I tried to ask this question:

Prove that $\{\gamma_\mu , \gamma_\nu\} = 0$,

but I was unable to resolve it. Can someone help me?

  • 3
    $\begingroup$ The definition of gamma matrices is that they satisfy $\{\gamma_\mu,\gamma_\nu\}=2\eta_{\mu\nu}$, where $\eta_{\mu\nu}$ is the usual Minkowski metric $\endgroup$ – WAH Apr 11 at 16:08
  • $\begingroup$ One could start with writing down 4-by-4 representations of the matrices and multiplying them. $\endgroup$ – Vadim Apr 11 at 16:11
  • $\begingroup$ Who asked you to prove this? $\endgroup$ – G. Smith Apr 11 at 17:09

One does not prove this relation (it is not even true). The defining conditions for the Dirac matrices are that they satisfy the anticommutation relations

$$\{\gamma_a, \gamma_b\} = 2g_{ab} $$

and the conjugacy relation $$\gamma^0\gamma^a\gamma^0=\gamma^{a\dagger}$$

The anticommutation relations are necessary and sufficient for solutions of the Dirac equation to satisfy the Klein-Gordon equation. To show this you factorise the Klein-Gordon equation (according to Dirac's original argument) and use Clairaut's theorem $$\partial_a\partial_b = \partial_b\partial_a $$

The conjugacy relation is needed to put the Dirac equation into the form of a Schrodinger equation (I do not give a full treatment, as this would be a full treatment of a homework and exercises question).

| cite | improve this answer | |
  • $\begingroup$ Rather than saying “one does not prove this relation”, I think it would be clearer to say “one cannot prove this relation because it isn't true”. $\endgroup$ – G. Smith Apr 11 at 17:11
  • $\begingroup$ @G.Smith Very true. $\endgroup$ – Charles Francis Apr 11 at 17:13
  • $\begingroup$ @G.Smith I think Charles Francis implied that even the correct relationship can't be proven, as it is somewhat axiomatic for relativistic QM. $\endgroup$ – TheoreticalMinimum Apr 11 at 17:17

Not the answer you're looking for? Browse other questions tagged or ask your own question.