The Naive Zeeman Splitting

The additional term we wish to consider in the Hamiltonian is \bgroup\color{black}${e\over 2\mu c}\vec{B}\cdot\vec{L}$\egroup. Choosing the z axis so that the constant field points in the z direction, we have

\begin{displaymath}\bgroup\color{black}H_{\mathrm{Zeeman}}={eB_z\over 2\mu c}L_z.\egroup\end{displaymath}

In general, the addition of a new term to the Hamiltonian will require us to use an approximation to solve the problem. In this case, however, the energy eigenstates we derived in the Hydrogen problem are still eigenstates of the full Hamiltonian \bgroup\color{black}$H=H_{\mathrm{hydrogen}}+H_{\mathrm{Zeeman}}$\egroup. Remember, our hydrogen states are eigenstates of H, \bgroup\color{black}$L^2$\egroup and \bgroup\color{black}$L_z$\egroup.

\begin{displaymath}\bgroup\color{black}(H_{\mathrm{hydrogen}}+H_{\mathrm{Zeeman}})\psi_{n\ell m}=(E_n+m\mu_BB)\psi_{n\ell m}\egroup\end{displaymath}

This would be a really nice tool to study the number of degenerate states in each hydrogen level. When the experiment was done, things did not work our according to plan at all. The magnetic moment of the electron' s spin greatly complicates the problem. We will solve this later.

Jim Branson 2013-04-22