According to Wikipedia's Dirac equation article, the Dirac equation can be written in form $$ i\hbar\gamma^{\mu}\partial_{\mu}\psi-mc\psi=0, $$ where $\gamma^{\mu}$ are gamma matrices which are $4 \times 4$ matrices. Now is the wavefunction $\psi$ still a scalar like in classical quantum mechanics? If so then isn't this equation a matrix equation meaning that the same scalar function should obey multiple different equations? Why would it be so? Isn't that overdetermination?

  • 1
    $\begingroup$ Did you also read the bit where it says $\psi$ is a four component spinor? (That WP page is a mess, it would be fairly easy to do this). $\endgroup$
    – jacob1729
    Jun 10, 2019 at 10:44
  • $\begingroup$ Well that explains. I tried to see something like that said in that WP page but didn't notice that. $\endgroup$
    – Kirby
    Jun 10, 2019 at 10:47

1 Answer 1


In this context, $\psi$ stands for a set of four wavefunctions, forming a Dirac spinor. Both sides of the equation transform as Dirac spinors.

This is not unusual or overdetermined. It's just like how $\mathbf{F} = m \mathbf{a}$ really has "three things" on each side, because it relates two objects that transform as vectors.


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