Proof of the Anti-Commutation Relation for Gamma Matrices from Dirac Equation 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?
 A: 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*}
A: 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.
A: 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. $\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=g^{\mu\nu}+g^{\nu\mu}=2g^{\mu\nu}$.
A: Just write
$$
\gamma^\mu\gamma^\nu=\frac{1}{2}\{\gamma^\mu,\gamma^\nu\}+\frac{1}{2}[\gamma^\mu,\gamma^\nu]
$$
and note that the last term is antisymmetric.
