Why is ⟨K⁰|V₃|K⁰⟩ = 1 in Ademollo-Gatto theorem?

  • Context: Graduate 
  • Thread starter Thread starter Einj
  • Start date Start date
  • Tags Tags
    Doubt Proof Theorem
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
Sign up Log in
3 replies · 3K views
Einj
Messages
464
Reaction score
59
I'm studying the semileptonic decay of the kaon and I'm currently reading about the Ademollo-Gatto theorem. The question is probably silly. However, here it is.
The proof starts by just considering that [itex][V_-,V_+]=V_3[/itex], where V is the V-spin. Now, we consider the matrix element [itex]\langle K^0|V_3|K^0\rangle=1[/itex].

I understand that the [itex]|K^0\rangle[/itex] state is an eigenstate of V3 (is the same thing that happens for the usual isospin), but why exactly 1? Consider for example the third component of isospin. As K0 has I3=1/2 I'd expect that matrix element to be 1/2. Why is 1?

Thanks
 

Discussion

Physics news on Phys.org
Actually isn't the V-spin of the K0 equal to 1? It belongs to a V-spin triplet along with K0 bar and some linear combination of π0 and η0.
 
I'm not really sure but I think that that is the U-spin. The V-spin should be the operator that allows the transition from K0 to π- and so the K0 should belong to the doublet (K0,π-), while the V-spin triplet should be composed by K+, K- and π0/η8. But, again, I'm not really sure.