# Units in the nonrelativistic free particle path integral

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?

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.

• 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?
– Zack
May 9, 2021 at 3:33
• 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?
– Zack
May 9, 2021 at 3:35
• @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...
– Buzz
May 9, 2021 at 4:21
• @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.
– Buzz
May 9, 2021 at 4:24