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

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

• What does "linear for all product states" mean? Can you write a formula? Jan 24 at 13:11
• 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. Jan 24 at 13:22
• You're right, I could have stated the question more carefully. Hope the edit helps clarify what I mean. Jan 24 at 13:42
• We do not require the map to be trace preserving Jan 24 at 13:54
• 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. Jan 24 at 14:16

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 $$p_i>0$$.

• 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. Jan 24 at 14:53
• ... or $\mathcal N(\rho) = f(\rho)\rho$, for that matter. Jan 24 at 16:55