Justifying the identification of eight gluons with the ${\bf 8}$ from ${\bf 3}\otimes{\bf 3}^*$

Question

When we combine the fundamental ${\bf 3}$ and antifundamental ${\bf 3}^*$ of color $SU(3)$ of QCD i.e. single quark of three colors and a single antiquark of 3 anticolors, nine states are obtained. One particular linear combination transforms as the singlet ${\bf 1}$ and the other ${\bf 8}$ transform in the adjoint of $SU(3)$. The latter either states are identified with eight gluons.

I find this a bit perplexing! Gluons (unlike mesons) are certainly not made up of quark-antiquark pair. If this identification of ${\bf 8}$ with eight gluons of $SU(3)$ is rigorous, how is this identification justified? In what sense?

Answer 1

The ${\bf 8}$ not directely connected to the ${\bf 3}\otimes {\bf 3}^*= {\bf 1}\oplus {\bf 8}$ decomposition. Instead it comes from the fact that the gluons are labelled by the generators of the Lie algebra, and these, in turn, form the basis of the adjoint representation. In ${\rm SU}(3)$ this happens to be the ${\bf 8}$.