If a global symmetry is anomalous, classically it is still possible to talk about spontaneous breakdown of that symmetry. In particle physics, do we have such an anomalous symmetry (or symmetries) which spontaneously breaks down and gives rise to pseudo Nambu-Goldstone bosons (pNGbs)?

I know that spontaneous breakdown of approximate global symmetries (i.e., those which are weakly broken explicitly but not quantum mechanically) give rise to pNGbs whose classic example are pions. I am not sure whether breakdown of anomalous symmetries would also produce pNGbs.

  • $\begingroup$ Why wouldn't it? $\endgroup$
    – knzhou
    Nov 23, 2019 at 17:59
  • $\begingroup$ I would expect that it might. But is it obvious that a deviation from $\partial_\mu j^\mu=0$ leads to light bosons? @knzhou $\endgroup$
    – SRS
    Nov 23, 2019 at 18:19

1 Answer 1


Yes. Consider $N_f=2$ QCD with a large number of colors. The $U(1)_A$ is anomalous, but also spontaneously broken by the quark condensate. Because of large $N_c$ the effects of the anomaly are suppressed, and the $\eta’$ is a pseudo-Goldstone boson, $m_{\eta’}^2\sim 1/N_c$.

Note that the real world is not that close to this limit, because $m_{\eta’}\simeq 958$ MeV, heavier than the rho meson (and heavier than the proton).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy