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?"

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


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.