[Physics] Why can omega meson not decay into two charged pions

Question

Why is the decay of the $\omega$ meson similiar to the neutral $\rho^{0}$ meson into two charged $\pi$ prohibited?

My first thought was that it has something to do with isospin conservation in the strong interaction, but obviously $\pi^{+}$ and $\pi^{-}$ can couple to an Isospin $I = 0$ with $I_{3} = 0$. So this does not seem to be a problem.

I then assumed it might be due to charge conjugation, but then I noticed I don't really know how this works for charged pions that are not an eigenstate to $C$.

So $\omega$ has $C=-1$ and $\pi^0$ has $C = 1$ while $C|\pi^+> = – |\pi^->$. But from there I don't know how to proceed.
Thank you very much.

Answer 1

Here are the quantum numbers for omega and rho.

They differ in G parity, and G parity is conserved in strong interactions ( am using this paper ), it is a generalization of C parity.

The strong interaction was not believed to allow an ω to decay to the pion pair, as to do so would violate G parity.

The table of decays of the omega:

actually showed that a pi+ pi- decay was apparent in the two pion resonance of the rho:

Glashow suggested in 1961 that EM effects mixed the two states of pure isospin, ρ_I and ω_I, resulting in the mass eigenstates,ρ and ω, being superpositions of the two initial fields.

It is a long story given in chapter 3 of the link and summarized in 3.5 .

The short answer is that the channel exists, and a series of studies in the 1990s tries to explain how G parity is not conserved because omega has a sizable decay width to two pions.