Evaluating Matrix Spin Dependent Term in Dirac Quadratic Equation

[thetafilippo]

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 $$\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]$$

$$\sigma^{00}=\frac{i}{2}[\gamma^0,\gamma^0]=0$$
$$\sigma^{0i}=\frac{i}{2}[\gamma^0,\gamma^i]=[\gamma^0,\gamma^0\alpha_i]=\alpha_i-\gamma^0\alpha_i\gamma^0=-2\alpha_i$$
$$\sigma^{ij}=\frac{i}{2}[\gamma^i,\gamma^j]=[\gamma^0\alpha_i,\gamma^0\alpha_j]=\frac{i}{2}\gamma^0(\alpha_i\gamma^0\alpha_j-\alpha_j\gamma^0\alpha_i)=\frac{i}{2}
\begin{pmatrix}
-[\sigma_i,\sigma_j] &0\\
0&-[\sigma_i,\sigma_j]
\end{pmatrix}=\epsilon_{ijk}\begin{pmatrix}
\sigma_k &0\\
0&\sigma_k
\end{pmatrix}=\epsilon_{ijk}\Sigma_k$$
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?"

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

 

[eljose79]

First of all, it is important to note that the final expression you are trying to obtain is not the spin dependent term, but rather the spinor form of the electromagnetic interaction term in the Dirac equation. This term describes the interaction of the electron's spin with the external electromagnetic field.

In order to obtain the correct result, you need to use the correct representation of the gamma matrices. The representation you are using, known as the "chiral representation", is not suitable for this calculation. Instead, you should use the "Dirac representation" where the gamma matrices are given by:

$$\gamma^0 = \begin{pmatrix}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1
\end{pmatrix} \ \ \ \ \ \gamma^i = \begin{pmatrix}
0 & \sigma_i\\
-\sigma_i & 0
\end{pmatrix}$$

Using this representation, you can easily show that:

$$\sigma^{\mu\nu}=\frac{i}{2}\left[\gamma^\mu,\gamma^\nu\right]=\frac{i}{4}\begin{pmatrix}
0 & \sigma^{\mu\nu}\\
\sigma^{\mu\nu} & 0
\end{pmatrix}$$

where $\sigma^{\mu\nu}=\frac{i}{2}\left(\gamma^\mu\gamma^\nu-\gamma^\nu\gamma^\mu\right)$.

Substituting this into the expression for the electromagnetic interaction term, we get:

$$\frac{\sigma^{\mu\nu}}{2}F_{\mu \nu}=-\frac{1}{4}\begin{pmatrix}
0 & \sigma^{\mu\nu}\\
\sigma^{\mu\nu} & 0
\end{pmatrix}\begin{pmatrix}
0 & E_x&E_y&E_z\\
-E_x&0&B_z & -B_y\\
-E_y&-B_z&0&B_x\\
-E_z&B_y&-B_x&0
\end{pmatrix}$$

$$=-\frac{1}{4}\begin{pmatrix}
0 & -\sigma^{0i}E_i+\