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I am almost certain I have seen the answer to this question on this site before, but for the life of me I cannot find it after significant searching. If someone can located another question with the answer, I would gladly close this one.

Consider the path integral for the finite-temperature partition function of free bosons or fermions: $$ \mathcal{Z}_0 = \int \mathcal{D}[\bar{\psi}, \psi] \, e^{-S[\bar{\psi}, \psi]}, \quad S[\bar{\psi}, \psi] = \int_0^{\beta} d\tau \int d^d\vec{x} \, \bar{\psi} \Big[\partial_{\tau} - \frac{\hbar^2}{2m} \nabla^2 - \mu \Big] \psi , $$ where $\psi$ is a complex field for bosons and a Grassmann field for fermions. Commonly, the free energy is evaluated from such a path integral as follows: $$ \mathcal{Z} = \Big[ \det \Big( \partial_{\tau} - \frac{\hbar^2}{2m} \nabla^2 - \mu \Big) \Big]^{-\zeta}, $$ $$ \beta F = -\log \mathcal{Z} = \zeta \text{Tr} \log \Big( \partial_{\tau} - \frac{\hbar^2}{2m} \nabla^2 - \mu \Big) = \zeta \sum_{\vec{k},\omega_n} \log \Big(-i \omega_n + \frac{\hbar^2 \vec{k}^2}{2m} - \mu \Big) $$ There are then various integral tricks which can be performed to evaluate this free energy and get the desired free boson/fermion answer. (Of course, note that this result needs to be regularized, the sum is ill-defined as written, etc.)

Here's the question: how do I make sense of the dimensionful argument of the logarithm? Is there another unit of energy which has been lurking around that I have forgotten about, say the discretized temperature? My usual understanding of these sort of path integrals is that the continuum form written down is a shorthand, and the correct answer can be obtained by a properly discretized path integral along with a limiting procedure (after all, there shouldn't be anything singular about free bosons/fermions). So if one were to perform this path integral more carefully, how do the units inside the logarithm go away?

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The regularization issue that you mention is exactly where the solution of the problem lies. Path integrals are always formally divergent. In order to get a physically meaningful result, it is always necessary to take a ratio of two path integral expressions. For the case of a thermal partition function, what you normally want is the ratio of the partition function a temperature $T=1/\beta$ over its value at $T\rightarrow 0$. Taking this ratio $${\cal Z}_{{\rm reduced}}=\frac{{\cal Z}(\beta)}{{\cal Z}(\beta\rightarrow\infty)}$$ will give you an expression in which the argument of the logarithm in the numerator is rescaled by a constant (meaning $\beta$-independent) factor from the denominator with the same dimensions.

Note that how you actually normalize the partition function is irrelevant; the denominator of ${\cal Z}_{{\rm reduced}}$ is infinite, but it is constant. A thermodynamic potential like $F$ is only defined up to a constant anyway. To get an actual observable, you must take a derivative, to which the infinite constant will never contribute. Note that this is already apparent in the final equation in the question; the logarithm itself is ill-defined, but its derivative will have the correct dimensional behavior.

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  • $\begingroup$ I agree that the naive path integral is formally divergent, but certainly this arises due to some naive continuum limit in the action, correct? If I properly re-discretize the path integral for some large but finite number of imaginary time slices, then certainly everything should be finite. If I do this correctly, how are the units inside the logarithm fixed? $\endgroup$
    – Zack
    May 9, 2021 at 3:33
  • $\begingroup$ If I had to guess, I I think the issue may have something to do with the Jacobian of the transformation from imaginary time to Matsubara frequency space. The fields $\psi(\tau)$ and $\psi_n$ have slightly different units, so presumably there's a Jacobian to worry about in the measure of the path integral. Is there a good way to deal with this while working in the continuum, or is it best to just normalize as you write it? $\endgroup$
    – Zack
    May 9, 2021 at 3:35
  • $\begingroup$ @Zack If you discretize it (and I think you have to discretize both the $\vec{x}$ and $\beta$ integrations), the discretization scale $\epsilon$ will have units, and the scale inside the logarithm will be provided by a function that combines the Lagrange density and $\epsilon$. Exactly how you do the discretization of the derivatives will affect the precise form you get, and some versions end up looking a little nicer than others. You are right that if you do the transformation between continuous imaginary time and discrete frequencies, there is also a nontrivial Jacobian, which... $\endgroup$
    – Buzz
    May 9, 2021 at 4:21
  • $\begingroup$ @Zack effectively takes up the infinity. If you are interested in the issues of going back and forth between the discretized and continuum versions of the integral, I recommend the book Path Integrals And Quantum Processes by Mark S. Swanson, which works out a lot of these issues very carefully, including both normalized and unnormalized path integrals. If the book doesn't cover exactly the expressions you are considering, it comes extremely close. $\endgroup$
    – Buzz
    May 9, 2021 at 4:24

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