〈 Berger | Dillon 〉

The neutral π-meson particle π_0 (very very very shortlived!) is the anti-symmetric linear combination of uu' and dd' biquark states - where u' = anti-quark of the "up" quark u, and likewise for the "down" quark d. That is, π_0 = (uu' - dd')/√2. #physics #algebra

in terms of quantum numbers yes! but since QCD is strongly coupled we run into huge problems when we think of mesons being a bound state of quarks and gluons! confinement

But for weak interactions such linear combinations (also for the charged pions, e.g. π^- = u'd) are still good, right? (Since they produce correct decay rates, cross ratios, for the pions.) What do you think?

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?