Prove that the singlet state of 3 quark flavors is fully antisymmetric

Question

I have three quark flavors, according to group theory: $$3\otimes3\otimes3=10\oplus8\oplus8\oplus1$$ I have to show that the 1 is antisymmentric.

My idea would be to try to construct the states in the two octuplets and show that they are not antisymmetric and since the singlet should have the same quark content as the "centermost" states in the octuplets and the "0" state of the decouplet, the singlet has to be antisymmetric to be distinguishable from the others. But that would mean applying lowering operators at least 8 times(there are 2 states in the center of those octuplets that can only be distinguished by taking different paths to the "center") which would be really time consuming. Is there any quicker way to show it?

Answer 1

The simple approach is to start by observing that your singlet state must be (by definition) in a 1-dimensional irrep and be constructed from triple products of states. A bit of reflection will show that taking a determinant of the type $$ \left\vert\begin{array}{ccc} \psi_u(1)&\psi_u(2)&\psi_u(3)\\ \psi_d(1)&\psi_d(2)&\psi_d(3)\\ \psi_s(1)&\psi_s(2)&\psi_s(3) \end{array}\right\vert $$ will carry a 1d irrep of SU(3) since the determinant is invariant under SU(3). It is clearly antisymmetric under exchange of any two particles since the determinant picks up a sign under exchange of any two columns. To boot, you can verify using Young diagram methods (see here or here) that the other 3 pieces of your decomposition correspond to the fully symmetric and the two partially symmetric irreps.

Answer 2

I might figured this one out: The singlet must be a combination of all three $u,d,s$ quarks. That means a permutation function $F(u,d,s)=\psi_{singlet}$. Now it is singlet, so applying all lowering/raising operators must yield zero.

In case of 3 quark flavours we have 3 pairs of operators:

1. $T$ that switches $u$ for $d$ or vice versa
2. $U$ that switches $d$ for $s$ or vice versa
3. $V$ that switches $u$ for $s$ or vice versa

Applying these operators to our function $F(u,d,s)$: $$T_+F(u,d,s)=F(u,u,s)=0$$ $$T_-F(u,d,s)=F(d,d,s)=0$$ And the same applies for the rest. This means, that the all quark flavours are in the wavefunction in pairs with different signs, that is the only nontrivial way to construct the wavefunction or in other words, the wavefunction has to be fully antisymmetric.