Units in the nonrelativistic free particle path integral

Question

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?

Answer 1

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.