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.


2 Answers 2


Here are the quantum numbers for omega and rho.


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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.

  • $\begingroup$ Thank you so much! We did not cover G-parity in the lecture, that's why I didn't think of it. $\endgroup$
    – MmeTautou
    Commented Mar 26, 2017 at 11:32

As you rightly pointed out, the composite $\pi^+ \pi^-$ is not a definite isospin state, but has some component along $I=0$, so thats not a problem.

However, the quantum number violated here is G parity, defined by $$G=C exp(i\pi I_2)$$ Where $C$ is charge conjugation and $I_2$ is the second isospin generator. Clearly $\omega$ has G-Parity -1, since its an isospin singlet and has charge conjugation eigenvalue of -1.

As you correctly pointed out, pions are not eigenstates of charge conjugation. However, they are eigenstates of G since, under charge conjugation $\pi^+ \rightarrow -\pi^-$ and $\pi^- \rightarrow -\pi^+$, and action of $exp(i\pi I_2)$ is a rotation about the $I_2$ axis by $\pi$ and therefore takes states of $I=1$ to $I=-1$ and vice versa, thereby making the combined action if $C$ and $exp(i\pi I_2)$(Notice that they commute) $\pi^+ \rightarrow -\pi^+$ and $\pi^- \rightarrow -\pi^-$.

Hence the products have a G value of 1 while the initial state has G value of -1. Remember, that both charge conjugation and isospin are symmetries of the system and hence any combination of the two are symmetries of the system.


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