# Relativistic Correction to the Hydrogen Atom and Spherically Symmetric Operators

We know, the lowest order relativistic correction to the Hamiltonian for the Hydrogen atom is $$H^{'}=-\frac{p^{4}}{8m^{3}c^{2}}$$ Where, $$p=-\frac{\hbar}{i}\nabla$$ So, is $$p^{2}=-\hbar^{2}\nabla^{2}$$ and $$p^{4}=\hbar^{4}\nabla^{2}(\nabla^{2})$$?

In various sources, for calculating the first order correction to the energy, a fact has been utilized that the perturbation is spherically symmetric.

a) First of all, what should a spherically symmetric operator look like?

b) Secondly, does this mean that the operator $$\nabla^{2}(\nabla^{2})$$ is spherically symmetric? If so, how to prove that? Or is there any easy way to intuitively understand that? Any help is appreciated.

• It is a good exercise to do it for arbitrary perturbation potential, starting from the Dirac equation. The resulting form is actually used to model spin-orbit coupling in semiconductors, although replacing $mc^2$ by $E_g$. Commented Feb 16, 2023 at 15:07