Em05-B Excitation Transfer and Referenced Information Recovery Complementary Leakage, Flagged Decoding, and Population-Preserving Noise in a Signed Fermionic Channel

We study a specified even-parity excitation-qubit extraction isometry, its accessible channel, and its complete complement. Exact formulas distinguish occupation transfer, occupation-basis coherence, subsystem purity, reference correlations, and deterministic quantum recovery.

For extraction probability $\eta$, we evaluate the complementary pair-distance leakage as $1-{\eta }^2$ and prove its recovery lower bound without importing another model. We restate and prove the constant-complement criterion for exact recovery. A specified decoder has half-diamond error $1-\eta$; its flagged instrument exposes off-code events without postselection. We also determine the maximal decoded Bell fidelity, $(1+4\eta -{\eta }^2)/4$, and construct a balanced decoder. Writing $t=1-\eta$, these give the improved deterministic recovery bounds

The endpoint optima are $3/4$ and $0$; an exact intermediate half-diamond optimum is not asserted.

A separately declared population-preserving dephasing experiment retains its additional environment. At full transfer its optimal recovery error is $(1-\lambda )/2$, although its Bell-input accessible marginal is independent of $\lambda$. Complete dephasing reaches the entanglement-breaking recovery floor $1/2$ while preserving perfect occupation delivery. Normalized-Choi and matrix-unit estimates provide explicit referenced certification interfaces. An independent continuum argument separates the abstract full-transfer endpoint from finite-time attainability. These are finite-channel results, not a gravitational evaporation model or a Page-curve construction.

Occupation transfer, coherence, entanglement, marginal purity, and recovery of an unknown quantum message are different diagnostics. A recovery task must specify both the accessible system and the required quantum output, including its correlations with an external reference.

This paper gives an exact comparison within one declared channel family. Its central separation is operational: a pure accessible marginal can contain no message information, whereas a mixed accessible marginal can support exact referenced recovery. Moreover, two experiments can have identical accessible marginals for a Bell input and nevertheless have different optimal recovery errors.

The framework is established. Exact recovery and error-correction conditions are developed by Knill and Laflamme KnillLaflamme; channel continuity and information–disturbance relations by Kretschmann, Schlingemann, and Werner KSW; and approximate recovery through an environmental fidelity optimization by Bény and Oreshkov BenyOreshkov. Our half-diamond metric is not identified with their fidelity metrics. Concurrence and negativity use the established definitions Wootters,VidalWerner.

Finite-dimensional evaporation benchmarks also have precedent Avery, and Hayden–Preskill recovery depends on its own dynamical and access assumptions HaydenPreskill. Neither construction is identified with the channel below.

The contribution here is the explicit signed channel calculation, its local recovery witnesses, a complete flagged decoder, improved intermediate recovery bounds, and a population-preserving noise comparison. All model-specific arguments are supplied in this source. Related manuscripts and the revision proposal motivate these questions but are not formal premises. No exhaustive novelty or first-in-literature claim is made.

scope. The parameter $\eta$ is an extraction probability, not time. The finite channel supplies no black-hole microstate identification, gravitational factorization, passive Hawking-emission mechanism, backreaction law, island construction, or gravitational Page curve. The continuum appendix concerns only its explicitly declared controlled model.

All finite-system states and channels use ordinary matrix representations. A channel is a completely positive, trace-preserving (CPTP) linear map. For matrices and states, respectively, set

For channels with the same input and output types, define

For differences of finite-dimensional channels, this is the supremum of the output trace distance over referenced input states; a reference of input dimension suffices Watrous (Chapter 3).

Let $M$ be the input logical qubit, $Q$ an external reference, $R$ the accessible two-mode output, $E$ the residual two-mode output, and $M^{\prime }$ the decoded qubit. Logical bases of $M$ and $M^{\prime }$ are identified when comparing a decoded channel with ${\mathrm{id}}_M$. All entropy logarithms are base two.

Set

For a channel $\Lambda :M\to M^{\prime }$, define

For a purification $|{\psi }_{\rho }{⟩}_{QM}$ of ${\rho }_M$, define

These are target overlaps, hence squared fidelities against pure targets.

For an accessible channel $\mathcal{N}:M\to R$, put

These are separate optimization problems. An optimal Bell-fidelity decoder need not minimize half-diamond error.

A replacement channel and a unitary channel are denoted by

The following tasks have different types.

1. Observable readout: for a specified POVM $\{F_y\}$, the target is the classical channel

   $$\rho ⟼\sum _y\mathrm{Tr}(F_y\rho )|y⟩⟨y|.$$

   No quantum preservation of the message is required.

2. Deterministic quantum recovery: the output is a qubit, and performance is measured by $d_{\diamond }(\mathcal{DN},{\mathrm{id}}_M)$.

3. Instrument output: for CP maps $\{{\mathcal{I}}_y\}$ whose sum is trace preserving, the output includes both a classical outcome and a conditional quantum state:

   $$\hat{\mathcal{I}}(X)=\sum _y|y⟩⟨y|\otimes {\mathcal{I}}_y(X).$$

   Matching only the outcome probabilities does not match this quantum–classical channel.

An instrument distance requires a target with the same output type. A flagged qubit output cannot be compared directly with an unflagged identity channel.

A CPTP map contracts trace norm on Hermitian inputs and therefore contracts trace distance. Half-diamond distance contracts under fixed CPTP pre- and postprocessing. For states and an effect $0\le F\le I$,

For normalized vectors $x,y$,

In particular,

Proof. For Hermitian $X=X_{+}-X_{-}$, positivity and trace preservation give

The same argument applies with a reference. Preprocessing restricts the referenced input states available in the supremum; postprocessing contracts their output distances.

The positive and negative parts of $\rho -\sigma$ have equal trace ${\Vert \rho -\sigma \Vert }_1/2$, which proves the effect inequality. The difference of two rank-one states has support in their two-dimensional span; its nonzero eigenvalues are $\pm \sqrt{1-|⟨x,y⟩{|}^2}$. Finally, test the Bell-input output against $|\Phi ⟩⟨\Phi |$.

Both extrema in (eq:optimalError) and (eq:optimalBell) are attained.

Proof. The Choi matrices of CPTP maps of the fixed type are positive, have a fixed partial trace, and form a nonempty compact set. Bell fidelity and the channel norm are continuous on this finite-dimensional set.

A channel is entanglement breaking if its output is separable from every external reference for every referenced input state.

For a decoded replacement qubit channel,

For an entanglement-breaking qubit channel $\Lambda$,

Consequently an entanglement-breaking accessible channel obeys ${\epsilon }_{\mathrm{rec}}(\mathcal{N})\ge 1/2$.

For qubit Pauli channels, with $P\in \{I,X,Y,Z\}$,

In particular, $d_{\diamond }(\Phi ,{\mathrm{id}}_M)=1-a_I$.

Proof. A replacement channel has Bell output $I_Q/2\otimes \tau$, with Bell overlap $1/4$. A normalized product vector satisfies

by Cauchy–Schwarz. Convexity extends this bound to all separable states. Local decoding preserves separability. Equation (eq:BellWitness) proves the lower bounds.

For (eq:PauliComparison), the channel triangle inequality and ${\Vert {\mathrm{Ad}}_P\Vert }_{\diamond }=1$ give the upper bound. The four Pauli outputs on $|\Phi ⟩$ are orthogonal Bell projectors; their trace distance gives the matching lower bound.

Take

with ordered occupation bases

The occupations label two distinct modes or species. The fixed tensor representation is accessible bath first, residual second.

Let

Throughout the finite-channel sections, $t$ is a residual probability, not a time variable.

Define $V_{\eta }:{\mathbb{C}}^2\to {\mathcal{H}}_R\otimes {\mathcal{H}}_E$ by

The accessible and complete complementary channels are

For residual creators ${e_1}^{†},{e_2}^{†}$ and bath creators ${b_1}^{†},{b_2}^{†}$, expand

Putting bath creators before residual creators gives the minus sign in (eq:Vone). Both logical basis vectors have even total fermionic parity. The reference is an abstract logical register, not another occupation mode.

All partial traces and entanglement cuts below refer to the declared tensor representation. No basis-independent fermionic mode-entanglement identification is asserted.

If $q=\sqrt{\eta }{e}^{\mathrm{i}\alpha }$, the uncorrected occupied column has coefficients $q^2,q\sqrt{t},-q\sqrt{t},t$. The accessible correction $e^{-\mathrm{i}\alpha N_R}$ yields Definition def:extractor. This phase correction is fixed independently of the logical input. Its implementation is a separate certification obligation.

The map $V_{\eta }$ is an isometry and both declared channels are CPTP. A Kraus family for ${\mathcal{N}}_{\eta }$, indexed by residual occupation, is

The complete complement is

These channel identities apply with arbitrary external references.

Proof. The columns are orthogonal and the occupied column has squared norm

Hence ${V_{\eta }}^{†}{V}_{\eta }=I_M$. Reading its residual coefficients gives (eq:kraus), with

Taking the other partial trace gives (eq:complement). Tensoring the isometry with a reference proves the referenced versions.

Write

In the declared bases,

\rho_R
 &=
 \begin{pmatrix}
 1-p+pt^2&0&0&\eta z\\
 0&p\eta t&0&0\\
 0&0&p\eta t&0\\
 \eta z^*&0&0&p\eta^2
 \end{pmatrix},
 \label{eq:rhoR}\\
 \rho_E
 &=
 \begin{pmatrix}
 1-p+p\eta^2&0&0&tz\\
 0&p\eta t&0&0\\
 0&0&p\eta t&0\\
 tz^*&0&0&pt^2
 \end{pmatrix}.
 \label{eq:rhoE}

For the respective occupation operators

one has

The accessible purity is

The residual purity is obtained by interchanging $\eta$ and $t$.

Proof. Insert (eq:input) into the Kraus expressions. Only $A_{00}$ contributes the accessible $00,11$ coherence. Equation (eq:complement) gives the residual coherence $tz$. The occupation formulas follow by tracing against $N_R,N_E$, using

Finally, for Hermitian $\rho$, $\mathrm{Tr}({\rho }^2)={\sum }_{i,j}|{\rho }_{ij}{|}^2$, giving (eq:purity).

For every logical state,

The nonzero eigenvalues of $V_{\eta }{\rho }_M{V}_{\eta }^{†}$ are those of ${\rho }_M$. Every purified input has a pure joint referenced output.

Proof. The first equality uses $\eta +t=1$. An isometry embeds a spectral decomposition without changing its nonzero eigenvalues and maps pure vectors to pure vectors.

Occupation depends only on $p$, while coherence and purity also depend on $z$. None of these single-input marginal diagnostics is an all-input referenced recovery criterion.

Set

Its input marginal is $I_M/2$, so $p=1/2,z=0$.

For this Bell input,

including zero multiplicities. The residual spectrum follows by interchanging $\eta ,t$.

With $0{\log }_{2}0=0$ and $I(X:Y)=S(X)+S(Y)-S(XY)$,

In particular,

These spectra and numerical values are input-specific.

Proof. The Bell-input marginal matrices are diagonal. The joint state is pure and $S(Q)=1$. The two marginals of a pure bipartite state have the same nonzero spectrum, as follows by comparing the nonzero eigenvalues of its coefficient matrices $A{A}^{†}$ and $A^{†}A$. Thus $S(QR)=S(E)$ and $S(QE)=S(R)$. This proves the information identities; the displayed values follow from the spectra.

Equation (eq:balance) includes classical correlations. It does not state that both outputs contain independently recoverable copies of the message.

For the declared cut $Q|R$, define negativity VidalWerner by

For the Bell-input accessible state,

It is entangled for every $\eta >0$.

Proof. The coherence connects $|0,00⟩$ and $|1,11⟩$, with coefficient $\eta /2$. After partial transpose it enters the block on $|0,11⟩,|1,00⟩$:

The eigenvalues are

All other eigenvalues are nonnegative. The absolute value of the negative eigenvalue gives (eq:negativity). A separable state has positive partial transpose, so strict negativity proves entanglement.

This is reference entanglement across $Q|R$, not entanglement between the two accessible occupation modes.

The next criterion is an established exact-error-correction statement KnillLaflamme. Its proof is included to fix its domain and environmental meaning.

Let $V:{\mathcal{H}}_M\to {\mathcal{H}}_A\otimes {\mathcal{H}}_E$ be a finite-dimensional isometry, with accessible channel $\mathcal{N}$ and complement $\mathcal{C}$. The following are equivalent:

1. There is a CPTP map $\mathcal{D}:A\to M$ with $\mathcal{DN}={\mathrm{id}}_M$.

2. For some density operator ${\tau }_E$,

   $$\mathcal{C}(X)=\mathrm{Tr}(X){\tau }_E\text{for every }X\in \mathcal{B}({\mathcal{H}}_M).$$

3. For a Kraus family $\{A_a\}$ read from $V$,

   $${A_a}^{†}{A}_b=c_{ab}{I}_M$$

   for scalar coefficients $c_{ab}$.

Proof. Suppose (i). Compose $V$ with a Stinespring isometry $T:A\to M\otimes G$ for $\mathcal{D}$. The composite dilates the identity channel.

Every Kraus operator $B_j$ of the identity channel is proportional to $I_M$: its Choi matrix is rank one and is the sum of the positive rank-one matrices formed from the vectorized $B_j$, so each vectorized $B_j$ lies in the span of the vectorized identity. Therefore the composite isometry factors as

for a fixed normalized $\chi$. Tracing $MG$ leaves the original complement unchanged, hence proves (ii).

Conversely, suppose (ii). Diagonalize ${\tau }_E={\sum }_j{\lambda }_j|j⟩⟨j|$, omitting zero eigenvalues. The isometry has no component in the kernel of ${\tau }_E$. Write

The constant-complement identity on every matrix unit gives

Let $P_S$ project onto their span, and define an isometry $U:S\to M\otimes J$ by $U{\varphi }_{ij}=|i⟩|j⟩$. For an arbitrary fixed state ${\sigma }_M$, the map

is CPTP. Its action on $\mathcal{N}(X)$ is $X$, proving (i).

Finally,

This is a scalar multiple of $\mathrm{Tr}(X)$ for every $X$ exactly when every ${A_b}^{†}{A}_a$ is scalar. Thus (ii) and (iii) are equivalent.

The extractor admits exact CPTP recovery if and only if $\eta =1$.

Proof. For $t>0$,

is not scalar. Equivalently, the complementary basis outputs are unequal by (eq:rhoE). At $\eta =1$, the complement is vacuum replacement and the accessible map is an isometric encoding.

Let $\mathcal{N}$ have a qubit input and complement $\mathcal{C}$. For any orthonormal input pair $|0⟩,|1⟩$, put

Every deterministic decoder satisfies

Proof. Write $\epsilon =d_{\diamond }(\mathcal{DN},{\mathrm{id}}_M)$. Dilate the decoder by an isometry $T:A\to M^{\prime }\otimes G$ and apply the combined isometry to the Bell input, obtaining a pure state $|\Omega {⟩}_{Q{M}^{\prime }GE}$. Its Bell overlap $f$ satisfies $f\ge 1-\epsilon$ by the effect bound.

If $f=0$, then $\epsilon =1$, which already proves the claim. Otherwise project $\Omega$ onto $|\Phi {⟩}_{Q{M}^{\prime }}$. After normalization this gives $|\Phi ⟩\otimes |\chi {⟩}_{GE}$, whose squared overlap with $\Omega$ is $f$. The pure-state distance and partial-trace contraction give

Local decoding leaves the original $QE$ reduction unchanged.

Pinching $Q$ in its input-pair basis yields

The triangle inequality makes the left side at least $\ell /2$, giving (eq:pairBound).

For the declared extractor, define

where the suprema are over logical density matrices. These are unreferenced pair-distance diameters, not diamond distances to replacement channels.

For a unit Bloch vector $n=(n_x,n_y,n_z)$, let ${\rho }_{\pm }=(I\pm n\cdot (X,Y,Z))/2$. Then

Consequently,

and

The equatorial input pair gives the additional elementary bound ${\epsilon }_{\mathrm{rec}}({\mathcal{N}}_{\eta })\ge t/2$.

Proof. The input difference is $n\cdot (X,Y,Z)$. Its complementary image has odd-sector diagonal entries $-\eta t{n}_z$, twice, and an even-sector block

The two eigenvalues of this block are $t(\eta n_z+1)$ and $t(\eta n_z-1)$. Its trace norm is $2t$; the odd entries add $2\eta t|n_z|$. This proves (eq:environmentPair). Interchanging $\eta ,t$ gives (eq:accessiblePair).

Any difference of two qubit states is $rn\cdot (X,Y,Z)$ with $0\le r\le 1$. Thus the largest pair distances occur at $r=1,|n_z|=1$, which proves (eq:pairDiameters).

The computational-basis accessible outputs have distance $1-t^2$. If a decoder has error $\epsilon$, contraction and the triangle inequality give

so $\epsilon \ge t^2/2$. The complementary pair in the same basis attains $L_E$; Lemma lem:leakageBell gives its lower bound. For $n_z=0$, the accessible distance is $\eta$, yielding $1\le 2\epsilon +\eta$, hence $\epsilon \ge t/2$.

The upper bounds follow from the specified decoder of Section sec:specified and from ignoring the input and preparing $I_{M^{\prime }}/2$. The latter is the uniform Pauli channel and has error $3/4$ by Lemma lem:witnesses.

The complementary term in (eq:basicSandwich) exceeds $t^2/2$ when $\eta >\sqrt{2}-1$. At $\eta =1/2$, those two terms are $9/64$ and $1/8$. Neither is the best local witness available here: the equatorial pair gives $1/4$, and the optimized Bell witness below gives $5/16$. All these are proved lower bounds, not intermediate recovery optima.

Define the code isometry and projector

The decoder is

It is defined on the entire accessible space, not only the code.

The map ${\mathcal{D}}_{\star }$ is CPTP and

Its performance is

Its decoded Bell state

has concurrence $C(Q:M^{\prime })=\eta$.

Proof. A Kraus family is $W^{†}$, together with

Its effects sum to $I_R$, proving CPTP completeness.

The odd occupations contribute $p\eta t{I}_{M^{\prime }}$. The decoded lower-right entry is $p{\eta }^2+p\eta t=p\eta$, and its coherence is $\eta z$. This proves (eq:specifiedChannel).

The identity and replacement by $|0⟩$ have half-diamond distance one, attained on input $|1⟩$. Homogeneity proves (eq:specifiedError). The replacement term has Bell overlap $1/4$, proving (eq:specifiedBell). Every purified target overlap is at least $\eta$, and input $|1⟩$ attains $\eta$; this proves (eq:specifiedWorst).

For concurrence, use the spin-flip definition Wootters, with

In the basis $00,01,10,11$,

The square-root eigenvalues of $\omega \tilde{\omega }$ are

The positive part of the largest minus the other three is $\eta$.

Let $K$ be a two-dimensional classical register with labels $\mathsf{c},\mathsf{o}$, meaning code and off-code. Define CP maps

and the complete quantum–classical channel

The map ${\mathfrak{D}}_{\star }:R\to K{M}^{\prime }$ is CPTP and

Writing $B_{\eta }=\mathrm{diag}(1,\eta )$, its logical branches are

For input (eq:input), their probabilities are

For a referenced input ${\rho }_{QM}$, put $H_Q=⟨1{|}_M{\rho }_{QM}|1{⟩}_M$. The unnormalized branch outputs are

Whenever $p_y>0$, the conditional state is ${\tau }^y/p_y$. The unconditional reference obeys

Proof. The branch Kraus operators are precisely the code and off-code groups used in Theorem thm:specified. Their effects sum to $I_R$; attaching orthogonal classical labels makes their sum CPTP. Equations (eq:codeBranch)–(eq:offBranch) follow by grouping (eq:rhoR). Applying the same operator identities with a reference gives (eq:referencedCodeBranch) and (eq:referencedOffBranch). Trace preservation yields the final identity.

The code branch contains both coherent code transfer and residual-vacuum ambiguity. At $\eta =0$ it occurs with probability one, although no message is recoverable from $R$. Thus $\mathsf{c}$ must not be relabeled “successful recovery.”

Adding the flag leaves the flag-forgotten error $t$ unchanged. This statement concerns the declared decoder, not the possible value of the flag for other processing. Conditional normalization is nonlinear and is not a deterministic recovery channel. The no-signaling identity holds after summing all outcomes; conditional reference states may differ.

For the declared extractor,

It is attained by the complete decoder

Consequently,

Proof. Let $\mathcal{D}$ be any CPTP decoder and abbreviate

Put

All lie in $[0,1]$. If

a Kraus family $\{K_{\ell }\}$ and Cauchy–Schwarz give

The Bell overlap is therefore bounded by

Each step uses nonnegative coefficients and $a,b,r_j\le 1$.

The decoder ${\mathcal{D}}_1$ is CPTP, with Kraus operators $W^{†},|1⟩⟨01|,|1⟩⟨10|$. It has $a=b=r_2=r_3=k=1$, attaining the upper bound. Equation (eq:BellWitness) gives

Theorem thm:optimalBell solves only (eq:optimalBell). It provides a lower bound for (eq:optimalError); it does not identify ${\mathcal{D}}_1$ as a half-diamond-optimal decoder.

For $0\le x\le 1$, define a qubit channel

Its effects sum to $I_{M^{\prime }}$.

Put

and define

The balanced decoder is CPTP. Its decoded channel is unital and Pauli diagonal, with Bloch multipliers

Its Pauli probabilities are

Its exact error is

Thus

The endpoints are exact:

Proof. A complete Kraus family for the balanced decoder is

whose effects sum to $I_R$.

For a general value $x$, its decoded excitation probability on input (eq:input) is

and its coherence is $\eta \sqrt{1-x}z$. At $x=t^2/(1+t^2)$, the populations become $x+p{s}_{\eta }$, with $x=(1-s_{\eta })/2$. The channel is therefore unital with the stated Bloch multipliers. Expanding a unital Pauli-diagonal qubit channel gives (eq:balancedPauliI) and its companions.

The probabilities are nonnegative. In particular,

which proves $p_Z\ge 0$. Lemma lem:witnesses gives its exact error $1-p_I=U(\eta )$. The lower bound is Theorem thm:optimalBell; the other upper bounds were proved above.

At $\eta =0$, both $B$ and $U$ equal $3/4$. At $\eta =1$, both equal zero.

The optimized Bell lower bound dominates all the elementary lower bounds stated in Section sec:complementary:

At $\eta =1/2$, the improved sandwich is

As $t\downarrow 0$,

Proof. The first two comparisons follow directly from $B=t/2+t^2/4$. For the complementary term,

For $0\le t\le 1$,

Substitution gives the half-transfer values. Taylor expansion of (eq:balancedUpper) gives the endpoint asymptotics.

The bounds control the intermediate deterministic recovery problem and determine its leading error near full transfer. They do not solve its exact half-diamond optimum on $0<\eta <1$. Neither decoder is assigned unproved intermediate optimality.

This section declares a different experiment: extraction followed by output noise.

On the entire accessible space, put

It is Hermitian, unitary, parity preserving, and commutes with all occupation projectors and $N_R$.

For $0\le \lambda \le 1$, define