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?


1 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.

  • $\begingroup$ 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. $\endgroup$
    – suresh
    Commented Mar 4, 2014 at 2:56

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