# Feynman propagator in curved spacetime?

It is known that in flat spacetime the Feynman propagator is the inverse of some operator. For instance, in $$\phi^4$$: $$$$\mathcal{L}=-\frac{1}{2}\phi(\partial^2+m^2)\phi-\frac{\lambda}{4!}\phi^4$$$$ The Feynman propagator is $$(\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 $$$$S=\int d^4x \sqrt{-g}\frac{-1}{2}\phi(\nabla_{\mu}\nabla^{\mu}+m^2)\phi$$$$

Which one should be the propagator?$$[\sqrt{-g}(\nabla_{\mu}\nabla^{\mu}+m^2)]^{-1}$$ or just $$(\nabla_{\mu}\nabla^{\mu}+m^2)^{-1}$$?

• @NiharKarve it is $\int d^4 x \sqrt{-g}$? Dec 6, 2020 at 14:36

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.