State Family No Broadcasting Iff Commuting

Suppose one channel broadcasts every member of $S$. For any $\rho ,\sigma \in S$, the same channel broadcasts the pair, so [L2] gives $[\rho ,\sigma ]=0$. If $S$ has fewer than two elements, pairwise commutation is vacuous or follows from $[\rho ,\rho ]=0$.

Conversely, suppose all members commute pairwise. Each is diagonalizable by [L4], so [L3] gives a common eigenbasis $(v_j)$ (not assumed orthogonal). Group this finite basis by its joint eigenvalue tuples across $S$. If two tuples differ, some self-adjoint $\rho \in S$ has distinct real eigenvalues on the corresponding vectors, and those vectors are orthogonal because $⟨\rho v,w⟩=⟨v,\rho w⟩$ forces $(\lambda -\mu )⟨v,w⟩=0$. Within each common eigenspace, choose an orthonormal basis by Gram–Schmidt; every $\rho \in S$ acts there as one scalar, so the resulting orthonormal basis $(e_j)$ still diagonalizes every member. If $S$ is empty, any orthonormal basis suffices.

Define $K_j=\mathrm{\mid }{e}_j\otimes e_j⟩⟨e_j\mathrm{\mid }$ and $B(X)={\sum }_j{K}_{j}X{K}_j^{\ast }$. The map is completely positive by [L5]. Also $K_j^{\ast }{K}_j=\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$, so ${\sum }_j{K}_j^{\ast }{K}_j=I$; cyclicity [F1] gives $\mathrm{tr}B(X)=\mathrm{tr}X$ for every $X$, hence $B$ is a channel. For any $\rho ={\sum }_j{\lambda }_j\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$ in $S$, $B(\rho )={\sum }_j{\lambda }_j\mathrm{\mid }{e}_j\otimes e_j⟩⟨e_j\otimes e_j\mathrm{\mid }$, and the partial-trace identity [L6] gives either marginal as $\rho$. The same channel works for every member, proving sufficiency.