Width of the J/$\psi$ resonance The J/$\psi$ resonance has a mass of ~3 GeV but its width is only of 100 keV, why is it so narrow compared to the total mass? How can the width of these resonances be calculated?
 A: The small width of the $J/\psi$ was an important discovery, sometimes called the "November Revolution", see "The Revolution That Shook Particle Physics" (2014-10-21).
It was quickly understood to be due to a new, heavy, quark flavor, called "charm". The $J/\psi$ is a charm-anti-charm bound state with the quantum numbers $J^{PC}=1^{--}$. The reason for the small width is 


*

*The $J/\psi$ is almost the ground state of the $c\bar{c}$ system. There is a lighter state, called $\eta_c$, but there are no hadronic decays of $J/\psi\to\eta_c$, and the electromagnetic decay width $J/\psi\to\eta_c\gamma$ is small.  

*The $J/\psi$ therefore predominantly decays by $c\bar{c}$ annihilation into light quark pairs (eventually, pions). Because the $J/\psi$ is heavy, this process can be analysed in QCD perturbation theory.

*The $J/\psi$ is a color singlet and cannot annihilate into a single gluon. It has $J^P=1^-$ and cannot decay into two gluons. The process is therefore $J/\psi\to 3g\to 3(q\bar{q})$, which is proportional to $\alpha_s^3$ (and the wave function of the $c\bar{c}$ pair at the origin). If we treat the $c\bar{c}$ as a color-Coulomb bound state the wave function would bring in extra powers of $\alpha_s$, but the $J/\psi$ is not quite heavy enough to fully justify a perturbative treatment of the wave function. In any case, the large power of $\alpha_s$ explains the small $J/\psi$ width (the $\eta_c$, which can decay into $2g$, is indeed broader).    
