According to https://arxiv.org/abs/1206.3794, the standard treatment models the time evolution of open quantum systems as follows: starting with a system in the initial state $\rho_0$ on the Hilbert space $\mathcal{H}_S$, we perform an "assignment map" $\Phi$ from states on $\mathcal{H}_S$ to states on an expanded Hilbert space $\mathcal{H}_S \otimes \mathcal{H}_R$. We then time-evolve the enlarged state $\rho_{SR} = \Phi(\rho_S)$ by the unitary operator $U$ and then trace out $\mathcal{H}_R$ to get a new state on the original Hilbert space $H_S$. That is:

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

(We often impose the "consistency" requirement on $\Phi$ that for all inputs $\rho_s$, $\mathrm{Tr}_R[\Phi(\rho_S)] \equiv \rho_S$, so that the time evolution is trivial and $\rho_S' \equiv \rho_S$ if $U$ is the identity.)

The paper claims that if the assignment map $\Phi$ is

  1. defined on the entire space of states on $\mathcal{H}_S$,
  2. linear, and
  3. consistent (as defined above),


  1. $\Phi$ must be a product map, i.e. $\Phi(\rho_S) \equiv \rho_S \otimes \rho_R$ for a single fixed density operator $\rho_R$ that does not depend on $\rho_S$, and
  2. the time-evolution map (1) must be CP. (I think it also must be TP as well, although the paper doesn't say that.)

The paper then claims that if we allow the output of the assignment map $\Phi$ to be entangled, then

forget complete positivity; [the map (1)] may not even be positive.

In light of the above result, this would require that the assignment map $\Phi$ abandon at least one of the three requirements numbered above. But the paper doesn't give any examples of such a "pathological" non-positive map of the form (1).

But I don't see how this can be possible. If $\rho_{SR}$ is a state on $\mathcal{H_S} \otimes \mathcal{H}_R$, then $U \rho_{SR} U^\dagger$ is as well, and in that case its partial trace $\rho_S'$ should be a valid state on $\mathcal{H}_S$. So it seems to me that the map (1) must indeed be positive (and trace-preserving), although perhaps not necessarily completely positive.

What's an explicit example of a map of form (1) - but with the assignment map $\Phi$ not a product map (and therefore violating at least one of the three assumptions above) - such that the map (1) is non-positive?

Edit: I've figured out the answer to this question, but can't cancel the bounty. But I don't know the answer to a closely related question, which I've asked here.

  • $\begingroup$ I feel like what they're trying to get at is the issue of CP (non)-divisibility: given a dynamical process $t\mapsto \Phi_t$, it is not in genereal possible to write things like $\Phi_t=\Phi_{t/2}\circ\Phi_{t/2}$ and have $\Phi_{t/2}$ be a channel, and this is to some degree due to entanglement arising in the dynamics. But the way this is stated here seems wrong to me. Regardless of the specifics of $\Phi$, if it is a (CPTP) channel, then it clearly follows that the map $\rho_S\mapsto {\rm Tr}_R[U\Phi(\rho_S)U^\dagger]$, being a composition of (CPTP) channels, is also a (CPTP) channel. $\endgroup$
    – glS
    Sep 28, 2022 at 5:56
  • $\begingroup$ @glS Yes, but the paper says that "whether or not [(1)] is CP is solely determined by the properties of $\Phi$". It says that $\Phi$ maps states to states, so it seems to me that it must be positive and TP. It may not necessarily be completely positive, but it seems to me positivity alone is enough to guarantee that the full map (1) must be at least positive. $\endgroup$
    – tparker
    Sep 28, 2022 at 16:11
  • $\begingroup$ journals.aps.org/prl/abstract/10.1103/PhysRevLett.73.1060 claims to give an explicit example, but it isn't presented very clearly and I don't understand it. I don't understand their notation when they specify the assignment operator, and instead of explicitly giving the unitary time-evolution operation, for some reason they only give the global Hamiltonian $H$ (again using not-very-clear notation) and the evolution time period $t$, and they leave it to the reader to calculate the operator $U = e^{-i H t}$ that we actually care about. Surely there is a simpler minimal example? $\endgroup$
    – tparker
    Sep 30, 2022 at 23:36

1 Answer 1


I went back to the original paper at https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.73.1060 and figured it out.

The proposed assignment map in that paper is consistent for all input states $\rho_S$, but for some input states it outputs a trace-1 Hermitian operator that is not positive semidefinite, and hence not a state. (I'm using the term "state" to mean "density operator" in this answer.) The author chooses to deal with this issue by (rather artificially, IMO) restricting the domain of the assignment map to the set of input states that get mapped to a positive-semidefinite output state, therefore violating assumption #1 in the list in the question. Another way that one could deal with this issue would be to remove the requirement that the output of the assignment map be a state, but instead allow it to be an arbitrary linear operation on the enlarged Hilbert space - but the physical significance of this option is dubious.

The result is then that if you start with a system state $\rho_S$ that the chosen assignment map doesn't map to a state, but for some reason choose to apply equation (1) anyway, then you can end up with a $\rho_S'$ that is not positive semidefinite, and so the map (1) is not positive. I don't see why this result has any physical significance, though. If $\rho_{SR}$ is not a state, then $U \rho_{SR} U^\dagger$ won't be a state either, and so there's no reason to expect that its partial trace would be positive semidefinite. I don't see any physical motivation for considering the reduced dynamics for a linear operator that isn't actually a state.

As long as the output of the assignment map is a state for any input, then the map (1) will be positive and trace-preserving. But I'm not sure if the map (1) must necessarily be completely positive if the assignment map is not a product map; that's an interesting follow-up question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.