- #1
karlzr
- 131
- 2
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 [itex]K^{+}K^{-}[/itex] should be an eigenstate of G, according to the first statement. But after applying [itex]G=e^{-i\pi I_y}C[/itex] to [itex]K^+=u\bar{s}[/itex], we obtain [itex]\bar{K^0}=\bar{d}s[/itex]. Similarly, [itex]K^-[/itex] changes into [itex]K^0[/itex](here [itex]e^{-i\pi I_y}=e^{-i \pi \sigma_y/2}[/itex] for SU(2)) . Then how can we say [itex]K^{+}K^{-}[/itex] is an eigenstate of G?
In addition, the composite state consisting of [itex]K^{+}K^{-}[/itex] should be an eigenstate of G, according to the first statement. But after applying [itex]G=e^{-i\pi I_y}C[/itex] to [itex]K^+=u\bar{s}[/itex], we obtain [itex]\bar{K^0}=\bar{d}s[/itex]. Similarly, [itex]K^-[/itex] changes into [itex]K^0[/itex](here [itex]e^{-i\pi I_y}=e^{-i \pi \sigma_y/2}[/itex] for SU(2)) . Then how can we say [itex]K^{+}K^{-}[/itex] is an eigenstate of G?