Let $\mathcal{H}:=V-\Delta$ be the Schrödinger operator, $V\in C^\infty_0(\mathbb{R})$ and let $u\in L^2(\mathbb{R})$ be a solution of the equation
$$\left(\Delta-\frac{\partial^2}{\partial t^2}\right)u(x,t)=V(x)u(x,t)\qquad(1)$$
How does it follow that $u(x,t)\xrightarrow{t\to\pm\infty} 0\iff\sigma_p(\mathcal{H})=\emptyset$ in $L^2(\mathbb{R})$, where $\sigma_p$ is the point spectrum?
Note that this is a result from Professor Tao's blog post
I have thought a little bit about the "$\Longleftarrow$" direction:
$$\begin{aligned} &\mathcal{H}u=\lambda u \\ \iff&(V-\Delta)u=\lambda u \\ \iff&(\Delta+\lambda)u=V u\qquad(2) \end{aligned}$$
This is equivalent to $(1)$ if $\lambda=-\frac{\partial^2}{\partial t^2}$. However, that doesn't make sense.
$\sigma_p(\mathcal{H})=\emptyset$ implies that $\nexists$ $u\in L^2(\mathbb{R})$ satisfying $(2)$