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: $$ E_n \approx -m_e c^2 + m_e c^2 \left ( 1 - \frac{e^4}{32 \pi^2 c^2 \varepsilon_0^2 \hbar^2 n^2} + \mathcal O \left ( \frac{1}{c^4} \right) \right) $$ $$ E_n = -\frac{m_e e^4}{32 \pi^2 c^2 \varepsilon_0^2 \hbar^2 n^2} $$$$ E_n = -\frac{m_e e^4}{32 \pi^2 \varepsilon_0^2 \hbar^2 n^2} $$ 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.
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