Theorem: Selected operational access map

For the ideal encoding $W$, the optimal unrestricted half-diamond errors for the following declared targets and access classes are:

No value is asserted here for an unlisted mixed access pattern.

Proof. For $|A|\le 2$, the channel is constant. The subsystem optimum is $3/4$ by the qubit replacement argument; the upper bound is unchanged by composing with ${\mathrm{Tr}}_S$. For full two-qubit recovery, a maximally entangled four-dimensional input has overlap $1/16$ with every replacement Choi state. The replacement $I_4/4$ attains error $15/16$, by the sixteen-term two-qubit Pauli expansion of ${\mathrm{Dep}}_4$.

For the hybrid target, identify the output with $L\otimes S$ and write ${\mathcal{T}}_{\mathrm{hyb}}={\mathrm{id}}_L\otimes {\Delta }_{Z,S}$. Its normalized Choi state is supported on

It has support probability one. A replacement Choi state $\tau \otimes I_4/4$ has support probability $1/8$, because ${\mathrm{Tr}}_{L^{\prime }{S}^{\prime }}{P}_{\mathrm{hyb}}=I_{LS}/2$. Thus the error is at least $7/8$. For the upper bound, ${\mathcal{T}}_{\mathrm{hyb}}$ is the equal mixture of ${\mathrm{id}}_{LS}$ and ${\mathrm{Ad}}_{I_L\otimes Z_S}$, whereas ${\mathrm{Dep}}_4$ is the equal mixture of all sixteen Pauli conjugations. Their difference is $7/8$ times a difference of two channels, so its half-diamond norm is at most $7/8$.

For left-only triple or four-leg access, $L$ is returned exactly and no information about $S$ is available. Replacing $S$ by $I_2/2$ gives full-bulk error $3/4$; a Bell input on $S$ gives the matching lower bound. For the hybrid target, returning a fair sector label gives error $1/2$. The lower bound follows by using the two orthogonal sector inputs with the same logical state: they have identical accessible outputs, so a single decoder cannot be closer than $1/2$ to both orthogonal target outputs. The two-triple case is Theorem thm:patch.