Hilbert space and Hamiltonians 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?
 A: No, I don't see why that should be the case. The notion of a Hilbertspace underlying a quantum-mechanical system is quite independant of the postulate that time evolution is generated by a Hamiltonian. 
The notion of a vectorspace enters QM, because fundamentaly QM should be a linear theory and thus allow for arbitrary superpositions. The more wonderous idea is, that it is a space over the complex numbers.
The fact that a proper wavefunction should be $L^2$ reflects the interpretation as a probability-/ or charge-density.
Any (sensible) Hilbertspace admits a countable orthonormal basis. Pick a real number for each of those basis states and define
$$ (H)_{mn} = \langle \psi_m,H\psi_n\rangle \equiv \epsilon_n$$
as matrix elements of an operator $H$. What kind of dynamics this "Hamiltonian" describes, depends on the choice of eigenvalues, e.g energies. You should also make sure, that the spectrum is not too pathological, e.g it should be bounded from below to be interpretable as a sensible physical spectrum.
Non-unitary time-evolutions come about in open quantum systems, where you forget about the environment which couples to the system of interest, but which is nevertheless there. Probability can "leak" into the environment, hence a non-unitary evolution. An example is e.g the Lindblatt-Master-Equation describing the Markovian limit of a system-environment coupling.
A: Any Hilbert space $\mathcal{H}$ with the notion of unitary time evolution also possesses the notion of Hamiltonian. 
If $\mathcal{U}(t) : \mathcal{H} \to \mathcal{H}$ is the time evolution operator for every $t \in \mathbb{R}$, then it forms a one-parameter Lie subgroup of the Lie group of unitary operators, which is generated by some distinct element $H$ in the Lie algebra of linear operators. This generator is the Hamiltonian, and as the generator of a unitary operator, it is necessarily self-adjoint by Stone's theorem, so you get a Hamiltonian whose eigenvectors span the space.
Since non-unitary time-evolution comes into play if you are only considering a subspace of the full space of states (e.g. when you don't track all decay products for decaying systems), one can always get a unitary evolution by embedding the subsystem into "the whole system", find the Hamiltonian there, and then project it back onto the subsystem to get the Hamiltonian for the subsystem. But now, since time evolution was non-unitary here, it cannot be that this Hamiltonian is self-adjoint (since the exponential of self-adjoint operators is unitary), therefore we are forced to conclude that the eigenvectors of a Hamiltonian cannot span a subspace on which time evolution is non-unitary.
So, you cannot get a Hamiltonian both spanning the space and producing non-unitary time evolution, one of these must necessarily fail.
