TITLE: How much information does the Hamiltonian contain in quantum mechanics?
QUESTION [0 upvotes]: Given a Hamiltonian, let's say of a many-body system, through the Schrodinger equation,in principle we can find the eigenfunctions and their corresponding eigenvalues (spectrum). Now given an arbitrary observable (not necessarily comutating
 with the Hamiltonian),  we wish to work out its eigenvalues (spectrum). In the latter case we have to average the observable over the Hamiltonian's eigenfunctions. In this respect it seems that all the information of the physical system flows from the Hamiltonian to its eigenfunctions. My question is then, if it is possible to find a mathematical relationship between the Hamiltonian and the observable's spectra without using the Hamiltonian wavefunctions, but only a functional  relation between the two observables.

REPLY [1 votes]: The spectrum of a generic observable, especially if it doesn't commute with the Hamiltonian, is not determined by averaging it on the eigenfunctions of the Hamiltonian. The way the spectrum is determined is by the usual means of spectral theory, that is you look for those values $\lambda\in\mathbb R$ (observables are assumed to be self-adjoint operators, hence the spectrum will be contained in $\mathbb R$) for which $O-\lambda I$ is not invertible in the set of bounded operators on the representation Hilbert space $H$, where $O$ is the observable in question in a certain representation and $I$ is the identity over $H$.
More abstractly, if you are given the full C*-algebra of observables then you can characterise the spectrum of an operator as the set of values $\omega(O)$, where $\omega$ is any state from the state space of the C*-algebra that is such that the uncertainty $\Delta_\omega(O)$ vanishes.
On the other hand it is true that one can encode the information of the Hamiltonian by means of its spectral decomposition, which gives spectrum and (generalised) eigenspaces.
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