En02-C Resource-Limited Black-Hole Information Recovery Radiation Access, Decoder Memory, and Operational Recovery Thresholds

We distinguish information-theoretic recovery from executable decoding in explicitly specified finite unitary radiation models. Our principal benchmark has one message qubit, four initially pure internal qubits, and three consistent emission steps. The first two radiation prefixes are independent of the message and have optimal half-diamond recovery error $3/4$. The third prefix permits exact quantum recovery. For an ordered, single-pass observer without auxiliary quantum storage, however, the optimal error at one-live-qubit capacity is exactly $1/2$, even with unrestricted classical memory and arbitrary single-qubit instruments. The lower bound is proved directly on the logical-reference state: consuming the first carrier without quantum memory makes every subsequent recovered output separable from the reference. It does not assume approximate uniqueness of an off-code decoder. A seven-gate Clifford programme attains exact recovery with two live qubits and two classical record bits. We separately compile the conjugate-evolution, Bell-projection construction of the idealized Hayden–Preskill decoder. Under exact complementary-channel decoupling, its success map is $2^{-2k}\mathrm{id}$ for a $k$-qubit message. Uniform-angle coherent amplification has an explicitly priced register and oracle ledger; one amplification iteration is exact for $k=1$. Laboratory occupancy and external radiation storage are recorded separately. Complete-channel implementation errors add, and a specified flagged-erasure model has optimal error $3\ell /4$. Operational thresholds and completion-time bounds are derived only for the declared interfaces and schedules. The results are finite quantum-information statements, not a derivation of evaporation dynamics, a Page curve, or a solution of the information paradox.

Hayden–Preskill recovery concerns information retrieval in an idealized unitary, rapidly mixing model with extensive radiation control HP. Constructive decoding, including conjugate-evolution postselection and coherent amplification, is available in specified idealized settings YK. Clifford dynamics admits additional structured recovery procedures YoshidaClifford. These results must be distinguished from the resources of a particular observer.

We separate three questions:

1. Availability: what message information is present in the accessible radiation?

2. Existence: does a recovery channel achieve the requested reference-preserving accuracy?

3. Execution: can a permitted programme realize such a channel within its memory, circuit-access, record, and time budgets?

The separation itself is not claimed to be new. Our contribution is an exact finite example with a quantitatively optimal memory-constrained error, together with a resource ledger for a recognized constructive recovery architecture.

The principal separation is

The middle equality optimizes over all permitted one-live-qubit decoders, not only a chosen circuit.

En02-A supplies the distinction between specified Born semantics and an unknown-strategy characterization, as well as one-shot versus adaptive accounting En02A. En02-B supplies finite decoder interfaces and a perfect-code authorized-triple construction En02B. Neither supplies the sequential radiation maps or the quantitative approximate-memory obstruction proved here.

All principal theorems below use specified Born semantics. The emission evolution, its initial state, and the observer interface are stated independently. We do not use unrestricted finite-capacity measurement uniqueness as a premise.

The primary benchmark has trivial early radiation. A separate Hayden–Preskill benchmark uses maximally entangled early radiation. The former is not presented as an old-black-hole or generic-scrambling model.

A finite unitary radiation model is not, merely by being unitary, a model of Hawking spectra, gravitational backreaction, or rapid scrambling. Our three-step benchmark is a partial-emission model with a two-qubit remnant. It is not a complete evaporation history.

No entanglement wedge, island, quantum extremal surface, Page curve, or replica wormhole is constructed. No result changes the status of a field-theoretic insertion or symmetry problem in another manuscript. En02-C and Em02-C, if similarly named documents exist, are not identified.

All Hilbert spaces are finite dimensional. Channels are completely positive and trace preserving, abbreviated CPTP.

Let $M$ be the message, $R$ an arbitrary external reference, $B$ the initial internal register, and $E$ accessible early radiation. The initial state ${\sigma }_{BE}$ is independent of the message-reference input.

An injection unitary identifies $MB$ with $C_0$. Thereafter,

is a specified dimension-preserving unitary identification, possibly preceded by gates on $C_{j-1}$. Emitted carriers never interact again with the internal evolution.

Set

The consistent evolution is

obtained by composing injection and the first $t$ emission maps. The accessible channel is

The observer receives $E{D}_{\le t}$, not $C_t$. We identify the decoder's output register $\hat{M}$ with $M$ through a fixed computational basis.

For a decoder ${\mathcal{D}}_t:E{D}_{\le t}\to \hat{M}$, define

The diamond norm includes arbitrary entanglement with an external reference Watrous (Chapters 2–3).

where the infimum is over all CPTP decoders without implementation restrictions.

A measurement of one specified message observable is a different task. A measurement instrument must reproduce both its classical outcomes and its specified conditional quantum outputs. Neither task is silently substituted for subsystem recovery.

If ${\sigma }_{BE}$ is mixed, purify it by a register $F$. The inaccessible complementary output is then $C_{t}F$, not necessarily $C_t$ alone.

A channel $\mathcal{N}:M\to A$ has an exact CPTP left inverse if and only if a complementary channel satisfies

for a fixed density operator $\tau$.

Proof. Suppose a recovery exists. Compose a Stinespring isometry for $\mathcal{N}$ with one for the recovery. This dilates the identity channel. Every Kraus operator in any Kraus family of the identity is proportional to $I_M$, since its Choi operator has rank one. The dilation therefore factors as the unchanged message tensored with a fixed environment vector. Tracing the recovery environment makes the original complement constant.

Conversely, diagonalize $\tau ={\sum }_j{\lambda }_j|j⟩⟨j|$, omitting zero eigenvalues. A Stinespring isometry can be written

The constant-complement condition gives $⟨{\varphi }_{i^{\prime }{j}^{\prime }}|{\varphi }_{ij}⟩={\delta }_{i{i}^{\prime }}{\delta }_{j{j}^{\prime }}$. An isometry on their span sends ${\varphi }_{ij}$ to $|i⟩|j⟩$. Trace the second factor and define an arbitrary trace-preserving action on the orthogonal complement. This is an exact recovery.

This is the exact information–disturbance principle. Channel-level continuity versions are established in KSW. One-shot decoupling gives other finite-size recovery tools Decoupling; it is not used to assign an unproved scrambling estimate to our specified circuit.

At an observer event $\tau$, let $q_{\mathrm{lab}}(\tau )$ count delivered radiation, working ancillas, retained quantum records, and recovered output not yet returned.

Let $q_{\mathrm{buf}}(\tau )$ count emitted accessible radiation held by passive external suppliers. Any other permitted auxiliary quantum storage is counted separately in $q_{\mathrm{aux}}(\tau )$.

Define

and

The inaccessible remnant and external correctness reference are not observer storage. Radiation not yet emitted is not counted as an external radiation buffer.

Thus a small laboratory peak does not establish small total radiation storage.

We record

Here $G$ is the worst-history elementary unitary gate count and $D$ the instruction depth including destructive readouts. Outcome storage $C_{\mathrm{cl}}$ excludes the fixed programme description.

When a construction requires $U^{\ast }$ or $U^T$, their counts $n_{U^{\ast }},n_{U^T}$ are reported additionally. These operations are not identified with $U$ or $U^{†}$. The external-storage ledger accompanies $\mathbf{C}$.

Accessible carriers are delivered once in their specified order. A passive supplier may hold undelivered carriers but performs no measurement, encoding, selection, or decoding.

The observer may pause delivery. It may not return an intermediate quantum register to the supplier, revisit consumed radiation, use an unpriced quantum buffer, share auxiliary entanglement, or export quantum output before the final return.

A delivered carrier is counted before any gate involving it. Receiving a new qubit while another remains live therefore costs two live slots, even if a gate immediately consumes one.

Every laboratory qubit is destructively measured or returned. Classical memory is unrestricted unless a bound is stated. The internal emission maps are fixed and are not controlled by the observer's measurement outcomes.

For the lower bound, arbitrary single-qubit instruments and arbitrary classical computation are permitted at one-live-qubit capacity. For the constructive benchmark, the exact library is

together with zero-state allocation, destructive computational readout, classical control, and return. Any two laboratory qubits may interact. A classically selected Pauli costs at most one gate. This is not a quantum-controlled Pauli gate convention.

Batch access places the whole accessible register in the laboratory initially. In particular, $e$ batch early-radiation qubits already force peak at least $e$.

The following interfaces are distinct:

1. a supplied explicit gate description;

2. a specified oracle for selected evolution operations;

3. radiation with a fixed prior description of the encoding;

4. radiation with unresolved uncertainty about the encoding.

The main streaming benchmark grants its fixed Clifford description, but no calls acting on the inaccessible remnant. The Hayden–Preskill construction later grants operations on observer-owned duplicate registers explicitly.

Memory cost with classical assistance is an established global optimization problem Bisio. Our live-name count specifies a particular access model and is not automatically identical to ancillary dimension in a general quantum strategy.

A certificate contains:

1. the injection, emission maps, and initial internal state;

2. the finite decoder instruction tree;

3. its branch path maps and reported classical coarse-graining;

4. an equality or error bound for the complete decoded channel;

5. a branchwise laboratory and external-storage ledger;

6. an explicit consumption and final-return specification.

No successful branch is substituted for the complete channel.

Under specified Born semantics and Assumption ass:stream, a legal finite programme with such a certificate implements its declared channel, including every reference, and obeys its stated storage ledgers.

Proof. Decoder operations act only on emitted names and fresh laboratory ancillas. Each subsequent emission acts on the remnant. These operations commute after identifying their distinct tensor factors. Semantically, one may therefore place the fixed emissions before the decoder operations, without claiming that this rearranged batch schedule has the same memory cost.

On each complete decoder history, compose allocation isometries, unitaries, and destructive measurement bras into a linear path map $K_h$. Branch normalizations cancel against their probabilities. Backward induction on the finite instruction tree gives

Reported selective maps are

Their sum is CPTP.

Equality on all input matrix units is equality of channels. Tensoring with a reference preserves it. A diamond-error certificate already controls every such extension.

The two ledgers count all named quantum factors at every actual event, including supplier holdings. Their maxima are precisely the declared resource bounds. Commutation in the correctness proof does not erase storage in the actual schedule.

For a fixed interface and library, define

The complete output is deterministic unless a success flag and its failure output are explicitly part of the task.

Increasing capacity or enlarging the implementation library cannot increase this infimum. Relaxing an error target cannot remove a feasible programme.

The message initially occupies position $2$. The four internal qubits occupy $1,3,4,5$ and begin in $|0000⟩$. Early radiation $E$ is one dimensional.

Write

Let $\mathcal{B}$ prepare Bell pairs on $1,4$ and $3,5$: chronologically,

On positions $1,2,3$, define

The rightmost operation is applied first. This is a nine-gate Clifford circuit.

Define the full encoding unitary

and its message isometry

The authorized-triple channel uses the decoder normal form of En02-B. The remnant purification and the following sequential dynamics are specified here.

Set the injection unitary to

The leftmost ${\mathrm{CZ}}_{2,3}$ cancels the last chronological gate of $J$. Thus injection uses twelve elementary gates.

After injection:

1. $W_1$ emits position $1$, leaving $C_1=(2,3,4,5)$;

2. $W_2$ applies ${\mathrm{CZ}}_{2,3}$ on the remnant, then emits position $2$, leaving $C_2=(3,4,5)$;

3. $W_3$ emits position $3$, leaving $C_3=(4,5)$.

Emission itself is a tensor-factor reassignment, not an allocation. The dimensions are

At the third step the joint message encoding is $V$.

There is a nontrivial gate during the emission sequence, acting only on not-yet-emitted carriers. The process has a fixed remnant and does not assert complete evaporation.

Direct Clifford conjugation gives

The last identity follows also from $Y=iXZ$. These words all have weight three.

For every message operator $X$,

Indeed,

For a density operator $\rho =(I+xX+yY+zZ)/2$,

Tracing any one of the three radiation factors removes all three nonidentity terms.

For $t=0,1,2$,

where $I_{2^0}=1$. At $t=3$, the channel is (eq:N3), equivalently (eq:normal). These are equalities on all operators, including off-diagonal message matrix units.

Proof. The first two final radiation marginals follow from (eq:N3). Future gates act only on the remnant, so they cannot change already emitted marginals. In particular, the remaining ${\mathrm{CZ}}_{2,3}$ after the first emission does not alter the first carrier's law or reference correlations. At $t=0$, there is no accessible quantum factor.

If a one-qubit input channel is constant, its optimal recovery error is $3/4$.

Proof. Any decoded channel is a constant preparation $X\mapsto \mathrm{tr}(X)\sigma$. On a normalized logical Bell state with reference, its output is $\sigma \otimes I_R/2$. Its Bell overlap is $1/4$. The Bell-projector test therefore gives trace distance at least $3/4$.

The bound is attained by the completely depolarizing channel

The triangle inequality gives

The Bell-state lower bound makes this an equality.

For the three-step benchmark,

Proof. Use Proposition prop:prefix and Lemma lem:constant for the first three values. At the third step, apply $U$, destructively read positions $1,3$, forget their outcomes, and return position $2$. Equation (eq:normal) proves exact recovery with every reference.

The batch decoder has peak three, nine gates, and at most eight instruction rounds. Its two syndrome bits are independent of the logical-reference state.

More generally, sequential consistency gives

Thus unrestricted ${\epsilon }_{\mathrm{IT}}(t)$ is non-increasing. No corresponding total-batch-memory monotonicity is inferred: a later accessible input can itself be larger.

The lower bound is stated for all permitted one-live-qubit protocols. It does not require approximate uniqueness of a full-input recovery channel.

Let

From (eq:N3),

For every positive measurement effect $E$ on carrier $1$, the conditional state of $R23$ obtained from ${\omega }_{R123}$ is separable across $R\mid 23$, whenever that outcome has positive probability.

Proof. Write

Its outcome probability is $\mathrm{tr}E/2$. The normalized conditional state is

where

The first two operators commute, are independent, and satisfy $AB=-C$. Their four rank-one joint projectors are

Equation (eq:conditional) becomes

with

Each Bloch vector has length ${\Vert v\Vert }_2\le 1$, so every ${{\rho }_{\alpha \beta }}^R$ is a density operator. This is an explicit separable decomposition.

Under Assumption ass:stream, every deterministic decoder with $Q_{\mathrm{peak}}\le 1$ has

The lower bound permits arbitrary classical storage and arbitrary single-qubit instruments. It is attained in ${\mathfrak{L}}_{\mathrm{Cl}}$. Hence

Proof. Before carrier $2$ can enter a one-slot laboratory, carrier $1$ must cease to be a live quantum system. No intermediate quantum output or external quantum scratch register is permitted. The complete influence of the first carrier is therefore a classical measurement history, described by a POVM effect $E_h$.

Lemma lem:firstmeasurement makes the conditional state of $R23$ separable for every such history. All subsequent processing acts on the observer side, conditioned on classical data. It preserves separability between $R$ and the final recovered output.

A separable two-qubit state has Bell overlap at most $1/2$: for a pure product it is at most $1/2$ by Cauchy–Schwarz, and convexity extends the bound. Testing the Bell projector gives recovery trace distance at least $1/2$, and therefore the same half-diamond lower bound.

For attainment, read $Z_1,X_2,Z_3$ sequentially and report the negative product of their eigenvalues. This is logical $X$-readout by (eq:logicalreps). After consuming all three carriers, allocate and return the corresponding $X$-eigenstate. The unreported quantum channel is

Thus $d_{\diamond }({\Delta }_X,\mathrm{id})\le 1/2$, and the Bell test gives equality. An $X$-readout needs one $H$; preparation needs $H$ and at most one $Z$. The implementation uses peak one and at most three unitary gates.

At step $3$, information-theoretic recovery is exact, while any one-slot observer has error at least $1/2$.

If another consistent emission model has the same access interface and

then every one-slot decoder obeys

Proof. The first assertion combines Theorems thm:C2 and thm:C3. The second follows by the triangle inequality and CPTP contraction under ${\Phi }_{\pi }$.

The separability proof is stronger than an exact-only obstruction: it supplies a positive optimized approximate-error bound directly.

Define

They commute and have rank-one joint eigenspaces. Let

Then

Let the destructive computational outcomes after $B_{12}$ be $(t,s)$. Their eigenvalues are

Set

The last product is one classically selected Pauli, up to phase.

The chronological programme is:

1. receive carriers $1,2$;

2. apply $H_2,{\mathrm{CX}}_{1\to 2},{S_1}^{†},H_1$;

3. destructively read both carriers and retain $t,s$;

4. receive carrier $3$;

5. apply ${S_3}^{†},H_3,C_{ts}$;

6. forget the two record bits and return carrier $3$.

This programme recovers the message exactly with every reference. Each of its four measurement histories has probability $1/4$, independent of the message-reference state.

Its resource bounds are

with no evolution-oracle calls, no auxiliary entanglement, and $p_{\mathrm{succ}}=1$. Consequently,

Proof. Put $⟨{\beta }_{ts}|=⟨t,s|B_{12}$. The full branch map, including the inaccessible remnant, is

The logical representatives are $L_X=-P_{12}{Z}_3$ and $L_Z=-Q_{12}{Y}_3$. Also,

The correction signs therefore give

The $Z$ equation makes the two logical basis inputs have their corresponding output basis values. The $X$ equation makes their remaining environment vectors equal. Hence

for a fixed branch vector. The reduction on $12$ is $I_4/4$, so ${\Vert {\chi }_{ts}\Vert }^2=1/4$. Linearity proves branchwise reference preservation.

There are two live input qubits before the early readout. Both are consumed before the third arrives. Thereafter one is live. Four early gates, two late basis-change gates, and one Pauli correction give seven gates. The simultaneous early readout adds one round.

For capacities at least one, with the broad one-qubit lower-bound library and the exact Clifford upper-bound library,

The early upper bounds consume supplied radiation and prepare a maximally mixed output using one recyclable slot.

If the observer starts at step $3$, a passive supplier initially holds all three radiation qubits. The two-slot programme then has

It has not compressed three stored radiation qubits into two.

With just-in-time emissions and sufficient spacing to finish the early readout before carrier $3$ is emitted, there need be no external buffer and $Q_{\mathrm{stored},\mathrm{total}}=2$. If carrier $3$ is emitted while the two early carriers remain live, it occupies a supplier slot and total storage reaches three. These are different delivery schedules.

Keeping only one outcome $(t,s)$ gives the selective recovered map

Its trace is $1/4$, not one. All four corrected histories together give deterministic recovery.

The peak-one attainment programme is logical readout followed by preparation. It has recovery error $1/2$; it is not an exact subsystem decoder merely because it implements the corresponding rank-one measurement instrument.

We now change the initial model explicitly.

Let the message have $k$ qubits and dimension $m=2^k$. Let $B,E$ each have $b$ qubits, initially in $|{\Phi }_{\beta }{⟩}_{BE}$, where $\beta =2^b$. Let

Thus

The observer is given $ED$ in batch. It knows an explicit circuit for $U$, or is explicitly granted observer-register operations $U^{\ast }$ and, for amplification, $U^T$. It has no access to the actual remnant $C$.

The construction is the conjugate-evolution and Bell-projection architecture of Yoshida–Kitaev YK (Sections 4–5). We prove an exact specialization and its ledger below rather than importing an unquantified scrambling approximation.

Allocate two $k$-qubit registers $F,\hat{M}$ and prepare $|{\Phi }_m{⟩}_{F\hat{M}}$. Apply

Project $D{D}^{\prime }$ onto $|{\Phi }_{\delta }{⟩}_{D{D}^{\prime }}$. This projection can be implemented by inverse Bell preparation and destructive computational measurement, declaring success only when all $2d$ outcomes are zero.

The corresponding linear map from $ED$ to $\hat{M}{C}^{\prime }$ is

The successful output channel traces $C^{\prime }$, operationally by destructive readout with unreported outcome.

The complementary channel is

Assume

This is a checkable equality on all message matrix units. It is a hypothesis, not a consequence of unitarity.

For each computational $C$-label $a$, define a Kraus map $E_a:M\to ED$ by

Then

The maps $V_a=\sqrt{\gamma }{E}_a$ are orthogonal isometries. Let

be their range projector in $ED$.

Under (eq:exactdecouple),

The successful recovered map is therefore

Its success probability is $p=m^{-2}=2^{-2k}$, independent of every message-reference input.

Proof. From (eq:L),

Thus

using $m\beta =\gamma \delta$.

Likewise, the $a$-th block of $L{E}_{a^{\prime }}$ is

Summing over $a^{\prime }$ proves

All identities are linear, so tensoring with a reference preserves them.

For a generic $U$, equation (eq:L) still defines a selective map, but neither constant success probability nor exact conditional recovery is presumed. A different fidelity or decoupling certificate must then be supplied.

Before destructive readout, define the isometry

Let

Then

The reflection about $P$ is executable as

The inverse of $U^{\ast }$ is $U^T$, not $U^{†}$ in general.

Assume (eq:exactdecouple) and access to the two reflections above. Put

After $r$ iterations of

followed by the final Bell test, the success probability is

The successful recovered map is $s_r{\mathrm{id}}_M$.

If the success flag is forgotten while the failure quantum output is retained, the complete recovered channel has error at most $1-s_r$.

For $k=1$, one iteration gives $s_1=1$, hence deterministic exact recovery. For general $k\ge 1$, choosing a nonnegative integer $r$ nearest to

gives $1-s_r\le m^{-2}$ and $r=O(m)$.

Proof. For a unit vector $v\in \mathrm{supp}S$, let $a_v=Tv$. Equation (eq:Lnorm) gives

Define

These are orthonormal, and

The two reflections preserve their span and their product rotates the angle by $2\theta$. The same angle applies to every $v$ in the encoded support. Therefore the success component after $r$ iterations is

Using Theorem thm:HPpost gives success map $s_r{\mathrm{id}}_M$, including arbitrary references.

The failure map has constant trace $1-s_r$. When this is nonzero, divide it by that scalar to obtain a CPTP channel ${\mathcal{F}}_r$. The complete output channel is

Half-diamond distance between channels is at most one, giving the error bound.

When $m=2$, $\theta =\pi /6$, and $r=1$ gives angle $\pi /2$. For the nearest-integer choice, the final angle differs from $\pi /2$ by at most $\theta$. Hence failure is at most ${\sin }^2\theta =m^{-2}$.

The last result is an exact-decoupling specialization. It is not a proof that arbitrary evolution has a uniform Grover angle or the displayed worst-case error.

The batch input $ED$ occupies $b+d$ qubits. Appending $F,\hat{M}$ adds $2k$. Applying $U^{\ast }$ merely reassigns the $F,E$ factors into $C^{\prime },D^{\prime }$; it does not allocate them again. At the coherent stage the live factors are

with total

In an ancilla-free reflection implementation, the postselected construction and every fixed-$r$ amplification construction have

There is no external buffer in this batch convention.

The postselected construction uses one $U^{\ast }$ call and no $U^T$ call. The amplified construction uses

No call acts on the actual remnant. Original-evolution calls satisfy $n_U=n_{U^{†}}=0$ when the stated conjugate operations are the native oracle services.

At the end, destructive readout consumes $D{D}^{\prime }{C}^{\prime }$, a total of $2d+c$ qubits, and returns the $k$-qubit $\hat{M}$. Retaining all raw readout bits and a success flag costs at most $2d+c+1$ classical bits.

Proof. The displayed live register list accounts for every supplied and allocated factor. Reflections preserve those factors. Each $P$-reflection uses one $U^T$ and one $U^{\ast }$; the initial preparation uses one $U^{\ast }$. The final readouts and output consume or return every factor.

Let $g_U,D_U$ price the supplied $U^{\ast }$ and $U^T$ circuits. Let

have a declared ancilla-free implementation of gate count $z(n)$ and depth $d_0(n)$.

This is an additional implementation assumption. It is not a constant-cost elementary gate for arbitrary $n$.

A Bell-subspace reflection on two $n$-qubit registers consists of inverse Bell preparation, $Z_0(2n)$, and Bell preparation. It has count $4n+z(2n)$ and depth at most $4+d_0(2n)$.

Under these library assumptions, a fixed-$r$ programme has

and

The final measurements of $D{D}^{\prime }$ and $C^{\prime }$ may share one round.

No additional coherent success-flag qubit is required for these direct reflections. If a different implementation computes a predicate into a quantum flag, that flag and all predicate workspace must be added to its ledger.

Proof. Initial Bell preparation costs $2k$ gates and two rounds. There is an initial conjugate-evolution circuit. Each iteration contains two evolution circuits and the two Bell-subspace reflections. The terminal Bell-basis change costs $2d$ gates and two rounds; the terminal destructive readouts add one round.

If only an unknown oracle for $U$ is supplied, conjugating its matrix entries is not an automatic permitted operation. For an explicit circuit, conjugate gates and reversed adjoints can be priced in the declared library.

Take $k=b=c=d=1$ and

Thus $U$ is SWAP, initially $BE$ is a Bell pair, and

The complement is $I_C/2$, so exact decoupling holds.

SWAP has a three-CNOT implementation. Also,

which costs five gates and no ancilla. Therefore:

Postselection succeeds with probability $1/4$; one iteration is deterministic and exact. Direct recovery simply consumes $E$ and returns $D$. The table prices the chosen published architecture, not an optimum for this deliberately simple evolution.

A success map $p\mathrm{id}$ does not determine what remains after failure. A destructive Bell test generally changes the unique available radiation state. A failed attempt cannot be retried on the original unknown message without an additional recovery or state-supply assumption.

The coherent amplification programme avoids this issue by preserving all registers until its final test. Its inverse calls, reflections, and retained registers are explicitly charged.

Suppose

Then

For compatible complete-channel steps with individual errors ${\delta }_j$, the implementation error is at most ${\sum }_j{\delta }_j$.

Proof. Insert $\tilde{\mathcal{D}}\mathcal{N}$ and $\mathcal{DN}$, use the triangle inequality, and contract under CPTP composition. The stepwise statement follows by telescoping the channel products.

For adaptive decoding, retain the classical history in the complete step channels. A uniform error bound for every controlled branch bounds the entire controlled channel. A bound for a normalized postselected branch is not a complete-channel error bound.

For the two-slot benchmark decoder, seven unitary gates and two single-qubit destructive readouts give

when these are uniform complete-channel error bounds. This assumes compatible interfaces and no unmodelled correlated device memory. Radiation loss and encoding uncertainty are separate channel errors, not automatically gate noise.

With probability $\ell$, independent of the message, all three accessible radiation carriers are lost and replaced by a fixed state. A classical flag reports the erasure:

This models complete flagged erasure, not arbitrary partial radiation loss.

The unrestricted optimal error for this model is

A two-slot decoder attains it: recover on the $\mathrm{ok}$ branch and prepare $I_2/2$ on the lost branch.

Proof. On a logical Bell input, the good branch has Bell overlap at most one. The erased branch produces a constant output with Bell overlap $1/4$. The total overlap is at most $1-3\ell /4$, giving the lower bound.

The stated decoder implements $(1-\ell )\mathrm{id}+\ell \mathcal{T}$. Lemma lem:constant and homogeneity give error $3\ell /4$.

Suppose the observer is given radiation from one of the four channels

but no information identifying $P$. For a single decoder required to work uniformly,

This remains true with unlimited laboratory memory.

Proof. Averaging the four input Paulis gives $\mathcal{T}$. The averaged decoded channel is constant, so its error is at least $3/4$ by Lemma lem:constant. Convexity of the diamond norm bounds that averaged error above by the maximum of the four individual errors. Ignoring radiation and preparing $I_2/2$ attains $3/4$.

Knowledge, coherent circuit access, radiation loss, and workspace therefore price different obstructions.

For a finite emission horizon $T$, use the convention $\inf \varnothing =+\infty$. Define

and

The access convention is part of both definitions.

Within the declared horizon $T=3$:

1. for $0\le ϵ<3/4$, $t_{\mathrm{IT}}(ϵ)=3$;

2. for $0\le ϵ<1/2$ and any gate budget, $t_{\mathrm{op}}(1,G_0,ϵ)=+\infty$;

3. for $N\ge 2$, $G_0\ge 7$, and $0\le ϵ<3/4$, $t_{\mathrm{op}}(N,G_0,ϵ)=3$;

4. for $G_0\ge 3$ and $1/2\le ϵ<3/4$, $t_{\mathrm{op}}(1,G_0,ϵ)=3$.

Proof. The early prefixes have unrestricted error $3/4$, so no programme can cross these targets before step $3$. At step $3$, use Theorems thm:C3 and thm:C4.

The infinite value means no permitted programme reaches the target within the specified horizon. It is not a statement about every possible extension of the emission model.

Let physical availability times be $a_1\le a_2\le a_3$. Assume: