Let $A$ be a selfadjoint operator on some Hilbert space $H$, let $U(t)=e^{itA}$ be the corresponding continuous group, and let $f\in H$ be orthogonal to all eigenvectors of $A$. Are there examples such that $U(t)f$ does not converge weakly to 0 as $t\to+\infty$?


(From the RAGE Theorem, if $K$ is a compact operator on $H$ and $f$ as above, then
$$
\lim\_{T\to+\infty}\frac1T\int\_0^T\|KU(t)f\|\_H^2dt=0.
$$
This implies $\liminf\_{t\to+\infty}\|KU(t)f\|\_H=0$ and suggests weak convergence of $U(t)f$ to 0, at least for suitable sequences of times.)



Counterexamples to weak dispersion for the Schrödinger group
The answer is yes.


A measure-preserving invertible shift $T: X \to X$ on a probability space $(X,\mu)$ is said to be [weakly mixing](https://en.wikipedia.org/wiki/Mixing_(mathematics)) if $\lim\_{N \to \infty} \frac{1}{N} \sum\_{n=1}^N |\langle f \circ T^{-n}, g \rangle|^2 = 0$ for all $f,g$ in the Hilbert space $L^2(X)\_0$ of mean zero square-integrable functions, and [strongly mixing](https://en.wikipedia.org/wiki/Mixing_(mathematics)) if $\lim\_{n \to \infty} |\langle f \circ T^{-n}, g \rangle| = 0$ for all such functions. (The minus sign is not important here, and one can replace $T^{-n}$ by $T^n$ if one wishes.) There are examples of systems that are weakly mixing but not strongly mixing; see for instance [this previous MathOverflow post](https://mathoverflow.net/questions/47080/examples-of-transformations-which-are-weak-mixing-but-not-strong-mixing) for some examples (indeed in certain technical senses a "generic" shift is of this form). Note that weakly mixing shifts have no eigenfunctions in $L^2(X)\_0$ (indeed this is an if and only if, by the discrete version of the RAGE theorem).


Such systems $(X,\mu,T)$ are discrete flows, but they can be converted into continuous flows by the standard device of taking a [suspension](https://en.wikipedia.org/wiki/Suspension_(topology)). Namely, let $\tilde X$ be $X \times {\bf R}/\sim$ where we quotient by the equivalence relation $(x,t) \sim (T^{-n} x, t+n)$ and endow this space with the product measure $\tilde \mu$ of $\mu$ and Lebesgue measure on the unit interval, and the continuous shift $\tilde T^t (x,s) := (x,s+t)$. If one then lets $H \equiv L^2([0,1]; L^2(X)\_0)$ be the Hilbert space of functions $f \in L^2(\tilde X)$ that are of mean zero on every time slice $X \times \{t\}$, and lets $U(t): H \to H$ be the [Koopman operator](https://en.wikipedia.org/wiki/Composition_operator) $U(t) f(x,s) := f \circ \tilde T^{-t}(x,s) = f(x,s-t)$, one can easily verify that $U(t)$ is a strongly continuous unitary flow (and thus of the form $e^{itA}$ by [Stone's theorem](https://en.wikipedia.org/wiki/Stone%27s_theorem_on_one-parameter_unitary_groups)) that has no eigenfunctions, but such that $U(t)$ fails to weakly converge to zero (even if we restrict $t$ to the integers, in which case the continuous flow basically collapses back to the discrete flow $T^n$).


UPDATE: Here is another example. Let $\mu$ be an atomless compactly supported probability measure on ${\bf R}$ whose Fourier transform does not decay to zero at infinity (for instance one can take the standard Cantor set measure). If we let $U(t): L^2(\mu) \to L^2(\mu)$ be the modulation flow $U(t) f(x) = e^{itx} f(x)$, then the flow has no eigenfunctions (because of the atomless condition) but $U(t) 1$ does not converge weakly to zero (because $\langle U(t) 1, 1 \rangle$ is basically the Fourier transform of $\mu$).


