Transition from 4-potential to E and B In my lecture notes there is a step that i cannot follow: 
$$\frac{i}{2}[\gamma^{\mu},\gamma^{\nu}] (\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})=: \sigma^{\mu\nu}F_{\mu\nu}=i\vec{\alpha}  \vec{E}-\vec{\sigma}\vec{B}$$
Can somebody please help me with this step? 
The $\gamma^{\mu}$ are the dirac- gamma matrices, A is the electromagnetic 4-potential.
$\alpha_i$ is a 2x2 block matrix with pauli matrices $\sigma_i$ in the off-diagonal elements. 
The same step can be found in Itzykson - Quantum Field Theory: Page 66 Eq. 2-73.  Without further explanation. 
 A: You don't define $\overrightarrow{\sigma}$. It looks like neither do Itzykson/Zuber, or, rather, it is defined there as a vector with Pauli matrices as components (if I am not mistaken). However, those matrices are 2x2, whereas all other matrices in your equation are 4x4. Therefore, such definition seems incompatible with your equation. I believe you should define new 4x4 matrices $\sigma_i$, $i=1,2,3$ (as all further Latin indices), as $\sigma_i=\epsilon_{ijk}\sigma^{jk}$ (I may omit some constant here). Then $\alpha_i$ coincides with [\gamma_0,gamma_i] (up to a constant), which coincides with $\sigma^{0i}$ (up to a constant). Then you use $B_i=\epsilon_{ijk}F^{jk}$ (again, up to a constant). In the above I ignore the distinction between upper and lower indices.
A: If you define your gamma matrices as the block matrices:
$$
\gamma^0:=\begin{bmatrix}
    1       & 0  \\
    0      & -1 
\end{bmatrix}; \quad \gamma^i:=\begin{bmatrix}
    0       & \sigma^i  \\
    -\sigma^i      & 0 
\end{bmatrix}
$$
Then you can also define the block matrices $\sigma^{\mu\nu}:=\frac{i}{2}\left(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu\right)$.  And given the scalar potential $(A_0,A_1,A_2,A_3)$ you can define the Faraday tensor $F_{\mu\nu}:=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$.
So the first equality $$\frac{i}{2}[\gamma^{\mu},\gamma^{\nu}] (\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}):= \sigma^{\mu\nu}F_{\mu\nu}$$ follows by definition.  And what it is really saying is a matrix equation.  For every $\mu$ and $\nu$ there is a 4x4 complex matrix $\sigma^{\mu\nu}$ and a real scalar $F_{\mu\nu}$, scale the matrix by that scalar, and then take each such matrix and add them all up, you get a matrix.  Same for the left hand side, and in fact they are the same matrix and the same scalar and the same sum.
As for the second equality we need to relate $F$ to $E$ and $B$.  We have $F_{0i}=E_i$ and $B_i=\frac{-1}{2}\epsilon_{ijk}F^{jk}$, where $F^{jk}=F_{jk}$ numerically.  So now we can break that sum into two parts.  First what if $\mu=0$, then $\nu\neq 0$ (else we get the zero matrix scaled by the zero scalar) so let's call $\nu$ by $i$ to indicate it is 1,2 or 3. Then we have (for each fixed i): $$\sigma^{0i}F_{0i}=\frac{i}{2}\left(\gamma^0\gamma^i-\gamma^i\gamma^0\right)E_i=\frac{i}{2}\left(\begin{bmatrix}
    1       & 0  \\
    0      & -1 
\end{bmatrix}\begin{bmatrix}
    0       & \sigma^i  \\
    -\sigma^i      & 0 
\end{bmatrix}-\begin{bmatrix}
    0       & \sigma^i  \\
    -\sigma^i      & 0 
\end{bmatrix}\begin{bmatrix}
    1       & 0  \\
    0      & -1 
\end{bmatrix}\right)E_i.$$
So doing the block matrix computation  (again for each fixed i) we get:
$$\sigma^{0i}F_{0i}=\frac{i}{2}\left(\begin{bmatrix}
    0       & \sigma^i  \\
    \sigma^i      & 0 
\end{bmatrix}-\begin{bmatrix}
    0       & -\sigma^i  \\
    -\sigma^i      & 0 
\end{bmatrix}\right)E_i=i\begin{bmatrix}
    0       & \sigma^i  \\
    \sigma^i      & 0 
\end{bmatrix}E_i=i\alpha^iE_i.$$
So that's the case where $\mu=0$ and $\nu=i$.  Similarly if $\mu=i$ and $\nu=0$.  (But seriously I think your result may be off by a factor of two since both of these terms come up in the sum $\sigma^{\mu\nu}F_{\mu\nu}$.) Next you can consider the case where $\mu=i$ and $\nu=j$ but this time you use the commutator result for $[\sigma^i,\sigma^j]$ and what you'll do is figure out what $\vec\sigma$ is supposed to be, since no one has said.  (It's not a common term in my quantum field theory books.)  But you can define it to be what you need it to be by just computing the left hand side.  Computing the left hand side where you know what it is, should give you an expression that you can make look like the right hand side (except maybe for that pesky factor of two).
