# 2 quick questions about the baryon octet and decuplet?

1. Jan 27, 2010

### jeebs

Hi,
I have been looking at the baryon decuplet for spin 3/2 baryons, and the octet for spin 1/2 baryons. Here is a picture of the baryon decuplet:
http://schools-wikipedia.org/images/392/39222.png
Here is a picture of the baryon octet:

As I understand it, the spin of the baryon is determined by the spins of its constituent quarks, ie. you can either have all 3 quark spins aligned, giving spin 1/2 + 1/2 + 1/2 = 3/2,
or you can have 2 of the spins in one direction and the other in the opposite direction, giving spin 1/2 + 1/2 - 1/2 = 1/2.

The differences between the decuplet and the octet appear to be that the quarks $$\Delta^-, \Delta^+^+ and \Omega^-$$ are on the decuplet but not the octet, and the octet has the quark $$\Lambda$$ but the decuplet does not.

Also, the proton and neutron on the octet are the same quark combination as the $$\Delta^0 and \Delta^+$$ do on the decuplet, but for some reason they are called different names depending on which diagram you look at.

Why do we have different names for particles with exactly the same quark content?
I thought the answer to this might be to do with the spin of the baryon, but the octet and the decuplet both include the $$\Sigma$$ quarks, with the same name and quark content. What's going on here?

My other question is why should there be ten particles in the decuplet but only eight in the octet? By that I mean, take the $$\Delta^-$$ quark that appears on the decuplet only. It has quark content ddd. Clearly if it only appears in the decuplet diagram, the ddd combination cannot form a spin 1/2 baryon.
What is the reason why these quarks ARE able align so that the baryon they form has a spin 3/2, but are NOT able to align so that they form a baryon with spin 1/2?

Or, to put it more simply, why are there 10 quarks in the one diagram but not the other?

Thanks.