This question is a follow-up to this one.

The setup is the same as in that question. We consider a map from a certain set of density operators $\rho_S$ on a Hilbert space $\mathcal{H}_S$ to the complete set of density operators on that $\mathcal{H}_S$ that is a composition of:

  1. An "assignment map" $\Phi$ from some set of density operators $\rho_S$ on $\mathcal{H}_S$ to the set of density operators on a larger Hilbert space $\mathcal{H}_S \otimes \mathcal{H}_R$. (For the purpose of this question, we require that the assignment map's image consist of actual density operators (i.e. positive semidefinite and trace-1 Hermitian operators - the paper linked from the previous question does not make this requirement), and that it be consistent in the sense defined in the previous question, but we do not require that the map be linear or defined on the whole space of density operators on $\mathcal{H}_S$.)
  2. Unitary time evolution on the enlarged Hilbert space $\mathcal{H}_S \otimes \mathcal{H}_R$ with some unitary operator $U$.
  3. The partial trace over $\mathcal{H}_R$.

That is: $$\rho_S \to \rho_S' = \mathrm{Tr}_R[U \Phi(\rho_S) U^\dagger]. \tag{1}$$

It is known that if the assignment map $\Phi$ is a product map (in the sense defined in the previous question) $\Phi(\rho_S) \equiv \rho_S \otimes \rho_R$, then the composed map (1) must be completely positive and trace-preserving (CPTP).

As discussed in the previous question, in general - whether or not the assignment map is a product map - since we require that the output of the assignment map be a density operator, the inputs and outputs of this composed map (1) must all be density operators, so the composed map is (trivially) positive and trace-preserving, since all possible inputs and outputs are positive semidefinite and trace-1.

But if the assignment map is consistent but not a product map, then does the composed map (1) need to be completely positive?

  • $\begingroup$ I'm a bit confused tbh. You defined "consistent" in the previous question as $\Phi$ st ${\rm Tr}_R\Phi(\rho)=\rho$ for all $\rho$, which implies $\Phi(\rho)=\rho\otimes\sigma$ for some $\sigma$. But the propreties of such maps are quite trivial: if $\sigma$ is a state, then $\Phi$ is CPTP. If $\sigma$ has unit trace, then $\Phi$ is trace-preserving, if $\sigma\ge0$ then $\Phi$ is CP, etc. So how can a map be consistent and not a product map? Or are you asking how to prove that "consistent" implies it being a product map? $\endgroup$
    – glS
    Commented Oct 3, 2022 at 10:56
  • $\begingroup$ @glS No, consistency alone does not require that the assignment map $\Phi$ be a product map. Equation (1) of journals.aps.org/prl/abstract/10.1103/PhysRevLett.73.1060 gives an explicit example of an assignment map that's consistent but generically outputs an entangled state. $\endgroup$
    – tparker
    Commented Oct 3, 2022 at 14:35
  • $\begingroup$ @glS Another way to see this is to count degrees of freedom: if $\mathcal{H}_S$ and $\mathcal{H}_R$ have dimensions $N_S$ and $N_R$ respectively, then the set of states $\rho_{SR}$ on $\mathcal{H}_S \otimes \mathcal{H}_R$ has $(N_S N_R)^2 - 1$ real parameters, and fixing $\mathrm{Tr}_R[\rho_{SR}]$ imposes $N_S^2 - 1$ constraints, leaving $N_S^2 (N_R^2 - 1)$ remaining real degrees of freedom for $\rho_{RS}$ even after fixing its partial trace - and these are independent for each choice of input $\rho_S$, for a total of $(N_S^2 - 1) N_S^2 (N_R^2 - 1)$ real degrees of freedom for the entire map.. $\endgroup$
    – tparker
    Commented Oct 3, 2022 at 14:42
  • $\begingroup$ ... Whereas a product map only has $N_R^2 - 1$ real degrees of freedom, which specify the state $\sigma$. $\endgroup$
    – tparker
    Commented Oct 3, 2022 at 14:42
  • 1
    $\begingroup$ ah, sure, if you don't assume linearity then most basic results don't apply. I would stress explicitly that you don't require linearity in the question then though, because the term "quantum map" is usually used to refer to linear functions between linear operators, and people in this context almost always consider linear maps, hence why I was implicitly assuming that (I see now that it is mentioned in the question, but I would give it more prominence as that parenthetical remark is easy to miss). I would not call this a "map", but rather a more generic "function" between states $\endgroup$
    – glS
    Commented Oct 4, 2022 at 6:44

1 Answer 1


This answer will be based on this paper of Schmid et al. (arXiv). Therein the authors give several examples of how the input-output relation of the standard formulation—that is, trying to model the change ${\rm tr}_E(\rho_{SE})\mapsto {\rm tr}_E(U \rho_{SE} U^\dagger)$ via some map—can result in something that is not completely positive and sometimes not even a map at all (since the same marginal can result in different outputs).

The underlying issue here is that, while the evolution map and the inference map—the latter answering the question whether "given a particular state of knowledge of the initial system, what is the appropriate state of knowledge to assign to the final system?"—coincide when the initial state is a product state $\rho_{SE}=\rho_S\otimes\rho_E$, there is a conceptional difference between the two if initial system-environment correlations are present. This highlights the problem with trying to assign a map to this marginal propagation: To again quote the paper, "simply listing a set of marginal states at its input and at its output" is not enough to even define such a map "since if there is a common cause acting on the input and the output, then such pairs of states do not constitute constraints on the evolution map".


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