Does quantum observables demand both Heisenberg and Schrodinger representations? I will consider an observable $\mathcal{O}\in\mathcal{L}(\mathscr{H})$ and, for simple, let me assume $\mathscr{H}$ is finite dimensional. Now, for some time independent hamiltonian $\mathcal{H}$ I can evolve $\mathcal{O}\to\mathcal{O}(t)=U_t^\dagger\mathcal{O}U_t$, where $U_t=e^{-it \mathcal{H}}$. I then define $\delta\mathcal{O}(t)=\mathcal{O}(t)-\mathcal{O}$, which is hermitian for $\mathcal{O}$ is hermitian. Something sounds weird to call this object an observable; I don't see how to represent it in a time independent frame, that is, to the best of my knowledge this can be defined only at the Heisenberg picture.
In this simple, finite dimensional case, does legit quantum observables demand a representation in both Heisenberg and Schrodinger picture? In particular, is $\delta\mathcal{O}(t)$ an observable and, if not, what is the catch?
At a more operational perspective, it also seems to me that acquiring the outcomes of such operator would demand measurements which are non-local in time. So, a related question: is $\delta \mathcal{O}(t)$ measurable? I am aware of cases in which one can assign a two-point-measurement protocol to obtain such outcomes, but I am interested in the general case in which I do not have to assume anything about the state of the system.
 A: Strictly speaking, this operator $\delta\mathcal{O}$ is ill-defined because you are taking the difference of operators defined on different (but isomorphic) Hilbert spaces.
This point is usually not emphasized, but if we were being careful, the stricture of quantum mechanics is to assign a Hilbert space to each value of time (constant-time slices if you're doing field theory). The time-evolution operator, with this understanding, can be interpreted as the linear map from the Hilbert space at one time to the Hilbert space at a different time. Since the evolution operator is invertible, all these Hilbert spaces are isomorphic, and hence the distinction is usually forgotten, but this is also the reason why canonical commutation relations are always between operators at the same time.
So, the operator $\delta\mathcal{O}$ is being built from operators defined as acting on different Hilbert spaces and hence is ill-defined as an operator. The loophole, however, is that its expectation value is still well-defined as being the difference of the expectation values.
