Measure in the Fourier Representation of the Coherent States Path Integral

Question

The problem I have arises in the context of condensed matter physics. I am largely following chapter 4 about functional integration in the book by Altland and Simon. Consider the coherent states path integral

$$\mathcal{Z}=\int\mathcal{D}\left[\bar{\psi}\left(\tau\right),\psi\left(\tau\right)\right]e^{-S\left[\bar{\psi}\left(\tau\right),\psi\left(\tau\right)\right]}$$

with the action of a non-interacting system (and also diagonal for simplicity)

$$S\left[\bar{\psi}\left(\tau\right),\psi\left(\tau\right)\right]=\int_{0}^{\beta}{\rm d}\tau\sum_{i}\bar{\psi}_{i}\left(\tau\right)\left(\partial_{\tau}+\xi_{i}\right)\psi_{i}\left(\tau\right)$$

Here $\xi_{i}=\epsilon_{i}-\mu$ where $\left\{\epsilon_{i}\right\}_{i}$ is the spectrum of the system and $\mu$ is the chemical potential. It is then common to make a change-of-variables into Fourier modes

$$\psi_{i}\left(\tau\right)=\dfrac{1}{\sqrt{\beta}}\sum_{\omega_{n}}\psi_{in}e^{-i\omega_{n}\tau},\qquad\qquad\qquad\psi_{in}=\dfrac{1}{\sqrt{\beta}}\int_{0}^{\beta}{\rm d}\tau\psi_{i}\left(\tau\right)e^{i\omega_{n}\tau}$$

and similarly for $\bar{\psi}$. The action in this new basis is

$$S\left[\bar{\psi}_{n},\psi_{n}\right]=\sum_{n}\sum_{i}\bar{\psi}_{in}\left(-i\omega_{n}+\xi_{i}\right)\psi_{in}$$

However, I am not sure about the measure. Naively, it seems like the transformations is unitary so that the Jacobian must be unity. However, the units don't work out. Moreover, if the measure is simply $\mathcal{D}\left[\bar{\psi}_{n},\psi_{n}\right]=\prod_{i,n}{\rm d}\bar{\psi}_{in}{\rm d}\psi_{in}$, then $\beta$ disappeared from the problem, which makes no sense.

I would appreciate any thoughts about this matter!

EDIT 1:

Thanks to mike stone. Let me be more precise about this $\beta$ I am looking for. Indeed the Matsubara frequencies depend on $\beta$ so it doesn't completely disappear, but according to Altland and Simon (p. 169, eq. 4.35)

$$\mathcal{Z}={\rm det}\left[{\color{blue}\beta}\left(-i\hat{\omega}+\hat{H}-\mu\hat{N}\right)\right]^{-\zeta}$$

It is not clear to me where does this $\beta$ in blue come from. Yet, I understand and expect it to be there because of the units. My guess is that there is a non-trivial Jacobian I am missing somewhere.

Answer 1

Why does $\beta$ drop out? Surely the allowed (Matsubara) frequencies are $\omega_n=(2n+1)\pi/\beta$? This makes $$ e^{i(2n+1)\tau\pi/\beta} $$ antperiodic as it should be. After doing the Grassman integral, the PI becomes the functional determinant
$$ {\rm det}[\partial_\tau+\xi]= \prod_{n\in {\mathbb Z}}\left(\frac {-i(2n+1)\pi}{\beta}+\xi\right) $$ We can regulate the divergent infinite product by dividing by the same expression with $\xi=0$ to get $$ \prod_{n\in {\mathbb Z}}\left(1+\frac{i\beta\xi}{(2n+1)\pi}\right)= \cosh(\beta\xi/2)= e^{\ln\cosh(\beta\xi/2)}=e^{-\beta F} $$ where $$ F=- \xi/2-\frac 1\beta \ln(1+ e^{-\beta \xi}) $$

Answer 2

I was a student at the university of Alexander Altland and this book is since more than 4 years my bible.

You are indeed correct with your assumption, that the factor of beta comes from the Jacobian. In order to understand this in very detail you must understand that the change of coordinates inside the functional integral, is not a Fourier-transformation in the traditional sense (of functions), but a discrete Fourier-transform (for the integration-variables). In this spirit, the r.h.s. of the defining equation for $\psi_{i n}$ is actually to be interpreted as a sum, although the continuum-notation is much more appealing even though it is nonsense in the fermionic case (and also in the bosonic case, if saddle-point-approximations fail). You should also notice, that this Fourier-transform is not unitary, but just unitary up to a factor. (This is the factor that you are looking for.)

If you are still interested I can guide you through the calculation.