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  Anomalous commutators and gauge anomaly of gauge current

+ 3 like - 0 dislike

Suppose we know, that the dynamics of theory with chiral fermions (say, left) and gauge field (for simplicity, abelian) leads us to presence of anomalous commutator of canonical momentum $\mathbf E(\mathbf x)$:
[E_{i}(\mathbf x), E_{j}(\mathbf y)] \sim \Delta (A_{i}(\mathbf x), A_{j}(\mathbf y)), \qquad (1)
where $\mathbf A$ is canonical coordinate.

Is this information the only one which we need to conclude that corresponding fermion current $J_{\mu}^{L}$ isn't conserved, i.e.,
\partial_{\mu}J^{\mu}_{L} \neq 0?

In particular, if the answer is "yes", for the left charge $Q_{L} = \int d^{3}\mathbf r J_{L}^{0}$ there must be

\frac{dQ_{L}}{dt} \sim [H, Q_{L}] \neq 0

due to anomalous commutator.

Another formultion of the question: does the presence of anomalous commutator $(1)$ guarantee the presence of gauge anomaly, i.e., current non-conservation?

An edit

It seems that the answer is obviously yes. First, even if I don't know precise structure of anomalous commutators, it is enough to know that they are non-zero. Thus, in particular, the "Gauss laws" $G(x) = \nabla \cdot \mathbf E - J^{0}$ don't commute with each other. This means, that physical states $|\psi\rangle$ of theory no longer satisfy the relation
G(x)|\psi\rangle = 0
This means violation of the unitarity, and hence the gauge anomaly.

In the case when we know the precise form $(1)$ of anomalous commutators, it's elementary to compute the anomalous conservation law: by using the Gauss law, we have
\frac{dQ_{L}}{dt} \sim [H,Q_L] =|Q_L = \int d^{3}\mathbf r \nabla \cdot \mathbf E| =  [H, \int d^{3}\mathbf r \nabla \cdot \mathbf E(\mathbf r)] =
= \left[\frac{1}{2}\int d^{3}\mathbf y \frac{\mathbf E^{2}(\mathbf y )}{2},\int d^{3}\mathbf r \nabla \cdot \mathbf E(\mathbf r)\right]= \int d^{3}\mathbf r E_{i}(\mathbf r)\partial_{j}\Delta^{ij}(\mathbf A,\mathbf r)

asked Oct 28, 2016 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Oct 29, 2016 by NAME_XXX

are you asking if the anomaly cancellation may come only from the spins ?

@igael : rather about following: does existence of anomalous commutator guarantee the presence of gauge anomaly? I.e., are they directly related to each other?

In the example there are same sided chiral fermions ... But you may get both anomalies "and" current conservation / gauge invariance , if by some tuning, you make anomalies cancel or else vanish ; it depends of the other specifications of the theory. Read Edward Witten in "Global Gravitational Anomalies" and similar articles like the chapter "Gravitational Anomalies" in "Global aspects of current algebra" ( with  Luis Alavarez-Gaumé, a book from S.Treiman ) where he illustrates both cases.

@igael : did You mean that I can interpret the zero gauge anomaly for $\psi_{L}\oplus \psi_{R}$ as the particular example of anomaly cancellation?

yes, in some theories, particularly if fermions are in real representations of the gauge group and not only ...

@igael : can You please clarify me this question, http://physicsoverflow.org/37559/from-gauge-anomaly-to-chiral-anomaly ? Now i think that the left and right fermions sector separately are anomalous (and provide non-zero anomalous commutator of EM field canonical variables), but their total contribution into the gauge anomaly is zero. However, by themselves they remain anomalous (the contribution into anomalous commutator is still non-zero), and thus the anomalous commutators provide non-conservation of chiral charge.

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