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I’m a bit confused on the effects of anomalous global symmetries. So take for instance the following theory $$\mathscr{L}=\partial_\mu\phi\partial^\mu\phi^*+i\bar{\psi}\gamma_\mu\partial^\mu\psi-y \phi\bar{\psi}\psi+\text{h.c}-V(\phi)$$ with $V(\phi)=m^2|\phi|^2+\lambda |\phi|^4$ It has two global symmetries $U_V(1)$ with $\psi\to e^{i\theta}\psi$ and $U_A(1)$ with $\psi\to e^{I\gamma_5\theta}\psi$ and $\phi\to e^{-2 i\theta}\phi$.

These symmetries have significant physical consequences; of course $U_A(1)$ forbids a mass for $\psi$,also the interplay of $U_V(1)$ and $U_A(1)$ forbid $\phi$ from decaying since decaying into two fermions is forbidden by helicity consideration, and other decays are forbidden by either $U_V(1)$ or $U_A(1)$.

However we would usually consider the $U_A(1)$ to be anomalous; certainly it can’t be gauged. But it's unclear to me what physical affect this anomaly actually has. If $U_V(1)$ was gauged, then we would have $$\partial_\mu J^A_\mu=-\frac{g^2}{16\pi^2}F_{\mu\nu}\tilde{F}^{\mu\nu},$$ which would allow violations of $U_A(1)$.

However, when $U_V(1)$ is just a global symmetry, it seems like there is no physical consequence of the “anomaly”.

So my question is: are $U_A(1)$ and $U_V(1)$ good symmetries of the theory I described? If not, what observable consequences does this have? I understand that anomalies come from regularisation ambiguities so perhaps a different way to phase my question is: Is there a regularisation scheme that respects both $U_A(1)$ and $U_V(1)$ and if not what observables are ambiguous?

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  • $\begingroup$ Write down the currents you believe are classically conserved. This will specify what you believe your scalar actually does. $\endgroup$ Sep 24, 2020 at 22:15

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There is no anomaly problem with this system --- except that as written it does not have a continuous $U_A(1)$ symmetry. You need to include a term $i\bar\psi \gamma^5 \psi$ term in addition to the $\bar\psi\psi$ term. With that included it is a simple model that can be be used for illustrating chiral symmetry breaking.

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  • $\begingroup$ Are you saying we can consider QCD with $U_A(1)$ symmetry simply by adding the $i\bar\psi \gamma^5 \psi$ term (a term that should be there any way by EFT logic). If that is true why don't we write it? Is it due to experimental data? $\endgroup$
    – Kvothe
    Nov 18, 2021 at 19:28

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