Take the 2-minute tour ×  Consider the field integral for the
partition function of a free non-relativistic electron in a
condensed matter setting, i.e.  $$ Z = ∫D\bar\psi D\psi
\exp\left(-\sum_{k,ω} \bar\psi_{k,ω} (-iω + \frac{k^2}{2m} -
\mu) \psi_{k,ω}\right) $$  where the action is written in
Fourier representation and $\mu$ denotes the chemical potential.
Now, this integral is well known to be the determinant  $$ Z =
\det\left(β\left(-iω+\frac{k^2}{2m} - \mu\right)\right) $$
which is equal to the product of all eigenvalues of the
quadratic form in brackets.  But here is my problem:  How to
calculate a quadratic path integral if the quadratic form has
some eigenvalues that are equal to zero?  If the chemical
potential $\mu$ is positive, then all all momenta with
$\frac{k^2}{2m} = \mu$ (the Fermi surface) will represent an
eigenvalue equal to zero and would force the determinant to
become zero.  Of course, I could just drop the problematic
eigenvalues from the determinant and call it a day, but
unfortunate