Lifetime of up quark

I am taking an introductory particle physics course, and from having studied about the breit wigner cross section and propagator terms:

$$\frac{1}{(E-E_0)^2+ \frac{\Gamma^2}{4}}$$

where $$E$$ is the energy in the centre of mass frame, $$E_0$$ is the rest energy of the propagator particle, and $$\Gamma$$ is its decay width. I was under the impression that divergences in transition rates are necessarily supressed because $$\Gamma$$ is always non-zero. However I have now come across a decay $$\pi^0 \rightarrow \gamma \gamma$$ where the intermediate particle is an up quark. From wat I am aware, these have a vanishing decay rate. How then does the rate for this process not diverge?

I am vaguiely reminded about the need for the introduction of the Z bososn and electroweak unification because otherwise there would be divergences from massless propagators. The propagator here isn't massless, but if the energy is tuned right then the propagator term could diverge?

• This is outside my area of expertise, but it seems to me that the mistake here is the idea that the decay involves the creation of an up quark which then decays. If you look at the Feynman diagram, there isn't a vertex where an up quark decays into two gammas. The quarks form a triangle. On any surface of simultaneity, there are either 2 quarks or none -- which makes sense in terms of baryon number conservation. – Ben Crowell May 2 at 15:21
• Are you fully aware of the neutral pion decay loop diagram and at peace with it? – Cosmas Zachos May 2 at 16:05
• @CosmasZachos Thank you for your reply. I have seen this diagram, bu thought that the breit wigner cross section applies for T/u channel diagrams (similar too this case) as well as s channel diagrams. Perhaps this is my mistake? – Meep May 4 at 9:24
• I'm not sure what you imagined the Breit-Wigner had to do with up and down participants in the triangle diagram of the decay. The propagator of the neutral pion has a Γ by dint of this decay, alright, but you compute the triangle loop diagram, (with Steinberger, 1949), just, and you get the correct width. – Cosmas Zachos May 4 at 15:34