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

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.

share|cite|improve this question
Note that the determinant is the product of eigenvalues. This, first of all, represents some challenge because the number of eigenvalues is infinite and may even include a continuous part of the spectrum. Second, the first eigenvalues near the "ground state(s)", are also special and affect your approximation which is really a semiclassical approximation of a sort. So you may want to try to remove the neighborhood of the phase space near the ground state from your approximate $Z(t)$ integral and treat it separately and discretely. – Luboš Motl Apr 7 '12 at 20:28

Your Answer


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

Browse other questions tagged or ask your own question.