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so apparently the this cool looking decay of the $J/\Psi$ particle is OZI suppressed, which was confusing to me, because the Particle Data Group (link) says that 64% of its decays are into 3 gluons (as does wikipedia). So my question is, do I misunderstand what the PDG is saying, or does it emit 3 real gluons which don't decay into hadrons? Don't gluons have a finite lifetime because they're not white? Any help would be appreciated, thanks. Cool looking but OZI suppressed,

SuperCiocia
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Derian
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1 Answers1

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Hot tip: read the PDG listing mindful that the partial widths given, $\Gamma_1-\Gamma_6$ are not mutually exclusive! (Your $\Gamma_{80}$ is in $\Gamma_{3}$ and $\Gamma_{1}$.)

enter image description here

That is, the BRs must sum to 100%.

So, roughly, you either look at real particles in your detector, $$ \Gamma_1 + \Gamma_5 + \Gamma _6 \sim \Gamma \\ 87.7+5.94+5.93 \sim 100, $$ or else inferred virtual particles coupling to the charm vertex maintainig symmetries, (C, zero color, etc...), $$ \Gamma_2/\Gamma + \Gamma_3/\Gamma + \Gamma_4/\Gamma \sim \Gamma_1/\Gamma \\ 13.50+64.1+8.8 \sim 87.7 ~. $$ All gluons and most photons are virtual.

The striking takeaway, which confused experts to no end when the $\psi/J$ was discovered, is that the OZI suppression in, notably, $\Gamma_3$ reduces a strong rate to the pitiful status of competing with an EM one, $\Gamma_2$; also note the anaemic strong/EM ratio $\Gamma_3/\Gamma_4$.

At the time, nobody had seen anything of the sort,$^*$ making the narrowness of the $\psi/J$ a bewildering scandal.


$^*$I was in the corridor when (1975) George Zweig failed to convince Feynman of the logic of his eponymous rule yet again, and had to leave his (RPF's) office dejected with the door angrily slammed behind him... being right is not everything all the time...

Cosmas Zachos
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