In the $m_q \rightarrow 0$ limit the QCD lagrangian has the symmetry $U(N)_V \times U(N)_A$. Including just the two lightest quarks, $N=2$, and looking at the $U(2)_A=SU(2)_A \times U(1)_A$ part, we have upon SSB 8 PNGB's from the $SU(2)_A$. However, no goldstone boson is observed that matches the $U(1)_A$-part ("The $U(1)_A$ problem"). The way I see this explained is that $U(1)_A$ was never a real symmetry (the associated axial current is anomalous), and thus there is no SSB or GB either.

QUESTION: In the Peccei-Quinn solution to the strong CP problem one adds a new anomalous U(1) symmetry to cancel the CP-violating theta-term. How come we can have axions as a PNGB of this anomalous symmetry, when in the case of the $U(1)_A$ problem anomalousness was used to explain why there should not be a PNGB?

Related: Spontaneous symmetry breaking of anomalous global abelian symmetries and $U(1)$ of QCD

The answer (I think) seems to say that for $U(1)_{PQ}$ the symmetry is exact in the $m_a \ll \lambda_{QCD}$ limit (the anomaly is small) and thus we get PNGBs from SSB, while for $U(1)_{A}$ the anomaly is not small so we get no goldstone bosons. Are anomalies then merely the reason for explicit symmetry breaking, and if this breaking is small enough (so that we can have SSB) we will have (pseudo-) goldstone bosons? What happens to the goldstone bosons between this limit where the anomaly is small and a limit where it isn't (maybe for example with time dependent axion mass around when the axion mass is "turned on" at $T=T_{QCD}$)?

In the $m_q \rightarrow 0$ limit the QCD lagrangian has the symmetry $U(N)_V \times U(N)_A$. Including just the two lightest quarks, $N=2$, and looking at the $U(2)_A=SU(2)_A \times U(1)_A$ part, we have upon SSB 8 PNGB's from the $SU(2)_A$. However, no goldstone boson is observed that matches the $U(1)_A$-part ("The $U(1)_A$ problem"). The way I see this explained is that $U(1)_A$ was never a real symmetry (the associated axial current is anomalous), and thus there is no SSB or GB either.

In the Peccei-Quinn solution to the strong CP problem one adds a new anomalous U(1) symmetry to cancel the CP-violating theta-term. How come we can have axions as a PNGB of this anomalous symmetry, when in the case of the $U(1)_A$ problem anomalousness was used to explain why there should not be a PNGB?

This question has been answered here: Spontaneous symmetry breaking of anomalous global abelian symmetries and $U(1)$ of QCD

The answer (I think) seems to say that for $U(1)_{PQ}$ the symmetry is exact in the $m_a \ll \lambda_{QCD}$ limit (the anomaly is small) and thus we get PNGBs from SSB, while for $U(1)_{A}$ the anomaly is not small so we get no goldstone bosons.

- Are anomalies then merely the reason for explicit symmetry breaking, and if this breaking is small enough (so that we can have SSB) we will have (pseudo-) goldstone bosons?

- What happens to the goldstone bosons between this limit where the anomaly is small and a limit where it isn't (maybe for example with time dependent axion mass around when the axion mass is "turned on" at $T=T_{QCD}$)?

In the $m_q \rightarrow 0$ limit the QCD lagrangian has the symmetry $U(N)_V \times U(N)_A$. Including just the two lightest quarks, $N=2$, and looking at the $U(2)_A=SU(2)_A \times U(1)_A$ part, we have upon SSB 8 PNGB's from the $SU(2)_A$. However, no goldstone boson is observed that matches the $U(1)_A$-part ("The $U(1)_A$ problem"). The way I see this explained is that $U(1)_A$ was never a real symmetry (the associated axial current is anomalous), and thus there is no SSB or GB either.

QUESTION: In the Peccei-Quinn solution to the strong CP problem one adds a new anomalous U(1) symmetry to cancel the CP-violating theta-term. How come we can have axions as a PNGB of this anomalous symmetry, when in the case of the $U(1)_A$ problem anomalousness was used to explain why there should not be a PNGB?

Related: Spontaneous symmetry breaking of anomalous global abelian symmetries and $U(1)$ of QCD

The answer (I think) seems to say that for $U(1)_{PQ}$ the symmetry is exact in the $m_a \ll \lambda_{QCD}$ limit (the anomaly is small) and thus we get PNGBs from SSB, while for $U(1)_{A}$ the anomaly is not small so we get no goldstone bosons. Are anomalies then merely the reason for explicit symmetry breaking, and if this breaking is small enough (so that we can have SSB) we will have (pseudo-) goldstone bosons? What happens to the goldstone bosons between this limit where the anomaly is small and a limit where it isn't (maybe for example with time dependent axion mass around when the axion mass is "turned on" at $T=T_{QCD}$)?

Thanks!

In the $m_q \rightarrow 0$ limit the QCD lagrangian has the symmetry $U(N)_V \times U(N)_A$. Including just the two lightest quarks, $N=2$, and looking at the $U(2)_A=SU(2)_A \times U(1)_A$ part, we have upon SSB 8 PNGB's from the $SU(2)_A$. However, no goldstone boson is observed that matches the $U(1)_A$-part ("The $U(1)_A$ problem"). The way I see this explained is that $U(1)_A$ was never a real symmetry (the associated axial current is anomalous), and thus there is no SSB or GB either.

QUESTION: In the Peccei-Quinn solution to the strong CP problem one adds a new anomalous U(1) symmetry to cancel the CP-violating theta-term. How come we can have axions as a PNGB of this anomalous symmetry, when in the case of the $U(1)_A$ problem anomalousness was used to explain why there should not be a PNGB?

Related: Spontaneous symmetry breaking of anomalous global abelian symmetries and $U(1)$ of QCD

The answer (I think) seems to say that for $U(1)_{PQ}$ the symmetry is exact in the $m_a \ll \lambda_{QCD}$ limit (the anomaly is small) and thus we get PNGBs from SSB, while for $U(1)_{A}$ the anomaly is not small so we get no goldstone bosons. Are anomalies then merely the reason for explicit symmetry breaking, and if this breaking is small enough (so that we can have SSB) we will have (pseudo-) goldstone bosons? What happens to the goldstone bosons between this limit where the anomaly is small and a limit where it isn't (maybe for example with time dependent axion mass around when the axion mass is "turned on" at $T=T_{QCD}$)?

In the $m_q \rightarrow 0$ limit the QCD lagrangian has the symmetry $U(N)_V \times U(N)_A$. Including just the two lightest quarks, N=2, and looking at the $U(2)_A=SU(2)_A \times U(1)_A$ part, we have upon SSB 8 PNGB's from the $SU(2)_A$. However no goldstone boson is observed that matches the $U(1)_A$-part ("The $U(1)_A$ problem"). The way I see this explained is that $U(1)_A$ was never a real symmetry (the associated axial current is anomalous), and thus there is no SSB or GB either.

QUESTION: In the Peccei-Quinn solution to the strong CP problem one adds a new anomalous U(1) symmetry to cancel the CP-violating theta-term. How come we can have axions as a PNGB of this anomalous symmetry when in the case of the $U(1)_A$ problem anomalousness was used to explain why there should not be a PNGB?

Related: Spontaneous symmetry breaking of anomalous global abelian symmetries and $U(1)$ of QCD

The answer (I think) seems to say that for $U(1)_{PQ}$ the symmetry is exact in the $m_a \ll \lambda_{QCD}$ limit (the anomaly is small) and thus we get PNGBs from SSB, while for $U(1)_{A}$ the anomaly is not small so we get no goldstone bosons. Are anomalies then merely the reason for explicit symmetry breaking, and if this breaking is small enough (so that we can have SSB) we will have (pseudo-) goldstone bosons? What happens to the goldstone bosons between this limit where the anomaly is small and a limit where it isn't (maybe for example with time dependent axion mass around when the axion mass "turned on" at $T=T_{QCD}$)?

Thanks!

In the $m_q \rightarrow 0$ limit the QCD lagrangian has the symmetry $U(N)_V \times U(N)_A$. Including just the two lightest quarks, $N=2$, and looking at the $U(2)_A=SU(2)_A \times U(1)_A$ part, we have upon SSB 8 PNGB's from the $SU(2)_A$. However, no goldstone boson is observed that matches the $U(1)_A$-part ("The $U(1)_A$ problem"). The way I see this explained is that $U(1)_A$ was never a real symmetry (the associated axial current is anomalous), and thus there is no SSB or GB either.

QUESTION: In the Peccei-Quinn solution to the strong CP problem one adds a new anomalous U(1) symmetry to cancel the CP-violating theta-term. How come we can have axions as a PNGB of this anomalous symmetry, when in the case of the $U(1)_A$ problem anomalousness was used to explain why there should not be a PNGB?

Related: Spontaneous symmetry breaking of anomalous global abelian symmetries and $U(1)$ of QCD

The answer (I think) seems to say that for $U(1)_{PQ}$ the symmetry is exact in the $m_a \ll \lambda_{QCD}$ limit (the anomaly is small) and thus we get PNGBs from SSB, while for $U(1)_{A}$ the anomaly is not small so we get no goldstone bosons. Are anomalies then merely the reason for explicit symmetry breaking, and if this breaking is small enough (so that we can have SSB) we will have (pseudo-) goldstone bosons? What happens to the goldstone bosons between this limit where the anomaly is small and a limit where it isn't (maybe for example with time dependent axion mass around when the axion mass is "turned on" at $T=T_{QCD}$)?

Thanks!