〈 Berger | Dillon 〉

well the thing is the thing is that QCD becomes strongly coupled at distances ~ 10 femtometer ~ 1/Λ_QCD s.. so we've never actually seen such a state at those energies .. indeed a bound state makes perfect sense there, but that's just not what we call a π^0 w. mass ~135MeV

i think about it in terms of distances cause it's easier for me to visualize .. at distances<1/Λ QCD is well defined in terms of quarks and gluons .. as E->Λ the coupling becomes strong and our perturbative expansion breaks down .. (cont)

So how do we predict the spectrum? .. well for the lighter mesons we can .. QCD has an approximate global SU(3)_L x SU(3)_R family symmetry which is also spontaneously broken to SU(3)_V by the quark condensate <qq> .. (cont)

So by Goldstone's theorem we will have 3^2 - 1 = 8 goldstone bosons .. these are the light pseudoscalar mesons K^\pm π^\pm etc Since the symmetry is Global this means that they should be massless, though which is not the case .. the small masses are treated as a pertbtn .. cont

The Goldstones always come out massless, so could approximately represent mesons (as I understand). Would symmetry breaking not give them mass thereafter?

breaking the symmetry doesn't give them mass, since the symmetry isn't broken by the Higgs mechanism the vev <qq> is a so called 'non perturbative' effect, it's role strictly breaks the global SU(3)xSU(3) chiral symmetry .. we can't expand a strongly coupled theory abt a vev

I knew (or read) that the symmetry breaking doesn't give mass to gluons since they're massless, but wondered about mesons. How QCD explains their having nonzero mass (?). The non-perturbative aspect of QCD is something rather intriguing and which one day I'll have to study.

I agree 100% about the non-pert aspect of QCD .. I will admit i'm not an expert on it, though would love to learn more about it Well the mass term you're referring to is just the mass they obtain via the SM higgs mechanism m_q ~ v*y_q ! Then ..(cont)