# Relation between magnetic quantum number and angular momentum quantum number

The energy levels of a quantum rigid rotator are given by $$E_{\ell}=\frac{\hbar^{2}}{2 I} \ell(\ell+1), \quad \text { for } \ell=0,1,2,3, \dots$$ It is also stated that the degeneracy of each level is given by $$2\ell +1$$ but I couldn't find a formula which expresses the energy level as a function of $$m_{\ell}$$ so I'm not sure how for $$\ell=1$$, $$E_{m_{\ell}=0} = E_{m_{\ell}=-1} = E_{m_{\ell}=1}$$ On a more general note, I would love to know more about the relation between magnetic quantum number and angular momentum quantum number.

The energy does not depend on $$m$$, which is why all $$m$$ values have the same energy. Basically the eigenvalues of your Hamiltonian depend only $$\ell$$, in the same way that all the $$Y_{\ell,m}(\theta,\varphi)$$ for fixed $$\ell$$ are eigenstates of $$\vec L\cdot\vec L$$ with eigenvalue $$\ell(\ell+1)$$.
The connection between $$m$$ and $$\ell$$ is found in any elementary textbook on quantum mechanics or quantum physics, from the theory of spherical harmonics. A good approach is to use the ladder operators $$\hat L_\pm$$ but this requires a little more machinery.