Past exam of the quantum physics course of the University of Oxford / 2016 / 9 / b / Solution

Ignoring spin and working in the fixed quarkonium internal sector, the orbital degeneracy is ${\sum }_{l=0}^{n-1}(2l+1)=n^2$. HenceNo extra spin or colour multiplicity is being counted. The degeneracy between different $l$ at fixed $n$ is special to the Coulomb potential.