Can we craft a Hamiltonian such that the measurement is consistent with the discrete measurement taught in Quantum physics? So the way I understand this, the way measurement is taught is that
you have a wave function $\Psi(t)$. It's evolution over time is :
$$i \hbar \frac{d}{d t}\vert\Psi(t)\rangle = \hat H(t)\vert\Psi(t)\rangle$$
If you have an observable $\hat{O}$ it defines a basis of the Hilbert space s.t.
$$ \Psi(t) = \sum e_k O_k(t)$$
where $O_k$ are the eigenstates, $e_k$ the eigenvalues.
If you measure $\hat{O}$ in the real world, you will read the results $O_k$ with probability $\frac{e_k^2}{\sum_j e_j^2}$.
The question is, can we craft $\hat{H}$ such that the evolution of $\psi$ converges towards $O_k$ as the $O_k$ become attractors of the solutions of Schrödinger's equation. If we craft $\hat H(t)$ correctly we can have the size of the regions of attractor $O_k$ be $\frac{e_k^2}{\sum_j e_j^2}$ to respect the "discrete" view of measurement.
Is such a Hamiltonian findable ? Have I missed something glaringly obvious ?
Ps: I'm not a physicist.
 A: The short answer is no, the attractor idea doesn't work, because time evolution is unitary. 
In more detail, the equation
$$
 i\hbar\frac{d}{dt}|\Psi(t)\rangle
 =\hat H(t)|\Psi(t)\rangle
\tag{1}
$$
implies that the inner product between any two given solutions is independent of time:
$$
 \frac{d}{dt}\langle\Psi_1(t)|\Psi_2(t)\rangle = 0.
\tag{2}
$$
This is called unitarity, and it rules out the attractor idea. 
Intuitively, unitarity says that time evolution can't make an initially uniform distribution of state-vectors in the Hilbert space cluster toward the basis vectors in any given orthonormal basis, not even slightly. Unitarity says that an initially uniform distribution remains uniform forever.
To put a little more detail behind this intuition, consider one of the eigenstates $O_k$ of the measured observable, and consider some small neighborhood of this eigenstate in the Hilbert space. By small neighborhood, I mean a neighborhood in which every state-vector $\psi$ is approximately proportional to $O_k$, in the sense that the component of $\psi$ orthogonal to $O_k$ has a much smaller norm than the component of $\psi$ that is proportional to $O_k$. Since unitarity preserves inner products, if we run this neighborhood backward in time arbitrarily far, all of those state-vectors will still be in a small neighborhood of (a past version of) the eigenstate $O_k$. But the overwhelming majority of the state-vectors in the Hilbert space are not in a small neighborhood of any of the basis vectors in any given orthogonal basis, so the overwhelming majority of the state-vectors in the Hilbert space are not in any basin of attraction.
Here's an analogy: A unitary transformation in Hilbert space is essentially the complex-valued version of a rigid rotation in (many-dimensional) Euclidean space. The fact that a rigid rotation preserves angles is not compatible with the attractor idea.
