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Consider pure QED with massless electrons. Due to the axial anomaly the axial current is not conserved:

$$ \tag 1 \partial_{\mu}J^{\mu}_{5} \sim F_{\mu\nu}\tilde{F}^{\mu\nu} $$ On the other hand, it seems that this non-conservation has nothing to do with the particle physics processes, as there is no axial field coupled to $J^{\mu}_{5}$ in the lagrangian.

What is then the physical manifestation of the equation (1)? What are its observable consequences?

Qmechanic
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Name YYY
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1 Answers1

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This question reminds you that $$\langle 0|J^5_{\mu,0}(x)|\pi^0(p)\rangle=-if_{\pi} e^{-ipx}p_{\mu}~,$$ the mother of PCAC. That is to say, you already know this axial current corresponds to a SSB generator, and so is linear in the Goldstone boson corresponding to it, $$ J^5_{\mu,0}\propto f_\pi \partial_\mu \pi^0 + ... $$ where the ellipsis represents terms of higher order in the fields.

The current is basically the goldston: The corresponding charge pumps such goldstons into and out of the chirally non-invariant vacuum!

As a consequence, the corresponding term of the effective Lagrangian which gives you the above current divergence is
$$ \frac{e^2 N_c \pi^0}{48 \pi^2 f_\pi} F_{\mu\nu}\tilde{F}^{\mu\nu}. $$ It therefore induces neutral pion decay to two photons, quite observable and physical, really. enter image description here

This term was an early reassurance of the genius of the WZWN term of flavor-chiral anomalous effective actions.

Cosmas Zachos
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