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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.

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?''

[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 $\eta'$.

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.

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?''

[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.

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.

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?''

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.

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?''

[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 $\eta'$.

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.

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?''

[words of caution & side remark]: 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 $\eta'$.

But are there counter examples?

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.

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?''