Gaussian integration of path integrals

Question

I have a question regarding the Gaussian integration of path integrals with quadratic action.

Lets say the path integral has the standard form

$$ \int D[\Phi^\dagger,\Phi]e^{iS}$$ with a quadratic action of the form

$$ S = \int d\omega \, \Phi(\omega)^\dagger M(\omega) \Phi(\omega) $$ Here $\Phi^{(\dagger)}$ denote bosonic fields. I know that in the case of real bosonic fields, the matrix should fullfill a symmetry relation $M(\omega) = M^\intercal(-\omega)$. If this is not the case, one can "symmetrize" the matrix by replacing it with $$N(\omega) :=\tfrac{1}{2} (M(\omega)+M^\intercal(-\omega)).$$

I have 2 questions related to this procedure:

1. What exactly is the reason for doing this symmetrization? Is this to ensure that the gaussian integral converges?

2. What happens if the bosonic fields are complex? Would I have to check if $$M(\omega) = M^\dagger(-\omega)$$ holds and if this is not the case perform a similar symmetrization like this? $$ N(\omega) := \tfrac{1}{2}(M(\omega)+M^\dagger(-\omega)).$$

Answer 1

I have 2 questions...

(I answer two questions below.)

What exactly is the reason for doing this symmetrization?

In part for convenience, and in part so you can be reassured that the matrix is diagonalizable, and thus the integrations can be performed with formal ease.

The matrix $N$ can (hopefully) be written as $$ N = U D U^T\;, $$ where $D$ is diagonal.

Then the integrations reduce to $$ \int d\phi_1 d\phi_2 \ldots d\phi_N e^{i\sum_j \phi_j^2 d_j}\;, $$ where $\phi = U \Phi$, $d_j$ is a diagonal element and where I replaced the integral over $\omega$ with a sum over $j$.

(That's one question answered.)

Is this to ensure that the gaussian integral converges?

No. The integration could still diverge. For example the gaussian integral $$ \int_{-\infty}^{\infty} dx e^{-\alpha x^2}\;, $$ diverges when $\alpha < 0$.

(That's a second question...)

Answer 2

1. OP's symmetrization seems related to the fact that for a real field $\Phi(x)$, the Fourier transformed field satisfies $\widetilde{\Phi}(k)^{\ast}=\widetilde{\Phi}(-k)$.

2. Concerning convergence of complex Gaussian integrals, see e.g. this related Phys.SE post.