yesterday i had a discussion with some friends about the decays of pions which produced a few questions that we couldn't resolve properly.
if anyone can shed some light on this matter, or point out flaws in the reasoning below, that would be great.
so starting with the decay of a neutral pion, and the question of why it is not simply an annihilation qqbar -> gamma -> e+e-; my understanding is that this diagram would not be allowed since the pion is spin 0 and the photon is spin 1. is this correct?
there is a non-zero (~10^-8) branching fraction for pi0 -> e+e- (with no final state photons); i'm assuming this proceeds by some higher order diagram and is hence supressed.
so anyway, if the tree diagram qqbar(pion) -> gamma -> e+e- is not allowed because of spin, how is it that the charged pion decays to muon+neutrino with very high branching fraction? i have seen in text books and on wikipedia ( http://en.wikipedia.org/wiki/Pion
) this decay being drawn as ud -> W -> mu nu, but as in the case for the neutral pion, the pion is spin zero and the W is spin 1, so how can this happen?
or is it that the vertex pion -> vector boson is allowed, but just supressed in some way? if so what is the nature of that supression?
i have looked at the matrix element calculation for pi+ -> mu+ nu and it seems that the current on the quark side of the diagram cannot be written with gamma^mu(1 - gamma^5) ; does this mean that the propagator is not a W? if so then what is it?
so, if anyone has any thoughts on this, or you think i'm just being retarded, please let me know.