Nambu-Goldstone bosons from a quantum anomaly symmetry breaking?

Question

We know that:

Nambu-Goldstone bosons come from Goldstone theorem: a spontaneous (continuous)-symmetry breaking of the system leads to massless scalar modes.

quantum anomaly: is the anomalous phenomena where the (classical Noether) symmetry G respected by the physical system in a classical limit, but this symmetry G is broken by the quantum effect. This is the case where the action $S$ preserves the symmetry, but the path integral partition function $Z=\int [D\Psi][D\Phi]\dots e^{iS}$ and the measures $[D\Psi][D\Phi]\dots$ do not preserves the symmetry.

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• Question: whether there is any example that Nambu-Goldstone bosons can be derived from a spontaneous symmetry breaking caused by quantum anomaly effect? Alternatively, whether there is a known theorem to prove ``No Nambu-Goldstone bosons from a quantum anomaly symmetry breaking?''

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[words of caution & side remark]: I offer some further thoughts if you wish you can skip it. There is a statement learned: Nambu-Goldstone bosons do not generally appear for a spontaneously broken symmetry if the relevant global symmetry is broken by the effects of the anomaly and instantons. This is regarded as a reason why we do not observe a light pseudo-scalar meson $\eta'$ in the QCD mesons. 1 among the 9 mesons is this $\eta'$, which stands for axial $U(1)_A$ anomalous symmetry breaking, there are still $SU(N_{flavlor})_A=SU(3)_A$ which is spontaneous broken by dynamical symmetry breaking, which induces 8 among the 9 mesons, such as three $\pi$, three $\kappa,$ and one $\eta$. There are also $SU(N_{flavlor})_V=SU(3)_V$ broken by the nonzero explicit quark masses($m_u\neq m_d \neq m_s$). There is also $U(1)_V$ broken by Sphaleron, such that numebr of baryon $N_{baryon}$ is not conserved, but which only conserved numbers of $N_{baryon}-N_{lepton}$. Anyway, altogether $$SU(N)_V \times SU(N)_A \times U(1)_V \times U(1)_A$$ makes it to be the full $U(N)_V \times U(N)_A$.

Simply that $U(1)_A$ symmetry is anomalous broken by the quantum effect, but we do not see its Goldstone boson $\eta'$.

ps. please one may read full comments below the questions.

Answer 1

There are two different kinds of symmetry breaking involved in your question. The first would be spontaneous symmetry breaking, in which case we are dealing with a theory that is invariant under a certain symmetry group, but its vacuum is not. The breaking of the symmetry corresponds to a specific choice of the vacuum, the freedom of choosing a vacuum results in a new degree of freedom: the Nambu-Goldstone- boson. Depending on the type of symmetry that is broken, one might get one or more of them. In the case of the chiral symmetry of QCD, the $SU(N_f)_A$ part is broken spontaneously, resulting in eight massless bosons. The other axial part, $U(1)_A$, however, is not broken spontaneously.

The second kind of symmetry breaking would be induced by quantum anomalies. We speak of such an anomaly when a theory is classically invariant under a certain symmetry operation but the corresponding quantum theory is not. In terms of the path integral formalism this is manifest in the fact that the measure transforms in a nontrivial way. An example would be the breaking of the axial $U(1)_A$ part of the chiral symmetry in QCD, which is not related to any choice of a vacuum, and there is no Nambu-Goldstone boson connected to it. In fact, it can be traced back to something entirely different, namely instantons. The occurence of an anomaly is related to the number of fermionic zero modes of the theory: the difference between fermionic and antifermionic zero modes is given by the Pontryagin number of the topological instanton configuration of the gauge fields. This is also known as the Atiyah-Singer theorem.

Since these two possible symmetry breaking mechanism are quite distinct, the concept of Nambu-Goldstone bosons arising from quantum anomalies can only be a result of sloppy, non-standard terminology.

Answer 2

First note that chiral symmetry breaking appears at the level of states of theory, where exist set of nontrivial vacuums which are connected with each other by symmmetry transformations, and hence which are not invariant under the gauge transformation. The theory itself on the level of path integral is gauge invariant. If the theory isn't invariant, then we may talk about SSB of approximately exact symmetry, if breaking effects are small compared to themscale of CSB, otherwise there is no even approximate symmetry.

As follow from comments to the question, you thought that, in the spirit of 't Hooft's anomaly matching conditions, anomalies, which break the symmetry explicitly, leads to existence of Nambu-Goldstone bosons. This is misunderstanding of conditions.

Really, 't Hooft anomaly matching says that if we initially have theory with chiral fermions and gauge bosons, and feemion sector has the global symmetry group $G$, which is gauge anomaly free (no $G_{\text{gauge}}^2G$ anomalies), but which is anomalous itself (there exist $G^3$ anomalies), then after the confinement there must exist massless fermion bound states which reproduce anomalous structure of symmetry group in initial theory; otherwise there must exist spontaneous symmetry breaking. But is there symmetry breaking (i.e., current nonconservation) by anomaly?

In fact, no. Presence of $G^3$ anomalies means only that there are symmetric tensor with group indices structure in the triangle diagram with three $G$ currents, namely $$ d_{abc} \equiv \text{Tr}[[t_{a},t_{b}]_{+}t_{c}] $$ This means that if we decide to gauge the $G$ group, we will face with current nonconservation. Otherwise there is no current nonconservation and hence no symetry breaking. For example, the pure QCD global symmetry group $G \sim SU_{L}(3)\times SU_{R}(3)\times U_{B}(1)$ is exact in the chiral limit, although it is anonalous itself.