Proposition: Authorized and unauthorized access

Every boundary set of at least three qubits can recover the logical qubit algebraically. Every set of at most two qubits is independent of the logical input and cannot recover it exactly.

For an unauthorized set, the optimal subsystem recovery error, over arbitrary channels, is

Proof. For an accessible triple, $\left|C\right|=2$. The eight vectors

are orthonormal by (eq:KL). They form a basis of ${\mathcal{H}}_A$. The unitary sending them to $|i⟩|e⟩$, followed by tracing the last two factors, is an exact decoder. For a larger accessible set, select a triple and explicitly consume the other accessible qubits by unreported destructive measurement.

For an unauthorized set, (eq:privacy) makes every decoded channel a constant preparation channel. On a logical Bell state with a reference, its output has the form $\sigma \otimes I_R/2$, whose Bell overlap is $1/4$. The Bell-projector test gives trace distance at least $3/4$.

This bound is achieved by the completely depolarizing logical output $\mathcal{T}$. Indeed,

so

The Bell-state lower bound supplies equality.