Is there a way to quantify decoherence in the Heisenberg picture? In the usual picture of decoherence:
$\bullet$ you start with a state $| \psi(0) \rangle = | \psi_{S}(0) \rangle \otimes | \psi_{E}(0) \rangle$ for the system and environment (assumed to be uncorrelated initially usually).
$\bullet$ One then time evolves the state with the hamiltonian $H$, getting some $| \psi(t) \rangle = e^{- i H t} | \psi(0) \rangle$.
$\bullet$ One can then partial trace over the environment to get a reduced density matrix for the system: $\rho_{S}(t) = \mathrm{Tr}_{E}[ | \psi(t) \rangle \langle \psi(t) |  ]$.
$\bullet$ From there, one usually can use some basis $| s_{j} \rangle$ for the system to show that the off-diagonals $\langle s_{j} | \rho_{S}(t) | s_{k} \rangle$ for the reduced density matrix fall-off to zero after some time evolution takes place (so that the system ends up in some mixed state, and so decoheres).
My question: Taking a different route: can we quantify decoherence in the Heisenberg picture? (without referring to the state?)
Suppose we specify some initial state $|\psi(0) \rangle$ as above,  if we have some operator $A(0)=A_{S}(0)\otimes I_{E}$ and then time-evolve it in the Heisenberg picture to get an operator $A(t) = e^{+ i H t} A(0) e^{- i H t}$, is there a way to quantify decoherence in the system sector by looking at the operator $A(t)$? Does there exist some methods for doing this?
 A: Tradition has a way of making things unnecessarily difficult to understand, so forget about the traditional formulation for a moment. Don't factorize the Hilbert space and don't take a partial trace, at least not yet. Consider the state of the whole system instead, including the environment.
To begin, work in the Heisenberg picture, so observables depend on time and the state does not. If the state is a superposition of orthogonal terms $|n\rangle$ that can't be significantly mixed with each other by any operators representing practically-measurable observables $O(t)$ at times $t>T$ in the forseeable future, then the relative phases between the terms $|n\rangle$ also can't significantly affect anything measurable in the forseeable future. That's decoherence.
We could stop right here, because that's really all there is to it. (The details are a little ambiguous, but that's unavoidable. It's part of the so-called measurement problem.) I'll take a few more steps, though, to help clarify the connection with the traditional formulation:

*

*Suppose that decoherence has occurred in the sense defined above, so that the relative phases between the terms $|n\rangle$ don't affect anything measureable for $t>T$. We might as well make this explicit by replacing the superposition
$$
 \sum_n c_n|n\rangle
\tag{A}
$$
with the density matrix
$$
 \rho = \sum_n |c_n|^2|n\rangle\langle n|,
\tag{B}
$$
because we've already determined that this replacement doesn't have any significant effect on any future predictions.  Notice that we're still in the Heisenberg picture, and notice the key word future. We should only replace (A) with (B) when considering observables $O(t)$ with $t>T$, with $T$ defined as above. (Remember: quantum theory's usual state-projection rule applies in the Heisenberg picture, too, even though all time-dependence is carried by the observables. We're doing the same kind of thing here.) By the way, a standard way to quantify decoherence is to use the von Neumann entropy, defined as $-\sum_n |c_n|^2\log |c_n|^2$. We can define this even for (A), because the terms in (A) are defined by the criterion described in the second paragraph of this answer — but remember that this is relevant only for observables at times $t>T$. Finally, notice that we're still describing the whole system, even though we've switched to a density matrix (B). We haven't taken any partial traces.


*If we want to, we can switch to the Schrödinger picture, where the density matrix becomes
$$
 \rho(t)=\sum_n |c_n|^2\big|n(t)\big\rangle\big\langle n(t)\big|.
\tag{C}
$$


*If we want to, we can suppose that the Hilbert space is factorized, and we can take a partial trace over the second factor. (Remember that the von Neumann entropy of the reduced density matrix cannot be less than the von Neumann entropy of the full density matrix.)
Steps 1,2,3 show how my earlier description of decoherence is related to the traditional one, but conceptually they're not necessary:

*

*Replacing the superposition (A) with the density matrix (B) is not necessary. In fact, it's an idealization: (A) and (B) are not quite equivalent, because of the word "significantly" in the second paragraph of this answer, which is part of the ambiguity that I acknowledged in the third paragraph. The idealization (B) makes the decoherence criterion look less ambiguous than it really is. I like to keep ambiguities in the forefront instead of sweeping them under the rug, which is why I prefer (A).


*Using the Schrödinger picture is not necessary. I tried to make this obvious.


*Taking a partial trace is not necessary. It can be convenient for calculations if we don't care about observables that only affect the second factor, but computational convenience and conceptual clarity are not the same thing. Despite tradition, the concepts are more clear if we don't take a partial trace.
