= Solution
Work in <temporal gauge>, with zero scalar potential, and use the charge sign exactly as in the given covariant spatial derivative. A spatially uniform <vector potential> acts on a Fourier mode as
$$
\left(\nabla-\frac{ie\mathbf A(t)}\hbar\right)
\{\psi_{\mathbf k}(t)e^{i\mathbf k\cdot\mathbf r}\}
=i\left(\mathbf k-\frac{e\mathbf A(t)}\hbar\right)
\psi_{\mathbf k}(t)e^{i\mathbf k\cdot\mathbf r}.
$$
There are no spatial derivatives of $\mathbf A$. Substitution into the field equation and multiplication by $c^2$ therefore gives
$$
\boxed{\ddot\psi_{\mathbf k}
+\omega_{\mathbf k-e\mathbf A(t)/\hbar}^2\psi_{\mathbf k}=0,
\qquad
\omega_{\mathbf q}=\sqrt{c^2|\mathbf q|^2+\frac{m^2c^4}{\hbar^2}}.}
$$
The canonical <wavevector> $\mathbf k$ is conserved by spatial translation symmetry; the <kinetic momentum> is $\hbar\mathbf k-e\mathbf A(t)$. The absence of a term involving $\dot{\mathbf A}$ in this mode equation follows from <temporal gauge> and the spatial minimal-coupling operator. The <electric field> still affects evolution through the changing oscillator <frequency>.
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