Timeline for What does $B+L$ anomaly have to do with a phase redefinition of the left-handed quark field?
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| Oct 11, 2018 at 8:31 | vote | accept | Nanashi No Gombe | ||
| Oct 10, 2018 at 17:20 | comment | added | knzhou | @NanashiNoGombe There are instantons and $\theta$-vacua regardless of anomalies. The anomaly just ensures the different $\theta$-vacua are physically equivalent. | |
| Oct 10, 2018 at 15:51 | comment | added | Nanashi No Gombe | 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, 2018 at 15:02 | comment | added | knzhou | @NanashiNoGombe I edited to make the answer a bit more clear, hopefully it helps. | |
| Oct 10, 2018 at 15:01 | history | edited | knzhou | CC BY-SA 4.0 |
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| Oct 10, 2018 at 13:51 | comment | added | Nanashi No Gombe | 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, 2018 at 10:58 | history | answered | knzhou | CC BY-SA 4.0 |