Theorem: Exact radiation, residual recovery, and complete leakage

For this realization,

The complete output $RB$ recovers the diary exactly.

If all reservoir modes are declared accessible, then

Proof. For source eigenvalue $a_i$, let $q_i=g{a}_i{h_B}^{-1}f$. The Weyl unitary $\mathcal{W}(q)=e^{a^{†}(q)-a(q)}$ gives

The vacuum evolves to $|{\eta }_i(t)⟩$, up to a phase. For $a_i=\pm 1$, the phases are common. Factoring the coherent state between collected and uncollected modes gives

Tracing the orthogonal source labels proves (eq:cq).

Write ${\sigma }_i=|{\eta }_{i,R}⟩⟨{\eta }_{i,R}|$. Their trace distance is $z_R$. For any effect $0\le F\le I$,

because the maximum of $\mathrm{Tr}(F({\sigma }_0-{\sigma }_1))$ is the trace of its positive part, namely $z_R$.

For any decoder, its Bell-input overlap with the ideal output is

Indeed, the adjoint image of $|0⟩⟨0|$ is an effect and its complement is the adjoint image of $|1⟩⟨1|$. The Bell-projector test therefore gives

Bosonic parity interchanges ${\sigma }_0$ and ${\sigma }_1$. The optimal discrimination measurement can thus be chosen with symmetric errors, each $(1-z_R)/2$, by symmetrizing an optimal measurement. Measuring and preparing the inferred logical basis state produces the channel

Its Pauli probabilities are

The four Pauli Bell outputs are orthogonal. Lemma lem:unitary-mixture gives error $1-p_I=(3-z_R)/4$, proving (eq:Rerror). The decoder is completed on the orthogonal complement of the two-state support.

For the residual, define

The source labels make $L_t$ an isometry. Tracing $R$ gives

Retracting the range of $L_t$ gives

with error $(1-c)/2$. For the lower bound, the residual outputs of $|+⟩$ and $|-⟩$ have trace distance $c$. If a decoder has error $\epsilon$, contraction and the triangle inequality give $1\le 2\epsilon +c$. This proves (eq:Berror).

By Lemma lem:dictionaries, residual leakage equals the leakage of ${\Delta }_c$. This channel is Pauli covariant. Conjugating a replacement state by a Pauli leaves its distance to ${\Delta }_c$ unchanged after the corresponding input conjugation. Convexity therefore shows that twirling the replacement state to $I_2/2$ cannot increase the distance; that replacement is optimal.

Relative to ${\mathcal{T}}_{I_2/2}$, the Pauli coefficient differences, in the order $I,Z,X,Y$, are

Lemma lem:unitary-mixture gives

Complete recovery follows from Theorem thm:global. Finally,

which proves (eq:fullbath-overlap).