Exponential switch-on of a driven quantum oscillator ($P_1^{(0)}\simeq {\eta }^2(\omega \tau {)}^2/[1+(\omega \tau {)}^2]$)

For $H(t)=\hslash \omega (a^{†}a+1/2)-Fx{e}^{t/\tau }$ on $t\le 0$, start in the unforced ground state at $t=-\infty$. Define $x_{01}=\sqrt{\hslash /(2m\omega )}$ and $\eta =F{x}_{01}/(\hslash \omega )$. Time-dependent perturbation theory givesThus the first excited-state probability in the original oscillator basis tends to zero for a rapid turn-on and to ${\eta }^2$ for slow turn-on, within the weak-force approximation $|\eta |\ll 1$.

This finite adiabatic limit is not a failure of the adiabatic theorem: the instantaneous ground state is displaced and has a nonzero projection onto the original excited states. In fact a linear force preserves a coherent state, whose exact original-basis displacement at $t=0$ is $\alpha =i\eta \omega \tau /(1+i\omega \tau )$. It follows either from $\dot{\alpha }=-i\omega \alpha +iF{x}_{01}{e}^{t/\tau }/\hslash$ or from the forced oscillator equations. Consequently the exact original-basis probability is $P_1^{(0)}=|\alpha {|}^2{e}^{-|\alpha {|}^2}$. The instantaneous oscillator basis has displacement $\eta$, so its mean excitation is instead $|\alpha -\eta {|}^2={\eta }^2/[1+(\omega \tau {)}^2]$, which vanishes adiabatically.