# 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?

• Could you give a reference for the second paragraph? About how the B+L anomaly singles out of the vacua? Jul 26 '19 at 4:04

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.
• First question. I was taught that the meaning of an anomalous symmetry is that it is NOT a true symmetry of the quantum theory because although classically conserved, quantum-mechanically the symmetry breaks. Over here, you seem to claim the opposite, namely that $G^2G'$ anomaly $\Rightarrow G'$ is a true symmetry. Secondly, what do you exactly mean by "the instantons associated with the gauge group $G$ change the $G'$ charge"? How? Lastly, what is so special about the combination $G^2G'$? I hope I am not asking unrelated questions. I am still trying to piece together all the ideas. Thanks! :) Oct 10 '18 at 13:51
• @NanashiNoGombe I edited to make the answer a bit more clear, hopefully it helps. Oct 10 '18 at 15:02
• Thanks. I get the gist. A $G^2G′$ anomaly is necessary to make the $\theta$-term unphysical. There is no global classical symmetry $G′$ with a $G^2G′$ anomaly, which is why the $SU(3)_C\ \theta$-term is physical. Ok. One question though. In your first two points, you seem to imply that a $G^2G′$ anomaly $\Rightarrow$ instantons $\Rightarrow$ non-trivial $\theta$-vacuum. But isn't that the opposite of what you try to say afterwards, namely that an anomaly trivialises the $\theta$-vacuum? Oct 10 '18 at 15:51
• @NanashiNoGombe There are instantons and $\theta$-vacua regardless of anomalies. The anomaly just ensures the different $\theta$-vacua are physically equivalent. Oct 10 '18 at 17:20