# Free Particle Path Integral Matsubara Frequency

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I am trying to calculate $$Z = \int\limits_{\phi(\beta) = \phi(0) =0} D \phi\ e^{-\frac{1}{2} \int_0^{\beta} d\tau \dot{\phi}^2}$$ without transforming it to the Matsubara frequency space, I can show that $Z = \sqrt{\frac{1}{2\pi \beta}}$. However, I have a problem in obtaining the same result in the Matsubara frequency space: $$\phi (\tau) = \frac{1}{\sqrt{\beta}} \left( \sum_{n} \phi_n \ e^{i\omega_n\tau} \right),$$ with $\sum_n \phi_n =0, \omega_n = \frac{2\pi n}{\beta}$. And $$Z = \int \prod_n D\phi_n\ \delta\left(\sum_n \phi_n\right)\ e^{-\frac{1}{2} \sum_n \phi_n \phi_{-n} \omega_n^2 }$$ which, I think, vanishes.

I guess the problem lies in the measure. Any comments?

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user L. Su
retagged Jul 11, 2014
Note that in Matsubara frequency space your integrand is a Gaussian. Hence you only need to evaluate the determinant of the bilinear, i.e., the product of its eigenvalues. This is divergent, but can be evaluated with a regulator.

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user Robin Ekman
@RobinEkman Correct me if I am wrong. Roughly speaking, $Z \propto \prod_n \frac{1}{n^2} \rightarrow 0$. Why do we need a regulator here?

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user L. Su
Shouldn't it be $\propto \beta^{2n}/n^2$ \to \infty?

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user Robin Ekman
Then it should be written as $\propto {\beta}^{2n}/(n!)^2 \rightarrow 0$.

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user L. Su
Yes, that's right.

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user Robin Ekman
On the other hand there is a zero eigenvalue.

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user Robin Ekman
For $n=0$, we must have $\phi_0=0$.

This post imported from StackExchange Physics at 2014-07-11 17:23 (UCT), posted by SE-user L. Su

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