Consider the free particle on the real line with standard Hamiltonian $$H = \frac{1}{2m}P^2\:.$$
Next consider a state $\psi=\psi(x)$ consisting of a compactly-supported smooth function attaining constantly the value $c\neq 0$ in the interval $[a,b]$ and smoothly vanishing outside that interval.
$\psi$ seems a good candidate for a physically meaningful wavefunction. I stress that it simultaneously belongs to the domains of $H$, $X$, and $P$.
(The Fourier transform $\hat{\psi}$ of $\psi$ belong in the Schwartz space so that $\psi$ is an element of $D(P^k)$ for every $k=1,2,\ldots$)
Now perform a measurement of position and suppose that the particle is found in $[a',b'] \subset [a,b]$.
In view of that standard projection postulate the wavefunction immediately after the measurement is
$$\psi'(x) = Nc \quad\mbox {if $x\in [a',b']$, $\quad \psi(x) =0\quad$ if $x\not\in [a',b']$}$$
where $N$ is the normalization constant.
This new function does not belong to the domain of $H$. Indeed,
$$D(H) = \left\{\psi \in L^2(\mathbb R)\:\left|\: \int_{\mathbb R} p^4|\hat{\psi}(p)|^2 dp <+\infty\right.\right\}$$
In the considered case, for some positive constant $C$,
$$p^4|\hat{\psi'}(p)|^2 = C p^2 (1-\cos(p(b'-a'))\:.$$
The integral of this function diverges (I used an online integrator to be sure completely). Thus $\psi' \not \in D(H)$.
This elementary and very idealized example shows how the existence of incompatible observables ($X$ and $H$ here) and the collapse postulate make very difficult to define a class of physically sensible states as belonging to a suitable class of domains of physically meaningful observables.
There are actually several ways out from my no-go example. One may argue that the collapse postulate is oversimplified and physical measurement procedures of continuous spectrum observables admit a more accurate description in terms of POVM and quantum operations. This is true, my intention was just to show how the problem is subtle.
