# Schrödinger's equation to Dirac's

1. Dec 4, 2011

### fluidistic

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?

2. Dec 4, 2011

### Staff: Mentor

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, $K = p^2 / 2m$.

3. Dec 4, 2011

### tom.stoer

In relativistic quantum mechanics the terms in an equation for a wave function must transform according to some representation of the Lorentz group (just like in relativistic mechanics). You can't achieve that simply by introducing some gamma-factors