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A Why is a glueball not massless, if it's a Goldstone boson?

  1. Feb 6, 2018 #1

    Wim

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    Scalar glueballs in QCD appear as a result of violation of global conformal (scale) symmetry - the energy-momentum tensor has a nonzero trace. According to the Goldstone theorem, this (violation of global symmetry) corresponds to the appearance of scalar massless bosons.
    Why, then, are the scalar glueballs massive, and the massless variant is not considered anywhere?
     
  2. jcsd
  3. Feb 9, 2018 #2
    The Goldstone theorem applies to spontaneously broken symmetry. Conformal symmetry in QCD is broken by an anomaly, which is something different.
     
  4. Feb 9, 2018 #3

    Wim

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    Could you specify any literature about this?
     
  5. Feb 9, 2018 #4
    Could you tell me first, where you got the idea that scalar glueballs exist because of conformal symmetry breaking?
     
  6. Feb 9, 2018 #5

    Wim

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    Ок
    M.A. Shifman, A.I. Vainshtein and V.I. Zakharov Nuclear Physics B147 (1979) for example.
    (in p. 391)
     
  7. Feb 9, 2018 #6

    Wim

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    V.A. Novikov M.A. Shifman, A.I. Vainshtein and V.I. Zakharov Nuclear Physics B174 (1980) 378-396
    About glueballs
     
  8. Feb 9, 2018 #7
    OK, I thought you might have been reading modern works on holographic QCD. There is a kind of five-dimensional approximation to QCD where conformal symmetry can be spontaneously broken in the fifth dimension, and it really does produce a massless goldstone "glueball" - but this is considered an incorrect approximation.
     
  9. Feb 9, 2018 #8

    Wim

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    No, I confine to the renormalizable d = 4 QCD.
     
  10. Feb 10, 2018 #9
    Here is a brief explanation of why spontaneous symmetry breaking, and symmetry breaking through anomalies, are qualitatively different things.
     
  11. Feb 10, 2018 #10

    Wim

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    Ok
    1. "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."
    2. " The second kind of symmetry breaking would be induced by quantum anomalies..... which is not related to any choice of a vacuum
    and there is no Nambu-Goldstone boson connected to it."

    But appearance of glueballs is associated with the breaking of the symmetry corresponds to a specific choice of the vacuum -
    nonzero gluon condensate. This condensate is associated with a violation of the scale symmetry and сonsequently
    Nambu-Goldstone boson.

    So the question remains.
     
  12. Feb 10, 2018 #11
    It's not a choice; QCD is "born that way". The existence and size of the gluon condensate is determined by the anomaly, which is an intrinsic property of the theory. There are no QCD vacua without it.
     
  13. Feb 11, 2018 #12

    Wim

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    Anomalies are instantons?
    Please specify which anomaly in this case.
     
  14. Feb 11, 2018 #13
    I wrote this with great confidence when I was half-awake (and I definitely meant the conformal anomaly), but it seems to be a baseless assertion. I'll do some homework and then I'll be back with, hopefully, a better-justified analysis.
     
  15. Feb 12, 2018 #14
    After a few days' literature search, I now know a lot more about all the relevant topics, but can't say I have hit bottom yet. However, I have found for you a paper in which they discuss exactly your scenario. See Section V of "Scale anomaly and the scalars".
     
  16. Feb 13, 2018 #15

    Wim

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    This is very interesting, thank you. I'll figure it out, go through the links in the article and comment later...
     
  17. Feb 15, 2018 #16

    Wim

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    With scalar glueball it became clearer. Still with a tensor glueball to understand, what there for anomalies...
     
  18. Feb 17, 2018 #17

    vanhees71

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    There is a profound difference between "spontaneous symmetry breaking" and "anomalies". A global symmetry of the Hamiltonian (or the action functional) is called spontaneously broken, if the ground state is not symmetric. This necessarily implies that the ground state is degenerate, and the excitations connecting different ground states are thus massless (hand-waving argument for the validity of the Nambu-Goldstone theorem).

    Note that local symmetries cannot get spontaneously broken, although many texbooks call the Higgs mechanism the spontaneous breaking of a local symmetry. That there is no spontaneous symmetry breaking is clear since here the ground state is not degenerate, because the apparent different ground states with the same energy are connected by a local gauge transformation, i.e., they are in fact the same state. That explains also why there are no massless Nambu-Goldstone bosons in this case but the degrees of freedom which would lead to Goldstone bosons if the symmetry were only global (the "would-be Goldstone modes") are eaten up by the corresponding gauge fields, providing the third component for each of these gauge bosons that refer to the corresponding part of the gauge group, and these gauge bosons get massless (e.g., the electroweak standard model is Higgsed from ##\mathrm{SU}(2) \times \mathrm{U}(1) \rightarrow \mathrm{U}_1##, i.e., of the four gauge boson degrees of freedom three get massive, and each of them eats thus up one of the would-be Goldstone modes, providing the third ##m=0## spin component for a massive vector boson.

    Anomalies occur, if a symmetry of the classical action (no matter whether it's global or local) is no longer a symmetry in the quantized theory. This implies that in the quantized case there is no symmetry to begin with, and thus the symmetry of the classical action is explicitly broken. If this happens to a local gauge symmetry you can forget the entire model since a gauge theory whose gauge invariance is broken is unphysical (involving unphysical degrees off freedom in the dynamics leading to a non-unitary S-matrix, and the whole theory lacks physical meaning). For global symmetries anomalies are often welcome: E.g., in neutral-pion decay to 2 Gammas the anomalous breaking of axial U(1) symmetry leads to the correct decay rate for ##\pi^0 \rightarrow \gamma \gamma##. Mathmatically there are also remnants of the classical symmetry left, in terms of socalled anomalous Ward identities.
     
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