Dirac equation

How to construct

Starting from the relativistic relation between energy and momentum:

$E^{2}={\mathbf {p}}^{2}c^{2}+m^{2}c^{4}$

or $E=c{\sqrt {{\mathbf {p}}^{2}+m^{2}c^{2}}}$

From this equation we can not directly replace $E,{\mathbf {p}}$ by the corresponding operators since we don't have the definition for the square root of an operator. Therefore, first we need to linearize this equation as follows:

$E=c{\sqrt {{\mathbf {p}}^{2}+m^{2}c^{2}}}=c{\sqrt {(p_{x}^{2}+p_{y}^{2}+p_{z}^{2})+m^{2}c^{2}}}=c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^{2}$

where ${\mathbf {\alpha }}_{x},\alpha _{y},\alpha _{z}$ and ${\mathbf {\beta }}$ are some operators independent of ${\mathbf {p}}$.

From this it follows that:

$c^{2}(p_{x}^{2}+p_{y}^{2}+p_{z}^{2}+m^{2}c^{2})=[c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^{2}].[c(\alpha _{x}p_{x}+\alpha _{y}p_{y}+\alpha _{z}p_{z})+\beta mc^{2}]$

Expanding the left hand side and comparing it with the right hand side, we obtain the following conditions for ${\mathbf {\alpha }}_{x},\alpha _{y},\alpha _{z}$ and ${\mathbf {\beta }}$ :

$\alpha _{i}^{2}=\beta ^{2}=1$ where $i=1,2,3$ corresponds to $x,y,z$

${\mathbf {\alpha }}_{i}\alpha _{j}+\alpha _{j}\alpha _{i}={\alpha _{i},\alpha _{j}}=0$