# What is the physical reason the angular quantum number appears in the relativistic correction to the Schrödinger equation with a Coulomb potential?

The energy levels derived for the Schrödinger equation for the Coulomb potential are $$$$E_{n}=-m c^{2}\frac{(Z \alpha)^{2}}{2 n^{2}}.$$$$ If you add the relativistic corrections $$H'_{rel} = -\frac{\hbar^4}{8m^3c^2}\nabla^2\nabla^2 = \frac{-1}{8m^3c^2}\hat{p}^4$$ to the Hamiltonian, one could calculate that you find a relativistic energy shifts of $$$$\Delta E_{\mathrm{rel, nlm}} = -\frac{m c^{2}}{2}(Z \alpha)^{4}\left(\frac{1}{n^{3}\left(l+\frac{1}{2}\right)}-\frac{3}{4 n^{4}}\right).$$$$ We assume a spinless particle so that only the correction for the kinetic energy must be taken into account, i.e. no Darwin term and no spin-orbit coupling. Now, what is the physical reason the angular quantum number $$l$$ appears when the relativistic correction to the Schrödinger equation with the Coulomb potential is considered?

The reason that the non-relativistic hydrogen atom is degenerate for different $$l$$ with the same $$n$$ is that the Hamiltonian comutes with the Runge-Lenz vector. This is sometimes referred to as an accidental degeneracy that is a special property of the non-relativistic hydrogen atom. The degeneracy arises because the non-relativistic hydrogen problem is mathematically equivalent to a particle moving on the surface of a four-dimensional sphere, and this problem is symmetric under certain rotations in four-dimensional space.