Theorem: Exact radiation-recovery error for the frozen realization

For this realization,

Its optimal deterministic radiation recovery error is

The complete residual obeys ${\delta }_{\mathrm{full}}(t)\ge 1/2$ at every time. The complete output nevertheless preserves the qubit exactly.

If all bath modes are accessible, then

These statements are finite-realization dynamical conclusions, not properties inferred from source size or chaos.

Proof. On source eigenvalue $a_i$, the bath Hamiltonian is $H_B+g{a}_i\Phi (f)$. Let $q_i=g{a}_i{h_B}^{-1}f$, and use the Weyl unitary

The CCR give

Thus vacuum evolves, up to a phase, to the coherent state with amplitude $(e^{-\mathrm{i}t{h}_B}-I)q_i={\eta }_i(t)$. The phase is common for $a_i=\pm 1$. Coherent states factor under the orthogonal packet decomposition. Tracing the orthogonal source states gives (eq:cq).

For two pure radiation states with positive overlap $c_R$, their trace distance is $z_R=\sqrt{1-{c_R}^2}$. The optimal equal-prior discrimination success probability is $(1+z_R)/2$. This follows directly by maximizing

over effects $0\le E\le I$, using the positive part of ${\sigma }_0-{\sigma }_1$.

For any decoder, the decoded Bell overlap is

The two diagonal decoder effects form a discrimination POVM. Hence $F_e\le (1+z_R)/4$, and the Bell-projector test gives error at least $(3-z_R)/4$.

The two states are interchanged by radiation parity. Their Helstrom measurement can therefore be chosen with symmetric conditional errors. Measuring it and preparing $|0⟩$ or $|1⟩$ produces the complete decoded channel

Its Pauli probabilities are

Its distance from the identity is $1-p_I=(3-z_R)/4$: the Bell-state outputs are orthogonal for the four Paulis, and the triangle inequality gives the same upper bound. The decoder is completed on the unused radiation subspace by fixed-state preparation. At $z_R=0$, use the constant decoder.

The complete residual retains orthogonal source states on the two logical basis inputs. Thus their residual trace distance is one. If both residual outputs were within distance $\delta$ of one replacement state, the triangle inequality would give $1\le 2\delta$. This proves ${\delta }_{\mathrm{full}}\ge 1/2$.

Global preservation follows from Theorem thm:model-unitarity. Finally,

which gives the full-bath overlap.