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?


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.

  • $\begingroup$ There is a problem with the definition of the action that should be, for the harmonic oscillator, $\partial_\tau^2+\omega^2$.Please note that $\int D(\Psi^*,\Psi)e^{-S}\Psi^*\Psi=G$. $\endgroup$
    – Jon
    Nov 19, 2019 at 18:01
  • $\begingroup$ To the first point: This is not the case for the coherent state representation. Your propagator is in the real space representation, so Feynman Path integral. To the second point: I used this in the last line to conclude that it is 1. $\endgroup$
    – M_kaj
    Nov 19, 2019 at 18:06
  • $\begingroup$ Sorry, I thought you was convinced that the last line was wrong. Maybe, you should state more clearly your problem entering into details. $\endgroup$
    – Jon
    Nov 19, 2019 at 18:09
  • $\begingroup$ I edited it a bit more. I hope it gets more clear what i meant. $\endgroup$
    – M_kaj
    Nov 19, 2019 at 18:25

1 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.

  • 1
    $\begingroup$ Why should they be identical? What is the meaning of the derivative term? Especially considering that when we divide by Z we should get the expectation value of the Hamiltonian. $\endgroup$
    – M_kaj
    Nov 19, 2019 at 14:40
  • $\begingroup$ Indeed, you get the Hamiltonian. This is the term that goes down. The idea is to derive with respect to the integration limit. The formula you wrote in the simplest case is rather general. When you move from time to $\beta$ you get a Hamiltonian back. $\endgroup$
    – Jon
    Nov 19, 2019 at 15:18

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