# Tag Info

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.