Suppose we have a Lagrangian invariant under Chiral symmetry, such as QED with massless fermions:

$$ \mathscr{L} = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \bar{\psi} i \gamma^{\mu} D_{\mu} \psi .$$

In all textbooks and other posts on this website it is said that Chiral symmetry protects against the generation of mass terms at loop level. But if Chiral symmetry is an anomalous symmetry and it is broken at quantum level (making the Ward identities incorrect) how can it protect against the generation of mass terms? Is maybe Chiral symmetry not broken for QED and other simple theories? What am I missing in this picture?


1 Answer 1


The $\eta'$ meson (associated to the "ninth axial current") gains its mass from the anomalous symmetry breaking. The other eight pseudo-goldstone's mesons are not affected by the axial anomaly.

  • $\begingroup$ So, referring to the example at hand (QED with one massless fermion) is Chiral simmetry broken? And if yes why does it not affect the mass of the fermion? $\endgroup$ Jan 7, 2022 at 8:03
  • $\begingroup$ With only one fermion, the symmetry is broken so it gains a mass. For notes and some refences see page 12-17 in my notes: people.physics.illinois.edu/stone/lichnerowicz.pdf $\endgroup$
    – mike stone
    Jan 7, 2022 at 13:26
  • $\begingroup$ I tried looking at your notes and they look way too advanced, but if the symmetry is broken as you now say, and a mass term is indeed generated, why for example here: physics.stackexchange.com/questions/287207/…, it is said that no such mass term can be created by renormalization? $\endgroup$ Jan 7, 2022 at 13:48
  • 1
    $\begingroup$ The axial anomaly means that the "symmetry" was not there in the first place. Renormalization itself does not break any symmetries. Dynamical symmetry breaking is a different thing,and does and can give fermons mass. Indeed most of the mass in the universe comes from the spontaneous breaking of chiral symmetry by QCD interactions. $\endgroup$
    – mike stone
    Jan 7, 2022 at 14:38

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