In Cheng&Li's book "Gauge theory of elementary particle physics" section 16.3, the $U(1)$ problem is presented. Here is the story:

Suppose we have a theoy which contains only two quarks $q_1=u$ and $q_2=d$. The symmetry group is given by $SU(2)_L\times SU(2)_R\times U(1)_V\times U(1)_A$.

$U(1)_A$ is axial symmery, i.e., $q_i\rightarrow e^{i\beta \gamma_5}q_i,i=1,2$. In the limit $m_{u,d} \rightarrow 0$, the chiral current \begin{equation} J^5_{\mu}=\bar{u}\gamma_{\mu}\gamma_5u+\bar{d}\gamma_{\mu}\gamma_5d \end{equation} does not seem to correspond any observed symmetry in hardon spectra. If the $U(1)_A$ is spontaneously broken, then a massles Goldstone boson will appear. But in experiment, no such particle is observed.

My confusion is: why do we expect a Goldstone boson? This $U(1)_A$ symmetry is based on the assumption "$m_{u,d} \rightarrow 0$". But in the real world, both $m_u$ and $m_d$ are nonzero, i.e., we might have mass term like $\bar{u}m_uu+\bar{d}m_dd$. Therefore the $U(1)_A$ is not a symmetry. What is the logic and whole story here? I think I misunderstood something.


1 Answer 1


The short answer is the symmetry is explicitly broken, and spontaneously broken, at the very same time. (It is also anomalous, but this is not part of your question.)

Your text explains the "real world" issue better than most. It has already lavished chapters on explaining spontaneous/dynamical chiral symmetry breaking of the 3 axial generators, whose Goldstone bosons, in the absence of quark masses, would have been the 3 flavored pseudoscalar mesons, πs. The corresponding Noether currents would be strictly conserved, in the ideal world.

In real life, the non-Abelian axial symmetries are approximate, PCAC, so $$ \partial ^\mu J_\mu^{a ~ 5}\approx m \bar q \tau^a ~\gamma_5q \qquad \sim f_\pi m_a^2 \phi ^a \neq 0, $$ where m denotes linear combinations of quark masses you may compute for each broken generator. The Goldstone sombrero vacuum has tilted a little, and there is a small mass about a non-degenerate vacuum, as σ-model fans illustrate with the σ term.

The 3 almost Goldstone bosons, then, are almost massless, being pseudogoldstone bosons, the dramatically light pions, the featherweights of the hadronic spectrum. It is all a perturbation in the small quark masses normalized by the much bigger chiral symmetry breaking scale, m/v ~ 1/60 , two orders of magnitude!

The authors assume you would then expect, mutatis mutandis, to echo this for the isosinglet axial current, $$ \partial ^\mu J_\mu^{5} =2i (m_u \bar u \gamma_5u + m_d \bar d \gamma_5 d )\qquad \sim f_\eta m_\eta^2 ~\eta \neq 0, $$ by an analog of GOR/Dashen's formula, so a comparably light pseudogoldstone boson pseudoscalar.

But, as they detail, this, dramatically, fails to occur, because of the anomaly, seizing of the vacuum, etc, the interesting issues outranging your question, which they discuss.

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    $\begingroup$ Pseudo-Goldstone bosons appear due to both spontaneous and explicit symmetry breaking. So the logic of $U(1)$ problem is: If $U(1)_A$ is spontaneous and explicit broken, then where is the corresponding Pseudo-Goldstone boson? $\endgroup$
    – Sven2009
    Jul 28, 2020 at 11:20
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    $\begingroup$ Of course; if that is all there is, the axial singlet would have an exceptionally light pseudoscalar associated with it, here, the notional η; in actuality (3 flavors) the η' , which is not as light as it "should" be. The reason is the QCD topological term anomaly, etc, outside the ambit of this discussion, and explained further in the text. $\endgroup$ Jul 28, 2020 at 12:45
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    $\begingroup$ The entire subsequent development is a sublime effort to explain the "heavier" mass of the η', not "just" a pseudogoldston ... $\endgroup$ Jul 28, 2020 at 12:59

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