Path integral in statistical physics and temperature derivatives

Question

I'm interested how to take derivatives w.r.t. to $\beta$, of the partition function.

Defining it in the usual way as $ Z =\text{Tr} \, e^{- \beta H}$ it is clear that one gets

$ \partial_\beta Z =- \text{Tr} ( H e^{- \beta H}) $

In the coherent state path integral representation for a single field the partition function has the form

$ Z = \int D(\Psi^*,\Psi) \exp\left\{ - \int_0^\beta d\tau \left( \Psi^* \partial_\tau \Psi + H(\Psi^*,\Psi) \right) \right\} $

With the boundary condition $\Psi(0)= \pm \Psi(\beta)$.

How is the derivative w.r.t. to $\beta$ performed in this case?

EDIT:

Let me specify my problem more and show you where my problem is.

The main point of confusion comes from the continuum representation.

Let us consider the harmonic oscillator and calculate the expectation value of the Hamiltonian. We see from statistical mechanics that we can get it by

$ <H> = \frac{-1}{Z} \partial_\beta Z $

The action for this case, in coherent state basis, is given by:

$ S= \int_0^\beta d\tau \Psi^*(\tau)(\partial_\tau + \omega) \Psi(\tau) = \int_0^\beta d\tau \Psi^*(\tau) G^{-1}(\tau) \Psi(\tau) $

Putting this into the expression for the average of the Hamiltonian and taking the derivative just to the exponential and not the measure one gets

$ \frac{1}{Z} \int D(\Psi^*,\Psi) \exp\left\{ - \int_0^\beta d\tau \left( \Psi^* G^{-1} \Psi \right) \right\} \partial_\beta \int_0^\beta d\tau \Psi^*(\tau) G^{-1}(\tau) \Psi(\tau) $ $ = \frac{1}{Z} \int D(\Psi^*,\Psi) \exp\left\{ - \int_0^\beta d\tau \left( \Psi^* G^{-1} \Psi \right) \right\} \Psi^*(\beta) G^{-1}(\beta) \Psi(\beta) $

The only terms which are not equal in numerator and denominator are the terms at $\beta$. Therefore they cancel everywhere except at the "time slice" $\beta$. Therefore one gets

$ = \frac{\int D(\Psi^*,\Psi)_\beta \exp\left\{ - \left( \Psi^* G^{-1} \Psi \right)(\beta) \right\} \Psi^*(\beta) G^{-1}(\beta) \Psi(\beta)}{\int D(\Psi^*,\Psi)_\beta \exp\left\{ - \left( \Psi^* G^{-1} \Psi \right) (\beta)\right\} } =1 $

In the last step the Gaussian integral's are performed.

This result is wrong. I can reproduce the right result in the time sliced version of the path integral, but i do not see how it works out in the continuum limit.

Answer 1

With respect to $\beta$ is $$ \partial_\beta Z= -\int D(\Psi^*,\Psi) \exp\left\{ - \int_0^\beta d\tau \left( \Psi^* \partial_\tau \Psi + H(\Psi^*,\Psi) \right) \right\}\left.\int d^Dx\left( \Psi^* \partial_\tau \Psi + H(\Psi^*,\Psi) \right)\right|_{\tau=\beta}. $$ Of course, the formulas are identical.