Theorem: All ideal access channels and their optimal errors

For the encoding (eq:Vmatrix):

1. If $m\ge 3$, then

   $${\mathcal{C}}_A(M)=\mathrm{Tr}(M)\frac{I_{\bar{A}}}{2^k},{T_A}^{†}{T}_A=I_{C\otimes K_A},$$

   and

   $$\begin{array}{rl}{\mathcal{N}}_A(M)& =T_A(M\otimes I_{K_A}/2^k){T_A}^{†},\\ D_A(Y)& ={\mathrm{Tr}}_{K_A}({T_A}^{†}Y{T}_A)+\mathrm{Tr}((I_A-T_A{T}_A^{†})Y)\frac{I_2}{2}\end{array}$$

   satisfies $D_A{\mathcal{N}}_A={\mathrm{id}}_2$.

2. If $m\le 2$, then

   $${\mathcal{N}}_A(M)=\mathrm{Tr}(M)\frac{I_A}{2^m},$$

   and

   $$\underset{D:\mathcal{B}(H_A)\to \mathcal{B}(C)\mathrm{CPTP}}{\inf }{d}_{\diamond }(D{\mathcal{N}}_A,{\mathrm{id}}_2)=\frac{3}{4}.$$

3. In the second case, the complement is nonconstant and is itself exactly recoverable. More explicitly,

   $${\mathcal{C}}_A(M)=T_{\bar{A}}(M\otimes I_A/2^m){T_{\bar{A}}}^{†}.$$

Every assertion concerns the full logical qubit, including its reference entanglement.

Proof. For $m\ge 3$, the omitted set has size $k\le 2$, so Lemma lem:privacy gives the complementary matrix-unit identities. Comparing their entries gives

Expanding $V|i⟩$ in the omitted computational basis proves the first line of (eq:DA). The second line is the complete decoder in Theorem thm:exact; alternatively its traces add to $\mathrm{Tr}Y$, and direct substitution returns $M$.

For $m\le 2$, privacy makes the accessible channel replacement. Every subsequent decoder is therefore replacement by some state $\omega$. On a Bell input its output is $I_Q/2\otimes \omega$, whose Bell overlap is $1/4$. The Bell-projector test gives error at least $3/4$.

The decoder returning $I_2/2$ attains the bound. Indeed the completely depolarizing channel is

The triangle inequality gives $d_{\diamond }(\mathcal{T},\mathrm{id})\le 3/4$, matching the Bell lower bound.

Finally $|\bar{A}|\ge 3$, so the first part, with the roles interchanged, gives the displayed complement formula and its exact recovery. It cannot be constant because it recovers two distinct logical inputs.