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Chiral symmetry 
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#1
Sep1609, 04:56 PM

P: 969

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. 


#2
Sep1609, 05:20 PM

P: 25

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... 


#3
Sep1609, 10:01 PM

Physics
Sci Advisor
PF Gold
P: 6,242

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 interactionactually, of course, it's SU(2) x U(1), for the electroweak interaction (and it's actually a lefthanded x a righthanded copy of this).
As I understand it, each "generation" of quarks (u/d, s/c, b/t)and each generation of leptons as wellhas 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 lefthanded (L) and righthanded (R) representations are not the same (lefthanded fermions are SU(2) doublets; righthanded 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 John Baez' page on elementary particles for a much more detailed discussion. 


#4
Sep1609, 11:39 PM

P: 969

Chiral symmetry
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: [tex]m^2_{\pi}=2(m_u+m_d)\nu^3/f^2_{\pi}[/tex] 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. 


#5
Sep1809, 11:26 AM

Physics
Sci Advisor
PF Gold
P: 6,242




#6
Sep1809, 02:43 PM

P: 969

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. 


#7
Sep1809, 04:48 PM

P: 136

Once you accept that we must have a broken approximate symmetry, there is a wellestablished 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 ongoing 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. 


#8
Sep1909, 12:34 AM

P: 969

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). 


#9
Sep1909, 02:34 PM

P: 136

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 2dim 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 nonperturbative 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. 


#10
Sep2209, 04:01 PM

Sci Advisor
P: 1,229

The theory of the pions as Goldstone bosons is in excellent agreement with experiment. See, e.g., http://arxiv.org/pdf/hepph/0107332v1



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