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?
Best Answer
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.