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What is the significance of the fact that the anomany term (calculated from the triangle diagram) is a total divergence? Or, in other words, what is the significance of $$\partial_\mu j^\mu_A\sim Tr(W\tilde{W}) =\text{a total divergence}$$ for global anomalies. I think this fact is related to why baryon number violation in standard model cannot be a perturbative process. Perhaps someone can illuminate.

SRS
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  • Um...if the symmetry were non-anomalous, this divergence would vanish, making $j_A$ a conserved Noether current. Since it is anomalous, this divergence doesn't vanish, and the Noether current is not conserved. Do you have reason to believe there is another significance to this, and if yes, why? – ACuriousMind Oct 04 '14 at 15:06
  • My question was different. I'm not asking what is anomaly or why is right hand side non-zero but what is the significance that this non-zero right hand side is a total divergence. I think this fact is related to why baryon number violation in standard model cannot be a perturbative process. Perhaps someone can illuminate. – SRS Oct 04 '14 at 15:17
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    Ah, I see, you're talking about the sentence "An important fact is that the anomalous current non-conservation is proportional to the total derivative of a vector operator" in the Wiki article – ACuriousMind Oct 04 '14 at 15:23
  • @ACuriousMind : In the Wikipedia article (baryonic charge violation), since $K_\mu$ is the Hodge dual of the Chern-Simons 3-form, then the anomaly could be considered as "topological". No ? – Trimok Oct 04 '14 at 15:37
  • The wiki article you mentioned, does not explain the point why the baryon number non-conservation cannot* be a perturbative violation*. After all, a non-zero current violation may be integrated directly to give a non-zero charge violation. Isn't it? The wiki article only says the violation is non-perturbative and induced by instanton effects. That is fine. But it misses the point, what is the problem if I say baryon violation is also possible perturbatively. I have no prolem with the process being non-perturbative. But I have problem understanding why it cannot be perturbative. – SRS Oct 04 '14 at 15:40
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    @Trimok: Indeed. And I've just seen that I've already said this once to a similar question by the OP. – ACuriousMind Oct 04 '14 at 15:41

1 Answers1

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Leaving out numerical factors, we have that

$$ \mathrm{d}j_A = \mathrm{Tr}(F \wedge F)$$

This already shows that we are dealing with a topological quantity, since the RHS is the second Chern character of the gauge field (or rather, the principal bundle associated to it). Now, there is also the (3D) Chern-Simons form

$$ \omega = \mathrm{Tr}(F \wedge A - \frac{1}{3}A \wedge A \wedge A)$$

and it is easily calculated that $\mathrm{d}\omega = \mathrm{Tr}(F \wedge F)$, and so, $\mathrm{d}j_A = \mathrm{d}\omega$. Now, we can obtain the Noether charge by taking a spacelike three-dimensional slice $\Sigma$ in our four-dimensional spacetime and integrating the current over it. Such a spacelike slice will, for ordinary Minkowski space, always be the boundary of some four-dimensional region $M$, and so we find:

$$ \int_\Sigma j_A = \int_{\partial M}j_A = \int_M \mathrm{d}j_A = \int_M \mathrm{d}\omega = \int_\Sigma \omega$$

The RHS of this is now a topological (and gauge) invariant quantity, since the Chern-Simons form does not depend on the choice of a metric on spacetime, and is well-known to produce a topological field theory.

Thus, the Noether charge does only depend on the topological structure of the gauge bundle over this slice, and the topological structure of the bundle is precisely what instantons describe (for more on instantons, topology and vacua, see my answer here). Instantons are non-perturbative because each of them is its own local minimum of the action, i.e. they are all vacua, while perturbative things are only ever stemming from fluctuations around a single vacuum.

Additionally, the integral $\int\mathrm{Tr}(F\wedge F)$ is a discrete term, taking on values $8\pi^2 k$ with integer $k \in \mathbb{Z}$, so it is not a smooth function of anything, but jumps discontinuously when the topology changes. In contrast, perturbative results should vary smoothly when sending the perturbatively small parameter to zero, which cannot be the case here.

Community
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ACuriousMind
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  • @ Acuriousmind- Thanks. This is all very informative and important. But with all humility, I didn't get an answer to the fact why a total divergence term cannot contribute to the perturbation theory, as given in Matthew Schwartz's book on "Quantum Field Theory and Standard Model". – SRS Oct 04 '14 at 16:20
  • @Roopam: Perturbation theory always acts as if we're on $\mathbb{R}^4$ with sufficient fall off at infinity, and so the integrals over total divergences always vanish since the boundary term is zero. – ACuriousMind Oct 04 '14 at 16:31