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Let $J$ be a certain Noether current $$ J=J[\phi] $$ where $\phi$ is a field. This object is classically conserved, although in the quantum-mechanical case it may be anomalous.

In the functional integral formalism, the failure of $J$ to be conserved is associated to a non-trivial Jacobian1. One typically finds $$ \langle \partial\cdot J\rangle=\langle A[\phi]\rangle $$ where $\langle\,\cdot\,\rangle$ denotes an expectation value, and $A[\phi]$ is the anomaly function (the trace of the logarithm of the Jacobian of the transformation).

In the operator formalism, the anomaly holds as an operator equation, $$ \partial\cdot \hat J=A[\hat \phi] $$ where $\hat J=J[\hat\phi]$.

The non-conservation of $\hat J$ is usually attributed to its singular nature2. Indeed, being a non-linear operator, the coinciding space-time points in its definition lead to an $0\times\infty$ indeterminate, and one must introduce a regulator. When the regulator is removed, one usually finds a non-zero finite contribution, which we identify with $A[\hat\phi]$.

There is a paradox here, because $A[\hat\phi]$ is usually non-linear in the fields too, and therefore it includes coinciding space-time points too. In other words, even if $\partial\cdot \hat J$ has been constructed using an explicit regulator and taking the limit carefully, a divergence remains. Take as an example the chiral anomaly, where $A\propto \hat F\wedge \hat F\cdots \wedge \hat F$. Here, $\hat F$ is a singular operator, so the anomaly $A$ is ill-defined. Still, $\partial\cdot\hat J$ was supposedly finite.

What's going on here? How is this paradox resolved? How can we make sense of the singularities in the anomaly function, when this object was precisely constructed by collecting the singularities in $\hat J$ and identifying the finite contribution as the regulator is removed?


1: See for example Weinberg's QFT, Vol.II, §22.2. See also Peskin & Shroeder, §19.1 (in particular, the discussion around equation 19.61 on page 664).

2: See for example Peskin & Shroeder, §19.1 (in particular, the discussion around equation 19.22 on page 655). See also Itzykson & Zuber, §11-5-3 (in particular, the discussion around equation 11-229 on page 559).

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  • $\begingroup$ Just a minor comment, In the chiral anomaly example the gauge fields don't need to be propagating fields; they can be (and are practically used as) external c-number fields. From the path integral perspective there is basically no issue as the integration over the gauge orbit can be done after the integration over the fermions that just see an external field $\endgroup$ – TwoBs Dec 28 '17 at 23:18
  • $\begingroup$ Aren't you mixing up two things? Yes, the path integral measure may pick up a nontrivial phase under an "anomalous" transformation. So far so good. But the discussion of the singularity is not transparent to me. According to your discussion, any local product of field operators is singular. Or why are you saying that $A$ is ill-defined? $\endgroup$ – user178876 Dec 28 '17 at 23:30
  • $\begingroup$ @TwoBs sure, from the path integral POV there is no issue in treating the gauge fields as external c-numbers, and integrating first over the fermionic fields. But from the operator POV, the gauge fields may contain both an external c-number contribution and an internal q-number contribution (e.g., in QCD, where the chiral anomaly is associated to the $\theta$ term; here both $\psi$ and $A^\mu$ are quantum operators). In this sense, the gauge fields may be propagating fields, in which case the paradox arises. This is precisely the situation I am interested in here. $\endgroup$ – AccidentalFourierTransform Dec 29 '17 at 9:42
  • $\begingroup$ @marmot I included some references. "any local product of field operators is singular": indeed, local product of field operators are always singular. Think of OPEs (or the fact that propagators are singular at $x=y$). $\endgroup$ – AccidentalFourierTransform Dec 29 '17 at 10:10
  • $\begingroup$ @AccidentalFourierTransform I suspect that your question is basically why is the anomaly local (both $F_{\mu\nu}$ at the same point) and yet finite. One way to see that's indeed finite is thinking of the other, IR, side of the anomaly. They don't run and have the same value at all scales, including the deep IR, i.e. at large distances (from where the t'Hooft anomaly matching comes from) where no divergence is expected. $\endgroup$ – TwoBs Dec 29 '17 at 15:08
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Although contact terms can arise when local operators are multiplied, but in the case of the anomaly term they don't. Qualitatively, the reason for that is that the contact terms are sensitive to the short distance limit, and depend on the regularization scheme.

Anomalies, on the other hand, survive the infrared limit when the contact terms become unimportant and are independent of the regularization scheme. More quantitatively, anomalies can be attributed to nonlocal counterterms of the type: $$\int d^4x \frac{\partial_{\mu} A^{\mu}}{\Box} F_{\alpha \beta}F^{\alpha \beta}$$ (This term is non-local, but its gauge variation is local and generates the correct anomaly in the current divergence). The inverse Laplacian suppresses the ultraviolet dependence.

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  • $\begingroup$ +1 Thanks. The picture is becoming clear for me. I'll do some more thinking and report back later on. $\endgroup$ – AccidentalFourierTransform Dec 31 '17 at 15:33

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