In "Quantum Computation and Quantum Information" by Nielsen & Chuang, on page 358, as well as in "Exploring the quantum" by S.Haroche & J.M Raimond on page 177, they consider the following.

We take a system $A$ that may interact with an environment.

$A$ is described by a density matrix $\rho_A$.

It evolves through an interaction with the environment $E$.

They say that, it is possible to express $\rho_A$ after its interaction with $E$ as a quantum map : $\rho'_A=\mathcal{L}_A(\rho_A)$ only if the $A$ and $E$ were not entangled at the beginning. Because else "It is then, in general, impossible to define a linear map deducing the state of $A$ after its interaction with E from the knowledge of $\rho_A$ alone" (S.Haroche book on this page 177).

But I don't understand this.

I could imagine a fictive transformation such that at the beginning I had $\rho_{AE}=\rho_A \otimes \rho_E$ and at the end $\rho'_{AE}=\rho'_A \otimes \rho_E$. Thus a quantum map could exist ?

Is it because that the map linking my $\rho_A$ and $\rho'_A$ wouldn't in practice verify all the postulates of quantum maps (like it wouldn't conserve the trace for example).

But actually, more generally, I don't see what the entanglement has to do here, indeed as we deal with density matrix, if $A$ and $E$ are entangled, it will just imply that $\rho_A$ is a mixed state. Why couldn't I describe its evolution with a quantum map ?

  • 1
    $\begingroup$ If system and environment are entangled, then $\rho_{AE}$ can't be written as $\rho_A \otimes \rho_E$. $\endgroup$
    – Noiralef
    Commented Oct 22, 2018 at 17:44
  • $\begingroup$ @Noiralef I know i considered a fictive transformation leading to the same final partial density matrix. Thus the same L_A if it existed $\endgroup$
    – StarBucK
    Commented Oct 22, 2018 at 17:45

1 Answer 1


The problem is related to the construction and the assumptions about quantum maps. This also raises some questions about the construction itself and it has been studied extensively in the literature, starting from a work of Pechukas in 1994. I also suggest you to read a work by Jordan and others (in particular the discussion) and the review by Shaji and Sudarshan.

Let suppose we have a system $A$ in the state $\rho_A\in\mathcal{D}(\mathcal{H})$, where $\mathcal{D}(\mathcal{H})$ is the space of density matrices over $\mathcal{H}$. The idea to construct a non-unitary evolution $\mathcal{L}$ is as follows.

  1. We first construct an assignment $T:\mathcal{D}(\mathcal{H})\to\mathcal{D}(\mathcal{H}\otimes\mathcal{H}_E)$ which assignes $\rho_A\to\rho_{AE}$ (a state of the system + environment) for every $\rho_A\in\mathcal{D}(\mathcal{H})$
  2. We let the overall system evolve through a unitary interaction: $\rho_{AE}\to U\rho_{AE}U^{\dagger}$
  3. We take the partial trace over $E$ to get the state of the initial system after the interatction, i.e., $\rho_A'=\mathcal{L}(\rho_A)=\text{Tr}\{U\rho_{AE}U^{\dagger}\}$

Now, first of all, note that after the first step we should have $\text{Tr}_E\{\rho_{AE}\}=\rho_A$. However it can be shown (see the first two works) that the only assignment $\rho_A\to\rho_{AE}$ which satisfies this constraint for all $\rho_A\in\mathcal{D}(\mathcal{H})$ is the product state assignment (This assignment makes the overall map to be completely positive). This does not mean that an initial entangled condition is not possible for a non-unitary evolution but, if we require the initial system+enironment state to be entangled, we need to restrict the domain of the mapping $T$ (think, for example, that if $\rho_{AE}$ is entangled, $\rho_A$ cannot be pure), and therefore, the domain of $\mathcal{L}$. This is the main point of this problem.

If you don't care about these restriction, there may be some problems with $T$ (You can, for example, define $T$ over a subspace of "allowed" initial states and then extend by linearity $T$ over the whole domain $\mathcal{D}(\mathcal{H})$). Citing Shaji and Sudarshan:

The assignment map applied blindly to all states of $A$ with the restriction that certain correlations in $A+E$ are to be kept fixed will map some of the states of $A$ to negative matrices.


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