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

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

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


\vec{J}&=&\vec{L}+\vec{S} \\
... 2}(J^2-L^2-S^2)\to {\hbar^2\over 2}(j(j+1)-\ell(\ell+1)-s(s+1))

Since our eigenstates of \bgroup\color{black}$J^2$\egroup and \bgroup\color{black}$J_z$\egroup are also eigenstates of \bgroup\color{black}$L^2$\egroup and \bgroup\color{black}$S^2$\egroup (but not \bgroup\color{black}$L_z$\egroup or \bgroup\color{black}$S_z$\egroup), 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.

Jim Branson 2013-04-22