What Are the Symmetry and Orthogonality Properties of Baryon Decuplet States?

[CAF123]

Homework Statement

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?

Homework Equations

SU(3) approximate flavour symmetry

The Attempt at a Solution

[/B]
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!

 

[mark726]

Hello there,

Thank you for your post. You have made some good observations about the symmetry and structure of the multiplet. Let me address your questions and provide some additional insights.

a) The symmetry of the multiplet under exchange of particles is not just the mass of the states, but also their quantum numbers such as spin, isospin, and strangeness. The states in the decuplet all have the same quantum numbers, except for their different masses. This is known as approximate SU(3) flavour symmetry, which means that the states are not exactly the same, but have similar properties due to their shared quantum numbers.

As for the spin state, we can use the raising and lowering operators to determine the possible spin states in the multiplet. The state $|uuu\rangle$ has spin 3/2, which means that the other states in the multiplet must have spin 1/2 or 5/2. This is because the raising operator increases the spin by 1/2, and the lowering operator decreases it by 1/2. So the multiplet contains states with spin 1/2, 3/2, and 5/2.

b) You are correct that the flavour state obtained by demanding orthogonality to the ten states in the multiplet is a flavour singlet. This means that it is invariant under exchange of particles, as you mentioned. As for the spin and colour structure, the spin state of the flavour singlet can be any of the possible spin states in the multiplet, but the colour structure must be the same as the other states in the multiplet. This is because the flavour singlet is orthogonal to all the states in the multiplet, so it cannot have a different colour structure.

I hope this helps clarify things for you. Keep up the good work with your studies!