# Schrödinger's equation to Dirac's

Gold Member
The Shrödinger's equation is $i \hbar \frac{\partial \Psi (\vec r, t) }{\partial t}=-\frac{\hbar ^2}{2m} \nabla ^2 \Psi (\vec r ,t ) + V(\vec r ) \Psi (\vec r ,t)$.
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 $\gamma m_0$.
Say, if instead of using "m" in the Schrödinger's equation, I use $\gamma m_0$ where $m_0$ 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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jtbell
Mentor
I would like to know why the pass to the relativistic equation isn't as simple as changing m for $\gamma m_0$.
It's related to why you can't substitute $\gamma m_0$ for m in the non-relativistic kinetic energy equation, and thereby get the relativistic kinetic energy.

The terms of the SE, as you've written it, correspond to the statement

total energy = kinetic energy + potential energy

using the classical relationship between kinetic energy and momentum, $K = p^2 / 2m$.

tom.stoer