# Anomalous commutators and gauge anomaly

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 $\mathbf E(\mathbf x)$: $$\tag 1 [A_{i}(\mathbf x), A_{j}(\mathbf y )] = 0, \\ [A_{i}(\mathbf x),E_{j}(\mathbf y)] =-i\delta_{ij}\delta(\mathbf x - \mathbf y), \\ [E_{i}(\mathbf x), E_{j}(\mathbf y)] = i\Delta_{ij} (\mathbf A, \mathbf x)\delta (\mathbf x- \mathbf y)$$ where $\mathbf A$ is the canonical coordinate

Is this information the only one which we need to conclude that corresponding fermion current $J_{\mu}^{L}$ isn't conserved due to gauge anomaly, 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 $$\frac{dQ_{L}}{dt} = -i[H,Q_L] =|Q_L = \int d^{3}\mathbf r \nabla \cdot \mathbf E| = [H, \int d^{3}\mathbf r \nabla \cdot \mathbf E(\mathbf r)] =$$ $$= -i\left[\frac{1}{2}\int d^{3}\mathbf y \frac{\mathbf E^{2}(\mathbf y )}{2},\int d^{3}\mathbf r \nabla \cdot \mathbf E(\mathbf r)\right]= \int d^{3}\mathbf r E_{i}(\mathbf r)\partial_{j}\Delta^{ij}(\mathbf A,\mathbf r)$$

• The canonical momentum is the electric field. So, by analogy with classical mechanics of a particle we have $p_{i}\sim E_{i}(x)$. The Poisson brackets are $[p_{i},p_{j}]_{PB}=0$ and these go over to $[E_{i}(x),E_{j}(y)]_{PB}=0$ for the electromagnetic field. So, how do you get the third line in equation (1)? – Stephen Blake Oct 29 '16 at 12:14
• @StephenBlake : I've obtained the quantum correction to the symplectic space due to adiabatic evolution of the EM 3-potential, which captures the effect of anomaly. – Name YYY Oct 29 '16 at 12:32
• I'm unfamiliar with this procedure, would you supply a reference? I don't understand why the commutator of the electric field does not follow from just studying the free electromagnetic field so that nothing unusual should come in from the chiral fermion field. – Stephen Blake Oct 29 '16 at 15:27
• @StephenBlake : if the gauge theory is anomalous, i.e., the charge isn't conserved, the Gauss law isn't realized on physical states. In general, this means that the Gauss operators $G(\mathbf x), G(\mathbf y)$ don't commute. This leads to the statement, that the canonical commutation relations, which follow from naive canonical quantization of theory, are broken by quantum effects. Since the commutators are connected directly to the Poisson brackets (the ones which define the so-called symplectic structure), we state that quantum corrections break the canonical Poisson brackets. – Name YYY Oct 29 '16 at 16:19
• @Stephen Blake: the simplest way to capture these quantum corrections is to assume adiabatic evolution of the gauge fields from $t=0$ to $t=T$, with $H(t=0) = H(t=T)$. With this assumption, there is generated additional phase to the action - the Berry phase. It's generated because the fermion hamiltonian eigenstates are defined up to the unremovable phase factor. By the definition, it modifies the symplectic structure of the action explicitly, and hence may modify the canonical Poisson brackets. – Name YYY Oct 29 '16 at 16:24