Why must a hadronic decay of the $J/\psi$ meson include (at least) three gluons? Why is the decay mediated by a single gluon allowed for the $\rho^0$ meson?
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$\begingroup$ Added 2nd part to answer. Is this what you had in mind? $\endgroup$– Cosmas ZachosCommented Feb 6, 2019 at 15:07
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Gluons carry color. So a colorless meson cannot annihilate its valence quarks to decay to one colored gluon and nothing else. You are flat wrong that the ρ decays thus through one gluon.
Two colored gluons, may combine into a colorless object, think $\bar R G$ and $\bar G R$, and the singlet has C=+. But the ψ has C=- , like the neutral ρ. Neither can decay through 2 gluons.
The ψ can, and does decay through ggγ : check your PDG.
Finally the (also charmonium, but scalar) χ has C=+ , and can and does decay through gg.
It's hard to see how you'd learn about the three gluon decay without a justification of the reason.
Edit in response to comment: The possibility arises that you are not focussed on Zweig's rule violation, but are interested in a two-meson decay mode involving the original valence quarks. This mode is available to the ρ, above threshold for a 2 π decay, but not to the J/Ψ(3097), below threshold for a 2 D(1865) decay; so that channel is closed, and you must decay via annihilation of your valence quarks.
By sharp contrast, the Ψ(3770) is above threshold, and so does very much decay to 2 D s, so, then, analogously to the decay of the ρ , somewhat misleadingly dubbed in the question as a "one gluon decay". The colored quark pair created by that gluon has to combine with the (color altered) valence quarks and hadronize through a resolutely nonperturbative process involving quintillions of soft gluons.
answered Feb 3, 2019 at 1:48
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$\begingroup$ Hi, thanks for the reply. I'm not completely sure about your answer. Why do you affirm that the $\rho^0$ cannot decay via one gluon without any violation of color conservation? There is no energy conservation violation (it has enough mass), it is in the PDG and here is the Feynman diagram physics.stackexchange.com/questions/163965/… $\endgroup$ Commented Feb 3, 2019 at 14:49
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$\begingroup$ Alas, that Feynman diagram is purely metaphorical/poetic: you'd never think of it as the leading term in a valid perturbative expansion at such energies. Throw in a few trillion gluons all over the place with comparable significance. This type of decay preserving initial valence quarks is unavailable to the J/Ψ for energy availability reasons--see edit to the answer. The way the question is articulated, it is apparently asking about Zweig's rule suppressed decays. $\endgroup$ Commented Feb 3, 2019 at 19:19
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$\begingroup$ Sorry if I misunderstand something, how to deduce that the two-gluon system (as a color singlet) has $C=+1$? I'm stuck on the step that $C=(-)^{L+S}$. Since $L$ and $S$ has many different values, so I suppose $C=-1$ is also possible? $\endgroup$– luyuwuliCommented Sep 18 at 7:14
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$\begingroup$ Could you be more elaborate on this point? For parity conservation, $(-)^{L}=-1$, which gives $L=1,3$, and C parity conservation, $(-)^{L+S}=-1$ can deduce that $S=0,2$. The combination of $L=1, S=0$, $L=1, S=2$ and $L=3, S=2$ can ensure the angular momentum conservation. $\endgroup$– luyuwuliCommented Sep 18 at 12:54
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1$\begingroup$ oops. Thanks for the update. I'll check on that paper $\endgroup$– luyuwuliCommented Sep 18 at 14:39