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According to the 2006 Particle Data Book:
For lambda-0 I(JP) = 0(1/2+)
For sigma-0 I(JP) = 1(1/2+)
They are very similar particles indeed. I think the I quantum number refers to isospin, and the arrangement of the quarks. I hope this means something to you Mehh
If it does, could you explain it?
Also, sigma-0 decays with a lifetime of ~10-20 s into a lamdba-0 and either a) a photon, b) two photons or c) an electron and positron. The lambda-0 has lifetime ~10-10.
For lambda-0 I(JP) = 0(1/2+)
For sigma-0 I(JP) = 1(1/2+)
They are very similar particles indeed. I think the I quantum number refers to isospin, and the arrangement of the quarks. I hope this means something to you Mehh
If it does, could you explain it?Also, sigma-0 decays with a lifetime of ~10-20 s into a lamdba-0 and either a) a photon, b) two photons or c) an electron and positron. The lambda-0 has lifetime ~10-10.
Worzo
According to the 2006 Particle Data Book:
For lambda-0 I(JP) = 0(1/2+)
For sigma-0 I(JP) = 1(1/2+)
They are very similar particles indeed. I think the I quantum number refers to isospin, and the arrangement of the quarks. I hope this means something to you Mehh If it does, could you explain it?
Also, sigma-0 decays with a lifetime of ~10-20 s into a lamdba-0 and either a) a photon, b) two photons or c) an electron and positron. The lambda-0 has lifetime ~10-10.
For lambda-0 I(JP) = 0(1/2+)
For sigma-0 I(JP) = 1(1/2+)
They are very similar particles indeed. I think the I quantum number refers to isospin, and the arrangement of the quarks. I hope this means something to you Mehh
If it does, could you explain it?Also, sigma-0 decays with a lifetime of ~10-20 s into a lamdba-0 and either a) a photon, b) two photons or c) an electron and positron. The lambda-0 has lifetime ~10-10.
Ain't got a clue. I know they are different particles and have very different properties, but its still weird...
During lectures we were introduced to all 'possible' combinations of uds quarks.
For some reason we were given the following combinations:-
First 2 quarks spin antisymmetric, thus L = 0. Thus addition of 3rd gives L = 1/2. Giving a lambda-0
Or first 2 quarks spin symmetric, thus L = 1. Thus addition of 3rd gives L = 1/2 or 3/2
Giving Sigma-0 and someother quark.
The only explaination I can see for this is that there is hyperfine structure from either a u/d-s spin coupling or a u-d spin coupling....sigh....
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