Quantum Field Theory – Path Integral as a Functional Determinant Explained

Question

In Peskin and Schroeder on pg. 304, the authors call the fermionic path integral:
\begin{equation}
\int {\cal D} \bar{\psi} {\cal D} \psi \exp \left[ i \int \,d^4x \bar{\psi} ( i \gamma_\mu D^\mu – m ) \psi \right]
\end{equation}
a functional determinant,
\begin{equation}
\det \left( i \gamma_\mu D^\mu – m \right).
\end{equation}
I've never heard this way of thinking about it. Why would the generating functional be a functional determinant?

Answer 1

This is because the path integral ${\cal Z}$ is an infinite-dimensional version of a Grassmann-odd Gaussian integral

$$\int \!\mathrm{d}^n \bar{\theta} ~\mathrm{d}^n\theta ~e^{\sum_{i,j=1}^n\bar{\theta}_i ~M^i{}_j ~\theta^j}~\propto~\det(M), $$

where the indices $i,j$ can be interpreted as DeWitt's condensed notation.