What's the real resolution of the $U(1)_A$ problem?

Question

To recap the problem, consider QCD with three massless quark flavors. There is a symmetry $$SU(3)_L \times SU(3)_R \times U(1)_L \times U(1)_R$$ corresponding to independent rotations of the left-chiral and right-chiral quark fields. Vector symmetries are the subset $$SU(3)_V \times U(1)_V$$ which rotate the left-chiral and right-chiral quark fields the same way, while the axial symmetry $U(1)_A$ rotates the fields in opposite directions. Finally, we define $SU(3)_A$ by $$SU(3)_A = SU(3)_L \times SU(3)_R / SU(3)_V$$ and conventionally call it an "axial symmetry group", though it's merely a coset. All of these symmetries except $U(1)_V$ are explicitly broken by the quark masses, but we can treat this as a small effect and ignore it below.

The formation of the chiral condensate spontaneously breaks the symmetry to $SU(3)_V \times U(1)_V$, so we should have $8 + 1$ Goldstone bosons due to $SU(3)_A$ and $U(1)_A$. The $U(1)_A$ problem is the fact that there is no Goldstone boson corresponding to it. The candidate is the $\eta'$, which is much heavier than the $8$ others.

According to most textbooks, the resolution of the $U(1)_A$ problem is that the $U(1)_A$ symmetry is anomalous by a $U(1)_A SU(3)^2$ triangle diagram, and hence not a true symmetry of the quantum field theory. Since it's not a symmetry, it can't be spontaneously broken.

I don't buy this. The problem is that $SU(3)_A$ is also anomalous, by about the same amount. For example, the $U(1)$ subgroup of $SU(3)_A$ corresponding to the pion $\pi^0$ has a $U(1) U(1)_{\text{EM}}^2$ anomaly which accounts for the fast decay $\pi^0 \to \gamma \gamma$. This is important because it's how anomalies were discovered in the first place. So by this reasoning the pion should be heavy as well, but it isn't.

What distinguishes $U(1)_A$ here? Is the anomaly alone really the solution to the $U(1)_A$ problem?

Answer 1

We are trying to understand why the $\eta'$ acquires a mass in pure QCD (no external fields). This is indeed explained by the $U(1)_A\times [SU(3)]^2$ anomaly. Note that the $SU(3)$ is the QCD gauge symmetry, so we cannot turn these fields off. The anomaly in the $U(1)_A$ current is proportional to $G_{\mu\nu}\tilde{G}^{\mu\nu}$, which is a total divergence. This is not an impediment, because QCD has topological $|n\rangle$ sectors, and the partition function in the presence of a topological $\theta$-term ${\cal L}_\theta\sim\theta G_{\mu\nu}\tilde{G}^{\mu\nu}$ has $\theta$ dependence. The topological term has a vanishing vacuum expectation value (QCD does not spontaneously break CP), but the $\eta'$ mass is controlled by the associated susceptibility (with a caveat, explained by Witten and Veneziano) $$ \chi_{top} \sim \frac{1}{V} \int d^4x\, \langle G_{\mu\nu}\tilde{G}^{\mu\nu}(0)G_{\alpha\beta}\tilde{G}^{\alpha\beta}(x)\rangle . $$ This quanitity is of order $\Lambda_{QCD}^4$, and as a result the $\eta'$ acquires a mass comparable to other non-Goldstone bosons.

Pure QCD has a $SU(3)_L\times SU(3)_R$ flavor symmetry, which I can try to (weakly) gauge. We find that this symmetry is anomalous, and in the chiral broken phase this anomaly can be represented by a Wess-Zumino term, which (among other things) reproduces the $\pi^0\to 2\gamma$ decay. I can ask whether this anomaly contributes to the mass of the $\pi^0$. Note that 1) EM certainly does contribute to the mass of charged Goldstone bosons, 2) the anomaly is again proportional to $F_{\mu\nu}\tilde{F}^{\mu\nu}$, which is a surface term (but QED does not have topological sectors). Finally, in the standard model $SU(2)_L\times U(1)_Y$ is gauged, but the QCD anomaly (represented by the Wess-Zumino term) is cancelled by leptons.

P.S.: An explanation of why the $\pi^0$ does not acquire a mass (in more modern language) is also given here.