Assume a system described by a Hamiltonian H, and assume that the eigenstates of H, $φ_i$(r) are integrable in absolute square. We say that these states belong to a Hilbert space (they can even form a base in that space).
But, is the opposite true? Let a system be described by a wave-function S(r, t), integrable in absolute square. Does that imply that the system behavior is also described by a Hamiltonian?
Remark: te evolution of a system does not always admit a Hamiltonian. E.g. if the evolution is non-unitary (or at least, if there is a Hamiltonian, it would take complex eigenvalues.) To be clear, I don't know if my system evolves unitarily or not. I just gave the example to show that the existence of a Hamiltonian is not guaranteed. Whatever I know of the function S(r, t) is that it belongs to a Hilbert space.
So the question is, absolute square-itegrability, ensures (as a sufficient condition) the existence of a Hamiltonian for the system?
An example: I expand S(r, t) in a quantum superposition of the eigenfunctions $φ_i$(r),
S(r, t) = $∑_i$ $C_i$(t) $φ_i$(r), with $C_i$(t) = $F_i$ (t) exp(-i$E_i$t/ħ).
Introducing this superposition in the Schrodinger equation with the Hamiltonian H, I obtain that iħ ∂S(r, t)/∂t is not equal with HS(r, t). But, could it be that another Hamiltonian H' may exist s.t. the Schrodinger eq. be satisfied?