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For a localized quantum harmonic oscillator along $z$, let the transverse magnetic vector potential be $A_z(t)=2A_0\cos (\theta -\omega t)$ for $t>0$, with zero potential before the switch. The Coulomb gauge form of minimal coupling gives $H_1=e{A}_z{p}_z/m$. The creation operator matrix element $⟨1|p_z|0⟩=i\sqrt{m\hslash {\omega }_0/2}$ means that only $|1⟩$ is reached at first order from the ground state. Its interaction picture amplitude isIn particular $F_0=t$, so resonant finite-time expressions are nonsingular. The induced electric dipole moment is $-2e\sqrt{\hslash /(2m{\omega }_0)}\mathrm{Re}(e^{-i{\omega }_{0}t}{b}_1)$. It contains a driven component with the dynamic electric polarizability and free-frequency transients required by the initial conditions. The discontinuity in $A_z$ gives an impulsive electric field and an initial mechanical velocity $e{A}_z(0^{+})/m$ even though the initial canonical momentum has zero mean.

For $\omega \ne {\omega }_0$, isolate the terms proportional to $e^{-i\omega t}$ in the previous answer. Their coefficient isHence the driven electric dipole moment component isFor $t>0$, away from the instantaneous switch, the radiation electric field isThus the dynamic electric polarizability isBelow 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.