Short version: effective actions, particularly ones obtained after integrating chiral fermions, are ambiguous up to the addition of local counterterms. Should we think of the counterterms as part of the path integral measure, or the classical action?

With some more background: Take a QFT with some Weyl fermions coupled to a gauge field, and define the effective action $W[A]$ as the action in terms of the gauge fields obtained taking the path integral over the fermions,

$$Z[A]=e^{iW[A]}=\int \mathcal{D}\psi \mathcal{D}\bar{\psi} e^{iS[\psi,\bar{\psi},A]}.$$

We say this theory is anomalous if, under a gauge variation, $W[A]$ has a transformation which cannot be cancelled by adding local counterterms to $W[A]$. My understanding of these local counterterms is that they reflect regularisation ambiguities in the path integral of the fermions. A very typical example is in 2d, where if we work with the two chiralities separately, we are free to (and indeed usually have to) add a counterterm proportional to $A_\mu A^\mu$ to make $W$ invariant under gauge transformations when the theory is non-anomalous.

In this context, I think of this counterterm as being part of the definition of the path integral measure, $\mathcal{D}\psi \mathcal{D}\bar{\psi}$, and whether it is possible to define it in a gauge-invariant way.

However, this seems to me to collide with the way I read about counterterms in introductory QFT texts, where counterterms are usually there to parametrise the difference between the bare and renormalised couplings, and therefore are already there as part of the classical action $S$, not coming from the measure. If I try to think of the above in this way, it seems to me that even for non-anomalous theories $\mathcal{D}\psi\mathcal{D}\bar{\psi}$ is not gauge-invariant, but its gauge variation can be cancelled by adding a local counterterm to $S$.

I believe the second interpretation is incorrect, but only because I'm used to think of anomalies as variations of the measure. Then a non-anomalous theory would necessarily have a gauge-invariant measure. Am I right here? Are the local counterterms part of the path integral measure? Is this also true for the usual counterterms parametrising the difference between bare and renormalised coupling constants?


1 Answer 1


My thoughts on counterterms are as follows, I can not say with certainty that they are absolutely correct. I would like to be corrected if something is not right.

My short answer: Counterterms are better thought of as part of the definition of the measure, not the classical action. and in response to the question:

Then a non-anomalous theory would necessarily have a gauge-invariant measure. Am I right here?

I would say yes. But I would like to emphasize that for a non-anomalous symmetry, though there must be an invariant measure, there can also be some measures that are not invariant. Different choices of measures correspond to different choices of regulators.

Long thoughts: First, I would like to make a distinction: the counterterms that can be added to the effective action depend on, like the effective action itself, the background fields, on the other hand, the counterterms that are added to the classical action related to the difference between the bare and the renormalized parameters depend on the dynamical fields. So far these two are not exactly the same, though adding terms to the classical action and then doing the path integration surely affects the effective action; but I don't think that they are responsible for the "counterterm ambiguity", for the following reasons.

Any field theory that's not asymptotically free is essentially effective, in other words, they are only defined for computing separated point correlation functions. So their partition functions are defined only upto modifications that don't affect separated point correlation functions. These modifications are precisely the additions of local background dependent functionals to the log of the partition functions. Schematically, if a theory has background field $J$ then the log of its partition function can be written as a generating function of connected correlation functions of operators (for which the background acts as source): $$ \log Z[J] = \log \int \mathcal{D}\phi\, e^{i S[\phi,J]} = \sum_{n=0}^\infty \int_M \prod_{i=1}^n \left(\mathrm{d}^dx_i\right) \left\langle \prod_{i=1}^n \left(\phi(x_i)\right) \right\rangle \prod_{i=1}^nJ(x_i)\,, \tag{*}$$ where $M$ is the space-time. Now, modification of $\log Z$ of the form: $$ \log Z[J] \to \log Z[J] + \int_M \mathrm{d}^d x\, f(J(x)) \,, $$ for arbitrary local function $f$ does not affect any of the separated point correlation functions so these modifications should be allowed in the effective theory, and this is the counterterm ambiguity you mentioned. On the other hand, the counterterms that often appear in QFT text books related to the difference between bare and renormalized parameters, do affect separated point correlation functions and they are not arbitrary, we have to fix their finite parts by imposing some renormalization conditions. These renormalization conditions are physical, they are fixed by imposing constraints that some scattering amplitude must take some certain values and/or some excitation must have some certain mass etc. It doesn't seem like these numbers should be arbitrary. The origin of all these counterterms is of course the fact that a straightforward computation of the various components of the series expansion (*) of $\log Z$ seems to diverge and we have to renormalize them. But the non-local part (involving multiple space-time integral) of the series expansion affects the low energy theory and is fixed from physical observation/constraints. The local part however is left arbitrary/ambiguous. Different choices of local counterterms correspond to different regulators.

Now, there can be some classical symmetry group $G$ that acts on the fields including the background: $$ S[g \cdot \phi, g^{-1} \cdot J] = S[\phi,J]\,, \quad \forall g \in G,. $$ This symmetry would be preserved after quantization if: $$ \log Z[g^{-1} \cdot J] = \log Z[J]\, \quad \mbox{or infinitesimally,} \quad \delta_\xi \log Z[J] = 0\,,$$ where $\xi \in \mathrm{Lie}(G)$. If it happens that the variation of $\log Z$ is not zero but equal to the variation of a local functional, $\delta_\xi \log Z[J] = \delta_\xi C[J]$, then the choice of regulator $\log Z \to \log Z + C$ preserves the symmetry. In other words the measure $\mathcal{D}\phi e^{C[J]}$ is $G$-invariant. It seems that not all regulators have to preserve the same amount of symmetry. I don't know if there is a reason to choose a regulator that preserves the maximal amount of symmetry in some sense. Except in cases where we want to gauge a symmetry, in which case we must find a regulator that preserves the symmetry.

  • $\begingroup$ Thanks for the answer, I think your reasoning makes sense for the most part. The only thing I would add is that the two cases no longer seem that different to me: even in the case where the counterterms parametrise the difference between bare and renormalised coupling constants, choosing a different regularisation scheme should be equivalent to choosing a different measure. Then you also have a different relationship between bare and renormalised couplings, so there I want to say you can change the measure and classical action in such a way that the correlation functions are not affected. $\endgroup$
    – user121664
    Dec 18, 2016 at 9:41
  • $\begingroup$ In (*), should the first identity read $\log Z[J]=\log \int \mathcal{D}\phi e^{iS[\phi]+J\phi}$? $\endgroup$
    – user121664
    Dec 18, 2016 at 9:42
  • 1
    $\begingroup$ @user121664 I agree, I also think any counterterm can be thought of as being part of the measure, it's just that the ones built purely out of the backgrounds are the only ones that are arbitrary and make the partition function ambiguous. And, in (*) I was imagining $J \phi$ to be part of the Lagrangian. $\endgroup$
    – nio
    Dec 18, 2016 at 12:27
  • $\begingroup$ I just realized that I hadn't mentioned $J$ in one of the terms in (*) at all, now I've changed it to make explicit the $J$-dependence of the action. $\endgroup$
    – nio
    Dec 18, 2016 at 12:47

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