Non-strange non-baryonic states are eigenstates of G-parity

by karlzr
Tags: eigenstates, gparity, nonbaryonic, nonstrange, states
 P: 60 It is said that all non-strange non-baryonic states are eigenstates of G-parity. And all members of an isospin multiplet have the same eigenvalue. Can anyone give me a proof to these two statements, or show me where I can find one? In addition, the composite state consisting of $K^{+}K^{-}$ should be an eigenstate of G, according to the first statement. But after applying $G=e^{-i\pi I_y}C$ to $K^+=u\bar{s}$, we obtain $\bar{K^0}=\bar{d}s$. Similarly, $K^-$ changes into $K^0$(here $e^{-i\pi I_y}=e^{-i \pi \sigma_y/2}$ for SU(2)) . Then how can we say $K^{+}K^{-}$ is an eigenstate of G?
 Sci Advisor P: 3,447 In order to talk about G-parity you have to start with an eigenstate of isospin. Whereas both K+K- and K0K0-bar are superpositions of I=0 and I=1.
P: 60
 Quote by Bill_K In order to talk about G-parity you have to start with an eigenstate of isospin. Whereas both K+K- and K0K0-bar are superpositions of I=0 and I=1.
Do you mean $K^+ K^-$ and $K^0\bar{K}^0$ are not eigenstates of G-parity? because I don't think so. Consider $\pi^+ \pi^-$, it can also be superposition of I=0,1,2, but the G-parity is $(-1)^2=1$, since $\pi$ are eigenstate of G with eigenvalue -1.
Actually this is about a problem from Bettini's elementary particle physics, 3.20 on page 107. Here is the link: