Theorem: Exact optimal deterministic packet recovery

For the declared finite packet output,

These are information-theoretic decoder optima, not local decoder-implementation claims.

At every finite $g$ and compact smooth source profile,

Thus the optimal recovery error is strictly greater than $1/2$.

Proof. For any decoder, let $E={\mathcal{D}}^{\ast }(|+⟩⟨+|)$. Its Bell overlap is

using the positive-part characterization of trace distance.

Let $\Pi =e^{\mathrm{i}\pi N_{\mathrm{tot}}}$. Lemma lem:gaussian gives $\Pi {\rho }_{+}{\Pi }^{†}={\rho }_{-}$. For $T={\rho }_{+}-{\rho }_{-}$, choose the Helstrom effect $E_{+}=(I+\mathrm{sign}T)/2$, with value $1/2$ on the kernel. Then $\Pi E_{+}{\Pi }^{†}=I-E_{+}$. Both conditional correct-decision probabilities equal $q=(1+d_{\mathbf{h}})/2$.

Measure this effect and prepare the corresponding source basis state. The decoded qubit channel has Pauli probabilities

For a Pauli channel its half-diamond distance from identity is $1-{\lambda }_I$: the Bell outputs are orthogonal and the triangle inequality gives the matching upper bound. This decoder attains both values in (eq:recovery). Every decoder has error at least $1-F_e$, proving optimality.

For the strict upper bound on $d_{\mathbf{h}}$, compare the two vector states $W(-a){\Omega }_{\mathrm{U}}$ and $W(a){\Omega }_{\mathrm{U}}$ in the regular quasifree realization. Their overlap modulus is $e^{-2\Vert a{\Vert }_{\mathrm{U}}^2}$. Their pure-state distance is $\sqrt{1-e^{-4\Vert a{\Vert }_{\mathrm{U}}^2}}$. Restriction to packets is contractive. Lemma lem:compact makes the exponent finite.