# Proof of the Anti-Commutation Relation for Gamma Matrices from Dirac Equation

To be honest I think in this case the best proof is by direct computation. The gamma matrices are

$$$$\gamma^{0}=\begin{pmatrix} 1 & 0 & 0 & 0\newline 0 & 1 & 0 & 0\newline 0 & 0 & -1 & 0\newline 0 & 0 & 0 & -1 \end{pmatrix}\, \quad \gamma^{1}=\begin{pmatrix} 0 & 0 & 0 & 1\newline 0 & 0 & 1 & 0\newline 0 & -1 & 0 & 0\newline -1 & 0 & 0 & 0 \end{pmatrix}\,$$$$ and $$$$\gamma^{2}=\begin{pmatrix} 0 & 0 & 0 & -i\newline 0 & 0 & i & 0\newline 0 & i & 0 & 0\newline -i & 0 & 0 & 0 \end{pmatrix}\, \quad \gamma^{3}=\begin{pmatrix} 0 & 0 & 1 & 0\newline 0 & 0 & 0 & -1\newline -1 & 0 & 0 & 0\newline 0 & 1 & 0 & 0 \end{pmatrix}.$$$$

Direct calculation shows that $$\{\gamma^{0},\gamma^{0}\} = \gamma^{0}\gamma^{0} + \gamma^{0}\gamma^{0} = 2\eta^{00}\mathbb{I}_{4}\,$$

where $$\eta^{00}=1$$ and $$\mathbb{I}_{4}$$ is the $$4\times 4$$ identity matrix. Furthermore, direct calculation shows that

$$\{\gamma^{0},\gamma^{i}\} = \gamma^{0}\gamma^{i} + \gamma^{i}\gamma^{0} = 2\eta^{0i}\mathbb{I}_{4}\, = 0_{4,4}\,$$

where $$\eta^{0i}=0$$ for $$i=1, 2, 3$$ and $$0_{4,4}$$ is the $$4\times 4$$ matrix with all zero entries. Additional calculations show that

$$\{\gamma^{i}, \gamma^{i}\} = 2\eta^{ii}\mathbb{I}_{4}$$

and that

$$\{\gamma^{i}, \gamma^{j}\} = 2\eta^{ij}\mathbb{I}_{4}=0_{4,4}\,,$$ where $$\eta^{ii}=-1$$ and $$\eta^{ij}=0$$ for $$i\ne j$$ with both $$i$$ and $$j$$ taking values from $$1, 2, 3\,.$$

The results $$\{\gamma^{0},\gamma^{0}\} = 2\eta^{00}\mathbb{I}_{4}$$ $$\{\gamma^{0},\gamma^{i}\} = 2\eta^{0i}\mathbb{I}_{4}=0_{4,4}$$ $$\{\gamma^{i},\gamma^{i}\} = 2\eta^{ii}\mathbb{I}_{4},$$ $$\{\gamma^{i},\gamma^{j}\} = 2\eta^{ij}\mathbb{I}_{4}=0_{4,4}$$ can be summarised into the single formula

$$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}_{4}$$

where $$\eta^{\mu\nu}$$ satisfies

$$$$\eta^{\mu\nu}=\begin{pmatrix} 1 & 0 & 0 & 0\newline 0 & -1 & 0 & 0\newline 0 & 0 & -1 & 0\newline 0 & 0 & 0 & -1 \end{pmatrix}\,.$$$$ This means $$\eta^{\mu\nu}$$ is the metric tensor of the Minkowski space-time of special relativity.

I prefer the expression $$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}_{4}$$ instead of $$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}$$ (or $$\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}$$ as the poster wrote) which gives the false impression that $$\{\gamma^{\mu}, \gamma^{\nu}\}$$ is just a number since for any chosen values of the pair ($$\mu,\nu)$$ the entry in $$g^{\mu\nu}=\eta^{\mu\nu}$$ is equal to 0 or $$\pm 1$$. Clearly this is not the case since $$\{\gamma^{\mu}, \gamma^{\nu}\}$$ involves the sum of products of $$4\times 4$$ matrices.

Declaration: I did not come up with the notation $$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}_{4}$$ myself. I saw it on the Wikipedia entry for gamma matrices (https://en.wikipedia.org/wiki/Gamma_matrices) earlier today. I do note though that the two QFT books I have to hand use the notation $$\{\gamma^{\mu},\gamma^{\nu}\} = 2g^{\mu\nu}$$ (Itzykson & Zuber) and $$\{\gamma^{\mu},\gamma^{\nu}\} = -2g^{\mu\nu}$$ (Srednicki, where $$g^{\mu\nu} = \mbox{diag}(-1,1,1,1)\,$$) but again I think this notation is confusing.

Even if this is similar, but 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*}

The second-order derivative is $g^{\mu\nu}\partial_\nu\partial_\mu$, but since $\partial_\nu\partial_\mu$ is symmetric the symmetrised coefficients match, viz. $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=g^{\mu\nu}+g^{\nu\mu}=2g^{\mu\nu}$.