Total Angular Momentum and The Spin Orbit Interaction

The spin-orbit interaction (between magnetic dipoles) will play a role in the fine structure of Hydrogen as well as in other problems. It is a good example of the need for states of total angular momentum. The additional term in the Hamiltonian is

$$\bgroup\color{black}H_{SO}={Ze^2\over 2m^2c^2}{\vec{L}\cdot\vec{S}\over r^3}\egroup$$

If we define the total angular momentum $\vec{J}$ in the obvious way we can write $\vec{L}\cdot\vec{S}$ in terms of quantum numbers.

audio

$$\begin{eqnarray*} \vec{J}&=&\vec{L}+\vec{S} \\ J^2&=&L^2+2\vec{L}\cdot\vec{S}... ... 2}(J^2-L^2-S^2)\to {\hbar^2\over 2}(j(j+1)-\ell(\ell+1)-s(s+1)) \end{eqnarray*}$$

Since our eigenstates of $J^2$ and $J_z$ are also eigenstates of $L^2$ and $S^2$ (but not $L_z$ or $S_z$), these are ideal for computing the spin orbit interaction. In fact, they are going to be the true energy eigenstates, as rotational symmetry tells us they must.