Dtjytyk

I have one problem in Quantum Information Theory, specially in Zero-error communication.

I am reading a paper :

Duan, Runyao; Severini, Simone; Winter, Andreas, Zero-error communication via quantum channels, noncommutative graphs, and a quantum Lovász number, IEEE Trans. Inf. Theory $59$, No. $2$, $1164-1174$ $(2013)$. ZBL$1364.81059$.

And in page $5$, there is a paragraph :

Let $N : L(A) to L(B)$ be a quantum channel, i.e. a linear completely positive, trace-preserving map with Kraus operators $E_j : A to B$, so that $N(rho)$ = $sum_j E_jρE^dagger_j$. Then to send messages $m$ one has to associate them with states $rho$ such that different states $rho, sigma$ lead to orthogonal channel output states: $N(rho) perp N(sigma)$, because it is precisely the orthogonal states that can be distinguished with certainty. Clearly, these states may, w.l.o.g., be taken as pure (rank 1 operator $textbf{vv}^*$), as the orthogonality is preserved when going to any states in the support (i.e., the range) of $rho, sigma$, etc.

I totally don't understand why the last sentence is clear.

I knew that $N(rho) perp N(sigma)Leftrightarrow operatorname{Tr}[N(rho)N(sigma)] = 0$,but didn't obtain more informations from $operatorname{Tr}[N(rho)N(sigma)]=0$

Please explain why $rho, sigma$ can be replaced by pure states.

We can see why by understanding the following claim.

For a density operator $rhoinmathcal{S}(A)$, one may denote the image (or range) of $rho$ as $mathrm{im}(rho)$. Note that, for any vector $uinmathrm{im}(rho)$, there exists a positive real number $x$ such that $uu^*leq xrho$. Moreover, for any positive linear mapping $mathcal{N}$, it holds that $mathcal{N}(uu^*)leq x mathcal{N}(rho)$.

Claim. Let $mathcal{N}:mathcal{L}(A)rightarrow mathcal{L}(B)$ be a channel and let $rho,sigmainmathcal{S}(A)$ be density operators satisfying $mathrm{Tr}(mathcal{N}(rho)mathcal{N}(sigma))=0$. For any vectors $uinmathrm{im}(rho)$ and $vinmathrm{im}(sigma)$, it holds that $mathrm{Tr}(mathcal{N}(uu^*)mathcal{N}(vv^*))=0$.

Proof. From the above observations, there exist positive real numbers $x$ and $y$ satisfying $uu^*leq xrho$ and $vv^*leq ysigma$. It follows that $mathcal{N}(uu^*)leq x mathcal{N}(rho)$ and $mathcal{N}(vv^*)leq y mathcal{N}(sigma)$. Moreover,
begin{align*}
0leqmathrm{Tr}(mathcal{N}(uu^*)mathcal{N}(vv^*)) leq xy,mathrm{Tr}(mathcal{N}(rho)mathcal{N}(sigma)) = 0,
end{align*}
which yields the desired result.

This idea can be extended to any state in the image of $rho$ and $sigma$. Namely, if $rho_0$ and $sigma_0$ are any other density operators satisfying $mathrm{im}(rho_0)subseteqmathrm{im}(rho)$ and $mathrm{im}(sigma_0)subseteqmathrm{im}(sigma)$, there must exist positive constants $x$ and $y$ such that $rho_0leq xrho$ and $sigma_0leq ysigma$ and we find that
begin{align*}
0leqmathrm{Tr}(mathcal{N}(rho_0)mathcal{N}(sigma_0)) leq xy,mathrm{Tr}(mathcal{N}(rho)mathcal{N}(sigma)) = 0.
end{align*}