Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Haha it seems you and I are studying the same stuff almost simultaneously... – Faq 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
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.