Exact quantum, private, and entanglement-assisted erasure capacities

By [F1] and [F8], the coherent information of an input state is $(1-2ϵ)S(\rho )$, with $S(\rho )\le \ln d$ and $S(\rho )\ge 0$. If $ϵ\le 1\mathrm{/}2$, its maximum is $(1-2ϵ)\ln d$, attained at $I\mathrm{/}d$.

Now suppose $ϵ\ge 1\mathrm{/}2$. By [F2], there is an antidegrading channel $\mathcal{A}:E\to B$ with $\mathcal{E}=\mathcal{A}\circ {\mathcal{E}}^c$. Its tensor powers are channels by [F9]. For any $n$-use input state and purification reference $R$, [F5] gives $I(R;B^n)\le I(R;E^n)$. The joint state $R{B}^n{E}^n$ from a product Stinespring isometry is pure, so Schmidt decomposition gives $S(R{B}^n)=S(E^n)$ and $S(R{E}^n)=S(B^n)$. By the mutual-information definition [F6], $I(R;B^n)-I(R;E^n)=S(B^n)-S(E^n)-S(R{B}^n)+S(R{E}^n)=2[S(B^n)-S(E^n)]=2I_c(\rho ,{\mathcal{E}}^{\otimes n}),$ where the last equality is [F10]. Hence the data-processing inequality forces $I_c(\rho ,{\mathcal{E}}^{\otimes n})\le 0$. A pure input attains zero, so the $n$-use coherent-information supremum is zero and [F4] gives $Q=0$.

For $ϵ\le 1\mathrm{/}2$, [F2] says the channel is degradable; applying [F3] to step 1.1 gives $Q=P=(1-2ϵ)\ln d$. This includes $ϵ=1\mathrm{/}2$, where the value is zero.

For any finite $n$-use ensemble, [F7] and [F6] express private information as $I(X;B^n)-I(X;E^n)$. The antidegrading channel on $E^n$ and [F5] make this difference nonpositive. The singleton ensemble has private information zero, so each $n$-use private-information supremum is zero; [F4] gives $P=0$.

By [F1] and [F4], the entanglement-assisted objective is $S(\rho )+I_c(\rho ,{\mathcal{E}}_{d,ϵ})=2(1-ϵ)S(\rho )$. Since $2(1-ϵ)\ge 0$, [F8] makes its maximum $2(1-ϵ)\ln d$, proving the final formula.