How does a particle passing an atom perturb the atom's QM system? Consider the quantum mechanical system of an atom with Hamiltonian $\hat{H}_0$ and assume that we know the solutions of the eigenvalue problem $\hat{H}_0|n\rangle=E_n|n\rangle$ for $n=0,1,2,\ldots$. Let now $\Psi_n(x)$ denote the position space vector corresponding to $|n\rangle$ where we are restricting ourselves to one dimension of position.
Assume now that the atom sits at $x=0$ and that a charged particle passes the atom with velocity $v$ such that the perturbing potential $V(x)=V_0e^{-|x-x_0(t)/\lambda|}$ is introduced where $\lambda\in\mathbb{R}$ is the reach of the potential $V(x)$ and $x_0(t):=vt$.
Assignment: If, in the distant past for $t\to-\infty$, the system is in the ground state $|0\rangle$, compute the probability to find the atom in a state $|n\rangle, n>0$ for $t\to\infty$ (Compute the relevant integral piecewise by considering the two cases $x<x_0(t)$ and $x\ge x_0(t)$ and use that $\Psi_n(x)$ is mostly located around the atom's position in order to expand the phase factor in the integral around $x=0$). What conditions do $\Psi_n(t)$ have to satisfy such that the transition probability is positive?

My try: If we denote by $V_i(t)$ the perturbing potential in the interaction picture, we get: $V_i(t)=V_0\cdot\left( e^{i\hat{H_0}t/\hbar} \circ e^{-|x-x_0(t)|/\lambda} \circ e^{-i\hat{H}_0t/\hbar} \right)$ where $\circ$ denotes composition.
We now plug this into the formula for the transmission probability
$P(|0\rangle\to|n\rangle)=\left| \frac{1}{\hbar}\int_{-\infty}^{\infty}dt' \langle n|V_i(t')|0\rangle \right|^2=\frac{1}{\hbar^2}\left|\int_{-\infty}^{\infty} dt  \,e^{i(E_n-E_0)t/\hbar}\cdot \langle n|V_0e^{-|x-x_0(t)|/\lambda}|0\rangle\right|^2$
and by distinguishing between by $x<x_0(t)$ and $x\ge x_0(t)$
and switching to position space, this is equal to
$\frac{1}{\hbar^2}\cdot \left| \int_{-\infty}^{\infty}dt \, e^{i(E_n-E_0)t\hbar}\cdot V_0\cdot\left( \langle\Psi_n(x)|e^{(x_0(t)-x)/\lambda}\cdot\mathbb{1}_{x\ge x_0(t)}|\Psi_0(x)\rangle + \langle\Psi_n(x)|e^{(x-x_0(t))/\lambda}\cdot\mathbb{1}_{x< x_0(t)}|\Psi_0(x)\rangle \right) \right|^{\,2}$
How do I continue? In particular, since the integral is wrt. time and not position, how do I use the two cases $x<x_0$ and $x\ge x_0$ and make a Taylor expansion?
 A: The relevant integral is $$\langle n|V|0\rangle=\int_{-\infty}^\infty dx \Psi_n^*(x) V(x)\Psi_0(x)=\int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x)e^{-|x-x_0(t)|}.$$ If we use the variable $\omega_n=(E_n-E_0)/\hbar$, the probability becomes
\begin{aligned}
P&=\frac{1}{\hbar^2}\left|\int_{-\infty}^\infty dt e^{i\omega_n t}\int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x)e^{-|x-x_0(t)|}\right|^2\\
&=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) \int_{-\infty}^\infty dt e^{i\omega_n t}e^{-|x-x_0(t)|}\right|^2\\
&=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) \left(e^{-x}\int_{-\infty}^{\lambda x/v} dt e^{i\omega_n t}e^{v t/\lambda}+
e^x\int_{\lambda x/v}^{\infty} dt e^{i\omega_n t}e^{-v t/\lambda}
\right)\right|^2\\
&=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) e^{i\lambda \omega_n x/v}\left(\frac{\lambda}{v+ i\omega_n\lambda}+
\frac{\lambda}{v- i\omega_n\lambda}
\right)\right|^2\\
&=\frac{1}{\hbar^2}\left| \int_{-\infty}^\infty dx \Psi_n^*(x) \Psi_0(x) e^{i\lambda \omega_n x/v}\frac{2\lambda v}{v^2+\omega_n^2\lambda^2}\right|^2\\
&\vdots
\end{aligned} This should get you on your way!
