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The baryon number B or lepton number L violation in the standard model arise from triangle anomaly. Right? Triangle diagrams are perturbative diagrams. Then why the B or L violation in Standard model is said to be a non-perturbative effect? I'm confused.

SRS
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1 Answers1

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It is a non-perturbative effect because it is 1-loop exact.

The triangle diagram is actually the least insightful method to think about this, in my opinion. The core of the matter is the anomaly of the chiral symmetry, which you can also, for example, calculate by the Fujikawa method examining the change of the path integral measure under the chiral transformation. You can obtain quite directly that the anomaly is proportional to

$$\int \mathrm{Tr} (F \wedge F)$$

which is manifestly a global, topological term, (modulo some intricacies) it is the so-called second Chern class and takes only values of $8\pi^2k$ for integer $k$. It is, by the Atiyah-Singer index theorem (this can also be seen by Fujikawa), essentially the difference between positive and negative chiral zero modes of the Dirac operator. This is obviously a discontinuous function of $A$ (or $F$), which is already bad for something which, if it were perturbative, should be a smooth correction to something, and it is also the number describing which instanton vacuum sector we are in, see my answer here. Since perturbation theory takes place around a fixed vacuum, this is not a perturbative effect, since it is effectively describing a tunneling between two different vacuum sectors.

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ACuriousMind
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  • ACuriousMind Why do you say that "The triangle diagram is actually the least insightful method to think about this"? Doesn't the chiral anomaly arise due to triangle diagrams? Is it the same case for baryonic or leptonic current anomaly? – SRS Aug 01 '14 at 13:12
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    @Roopam: "Due" is a difficult word here. If you insist on thinking about everything in terms of diagram, then yes, it arises due to them. But it is an anomaly of a symmetry, an effect produced by the non-invariance of the path integral measure under the symmetry, perfectly derivable without any reference to Feynman diagrams. I think it is not insightful because the triangle is "just another diagram" when the effect product is fundamentally different from "just another perturbative correction". It's just my opinion though, if you like the diagram best, it's your choice. – ACuriousMind Aug 01 '14 at 13:53
  • @ACuriousMind- can you please explain what is meant by one-loop exact? – SRS Sep 06 '14 at 19:23
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    @Roopam: one-loop exact means that there are no higher order diagrams that would contribute to the effect – ACuriousMind Sep 07 '14 at 14:02
  • could anyone please explain to me WHY the 1 loop contribution is exact? –  Oct 02 '15 at 10:45
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    @DancehallWashington: The one-loop result is already the same as the full non-perturbative result of e.g. the Fujikawa method. Hence there are no other loop orders which contribute. – ACuriousMind Oct 02 '15 at 12:23