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Nonstrange nonbaryonic states are eigenstates of Gparity 
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#1
Nov1612, 01:49 AM

P: 86

It is said that all nonstrange nonbaryonic states are eigenstates of Gparity. 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? 


#2
Nov1612, 05:57 AM

Sci Advisor
Thanks
P: 4,160

In order to talk about Gparity you have to start with an eigenstate of isospin. Whereas both K^{+}K^{} and K^{0}K^{0}bar are superpositions of I=0 and I=1.



#3
Nov1612, 02:41 PM

P: 86

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=HNc...page&q&f=false 


#4
Nov1612, 03:52 PM

Sci Advisor
Thanks
P: 4,160

Nonstrange nonbaryonic states are eigenstates of Gparity
Sorry, my comment was incorrect. This paper might be of help to you, especially on p 9 where they work out the Gparity of K Kbar systems.



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