Imagine you want to solve the Dirac equation: $$ (i\gamma^\mu \partial_\mu - m )\psi=0 \\ $$ Where $\gamma^\mu$ are the $4 \times 4$ gamma matrices, and $\psi$ is a 4 component spinor. $$ \psi = \begin{pmatrix} \psi_1 \\ \psi_2 \\ \psi_3 \\ \psi_4 \end{pmatrix} $$ When you trying to solve this differential equation, the derivatives of the different spinor components would be mixed by the matrices, and appear in all equations. You will finally get, a system of 4 equations, all of which depends on all the 4 spinor components.

My question is, how would you even solve this? It is possible to arrive to a differential equation that depends, only, on one function at a time?


1 Answer 1


Solving the Dirac equation for a free particle (i.e. without any external fields) $$(i\gamma^\mu \partial_\mu - m )\psi=0 \\$$ which you have written down, is actually not that hard. Searching for "Dirac equation free particle" will give many hits.

You begin with the approach where the wave spinor field $\psi(\mathbf{x},t)$ is the product of a constant spinor $u$ and a plane wave $$\begin{pmatrix} \psi_1(\mathbf{x},t) \\ \psi_2(\mathbf{x},t) \\ \psi_3(\mathbf{x},t) \\ \psi_4(\mathbf{x},t) \end{pmatrix} =\begin{pmatrix} u_1 \\ u_2 \\ u_3 \\ u_4 \end{pmatrix} e^{i(\mathbf{p}\mathbf{x}-Et)}$$ and $\mathbf{p}$ and $E$ are still unknown constants (the momentum and energy of the particle). This will lead you to the constraint $$E^2=\mathbf{p}^2+m^2$$ (which of course is the relativistic energy-momentum relation) and 3 algebraic relations between the 4 numbers $u_1,u_2,u_3,u_4$. So you don't have differential equations anymore.

  • $\begingroup$ What if you want to add a potential? $\endgroup$ Sep 20, 2023 at 18:54
  • $\begingroup$ @ÁlvaroRodrigo Then this simple approach doesn't work anymore. $\endgroup$ Sep 20, 2023 at 18:58
  • $\begingroup$ And what is the correct approach? $\endgroup$ Sep 20, 2023 at 19:35

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