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Decay widths for $\rho$ meson is $149 MeV$ while for the $\phi$ meson it is $4MeV$. Why is there such a difference?

I know that the phi meson decays primarily to $K \bar K$ states as the $\pi^+ \pi^- \pi^0$ states are OZI suppressed. The $\rho$ meson decays predominantly to $\pi^+\pi^-$ via strong interaction.

Is OZI suppression the culprit?

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Here is the phi mass 1019.445±0.020 MeV/c2. Here is the K mass 493.667±0.013 MeV/c2 , times two 987

1019-987~32 MeV/c2 are left over to be shared as kinetic energy of the two kaons.

It is a matter of phase space. The two pions of the rho are ~280 MeV/c, leaving a lot of phase space to facilitate the decay, i.e. give larger probability because of larger integration scope. enter image description here

In a e+e-scattering experiment: the probability of generating light quarks at a given energy is higher than heavy quarks to make up the phi, because of phase space.

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  • $\begingroup$ I don't see why the phase space argument answers the question. The fact that the phase space is small for the $\phi$ to decay into 2 kaons is an argument to explain why this partial decay width is small. It doesn't explain why the total decay width of the $\phi$ is small. One should explain why in spite of such a small phase space, this decay channel remains the one having the largest branching ratio. $\endgroup$ – Paganini Jun 15 '15 at 14:27
  • $\begingroup$ @Paganini The phi is made up with the strange antistrange pair of quarks, therefore kaons . The other decay mode of e+e- is 10^-4 branching ratio to K+K- journals.aps.org/prl/abstract/10.1103/PhysRevLett.21.1504 $\endgroup$ – anna v Jun 15 '15 at 15:35
  • $\begingroup$ ok, thanks. I thought the $\phi$ meson was a more complicated objects made of a superposition of $u\bar{u}, d\bar{d}, s\bar{s}$. If it's only $s\bar{s}$, then your explanation is clear. $\endgroup$ – Paganini Jun 15 '15 at 18:44

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