# Magnetism and Lande factor

When an atom is placed in a magnetic field region, we can have two possibilities: $B_{ext}\gg B_{int}$ , when the external field is way greater than the internal field produced by the motion of electrons, and viceversa $B_{int}\gg B_{ext}$ . These two cases give two different hamiltonians of perturbation. Hence we do distinguish the normal and anomalous Zeeman effects:

• Normal Zeeman effect:

${\hat {H}}={\frac {eB_{ext}}{2m}}\left(L_{z}+2S_{z}\right)\;\Rightarrow E_{M_{L},M_{S}}={\frac {eB_{ext}\hbar }{2m}}\left(M_{L}+2M_{S}\right)$ • Anomalous Zeeman effect:

$\left\langle {\hat {H}}\right\rangle ={\frac {eB}{2m}}\left\langle J_{z}\right\rangle {\underset {Lande\;factor}{\underbrace {\left[1+{\frac {\left\langle {\hat {S}}{\hat {J}}\right\rangle }{\left\langle {\hat {J}}^{2}\right\rangle }}\right]} }}$ $\Rightarrow E_{J^{2},L^{2},S^{2}}={\frac {eB\hbar ^{2}}{2m}}M_{J}\left[1+{\frac {J(J+1)-L(L+1)+S(S+1)}{2J(J+1)}}\right]$ In exercises, unless it is specified, you can not determin whether a given $B_{ext}$ is greater than $B_{int}$ or not and so you have to evaluate both energy split terms and . In conclusion we want to remember the hamiltonian of the spin-orbit coupling:

${\hat {H}}_{SO}={\frac {Ze^{2}}{4\pi \epsilon _{0}}}{\frac {1}{m_{e}^{2}c^{2}}}{\frac {{\boldsymbol {S}}{\boldsymbol {L}}}{r^{3}}}$ And its energy spectrum, in the $n,l,m$ basis, is:

$E_{n,l}=-{\frac {1}{2}}m_{e}c^{2}\left(Z\alpha \right)^{2}\left[{\frac {2\left(Z\alpha \right)^{2}}{n^{2}(2l+1)}}-{\frac {3(Z\alpha )^{2}}{4n^{4}}}\right]$ 1. Spin-orbit effect can be neglected. Hence $H_{B}$ is a perturbation for a normal hydrogen-like hamiltonian. Hence here we work in the $\left|L^{2},M_{L}\right\rangle \left|S^{2},M_{S}\right\rangle$ basis.
2. Spin-orbit can not be neglected, hence it has to be considered as part of the global hamiltonian to which $H_{B}$ is a perturbation.Hence here we work in the $\left|J^{2},M_{J},L^{2},S^{2}\right\rangle$ basis.