Electron magnetic moment in coupled squared Dirac equation

Question

Given a 4-potential $A^\mu=(\phi,\vec{A})$, the Dirac equation with minimal coupling reads $$[i\gamma^\mu(\partial_\mu+ieA_\mu)-m]\psi=0\tag{1}\label{1}$$ By means of a non relativistic reduction, one can prove that in the non relativistic limit we are left with $$\left\{\frac{(\vec{p}-e\vec{A})^2}{2m}-\underbrace{\frac{e}{m}\frac{\vec{\sigma}}{2}}_{\ g=2}\cdot\vec{B}+e\phi\right\}\psi=0\tag{2}\label{2}$$ which is Pauli-Schrödinger equation with the correct gyromagnetic factor.

Let

$$D_\mu:=\partial_\mu+ieA_\mu\qquad\text{and}\qquad \sigma^{{\mu\nu}}:=[\gamma^\mu,\gamma^\nu]$$

As Ref. 1 observes, the non relativistic reduction procedure can also be performed on the squared Dirac equation, the equation we get applying $[i\gamma^\mu(\partial_\mu+ieA_\mu)+m]$ to \eqref{1}, that is

$$[-D_\mu D^\mu-m^2\color{red}{-\frac{e}{2}\sigma^{\mu\nu}F_{\mu\nu}}]\psi=0\tag{3}\label{3}$$ That means that each component of the Dirac spinor satisfies the KG equation with the additional term in red. As pointed out in Ref. 2, such term only appears because we imposed the minimal coupling in \eqref{1} and then we squared. It is straightforward to prove that using the Dirac representation for the $\gamma$ matrices $$\frac{1}{2}\sigma^{\mu\nu}F_{\mu\nu}=(\color{blue}{i\vec{\alpha}\cdot\vec{E}}+\color{green}{\vec{\Sigma}\cdot\vec{B})}\qquad \alpha^i=\gamma^0\gamma^i=\begin{pmatrix}0 &\sigma^i \\ \sigma_i & 0\end{pmatrix}, \Sigma^i=\begin{pmatrix}\sigma^i &0 \\0 & \sigma_i \end{pmatrix}.\tag{4}\label{4}$$ The calculations are quite straightforward, now the problem comes. Then, Ref. 1 and Ref.3 (which is using the chiral representation though) observe that this is a magnetic interaction term, containing the correct (without considering non minimal prescriptions) gyromagnetic factor of Dirac theory $g=2$. I can see why the green piece is a magnetic interaction term and in fact this is the $4\times 4$ version of the term in \eqref{2}. Why is the piece containing the electric - the one in blue - considered as a part of the magnetic dipole term? Also, if I perform the non relativistic reduction in the usual way, like I would do with the KG equation, the electric term will still be there. On the other hand, there is no such term in \eqref{2}, so what does this electric piece represent and why does it only appear if we follow this path?

Addendum: this answer does not have the electric term. I'm not sure this is right, unless they are also assuming the electric field is zero.

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References:

1. Quantum Field Theory, 1987. Itsykson and Zuber. Section $2.2.3$, page 66 eqn $(2.73)$.
2. QED, Landau&Lifshitz. Section 32, eqn $(32.7a)$.
3. Quantum Field Theory and the Standard Model, 2013. Matthew D. Schwartz Section 10.4, eqn $(10.4.109)$

Answer 1

The coupling to the electric field is there because of Lorentz invariance. A magnetic dipole aquires an electric dipole component when seen from a moving frame. Note that the expectations $\psi^\dagger \alpha_i \psi$ of the $\alpha$ matrices are proportional to the current, and so are effectively zero when the Dirac particle is stationary.

And yes, my old answer had no electric field. My excuse is the "something like".