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

  1. Nov 16, 2012 #1
    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?
     
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  3. Nov 16, 2012 #2

    Bill_K

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    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.
     
  4. Nov 16, 2012 #3
    Do you mean [itex]K^+ K^-[/itex] and [itex]K^0\bar{K}^0[/itex] are not eigenstates of G-parity? because I don't think so. Consider [itex]\pi^+ \pi^-[/itex], it can also be superposition of I=0,1,2, but the G-parity is [itex](-1)^2=1[/itex], since [itex]\pi[/itex] 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:
    http://books.google.com/books?id=HNcQ_EiuTxcC&printsec=frontcover#v=onepage&q&f=false
     
  5. Nov 16, 2012 #4

    Bill_K

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    Sorry, my comment was incorrect. This paper might be of help to you, especially on p 9 where they work out the G-parity of K K-bar systems.
     
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