We have a Lagrangian $$\mathcal{L}=\frac{1}{2} \partial_\mu \Phi \partial^\mu \Phi - \frac{1}{2} M^2 \Phi^2- \frac{\lambda}{4}\phi^2 \Phi^2 - \frac{g}{2} \Phi \phi^2+\cdots $$ where $\Phi$ denotes a real scalar field of a heavy particle and $\phi$ denotes a light particle. Now we want to integrate out the heavy particles such that only light particle configurations remain or mathematically $$\displaystyle \int D\Phi \, e^{iS[\phi,\Phi]} \sim e^{iS_{eff}[\phi]}.$$ We also know that \begin{multline*} \int D\Phi \, \exp \left \{-\frac{1}{2} \int d^4x\, d^4y \, \, \Phi(x) G(x,y) \Phi(y) \right. \left. - \int d^4x H(x)\Phi(x) \right \} \\ = \left( \det \frac{G(x,y)}{2 \pi} \right)^{-\frac{1}{2}} \exp \left \{ \frac{1}{2} \int d^4x \, d^4y H(x) G^{-1}(x,y) H(y) \right \}, \end{multline*} for bosonic fields. Now we can rewrite our Lagrangian and obtain $$G(x,y) =\delta^{(4)}(x-y) \left[ \partial_{\mu} \partial^{\mu}+M^2 +\frac{1}{2}g\cdot \phi^2(x) \right]$$ and $$H(x)=\frac{g}{2}\phi (x)^2.$$ Furthermore we know that $$\int d^4z G(x,z) G^{-1}(z,y)=\delta^{(4)}(x-y).$$ If we plug $G(x,y)$ in our relation we obtain \begin{align} \int \delta^{(4)}(x-z) \left[ \partial_{\mu} \partial^{\mu}+M^2 +\frac{1}{2}g\phi^2(z) \right]G^{-1}(z,y) &= \left[ \partial_{\mu} \partial^{\mu}+M^2 +\frac{1}{2}g\phi^2(x) \right]G^{-1}(x,y) \\ &=\delta^{(4)}(x-y). \end{align} My question is how can we determine $G^{-1}(x,y)$? In the free theory we could work in the Fourier-space. But now we have an explicit $x$-dependence in our operator $G(x,y)$.
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$\begingroup$ I'm just guessing, but I would try the Fourier method anyway. Consider the Fourier transform of $\phi$, make the necessary manipulations, then maybe try to make a series expansion in $\phi$ at the end to try to undo the Fourier transforms term by term. I expect something like this should happen because I expect that integrating out that interaction term will lead to infinitely many interaction terms with even powers of $\phi$ $\endgroup$– Níckolas AlvesCommented Apr 27 at 19:24
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It is not important to calculate the explicit form of $G^{-1}(x,y)$. Thus write the operator $ M^2+\partial_{\mu} \partial^{\mu}+\frac{1}{2}g \phi(x)^2 $ in the denominator with $\displaystyle G^{-1}(x,y) = \frac{\delta^{(4)}(x-y)}{M^2+\partial_{\mu} \partial^{\mu}+\frac{1}{2}g \phi(x)^2}$. As $M^2 \gg \partial_{\mu} \partial^{\mu}+\frac{1}{2}g \phi(x)^2 $ we can Taylor expand the denominator to considered order $G^{-1}(x,y)= \delta^{(4)}(x-y) \left[ \frac{1}{M^2} -\frac{1}{M^4} (\partial_{\mu} \partial^{\mu}+\frac{1}{2}g \phi(x)^2)+ \dots \right]$ and now plug this in our former relation and thus only light particle configurations remain in the path integral.