In my quantum mechanics class we proved the Ehrenfest theorem for a conservative physical system with time-independent potential $V:\mathbb{R}^N\to \mathbb{R}$. The issue is that many steps in the proof, as well as the statement itself, were done without proper (if any) mathematical justification. I am trying to fill those holes but it seems a really hard task. The Schroedinger equation here is $$i\hbar \frac{\partial \psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\Delta\psi(x,t)+V(x)\psi(x,t),\qquad x\in \mathbb{R}^N $$ The statement is the following. Let $\psi\in L^2(\mathbb{R}^N)$ be a solution to Schroedinger's equation. Then denoting with $\left\langle x\right\rangle,\left\langle p\right\rangle,\left\langle\nabla V\right\rangle$ the expected values of the observables position, momentum and gradient of the potential respectively on the state $\psi$, which are given by \begin{gather*}\left\langle x\right\rangle(t)=\int_{\mathbb{R}^N}x|\psi|^2(x,t)dx,\qquad\left\langle p\right\rangle(t)=-i\hbar\int_{\mathbb{R}^N}\psi^*(x,t)\nabla \psi(x,t)dx\\ \left\langle\nabla V\right\rangle(t)=\int_{\mathbb{R}^N}\nabla V(x)|\psi|^2(x,t)dx \end{gather*} ,then $$\left\langle x\right\rangle'(t)=\frac{\left\langle p\right\rangle(t)}{m},\qquad \left\langle p\right\rangle'(t)=-\left\langle \nabla V\right\rangle(t) $$
We didn't mention this in class but the potential $V$ needs to be differentiable in some sense in order to make sense of the expression $\nabla V$. Perhaps almost everywhere differentiability is best, since for weak differentiability I would need to put $V$ in a Sobolev space, which I don't want to since the potential might not be in $L^p(\mathbb{R}^N)$ for any $p$? Moreover, I need $\psi \in D(\nabla V)$, since $\nabla V$ needs to be evaluated on the state $\psi$. Since $\nabla V$ is the multiplication operator by the function $\nabla V$, the requirement on the domain would be $$\psi\nabla V\in L^2(\mathbb{R}^N) $$
The first thing we do in the proof is differentiate $x(t)$ and $p(t)$ inside the integral sign. I'm not sure how to justify this. Take for instance the case of $x(t)$. Using Leibniz's integral rule, I need a time interval $(t_0-\varepsilon,t_0+\varepsilon)$ and a function $g(x)\in L^1(\mathbb{R}^N)$ such that $$|f(x,t)|:= \left|\frac{i\hbar}{2m}x\left(\psi^*(x,t)\Delta \psi(x,t)-\psi(x,t)\Delta \psi^*(x,t)\right)\right|\leq g(x)$$ for all $t\in (t_0-\varepsilon,t_0+\varepsilon)$ and for a.e. $x\in \mathbb{R}^N$. Now, if I assume $\psi(\cdot,t)\in \mathcal{S}(\mathbb{R}^N)$ for all $t$ then $f(\cdot,t)\in \mathcal{S}(\mathbb{R}^N)\subset L^1(\mathbb{R}^N)$. But I know nothing about the behavior in time, even in a small interval, so I don't know how to proceed.
To integrate by parts/apply divergence theorem and have the terms at the boundary vanish, i need rapid decrease for $|x|\to \infty$ of the solution to Schroedinger's equation $\psi$. The easiest way to guarantee this is to require $\psi(\cdot,t)\in \mathcal{S}(\mathbb{R}^N)$ for all $t$. Perhaps this could be simplified (I wonder if $\psi(\cdot,0)\in \mathcal{S}(\mathbb{R}^N)$ is enough to guarantee $\psi(\cdot,t)\in \mathcal{S}(\mathbb{R}^N)$ for all $t$, for instance).
In the proof of $\left\langle p\right\rangle'(t)=-\left\langle\nabla V\right \rangle(t)$ at some point I need to integrate by parts $V\nabla \psi$, which means that I need $V(x)\psi(x,t)\to 0$ as $|x|\to \infty$. Perhaps the right condition on $V$ would be to require that $V(x)\leq |x|^N$ as $|x|\to \infty$ for some $N>0$? Then if $\psi(\cdot,t)\in \mathcal{S}(\mathbb{R}^N)$ we would have $V\psi \to 0$.
I'd appreciate some tips on how to work out these issues by myself but I also wouldn't mind simply being given a reference as long as it doesn't have too many prerequisites.