 Solution

To first order in $F$, time-dependent perturbation theory replaces the state on the right-hand side by its unperturbed value $e^{-i{E}_{0}t/\hslash }|0⟩$. The bound state is even, while the position operator $x$ is odd. The parity selection rule therefore givesso ${\dot{b}}_k=0$ and, with no initial continuum component,

Projection onto the odd state givesWriting ${\Omega }_k=(E_k-E_0-\hslash \omega )/\hslash$ and integrating from zero to $t$ yields