I am trying to understand an argument for why anomalies are one-loop exact, given by Bilal in Lectures on Anomalies. The relevant paragraph is reproduced here:

Let us first explain why the anomaly we have just computed corresponds to a one-loop effect. One can introduce a formal loop-counting parameter by rescaling the action as $S \rightarrow \frac{1}{\lambda}S$. Then in computing Feynman diagrams, the propagators get an extra factor $λ$ while the vertices get an extra $\frac{1}{\lambda}$. Thus every Feynman diagram comes with a factor $λ^{ I−V}$ , where $I$ is the number of internal lines and $V$ the number of vertices. By a well-known relation, one has $I − V = L − 1$ with $ L$ being the number of loops, and one sees that $\lambda$ is a loop counting parameter. Since the classical action comes with a $\frac{1}{\lambda}$ it is the tree-level contribution ($L = 0$) to the effective action, while the anomaly, like any determinant, has no factor of $\lambda$ and corresponds to a one-loop contribution ($L = 1$).

I understand everything up until the section in italics. What does it mean that any determinant has no factor of $\lambda$? Am I meant to somehow be inferring that the anomaly is invariant under such a rescaling and so it is a determinant, or is it the other way around (i.e. determinant implies invariance)?

  • $\begingroup$ Both are consistent. The Jacobean is invariant, and hence the anomaly, which he also computes perturbatively. $\endgroup$ Jan 12 '20 at 19:43
  1. From the Boltzmann factor perspective $e^{\frac{i}{\hbar}S}$, if we exponentiate the path integral measure $\rho= e^{\frac{i}{\hbar}\frac{\hbar}{i}\ln\rho}$ in the path integral $Z=\int\!{\cal D}\phi~\rho~e^{\frac{i}{\hbar}S}$, the anomaly in the measure is formally a 1-loop effect, cf. the $\hbar$/loop-expansion. This seems to be essentially Bilal's argument in a nutshell, with $\lambda$ playing the role of $\hbar$.

  2. A more rigorous argument can be given by the fact that boson- and massive fermion-lines have a parity-conserving gauge-invariant UV regularization. Also recall that anomalies only appear when the regulator fails to conserve the pertinent symmetries. Therefore any loop diagram (without divergent subdiagrams) that contains a loop-participating boson- and massive fermion-line cannot be anomalous. In other words, the anomalous loops must be made purely from massless fermions. It turns out that a higher-loop diagram (without divergent subdiagrams) must contain a boson loop-line, and is hence anomaly-free. We should stress that in the presence of a one-loop anomaly, it will create anomalous higher-loop diagrams with divergent subdiagrams. On the other hand, 0-loop/tree diagrams are always finite. So in that sense, anomalies are a 1-loop effect.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.