Is the following map linear over the space of density matrices?

Question

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 $\mathcal{S}(\mathcal{H}_2 \otimes \mathcal{H}_2)$?

Answer 1

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 $p_i>0$.