# Decay of solutions to Schrodinger equation with local minimum in potential

Consider the one-dimensional Schrodinger operator on the real line $\mathbb{R}$ given by $$L = - \partial_x^2 + V$$ where $V$ is a potential with the following properties:

• $V$ is non-negative, and infinitely differentiable
• $|V| = \frac{1}{|x|^2}$ for $|x| \gg 1$ (so in particular it is in $L^p$ for any $p\geq 1$).

Question: Are there general theorems concerning the (pointwise or local energy) decay of the Schrodinger ($i\partial_t \Phi = L\Phi$) equations for such operators? I am particularly interested in the case where $V$ has more than one local maximum.

Motivation:

First let us consider the case where $V$ has exactly one local (and hence global) maximum. In the classical particle picture we see that the global maximum corresponds to an unstable fixed point of the dynamics, and for most energies a particle either has large energy so it flies over the hump, or has small energy so it comes in, turns around, and bounces off. Only at a very specific energy can you achieve the balancing act of sending in a particle that comes eventually to rest on the top of the hump.

This classical picture is essentially sufficient for analyzing the quantum picture. If we assume that the local maximum is non-degenerate, then we can prove relatively simply an integrated local energy decay estimate for solutions. One version of which states that, assuming the local maximum is at the origin, for every positive $\delta$ there exists some $b$ such that $$\int_0^\infty \int_{\mathbb{R}} \frac{1}{(1 + (bx)^2)^{\frac12 + \delta}} \Phi_x^2 ~\mathrm{d}x ~\mathrm{d}t \tag{*}$$ is bounded by some universal constant times some quantity depending only on the initial data.

In the case where $V$ has more than one local maximum, the classical picture is drastically different. We see that between two consecutive local maxima of the potential, we expect classically there to be stable trapped particles, since classical particles cannot jump over the hump. So the classical picture would contradict local energy decay, since a purely classical picture would admit spatially localised solutions over a non-trivial range of energies.

On the other hand, when dealing with quantum phenomenon there should be tunneling where particles can escape through finite barriers. So I expect the intuition to be that in the quantum case some (possibly weaker) energy decay is still available.

Related question: At the present I just don't have a starting point to search from. So if someone can give me a few names and/or papers to start looking, or some keywords to search for, that would be appreciated.

Edit: As Christian Remling pointed out, there is the standard RAGE theorem which in particular implies that energy will escape from any compact set. What I seek is something a bit stronger. The RAGE theorem (as far as I know) does not give explicit rates, and I am hoping for some sort of result giving either the localised energy has an explicit rate of decay $\leq t^{-\alpha}$ for some $\alpha > 0$ or that the decay can be made explicit in the integral sense (something like equation (*) above but with possibly the $L^1$ integration in $t$ replaced by some higher $L^p$ for $p < \infty$).

Your assumptions on $V$ imply that $L=-d^2/dx^2 + V(x)$ on $L^2(\mathbb R)$ has purely absolutely continuous spectrum. This implies decay estimates of the type $\| Ke^{-itL}\psi\|_2 \to 0$ as $|t|\to\infty$ for relatively compact $K$; the case of interest here is $K=$ multiplication by $\chi_{(-R,R)}(x)$. (Some people call this the RAGE Theorem; it's really the Riemann-Lebesgue lemma in disguise.)
This argument is rather general; I only used that $V\ge 0$ and $V$ has sufficiently rapid decay at $\pm\infty$.
• My fault for not having mentioned it in the original statement. Yes I am aware of the RAGE type results, but I am looking for something slightly stronger. For example, something like RAGE cannot be directly used to conclude $t^{-\alpha}$ decay for any $\alpha > 0$, nor that $\|K e^{-itL}\psi\|_{L^p_t L^2_x} < \infty$ for any $p < \infty$. – Willie Wong Apr 1 '15 at 14:31