Evaluating $\sigma^{\mu\nu}F_{\mu\nu}=i\alpha \cdot E+\Sigma\cdot B$ matrix, spin dependent term in quadratic Dirac equation

Question

I derive the quadratic form of Dirac equation as follows $$\lbrace[i\not \partial-e\not A]^2-m^2\rbrace\psi=\lbrace\left( i\partial-e A\right)^2 + \frac{1}{2i} \sigma^{\mu\nu}F_{\mu \nu}-m^2\rbrace\psi=0$$ And I need to find the form of the spin dependent term to get the final expression $$g \frac{e}{2} \frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-g\frac{e}{2}\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\Sigma}\cdot\mathbf{B}\right)$$ But I don't get this expression.

I'm using the Dirac representation with these quantities $$\vec{\alpha}=\begin{pmatrix} 0 & \vec{\sigma}\\ \vec{\sigma} & 0 \end{pmatrix} \ \ \ \ \ \vec{\Sigma}=\begin{pmatrix} \vec{\sigma}& 0\\ 0&\vec{\sigma} \end{pmatrix}$$ Where $\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$ is the Pauli matrix vector.

I constructed the electromagnetic tensor term by term, using the definition $F_{\mu\nu}=\partial_{\mu} A_{\nu}-\partial_{\nu} A_{\mu}$ with the metric tensor $g^{\mu\nu}=\textrm{diag}(+1,-1,-1,-1)$ and I get $$F_{\mu\nu}=\begin{pmatrix} 0 & E_x&E_y&E_z\\ -E_x&0&B_z & -B_y\\ -E_y&-B_z&0&B_x\\ -E_z&B_y&-B_x&0 \end{pmatrix}$$

I evaluate the $\sigma^{\mu\nu}$ matrix starting from its definition in terms of gamma matrices $\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]$

$$\sigma^{00}=\frac{i}{2}[\gamma^0,\gamma^0]=0$$ $$\sigma^{0i}=\frac{i}{2}[\gamma^0,\gamma^i]=\frac{i}{2}[\gamma^0,\gamma^0\alpha_i]=\frac{i}{2}[\alpha_i-\gamma^0\alpha_i\gamma^0]=\frac{i}{2}2\alpha_i=i\alpha_i$$ $$\sigma^{ij}=\frac{i}{2}[\gamma^i,\gamma^j]=[\gamma^0\alpha_i,\gamma^0\alpha_j]=\frac{i}{2}\gamma^0(\alpha_i\gamma^0\alpha_j-\alpha_j\gamma^0\alpha_i)=\frac{i}{2} \begin{pmatrix} -[\sigma_i,\sigma_j] &0\\ 0&-[\sigma_i,\sigma_j] \end{pmatrix}=\epsilon_{ijk}\begin{pmatrix} \sigma_k &0\\ 0&\sigma_k \end{pmatrix}=\epsilon_{ijk}\Sigma_k$$ And the remaining terms follow by the antisymmetry property $\sigma^{\mu\nu}=-\sigma^{\nu\mu}$

$$\sigma^{\mu\nu}=\begin{pmatrix} 0 & 2\alpha_x & 2\alpha_y & 2\alpha_z\\ -2\alpha_x&0&\Sigma_z & -\Sigma_y\\ -2\alpha_x&-\Sigma_z&0&\Sigma_x\\ -2\alpha_x&\Sigma_y&-\Sigma_x&0 \end{pmatrix}$$

Now, my questions are:

"Why these calculations do not yield the correct result?"

"What I should do to obtain the correct result? What I'm missing?"

$$\frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-\left(i\vec{\alpha}\cdot\mathbf{E}+\vec{\sigma}\cdot\mathbf{B}\right)$$

Answer 1

You did not quite explain how you failed to obtain the target result. I would not like to spoil the fun of catching your factors and signs involved, so I will strictly deal with significant proportionalities.

$$ \sigma^{\mu\nu} F_{\mu \nu}= \sigma^{0i} F_{0 i}+\sigma^{i0} F_{i 0}+ \sigma^{ij} F_{ij}=2\sigma^{0i} F_{0 i} + \sigma^{ij} F_{ij} . $$

Now, $$ \sigma^{0i} F_{0 i} \propto \alpha_i E_i, $$ and $$ \sigma^{ij} F_{ij} \propto \epsilon_{ijk}\Sigma_k ~~ \epsilon_{ijm} B_m =2 \Sigma_k B_k, $$ by virtue of the 2-index Levi-Civita contraction identity.

Proceed to fix numerical normalizations, if needs be by assuming sparse special constant EM fields.