Feynman propagator in curved spacetime? It is known that in flat spacetime the Feynman propagator is the inverse of some operator. For instance, in $\phi^4$:
\begin{equation}
\mathcal{L}=-\frac{1}{2}\phi(\partial^2+m^2)\phi-\frac{\lambda}{4!}\phi^4
\end{equation}
The Feynman propagator is $(\partial^2+m^2)^{-1}$.
But in curved spacetime, the story confuses me, because of the appearence of $\sqrt{-g}$. For example, I have a theory
\begin{equation}
S=\int d^4x \sqrt{-g}\frac{-1}{2}\phi(\nabla_{\mu}\nabla^{\mu}+m^2)\phi
\end{equation}
Which one should be the propagator?$[\sqrt{-g}(\nabla_{\mu}\nabla^{\mu}+m^2)]^{-1} $ or just $(\nabla_{\mu}\nabla^{\mu}+m^2)^{-1}$?
 A: The perturbation series can be organized in different ways. This answer derives one way.
Recall how the Feynman-diagram series is derived from the functional integral formulation. Use the abbreviation
$$
 d\omega\equiv d^4x\ \sqrt{|g|}
$$
and write
$$
 S = \int d\omega\ \left(\frac{\phi K\phi}{2}+V(\phi)\right)
\tag{1}
$$
with
$$
 K\equiv -(\nabla_\mu\nabla^\mu+m^2)
\tag{2}
$$
and
$$
 V(\phi)\equiv -\frac{\lambda}{4!}\phi^4.
\tag{3}
$$
Correlation functions are generated by taking derivatives of
$$
 Z[J]\propto
 \int [d\phi]\ \left(iS[\phi]+i\int d\omega\ \phi J\right)
\tag{4}
$$
with respect to the source $J(x)$. More explicitly, applying
$$
 \frac{-i}{\sqrt{|g|}}
  \frac{\delta}{\delta J(x)}
\tag{5}
$$
to $Z[J]$ is the same as inserting a factor of $\phi(x)$ in the integrand. The small-coupling expansion comes from rewriting (4) as
$$
 Z[J]\propto
 \exp\left(i\int d\omega\ V\left(\frac{-i}{\sqrt{|g|}}
  \frac{\delta}{\delta J(x)}\right)\right)
 Z_0[J]
\tag{6}
$$
and then expanding in powers of $V$, with
\begin{align*}
 Z_0[J]
 &\propto
 \int [d\phi]\ \left(i\int d\omega\ \frac{\phi K\phi}{2}
  +i\int d\omega\ \phi J\right)
\\
 &\propto
 \int [d\phi]\ \left(i\int d\omega\ 
  \frac{(\phi+K^{-1}J) K(\phi+K^{-1}J)}{2}
  -i\int d\omega\ \frac{J K^{-1}J}{2}\right)
\\
 &\propto
 \int [d\phi]\ \left(-i\int d\omega\ \frac{J K^{-1}J}{2}\right).
\tag{7}
\end{align*}
This shows that we can account for the required factors of $\sqrt{|g|}$ by

*

*writing the measure of each spacetime integral as $d\omega\equiv d^4x\ \sqrt{|g|}$ instead of $d^4x$,


*using $K^{-1}=-(\nabla_\mu\nabla^\mu+m^2)^{-1}$ as the propagator, without any factor of $\sqrt{|g|}$,


*including an extra factor of $\left(\sqrt{|g|}\right)^{-4}$ for each vertex (from equation (6)).
This clearly isn't the only way to organize the series, and it may or may not be the most convenient way. The two-point correlation function is not equal to $K^{-1}$ in this approach, because applying two factors of (5) to $Z[J]$ gives $|g|^{-1/2}K^{-1}+O(\lambda)$. If we want the propagator in the perturbation series to be the same as the $\lambda=0$ version of the two-point correlation functions, then we can organize things differently.
