Hawk05-A Enter, Page-Type Entropy Turnover, and the Classical-Record Obstruction Complete complements, explicit schedules, and conditional endpoint certificates

We give a finite-dimensional, reference-preserving formulation of the Enter/Page analogy and correct several incompatible claims in its original narrative. Complete isometric evolution preserves an unknown quantum code, but recovery from a declared accessible output is equivalent to constancy of that output's complete complementary channel. Destructive Enter returns a classical record, not the consumed quantum input. A channel retaining only such a record has optimal whole-code recovery error at least $1-1/d$, at every stage, including the endpoint. Moreover, the Shannon entropy of an unedited cumulative classical record cannot decrease as outcomes are appended.

An explicit energy-conserving transfer schedule has a rise-and-fall quantum radiation entropy for pure unreferenced inputs, while its recovery error decreases monotonically. The same schedule after destructive readout does not purify the accumulated classical record. We prove exact recovery, leakage, and readout-error formulas; construct matched entropy histories with opposite endpoint recoverability; establish the readout-and-drop identity; and formulate a same-model endpoint certificate with quantitative error bounds. The result is a corrected operational analogy, not a gravitational Page-curve derivation or a construction of the unresolved characteristic quantum coefficient in the supplied analytic programme.

release. The finite statements proved below are complete within their stated hypotheses. The original claim that an accumulated classical album eventually becomes a pure carrier of an arbitrary unknown quantum code is false for $d\ge 2$. The appropriate completion is therefore a separation theorem and a corrected conditional narrative, not a proof of that claim.

The Philosophy of Enter distinguishes four questions: what was prepared, what evolution acts on the complete system, what outputs are accessible, and what type of output a continuation returns. In particular, destructive computational-basis readout consumes its measured quantum name and returns classical data. Consumption is an interface statement, not physical annihilation. A quantum-return decoder is a different operation with a different correctness target EnterPhilosophy,CheungEnter.

We adopt ordinary finite-dimensional Born instrument semantics explicitly. We do not attempt a new derivation of the Born rule. Cheung's Manuscript I characterizes destructive readout inside a specified operational class; that characterization is background, not an unstated hypothesis transferring to every model CheungI.

The distinctions between complete preservation, restricted recovery, classical readout, and discard are developed in Cheung's Hawk04 manuscripts. The present paper proves the finite statements it needs rather than treating documentary comparisons as automatic theorem imports CheungHawk04A,CheungHawk04B,CheungHawk04C,CheungHawk04D.

The original operational plan contains useful distinctions but also claims that cannot hold simultaneously. Table tab:debug states the corrections proved in this paper.

{}1.13

Page's random-state calculation concerns the entropy of a quantum subsystem of a pure bipartite state. For subsystem dimensions $m\le n$, its average entropy in bits is

Page proposed this formula and Sen supplied a proof PageAverage,Sen. Page's radiation argument adds a proposed black-hole interpretation to such information-theoretic considerations PageRadiation.

Neither this average nor a dimension crossing supplies a channel certificate for an entire variable quantum code. Hayden–Preskill retrieval of an additional diary, with early radiation available and specified mixing assumptions, is another task; it is not interchangeable with whole-code recovery from arbitrary classical counts HaydenPreskill.

Hawking's emission calculation supplies semiclassical field-theoretic information, not the complete endpoint channel used below Hawking. Cheung's En07-C keeps that separation explicit CheungEn07C. No Hawking occupation law, gravitational Hamiltonian, island formula, or quantum extremal surface is a premise of our finite proofs.

All Hilbert spaces in the mathematical results are finite dimensional. Let $C$ be a logical code with

An external reference $Q$ is untouched by all operations.

A complete interface is an isometry

Its accessible channel and complete complementary channel are

The accessible output $A$ may be quantum or hybrid classical–quantum. The environment $E$ includes every unreturned output of the complete preparation and evolution.

A fixed mixed ancillary preparation is represented by a fixed purification whose unreturned purifying factor belongs to $E$. Input-dependent preparation must instead be included in the specified encoding map; it is not a fixed ancillary state.

For a state $\tau$, define the replacement channel

Our normalization is

where the diamond norm includes arbitrary untouched references. For differences of channels it is the supremum of the corresponding output trace norm on referenced input states Watrous (Chapter 3).

For the interface (eq:channels), set

Define

The infima are attained: the finite-dimensional sets of channels and states are compact, and the displayed objectives are continuous.

All finite entropies are in bits:

with the continuous endpoint convention.

CPTP pre- and postprocessing do not increase $d_{\diamond }$. Consequently

Proof. For referenced states, trace distance contracts under a CPTP map. Taking the defining supremum gives channel contraction. For every decoder $\mathcal{D}$, the triangle inequality gives

Take infima and interchange the two channels.

For every isometry $V:C\to AE$, the complete channel ${\mathrm{Ad}}_V(X)=VX{V}^{†}$ has a deterministic CPTP left inverse. Moreover, for every operator $X$ on $QC$,

Proof. For any fixed code state ${\omega }_C$, define

Both terms are completely positive, their traces sum to $\mathrm{Tr}Y$, and $\mathcal{L}(VX{V}^{†})=X$. An isometric embedding preserves all nonzero singular values. Apply this observation also to $I_Q\otimes V$.

Complete preservation does not determine which output factor supports recovery. The following standard error-correction equivalence is proved to fix the complete complement and the decoder domain KnillLaflamme,CheungHawk04B,CheungEm05B.

For $V:C\to AE$, the following are equivalent:

1. There is a CPTP decoder $\mathcal{D}:A\to C$ with $\mathcal{DN}={\mathrm{id}}_C$.

2. $\mathcal{C}={\mathcal{T}}_{\tau }$ for a state ${\tau }_E$.

3. In a code basis, $\mathcal{C}(|i⟩⟨j|)={\delta }_{ij}{\tau }_E$.

If $r=\mathrm{rank}{\tau }_E$, there are a state ${\sigma }_K$, $\dim K=r$, and an isometry $T:C\otimes K\to A$ such that

An exact decoder on the entire accessible input space is

All equalities preserve arbitrary references.

Proof. Matrix units span $\mathcal{B}({\mathcal{H}}_C)$, proving (ii)–(iii).

Assume (ii), and diagonalize the nonzero part of the residual state:

Write

The complementary matrix-unit equations imply

Thus $T(|i⟩|\alpha ⟩)=|u_{i\alpha }⟩$ is an isometry. Taking ${\sigma }_K={\sum }_{\alpha }{\lambda }_{\alpha }|\alpha ⟩⟨\alpha |$ gives (eq:normalform). Formula (eq:exactdecoder) is CP, trace preserving, and recovers each matrix unit.

Conversely, dilate an exact decoder by an isometry $W:A\to CG$, and set

This is a dilation of the identity channel. The identity channel has rank-one Choi operator. Therefore every Kraus operator obtained by expanding $Z$ in an environment basis is a scalar multiple of $I_C$: a sum of positive rank-one Choi operators can have rank one only if each summand has its vector in that same one-dimensional range. Consequently

for a fixed normalized vector $\chi$. Tracing $CG$ gives a fixed state on $E$, proving (ii).

Equality of linear maps on all matrix units remains equality after tensoring with ${\mathrm{id}}_Q$.

If $\mathcal{N}$ is replacement, $E$ exactly recovers the code. If $\mathcal{C}$ is replacement, $A$ exactly recovers it. For $d\ge 2$, both complementary outputs cannot independently recover the entire code. At exact recovery,

Proof. Interchange $A$ and $E$ in Theorem thm:exact. If $A$ recovers, its complement is constant and cannot itself recover two distinct states. The rank bound follows from the orthonormal family $\{u_{i\alpha }\}$.

This is the relevant no-hiding conclusion BraunsteinPati. It does not assert that every nonconstant complement independently contains a recoverable copy of the code.

For a pure logical input, (eq:normalform) gives

Thus exact recovery can coexist with positive raw radiation entropy. The decoded logical output is pure; the accessible syndrome need not be. A pure fixed complement is the special case $r=1$.

For a qubit code, let

Both channels are computational dephasing, not replacement. For any replacement state ${\tau }_A$,

Indeed, the two accessible basis outputs have trace distance one; at least one has distance at least $1/2$ from ${\tau }_A$. Appending an arbitrarily large fixed residual spectator changes neither channel nor this bound. Conversely, an arbitrarily large fixed residual can accompany the identity accessible channel.

Let

Every replacement accessible channel has

Also,

Consequently every accessible channel satisfies ${\epsilon }_{\mathrm{rec}}(\mathcal{N})\le a_d$.

Proof. A decoder following a replacement channel prepares a fixed state ${\omega }_C$. Its output on ${\Phi }_d$ is $I_Q/d\otimes {\omega }_C$, with Bell overlap $1/d^2$. Testing the projector ${\Phi }_d$ gives trace distance at least $a_d$.

For the matching upper bound, let $W_{ab}$, $0\le a,b<d$, be the generalized Weyl unitaries, with $W_{00}=I$. Their uniform twirl is

Writing the twirl as its identity term plus its remaining CPTP average gives

for a channel $\mathcal{M}$. The diamond distance between two channels is at most one in our normalization, so the upper bound is $a_d$. The Bell test gives equality. Ignoring any accessible input and preparing $I_C/d$ proves the universal upper bound.

If

then

The replacement state may be pure or mixed.

Proof. For every decoder, contraction and the triangle inequality give

Take the infimum.

A nearly replacement channel need not have high entropy: replacement by a fixed pure state has zero output entropy. Thus “featureless”, “thermal”, and “high entropy” are not synonyms.

A purely classical output channel has the form

A quantum-return decoder receiving only this record can prepare states ${\omega }_y$, giving

No input-correlated quantum side resource is supplied to this decoder.

For every purely classical output channel,

The bound is tight: complete computational-basis measurement has optimal recovery error exactly $b_d$. The bound applies to an arbitrarily long adaptive classical record, not merely to one measured cell.

Proof. On a Bell input, every decoded classical-output channel produces a separable state on $QC$, as is immediate from its measure-and-prepare form. For a product state,

Convexity gives the same bound for every separable state. The Bell-projector test therefore gives recovery error at least $1-1/d$. This is the entanglement-breaking mechanism HorodeckiShorRuskai; compare the local bounds in CheungHawk04B,CheungHawk04C.

For tightness, measure the basis and prepare the observed basis state. The composite is dephasing:

For $Z={\sum }_i{e}^{2\pi ii/d}|i⟩⟨i|$,

Separating the identity term gives $d_{\diamond }(\Delta ,{\mathrm{id}}_C)\le 1-1/d$. The Bell lower bound gives equality.

Any finite adaptive protocol returning only classical data has a POVM indexed by its complete reported histories. It therefore falls under the same argument, regardless of the number of outcomes.

For $d\ge 2$, a purely classical accessible output cannot have a constant complete complement. This remains true after the final named residual register has been consumed.

Proof. A constant complete complement would imply exact recovery by Theorem thm:exact, contradicting Theorem thm:classical.

The word “until” must therefore be removed from the proposed photograph and photocopy recovery bounds. No later classical collection size removes the whole-code obstruction.

Let a measured register $M$ have basis $\{|y⟩\}$, and let $K$ be the quantum factor explicitly retained. Define

The measured name $M$ is absent from the returned interface.

An isometric dilation of destructive basis readout is

where $L$ is unreturned. For a superposition,

The complete channel preserves the superposition, while the returned record is dephased.

Proof. The columns of $W_M$ are orthonormal. Tracing $L$ gives (eq:enter) without $K$; tensoring with $K$ gives the general statement.

The returned register $Y$ is classical precisely because its dephasing environment is unreturned. A coherent copy that remains jointly accessible is a different interface.

A finite adaptive tree of isometries, unitaries, destructive basis readouts, and declared discards defines a reference-preserving CPTP channel. Every unreturned dilation factor belongs to its complete complement.

Proof. Compose the linear operators along each complete history $h$ to obtain a Kraus map $K_h$. Completeness of each readout and unitarity or isometricity of each coherent step imply, by backward induction,

The reported channel is a sum of the maps $X\mapsto K_{h}X{K}_h^{†}$, with orthogonal classical labels for reported histories and with any additional unreported indices traced out. This is CPTP. A dilation retaining every index is isometric. Tensoring with ${\mathrm{id}}_Q$ proves reference preservation.

This is the finite instrument architecture used in Cheung's sequential and hybrid formulations CheungEn02C,CheungVIIG. It does not supply a gravitational or thermodynamic implementation.

Let

where the ${\mathcal{F}}_y$ are CP and ${\sum }_y{\mathcal{F}}_y$ is TP. Then $\mathcal{H}$ exactly recovers the entire quantum code if and only if:

1. there are input-independent probabilities $p_y$ with

   $$\mathrm{Tr}{\mathcal{F}}_y(X)=p_y\mathrm{Tr}X;$$

2. for every $p_y>0$, the normalized channel ${\mathcal{N}}_y={\mathcal{F}}_y/p_y$ has an exact quantum-return decoder on $K$.

Thus an input-informative retained classical record is incompatible with exact whole-code recovery, even when a quantum factor is also returned.

Proof. A complete dilation can retain a copy of the outcome $y$ in its environment. If $\mathcal{H}$ is exactly recoverable, its complete complement is constant by Theorem thm:exact. Measuring the environment's outcome label proves (i).

Restrict an exact decoder to the $y$-th diagonal input block to obtain a CPTP map ${\mathcal{D}}_y:K\to C$. Then

The identity has rank-one Choi operator. Positivity implies that every normalized summand has that same Choi operator, hence ${\mathcal{D}}_y{\mathcal{N}}_y={\mathrm{id}}_C$. This proves (ii).

Conversely, read the classical label and apply its branch decoder. The resulting channel is ${\sum }_y{p}_y{\mathrm{id}}_C={\mathrm{id}}_C$. If $p_y=0$, its CP map has zero effect and is identically zero, so such branches require no decoder.

Classical correction labels can therefore accompany quantum recovery when a genuine quantum carrier remains accessible. This theorem does not identify the correction labels themselves with the quantum code.

For a classical register, let ${\mathsf{dropBit}}_Y={\mathrm{Tr}}_Y$.

For every operator $X_{AM}$,

The identity remains valid with every untouched reference. An additionally authorized residual Enter followed by drop leaves the pre-existing channel to $A$, and its recovery error, unchanged.

Proof. Write

Readout retains only the diagonal blocks:

Dropping $Y$ gives ${\sum }_y{X}_{yy}={\mathrm{Tr}}_M{X}_{AM}$. The calculation holds for operator-valued blocks on $QA$. Applied to the output of a complete process, it gives exactly the same channel to $A$ as tracing $M$ without the extra readout.

This identity is also recorded in CheungHawk04B,CheungEn07A,CheungEn07B. It does not imply physical annihilation, a final energetic burst, or a successful decoder.

If the terminal outcome is kept, the accessible output becomes $AY$, which is a different channel from $A$. Theorem thm:drop concerns keeping $A$ and discarding the additional outcome. The two tasks must not be conflated.

An append-only record at stage $n$ is $Y_{\le n}=(Y_1,\dots ,Y_n)$. Earlier values are neither erased nor relabelled by subsequent operations. Later physical operations are trace preserving, so the earlier record law remains its marginal law.

For every fixed input state and append-only classical record,

If a quantum factor $K_n$ is also retained, its hybrid output has

In particular, once the record has positive entropy, the raw classical–quantum output cannot later become pure while that record remains unchanged.

Proof. Classical conditional Shannon entropy is nonnegative, giving (eq:recordmonotonicity). For a block-diagonal state ${\sum }_y{p}_y|y⟩⟨y|\otimes {\rho }_y$, its eigenvalues are $p_y$ times those of ${\rho }_y$. Substitution in the entropy definition gives (eq:hybridentropy).

For a separate classical message $L$,

Additional photographs may reduce uncertainty {about $L$} while the entropy of the complete album increases or stays constant. Entropy of an erased, compressed, postselected, or relabelled summary is another quantity. It is not the entropy of the original complete append-only record.

Conditioning on one observed history makes its classical value definite. It does not prove that the averaged channel returns a pure state or preserves entanglement with a reference.

We now construct a quantitative model rather than infer a Page-type shape from dimensional language. It reproduces the flagged-transfer calibration of Cheung's Hawk02-B in a fully specified finite setting CheungHawk02B. It is not a microscopic black-hole model.

Let

Initially the code occupies $M$, and every cell is vacant. For $0\le {\theta }_j\le \pi /2$, define a unitary $U_j$ on $M{R}_j$ by the rotations

It is identity on the orthogonal complement of these mutually orthogonal two-dimensional subspaces.

Let $P_M$ and $P_{R_j}$ project onto the occupied logical sectors. For $\omega >0$, declare

Each $U_j$ is unitary and commutes with $H_0$. After $n$ transfers, let

For $q_n<1$, there is an isometry $J_n:C\to R_{\le n}$ such that the complete prepared evolution is

Here $|\varnothing {⟩}_R$ means that all emitted cells are vacant. Future vacant cells are fixed spectators in the complete residual.

Proof. The rotations are orthogonal unitary blocks. Their two basis vectors have the same $H_0$ eigenvalue $\omega$, so the rotations commute with $H_0$.

Let $|j,i{⟩}_R$ denote label $i$ in cell $j$, with every other emitted cell vacant. Induction gives the emitted amplitudes

They satisfy

Thus

has orthonormal columns and proves (eq:flaggedisometry). At $q_n=1$, the emitted term is zero and no $J_n$ is required.

The pulse parameters are externally scheduled. Commutation with $H_0$ proves conservation of this declared Hamiltonian, including its probability distribution. It does not account for every actuator, pulse-generation resource, or cyclic control reset.

Suppress the stage index and write $q=q_n$, $J=J_n$. The channels are

The occupied and vacant blocks are orthogonal.

For the schedule above,

For a pure referenced input ${\rho }_{QC}$, with marginal ${\rho }_C$,

All recovery claims are deterministic and reference preserving.

Proof. Let $P=J{J}^{†}$. The decoder

is CPTP and satisfies

Lemma lem:replacement gives the upper bound $q{a}_d$. On a Bell input, the occupied branch contributes at most $1-q$ to decoded Bell overlap. The vacant branch is input independent and contributes exactly $q/d^2$. The Bell test proves the matching lower bound. Endpoint $q=1$ is replacement.

For leakage, compare ${\mathcal{C}}_q$ to the replacement state

The difference is $q({\mathrm{id}}_C-{\mathcal{T}}_{I_C/d})$ on the occupied block, so the distance is $q{a}_d$, but this concerns the channel with occupied weight $q$, namely ${\mathcal{C}}_q$. Its distance to replacement is therefore $q{a}_d$. Accordingly, the asserted expression (eq:flagleak) would have the weights reversed. The correct complete leakage is

To prove the lower bound for (eq:correctflagleak), use the Bell input and the following effect: Bell projection on the occupied reference–memory block, together with the full vacant reference block. The actual output has probability one. A replacement output whose occupied probability is $s$ has probability

This alone gives a lower bound $s{a}_d$, not $q{a}_d$ uniformly in $s$, so we use a more precise block trace-norm calculation.

Pinching a candidate replacement state into the occupied and vacant blocks cannot increase distance. Write it as $s\sigma \oplus (1-s)$, and twirl its occupied state using logical Weyl unitaries. Covariance of the channel and convexity show that one can take $\sigma =I_C/d$. On the Bell input, the occupied block difference is

Its trace norm is

The vacant block difference has trace norm $|s-q|$. If $s\le q$, the sum is $2q-2s/d^2\ge 2q{a}_d$. If $q\le s\le q{d}^2$, it is $2s{a}_d\ge 2q{a}_d$. If $s\ge q{d}^2$, it is $2(s-q)\ge 2q{a}_d$. Thus every replacement has half-trace distance at least $q{a}_d$, and the upper bound proves (eq:correctflagleak).

The entropy formulas follow from orthogonal blocks. For a pure referenced input, the occupied $QR$ branch is pure, whereas its vacant branch has state ${\rho }_Q$, giving $S(QR)=h_2(q)+qS({\rho }_C)$. Substitution in mutual information gives (eq:referenceinformation).

theorem statement. The leakage formula used throughout the rest of the paper is

Equation (eq:flagleak) is retained only as an explicitly debugged reversed-weight formula. The residual recovery error, which is a different quantity, is

This follows by interchanging the occupied and vacant weights in the recovery proof. It must not be substituted for residual leakage.

For pure unreferenced inputs,

A schedule taking $q$ from one to zero exhibits an entropy peak at $q=1/2$. Nevertheless,

both decrease with transfer. Their maximum occurs initially, not at the entropy peak.

For a Bell input, the final radiation entropy is ${\log }_{2}d$, not zero. Exact recovery preserves its entanglement with $Q$; it does not make the referenced radiation marginal pure.

Measure the accessible occupied basis $\{J|i⟩\}$ and its vacant flag. Complete this to a basis on unused accessible subspaces. The resulting classical channel is

The classical flagged channel has

Thus its endpoint error at $q=0$ is $b_d$, not zero. For a pure input $|\psi ⟩={\sum }_i{\alpha }_i|i⟩$, its snapshot record entropy is

A family of snapshot entropies is not an append-only history entropy.

Proof. Prepare the observed basis state on an occupied outcome and prepare $I_C/d$ on the erasure outcome. The composite channel is

The triangle inequality gives the upper bound in (eq:measuredflag).

On the Bell input, every occupied classical branch has conditional reference state $|i⟩⟨i|$, so its contribution to Bell overlap is at most $(1-q)/d$, summed over all occupied outcomes. The erasure contribution is $q/d^2$. The Bell test gives

proving optimality. The entropy formula follows from the classical orthogonal flags.

Take

Then $q_0=1$, $q_1=1/2$, and $q_2=0$. For a basis input, the coherent radiation entropies are

The final coherent radiation is a superposition of emission in the first cell and emission in the second cell.

If each cell is destructively read out when returned, the final record has two equally likely histories:

Its cumulative record entropies are

For an arbitrary code input, the final classical record additionally reports the measured logical basis label. Its quantum recovery error is $b_d$. The coherent time superposition has not survived as a classical album.

Suppose each later transfer acts only on the complete residual and returns an additional quantum factor, leaving earlier accessible factors untouched. Then

Proof. The earlier accessible channel is obtained from the later one by tracing the newly returned factor. A later decoder may therefore discard that factor and use an earlier decoder.

The later complete residual channel is a CPTP postprocessing ${\Lambda }_n{\mathcal{C}}_n$. For every replacement comparison,

Take the infimum.

If earlier accessible quantum factors are destructively measured, edited, or lost, the required marginal relation can fail. The proposition does not silently preserve factors that the program has consumed.

Fix a state ${\sigma }_K$, a purification $|\chi {⟩}_{KP}$, and $L\simeq C$. Use the same output spaces

The isometries

have, for every pure logical input, identical accessible spectra and entropy $S({\sigma }_K)$, but

For $H_A=I_{C_A}\otimes H_K$, their energy-measurement laws also agree for every logical input. One may choose ${\sigma }_K$ Gibbs.

Proof. Their accessible channels are

For pure $X$, both outputs have the nonzero spectrum of ${\sigma }_K$. Tracing $K$ decodes the first channel. The second is replacement, so Lemma lem:replacement applies. Every spectral effect of $H_A$ acts trivially on $C_A$, proving equality of the selected energy laws.

This construction is the finite separation recorded in Cheung's En07-C and related comparisons CheungEn07C,CheungOai265A. It does not claim equality of the channels on all inputs.

There are two energy-conserving sequential complete isometries with the same accessible spaces and complete residual spaces such that:

1. every pure logical input has the same radiation entropy at every stage in both models;

2. for one fixed logical input, their accessible density operators agree at every stage;

3. at complete transfer, one model has recovery error zero and complete leakage zero, while the other has recovery error $a_d$ and complete leakage $a_d$.

Proof. Use the schedule of Section sec:schedule, append a degenerate register $L\simeq C$, and keep $L$ inaccessible.

In model $G$, run the schedule on the variable input and leave $L$ fixed in $|0⟩$. In model $L_{\mathrm{ret}}$, retain the variable input in $L$ and run the schedule on a dummy input fixed in $|0⟩$. Both are isometries into the same output spaces. The transfer unitaries conserve (eq:energy); the additional logical labels in $L$ are degenerate and do not alter that ledger.

For pure inputs, the radiation entropy in $G$ is $h_2(q_n)$. The dummy schedule has the same entropy. For input $|0⟩$, the radiation states agree exactly at every stage.

At $q_N=0$, model $G$ returns $J_N|\psi ⟩$ and has fixed complete residual. Model $L_{\mathrm{ret}}$ returns the fixed state $J_N|0⟩$, while its complete residual retains $|\psi {⟩}_L$. Replacement recovery has error $a_d$. The residual channel is an isometric copy of the code with fixed spectators, whose distance to replacement is $a_d$ by Lemma lem:replacement and reversible compression.

Cheung's Hawk01-B gives a sequential entropy-history separation of this kind CheungHawk01B. The proof above displays the hidden memory explicitly. It does not compare two models under identical assumptions of complete access and a unique empty endpoint.

An entropy bound in a separately specified independent-factor model does not estimate the diamond distance of an unspecified complete complement. The independent-factor hypotheses in the supplied Hawk03-A comparison must remain visible CheungHawk03A. We use no such bound to identify a time of maximal leakage or to discharge a recovery certificate.

The only dimension-independent approximation theorem imported below is the continuity of Stinespring dilations and its information–disturbance consequence KSW. Its finite-dimensional form implies

The norm convention is important.

For a complete interface,

In particular, a purely classical accessible channel satisfies

Proof. Substitute ${\inf }_{\mathcal{D}}{\Vert \mathcal{DN}-\mathrm{id}\Vert }_{\diamond }=2{\epsilon }_{\mathrm{rec}}$ and ${\inf }_{\tau }{\Vert \mathcal{C}-{\mathcal{T}}_{\tau }\Vert }_{\diamond }=2{\delta }_{\mathrm{full}}$ into the displayed information–disturbance inequalities. The universal cap $a_d$ is Lemma lem:replacement. The classical lower bound follows from ${{\epsilon }_{\mathrm{rec}}}^2\le 2{\delta }_{\mathrm{full}}$ and Theorem thm:classical.

Thus a nearly constant complete complement supplies a recovery upper bound, while a nearly constant accessible channel supplies the lower bound in Proposition prop:featureless. The directions cannot be exchanged.

Define the normalized Choi operator of a channel with code input by

For channels $\mathcal{F},\mathcal{Q}$ with input dimension $d$,

Proof. The first inequality uses the Bell input. For a pure referenced input, write

allowing $T$ to be rectangular for a larger reference. Then ${\Vert T\Vert }_{\infty }^2\le d$, and its output difference is

Its trace norm is at most $d$ times the Choi trace norm. Convexity handles mixed referenced states. The state characterization of channel diamond distance gives the result Watrous (Chapter 3).

An endpoint certificate supplies:

1. Code and complete preparation. A justified finite code, a specified preparation map, all ancillary purifications, and the external-reference task.

2. Complete evolution. A complete unitary or isometric description including every lost output and measurement environment.

3. Access-preserving output identification. A declared accessible output, its complete complement, and any proxy identifications. These identifications preserve the two sides separately; they do not mix accessible and inaccessible outputs.

4. Complete-complement certificate. Exact replacement of the complete complement, or an explicit norm/Choi comparison proving approximate replacement on the whole code.

These obligations express the same-model discipline in CheungEn07B,CheungHawk04D. They are not supplied merely by termination, a selected spectrum, or a mass-loss estimate.

Suppose Definition def:four is satisfied. After the specified access-preserving identifications, let $\mathcal{V}$ and $\hat{\mathcal{V}}$ be actual and proxy complete channels with

Let $\hat{\mathcal{C}}$ be the proxy complete complement, and suppose a feasible state ${\tau }_E$ satisfies

Then

The recovery bound is realized by a fixed proxy decoder, unless the constant-decoder cap is smaller.

If a separately supplied proxy decoder has normalized decoded Choi discrepancy

the same decoder on the actual accessible output has error at most

Terminal residual Enter followed by drop changes none of these pre-existing accessible channels.

Proof. Partial traces contract the complete-channel comparison, giving accessible and complementary distances at most $b$. Lemma lem:choi gives

The triangle inequality proves (eq:endleak). Proposition prop:approx supplies a proxy decoder with error at most $\sqrt{2de}$. Applying that fixed decoder to the actual accessible channel adds at most $b$. The constant decoder gives the alternative cap $a_d$. The same Choi comparison and contraction prove the specified-decoder bound. The final assertion is Theorem thm:drop.

A certified constant complete complement proves quantum recovery, not purity of an arbitrary raw record. A purely classical endpoint with $d\ge 2$ cannot satisfy exact obligation (E4).

An information-free residual may still have positive dimension and positive energy. For example,

has zero leakage and residual energy $ϵ$.

Proof. Use Remark rem:purity and Corollary cor:classicalendpoint. The displayed residual channel is replacement by $|1⟩⟨1|$, so Theorem thm:exact applies directly.

No gravitational instance of Definition def:four has been constructed in this paper.

Let $H$ be a fixed finite-dimensional Hamiltonian and let $V_t=e^{-itH}{V}_0$ describe a fixed prepared interface. There are arbitrarily late times with complete-channel distance arbitrarily close to that at $t=0$. If the initial accessible channel is replacement, then

Proof. There are finitely many eigenvalues. Simultaneous recurrence of their phases gives arbitrarily late times for which $e^{-itH}$ is arbitrarily close to the identity in operator norm. For isometries,

Indeed, apply the pure-state trace-distance bound to every referenced input and then use convexity. Partial trace gives accessible-channel recurrence. Lemma lem:stability gives recovery-error recurrence. The initial error and the universal upper bound are both $a_d$.

The proposition concerns a fixed finite autonomous closed proxy, not the externally controlled terminal schedule in Section sec:schedule. An outgoing limit or a changing access model needs its own justification CheungHawk04A,CheungHawk04D.

The ledgers are:

These are distinct mathematical objects. Their numerical values are not being compared by a universal inequality.

Equality of reduced channels, including a reset channel or partial trace, does not determine heat, work, or cyclic reusable-memory reset cost.

Proof. Let $M$ be a qubit and $L$ an independent blank qubit. For every referenced input,