In quantum electrodynamics (QED), the chiral current is not conserved: $$ \partial_\mu(\overline\psi\gamma^\mu\gamma_5\psi)\sim F\wedge F + \text{mass term}. \tag{1} $$ If the current were conserved, then the chiral rotation $$ \psi\to \exp(i\alpha\gamma_5)\psi \tag{2} $$ with constant $\alpha$ would be a symmetry, but the current isn't conserved (not even if the mass term is zero), so (2) may or may not be a symmetry. According to section 2.5 in ref 4:

A. It is a symmetry of noncompact massless QED in topologically trivial spacetime.

B. It's not a symmetry of compact massless QED, not even in topologically trivial spacetime. (As emphasized in ref 4 at the end of section 3.4, compact QED is favored by the fact that the charges of electron and proton have equal magnitude as far as our most precise measurements can tell.)

In contexts with massless fermions, many sources (refs 1,2,3) seem to say that for finite $\alpha$, the only effect of the chiral rotation (2) is to change the coefficient of the so-called theta term $\int F\wedge F$ in the action. But this term is zero for an abelian gauge field in topologically trivial spacetime, so if that really were the only effect, then (2) would be a symmetry of massless QED. That would be correct for noncompact QED in topologically trivial spacetime (statement A) but not for compact QED (statement B).

This seems to imply that the chiral rotation (2) has an additional effect in compact QED. If this reasoning is correct, then what is that additional effect?


  1. Page 457 in Weinberg's The Quantum Theory of Fields Volume II

  2. "The Electroweak Vacuum Angle" (arXiv:1402.6340)

  3. Chapter 3 in Tong's Lectures on Gauge Theory (http://www.damtp.cam.ac.uk/user/tong/gaugetheory.html)

  4. Harlow and Ooguri, Symmetries in Quantum Field Theory and Quantum Gravity (https://arxiv.org/abs/1810.05338)


1 Answer 1


It is only the global rotation effects that is zero for topologically trivial spacetimes. There are still local effects that are captured by your eq 1.

  • $\begingroup$ Are you assuming that $\alpha$ is independent of $x$ and $t$? If $\alpha$ is local, i.e. depends on $x$ and $t$, then there are effects. $\endgroup$
    – mike stone
    Jun 27, 2021 at 0:23
  • 3
    $\begingroup$ Surely you are an expert on this! If $\int F^2=0$ then when $m=0$there is a global symmetry in which the *total* (but not *local*) chiral charge is conserved. Computie the determinant of $D_{Dirac} +m\cos \alpha + m\gamma^5 \sin\alpha $ in the absence of zero modes. Ypu will see that thia is independent of $\phi$. Consequently all correlators that would be changed by the transformation must vanish. $\endgroup$
    – mike stone
    Jun 27, 2021 at 0:30
  • $\begingroup$ @ChiralAnomaly I believe if you do the calculation correctly, you will end up not with the $\theta$-term, but with $\int \alpha F \wedge F$. When $\alpha$ is constant (global transformation), it is equivalent to the $\theta$ term with $\theta = \alpha$. When $\alpha$ isn't constant, it is not. $\endgroup$ Jun 27, 2021 at 7:37
  • $\begingroup$ @ChiralAnomaly also I believe your second reference says this is the only affect for a theory of massless particles and for global transformations. If you have a mass term in your action obviously it also changes under chiral transformations. So your argument seems to be a tautology: you've proven that the theory of massless particles must have massless particles. $\endgroup$ Jun 27, 2021 at 7:40
  • $\begingroup$ I have a toy QM meodel that illustrates a lot of these issues. people.physics.illinois.edu/stone/toy_instanton.pdf Ithas the advanatge that the functional integrals are analytically doable, and yet contains many of the features. Perhaps this will help? $\endgroup$
    – mike stone
    Jun 27, 2021 at 13:30

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