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.
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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.
answered Oct 13, 2019 at 19:35