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

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

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

$$ \begin{equation} \eta^{\mu\nu}=\begin{pmatrix} 1 & 0 & 0 & 0\newline 0 & -1 & 0 & 0\newline 0 & 0 & -1 & 0\newline 0 & 0 & 0 & -1 \end{pmatrix}\,. \end{equation} $$ 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. $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=g^{\mu\nu}+g^{\nu\mu}=2g^{\mu\nu}$.