Hawk04-A Unitary Information Preservation and Radiation Recovery: A Complete-Complement Formulation of the Black-Hole Information Problem

We prove the precise conditional statement that quantum information cannot be destroyed by unitary evolution of a complete closed system. For a finite logical code, a fixed pure ancillary preparation, and a unitary evolution, the complete output channel has a CPTP left inverse, including every external reference. Every logical matrix unit and every input-state trace distance are preserved.

This global statement does not imply recovery from radiation alone. For a complete isometry $V:C\to R\otimes B$, we prove that one decoder recovers the entire code from $R$ exactly if and only if the complete complementary channel to $B$ is constant. A Schmidt-space construction provides the decoder and the exact no-hiding normal form. We distinguish this result from partial decoherence, restricted input ensembles, and mixed-output descriptions with omitted preparation purifiers.

For the regulated quartic SYK–scalar architecture of Hawk02-E, we reconstruct the self-adjoint dynamics and a reference-preserving finite approximation. A finite complementary Choi certificate $e$, input dimension $d$, and complete-channel regulator allowance $b$ yield ${\delta }_{\mathrm{full}}\le de+b$. A finite decoder can be lifted to the continuum with recovery error at most $\sqrt{2de}+b$, or certified directly using its decoded Choi operator. We also give an exact four-Majorana realization with nonzero coupling for which the optimal radiation recovery error is $(3-z(t))/4\ge 1/2$, although global recovery is exact. This is a dynamical negative control, not a scrambling argument.

Finally, we prove endpoint continuity with access-preserving output identifications and state the separate gravitational obligations. The results establish unitary preservation and operational recovery criteria. They do not establish unitarity of quantum gravity, a generic interacting evaporation-recovery window, or a gravitational endpoint.

{{0.93} Central distinction.

The first assertion is proved from unitarity. The second requires a bound on the complete complementary channel. Assuming global unitarity does not prove that physical black-hole evaporation has that property. }

The assertion

is a theorem about reversible quantum dynamics. It is not, by itself, a derivation of the microscopic dynamics of an evaporating black hole.

Hawking's radiation calculation and his subsequent information-loss proposal provide the historical setting Hawking1975,Hawking1976. The relevant tension is between a complete reversible description and a proposed final description that fails to preserve arbitrary initial quantum states. Thermal occupation data, local radiation marginals, and a complete input-to-output quantum channel are different objects.

The exact no-hiding principle is due to Braunstein and Pati BraunsteinPati. Exact error correction and its syndrome structure are established in the work of Knill and Laflamme KnillLaflamme. Approximate information–disturbance relations are supplied by Kretschmann, Schlingemann, and Werner KSW,KSWContinuity. We do not claim these principles as new results. We give local proofs of the exact statements needed here and make the approximate external input explicit.

The supplied Cheung manuscripts motivate the complete-output discipline adopted in this paper. Hawk01-B distinguishes entropy histories from reference-preserving recovery. Hawk02-D supplies an autonomous quadratic benchmark with a complete scattering endpoint. Hawk02-E specifies an interacting regulated source–reservoir interface but does not establish its small dynamical residual bound CheungHawk01B,CheungHawk02D,CheungHawk02E. The operational manuscripts separately identify decoder execution, coherent collection, and channel-level error composition CheungEn02B,CheungEn02C,CheungEn02E,CheungEm03C. The dephasing comparisons in Em05-A and Em05-B distinguish delivered occupation from recovered quantum coherence CheungEm05A,CheungEm05B.

. This paper supplies:

1. a complete proof of global code preservation under unitary evolution;

2. an exact complete-complement recovery theorem and decoder;

3. counterexamples repairing overly broad readings of no-hiding;

4. a self-contained finite-regulator certification interface for the specified interacting architecture;

5. an exact, nonzero-coupling interacting-family negative control;

6. an access-preserving endpoint continuity theorem.

It is a theorem-and-certification paper, not a claim of a new universal resolution of black-hole evaporation.

policy. Cheung citations identify the supplied versions or excerpts. They are comparisons, not unprovided theorem premises. The model-specific arguments used below are proved in this source. No external mathematics catalogue is a premise.

Let $C\simeq {\mathbb{C}}^d$, with $d\ge 2$, be the logical code. Output Hilbert spaces are separable. States are normal density operators, and channels act on trace-class operators. CPTP means completely positive and trace preserving. Finite-dimensional channel facts and conventions are standard Watrous.

We use

The diamond distance retains arbitrary external references. For channels with input $C$, a reference of dimension $d$ suffices.

For an isometry

define

Here $B$ is the complete inaccessible output. It includes uncollected modes, retained source degrees of freedom, inaccessible apparatus outputs, and preparation purifiers.

For a residual state $\sigma$, set

Define

A channel $\mathcal{V}:C\to {\mathcal{H}}_{\mathrm{out}}$ preserves the code globally if a single CPTP map $\mathcal{L}$ satisfies

This equality includes every code-reference input. It does not assert access to, or efficient implementation on, the complete output.

For a declared partition $R\otimes B$, radiation recovery means a single CPTP decoder $\mathcal{D}:R\to C$ satisfying

exactly, or the corresponding approximate diamond bound. The decoder is not allowed to act on $B$.

preparations. If an ancillary state is mixed, choose a fixed purification and retain its purifying system in the complete dilation. A code-dependent initial environment is a different task: the code-to-total-state preparation must then itself be specified and tested for preservation.

coherences. With $E_{ij}=|i⟩⟨j|$, the complete channel is determined by

Agreement on $E_{ii}$ alone is not agreement on a quantum channel. For a qubit, both $E_{01}$ and $E_{10}={E_{01}}^{†}$ are required.

Let $J:C\to {\mathcal{H}}_{\mathrm{in}}$ be an isometry, let $|\Omega ⟩$ be a fixed normalized ancillary state, and let $U$ be unitary on the complete physical input space. After any fixed unitary regrouping of the output, set

Then:

1. $V^{†}V=I_C$;

2. the complete channel $\mathcal{V}(X)=VX{V}^{†}$ has a CPTP left inverse;

3. for every trace-class operator $X$, including operators on the code and an arbitrary reference,

   $${\Vert (V\otimes I)X(V^{†}\otimes I)\Vert }_1={\Vert X\Vert }_1;$$

4. every logical coherence and every code-state trace distance are preserved;

5. for a Bell input, the reference and the complete output have mutual information $2{\log }_{2}d$.

These assertions are conditional on the stated unitary dynamics and complete output space.

Proof. The composition of an isometry, a normalized pure preparation, and a unitary is an isometry, proving the first item.

Let $P=V{V}^{†}$, and choose any code density operator $\tau$. Define

Both terms are completely positive. Their traces sum to $\mathrm{Tr}Y$. Thus $\mathcal{L}$ is CPTP, and

An isometric embedding leaves the nonzero singular values of an operator unchanged. Equivalently, its modulus is the isometric embedding of the original modulus on the range, with zero on the orthogonal complement. Taking its trace proves the third item, also after tensoring with a reference. Applying this identity to differences of density operators proves trace-distance preservation. Applying the defining isometry to $E_{ij}$ proves preservation of all matrix units.

Finally,

is pure, and both reduced states have nonzero eigenvalues $1/d$. Therefore

All three entropies in this equality are finite, even if the complete output space is infinite dimensional.

Distinct code states cannot have identical complete outputs under the hypotheses of Theorem thm:global. In particular, a complete replacement channel cannot describe that evolution on a nontrivial code.

Proof. Apply the CPTP left inverse, or use preserved trace distance.

The theorem concerns reversible encoding of a quantum state. It is not a conserved scalar particle number or a thermodynamic entropy-production law. Subsystem entropy may increase after a partial trace while the complete evolution remains reversible.

If $U(t)=e^{-\mathrm{i}tH}$ for a self-adjoint Hamiltonian, the theorem applies at every finite time. The existence of that Hamiltonian is a separate analytical obligation. It is established for the regulated microscopic model in Section sec:model.

Let $V:C\to R\otimes B$ be an isometry. The following are equivalent:

1. a CPTP decoder on $R$ exactly recovers the code;

2. $\mathcal{C}={\mathcal{T}}_{\sigma }$ for some residual density operator $\sigma$;

3. for a code basis,

   $$\mathcal{C}(E_{ij})={\delta }_{ij}\sigma ;$$

4. for the Bell input,

   $$({\mathrm{id}}_Q\otimes \mathcal{C})(|{\Phi }_d⟩⟨{\Phi }_d|)=\frac{I_Q}{d}\otimes \sigma .$$

When these statements hold, there is an auxiliary Hilbert space $K$, a state ${\sigma }_K$, and an isometry $A:C\otimes K\to R$ such that

The decoder can be chosen as

Proof. The equivalence of (ii) and (iii) follows from linearity. Expanding the normalized Bell state in matrix units proves the equivalence of (iii) and (iv).

Assume (ii). Write the positive spectral decomposition

The sum is finite or countable. Each code basis vector has an expansion

The equality of all complementary matrix units gives

Thus the rule

extends to an isometry. Set ${\sigma }_K={\sum }_{\alpha }{p}_{\alpha }|\alpha ⟩⟨\alpha |$. Tracing $B$ gives (eq:normal-form).

The first term of (eq:exact-decoder) is completely positive. The second is a positive-effect measurement followed by preparation. Their traces add to $\mathrm{Tr}Y$, so the map is CPTP. Substitution into (eq:normal-form) proves exact recovery.

Conversely, suppose (i). Use the normal Stinespring representation of the decoder, with an isometry

The complete decoded isometry is

Since its $C$ output is exactly $|i⟩$ on input $|i⟩$,

for normalized ${\eta }_i$. The decoded off-diagonal matrix unit $E_{ij}$ must also be exact. Tracing $G\otimes B$ therefore gives

Hence all ${\eta }_i$ are one common vector $\eta$. Tracing $C\otimes G$ shows

This proves (ii).

The theorem is the complete-dilation error-correction/no-hiding structure associated with BraunsteinPati,KnillLaflamme. The proof above is included to make the required output ledger and decoder explicit. Stinespring's representation is an external standard mathematical input KSWContinuity,Watrous.

If the accessible channel $\mathcal{N}$ is constant on the entire code, then the complete complementary output $B$ exactly recovers that code. The arbitrary quantum state cannot reside only in inaccessible correlations between $R$ and $B$.

Proof. Interchange $R$ and $B$ in Theorem thm:exact.

For $d\ge 2$, the two complementary channels of an isometry cannot both be constant on the entire code. Nor can both outputs separately recover the entire unknown code.

Proof. If one is constant, the other is exactly recoverable. A constant channel cannot recover two distinct inputs. If one output is exactly recoverable, its complement is constant by Theorem thm:exact, and therefore is not itself recoverable.

Corollary cor:nohiding requires:

1. a complete isometric dilation;

2. independence of the entire output channel on every code state;

3. retention of all preparation purifiers.

It does not say that correlations are irrelevant to recovery. It does not apply merely because one mean, one marginal for one input, or one selected input ensemble is unchanged.

If the complete endpoint residual is a fixed pure state, then the radiation channel is an isometric encoding:

In particular, a pure code input has a pure complete radiation output.

If the fixed residual is mixed, radiation recovery is still exact, but a pure code input can have mixed radiation of the form (eq:normal-form).

Proof. In Theorem thm:exact, a rank-one residual has a one-dimensional $K$. The general mixed case retains the input-independent syndrome state ${\sigma }_K$.

A horizonless effective geometry, by itself, establishes neither hypothesis of this proposition.

Let $S$ and $E$ both contain a $d$-dimensional register, and prepare $E$ in $|0⟩$. The swap unitary gives

The local channel on $S$ is a reset, but the environmental channel is the identity on the input state. An entangled diary obeys the same identity with $I_Q$ tensored onto the swap.

The example proves a transfer mechanism, not a universal account of practical reset hardware. A measurement record containing only a classical outcome does not, on its own, store an arbitrary qubit's phase.

The isometry

has both reduced channels equal to computational dephasing:

Neither output alone recovers an arbitrary qubit. Its optimal local recovery error is $1/2$, whereas the complete output is exactly recoverable.

Proof. Tracing either output annihilates $E_{01}$ and $E_{10}$. The two inputs $|+⟩$ and $|-⟩$ have identical dephased outputs but orthogonal ideal targets. If a decoder has error $\epsilon$, the triangle inequality therefore gives $1\le 2\epsilon$.

The identity decoder attains $1/2$, since

and $d_{\diamond }({\mathrm{Ad}}_Z,\mathrm{id})=1$, witnessed by $|+⟩$. Global recovery follows from Theorem thm:global.

Both marginals in this example depend on the logical populations. Consequently, complete no-hiding does not assert that either marginal must carry the whole qubit. This is the distinction also illustrated by the supplied Em05-A noise comparison CheungEm05A.

Define

The two basis inputs have identical maximally mixed marginals on both outputs, yet orthogonal complete outputs. However,

The reduced channels are therefore not constant on the full code. There is no contradiction with Corollary cor:nomasking.

Let $U_k$, $k=1,\dots ,d^2$, be a Weyl unitary basis. The channel

has constant marginals on the ciphertext $R$ and key $K$, but exact recovery from $RK$. This does not contradict no-hiding.

Proof. The Weyl twirl gives the ciphertext marginal $I_R/d$; the key marginal is uniformly distributed. Controlled application of ${U_k}^{†}$ recovers $\rho$.

A pure complete dilation is

The complete complement of $R$ is $KF$, not $K$. Its off-diagonal key-purifier matrix entries contain

They are generally input dependent. Since the ciphertext channel is constant, Corollary cor:nohiding proves exact recoverability from the complete $KF$ output.

These examples repair the unsupported statement that every form of decoherence transfers the entire unknown state into a measurement record or into one environmental marginal.

For every code density operator $\sigma$,

Thus a constant accessible channel has optimal recovery error $a_d$.

Proof. For the Bell input, a replacement output is $(I_Q/d)\otimes \sigma$, whose overlap with $|{\Phi }_d⟩$ is $d^{-2}$. The Bell-projector test proves the lower bound.

The Weyl twirl represents

Separating the identity term and using that unitary channels have diamond norm one gives the matching upper bound. A decoder after a constant channel remains constant.

For isometries $V_1,V_2:C\to R\otimes B$,

The same bound holds for either reduced channel.

Proof. For a pure referenced input, the two output vectors differ by at most $\Vert V_1-V_2\Vert$. Their pure-state trace distance is no larger than that vector distance. Convexity handles mixed inputs, and partial trace is contractive.

For a channel $\Phi$, use the normalized Choi state

For channels $\Phi ,\Psi$ with input dimension $d$,

If

then

Proof. A pure referenced input can be written as

Multiplication on both sides of the Choi difference therefore bounds its trace norm by $d$ times the Choi trace norm.

Alternatively, write

Its output difference has trace norm at most

Divide by two, use convexity, and take the input supremum.

The dimension factor is indispensable to these conversions. One off-diagonal test alone is not enough for a growing code; a full operator basis is required.

For finite-dimensional complementary channels, the KSW information–disturbance theorem gives

Indeed its full-norm inequality

becomes the first inequality after substituting $2{\epsilon }_{\mathrm{rec}}$ and $2{\delta }_{\mathrm{full}}$. The corresponding converse gives the second KSW. We apply these bounds to finite proxies below; no unstated infinite-dimensional attainment assertion is needed.

Let $\hat{V}:C\to \hat{R}\otimes \hat{B}$ be a finite-dimensional isometry embedded into a common complete output space. Suppose

For a finite residual density operator $\hat{\sigma }$, put

Then

There exists a finite CPTP decoder, completed on the full radiation space, for which

Accordingly,

Independently, if a specified finite CPTP decoder $\hat{\mathcal{D}}$ has

its complete extension satisfies

Proof. Partial trace of the complete-channel comparison gives

The triangle inequality and Lemma lem:choi prove (eq:delta-certificate).

For the finite channel, $\widehat{{\delta }_{\mathrm{full}}}\le de$. The finite decoder set is compact, so an optimal decoder exists. Equation (eq:KSW) gives its recovery error at most $\sqrt{2de}$.

If $P_R$ is the projection onto the embedded $\hat{R}$, complete this decoder by

The map is CP, and its two traces sum to $\mathrm{Tr}Y$. Postprocessing contraction adds at most $b$, proving (eq:epsilon-certificate). The constant decoder supplies the additional bound $a_d$.

For the specified decoder, Lemma lem:choi gives the finite error $d{r}_{\mathrm{dec}}$. The same extension and comparison prove (eq:direct-certificate).

The regulator allowance $b$ already compares complete channels in diamond distance. It is not multiplied by $d$. Moreover, applying finite recovery first and then lifting the decoder gives $\sqrt{2de}+b$, rather than putting that comparison allowance inside the square root. Neither expression estimates $e$ from dynamics.

If ${\mathcal{N}}_{\mathrm{rep}}$ is a replacement channel and

then

For finite-dimensional complete outputs,

Proof. For every decoder, postprocessing contraction gives

The decoded replacement is constant, so Lemma lem:replacement gives the lower bound. Combine it with (eq:KSW).

This proposition concerns a uniform complete channel comparison, not agreement with a thermal spectrum on selected observables.

We now specify the architecture of Hawk02-E CheungHawk02E and prove the analytical inputs used here. The proof applies to every frozen coupling array. It does not assume that the array has a useful radiation-recovery regime.

Let $N=2n\ge 4$, with Majoranas

acting on ${\mathcal{H}}_S={\mathcal{F}}_{\mathrm{f}}({\mathbb{C}}^n)$. Freeze real quartic coefficients and set

Every quartic monomial is self-adjoint with norm one, so

Write

The $S_j$ commute, and $\Vert A\Vert \le |\alpha |+|\zeta |$. For the estimates, one may use a fixed finite family of even self-adjoint $A_a$, including the choice $A_a=A$.

Fix

On ${\mathcal{H}}_{\mathrm{bath}}={\mathcal{F}}_{\mathrm{s}}(\mathfrak{h})$, let

with annihilation conjugate-linear in $f$. The inner product is conjugate-linear in its first argument.

For frozen real $g_a$, define

Set

Here

This is a fixed-realization quartic SYK–scalar-reservoir specification. The normalization is printed rather than inferred from a different SYK convention. The bath is initially vacuum. No gravitational geometry is assigned to these objects.

The Hamiltonian (eq:H) is self-adjoint on ${\mathcal{H}}_S\otimes D(H_B)$, bounded below, and satisfies the form inequality

It generates $U(t)=e^{-\mathrm{i}tH}$. For every isometric diary loading and fixed pure reservoir preparation, the complete diary channel at every finite time is exactly recoverable.

Proof. The annihilation estimate is

The CCR give

Therefore

For every $b>0$, the spectral theorem bounds the square-root term by $b\Vert H_B\psi \Vert +c_b\Vert \psi \Vert$. Thus the interaction has relative operator bound zero. Kato–Rellich proves self-adjointness on the stated domain; the source term is bounded. These are the standard field estimates used in spin–boson Hamiltonian constructions HaslerHerbst.

For normalized vectors in the form domain,

Since $H_S\ge 0$, the lower bound follows. The spectral theorem supplies the unitary group. Apply Theorem thm:global to the complete prepared evolution.

This is an actual preservation theorem for the printed microscopic Hamiltonian, not a proof that quantum gravity uses that Hamiltonian.

Choose a simultaneous spin eigenvector $|B_s⟩$ of fixed source parity, and prepare

This Euclidean-filtered preparation is motivated by Kourkoulou and Maldacena KourkoulouMaldacena. The exact normalization here follows from $e^{-\beta M_N}\le Z_s(\beta )\le 1$.

The source preparation is completed before the diary arrives. If it is heralded, its success decision is independent of the diary. An attempted deterministic preparation protocol must retain failure outputs in its own complete channel.

Encode a qubit in two carrier modes:

Both basis vectors are even. Choose two source modes $T$, fix a fermionic occupation ordering, and let $U_{\mathrm{sw}}$ be the second quantization of their one-particle exchange with the carrier. Define

This is one message-independent, parity-preserving isometry.

The even diary introduces no logical-input-dependent permutation sign. The old source two-mode state is displaced into the carrier, including its correlations with the remaining source. The carrier marginal is diary independent, but it is generally neither vacuum nor uncorrelated with the source. It remains in the complete residual. This differs from the vacuum destination used in Hawk02-D CheungHawk02D.

Fix an orthonormal packet family $p_1,\dots ,p_q$ and

The declared access partition is

After the fixed regrouping,

Its complete matrix units are

For every logical input, the prepared evolution obeys

where $E_I=⟨V_{\mathrm{int}}⟩$. Moreover,

On the prepared finite-time domain,

Proof. The initial bath and interaction means vanish. Autonomous unitary evolution conserves the total Hamiltonian expectation. The form bound in Theorem thm:model-unitarity gives the bath-energy bound.

The occupation estimates in the next subsection give finite moments of every fixed particle-number order on bounded time intervals. Since the frequency band is bounded, differentiation can be justified by finite-particle approximation. The commutator identity and differentiated conservation law then give the currents.

Neither positive bath energy nor vanishing source energy establishes a small complete residual. Interaction energy cannot be omitted or renamed as black-hole mass.

Let $P_m$ project onto total bath occupation at most $m$. For $x\ge 0$, put

At fixed $\mu ,\Lambda$, every finite observation interval $[0,T]$ and finite packet family admit a finite-dimensional prepared isometry ${\hat{V}}_t$ such that

uniformly for $t\in [0,T]$. The same bound holds for both complete complementary outputs after their common embeddings.

For $0<b<1$, it suffices to choose $m\ge 1$ and a mesh with

when $T>0$. Then $b_{\mathrm{reg}}\le b$.

Proof. In interaction picture,

On sectors containing at most $k$ particles,

An $\ell$-fold product acting on bath vacuum has norm at most $(2G{)}^{\ell }\sqrt{\ell !}$. The ordered integration simplex has volume $t^{\ell }/\ell !$. The vacuum Dyson term is consequently bounded by

The bound is uniform over source, carrier, and reference inputs.

The series converges, also after multiplication by any fixed power of $\ell$. Since each term has occupation at most $\ell$, it converges in every fixed particle-number graph norm. On these vectors the differentiated series satisfies the Schrödinger equation; uniqueness for the self-adjoint Hamiltonian identifies it with the actual evolution.

The evolution generated by $P_{m}H{P}_m$, on its projected space, has the same Dyson terms through order $m$. Inserted projections are contractions, so the difference of the two prepared isometries is at most $2Z_m(2GT)$.

Partition $[\mu ,\Lambda ]$ into intervals of width at most $\Delta$, and replace frequency multiplication by its bin midpoint value ${\tilde{h}}_B$. Then

Duhamel's formula adds at most $Tm\Delta /2$.

For an exact finite active-mode realization of this discretized model, take

It contains all coupling vectors and the complete packet space, and is invariant under ${\tilde{h}}_B$. Its orthogonal complement remains vacuum. Thus the occupation-truncated dynamics on ${\mathcal{F}}_{\mathrm{s}}(\mathfrak{f})$ is finite dimensional.

The total-occupation subspace is embedded into the product

This is an embedding, not an assertion that the total cutoff itself factorizes. The source and carrier are retained. Apply Lemma lem:isometry and partial trace.

For the explicit selection, let $x=2GT$. The first condition gives

The subsequent tail-term ratios are at most $1/2$, so $Z_m(x)\le 2^{-m}$. The second condition makes $2Z_m(x)\le b/2$; the mesh condition bounds the remaining term by $b/2$.

The theorem refines finite occupation and frequency proxies. It does not remove the fixed infrared or ultraviolet cutoffs. The constants displayed above are not uniform as $\mu \downarrow 0$ or $\Lambda \uparrow \infty$.

If

for all $a,b$, the reservoir channel from any inserted finite code is entanglement breaking. Every deterministic radiation decoder has error at least

Proof. Choose common spectral projectors $P_{\lambda }$ of the source operators. The propagator is block diagonal:

Tracing the source kills cross-projector terms. The accessible channel therefore has the form

The $F_{\lambda }$ form a POVM. Thus the channel is measure-and-prepare.

A decoder preserves separability of its Bell-input output. Every separable state on $Q\otimes C$ has Bell overlap at most $1/d$, by Cauchy–Schwarz for product vectors and convexity. The Bell-projector test proves the lower bound.

If ${\mathcal{N}}_{\mathrm{cl}}$ is entanglement breaking and $d_{\diamond }(\mathcal{N},{\mathcal{N}}_{\mathrm{cl}})\le \eta$, then

Proof. Use postprocessing contraction and Theorem thm:QND's separable Bell-state argument.

This repairs the proposed shortcut of replacing source operators by classical histories while assuming that quantum recovery survives. A classical-history approximation needs its own complete-channel error; if that error is small, it gives an obstruction rather than a high-accuracy decoder.

We can evaluate a member of the interacting family exactly, without a disorder average or an approximate scrambling premise.

Fix

one bath species, $g\ne 0$, and any fixed $0<\mu <\Lambda$. Then

Take $c_j=({\chi }_{2j-1}+\mathrm{i}{\chi }_{2j})/2$, so $S_j=2{c_j}^{†}{c}_j-I$.

Prepare $|B_s⟩=|11{⟩}_S$. Its filtered state (eq:filtered) is still $|11{⟩}_S$. Swap the entire two-mode diary carrier with the source. The carrier becomes the fixed state $|11{⟩}_D$, and the logical source states are $|00{⟩}_S,|11{⟩}_S$. Both have source energy zero, and their coupling eigenvalues are $-1,+1$, respectively.

For

define

These are square-integrable coherent amplitudes. For the declared packet projection $P_{\mathfrak{r}}$, put

and

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.