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.



Vacuum region with positive measure for the Schrödinger equation
The purpose of this answer is to extend [Christian Remling's answer](https://mathoverflow.net/a/391661/3948) 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.




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### 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.


