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

Question

Let's say we have a complex scalar field in a curved background whose action is: \begin{equation} S=-\int d^4x \sqrt{-g}\phi^\ast(\square_g+m^2) \phi \end{equation} For some purpose I want to calculate its partition function $Z$: \begin{equation} Z=\int \mathcal{D}\phi^\ast \mathcal{D}\phi\,e^{iS[\phi^\ast,\phi]} \end{equation} 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?

Answer 1

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.