Getting units right when computing the effective action

Question

In QFT in Euclidean signature, the one-loop effective action is given by $$\Gamma[\Phi] = S[\Phi] + \frac{1}{2} \mathrm{STr}\log S^{(2)}, \tag{1}$$ where $S[\Phi]$ is the theory's classical action, $\Phi$ stands for the collection of all fields of the theory, $\mathrm{STr}$ is the supertrace (which is analogous to the trace, but adds a minus sign for fermionic variables), and $S^{(2)}$ is the Hessian of the action with respect to the fields. Eq. (1) is given, for example, in Eq. (15) of arXiv: 1108.4449 [hep-ph] (from now on I'll refer to this paper as "Braun").

Many QFT books will deal with this expression, or equivalent ones, in the case of scalar fields. Weinberg writes the Lorentzian version on Sec. 16.2 (Vol. 2), Peskin & Schroeder do the same on Chap. 11. However, I'm confused about what happens when dealing with multiple fields with different mass dimensions. For example, this might be the case in Yukawa theory.

In this situations, it seems to me that different entries of $S^{(2)}$ will have different dimensions. For example, we get the entry $\frac{\delta^2 S}{\delta \phi \delta \phi}$, which does not have the same dimension as $\frac{\delta^2 S}{\delta \bar{\psi} \delta \psi}$. This seems troublesome to me when we are interested in computing $\log S^{(2)}$. If all entries of $S^{(2)}$ had the same dimension, then $\log S^{(2)}$ would actually just mean $\log \frac{S^{(2)}}{\sigma}$ for some constant $\sigma$ with the appropriate dimension to render all entries of $\frac{S^{(2)}}{\sigma}$ dimensionless. $\sigma$ itself would then just add a negligible shift on the definition of $\Gamma$. Nevertheless, this does not seem to be the case since different entries of $S^{(2)}$ seem to have different dimensionalities, and hence I can't see how to make sense of $\log S^{(2)}$.

Notice this does yield issues. A way of computing the supertrace on Eq. (1) is to follow the trick Braun gives on Eqs. (17) and (18). We split $S^{(2)}$ as $S^{(2)} = \mathcal{P} + \mathcal{F}$, where $\mathcal{P}$ doesn't depend on fields but $\mathcal{F}$ does. Then we proceed to write $$\mathrm{STr}\log S^{(2)} = \mathrm{STr}\left[\mathcal{P}^{-1} \mathcal{F}\right] - \frac{1}{2}\mathrm{STr}\left[(\mathcal{P}^{-1} \mathcal{F})^2\right] + \frac{1}{3}\mathrm{STr}\left[(\mathcal{P}^{-1} \mathcal{F})^3\right] + \cdots, \tag{2}$$ where the dots represent other terms on the expansion (including $\mathrm{STr}\log \mathcal{P}$). If the dimension of all entries of $S^{(2)}$ were the same, Eq. (2) would make perfect sense. However, if they are not, then Eq. (2) will eventually lead to summing terms with different dimensions.

How do we make sense of these issues? Do the entries of $S^{(2)}$ actually have different dimensions or did I get something wrong? If they do, how can Eqs. (1) and (2) make sense?

Answer 1

1. First of all, the last term in OP's eq. (1) is a 1-loop effect and therefore there is an implicitly written power of Planck's constant $\hbar$ in front, which provides the correct dimension for an action term.

2. Secondly, the functional determinant (FD) must be dimensionless.

   • If the FD is defined via a Gaussian path integral with infinitely many integrations, the only meaningful outcome is if dimensionality is somehow cancelled in the normalization of the path integral measure.

   • Alternatively, a FD is often defined relatively to fiducial/reference/background FD, which in turn is normalized to 1. In OP's example $\cal P$ has the fiducial/reference/background role.