Dirac equation

Remarks

The matrices $\gamma^\mu$ generate a representation of the Clifford algebra of $(\R^4, \eta)$. A standard choice is the Dirac representation, in which $\gamma^0 = \mathrm{diag}(1, 1, -1, -1)$.

Example: in the Dirac representation, $\psi(x) = e^{-imt} (1, 0, 0, 0)$ is a solution. Indeed $\psi$ depends only on $t$, and $i \gamma^0 \partial_t \psi = i \gamma^0 (-im) \psi = m \gamma^0 \psi = m \psi$, since $(1, 0, 0, 0)$ is an eigenvector of $\gamma^0$ for the eigenvalue $1$.

Non-example: for $m > 0$, $\psi(x) = e^{imt} (1, 0, 0, 0)$ is not a solution, because $i \gamma^0 \partial_t \psi - m \psi = -2m \psi \ne 0$.

Applying the operator $i \gamma^\nu \partial_\nu + m$ to the equation and using the anticommutation relations gives $(\partial_t^2 - \Delta + m^2) \psi = 0$, the Klein–Gordon equation for each component.