Question 6.3: Consider an eigenstate of a central potential Ψnlm with ener...

Since the Hamiltonian commutes with \hat{L} _{\pm} we have

\left(\hat{H}\hat{L} _{\pm }-\hat{L} _{\pm }\hat{H}\right) \psi _{n\ell m}=0,

so

\hat{H}\hat{L} _{\pm }\psi _{n\ell m}= \hat {L} _{\pm }\hat{H} \psi _{n\ell m}=E_{n} \hat{L} _{\pm }\psi _{n\ell m}

or

\hat{H} \psi _{n\ell m\pm 1}=E_{n}\psi _{n\ell m\pm 1}

(I canceled the constant \hbar \sqrt{\ell(\ell +1)-m(m\pm 1)} from both sides in the last expression). This argument could obviously be repeated to show that \psi _{n\ell m\pm 2} has the same energy as \psi _{n\ell m\pm 1}, and so on until you’ve exhausted the ladder of states. Therefore, rotational invariance explains why states which differ only in the quantum number m have the same energy, and since there are 2\ell+1  different values of m, 2\ell+1  is the “normal” degeneracy for energies in a central potential.