1
$\begingroup$

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 away from the support of the potential or just in $C^\infty_c(\mathbb{R}^n)$ with $\text{supp } \phi \cap \text{supp } V = \emptyset $, 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 $$ i\partial_t \psi = -\Delta\psi, \quad \psi(0,x) = \psi_0(x) $$ for small times? 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.

$\endgroup$
  • $\begingroup$ Whether or not the $L^2$ norm spreads out at large times is closely related to the spectral properties of $-\Delta+V$ (sometimes grandiosely called the RAGE thm, but it's really just basic facts about the Fourier transform in disguise). $\endgroup$ – Christian Remling Mar 5 at 17:58
  • $\begingroup$ It is perfectly reasonable to study something like $\langle \psi, x\psi\rangle$ also (assuming it exists), and there is a large literature on this too (key word "quantum dynamics" perhaps). $\endgroup$ – Christian Remling Mar 5 at 17:59
  • $\begingroup$ About your second question: are you actually asking whether the solution for an arbitrary potential $V(x)$ is always "close" to the free particle solution? If this is what you are asking, i think the answer is evidently no. $\endgroup$ – Konstantinos Kanakoglou Mar 5 at 21:27
  • $\begingroup$ I am also a little confused by what you mean "in some norm". Aren't we speaking about $L^2$ - wave functions? $\endgroup$ – Konstantinos Kanakoglou Mar 5 at 21:28
  • $\begingroup$ I edited slightly the question. Basically I would like to see that as long as the particle is far away from the barrier, it does not "see it", which means that there is some property of some particular solution that holds independently of the potential, but maybe just for small times. @KonstantinosKanakoglou L^2-norm is perfectly fine, any observable would do fine as long as it remains finite for small times. I was trying to prove this result for some weighted norms or for averages of some simple compactly supported observables, but I was not able too. $\endgroup$ – Ivan Mar 6 at 15:10
1
$\begingroup$

Here are some remarks to put Q2 into context, though I'm not answering the actual question. Basically, I'm going to give a quick summary of my 2007 paper Finite propagation speed and kernel estimates for Schrodinger operators; see reference 17 here.

If $H_1=-\Delta$ and $H_2=-\Delta+V$, and the support of $\psi$ is disjoint from the support of $V$, then the solutions of the wave equations $\psi_{tt}=H\psi$ will agree for a while since this equation has finite speed of propagation. In other words, $(\cos t\sqrt{H_j})\psi$ is independent of $j$ for a while.

This will imply the desired kind of statement for certain other functions $f(H_j)\psi$, by writing $f(\lambda)=\int g(x)\cos x\sqrt{\lambda}\, dx$ (essentially a Fourier transform if the spectra are positive), and if $g$ decays, then we can approximately cut off the integrals. This would work for example for the semigroups $e^{-tH_j}$, or in the discrete case, but for the actual operator $e^{-iH_j}$ we would need to work with the function $e^{-i\lambda^2}$, which doesn't have a decaying Fourier transform.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.