# Why is the phi meson decay width much smaller than rho meson?

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?

• – dmckee --- ex-moderator kitten Jun 13 '15 at 16:02
• I know the relation : $\tau \sim \frac{1}{\Gamma}$. My question is why is the decay width much smaller – user44840 Jun 13 '15 at 17:12

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. • 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. – Paganini Jun 15 '15 at 14:27
• 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. – Paganini Jun 15 '15 at 18:44