As far as I understand, typical decay for $\omega$ meson is into $3\pi$, while for $\rho^0$ is into $2\pi$. In fact they are quite similar particles (same spin, parity, similar masses). Why this difference in their decay?
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1$\begingroup$ See en.wikipedia.org/wiki/G-parity $\endgroup$– rob ♦Commented May 8, 2022 at 21:00
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2 Answers
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The $\rho$ forms an isospin triplet ($I=1$)and the $\omega$ is an isospin singlet ($I=0$). The $g$-parity $$ g= Ce^{i\pi I_2} $$ that determines the odd versus number of pions in the decay goes as $(-1)^I$.
answered May 8, 2022 at 21:05
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In particle physics, a rho meson is a short-lived hadronic particle that is an isospin triplet
The isospin of the omega meson is zero
In nuclear physics and particle physics, isospin (I) is a quantum number related to the up- and down quark content of the particle. More specifically, isospin symmetry is a subset of the flavour symmetry seen more broadly in the interactions of baryons and mesons.
Isospin is conserved in the strong interactions, so the difference in the decays is due to this .
