〈 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