Hydrogen Atom in a Weak Magnetic Field

One way to break the spherical symmetry is to apply an external B field. Lets assume that the field is weak enough that the energy shifts due to it are smaller than the fine structure corrections. Our Hamiltonian can now be written as $H=H_0+(H_1+H_2)+H_3$, where $H_0={p^2\over{2\mu}}-{Ze^2\over r}$ is the normal Hydrogen problem, $H_1+H_2$ is the fine structure correction, and

$$\bgroup\color{black}H_3={e\vec B\over{2mc}}\cdot(\vec L +2\vec S)={eB\over{2mc}}(L_z +2S_z)\egroup$$

is the term due to the weak magnetic field.

We now run into a problem because $H_1+H_2$ picks eigenstates of $J^2$ and $J_z$ while $H_3$ picks eigenstates of $L_z$ and $S_z$. In the weak field limit, we can do perturbation theory using the states of definite $j$. A direct calculation of the Anomalous Zeeman Effect gives the energy shifts in a weak B field.

$\Delta E=\left<\psi_{n\ell jm_j}\left\vert{eB\over{2mc}}(L_... ...j}\right> ={e\hbar B\over {2mc}}m_j \left(1\pm {1\over{2\ell +1}}\right)$

This is the correction, due to a weak magnetic field, which we should add to the fine structure energies.

$$\bgroup\color{black}E_{njm_j\ell s}=-{1\over 2}\alpha^2mc^2\l... ...^3}\left[{1\over j+{1\over 2}}-{3\over 4n}\right]\right)\egroup$$

Thus, in a weak field, the the degeneracy is completely broken for the states $\psi_{njm_j\ell s}$. All the states can be detected spectroscopically.

The factor $\left(1\pm {1\over{2\ell +1}}\right)$ is known as the Lande $g$ Factorbecause the state splits as if it had this gyromagnetic ratio. We know that it is in fact a combination of the orbital and spin g factors in a state of definite $j$.

In the strong field limit we could use states of definite $m_\ell$ and $m_s$ and calculate the effects of the fine structure, $H_1+H_2$, as a correction. We will not calculate this here. If the field is very strong, we can neglect the fine structure entirely. Then the calculation is easy.

$$\bgroup\color{black}E=E^0_n+{eB\hbar\over{2mc}}(m_\ell +2m_s)\egroup$$

In this limit, the field has partially removed the degeneracy in $m_\ell$ and $m_s$, but not $\ell$. For example, the energies of all these $n=3$ states are the same.

$$\bgroup\color{black}\matrix{ \ell=2 & m_\ell=0 & m_s={1\over ... ...s={1\over 2} \cr \ell=2 & m_\ell=2 & m_s=-{1\over 2} } \egroup$$