 Solution

ID: 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} (2 l + 1) = n^{2}

. Hence

g_{1} = 1 (1 s), g_{2} = 4 (2 s and three 2 p states) .

No extra spin or colour multiplicity is being counted. The degeneracy between different

l

at fixed

n

is special to the Coulomb potential.

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