Singular Pair No Broadcasting

Let $S=\mathrm{supp}({\rho }_0)+\mathrm{supp}({\rho }_1)$, the linear span of the two supports. It is nonzero. Set $\sigma =({\rho }_0+{\rho }_1)\mathrm{/}2$. For $x\in S$, $⟨({\rho }_0+{\rho }_1)x,x⟩=0$ implies both nonnegative terms vanish, hence $x$ lies in both kernels and is orthogonal to $S$, so $x=0$. Therefore $\sigma$ is positive definite on $S$; it has trace one and support $S$. By linearity, $B(\sigma )$ has both marginals equal to $\sigma$.

For any nonnegative $X$ on $S$, faithfulness in step 1.1 and [L2] give $0\le X\le c\sigma$ for some finite $c$. Positivity of $B$ yields $0\le B(X)\le cB(\sigma )$. By [L3], $B(\sigma )$ is supported on $S\otimes S$. If $0\le A\le C$ and $x\in \ker C$, then $0\le ⟨Ax,x⟩\le ⟨Cx,x⟩=0$. Thus $\mathrm{\parallel }{A}^{1\mathrm{/}2}x{\mathrm{\parallel }}^2=0$, so $Ax=0$ and $\ker C\subseteq \ker A$; taking orthogonal complements gives $\mathrm{supp}A\subseteq \mathrm{supp}C$. Applying this to $A=B(X)$ shows $B(X)$ is supported on $S\otimes S$.

Nonnegative operators span $\mathcal{L}(S)$ over $\mathbb{C}$ by [F3]. Therefore step 2.1 shows $B(X)$ is supported on $S\otimes S$ for every $X\in \mathcal{L}(S)$. Define $\tilde{B}:\mathcal{L}(S)\to \mathcal{L}(S\otimes S)$ by restricting the input and identifying each supported output with its compression to $S\otimes S$. For each finite amplification, the input inclusion and output compression preserve nonnegativity, so $\tilde{B}$ is completely positive by [L4]. In an orthonormal basis of $H$ extending one of $S$, the embedded input has $X$ as its $S$-block and zero elsewhere, so its trace is $\mathrm{tr}X$; output compression leaves the supported $B(X)$ unchanged. Thus $\tilde{B}$ is trace-preserving and a channel by [L1]. It broadcasts the restrictions of ${\rho }_0,{\rho }_1$ on $S$. For fixed $0<\epsilon <1$, the states ${\tau }_i=(1-\epsilon ){\rho }_i+\epsilon \sigma$ are faithful on $S$ and are broadcast by $\tilde{B}$. By [L5], $[{\tau }_0,{\tau }_1]=0$. Direct expansion using $\sigma =({\rho }_0+{\rho }_1)\mathrm{/}2$ gives $[{\tau }_0,{\tau }_1]=(1-\epsilon )[{\rho }_0,{\rho }_1]$; since $1-\epsilon \mathrm{\ne }0$, the original states commute.

Suppose $[{\rho }_0,{\rho }_1]=0$. Choose an orthonormal eigenbasis of ${\rho }_0$ by [F1] and group it into its finitely many eigenspaces $S_{\lambda }$. For $x\in S_{\lambda }$, ${\rho }_0({\rho }_{1}x)={\rho }_1({\rho }_{0}x)=\lambda {\rho }_{1}x$, so $S_{\lambda }$ is invariant under ${\rho }_1$. The restriction of ${\rho }_1$ to $S_{\lambda }$ is self-adjoint, since $⟨{\rho }_{1}x,y⟩=⟨x,{\rho }_{1}y⟩$ for $x,y\in S_{\lambda }$. Apply [F1] on each $S_{\lambda }$ and combine these orthonormal eigenbases to obtain a common orthonormal basis $(e_j)$. Write ${\rho }_i={\sum }_j{\lambda }_{i,j}\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$. The Kraus operators $K_j=\mathrm{\mid }{e}_j\otimes e_j⟩⟨e_j\mathrm{\mid }$ obey ${\sum }_j{K}_j^{\ast }{K}_j=I$. Their map $B(X)={\sum }_j{K}_{j}X{K}_j^{\ast }$ is a channel by [F2]. It sends ${\rho }_i$ to ${\sum }_j{\lambda }_{i,j}\mathrm{\mid }{e}_j\otimes e_j⟩⟨e_j\otimes e_j\mathrm{\mid }$, whose two marginals are ${\rho }_i$. Thus the pair is broadcastable.