We are considering the Schrödinger equation on $\mathbb{R}^d \times [0,T]$

$$i \partial_t \psi(x,t)=-\Delta \psi(x,t) + u(t)V(x) \psi(x,t), t>0$$ $$\psi(x,0):=\psi(x_0) \in L^2(\mathbb{R}^d)$$ with $u\in C([0,T])$ and $V \in C_c(\mathbb{R}^d).$

Standard picard iteration shows that this equation has a unique solution.

Are there any sufficient conditions that the flow of this evolution equation leaves closed convex sets invariant?

I could for example imagine that if $X$ is such a set and $\psi(x_0)\in X$ and for all $y \in X$ and $t>0$ $$\langle i \partial_t(x,t), y \rangle =0$$ that this condition is true. I do not know if this is indeed true, but this was a possible condition that came first to my mind.

We are considering the Schrödinger equation on $\mathbb{R}^d \times [0,T]$

$$i \partial_t \psi(x,t)=-\Delta \psi(x,t) + u(t)V(x) \psi(x,t), t>0$$ $$\psi(x,0):=\psi(x_0) \in L^2(\mathbb{R}^d)$$ with $u\in C([0,T])$ and $V \in C_c(\mathbb{R}^d).$

Standard picard iteration shows that this equation has a unique solution.

Are there any sufficient conditions that the flow of this evolution equation leaves closed convex sets invariant?

I could for example imagine that if $X$ is such a set and $\psi(x_0)\in X$ and for all $y \in X$ and $t>0$ $$\langle \partial_t\psi(x,t), y \rangle =0$$ that this condition is true. I do not know if this is indeed true, but this was a possible condition that came first to my mind.