Let me restate the $U(1)$ problem of QCD:

Let us forget about the $s$ quark, and consider the $u$ and $d$ massless. This is a good approximation since $m_{u,d} \ll \Lambda_{QCD}$. Then $\mathscr{L}_{QCD}$ has a $$U(2)_L \otimes U(2)_R ~=~ U(1)_A \otimes U(1)_V \otimes SU(2)_{A} \otimes SU(2)_{V}$$ symmetry. Where $A$ refers to axial and $V$ to vectorial. The $SU(2)_{L-R}$ is spontaneously broken leading to the emergence of a triplet of pseudoscalar Goldstone bosons.

The puzzle regards $U(1)_{A}$ which turns out cannot be spontaneously broken since (according say to Schwarz p. 637) it is anomalous and hence it is not a symmetry in the first place.

The suggested solution to the above problem is the one due to Peccei and Quinn. They introduced a $U(1)_{PQ}$ which is subject to axial anomaly as well. However, even though the symmetry is anomalous, we know its spontaneous symmetry breaking leads to axions. How come this symmetry - even though it's anomalous - can be spontaneously broken?

  • $\begingroup$ I think you are confusing the U(1) puzzle and the strong CP problem. There is no problem with the anomalous U(1)_A. Because of the anomaly no Goldstone boson is expected, and indeed none is observed. $\endgroup$
    – Thomas
    Oct 21, 2015 at 3:08
  • $\begingroup$ There is (somewhat related, but different) puzzle called the strong CP problem. The existence of the amomaly makes it clear that the $\theta$ parameter is physical, and that observables do depend on it. But $\theta$ violates CP, so it must be very small. The $U(1)_{PQ}$ is a symmetry of a proposed extension of QCD, in which $\theta=0$ dynamically. However, you should keep in mind that i) there is nothing wrong with setting $\theta=0$ by hand, ii) there is no evidence for axions. $\endgroup$
    – Thomas
    Oct 21, 2015 at 3:11
  • 1
    $\begingroup$ Isn't $U(1)_{PQ}$ itself anomalous? The dynamical solution (axion) is related to the spontaneous symmetry breaking of $U(1)_{PQ}$. How can an anomalous $U(1)_{PQ}$ lead to the appearance of a Goldstone boson? Judging by the fact that indeed $U(1)_A$ because it is anomalous it doesn't lead to Goldstones in the spectrum. $\endgroup$
    – EEEB
    Oct 21, 2015 at 11:04
  • 1
    $\begingroup$ I see what you are asking. $\endgroup$
    – Thomas
    Oct 21, 2015 at 13:58

1 Answer 1


Yes, $U(1)_{PQ}$ has standard model anomalies. The difference is that in QCD $$ {\cal L}=f_\pi^2(\partial_\mu \eta')^2 - V(\eta'-\theta) $$ where $f_\pi$ and $V^{1/4}$ are both $O(\Lambda_{QCD})$, so there is no sense in which the anomaly is a small effect. In particular, you cannot argue that there is a spontaneously broken $U(1)$ which is weakly broken by the anomaly. Indeed, the mass of the $\eta'$ is $m\sim 1$ GeV, comparable to the mass of the proton.

[There are exceptions to this statement, even in QCD. You can consider QCD at very high baryon density. Then $f_\pi\sim \mu \gg \Lambda_{QCD}$ and $V\sim \Lambda_{QCD}^4 (\Lambda_{QCD}/\mu)^c$ where $c$ is a positive power. Then the $\eta'$ is light and we can talk about spontaneous $U(1)$ breaking, weakly broken by the anomaly. A similar situation exists in QCD at large N.]

In axion extensions of the SM we have $$ {\cal L}=f_a^2(\partial_\mu a)^2 - V_a(a-\theta) $$ where $V_a$ comes from SM anomalies (mostly QCD), so $V\sim \Lambda_{QCD}^4$. But $f_a$ is a free parameter, so we can make it as large as we want (phenomenologically it has to be much bigger than $\Lambda_{QCD}$). Then $m_a^2=\chi/f_a^2$ where $\chi\sim\Lambda_{QCD}^4$ and $m_a\ll \Lambda_{QCD}$, and we have an approximate $U(1)_{PQ}$.

Final Remark: What do we mean by approximate spontaneous symmetry breaking? The standard example is the $SU(2)_L\times SU(2)_R$ symmetry of QCD. This symmetry is spontaneously brokken by the quark condensate, $\langle \bar{\psi}_L\psi_R\rangle \simeq -\Lambda^3_{QCD}$, and explicitly broken by the quark masses $m_q$. If $m_q=0$, then chiral symmetry is exact amd sponatenous breaking leads to massless Goldstone bosons. If $m_q\neq 0$ then Goldstone bosons acquire a mass $m_{GB}^2\sim B/f_\pi^2$ with $B=m_q\langle\bar\psi_L\psi_R\rangle$. As long as $m_{GB}\ll \Lambda_{QCD}$ we can view the effect of the quark mass as a small perturbation, and talk about the spontaneous breaking of an approximate symmetry. If $m_{GB}\sim \Lambda_{QCD}$ there is no sense in which an approximate symmetry is broken.

  • $\begingroup$ Thanks for the answer, however I won't lie claiming I completely understood. What do you mean by ' In particular, you cannot argue that there is a spontaneously broken U(1) which is weakly broken by the anomaly' $\endgroup$
    – EEEB
    Oct 21, 2015 at 21:16
  • $\begingroup$ Added a postcsript. $\endgroup$
    – Thomas
    Oct 21, 2015 at 22:43

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