How to deal with path integral in curved space-time for a free scalar field?

Let's say we have a complex scalar field in a curved background whose action is: $$$$S=-\int d^4x \sqrt{-g}\phi^\ast(\square_g+m^2) \phi$$$$ For some purpose I want to calculate its partition function $$Z$$: $$$$Z=\int \mathcal{D}\phi^\ast \mathcal{D}\phi\,e^{iS[\phi^\ast,\phi]}$$$$ I know that in Minkowski space-time this partition function is just $$\propto [\det(\square+m^2)]^{-1}$$. But how should I do with the square root of the determinant of the metric? I am not sure of what I do next:

To evaluate this integral I split it into an infinite number of integrals: \begin{align} Z &\propto \prod_{i}\int d\phi^\ast_id\phi_i\,e^{-ia^4\sqrt{-g_i}\phi^\ast_i [(\square_g+m^2)\phi]_i} \\ &\propto\prod_i \frac{1}{\sqrt{-g_i} \lambda_i} =: \frac{1}{\det(\square_g+m^2)}\prod_i \frac{1}{\sqrt{-g_i}} \end{align} Is this a good way of doing this?

Simply put, if you want to perform a Gaussian integral, you need to redefine the operator to include the factor $$\sqrt{-g}$$. That is, for an integral $$\int dx\,\exp\left(-x^\dagger A x\right) = \frac{1}{\sqrt{\det A}}$$

The operator $$A$$ now includes the factor $$\sqrt{-g}$$, that is $$A = \sqrt{-g}(\square + m^2)$$

You should now compute the determinant of that new object. However functional determinants make little sense alone, you will need to regularize it against something otherwise the thing just diverges. One usually takes what you are computing, that is the free part, as the reference:

$$\log\frac{\det A_{\rm full}}{\det A_{\rm free}}$$ The object above for example a common term appearing in effective actions and makes sense, but needs still renormalization.

P.D. In the case of complex Gaussian integral on $$\phi,\phi^*$$, there might be factors of two you have be careful about.

• I figured out that since $\sqrt{-g}$ is just the determinant of the jacobian of a coordinate transformation, this would mean that the result of the path integral depends on the coordinate system I use. But since there is an integral in the exponential, the result should not depend upon my coordinate transformation. So shouldn't I change the measure of the path integral so the result doesn't depend on this transformation? May 26, 2021 at 16:48
• The expression for $\sqrt{-g}$ is coordinate dependent, however it appears within a functional determinant, the result does not depend on your choice of coord. The integral you have in the exponential does not change anything, you can imagine you are summing over several points in a lattice, with a modified operator as I wrote above. May 26, 2021 at 17:28
• You are writing the same thing by the way $\prod \sqrt{-g_i}$ is just the determinant of for a function which is diagonal in $i$, which is the case. May 26, 2021 at 17:32
• It is a bit out of the scope of my question but, if $\det(\sqrt{-g})$ is indeed coordinate independent, does this mean that it is a constant with respect to an eventual summation over some metrics? May 26, 2021 at 18:46
• As you write it it should be understood as $\det(\sqrt{-g}\delta(x-y))$. I would guess no, it is in principle metric dependent. Coordinate independence does not imply in any way metric independent. How are you defining a sum over metrics? May 26, 2021 at 19:21