= Solution
For $\omega\ne\omega_0$, isolate the terms proportional to $e^{-i\omega t}$ in the previous answer. Their coefficient is
$$
\frac1{i(\omega_0-\omega)}-\frac1{i(\omega_0+\omega)}
=-\frac{2i\omega}{\omega_0^2-\omega^2}.
$$
Hence the driven <electric dipole moment> component is
$$
\mathbf P_\omega(t)=-\frac{2e^2A_0\omega}{m(\omega_0^2-\omega^2)}
\sin(ky-\omega t)\hat{\mathbf z}.
$$
For $t>0$, away from the instantaneous switch, the radiation <electric field> is
$$
\mathbf E(t)=-\partial_t\mathbf A
=-2A_0\omega\sin(ky-\omega t)\hat{\mathbf z}.
$$
Thus the <dynamic electric polarizability> is
$$
\boxed{\mathbf P_\omega=\alpha_{\mathrm{dyn}}(\omega)\mathbf E,
\qquad\alpha_{\mathrm{dyn}}(\omega)=\frac{e^2}{m(\omega_0^2-\omega^2)}.}
$$
Below resonance the driven <electric dipole moment> is in phase with the <electric field>; above resonance its sign reverses. The same susceptibility follows from the classical <Lorentz oscillator model> without damping. At exact resonance the separation into bounded driven and transient terms fails: their apparent poles combine into the finite-time secular term identified in part (c). Without damping there is no bounded steady resonant response, and the first-order state expansion eventually ceases to apply.
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