= Exponential switch-on of a driven quantum oscillator
{title2=$P_1^{(0)}\simeq\eta^2(\omega\tau)^2/[1+(\omega\tau)^2]$}
For $H(t)=\hbar\omega(a^\dagger a+1/2)-Fx e^{t/\tau}$ on $t\leq0$, start in the unforced <ground state> at $t=-\infty$. Define $x_{01}=\sqrt{\hbar/(2m\omega)}$ and $\eta=Fx_{01}/(\hbar\omega)$. <Time-dependent perturbation theory> gives
$$
c_1(0)=\frac{iFx_{01}}\hbar\int_{-\infty}^0 e^{(1/\tau+i\omega)t}\,dt=\frac{i\eta\omega\tau}{1+i\omega\tau}.
$$
Thus 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 $|\eta|\ll1$.
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 $\dot\alpha=-i\omega\alpha+iFx_{01}e^{t/\tau}/\hbar$ or from the forced oscillator equations. Consequently the exact original-basis probability is $P_1^{(0)}=|\alpha|^2e^{-|\alpha|^2}$. 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.
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