One of the postulates of quantum mechanics is that, given a quantum state $\psi_{0}$ at time $t=0$, the state of the system at a posterior time $t > 0$ is given by $\psi_{t} = e^{-iHt}\psi_{0}$, where $H$ is the Hamiltonian operator of the system.
The operator-valued function $\mathbb{R}^{+}\ni t \mapsto U(t) = e^{-iHt}$ is an example of what mathematicians call a strongly continuous one-parameter unitary group. As a matter of fact, I have seen in many mathematical physics books the above postulate being restated as follows: given an initial quantum state $\psi_{0}$, the state at a posterior time $t > 0$ is given by $\psi_{t} = U(t)\psi_{0}$, where $U(t)$ is a strongly continuous one-parameter unitary group. It turns out that Stone's Theorem ensures that for each strongly continuous one-parameter unitary group $U(t)$, there exists a self-adjoint operator $H$ such that $U(t) = e^{-iHt}$, so we are now back to the previous formulation by taking $H$ to be the Hamiltonian of the system.
This second approach, however, made me think about the nature of this postulate. Stone's Theorem ensures the existence of some self-adjoint operator $H$, which is not necessarily the Hamiltonian of the system. So, my question is pretty basic and straightfoward: is there any reason or interpretation why the time-evoution is defined in terms of the Hamiltonian $e^{-iHt}$, and not any other self-adjoint operator/observable? Or is it just another postulate of quantum mechanics that we should accept without any apparent reason?
Note that, by setting $U(t) = e^{-iHt}$, the associated Schrödinger equation is then reduced to finding eigenvalues of $H$; of course, if another observable, say $A$, was used instead of $H$, one would find eigenvalues of $A$ instead. Hence, I would expect that this postulate has some sort of "we want to find a basis of eigenvectors of $H$" explanation, but I am not quite sure if this is the reason behind the postulate, since it is not always possible to find such a basis of eigenstates anyways.
