Full Rank Pair No Broadcasting

Since ${\rho }_0,{\rho }_1$ are faithful, the measurement-minimum theorem [L2] supplies a rank-one PVM $(E_b)$ whose classical overlap is $F({\rho }_0,{\rho }_1)$. Every $p_{s,b}=\mathrm{tr}({\rho }_s{E}_b)$ is positive: if $E_b=\mathrm{\mid }b⟩⟨b\mathrm{\mid }$ and ${\rho }_s$ is positive definite, then $p_{s,b}=⟨{\rho }_{s}b,b⟩>0$.

Put ${\tilde{\rho }}_s=B({\rho }_s)$. Data processing under $B$ gives $F({\rho }_0,{\rho }_1)\le F({\tilde{\rho }}_0,{\tilde{\rho }}_1)$. Data processing under either partial trace, whose outputs are ${\rho }_s$ by [L1], gives the reverse inequality. Hence equality holds. Measuring $E_b\otimes I$ on the first output factor, or $I\otimes E_b$ on the second, produces the same probabilities as measuring $E_b$ on the input; each classical overlap is $F({\tilde{\rho }}_0,{\tilde{\rho }}_1)$ and so is optimal by [L2].

For any two states $R,S$ and a finite POVM $(Q_b)$ attaining the minimum overlap, choose a polar unitary $U$ with $U\sqrt{R}\sqrt{S}=\sqrt{\sqrt{S}R\sqrt{S}}$. For each outcome let $A_b=U\sqrt{R}{Q}_b^{1\mathrm{/}2}$ and $C_b=\sqrt{S}{Q}_b^{1\mathrm{/}2}$. The Hilbert–Schmidt norms satisfy $\mathrm{\parallel }{A}_b{\mathrm{\parallel }}_2^2=\mathrm{tr}(R{Q}_b)$ and $\mathrm{\parallel }{C}_b{\mathrm{\parallel }}_2^2=\mathrm{tr}(S{Q}_b)$, while their inner product is $\mathrm{tr}(U\sqrt{R}{Q}_b\sqrt{S})$. Thus Cauchy–Schwarz and the triangle inequality give ${\sum }_b\sqrt{\mathrm{tr}(R{Q}_b)\mathrm{tr}(S{Q}_b)}\ge {\sum }_b\mathrm{\mid }\mathrm{tr}(U\sqrt{R}{Q}_b\sqrt{S})\mathrm{\mid }\ge \mathrm{\mid }\mathrm{tr}(U\sqrt{R}\sqrt{S})\mathrm{\mid }=F(R,S)\mathrm{.}$ Equality of the endpoints forces equality term by term in Cauchy–Schwarz and common nonnegative phase in the triangle inequality. For each outcome with $\mathrm{tr}(S{Q}_b)>0$, this gives a scalar ${\gamma }_b\ge 0$ with $U\sqrt{R}{Q}_b^{1\mathrm{/}2}={\gamma }_b\sqrt{S}{Q}_b^{1\mathrm{/}2}$. Applying this to the two optimal PVMs of step 2.1, there are unitaries $U,V$ and nonnegative ${\alpha }_b,{\beta }_b$ such that $U\sqrt{{\tilde{\rho }}_0}(I\otimes E_b)={\alpha }_b\sqrt{{\tilde{\rho }}_1}(I\otimes E_b),V\sqrt{{\tilde{\rho }}_0}(E_b\otimes I)={\beta }_b\sqrt{{\tilde{\rho }}_1}(E_b\otimes I),$ and both $U,V$ satisfy $U\sqrt{{\tilde{\rho }}_0}\sqrt{{\tilde{\rho }}_1}=V\sqrt{{\tilde{\rho }}_0}\sqrt{{\tilde{\rho }}_1}=\sqrt{\sqrt{{\tilde{\rho }}_1}{\tilde{\rho }}_0\sqrt{{\tilde{\rho }}_1}}$. For these two PVMs every outcome probability is positive by step 1.1 and the broadcaster marginals, so neither Cauchy–Schwarz vector is zero; dependence therefore gives the displayed scalar equations.

Set $G={\sum }_b{\alpha }_b{E}_b$ and $J={\sum }_b{\beta }_b{E}_b$. Summing the equations in step 3.1 gives $U\sqrt{{\tilde{\rho }}_0}=\sqrt{{\tilde{\rho }}_1}(I\otimes G)$ and $V\sqrt{{\tilde{\rho }}_0}=\sqrt{{\tilde{\rho }}_1}(J\otimes I)$. Multiply the first by $\sqrt{{\tilde{\rho }}_0}{U}^{\ast }$ on the left and trace out the first factor; multiply the second by $\sqrt{{\tilde{\rho }}_0}{V}^{\ast }$ and trace out the second. The marginal identities yield ${\rho }_0=AG=CJ$, where $A={\mathrm{Tr}}_1(\sqrt{{\tilde{\rho }}_0}{U}^{\ast }\sqrt{{\tilde{\rho }}_1})$ and $C={\mathrm{Tr}}_2(\sqrt{{\tilde{\rho }}_0}{V}^{\ast }\sqrt{{\tilde{\rho }}_1})$. Since ${\rho }_0$ is invertible, both $G$ and $J$ are invertible.

Left-multiply the two summed equations of step 4.1 by $\sqrt{{\tilde{\rho }}_1}$ and trace out the first factor in the first equation and the second in the second. The partial-trace adjoint identity and the local-factor multiplication rule in [L1] give $A^{\ast }={\rho }_{1}G$ and $C^{\ast }={\rho }_{1}J$. Taking adjoints in ${\rho }_0=AG=CJ$ gives ${\rho }_0=G{\rho }_{1}G=J{\rho }_{1}J$. For either positive operator $G$ or $J$, conjugating this equation by ${\rho }_1^{1\mathrm{/}2}$ shows $({\rho }_1^{1\mathrm{/}2}G{\rho }_1^{1\mathrm{/}2}{)}^2={\rho }_1^{1\mathrm{/}2}{\rho }_0{\rho }_1^{1\mathrm{/}2}$, and likewise for $J$. Uniqueness of the positive square root [L4] therefore yields $G=J$.

The operators $G=J={\sum }_b{\alpha }_b{E}_b$ are positive and invertible by step 4.1, so write $G\mathrm{\mid }b⟩=g_b\mathrm{\mid }b⟩$ with each $g_b>0$. Substituting $G=J$ into step 4.1 and eliminating $\sqrt{{\tilde{\rho }}_1}$ gives $V^{\ast }U\sqrt{{\tilde{\rho }}_0}=\sqrt{{\tilde{\rho }}_0}(G^{-1}\otimes G)\mathrm{.}$ For $w=\sqrt{{\tilde{\rho }}_0}(\mathrm{\mid }b⟩\otimes \mathrm{\mid }c⟩)\mathrm{\ne }0$, this equation says the unitary $V^{\ast }U$ has eigenvalue $g_c\mathrm{/}{g}_b>0$ on $w$, whose modulus must be one. Hence $g_b=g_c$; when they differ, $w=0$.

For any $b,b^{\mathrm{\prime }}$ with $g_b\mathrm{\ne }{g}_{b^{\mathrm{\prime }}}$ and any $c$, either $g_c\mathrm{\ne }{g}_b$, in which case step 5.2 gives $\sqrt{{\tilde{\rho }}_0}\mathrm{\mid }b,c⟩=0$, or $g_c=g_b$, in which case it gives $\sqrt{{\tilde{\rho }}_0}\mathrm{\mid }{b}^{\mathrm{\prime }},c⟩=0$. In the first case the column of ${\tilde{\rho }}_0$ vanishes; in the second its corresponding row vanishes because $\sqrt{{\tilde{\rho }}_0}$ is self-adjoint. Therefore every summand in $⟨b^{\mathrm{\prime }}\mathrm{\mid }{\rho }_0\mathrm{\mid }b⟩={\sum }_c⟨b^{\mathrm{\prime }},c\mathrm{\mid }{\tilde{\rho }}_0\mathrm{\mid }b,c⟩$ is zero. Thus $[G,{\rho }_0]=0$. Since ${\rho }_0=G{\rho }_{1}G$ from step 5.1, ${\rho }_1=G^{-1}{\rho }_0{G}^{-1}$ also commutes with ${\rho }_0$.

Conversely, diagonalize ${\rho }_0$ in an orthonormal basis by [L6]. Commutation makes each eigenspace invariant under ${\rho }_1$; the restriction of the self-adjoint ${\rho }_1$ to each eigenspace has an orthonormal eigenbasis by [L6]. Combining them gives a common orthonormal eigenbasis $(\mathrm{\mid }j⟩)$. Define $K_j=\mathrm{\mid }j,j⟩⟨j\mathrm{\mid }$ and $B(X)={\sum }_j{K}_{j}X{K}_j^{\ast }$. By [L7], $K_j^{\ast }{K}_j=\mathrm{\mid }j⟩⟨j\mathrm{\mid }$, so the basis identity gives ${\sum }_j{K}_j^{\ast }{K}_j=I$; [L6] therefore makes $B$ completely positive and trace-preserving, hence a channel. For either diagonal state ${\rho }_s={\sum }_j{\lambda }_{s,j}\mathrm{\mid }j⟩⟨j\mathrm{\mid }$, $B({\rho }_s)={\sum }_j{\lambda }_{s,j}\mathrm{\mid }j,j⟩⟨j,j\mathrm{\mid }$, whose two marginals are both ${\rho }_s$. This proves the reverse direction.