# Eigenstates of rotor with magnetic field (Zeeman effect)

The representation of angular momentum (of $L^2$ and $L_z$) in the space is given by

$$\langle\vec{r}|L_z|\psi\rangle = \frac{\hbar}{i}\frac{\partial}{\partial\phi}\langle\vec{r}|\psi\rangle$$

and $$\langle\vec{r}|L^2|\psi\rangle = -\hbar^2\left[\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2} + \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)\right]\langle\vec{r}|\psi\rangle$$

I want to solve the Zeeman's effect problem, consisting of a Hamiltonian of the form

$$H = \frac{L^2}{2\mu R^2} - \frac{eBL_z}{2\mu},$$ where $\mu,R,e,B$ are constants. I have to solve the equation

$$H|\psi\rangle = E|\psi\rangle.$$

Clearly the eigenstates must be eigenstates of $L^2$ and $L_z$ simultaneously. I write this as $|\psi\rangle = |\ell, m \rangle$. Taking into account that we define

$$L^2|\ell, m\rangle = \hbar^2\ell(\ell+1)|\ell, m\rangle, \quad\quad L_z|\ell, m\rangle = \hbar m |\ell, m\rangle.$$

Then we can find the eigenenergies: $$E_{\ell, m} = \frac{\hbar^2\ell(\ell+1)}{r\mu R^2}-\frac{e\hbar B}{2\mu}m$$

I want to find the eigenstates, therefore I do the following: $$\langle\theta, \phi | H |\ell,m\rangle = E_{\ell,m}\langle\theta, \phi |\ell,m\rangle.$$

This differential equation is difficult to solve. Is my approach correct to find the eigenstates?

In the standard angular momentum (=spherical harmonic) basis, your Hamiltonian is already diagonal, so no further diagonalisation is needed. In other words, the usual basis is $|\ell,m\rangle$, defined as \begin{aligned} L^2|\ell,m\rangle&=\ell(\ell+1)|\ell,m\rangle\\ L_z|\ell,m\rangle&=m|\ell,m\rangle \end{aligned} where $\langle \theta,\phi|\ell,m\rangle\equiv Y^m_\ell(\theta,\phi)$ are the spherical harmonics (modulo, perhaps, the normalisation).
In this basis, $E_{\ell,m}$ is given by the formula you wrote, while the eigenvectors are just spherical harmonics. You don't need to solve any eigenvalue (or differential) equation, mathematicians solved it for you long ago!