[\phi(x)\phi(y)] | \Omega \rangle = \frac{ \int \mathcal{D}\phi \phi(x)\phi(y) \exp[i(S_0+S_I)] }{ \int \mathcal{D}\phi \exp[i(S_0+S_I)] }.$$
My question is:
Would I have to do the calculations again for the new expression? Or is there some short cut, such as factoring out $\exp[iS_I]$ so that the new expression is a product of the old expression and $\exp[iS_I]$?

2. Sep 6, 2014

### dextercioby

Sure, the new 2-point correlation functions will be some nonlinear function of the 'free one'. Therefore you need a series expansion for it, in the same way you need it for the generating functional(s).