# Why is $\psi$ meson decay over a virtual photon so likely?

So a $\psi$ is a meson with a charm and an anticharm quark. The particle data group lists the following branching fractions for the following decay modes of the $\psi$:

• virtual $\gamma\rightarrow$ hadrons: 13.5%
• g g g: 64.1%
• $\gamma$ g g: 8.8%

Now if I understood it correctly, the virtual photon should just contribute $\alpha$ to the probability amplitude (because it introduces two new vertices in the feynman diagram with each having $\sqrt{\alpha}$). So I would have assumed that the branching fraction for the virtual photon decay mode, which is proportional to the square of the probability amplitude, would have been

$BF(\gamma\rightarrow h) = (BF(ggg)+BF(\gamma gg))\cdot\alpha^2$

Now, obviously, this is way less than 13.5%, so where have I gone wrong?

EDIT:

Here the feynman diagrams I thought of for the "g g g" decay mode and the equivalent with a virtual photon:

From this, I thought that the virtual photon just added two vertices, and that I have to multiply $\sqrt{\alpha}$ to the probability amplitude for each of them. Are these even correct?

• Have you compared the first with the third mode listed? Have you back-of-the-envelope estimated the strength of the "strong" coupling at these energies? Commented Feb 2, 2018 at 14:54
• @CosmasZachos No I haven't. I'll try to draw the feynman diagrams I thought of and share them to explain better what I did.
– Keno
Commented Feb 2, 2018 at 15:01
• I may be misrecalling, but the strong α could be only 35 times larger than the EM one... Which is why it can compete so effectively against purely leptonic decays.... and you have 3 gluons, to boot.... There are superb reviews on the subject. You don't need to re-invent the OZI "wheel"... Commented Feb 2, 2018 at 15:17
• @CosmasZachos I updated my post to show my feynman diagrams. Based of these, I didn't think the $\alpha$ for the strong force mattered at all.
– Keno
Commented Feb 2, 2018 at 15:25
• No, once the photon has produced quarks, they dress-themselves softly and need not disappear to produce gluons--which are unobservable. Conversely, the 3 gluons in the hadronic decay mode resolve to quark-antiquark pairs and ... The whole point of the OZI rule explaining the narrowness better than that of the φ 's is that the strong coupling does matter, a lot, as you go up in energy. You are clearly asking for a review, of which there are dozens. Commented Feb 2, 2018 at 15:34