- #1
Roumpedakis
- 3
- 0
As we know the algebra of SU(3) consist of two Cartan generators and 6 raising and lowering operators. We define the eigenstates of the Cartan operators as u,d,s, correspoding to the three lightest quarks.
Now when we study the [itex] 3\otimes 3[/itex] tensor product we can show that the Hilbert space of these states decompose as
[itex]3\otimes 3 = 8\oplus 1[/itex]
Which means that we can divide this Hilbert into two invariant subspaces.
And similar for baryons
[itex]3\otimes 3\otimes 3=10\oplus 8 \oplus 8\oplus 1[/itex]
My question is following. We can categorize the mesons in two multiplets of 8 particles and 1 singlet, and the baryons in one multiplet of 10 particles, one of 8 particles and one singlet. So is not obvious to me which is the correspondence between these groups of particles and the above Hilbert spaces. If the decomposition of [itex] 3\otimes 3[/itex] has one 8-plet why we have two mesons diagrams and for baryons why we have only one 8-plet of particles?
Now when we study the [itex] 3\otimes 3[/itex] tensor product we can show that the Hilbert space of these states decompose as
[itex]3\otimes 3 = 8\oplus 1[/itex]
Which means that we can divide this Hilbert into two invariant subspaces.
And similar for baryons
[itex]3\otimes 3\otimes 3=10\oplus 8 \oplus 8\oplus 1[/itex]
My question is following. We can categorize the mesons in two multiplets of 8 particles and 1 singlet, and the baryons in one multiplet of 10 particles, one of 8 particles and one singlet. So is not obvious to me which is the correspondence between these groups of particles and the above Hilbert spaces. If the decomposition of [itex] 3\otimes 3[/itex] has one 8-plet why we have two mesons diagrams and for baryons why we have only one 8-plet of particles?