Schrödinger's equation to Dirac's

  • Thread starter fluidistic
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  • #1
fluidistic
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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].
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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  • #2
jtbell
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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].
It's related to why you can't substitute [itex]\gamma m_0[/itex] 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, [itex]K = p^2 / 2m[/itex].
 
  • #3
tom.stoer
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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
 

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