A better Schrödinger Equation with Relativistic Effects?

Question

When you derive the Schrödinger Equation from the Hamiltonian, you perform the following approximation: $$ E^2 = (pc)^2 + (mc^2)^2 \; \; \; \Rightarrow \; \; \; E = \sqrt{(pc)^2 + (mc^2)^2} $$ $$ E = mc^2 \sqrt{1 + \frac{p^2}{m^2c^2}} \approx mc^2\left ( 1 + \frac{p^2}{2m^2 c^2} + \mathcal O(p^4)\right ) $$ $$ E_k =E - E_0 = E - mc^2 = \frac{p^2}{2m} $$ But recently I've heard about the Pade Approximation, which is more precise until a certain degree: $$ E = mc^2 \sqrt{1 + \frac{p^2}{m^2c^2}} \approx mc^2 \frac{4 + 3\frac{p^2}{m^2c^2}}{4 + \frac{p^2}{m^2c^2}} $$ $$ E_k =E - E_0 = \frac{2mc^2p^2}{4m^2 c^2 - p^2} $$ $$ E_k \approx \frac{2mc^2p^2}{4m^2 c^2 - p^2} $$ The Time-Independent Schrödinger Equation is: $$ \hat H \psi = E \psi $$ $$ \left (V(\mathbf r) + \frac{2mc^2 \hat p^2}{4m^2 c^2 - \hat p^2} \right ) \psi = E \psi $$ Rearranging we get: $$ \nabla^2 V(\mathbf r) \psi - [2mc^2 + E] \nabla^2 \psi = \frac{4m^2 c^2}{h^2} \left [ E - V(\mathbf r) \right ] \psi $$ Which is a linear second order partial differential equation. If we solve it for the Hydrogen Atom, we get: $$ E_n = -m_ec^2 + m_ec^2\sqrt{1 - \frac{e^4}{16 \pi^2 c^2 \varepsilon_0^2 \hbar^2 n^2}} $$ If we perform a Taylor approximation: $$ E_n \approx -m_e c^2 + m_e c^2 \left ( 1 - \frac{e^4}{32 \pi^2 c^2 \varepsilon_0^2 \hbar^2 n^2} + \mathcal O \left ( \frac{1}{c^4} \right) \right) $$ $$ E_n = -\frac{m_e e^4}{32 \pi^2 \varepsilon_0^2 \hbar^2 n^2} $$ which is precisely the Schrödinger Energy Levels for Hydrogen. My Question is: Is this a relativistic corrected form of the Schrödinger Equation? Does this equation takes into account higer order terms in the Taylor Approximation? Which effects does this takes into account? Do spin will arise in this equation (as in Dirac Equation due to relativistic effects)? Is this approximation even correct? Is there any papers written about this? If so, could you provide any of the links, please.

Answer 1

Your spectrum does not contain relativistic corrections, it is the "gross structure." It is the correct non relavistic limit (gross structure), since your equation contains Schrödinger in the nonrelativstic limit. You drowned your Padé approximant with too crude approximations to recover relativistic corrections in the final result. A more careful treatment would reveal a "fine structure" (relativistic correction).

Even then, your method would not predict the correct result. Your starting point is still to replace Schrödinger's equation by Klein Gordon, which you then approximate. KG is a legitimate relativistic model, and it turns out that its spectrum can be computed exactly. However, it does not accurately capture the fine structure of hydrogen. Historically, Schrödinger first started with this approach and then settled for his non relativistic equation as it better modelled hydrogen.

To get the correct fine structure, you only focused on the higher order kinetic terms, which your Padé approximant accurately models. You'll also need the spin orbit coupling and the Darwin term which are absent in your model. On top of that, to get the Lamb shift, you need QED and for the hyperfine structure, you need nuclear spin as well.

To summarise, it does contain some relativistic corrections. It does take into account higher Taylor terms of the kinetic energy, but neglects other corrections of similar magnitude. By construct, there is no spin, you need to add extra internal degrees of freedom for this. You can start by looking at the extensive literature of the fine structure.