The Shrödinger's equation is [itex]i \hbar \frac{\partial \Psi (\vec r, t) }{\partial t}=-\frac{\hbar ^2}{2m} \nabla ^2 \Psi (\vec r ,t ) + V(\vec r ) \Psi (\vec r ,t)[/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Where m is the mass of the considered particle at rest. I would like to know why the pass to the relativistic equation isn't as simple as changing m for [itex]\gamma m_0[/itex].

Say, if instead of using "m" in the Schrödinger's equation, I use [itex]\gamma m_0[/itex] where [itex]m_0[/itex] is the mass of the particle at rest and gamma is Lorentz factor, what would I obtain? Wouldn't this be a more accurate equation than Schrödinger's?

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# Schrödinger's equation to Dirac's

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