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


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


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!


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.

  • $\begingroup$ I believe the footnote 9 on page 169 (I'm using 2nd edition) answers your question in Edit 1, it's a matter of dimension and normalization. $\endgroup$
    – S. White
    Nov 30, 2021 at 7:38

2 Answers 2


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 $$ F=- \xi/2-\frac 1\beta \ln(1+ e^{-\beta \xi}) $$

  • $\begingroup$ You are right. I was't careful in my wording. What I meant to say is that the resulting partition function according to Altland and Simon (p. 169, eq. 4.35 if relevant) is $\mathcal{Z}={\rm det}\left[{\color{blue}\beta}\left(-i\hat{\omega}+\hat{H}-\mu\hat{N}\right)\right]^{-\zeta}$ where $\zeta=\pm 1$ depends on the statistics. This $\beta$ in blue is the one I am looking for. This makes $\mathcal{Z}$ dimensionless and gives a result that agrees with other calculations. $\endgroup$
    – eranreches
    Apr 3, 2020 at 17:47
  • $\begingroup$ The measure is always ill defined up to a regularization dependent constants. In particlular the Jacobian from Fourier transforming from a continuous interval $t\in [0,\beta]$ to a discrete set $n\in {\mathbb Z}$ is rather ill defined unless you restrict to a finite discrete set of points $t_n$ and an equally numerous finite set of $\omega_n$ and then take the limit. But then there is no good discrete definition of $\partial_\tau$ (the notorious Fermion doubling problem) $\endgroup$
    – mike stone
    Apr 3, 2020 at 18:01
  • $\begingroup$ Thanks for the follow up! The derivation of this path integral in Altland and Simon is based on taking the continuum limit of a discrete version. I understand there is a problem with the derivative in the fermionic case, but at least for bosons it should be fine. Restricting ourselves to the latter, is there a Jacobian contribution in the discrete version where the measure is 'better'-defined? This $\beta$ must come from somewhere. Maybe not from the measure, but where from then? $\endgroup$
    – eranreches
    Apr 3, 2020 at 18:24
  • $\begingroup$ I'd say that it comes from regulating the determinant. I've amended my answer. $\endgroup$
    – mike stone
    Apr 3, 2020 at 20:35

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.


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