# 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.

• I'm sure somebody will be able to give you the answer, but if you don't want to wait you can try to look up the terms 'unique continuation'. Commented Apr 26, 2021 at 18:11
• Thank you for your comment Leo. I know the basic result that a solution can not have compact support for two distinct times (this basically follows by the Paly-Wiener theorem), and I know that there are more refined unique continuation results of this type, but still I can't find a reference which answers my question. Commented Apr 26, 2021 at 18:27

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.

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.

• Thank you, very nice argument! It remains to understand whether $\Omega$ can have positive measure, but probably this is much harder. Commented May 5, 2021 at 15:31
• Yeah, I have absolutely zero idea for that one. Commented May 5, 2021 at 16:08

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 $$|t|, then we can set $$\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$$.

• Interesting, thank you! Do you think that a similar argument (with suitable modifications) could work also in higher dimensions? Commented May 3, 2021 at 8:24
• @RaffaeleScandone: The statement feels to me like it should be true in higher dimensions as well (but I don't have any expertise here), but this argument seems to depend crucially on the fact that an open set separates the space into two components in $d=1$, which of course isn't true for $d>1$. Commented May 3, 2021 at 14:48
• Minor remark: The usual spherical mean argument extends this also to $d = 3$. Commented May 3, 2021 at 21:16
• @WillieWong I agree, but it seems to me that you need that $\psi$ has a non trivial radial component, or am I missing something? Commented May 3, 2021 at 22:57
• @RaffaeleScandone: there's a second more technical step. Let me write an answer. Commented May 4, 2021 at 13:28