Timeline for Spontaneous symmetry breaking of anomalous global abelian symmetries and $U(1)$ of QCD
Current License: CC BY-SA 3.0
10 events
when toggle format | what | by | license | comment | |
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Mar 9, 2017 at 11:42 | vote | accept | EEEB | ||
Aug 21, 2016 at 5:15 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
Oct 23, 2015 at 2:13 | history | tweeted | twitter.com/StackPhysics/status/657379206192590848 | ||
Oct 21, 2015 at 13:58 | comment | added | Thomas | I see what you are asking. | |
Oct 21, 2015 at 13:42 | answer | added | Thomas | timeline score: 4 | |
Oct 21, 2015 at 11:04 | comment | added | EEEB | Isn't $U(1)_{PQ}$ itself anomalous? The dynamical solution (axion) is related to the spontaneous symmetry breaking of $U(1)_{PQ}$. How can an anomalous $U(1)_{PQ}$ lead to the appearance of a Goldstone boson? Judging by the fact that indeed $U(1)_A$ because it is anomalous it doesn't lead to Goldstones in the spectrum. | |
Oct 21, 2015 at 3:11 | comment | added | Thomas | There is (somewhat related, but different) puzzle called the strong CP problem. The existence of the amomaly makes it clear that the $\theta$ parameter is physical, and that observables do depend on it. But $\theta$ violates CP, so it must be very small. The $U(1)_{PQ}$ is a symmetry of a proposed extension of QCD, in which $\theta=0$ dynamically. However, you should keep in mind that i) there is nothing wrong with setting $\theta=0$ by hand, ii) there is no evidence for axions. | |
Oct 21, 2015 at 3:08 | comment | added | Thomas | I think you are confusing the U(1) puzzle and the strong CP problem. There is no problem with the anomalous U(1)_A. Because of the anomaly no Goldstone boson is expected, and indeed none is observed. | |
Oct 20, 2015 at 22:53 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 5 characters in body; edited tags; edited title
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Oct 20, 2015 at 22:16 | history | asked | EEEB | CC BY-SA 3.0 |