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Chiral symmetry breaking and approximate flavour symmetry

  1. Jul 7, 2009 #1
    I have 2 questions:

    1. When there are no fermion mass terms, the Dirac part of the Lagrangian posseses an SU(N) left X SU(N) right flavour symmetry for N flavours of fermions. This can be "re-arranged" as an SU(N) vector X SU(N) axial symmetry. The axial part is spontaneously broken by the quark condensate. People often talk about spontaneous breaking of the SU(2) axial symmetry of up and down quarks, and how the pions are the associated goldstone bosons; however, when there are no quark masses in the Lagrangian it actually possessed and SU(6) vector X SU(6) axial symmetry. Why does no one talk about this symmetry, and why dont people study all the mesons as the goldstone bosons of this broken symmetry as they do with broken SU(2) symmetry?

    2. Why do people talk about there being an approximate SU(3) flavour symmetry amongst the u,d and s quarks when actually there is a U(3) symmetry? There is no reason why the transformation has to be a special one, the Dirac Lagrangian is invariant under a global unitary transformation.
     
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  3. Jul 7, 2009 #2

    Haelfix

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    Practical reasons. Even at the level of SU(3) * SU(3), the quark mass terms badly break the symmetry. For the full flavor hierarchy, the approximation is even worse and so not really used in practise.

    For the second question, I believe is a bit of a physics group theory convention issue in how we are actually naming things. The centralizers of the unitary group have these factors of U(1)/Z2 floating around and its a little confusing. I'll try to explain it if I get the chance and if I remember.

    edit:I think the general gist is that for instance U(2) (the global flavor symmetry group for u and d quarks) is locally isomorphic to something like SU(2) * U(1)/Z2 where the latter is something like a baryon number.

    Its a group theory vs lie algebra sloppy physicist notation. For instance the real standard model group is not SU(3)*SU(2)*U(1), but actually something like S(U(3)*U(2))
     
    Last edited: Jul 8, 2009
  4. Jul 8, 2009 #3
    The other axial U(1) is spoiled by anomaly even with massless quarks so it does not help chiral perturbation.
     
  5. Jul 8, 2009 #4

    Haelfix

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    Quite so, the axial U(1) is violated by instantons and this gives mass to one of the Goldstone bosons.
     
  6. Jul 8, 2009 #5
    But I was thinking about an energy region before the electroweak symmetry has been broken. Then there truly are no mass terms for the fermions, and chiral symmetry is an exact SU(6) left X SU(6) right symmetry, which is spontaneously broken by the quark condensate. All the mesons would then presumably be the goldstone bosons of this broken symmetry.

    Thanks for taking the time to reply by the way. I know that U(N) is homomorphic to SU(N) X U(1). Maybe we talk about approximate SU(3) flavour symmetry simply because we are used to working with SU(3) colour , when in actually fact it should be U(3).
     
  7. Jul 8, 2009 #6

    Haelfix

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    So lets see.
    we have U(3)*U(3) ~ SU(3) * SU(3) * U(1) * U(1).

    The First U(1) is a vector current and associated with baryon number, which is exact and not approximate. The second U(1) is an axial current and anomalous.

    The part of that decomposition that is both approximate and spontaneously broken is thus the special unitary groups which breaks down to its diagonal subgroup, so it makes sense to talk about them.
     
  8. Jul 8, 2009 #7

    blechman

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    In such a regime, chiral symmetry is restored, QCD is perturbative, and it no longer makes sense to talk about mesons and baryons!
     
  9. Jul 8, 2009 #8
    Ok so at such high energies QCD matter is in the quark gluon plasma state, but surely that doesnt mean that the meson states dont exist in the theory, just that they arent excited. Or if they are, they quickly get torn apart. The mesons should still be the goldstone bosons of the broken symmetry, youre just not likely to see them at such high energies.
     
  10. Jul 8, 2009 #9
    Actually never mind Im being an eejit
     
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