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I have a map $\mathcal{N}$ from the space of two-qubit subnormalised density matrices $\mathcal{S}(\mathcal{H}_2 \otimes \mathcal{H}_2)$ to itself (positive operators with trace between 0 and 1). Notice that it is not necessarily a physical one (e.g. it may be nonlinear, that's perfectly fine). I know $\mathcal{N}$ is

  1. linear for all product states $\rho_A \otimes \rho_B$. By this I mean $\mathcal{N} \left( \sum_{ijkl} c_{ik}d_{jl} |i\rangle \langle k| \otimes |j \rangle \langle l | \right)= \sum_{ijkl} c_{ik}d_{jl} \mathcal{N} \left( |i\rangle \langle k| \otimes |j \rangle \langle l | \right)$ for every $\rho_A = \sum_{ik}c_{ik}|i\rangle \langle k|, \rho_B = \sum_{jl}d_{jl}|j \rangle \langle l| \in \mathcal{S}(\mathcal{H}_2)$

  2. continuous on the whole space of two-qubit density matrices $\mathcal{S}(\mathcal{H}_2 \otimes \mathcal{H}_2)$.

Does it automatically follow that $\mathcal{N}$ has to be linear over all $\mathcal{S}(\mathcal{H}_2 \otimes \mathcal{H}_2)$?

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    $\begingroup$ What does "linear for all product states" mean? Can you write a formula? $\endgroup$ Jan 24, 2023 at 13:11
  • $\begingroup$ Do you require the map to be trace-preserving (i.e. $\mathcal S$ are positive operators with trace one)? Generally, I think being more specific (formulas!) in the question would be rather helpful. $\endgroup$ Jan 24, 2023 at 13:22
  • $\begingroup$ You're right, I could have stated the question more carefully. Hope the edit helps clarify what I mean. $\endgroup$
    – bb2002
    Jan 24, 2023 at 13:42
  • $\begingroup$ We do not require the map to be trace preserving $\endgroup$
    – bb2002
    Jan 24, 2023 at 13:54
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    $\begingroup$ Well, but then $\mathcal N$cannot be defined on $|i\rangle\langle k|\otimes |j\rangle\langle l|$, no? What I mean is that you write " map [...] from the space of two-qubit subnormalised density matrices [...]to itself". But if I am not completely mistaken, then $|i\rangle\langle k| \otimes |j\rangle\langle l|$ is not from the domain of the map. $\endgroup$ Jan 24, 2023 at 14:16

1 Answer 1

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No.

First, choose a continuous map $f(\rho)$ into the reals, such that

  1. $f(\rho)$ is zero on all separable states (the convex hull of product states)

  2. $f(\rho)>0$ for all other states.

  3. $f(\rho)\le 1$.

Such a map exists -- e.g., it could be the (trace norm) distance to the set of separable states, or some nice entanglement measure (for two qubits: negativity, entanglement of formation, ... ), suitably normalized.

Then, construct $$\mathcal N(\rho) = f(\rho) \sigma_1 + (1-f(\rho))\sigma_2$$ for any two states $\sigma_1$ and $\sigma_2$.

This map will be linear on the separable states (in fact, constant, namely $\sigma_1$), and thus on all product states. On the other hand, it will not be linear in general. In particular, any separable state can also be decomposed as a convex combination which contains entangled states, $\rho=\sum p_i \rho_i$ with some $\rho$ entangled, and $\mathcal N$ will not be linear over such a decomposition, as $\mathcal N(\rho)=0$, while $\mathcal N(\rho_i)\ne0$ for some $i$, and $p_i>0$.

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  • $\begingroup$ I started writing this answer before the OP clarified in their comments. If the map needs not be trace preserving $\mathcal N(\rho) = f(\rho)\sigma_1$ will do. $\endgroup$ Jan 24, 2023 at 14:53
  • $\begingroup$ ... or $\mathcal N(\rho) = f(\rho)\rho$, for that matter. $\endgroup$ Jan 24, 2023 at 16:55

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