vacuum - What does $B+L$ anomaly have to do with a phase redefinition of the left-handed quark field?

According to this answer, the reason why $SU(2)_L$ weak theory does not have a theta vacuum is because any theta term can be reabsorbed with a phase redefinition of the left-handed quark field.

However, I heard another equivalent argument saying that there is a $B+L$ anomaly in the Standard Model which uniquely selects one of the many topologically distinct vacua as the one true vacuum. An instanton of $SU(2)$ lets one initial state tunnel into another in a different $B+L$ sector and, therefore, does not represent a vacuum to vacuum process.

I am trying to get the overall picture and understand how these two arguments are related. Could someone elucidate on that and explain what the $B+L$ anomaly is and how it selects a unique vacuum and how that is equivalent to a phase redefinition?

Answer

Let $G$ be a nonabelian gauge group and $G'$ be a global, classical symmetry.

• Suppose there exists a $G^2 G'$ anomaly, which is to say that the triangle diagram with two $G$ currents and one $G'$ current does not vanish. We say the $G'$ symmetry is anomalous.


• This implies the instantons associated with the gauge group $G$ may change the $G'$ charge. Also, the possibility of instantons implies the energy eigenstates are the $|\theta \rangle$ vacua. The specific $|\theta \rangle$ vacuum we live in corresponds to the value of the $\theta$-term.


• Redefining the fields using the symmetry generated by $G'$ is equivalent to shifting the value of the $\theta$-term associated with the gauge group $G$.



• Since $G'$ is a symmetry, that means all values of $\theta$ are equivalent, so we don't have to worry about $\theta$ terms or any effects of $|\theta \rangle$ vacua.


• In the case of $G = SU(2)_L$, we have $G' = U(1)_{B+L}$ which is why we don't have to worry about the $SU(2)_L$ $\theta$-term. One could also choose $U(1)_B$ or $U(1)_L$ for $G'$. In the former case, we get a non-chiral rotation of all the quarks.


• If there were massless fermions in the theory, we could have a global symmetry $G'$ from chiral rotations. Very often, such a chiral symmetry has a $G^2 G'$ anomaly, called the chiral anomaly. This is very important historically, but not relevant to this question because the Standard Model doesn't have chiral symmetry due to the fermion masses.


• Focusing on the Standard Model, the reason the $SU(2)_L^2 U(1)_{B+L}$ anomaly can exist is because the electroweak force is chiral. If it were not chiral, the effects of rotating the left-handed fermions would be exactly cancelled by those of rotating the right-handed fermions.


• Now consider the strong force, $G = SU(3)_C$. Since the strong force is not chiral, it behaves rather differently. It turns out that there is no global classical symmetry $G'$ with a $G^2 G'$ anomaly. That's why the $SU(3)_C$ $\theta$-term can have physical effects.


• On the other hand, if the up quark were massless, than we could let $G'$ correspond to chiral rotations of that quark alone, so the $\theta$-term would have no effects. That's an old proposed solution to the strong CP problem.