SU(3) Transformation behavior of hadronic wave function

Question

We were given a “baryon” wave function $$ \sum_{i,j,k=1}^3 \epsilon_{ijk} q_i q_j q_k$$ which is supposedly invariant under SU(3) gauge transformations. The structure of those transformations was given as $$\psi \to \psi' = U(x) \psi \,\quad U(x) = \exp(\mathrm i g \chi_a(x) T_a) \,.$$

Then we had to show that this wave function is invariant under infinitesimal transformations. In later problems it was asked to show that a “meson” wave function $\sum_{i=1}^3 \bar q_i q_i$ is invariant whereas the wave function $\bar q_1 q_1$ is not.

The best answer that I was able to come up with is that in the “baryon” case the $\epsilon$-symbol is a top-dimensional form. The space of top-dimensional forms only has one dimension, therefore it cannot transform into anything else, it is a singlet.

For the “mesons” in the first case I have (summation implicit) $$\bar q_i q_i \to \bar q_j U^\dagger(x)_{ji} U(x)_{ik} q_k = \bar q_j \delta_{jk} q_k = \bar q_j q_j$$ due to the unitarity of the transformation. In the second argument this breaks because the summation between the unitary transformation is not completely, therefore not yielding an inverse: $$\bar q_1 q_1 \to \bar q_j U^\dagger(x)_{j1} U(x)_{1k} q_k \,.$$

My tutor and the tutor of the other group were not able to really give some convincing argument, so I am asking here. Are those arguments the correct ones for the invariance and non-invariance under SU(3) color transformations?

Answer 1

In the baryon case you can use a direct proof similar to the one you used for the meson. If you make a $SU(3)$ transformation you get (from now on a summation over repeated indices is assumed) $$ \varepsilon_{ijk}q_i q_j q_k \rightarrow \varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} q_\ell q_m q_p $$ The expression $$ \varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} $$ is completely antisymmetric in the indices $\ell,m,p$, so it must be proportional to $\varepsilon_{\ell m p}$ and we obtain $$ \varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} = \lambda \varepsilon_{\ell m p} $$ By contracting both sides with $\varepsilon_{\ell m p}$ we get on the left side the determinant of $U$, on the right side the number of nonzero elements of $\varepsilon$, $$ 3!\det U = \varepsilon_{\ell m p} \varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} = \varepsilon_{\ell m p}\lambda \varepsilon_{\ell m p} = 3! \lambda $$ and as in $SU(3)$ the determinant is one, $\lambda=1$, so $ \varepsilon_{ijk} U_{i\ell} U_{jm} U_{kp} =\varepsilon_{\ell m p}$. Inserting this expression in the first one we wrote we get the proof of the invariance.