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The energy levels derived for the Schrödinger equation for the Coulomb potential are \begin{equation} E_{n}=-m c^{2}\frac{(Z \alpha)^{2}}{2 n^{2}}. \end{equation} 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 \begin{equation} \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). \end{equation} 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?

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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.

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  • $\begingroup$ Thanks was great help. $\endgroup$
    – drandran12
    May 3, 2020 at 10:14

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