Let the Hamiltonian in one dimension be $H+z$, then I would like to evaluate $\det(H+z)$.
I have thought that if I know the function $Z(t) = \sum_{n>0}\exp(-tE_{n})$ I can use
$$\sum_{n} (z+E_{n})^{-s} \Gamma (s) = \int_{0}^{\infty}\mathrm{d}t\ Z(t)\exp(-zt)t^{s-1}$$
so i can use the zeta regularization to define the quotient
$$\log\det(H+z) - \log\det(H)= -\zeta'(0,z)+ \zeta'(0,0)$$
with $\sum_{n} (z+E_{n})^{-s}= \zeta (s,z)$. The derivative is taken on the variable $s$ ... at $s=0$.
However if $Z(t)$ is too hard to evaluate, can I make $Z(t)= \int \mathrm{d}x\mathrm{d}p\exp(-tp^{2}-tV(x))$ as an approximation to evaluate the functional determinant?? Thanks.
