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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?

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  • $\begingroup$ Could you give a reference for the second paragraph? About how the B+L anomaly singles out of the vacua? $\endgroup$
    – Optimus Prime
    Commented Jul 26, 2019 at 4:04

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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.
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  • $\begingroup$ 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! :) $\endgroup$
    – Nanashi No Gombe
    Commented Oct 10, 2018 at 13:51
  • $\begingroup$ @NanashiNoGombe I edited to make the answer a bit more clear, hopefully it helps. $\endgroup$
    – knzhou
    Commented Oct 10, 2018 at 15:02
  • $\begingroup$ 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? $\endgroup$
    – Nanashi No Gombe
    Commented Oct 10, 2018 at 15:51
  • $\begingroup$ @NanashiNoGombe There are instantons and $\theta$-vacua regardless of anomalies. The anomaly just ensures the different $\theta$-vacua are physically equivalent. $\endgroup$
    – knzhou
    Commented Oct 10, 2018 at 17:20

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