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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.

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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.

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  • $\begingroup$ Awesome. Thanks. Could you provide some reference that emphasizes such point? Does this also implies that higher moments $<\delta\mathcal{O}^n(t)>, n\geq 2$ are well-defined? $\endgroup$
    – Janov
    Commented Dec 9, 2020 at 21:36
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    $\begingroup$ @MarceloBroinizi They can be defined at any rate. The standard prescription is the path integral which enforces a time-ordering of the operators. The only source I can think of off the top of my head would be Hartmann's quantum gravity lectures. There's a chapter related to these things in there. $\endgroup$
    – Richard Myers
    Commented Dec 9, 2020 at 21:44
  • $\begingroup$ @RichardMyers Which definition of QFT do you have in mind? The definition as a functor from a category of bordisms to the category of Hilbert spaces? In Hamiltonian lattice QFT, all of the observables act on the same Hilbert space, by construction, so it might be worth specifying which flavor of QFT your answer assumes. (I mention lattice QFT because it's mathematically unambiguous and relatively broadly applicable, even though it's not as aesthetically appealing as, say, the functorial definition.) $\endgroup$
    – Chiral Anomaly
    Commented Dec 10, 2020 at 0:50
  • $\begingroup$ @ChiralAnomaly I suppose I have in mind the functor definition, but am not intimately familiar with all the details of lattice models. I was under the impression that it took the perspective that operators are only defined insofar as they produce correlators, not that they involved a different definition of the Hilbert space. $\endgroup$
    – Richard Myers
    Commented Dec 10, 2020 at 4:32
  • $\begingroup$ @RichardMyers, I've checked the reference you mentioned. It indeed provide some insight to the question but it does not mention that (i)"$\mathcal{O}$ and $\mathcal{O}(t)$ live in different Hilbert spaces", in fact it seems to claim the contrary: it says that ordering of operators matter due to the lack o analicity on time slices, but this translates in non-commuativity at the level of operators, but assuming (i), $[\mathcal{O}(t),\mathcal{O}]=0$. If you remember some further reference on the claim (i) please let me know. $\endgroup$
    – Janov
    Commented Dec 10, 2020 at 14:30

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