Vacuum region with positive measure for the Schrödinger equation Let us consider the free Schrödinger equation $(i\partial_t+\Delta_x)\psi=0$ in $\mathbb{R}_t\times\mathbb{R}_x^d$.  I'm trying to understand the structure of the vacuum region
$$\Omega(\psi):=\{(t,x)\in \mathbb{R}_t\times\mathbb{R}_x^d \;\;s.t.\;\,\psi(t,x)=0\}$$
for solutions with finite energy. In particular, my question is the following: does there exist a non-zero solution $\psi\in \mathcal{C}(\mathbb{R},H^1(\mathbb{R}^d))$ of the free Schröodinger equation such that $\Omega(\psi)\subseteq\mathbb{R}^{1+d}$ has positive measure? If yes, is it even possible that $\Omega(\psi)$ contains an open ball? Thank you for any suggestion.
 A: The purpose of this answer is to extend Christian Remling's answer to dimension $d = 3$. There are two steps. (N.B. below the cut I show how to replace part 1 by a different argument that works in all dimensions, and so this should answer the question posed.)
We assume that we have a solution $\phi$ to the Schrodinger equation such that it vanishes on a (WLOG) $[-a,a]\times B(0,R)\subset \mathbb{R}\times\mathbb{R}^3$
1. Controlling the radial parts
Let $\psi:\mathbb{R}\times\mathbb{R}^+ \to \mathbb{C}$ be the function
$$ \psi(t,r) = r \cdot \frac{1}{4\pi} \int_{\mathbb{S}^2} \phi(t, r\omega) ~d \omega $$
the spherical mean of $\phi$ multiplied by the radius. We have that
$$ \psi(t,r) \equiv 0, \text{ when } r \leq R , |t| \leq a $$
and that $\psi$ is a solution to the one dimensional Schrodinger equation
$$ i \partial_t \psi = \partial^2_{rr} \psi $$
Thus Christian Remling's answer sufficies to imply that $\psi \equiv 0$ everywhere.
(Note that if $\phi(t,\bullet)\in L^2(\mathbb{R}^3)$, then $\psi(t,\bullet)\in L^2(\mathbb{R}_+)$).
Remark: when dimension $d \neq 3$, or when $d = 3$ but considering other spherical harmonics, we get that the equation being satisfied is $i \partial_t \psi = \partial^2_{rr} \psi + \alpha r^{-2} \psi$; so if Christian Remling's answer can be extended to the one dimensional Schrodinger equation with inverse square potentials, then this would also give a general answer.
2. Controlling the rest
Absent the needed result from the previous remark, we can argue thus in $d = 3$.
Applying the same argument, but now with center at $x_0 \in B(0,R/2)$, we see that the spherical means of $\phi(t,x)$ vanishes for all radii if we center it at $x_0$. I claim that this is enough to guarantee that $\phi(t,x) \equiv 0$.
The following proof is probably not the most straightforward, but that's the one I know.
Let $u$ be the solution to the linear wave equation $(-\partial^2_s + \triangle)u = 0$ (I use $s$ for the time parameter to disambiguate from the time parameter in the Schrodinger equation) in $d = 3$ with initial data $u(0,x) = 0$ and $\partial_su(0,x) = \phi(t,x)$ for any fixed $t$. Using the fundamental solution and the spherical mean property, we have that $u(s,x) = 0$ for all $x \in B(0,R/2)$. By the strong version of the Holmgren uniqueness theorem (see chapter IV, section 3 of F. John Partial Differential Equations) we have that $u \equiv 0$. This implies that $\phi(t,x) = 0$ for all $x$. Since $t$ is arbitrary: we are done.

Radial part: redux
Let's now prove the spherical mean property directly using the fundamental solution of the Schrodinger equation; this argument holds for all dimensions.
Let $\phi_0(x) = \phi(0,x)$. Suppose for $(t,x) \in [-a,a]\times B(0,R)$ we have that $\phi(t,x) = 0$, this implies that, using the fundamental solution to the Schrodinger equation, evaluated at $x = 0$, that
$$ \int e^{i|x|^2/4t} \phi_0(x) ~dx = 0 $$
for all $t \in [-a,a]\setminus \{0\}$. Let $\tilde{\phi}_0$ be the spherical mean of $\phi_0$. We note that by assumption $\tilde{\phi}_0(r) = 0$ for all $r \leq R$.
Our integral identity above implies
$$ \int_0^\infty e^{i r^2 / 4t} \tilde{\phi}_0(r) r^{d-1} ~ dr = 0 $$
Change variables you get
$$ \int_0^\infty e^{i \rho / 4t} \underbrace{\tilde{\phi}_0(\sqrt{\rho}) \rho^{d/2-1}}_{g(\rho)} ~ d\rho = 0 $$
Since $\tilde{\phi}_0$ is supported away from $r = 0$, we can extend $g(\rho)$ by zero to the negative half line. The vanishing of the above quantity for all $|t| \leq a$ shows that $g$ has compact Fourier support and hence is analytic; but its vanishing for $\rho < R^2$ implies that $g \equiv 0$, and hence the spherical mean vanishes.
Combining this with part 2 from above, we extend the uniqueness property to all dimensions.
A: This is only a very partial answer. In dimension $d=1$, the Paley-Wiener argument you refer to in your comment shows that $\psi(x,t)$ can not be zero on an open set: If $\psi(x,t)=0$ for $0\le x\le a$, say, for all $|t|<d$, then we can set $\psi(x,t)=0$ for all $x\ge 0$ for these $t$ and this will still solve the equation. Vanishing Fourier transform on a half line means that $\widehat{\psi}(k,t)$ is in the Hardy space $H^2$, but clearly this can not hold for all $t$ from an interval: if $\widehat{\psi}(k,0)\in H^2$, then
$\widehat{\psi}(k,t)=e^{-itk^2}\widehat{\psi}(k,0)$ has completely the wrong asymptotics for large $k\in\mathbb C^+$ when $t\not= 0$.
