Decay of the Meson

Question

I'm trying to draw Feynman diagrams for 'The decay of the pseudoscalar meson $\phi = |s \bar{s}\rangle $ into these final states:

i) $\phi \rightarrow \pi^{+} \pi^{-},$ where $\pi^{+}=|u \bar{d}\rangle$ and $\pi^{-}=|d \bar{u}\rangle$

ii) $\phi \rightarrow \pi^{+} \pi^{-} \gamma$ '

but when I was reading up how to do them, I came across this from David Griffin's 'Introduction to Elementary Particles' which seems to explicitily say that the reaction isn't possible. This tarries with everything else I've found on the web: the ϕ doesn't decay into just two two pions. Is there an error in the question ? If not, how I can go about attacking how to draw the Feynman diagrams ?

Answer 1

A quick trip to the PDG will reassure you that reaction i) goes at a BR of $7\cdot 10^{-5}$ and reaction ii) at $4\cdot 10^{-5}$.

What Griffiths' book calls "forbidden" means "greatly suppressed", and I assume you went where he is sending you — but I am not familiar with that text.

In the strong interactions, I and G are pretty good symmetries, so the φ having negative G cannot go strongly to two pions each with G=-, so collectively with G=+.

Nevertheless, electromagnetism breaks I and hence G, and an exchanged photon costs an α in the amplitude and the square of that in the rate. Do you now see where the suppression factor in the BR must be coming from? What do your Feynman diagrams look like?