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Sep 11, 2013 at 17:08 comment added wonderich Read the statement of t 'Hooft PRL 37,8(1976): "When one attempts to construct a realistic model of nature one is often confronted with the difficulty that most simple models have too much symmetry. Many symmetries in nature are slightly broken, which leads to, for instance, the lepton and quark masses, and CP violation. Here I propose to consider a new source of symmetry breaking: the Bell-Jackiw anomaly." So even 't Hooft regards ABJ anomaly as a way of symmetry breaking. So it is fair for me to ask the possibility of Nambu-Goldstone bosons from a quantum anomaly symmetry breaking.
Sep 11, 2013 at 17:05 comment added wonderich To Lubos and Heidar again: you are answering the standard textbook statement (which everyone can read from standard QFT textbook, and I had learnt), I am not asking this level of question. I am asking a more subtle (or deeper) level question. Please DO digest my question before attempt to answer it. Thank you.
Sep 11, 2013 at 2:50 comment added wonderich To Lubos and Heidar: look into the original study of tHooft, the title of the paper: Phys. Rev. Lett. 37, 8–11 (1976) - Symmetry Breaking Through Bell-Jackiw Anomalies-Gerard 't Hooft, it may be fair to give some credits to question the relation between the anomaly and symmetry breaking. (as I said I agree the symmetry is not even there originally, but does it matter how we interpret why there is no symmetry there?) You are using the canonical way of thinking, I know, but I still question an alternate way of thinking.
Sep 5, 2013 at 16:57 comment added Luboš Motl Dear @Idear, I agree with Heider, too. You seem to be manipulating yourself into thinking strange (wrong) things by the misleading vocabulary. Why do you keep on saying "a symmetry is broken by an anomaly" if you accepted (in a comment above) that an anomaly is not (spontaneous) symmetry breaking? An anomaly is a proof that the apparent symmetry has been an illusion from the beginning. The current is not conserved. There is a right-hand side just like there would be from an explicit symmetry breaking.
Aug 22, 2013 at 8:11 history edited wonderich CC BY-SA 3.0
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Aug 22, 2013 at 8:05 comment added wonderich Continue: I mean, for example,can one do an explicit analytic exercise using Fujikawa's path integral approach, where a global symmetry transformation G (of underlying fermionic) is broken by a quantum anomaly, but this symmetry G is explicitly the same symmetry of a bosonic fields potential(say G broken down to N, such as O(N) broken down to O(N-1), or $U(N)$ broken down to $SU(N)$)? I wonder whether it makes sense to do this analysis to see there are/aren't |G/N| number of Goldstone bosons?
Aug 22, 2013 at 7:52 comment added wonderich continue: and Weinberg's book QFT II, Sec. 22.7 Anomalies and Goldstone bosons. " the gauged effective field theory of Goldstone bosons must have an anomaly for the fictions symmetry which is equal to what has been produced from the underlying (fermion) theory." I thought there may be some room to think a bit more for this question?
Aug 22, 2013 at 7:47 comment added wonderich TO Heidar: I do agree with what Heidar said. My original understanding is that the quantum anomaly for a current non-conservation respect to G is a fact that there is no that G symmetry at all for that system from the beginning. So that is why we do not have $\eta'$ meson, as I gave the example. However, I read this statement from Fuijikawa's book "Path Integral and Quantum Anomaly": Sec 5.6.2 "Nambu-Goldstone bosons do not generally appear for a spontaneously broken symmetry if the relevant global symmetry is broken by the effects of the anomaly and and instantons"
Aug 21, 2013 at 19:32 comment added BebopButUnsteady In support of @Heidar, the anomaly is not like spontaneous breaking at all. If anything, it is like just adding a term in the lagrangian which violates the symmetry. If a system has a $SU(2)$ spin symmetry, you don't expect a goldstone boson when you turn on an external magnetic field. If in zero field there is SSB, then you expect turning on a magnetic will cause the goldstone boson (magnon) to develop a gap. Its the same with the $\eta'$ (and also with the other pions with respect to quark mass to a lesser extent).
Aug 21, 2013 at 18:37 comment added Heidar However, as you very likely know well, the 't Hooft anomaly matching condition says that the anomaly is scale independent, the UV and IR anomalies must match. For me, this seems to imply that the anomalous symmetry is not a symmetry of the theory at all in any limit. It just appears to be a symmetry in the naive classical limit, but that's just an illusion. Thus its not a spontaneous nor explicit breaking of symmetry, but absence of symmetry. This line of reasoning might very likely be wrong, though.
Aug 21, 2013 at 18:37 comment added Heidar I am a little confused by the statement "...a spontaneous symmetry breaking caused by quantum anomaly effect". I am not sure I understand why the quantum anomaly can be seen as 'spontaneous symmetry breaking' of symmetry...? Let me make a few naive comments on this, which might very well be wrong. Usually anomalies are described as the impossibility to find a regularization scheme in which the anomalous symmetry is explicitly preserved, and thus it implies that the anomaly is a problem of the UV completion.
Aug 21, 2013 at 18:11 history tweeted twitter.com/#!/StackPhysics/status/370246887859032064
Aug 21, 2013 at 15:40 comment added Trimok I wonder if it is not related to this 'Hooft paper, that is : "chiral U(1) is explicitly broken by instantons" ?
Aug 21, 2013 at 12:01 history edited wonderich CC BY-SA 3.0
combine the question with a thought on a QCD example.
Aug 21, 2013 at 11:43 history edited wonderich CC BY-SA 3.0
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Aug 21, 2013 at 11:37 history edited wonderich CC BY-SA 3.0
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Aug 21, 2013 at 10:45 history asked wonderich CC BY-SA 3.0