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Chiral symmetry

  1. Sep 16, 2009 #1
    If you model the quarks as massless, there should be no flavor mixing, because flavor mixing is achieved through the CKM matrix, which is a mass matrix.

    However, if quarks are massless, there ought to be an axial flavor symmetry, but there isn't.

    So to reconcile this, we must spontaneously break the axial flavor symmetry. This is done by introducing a "fermion condensate" flavor mixing field: [tex]\bar{\Psi}^{\alpha j}P_L \Psi_{\alpha i} [/tex] where [tex]\alpha[/tex] is the color index and 'i' the flavor index (refers to the flavor of quark) and 'j' is also the flavor index (refers to flavor of the antiquark).

    How do you spontaneously break this field? You need to add some terms to the Lagrangian involving this composite field, to spontaneously break it, right? And how come when you spontaneously break it, you don't set it equal to some constant VEV (as you do with the Higgs), but allow the VEV to vary in spacetime, and call this variation the pion? Then you treat the vacuum expectation value as separate, independent field, and give the pion field its own Lagrangian?

    The flavor symmetry in the book is called [tex] SU(2)_LxSU(2)_R [/tex] because the book is assuming only two flavors of quarks (up and down) for simplicity. So since there are 6 quarks, is the real flavor symmetry [tex] SU(6)_LxSU(6)_R [/tex]? The problem I have with this is some of the arguments of the book rely on the fact that SO(4)=SU(2)xSU(2), which is no longer true if 2=6.
    Last edited: Sep 16, 2009
  2. jcsd
  3. Sep 16, 2009 #2
    In QCD you have an approximate chiral simmetry because 3 of the quarks are light (the mass of u,d,s is less than the typical energy[tex] \Lambda_{qcd}[\tex]). So you have an SU(3)xSU(3) symmetry. The SU(2)xSU(2) is "more" exact because u and d are lighter.
    This isn't true for the others quarks and there isn't any evidence of SU(6)xSU(6) symmetry...
  4. Sep 16, 2009 #3


    Staff: Mentor

    The strong interaction symmetry group is SU(3)_L x SU(3)_R just for the first generation (up/down) quarks. The SU(2) symmetry is for the weak interaction--actually, of course, it's SU(2) x U(1), for the electroweak interaction (and it's actually a left-handed x a right-handed copy of this).

    As I understand it, each "generation" of quarks (u/d, s/c, b/t)--and each generation of leptons as well--has its own "copy" of each of the above (of course the leptons are SU(3) singlets because they don't participate in the strong interaction), and the left-handed (L) and right-handed (R) representations are not the same (left-handed fermions are SU(2) doublets; right-handed fermions are SU(2) singlets). So there's no SU(6)--it's more like three "copies" of [SU(3) x SU(2) x U(1)]_L x [SU(3) x SU(2) x U(1)]_R.

    Check out http://math.ucr.edu/home/baez/qg-spring2003/elementary/" [Broken] for a much more detailed discussion.
    Last edited by a moderator: May 4, 2017
  5. Sep 16, 2009 #4
    The strong interaction group is just SU(3). With SU(3)_L x SU(3)_R you have much more freedom than SU(3), because you are then allowed to choose different angles for left and right chiral quarks. I'll check out the website - looks like a good one.

    A pion is made of a quark and antiquark, each a different flavor. So in principle, to understand the pion, you only need to know about quarks. However, this is not quite true. Something weird happens. You create a new field to represent the pion, constructed from the quark and antiquark field! This is totally different from representing two particles as 2 numbers in a Fock space. You're creating a new field operator, albeit a composite one. What's even stranger is that this new pion field is not even just the combination of quark and antiquark fields: it is the vacuum expectation value of the combination, so now you have to wonder about how symmetry breaks. In fact, using these weird concepts, we can calculate the pion mass in terms of the quark mass, and it comes out to be:


    So it shows that you simply don't add the mass of the quark and the antiquark to get the mass of the pion. I don't even know if the mass is higher or lower than just adding the quark masses.

    I'm just having a hard time justifying any of this. It seems pretty ad hoc.
    Last edited by a moderator: May 4, 2017
  6. Sep 18, 2009 #5


    Staff: Mentor

    Oops, you're right--I was mis-stating it. Both left and right-handed quarks have the same irrep of SU(3); the difference is just that the left-handed quarks are SU(2) doublets while the right-handed quarks are SU(2) singlets.

    Well, it seems to match experiments--if you fit all the free parameters. I think I've read that there are seventeen free parameters that need to be fit to experimental data in order to use the Standard Model. I think particle physicists would agree that a lot of this is "ad hoc"--in the sense that we don't yet understand why these particular patterns and these particular values of the parameters are the ones that Nature is using. String theory is supposed to help with that.
  7. Sep 18, 2009 #6
    Most of the free parameters in the Standard Model have to do with mass. After all, there are only 3 forces, so there are only 3 coupling constants you have to specify. There is also a mixing angle between photons and Z's. So 4 parameters. I think that's it, unless I'm forgetting some others. Addings the Higgs and stuff and the mixing angles for quarks and neutrinos pushes it up to 17 or thereabouts.

    So the Standard Model doesn't feel ad hoc until you give the particles mass.

    With QCD, you are giving a product of quark fields a vacuum expectation value, but the quark fields themselves don't have vacuum expectation values. That just seems weird.
  8. Sep 18, 2009 #7
    The global SU(2)xSU(2) is an approximate symmetry of the QCD action due to the non-zero masses of the quarks (similarly for SU(3)xSU(3) when all quarks are considered). If this approximate symmetry were true at low energies, any hadron that appears from quark groupings would come with a parity-flipped cousin. Because these aren’t observed, you can conclude that the SU(2)xSU(2) (and SU(3)xSU(3) in the general case) must be broken in a spontaneous way. Granted, that’s not a completely satisfactory picture because we haven’t shown the dynamical reason for such a spontaneous breakdown. The reason for that is: QCD is hard. But, it so happens that nature gave us SU(3)xSU(3) (3 families of quarks) and there is a result based on anomaly requirements that says this group cannot be an unbroken symmetry, which means it must be broken spontaneously (i.e., by a background of fields).
    Once you accept that we must have a broken approximate symmetry, there is a well-established theorem that says there must be a light Goldstone boson. Furthermore, we know it must be a colorless combination of quarks. As I mentioned already, QCD is hard, by which I mean having a complete calculational description of hadrons based on elementary quarks is a very rich and on-going subject. But we can go a long way using effective field theories. We do these kinds of things all over physics, such as replacing a complicated quantum mechanical problem with a classical approximation and we don’t blink thrice. In this case, we know the approximate theory contains a light boson field (which is naturally just a product of quark fields). The resulting situation is not unlike the Higgs field, since this scalar [tex]\sigma[/tex] now has a background vacuum expectation value[tex]<\sigma>[/tex], and “fluctuations” about that background are represented by the quantum field [tex]\sigma_{0}[/tex]. That is, [tex]\sigma(x)=<\sigma>+\sigma_{0}(x)[/tex]. Don’t confuse the background with the effective quantum field.

    Think quantum mechanically, for identical spin-1/2 particles: They can act like a single object of spin 0, without being able to resolve them. It is this quantum pairing that then forms a condensate. You couldn’t get such a condensate from resolved quarks. Said a slightly different way, you can create quantum states out of the vacuum that, in the framework of quantum mechanics, cannot be described as a quantum state of individual quarks. Hence you can have a v.e.v. of the former and not the latter.
  9. Sep 19, 2009 #8
    If all quarks are massless it would be [tex] SU(6)_L \times SU(6)_R [/tex] flavor symmetry. But I guess only the u, d, and s quarks have small mass and can be approximated as massless, so it would be [tex] SU(3)_L \times SU(3)_R [/tex]. Also the electric charge difference between u and d quarks also doesn't allow [tex] SU(2)_L \times SU(2)_R [/tex] to be an exact symmetry.

    Just from the form of the Lagrangian the symmetry would actually be [tex] U(2)_L \times U(2)_R [/tex], but due to the axial [tex] U(1)_A [/tex] anomaly there is no [tex] U(1)_A[/tex] axial symmetry, so [tex] U(1)_L \times U(1)_R =U(1)_V \times U(1)_A[/tex] reduces to just [tex] U(1)_V [/tex] so the total symmetry would be [tex] SU(2)_L \times SU(2)_R \times U(1)_V[/tex]. So because of spontaneous breakdown [tex] SU(2)_A [/tex] gets broken and the full symmetry is [tex] SU(2)_V \times U(1)_V[/tex]. That's a big drop right there, from [tex] U(2)_L \times U(2)_R [/tex] to [tex] SU(2)_V \times U(1)_V[/tex]. But instead of spontaneous breakdown, if there is an [tex] SU(2)_A [/tex] anomaly, then you can achieve the same thing, and there would be no goldstone bosons! So couldn't the reason we don't observe the parity-different cousins be because of another anomaly and not spontaneous symmetry breaking?

    I can actually accept everything now. It makes intuitive sense. But is there any support from computer calculations that show this is the right track, that starting from the pure QCD Lagrangian (in terms of quarks and gluons), the ground state of a certain quark and antiquark pair forms a pion? I guess the only way to calculate with QCD at large coupling is lattice calculations, so I'm wondering if the computer calculations show we're on the right track.

    But I can totally accept everything now. Nucleons exchange pions and exchange of spin zero particles are always attractive, so nucleus is held together - the only assumption is spontaneous breaking of axial [tex] SU(2)_A [/tex] flavor symmetry (and some other small ones).
  10. Sep 19, 2009 #9
    Let's just talk about the SU(2)xSU(2), because the abelian parts are another story, and for that matter let's leave out QED...we're just looking at QCD. (1) If the quarks are taken to be massless so that this rigid symmetry is exact, then there can't be any local anomalies in the SU(2) factors since there aren't any complex representations of that group. (2) If the quark masses are introduced, the symmetry is only approximate, but is broken anyway. But massive fermions don't contribute to any anomalies in the remaining symmetry (nor in the approximate symmetry).

    SU(2) factors can have *global* anomalies if there are an odd number of 2-dim representation fermions in the given SU(2).

    It's known from computational models that quarks seem to indeed be confined starting from QCD. But not only are the composites *quantum* pairings of elementary particles, but the strong interaction makes non-perturbative effects important for understanding these objects. In the end, the picture of a quark bound state like those we first learn about in quantum mechanics is a very loosely true picture. Still lots of work to be done.
  11. Sep 22, 2009 #10


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