My textbook on QFT says that the Dirac equation can be used to show the following relation:
$$\{\gamma^{\mu},\gamma^{\nu}\}=2g^{\mu\nu}$$
I have searched around and unable to find how to prove this as it seems like it has to be assumed at some point by definition. My understanding was that this relation is a fundamental one and that it is assumed in order that the gamma matrices generate a matrix representation of the Clifford algebra, so it is a mathematical assumption rather than something which you derive from a physical equation. One approach I started is to take the Dirac equation and then multiply as follows:
$$(i\gamma^{\nu}\partial_{\nu}-m)\psi=0$$
$$(i\gamma^{\mu}\partial_{\mu}+m)(i\gamma^{\nu}\partial_{\nu}-m)\psi=0$$
$$-(\gamma^{\nu}\gamma^{\mu}\partial_{\nu}\partial_{\mu}+m^2)\psi=0$$
Is there some way to use this to show the given identity?
Best Answer
Even if this is similar, this answer should be clearer, as it was to me.
We are here. \begin{eqnarray*} (\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu + m^2)\psi &=& 0\\ (\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu + m^2)\psi &=& 0 \end{eqnarray*} Adding both the equations, \begin{eqnarray*} [(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu) + 2m^2]\psi &=& 0\\ \end{eqnarray*} Dividing by 2, \begin{eqnarray*} \left[\frac{1}{2}(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu) + m^2\right]\psi &=& 0\\ \end{eqnarray*} and comparing with the Klein Gordon equation, \begin{eqnarray*} (\partial^\mu \partial_\mu+ m^2)\psi &=& 0\\ \Rightarrow (g^{\mu\nu} \partial_\nu \partial_\mu+ m^2)\psi &=& 0\\ \end{eqnarray*} we get, \begin{eqnarray*} g^{\mu\nu} \partial_\nu \partial_\mu &=& \frac{1}{2}(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu)\\ &=& \frac{1}{2}(\gamma^\nu \gamma^\mu \partial_\nu \partial_\mu+\gamma^\mu \gamma^\nu \partial_\nu \partial_\mu) {\text{ :as $\partial_\nu \partial_\mu =\partial_\mu \partial_\nu$},}\\ &=& \frac{1}{2}(\gamma^\nu \gamma^\mu +\gamma^\mu \gamma^\nu )\partial_\nu \partial_\mu\\ \end{eqnarray*} So, we have \begin{eqnarray*} (\gamma^\nu \gamma^\mu +\gamma^\mu \gamma^\nu ) &=& 2 g^{\mu\nu}\\ \Rightarrow \{\gamma^\nu, \gamma^\mu \} &=& 2 g^{\mu\nu}\\ \end{eqnarray*}