= Solution
For this mode, abbreviate
$$
\omega=\omega_{\mathbf k}>0,\qquad
\Omega=\omega_{\mathbf k-e\mathbf A_0/\hbar}>0.
$$
The electric pulse is $\mathbf E(t)=-\mathbf A_0\delta(t)$. In <temporal gauge> the mode equation has a finite step in its <frequency> squared, with no delta multiplying the mode. Thus \b[both $\psi_{\mathbf k}$ and its first time derivative are continuous at zero].
Writing the positive-frequency incident amplitude as $a_{\mathbf k}^{\rm in}$, the matched solution is
$$
\boxed{\psi_{\mathbf k}(t>0)
=\frac{a_{\mathbf k}^{\rm in}}{\sqrt{2\omega}}
\left(\cos\Omega t-i\frac{\omega}{\Omega}\sin\Omega t\right).}
$$
Its initial value is $a_{\mathbf k}^{\rm in}/\sqrt{2\omega}$ and its initial derivative is $-i\omega a_{\mathbf k}^{\rm in}/\sqrt{2\omega}$, as required. In the normalized outgoing-frequency basis it is
$$
\psi_{\mathbf k}(t>0)
=\frac{a_{\mathbf k}^{\rm in}}{\sqrt{2\Omega}}
\left(\alpha_{\mathbf k}e^{-i\Omega t}
+\beta_{\mathbf k}e^{i\Omega t}\right),
$$
$$
\boxed{\alpha_{\mathbf k}=\frac{\Omega+\omega}{2\sqrt{\Omega\omega}},
\qquad
\beta_{\mathbf k}=\frac{\Omega-\omega}{2\sqrt{\Omega\omega}},
\qquad \alpha_{\mathbf k}^2-\beta_{\mathbf k}^2=1.}
$$
The negative-frequency contribution appears even though only <positive frequency> was incident. These are temporal-mode matching coefficients, so their conserved difference is hyperbolic rather than the sum-one rule of ordinary spatial reflection and transmission <probabilities>.
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