# Why does a triangle anomaly appear in a gauge theory?

I have read that when we construct a theory with abelian gauge symmetry there will appear some anomalies when we do the quantum correction to the theory. In 4D space such anomalies are explained by triangular loop feynman diagrams. But I don't understand how this anomaly is explained by the triangular diagram. And does the theory naturally give the anomaly, or do we assume that there should be an anomaly and accordingly we set some conditions to cancel the anomaly? In layman's terms, does the theory naturally predict this Feynman triangular diagram?

• Here you go diarium.usal.es/vazquez/files/2015/02/UAM_slides2.pdf, 'Introduction to anomalies in QFT' by Miguel Á. Vázquez-Mozo – Vicky May 9 '19 at 4:44
• Yes, the rules of QFT tell us that we need to compute various diagrams, in particular the triangular diagram, and the triangular diagram gives the anomaly. The other diagrams don't. It is hard to answer "I don't understand how this anomaly is explained by the triangular diagram" without further details. The best I can suggest is to take a QFT textbook you have studied and read the explanation there. If you haven't studied a textbook where this is explained, it is highly unlikely someone on Q&A site can explain it properly from zero prior knowledge. – Peter Kravchuk May 9 '19 at 6:22
• – AccidentalFourierTransform May 9 '19 at 22:47
• The triangle diagrams are the one-loop correction to a two-point function with a vector current or an axial current inserted. The Ward identity then says that if you multiply this by the momentum associated with the current the result should be zero. This is the quantum version of Noether's theorem that the currents are conserved classically. When you work out the details you find that you cannot have both the Ward identity satisfied for the vector current and the axial current. This gives the anomaly. Zee and Cheng & Li explain this very well. Peskin & Schroder too. – Oбжорoв May 10 '19 at 11:34

## Perspective

We don't assume that there should be an anomaly. For a given pattern of interactions between gauge fields and matter fields, the quantum theory either does or does not have a chiral anomaly. Calculating the triangle diagram in four-dimensional spacetime, or the $$(n+1)$$-gon diagram in $$2n$$-dimensional spacetime, is one way to find out whether or not the quantum theory has a chiral anomaly — at least for a local chiral anomaly. There can also be a global chiral anomaly, but that's another story.

The only way to cancel chiral anomalies is to change the pattern of interactions -- that is, to change the theory. This is how we knew that the top quark must exist, long before it was directly observed: with an incomplete third generation of quarks and leptons, the Standard Model would have had a chiral anomaly that makes it mathematically inconsistent. To cancel this bad chiral anomaly, we needed to include the top quark in the model.

(Not all chiral anomalies are bad. As explained below, only the ones that obstruct the quantization of the gauge fields are bad.)

## Why do chiral anomalies appear in triangle diagrams?

To help explain why the triangle diagram is relevant to the chiral anomaly, think of constructing the quantum theory in two steps:

• First we construct the theory using quantum fermions and classical gauge fields, which gives the semiclassical model.

• Then we try to promote the classical gauge fields to quantum gauge fields, to get a fully quantum model.

Chiral anomalies don't obstruct the first step, but they can obstruct the second step. To understand chiral anomalies, we can focus on the semiclassical model that comes from the first step, specifically on how it depends on the classical gauge field.

In the semiclassical model, only the fermions are quantized, so only the fermions can appear as internal lines in Feynman diagrams. In a model without any 4-fermion interactions (which would be non-renormalizable in four-dimensional spacetime), this means that Feynman diagrams in the semiclassical model can't have any more than one loop, and it must be a fermion loop. Diagrams with no loops ("tree" diagrams) would already be present for non-quantized fermions, so chiral anomalies can must come from the one-loop diagrams.

That explains why the relvant diagrams have one loop. To see why the loop is a triangle, consider the generalization to $$2n$$-dimensional spacetime. Let $$\psi$$ denote a fermion field, and let $$\Gamma$$ denote the product of the $$2n$$ Dirac matrices $$\gamma^\mu$$, normalized to be hermitian. (The traditional notation $$\Gamma=\gamma_5$$ is only appropriate in the four-dimensioanl case.) If the semiclassical model were invariant under the chiral symmetry transformation $$\psi\to\exp(i\Gamma\theta)\psi$$, then it would have a corresponding conserved current $$j_\Gamma^\mu\sim\overline\psi\Gamma\gamma^\mu\psi. \tag{1}$$ A chiral anomaly spoils this conservation law, so that $$\partial_\mu j_\Gamma^\mu \tag{2}$$ is equal to some non-zero expression involving the classical gauge fields. This expression must be gauge-invariant and invariant under proper Lorentz transformations, and it should change sign when any single spacetime direction is reversed. A simple guess with these properties consists of $$n$$ factors of the gauge-invariant field-strengh tensor $$F_{ab}$$ (the Faraday tensor), antisymmetrized over the whole set of $$2n$$ indices to make it invariant under proper Lorentz transformations. This plausible guess turns out to be correct, as demonstrated explicitly in AccidentalFourierTransform's nice answer to this question:

How to calculate an axial anomaly in 1+1 dimensions?

To relate this to Feynman diagrams, note that if the quantity (2) is non-zero and proportional to a product of $$n$$ copies of the field-strength tensor, then the semiclassical model must have a non-zero Feynman diagram involving a factor of (2) along with $$n$$ external gauge-field lines. Since an external gauge-field line can enter only through the interaction term $$\sim\overline\psi\gamma^\mu A_\mu\psi$$, such a Feynman diagram must correspond to the expectation value of the product of $$n$$ factors of $$\overline\psi\gamma^\mu A_\mu\psi$$ along with the one factor of (2). Such product involves $$n+1$$ fermion-field pairs altogether, and each of these pairs contributes one fermion line-segment to the diagram. We already know that the relevant diagram must involve a single fermion loop, and now we know that the loop must be made of $$n+1$$ segments. In the case of four-dimensional spacetime, we have $$2n=4$$ and therefore $$n+1=3$$, so it's a triangle diagram.