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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.


I know this is a year old question, but I am going to attempt an answer. As far as I can tell, this is not really a caveat. The reason for this is that I can always set the overall phase of the quark mass determinant to be zero with a chiral U(1) transformation. For a discussion of this see for example the chapter on theta vacua in Weinberg's QFT book. The ...

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