# Fine Structure Correction

The fine structure correction is composed of the relativistic correction and spin-orbit coupling. The lowest-order relativistic correction to the Hamiltonian is

$$H_r' = -\frac{p^4}{8m^3c^2}$$

According to Griffiths, this perturbation is spherically symmetric, so it commutes with $L^2$ and $L_z$. He uses this to justify the use of nondegenerate perturbation theory for the relativistic correction, even though the hydrogen atom is very degenerate.

$p = -i\hbar\vec \nabla$. With this and $H_r'$ above, how can you tell $H_r'$ is spherically symmetric? $\vec \nabla ^4$ won't depend on $\theta$ or $\phi$?

Also, I know that $H$, $L^2$, and $L_z$ share common eigenfunctions, which are the spherical harmonics. So $[H,L_z] = 0$ etc, but how do we know that $[L_z,$anything spherically symmetric$] = 0$? The hydrogen atom is degenerate in $n$, but does does knowing that $L_z$ and $L^2$ commuting with the perturbation guarantees that $n,l,m$ are the good quantum numbers?

The easiest way to see that $p^4$ is spherically symmetric is to view it in momentum space. If you apply $p^4$ to a momentum eigenstate $|p\rangle$ the result clearly only depends on the magnitude of the momentum vector of the state and not on its orientation, so if $R$ is a rotation operator we have $$p^4 R|p\rangle = Rp^4|p\rangle$$ Since the momentum eigenstates form a basis we can extend this result to a general state by linearity, so we have $p^4R = Rp^4$, in other words $H_r^\prime$ is spherically symmetric.
The angular momentum operators are defined to be the generators for rotations, so if $R_z(\delta\theta)$ is an infinitesimal rotation around the $z$ axis, $L_z$ is given by $$R_z(\delta\theta) = 1 -\imath\delta\theta L_z + O(\delta\theta^2)$$ If we have a spherically symmetric operator, $Q$, then $[R,Q] = 0$ for all rotations $R$. In particular \begin{align*} 0 & = [R_z(\delta\theta),Q]\\ & = [1- \imath\delta\theta L_z, Q]\\ \Rightarrow 0 &= [L_z, Q]\end{align*}
When we are doing degenerate perturbation theory the problem is essentially to find a basis in which the perturbation Hamiltonian, $H_r^\prime$ is diagonal. We can do this the old fashioned way by finding the eigenvectors, but this is long and boring. The trick to getting around this is to find some operator, $S$, we already understand which commutes with the perturbation. Since $H_r^\prime$ and $S$ commute they have a basis of mutual eigenvectors, so if we use this basis $H_r^\prime$ will already be diagonal and we will have saved a lot of work.