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A symmetry is anomalous when the path-integral measure does not respect it. One way this manifests itself is in the inability to regularize certain diagrams containing fermion loops in a way compatible with the symmetry. Specifically, it seems that the effect is completely determined by studying 1-loop diagrams. Can someone give a heuristic explanation as to why this is the case? And is there a more rigorous derivation that "I just can't find any good way to regularize this thing."?

An alternative approach, due to to Fujikawa, is to study the path integral of the fermions in an instanton background. Then one sees that the zero modes are not balanced with respect to their transformation under the symmetries, leading to an anomalous transformation of the measure under this symmetry. Specifically, the violation is proportional to the instanton number, and thus one finds the non-conservation of the current is proportional to the instanton density. This is also found by the perturbative method above.

My question, which is a little heuristic, is how is it that the effect seems perturbative (and exact at 1-loop) on the one hand, and yet related to instantons, which are non-perturbative, on the other?

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These are all good questions. Perhaps I can answer a few of them at once. The equation describing the violation of current conservation is

$$\partial^\mu j_\mu=f(g)\epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$$

where f(g) is some function of the coupling constant. It is not possible to write any other candidate answer by dimensional analysis and by parity (assuming the current is the ordinary axial current...)

Now we integrate both sides over $\int d^4x$, and we find on the left hand side $\Delta Q$, meaning, now that the current is violated, the charge can change while the system evolves, while the right hand side is $$f(g)\int d^4x \epsilon^{\mu\nu\rho\sigma}F_{\mu\nu}F_{\rho\sigma}$$

The object on the right hand side is a known topological invariant of the gauge bundle, and it is an integer (if all the charges are appropriately quantized). So on the left hand side we get $\Delta Q$, which must be an integer (if all fundamental particles carry integer charge) and the right hand side is an integer too, up to the function $f(g)$.

This means that the function $f(g)$ cannot, in fact, depend on $g$. (More precisely, there is a scheme where it does not.) Hence, it is exact at one loop. This is the modern proof (without any computation) of the ABJ theorem about one-loop exactness of the anomaly.

So you see the deep connection between one loop and instantons... The violation of the conservation equation is at one loop, but to lead to interesting consequences we need to have a nontrivial gauge bundle.

About some of the other comments you made: ANY regularization scheme that respects Bose symmetry will lead to the anomaly, it is totally unavoidable. This is proven in http://inspirehep.net/record/154341?ln=en.

Another comment: anomalies can also arise from boson loops, for example, the trace anomaly. (It is not one-loop exact in any sense I am aware of.)

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I really wish this site was not about to be shut down. There is nowhere else you can reliably expect to see discussion of QFT at this level. –  Mitchell Porter May 2 '12 at 2:03
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I noticed this just now ;) too bad. But it seems all the material won't disappear into thin air, which is good. –  Zohar Ko May 2 '12 at 2:31
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I like this short argument. But if f(g) is independent of g, shouldn't the effect already arise at 0 loop, hence classically? –  Arnold Neumaier May 2 '12 at 18:46
    
No, the charges are normalized to one when there is a 1/g^2 in the expression for the current, so a constant $f$ is one-loop. –  Zohar Ko May 2 '12 at 19:10
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