I've been playing around with numerical solutions to the Schroedinger equation and I came across something that feels very natural, but I was not able to prove it with the math I know.

The motivation comes from the following physical situation. Consider scattering of a wave-packet over a square potential barrier. If the wave-packet is far enough from the barrier it looks like it evolves according to the free equation, like there is no barrier at all (you can find a video in the post here). I would like to understand how one could see this mathematically. So:

1. What does it mean in this context for a $L^2$-function $\psi$ to be localized somewhere? It is clear that just taking the average $<x>= (\psi,x\psi)$ is not a good choice (consider, for example, a superposition of two Gaussian states). On the other hand simply saying that $\psi$ has compact support excludes all the Gaussians.

2. Given a potential $V(x) \in C^\infty_c(\mathbb{R}^n)$ and initial state $\psi_0(x)$ either localized in some sense or just in $C^\infty_c(\mathbb{R}^n)$, is it true that the solution of the Schroedinger equation $$ i\partial_t \psi = (-\Delta+V(x))\psi, \quad \psi(0,x) = \psi_0(x) $$ is close to the corresponding solution of the free equation $$ i\partial_t \psi = -\Delta\psi, \quad \psi(0,x) = \psi_0(x) $$ in some norm? I don't assume that $V$ is small or something, so I don't think that perturbation theory will be enough, since Schroedinger equation is not local.

Thank you for your help.