# Baryon decuplet properties

1. May 18, 2016

### CAF123

1. The problem statement, all variables and given/known data
Consider composite states of three quarks, that transform as 3 ⊗ 3 ⊗ 3. The action of the corresponding raising and lowering operators on a typical state in this tensor product representation is $$I_+^{3 ⊗ 3 ⊗ 3} = I_+^3 \otimes \text{Id} \otimes \text{Id} + \text{perms}$$ The state $|H \rangle$for which $I_+|H \rangle = U_+|H\rangle = V_+|H \rangle = 0$ is $|uuu \rangle$. The lowering operators may be applied to determine the ten states in this multiplet.

a) What is the symmetry of this multiplet under exchange of particles? If the states are colour singlets, what does this imply for the spin state in the multiplet?

b) Find a flavour state orthogonal to all the states in the above multiplet. Comment on its symmetry properties under exchange of particles. What is the explicit spin and colour structure of your state?

2. Relevant equations
SU(3) approximate flavour symmetry

3. The attempt at a solution

a) The symmetry of the multiplet under exchange of particles is the mass of the states. States within the decuplet are supposed to assume the same masses, while in reality the states do not so this is only approximate symmetry. The states are all of the form $|qqq \rangle$, baryons, which are colourless states (or colour SU(3) singlet). I'm not sure how to deduce what the means for the spin though. A generic wavefunction of the state is composed of flavour, (this component obtained through the analysis above via the raising and lowering operators), colour (I think the colour component can generally be written as $\epsilon_{ijk} q^i q^j q^k$ so that is transforms trivially under SU(3) colour), space and spin. The collection of all must be antisymmetric under exchange by Pauli principle.

b) I suppose the flavour state obtained through demanding it is orthogonal to the ten states obtained is a flavour singlet? It is therefore invarant under exchange of the particles. The colour structure is as given above with the epsilon but I am not sure how the spin factors in.

Thanks!

2. May 23, 2016