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Nambu-Goldstone bosons from a quantum anomaly symmetry breaking?

+ 6 like - 0 dislike
73 views

We know that:

Nambu-Goldstone bosons come from Goldstone theorem: a spontaneous (continuous)-symmetry breaking of the system leads to massless scalar modes.

quantum anomaly: is the anomalous phenomena where the (classical Noether) symmetry G respected by the physical system in a classical limit, but this symmetry G is broken by the quantum effect. This is the case where the action $S$ preserves the symmetry, but the path integral partition function $Z=\int [D\Psi][D\Phi]\dots e^{iS}$ and the measures $[D\Psi][D\Phi]\dots$ do not preserves the symmetry.


  • Question: whether there is any example that Nambu-Goldstone bosons can be derived from a spontaneous symmetry breaking caused by quantum anomaly effect? Alternatively, whether there is a known theorem to prove ``No Nambu-Goldstone bosons from a quantum anomaly symmetry breaking?''

[words of caution & side remark]: I offer some further thoughts if you wish you can skip it. There is a statement learned: 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 instantons. This is regarded as a reason why we do not observe a light pseudo-scalar meson $\eta'$ in the QCD mesons. 1 among the 9 mesons is this $\eta'$, which stands for axial $U(1)_A$ anomalous symmetry breaking, there are still $SU(N_{flavlor})_A=SU(3)_A$ which is spontaneous broken by dynamical symmetry breaking, which induces 8 among the 9 mesons, such as three $\pi$, three $\kappa,$ and one $\eta$. There are also $SU(N_{flavlor})_V=SU(3)_V$ broken by the nonzero explicit quark masses($m_u\neq m_d \neq m_s$). There is also $U(1)_V$ broken by Sphaleron, such that numebr of baryon $N_{baryon}$ is not conserved, but which only conserved numbers of $N_{baryon}-N_{lepton}$. Anyway, altogether $$SU(N)_V \times SU(N)_A \times U(1)_V \times U(1)_A$$ makes it to be the full $U(N)_V \times U(N)_A$.

Simply that $U(1)_A$ symmetry is anomalous broken by the quantum effect, but we do not see its Goldstone boson $\eta'$.

ps. please one may read full comments below the questions.

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user Idear
asked Aug 21, 2013 in Theoretical Physics by wonderich (1,400 points) [ no revision ]
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I wonder if it is not related to this 'Hooft paper, that is : "chiral U(1) is explicitly broken by instantons" ?

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user Trimok
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.

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user 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.

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user Heidar
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).

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user BebopButUnsteady
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"

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user Idear
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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.

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user Idear
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.

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user Idear

1 Answer

+ 3 like - 0 dislike

There are two different kinds of symmetry breaking involved in your question. The first would be spontaneous symmetry breaking, in which case we are dealing with a theory that is invariant under a certain symmetry group, but its vacuum is not. The breaking of the symmetry corresponds to a specific choice of the vacuum, the freedom of choosing a vacuum results in a new degree of freedom: the Nambu-Goldstone- boson. Depending on the type of symmetry that is broken, one might get one or more of them. In the case of the chiral symmetry of QCD, the $SU(N_f)_A$ part is broken spontaneously, resulting in eight massless bosons. The other axial part, $U(1)_A$, however, is not broken spontaneously.

The second kind of symmetry breaking would be induced by quantum anomalies. We speak of such an anomaly when a theory is classically invariant under a certain symmetry operation but the corresponding quantum theory is not. In terms of the path integral formalism this is manifest in the fact that the measure transforms in a nontrivial way. An example would be the breaking of the axial $U(1)_A$ part of the chiral symmetry in QCD, which is not related to any choice of a vacuum, and there is no Nambu-Goldstone boson connected to it. In fact, it can be traced back to something entirely different, namely instantons. The occurence of an anomaly is related to the number of fermionic zero modes of the theory: the difference between fermionic and antifermionic zero modes is given by the Pontryagin number of the topological instanton configuration of the gauge fields. This is also known as the Atiyah-Singer theorem.

Since these two possible symmetry breaking mechanism are quite distinct, the concept of Nambu-Goldstone bosons arising from quantum anomalies can only be a result of sloppy, non-standard terminology.

This post imported from StackExchange Physics at 2014-06-04 11:33 (UCT), posted by SE-user Frederic Brünner
answered Mar 5, 2014 by Frederic Brünner (1,060 points) [ no revision ]

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