# Removing “electric dipole moment” from non-relativistic Dirac equation

I have found many sources (c.f. Schwartz's QFT book section 10.4) that try to obtain the non-relativistic limit of the Dirac equation by first "squaring it" so that it looks somewhat like the Klein-Gordon equation.

Kleiner eqn 6.113 for examples shows that in the chiral basis, this "squared" Dirac equation becomes

$$\left[-(\hbar \partial_\mu + i\frac{e}{c}A_\mu)^2+\frac{e\hbar}{c}\vec{\sigma}\cdot(\vec B \pm i \vec E)-M^2c^2\right]\left\{{\xi(x)\atop \eta(x)}\right\} = 0$$,

where $$\xi(x)$$ and $$\eta(x)$$ are the two-components chiral spinors. Kleiner then claims that we remove the fast oscillations from those spinors by defining $$\xi(x)\equiv e^{-iMc^2t/\hbar }\psi(\vec{x},t)$$ to obtain the non-relativistic Pauli equation

$$\left(i\partial_t + \hbar^2/2M\left(\vec \nabla-i \frac{e}{c \hbar}\vec A\right)^2+\frac{e}{2Mc}\vec\sigma \cdot \vec B-e \phi\right)\psi(\vec x,t)=0$$.

Same thing for $$\eta$$ presumably.

My question is this: How was the $$\vec \sigma\cdot \vec E$$ term removed when taking the non-relativistic limit in this way? As far as I can tell, the removal of the rapid oscillations is only going to affect the part of the equation involving covariant derivatives.

I know there are other ways to get the non-relativistic limit, but I would like to understand this one, since it comes up a lot.