The general formulation of quantum mechanics is done by describing quantum mechanical states by vectors $|\psi_t(x)\rangle$ in some Hilbert space $\mathcal{H}$ and describes their time evolution by the Schrödinger equation
$$i\hbar\frac{\partial}{\partial t}|\psi_t\rangle = H|\psi_t\rangle$$
where $H$ is the Hamilton operator (for the free particle we have $H=-\frac{\hbar^2}{2m}\Delta$).
Now I have often seen used spaces like $\mathcal{H}=L^2(\mathbb{R}^3)$ (in the case of a single particle), but I was wondering whether this is correct or not.
In fact shouldn't we require to be able to derivate $\left|\psi_t\right>$ twice in $x$ and thus choose something like $\mathcal{H} = H^2(\mathbb{R}^3)$?
If we treat directly $\psi(t,x) := \psi_t(x)$, shouldn't we require them to be in something like $H^1(\mathbb{R};H^2(\mathbb{R}^3))$? i.e., functions in $H^1(\mathbb{R})$ with values in $H^2(\mathbb{R}^3)$, e.g. the function $t\mapsto\psi_t$.
Best Answer
Unfortunately, taking $\mathcal{H}=H^2(\mathbb{R}^3)$ won't work. If $f\in H^2(\mathbb{R}^3)$ then, in general, it is not true that $\Delta f\in H^2(\mathbb{R}^3)$. This is a problem, because the Hamiltonian must be an operator from $\mathcal{H}$ to $\mathcal{H}$.
The only mathematically sound way out is taking $\mathcal{H}=L^2(\mathbb{R}^3)$ and considering the Laplacian as an unbounded operator, meaning that it is not defined on the whole $L^2(\mathbb{R}^3)$ space but only on a dense subspace. This idea dates back to von Neumann.