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  From gauge anomaly to chiral anomaly

+ 3 like - 0 dislike

Suppose the theory of chiral Weyl fermion (say, left) $\psi_{L}$, which interacts with abelian gauge field. This theory contains the gauge anomaly, which I write in the form
\tag 1 \frac{dQ_{L}}{dt} = \text{A},
where $Q_{L}$ is the left charge and $A$ is anomaly function.

The same thing is true about right fermion $\psi_{R}$. If the gauge field is vector (not axial vector), then
\tag 2 \frac{dQ_{R}}{dt} = -\text{A},
The underlying reason for this is that the dynamics of theory generates anomalous commutator between canonical momentums of EM field (the electric field $\mathbf E$): pcecisely,
\tag 3 [E_{i}(\mathbf x), E_{j}(\mathbf y)]_{L/R} = -i\Delta^{ij}_{L/R}(\mathbf A, \mathbf y)\delta (\mathbf x - \mathbf y),
where $L,R$ denotes the subspaces of left and right fermions. This gives (see the question) the anomaly $\text{A}$:
\tag 4 \frac{dQ_{L/R}}{dt} = \text{A} = \int d^{3}\mathbf r E_{i}(\mathbf r)\partial_{j}\Delta^{ij}_{L/R}(\mathbf A, \mathbf r)

Suppose now we take the "direct sum" of left and right representations:
\psi = \psi_{L} \oplus \psi_{R}
In this case, by using $(1), (2)$, we see, that there is no gauge anomaly of vector charge $Q_{\text{vector}}$,
\tag 5 \frac{dQ_{\text{vector}}}{dt} = \frac{dQ_{L}}{dt} + \frac{dQ_{R}}{dt} = \text{A} - \text{A} = 0,
but there is the chiral anomaly of axial charge $Q_{\text{axial}}$,
\tag 6 \frac{dQ_{\text{axial}}}{dt} =\frac{dQ_{L}}{dt} - \frac{dQ_{R}}{dt} =\text{A}+\text{A}= 2\text{A}
Since the vector current is the gauge current, then the total contribution into anomalous commutator from the left and right particles must vanish:
\tag 7 [E_{i}(\mathbf x), E_{j}(\mathbf y)]_{L\oplus R} = -i\Delta^{ij}_{L}(\mathbf A, \mathbf y)\delta (\mathbf x - \mathbf y)-i\Delta^{ij}_{R}(\mathbf A, \mathbf y)\delta (\mathbf x - \mathbf y) = 0

Although we assume the left and right particles with the same charge and mass, this looks like anomaly cancellation. The left and right fermions remain anomalous separately.

My question is following. Although the anomalous commutator $(3)$ exists on subspaces $L$ and $R$, it vanishes for their direct sum, as is shown by $(7)$. But the chiral anomaly $(6)$ exists. In terms of broken canonical commutator $(3)$ I can understand this phenomena as the fact that this commutator violates chiral symmetry. This is the direct consequence of Eqs. $(4)$, $(6)$. But to me is very strange that the gauge anomaly of left and right fermions give the ungauged anomaly for the difference of their fermion number. Is my understanding correct?

asked Oct 29, 2016 in Theoretical Physics by NAME_XXX (1,060 points) [ revision history ]
edited Oct 30, 2016 by NAME_XXX

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