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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?

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Haha it seems you and I are studying the same stuff almost simultaneously... –  Love Learning Mar 4 '14 at 1:17
@LoveLearning: Haha kind of, but I'm not really studying this stuff now so its not really a coincidence. I just stumbled upon this when trying to answer your earlier question. –  JeffDror Mar 4 '14 at 1:19

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up vote 5 down vote accepted

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.

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Just to add to @Qmechanic's perfect answer. You can also see the discussion in the appendix of Ramond's QFT book. It is also interesting that the integral for real Grassmann variables gives the Pfaffian. –  suresh Mar 4 '14 at 2:56

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