In quantum electrodynamics (QED), the chiral current is not conserved: $$ \partial_\mu(\overline\psi\gamma^\mu\gamma_5\psi)\sim F\wedge F + \text{mass term}. \tag{1} $$ It would be conserved if the chiral transformation $$ \psi\to \exp(i\alpha\gamma_5)\psi \tag{2} $$ were a symmetry, so we infer that (2) is not a symmetry, not even when the fermions are massless, at least not when the chiral rotation angle $\alpha$ is close to zero. This conclusion holds for any spacetime topology, because (1) is a local relationship.

In other contexts with massless fermions, many sources (refs 1,2,3,4) 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, it would imply that (2) is a symmetry of QED — when spacetime is topologically trivial and the fermions are massless.

The first argument says that (2) is not a symmetry, but the second argument says it is. What's the resolution of this apparent contradiction? I must be making a mistake somewhere, but where?

(In case this isn't clear: The question is about the case where $\alpha$ is independent of the coordinates and the fermions are massless. Otherwise, (2) wouldn't even be a symmetry of the action, much less of the quantum field theory.)


  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. This answer to the question Why is there no theta-angle (topological term) for the weak interactions?

Related: Is it the chiral anomaly which is solely responsible for having instanton effects (and therefore, the $\theta-$term) in the QCD action?


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.

  • $\begingroup$ Thank you for the response! I don't see how this answers the question, though. Consider the generating function $Z[\eta,\overline\eta]$, which is defined as the partition function with $\exp(\int \overline\psi\eta + \overline\eta\psi)$ inserted. This encodes all of the theory's correlation functions for the spinor field, and the second argument says that if I apply a chiral rotation to $\eta$ and $\overline\eta$, then I can change the integration variables ($\psi$ and $\overline\psi$) to show that the generating function is invariant. Where is the flaw in the second argument? $\endgroup$ Jun 27 at 0:20
  • $\begingroup$ 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 at 0:23
  • $\begingroup$ Right. I understand that there are effects when $\alpha$ depends on the coordinates. The question is about the global rotation, for which $\alpha$ is independent of the spacetime coordinates. (That's what I meant by symmetry.) $\endgroup$ Jun 27 at 0:25
  • 2
    $\begingroup$ 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 at 0:30
  • $\begingroup$ Even though $\int F\wedge F=0$ when integrated over the whole spacetime (if the action is finite), we can still have $\int F\wedge F\neq 0$ when integrated over only part of the spacetime, such as the part bounded by two Cauchy surfaces $\Sigma$ and $\Sigma'$. This says that the charge $Q\equiv\int_\Sigma n\cdot j_5$ is not necessarily invariant under changes of the Cauchy surface (where $n$ denotes the timelike normal to the Cauchy surface). In other words, it says that the chiral charge isn't conserved. What am I missing? $\endgroup$ Jun 27 at 1:24

I found at least two mistakes in my reasoning. First, I wasn't paying close enough attention to the difference between compact QED, whose gauge field is based on the compact group $U(1)$, and noncompact QED, whose gauge field is based on the noncompact group $\mathbb{R}$. Most texts gloss over this because they look the same in perturbation theory, but nonperturbatively it does make a difference. According to section 2.5 in ref 1,

  • The global chiral rotation is not a symmetry of compact QED, not even when spacetime is topologically $\mathbb{R}^4$.

  • The global chiral rotation is a symmetry of noncompact QED when spacetime is topologically $\mathbb{R}^4$.

Second, I wasn't paying close enough attention to the difference between a dynamic gauge field, which is what QED has, and a background gauge field, which is often used as a tool for studying chiral anomalies. According to section 2.5 in ref 1 again,

  • The global chiral rotation is a symmetry when the gauge field is just a background field (if spacetime is $\mathbb{R}^4$), even when the gauge field is based on the compact group $U(1)$.

The details behind these assertions are spelled out in the cited paper. The basic idea is that in compact QED, the would-be symmetry is obstructed by the presence of additional observables that are constructed from the compact-$U(1)$ gauge field itself. Those observables are absent when the gauge field is merely a background field, or when it's a dynamic field based on the noncompact group.

So I was making at least two different mistakes, namely conflating the compact and noncompact cases and conflating the dynamic and background gauge field cases. This answers my question, in the sense that it identifies mistakes in my reasoning. (Thanks to users mike stone and Prof. Legolasov for comments/chat that helped direct my attention toward these issues.)

My understanding is still incomplete, though. If a chiral rotation is a good symmetry in the compact $U(1)$ case when the gauge field is only a background field, then a nonperturbative regulator that preserves that symmetry ought to exist. When the background gauge field is absent, we can use a lattice formulation witih the overlap Dirac operator (reviewed in ref 2), which preserves an exact chiral-rotation symmetry, without doublers. When the gauge field is nonzero, we can generalize that transformation to one that is still an exact symmetry of the lattice action, but now that transformation depends explicitly on the gauge field, so its effect on the path-integral measure must be considered carefully. That's where the chiral anomaly is hiding. This lattice formulation of compact QED is actually what I had in mind when writing the question, even though I didn't say that out loud. I still don't fully understand how it all plays out in the lattice formulation (I still have a lot to learn about the subject that inspired my username!), but that's beyond the scope of the question I asked here. One step at a time.

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

  2. Creutz, Confinement, chiral symmetry, and the lattice (https://arxiv.org/abs/1103.3304)


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