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?
Path integral as a functional determinant
JeffDror
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