Feynman propagator in curved spacetime?

Question

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 $(\nabla_{\mu}\nabla^{\mu}+m^2)^{-1}$?

Answer 1

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 $$ Z[J]\propto \int [d\phi]\ \exp \left(iS[\phi]+i\int d\omega\ \phi J\right) \tag{4} $$ 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]\ \exp \left(i\int d\omega\ \frac{\phi K\phi}{2} +i\int d\omega\ \phi J\right) \\ &\propto \int [d\phi]\ \exp \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]\ \exp \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.

Answer 2

Your notation is confusing because you have $\nabla$ and $\partial$'s . Remember how $\sqrt g$ plays with the covariant derivative. For a scalar field the Laplacian is $$ -\nabla^2\varphi\equiv - \nabla_\mu \nabla^\mu \varphi =-\frac 1{\sqrt g}\nabla_\mu g^{\mu \nu} \sqrt g \partial_\nu\varphi, $$
where we are using $$ \Gamma^{\nu}_{\nu\mu}= \frac 1{\sqrt g} \partial_\mu \sqrt g, $$ to supply the necessary connection terms for the second $\nabla_\mu$.

The Laplacian is self-adjoint wrt to the scalar product $$ \langle f, g\rangle = \int \sqrt g f^*g d^dx $$ The propagator is the inverse of the Laplacian $$ (- \nabla^2)^{-1}_{xx'}=G(x,x')= \sum_n \frac 1 {\lambda_n} \chi_n(x) \chi_n^*(x') $$ where the $\chi_n$ are a complete orthonormal set of eigenfunctions.