# 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 $p < \infty$).

There are several decay results in the 1D case, but probably they are not enough for you. Goldberg and Schlag (Comm. Math. Phys. 251 (2004) 157–178) proved pointwise decay of the $$L^\infty$$ norm as $$t^{-1/2}$$ like in the free case, for any potential such that $$(1+|x|)V\in L^1$$, plus some spectral assumptions. Of course this rules out the case $$V\sim|x|^{-2}$$ that is of interest to you. Anyway, in these results the local structure of $$V$$ is irrelevant. Maybe this suggests that either localization is so strong to produce a bound state, or you have dispersion like in the free case.

• This is actually very helpful. When I asked the question I think I picked the asymptotic because I thought it is "nice enough". (This was just something I was wondering about and not coming from a specific problem.) So knowing that with slightly better asymptotics you get a good decay regardless of local structure is helpful. (Note For Self: Goldberg-Schlag required non-resonance at zero, but this is automatic if $V$ is signed and non-trivial.) Jun 6 at 13:07
• Maybe I should add that you get even better decay ($t^{-3/2}$) in weighted spaces. This is a remark by Schlag, later improved by Mizutani J. Math. Soc. Japan Vol. 63, No. 1 (2011) pp. 239–261, probably the best reference Jun 6 at 13:52
• Yeah, I actually became aware of the $t^{-3/2}$ decay (I wrote up some hand-wavy notes for the case of barrier potentials) when non-resonant about three years ago, when a physicist mentioned off-handedly that generic Schrodinger problems should decay at $t^{-3/2}$ and not $t^{-1/2}$ but wasn't able to explain why. Jun 6 at 14:53
• I am going to accept this answer: it doesn't strictly speaking answer the question I posed, but it does get me toward thinking about the right kind of stuff. Jun 6 at 14:55

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 $\pm\infty$.

There is also much work on more specific estimates in more specialized situations. I'm not very familiar with this, but this paper might provide an entry point to the literature.

• 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 $p < \infty$. Apr 1, 2015 at 14:31
• That's a good starting point; do you mind editing that into your answer for future reference? Apr 2, 2015 at 7:35
• @WillieWong: Sure, no problem, I just did it. Apr 2, 2015 at 16:00