Hawk07-B Operational Functions and Conditional Predictions for Finite Wormhole Certificates Payload scaling, private recovery, hybrid communication, and resource-priced composition in the Enter framework

We extrapolate the finite model of Hawk07-A by proving dimension-dependent recovery, operational-function, and composition theorems. The word “wormhole” denotes a declared quantum-return certificate, not a spacetime tunnel.

For a logical payload of dimension $d$, an independent maximally entangled early resource, one known Haar-sampled vault unitary, and nested release, we rederive the complete complementary moment and obtain a sufficient full-diamond recovery threshold

Here $r_{\ast }$ is the released quantum dimension, not the entire receiver dimension; $f$ is the blank-factor dimension; and $p$ is the allowed measure of uncertified devices. A positive-residual, multi-cut design is given explicitly. The proposed inverse-linear error scaling is corrected to the inverse-square scaling actually supplied by this certificate. No optimality claim is made.

The certified operational functions include referenced state return, entanglement transfer, classical message transmission, logical postprocessing, and privacy against the declared complete residual. For square-power payload dimensions, a new calculation proves coexistence of recovery and coherent entropy turnover on a family of Haar measure at least $1/(4d)$, with one physical residual qubit at the final cutoff. This extends, rather than silently reuses, Hawk07-A's dimension-$16$ calculation.

A finite convoy is constructed using separately prepared resources. Its joint error is bounded even for mutually entangled payloads. A different theorem states a sufficient complete memory-reset contract for reuse; small previous leakage alone does not supply that contract. Noise, batch workspace, classical communication, terminal discard, and the distinction between a quantum payload and a classical log are accounted for separately.

These are conditional predictions of finite quantum-information models. They predict neither transport of a material spacecraft through spacetime nor a metric, travel time, energy condition, fuel cost, gravitational Page curve, or geometric wormhole.

of “prediction.” A prediction below is a consequence of explicitly stated preparation, channel, access, and implementation hypotheses. It is not an observation, an empirical forecast of gravitational wormholes, or a claim that those hypotheses have been implemented.

The Philosophy of Enter separates destructive readout from quantum return, and channel descriptions from thermodynamic or geometric identifications. An interface must specify its input, accessible systems, returned output, retained references, and permitted continuations CheungPhilosophy.

In the Enter calculus, a runtime qubit name grants access to a tensor factor. Destructive Enter consumes the measured name and returns a classical record. The Born instrument is a declared semantics; the operational characterization of that semantics is a separate result within a specified strategy class CheungEnter,CheungI. This manuscript does not rederive measurement probabilities from wormhole terminology.

The mathematical setting here is ordinary finite-dimensional quantum theory. Enter contributes the discipline of stating access and return types explicitly. Entanglement, scrambling, release, decoding, and communication are additional resources, not consequences of naming an operation.

The geometric ER $=$ EPR proposal and gravitational traversability constructions concern additional geometric and dynamical data MaldacenaSusskind,GaoJafferisWall,MaldacenaStanfordYang. They motivate comparisons, but provide no automatic dictionary for the finite factors used here.

A “ship” will mean a finite logical factor $L$, possibly entangled with an untouched reference. It does not mean a specified mass distribution, material body, world tube, or gravitational source.

The receiver $R$ and complete residual $B$ are the two operational mouths. The complete complementary channel is the object called a throat in the overlay. It is a channel, not a third register or a tunnel.

“Travel” means returning the logical payload through a specified receiver with a quantified error. The model does not assert motion of its laboratory qubits through a spacetime bridge.

{}1.13

All companion sheets are comparisons with their own models. The calculations below are stated and proved locally. In particular, the attached Hawk07-A is a predecessor CheungHawk07A, not permission to change its model without rechecking the hypotheses.

All Hilbert spaces in the main model are finite dimensional. Channels act on matrix algebras in the Schrödinger picture. For states and channels set

The diamond norm includes arbitrary untouched references WatrousBook.

A complete interface is an isometry

Its accessible and complementary channels are

Every unreturned purifier, lost mode, apparatus environment, and record store required by the declared realization belongs to the complete residual. An untouched external reference is not granted to the decoder.

For a state $\tau$, define

Write $a_d=1-d^{-2}$ and $b_d=1-d^{-1}$.

A tuple $(V,R,B,\mathcal{D},\epsilon )$ is an operational wormhole certificate when:

1. $V$ is a complete isometry and the receiver accesses only $R$;

2. $\mathcal{D}:\mathcal{B}(R)\to \mathcal{B}(L_{\mathrm{out}})$ is a specified deterministic CPTP quantum-return map;

3. ${\Vert \mathcal{DN}-{\mathrm{id}}_L\Vert }_{\diamond }\le \epsilon$;

4. physical $B$ remains a declared output factor;

5. no two commuting code-preserving star representations both implement the same logical algebra through $V$.

The terms “mouth” and “throat” add no geometric hypothesis.

For a complete interface:

1. exact full-code recovery from $R$ is equivalent to $\mathcal{C}={\mathcal{T}}_{{\tau }_B}$;

2. if exact recovery holds and $s=\mathrm{rank}{\tau }_B$, then $\dim R\ge ds$;

3. $${\epsilon }_{\mathrm{rec}}(\mathcal{N})\le \min \{a_d,\sqrt{2{\delta }_{\mathrm{full}}(\mathcal{C})}\};$$

4. if $d_{\diamond }(\mathcal{N},{\mathcal{T}}_{{\omega }_R})\le \eta$, then

   $${\epsilon }_{\mathrm{rec}}(\mathcal{N})\ge \max \{0,a_d-\eta \}.$$

Proof. For exact recovery, dilate a decoder by $W:R\to L_{\mathrm{out}}G$. The composite $(W\otimes I_B)V$ dilates identity. The rank-one Choi operator of identity implies

for fixed $\chi$, proving complement constancy.

Conversely, diagonalize ${\tau }_B={{\sum }_{\alpha =1}}^s{\lambda }_{\alpha }|\alpha ⟩⟨\alpha |$ on its nonzero support. Write

Constancy on matrix units gives $⟨u_{j\beta }\mid u_{i\alpha }⟩={\delta }_{ij}{\delta }_{\alpha \beta }$. The resulting $ds$ orthonormal vectors define an isometry $T:L{\mathbb{C}}^s\to R$ and

The map

is CPTP and exactly recovers the code. This is the Knill–Laflamme structure KnillLaflamme.

The information–disturbance estimate gives

Changing normalization gives the square-root bound.

A decoded replacement has Bell overlap $d^{-2}$, hence half-diamond error at least $a_d$. Preparing $I_L/d$ attains this value: depolarization is a uniform average of $d^2$ Weyl conjugations, including identity. It also supplies the universal constant-decoder cap. Contraction and the triangle inequality prove (eq:replacementlower).

The criterion is for the entire declared input, not a single effect or an unspecified subsystem CheungOai290280,CheungHawk04B,CheungEn07B.

For $d\ge 2$, two commuting star representations ${\pi }_1,{\pi }_2$ cannot both satisfy

Proof. Commutation would give

Injectivity of $V$ would make $\mathcal{B}(L)$ commutative.

This excludes duplicate actions on the same code, not commuting algebras of independent degrees of freedom CheungHawk03A.

Declare

Before the unknown payload is used, prepare the independent resource

The vault $LHF$ has dimension $D=dhf$. Sample one Haar unitary $U\in U(D)$, retain its device description, and use that same $U$ at every cut.

At cut $k$, regroup the vault as $P_k{B}_k$, with

The receiver is the entire factor

The reference $Q$, possibly entangled with $L$, is untouched. The complete isometry is

with the indicated output regrouping. Later release does not act on already released cells.

This is a finite entanglement-assisted recovery task, comparable in structure to Hayden–Preskill retrieval HaydenPreskill. It is not a model of its gravitational dynamics.

For the normalized Bell input on $QL$, put

At a cut $D=rb$,

Proof. The two-copy Haar twirl of $I_{P{P}^{\prime }}\otimes {\mathsf{S}}_{B{B}^{\prime }}$ is $\alpha I+\beta \mathsf{S}$, where

Indeed, the twirl commutes with all $U\otimes U$, and its identity and swap coefficients are fixed by the traces $r^{2}b$ and $r{b}^2$.

After tracing $E$, the initial reference–vault state is

Its purity is $1/h$, its vault marginal has purity $1/(dh)$, and its reference marginal has purity $1/d$. The swap identity therefore gives

Subtracting the second expression divided by $d$ proves (eq:moment).

For an input dimension $d$, normalized Choi operators obey

To see the factor, write any pure referenced input as $(T\otimes I)|{\Phi }_d⟩$, with $\mathrm{Tr}(T^{†}T)=d$ and ${\Vert T\Vert }_{\infty }^2\le d$. Apply trace-norm multiplication bounds and then convexity WatrousBook. Since ${X_{QB}}^U$ acts on dimension $db$,

Taking expectations gives $\mathsf{K}$. Finally use

This repeats the general calculation on the newly declared dimensions; it does not infer dimension scaling from Hawk07-A's special turnover theorem.

For the specified nested schedule,

Proof. Discarding the additional received cells takes ${{\mathcal{N}}_{k+1}}^U$ to ${{\mathcal{N}}_k}^U$, so an earlier decoder remains an available policy. The later complement is a partial trace of the earlier complement. That partial trace maps replacement channels to replacement channels and maps the named center ${{\sigma }_{B_k}}^U$ to ${{\sigma }_{B_{k+1}}}^U$. Use diamond contraction.

Fix $0<p<1$, $\zeta >0$, a positive-residual cutoff $K$, and a retained cut $k_{\ast }\le K$. If

then, with Haar probability at least $1-p$, the same sampled device has

A sufficient condition is

For full-diamond target $0<\epsilon <1$, the choice $\zeta ={\epsilon }^2/8$ gives the sufficient condition

Proof. Markov's inequality bounds

Proposition prop:nested propagates this one event to every later retained cut. Apply Theorem thm:correction. The full-diamond error is at most $2\sqrt{2\zeta }$.

The probability concerns the selection of a device $U$. It is not a decoder success flag or the probability that a particular input survives one use.

At cut zero, for every $U$,

No comparable lower bound at an arbitrary nonzero cut follows merely because (eq:fullthreshold) has not been met.

Proof. A unitary confined to the vault leaves its inaccessible reference marginal $E$ unchanged. The complement is

Undoing $U$ and tracing $HF$ recovers $X$. Contraction through this inverse gives a leakage lower bound $a_d$; comparison with the output on $I_L/d$ gives the matching upper bound. Theorem thm:threshold is sufficient, not necessary, so its failure supplies no converse.

At a cut with specified residual dimension $b$, (eq:momentbound) becomes

Thus its exact second-moment sufficient condition is

With $\zeta ={\epsilon }^2/8$, this reads

These are design allowances, not a converse capacity theorem.

Let $d=2^{\ell }$, $\ell \ge 1$, take $f=1$, choose $0<\epsilon <1$, $0<p<1$, and integers $q\ge 1$, $w\ge 0$. Set

Let the vault have $n=\ell +m$ qubits and declare

Then $0\le k_{\ast }\le K<n$, there are $w+1$ retained cuts, and with Haar probability at least $1-p$ all those cuts have full-code recovery error at most $\epsilon$. The cutoff retains $q$ physical vault qubits.

At the first retained cut,

Proof. The choice of $h$ gives $H_{\ast }\le h<2H_{\ast }$. In particular $dh/b_{\ast }\ge 4d^2/({\epsilon }^2\sqrt{p})$; hence $k_{\ast }$ is a valid nonnegative integer and Theorem thm:threshold applies. Nested release reaches residual dimension $2^q$ at $K$. Substitution proves (eq:rwindow)–(eq:receivercost).

For fixed $b_{\ast }$ and $p$, this construction uses

The logarithms of these dimensions count qubits. A polynomial Hilbert-space dimension does not imply a polynomial algorithm in the payload qubit count $\ell ={\log }_{2}d$.

For a declared released dimension $r_{\ast }$, rearranging (eq:fullthreshold) gives the sufficient range

For qubit payloads, one may choose $\ell$ with $2^{\ell }$ below this allowance.

Changing $d$ changes the model or requires an explicitly fixed larger code. For a single device intended to accept every state of a smaller subcode, certify the largest declared code once, then use a fixed isometric subcode embedding and its CPTP left inverse. Contractivity transfers the same error allowance. This is not repeated independent certification of arbitrary unannounced code spaces.

With $U=I$, release the diary $L$, retain $H$, and take $h\ge 2$. Then

Tracing $E$ is an exact decoder even though $r=d$. Consequently the $d^2{\epsilon }^{-2}$ law is not a lower bound on every operational certificate. It is the sufficient scaling established for the Haar estimate.

No equation here relates $d$ to kilograms, physical volume, or a spacecraft's mechanical complexity.

Let $A_{\ast }=\dim R_{\ast }$. Use the unnormalized input-first Choi convention

A candidate decoder Choi matrix $C$ on $R_{\ast }^{\prime }\otimes L_{\mathrm{out}}$ defines

Let

Specify the receiver by

Among minimizing decoder matrices choose the one of least squared Frobenius norm, and denote its channel by ${\mathcal{D}}_{\ast }$.

Program (eq:sdp) defines a unique selected CPTP receiver. Its objective value is ${\epsilon }_{\mathrm{rec}}({\mathcal{N}}_{\ast }^U)$. For $k\ge k_{\ast }$, define

where ${\Pi }_{k\to k_{\ast }}$ discards the subsequently received cells. Then

Thus one decoder core supplies all later retained certificates.

Proof. For a Hermiticity-preserving map $\Delta$, diamond-norm duality gives

The Choi constraints on $C$ are exactly CPTP, and composition is linear in $C$, proving the objective assertion.

Decoder Choi matrices form a compact convex set; the continuous error attains its minimum. The minimizing decoder set is convex, and the strictly convex secondary norm selects a unique matrix. Finally ${\Pi }_{k\to k_{\ast }}{{\mathcal{N}}_k}^U={\mathcal{N}}_{\ast }^U$ by the stipulated schedule.

The decoder is selected from the known device matrix, not from a favorable unknown-input record. The optimization is not reported as executed.

Write

For qubit dimensions, a decoder of Choi rank $g\le d{A}_{\ast }$ has a Stinespring realization with at most

environment qubits.

Allocate a fresh output and environment, extend the Stinespring isometry on its prepared subspace to a unitary, and discard unused factors. This gives

The inaccessible active supplier register contains $n-k$ qubits. Keeping it separate, the corresponding upper bound on live named processing qubits is

Neither count includes an untouched test reference, retired physical record stores, or an unspecified thermal bath.

A useful ledger is therefore

with

If discarded decoder factors are implemented by Enter and their outcomes are retained, an elementary realization records at most $q_k+q_G$ such local bits. Its recovery need not send those bits between sites. Preparation distribution and quantum-cell transport are separate communication resources.

Arbitrary finite unitaries admit finite decompositions using continuous one-qubit gates and CNOT Barenco. We do not supply numerical values for $G_{\mathrm{gates}}$ or $D_{\mathrm{circuit}}$, nor an efficient synthesis algorithm. The matrix variable $C$ alone has $(d{A}_{\ast }{)}^2$ real Hermitian parameters.

Reordering operations on disjoint factors can preserve a channel without preserving the maximum storage occupied during execution. A finite sequential program must price every arrival, allocation, retained early register, and workspace factor on every branch. Branch Kraus maps $K_{\gamma }$ must satisfy

and its complete returned channel, not one normalized branch, must meet the error target.

This is the distinction illustrated in Cheung's batch and streaming benchmarks CheungEn02BC. The batch construction above is not a streaming optimum.

A Yoshida–Kitaev-type decoder requires its own oracle access, reflections, success handling, and complexity hypotheses YoshidaKitaev. Neither Haar typicality nor (eq:sdp) automatically installs that algorithm.

Suppose

Then:

1. Referenced payload return. For every state ${\rho }_{QL}$,

   $$d_{\mathrm{tr}}(({\mathrm{id}}_Q\otimes \Lambda )({\rho }_{QL}),{\rho }_{QL})\le e.$$

2. Entanglement transfer. For a Bell input, the decoded Bell overlap satisfies

   $$⟨{\Phi }_d|({\mathrm{id}}_Q\otimes \Lambda )(|{\Phi }_d⟩⟨{\Phi }_d|)|{\Phi }_d⟩\ge 1-e.$$

3. Classical message transmission. Encoding the $d$ orthonormal basis messages and measuring the decoded output in that basis has error at most $e$ for every message.

4. Logical processing after return. For every declared CPTP task $\mathcal{A}:L\to T$,

   $$d_{\diamond }(\mathcal{A}\Lambda ,\mathcal{A})\le e.$$

   In particular, any specified logical unitary may be applied after return without increasing this allowance.

5. Measurement-task transfer. Every effect $0\le F\le I$, including a joint reference–output effect, changes probability by at most $e$. A bounded Hermitian observable $O$ changes expectation by at most $2e{\Vert O\Vert }_{\infty }$.

Proof. The first assertion is the definition of diamond control. The Bell projector and each correct-message projector are effects, which proves the next two assertions. CPTP postprocessing contracts the diamond norm. The final claims follow from the operational characterization of trace distance and trace/operator-norm duality.

Thus possible functions of the operational construction are a quantum communication interface, a decoder for stored referenced information, an entanglement-distribution interface, and an input wire for subsequent logical processing. These functions do not require a geometric throat.

If $d_{\diamond }(\mathcal{C},{\mathcal{T}}_{{\tau }_B})\le \zeta$, then

for every referenced input. For any two unreferenced inputs,

An observer restricted to $B$, with equal prior probabilities for these inputs, has discrimination success probability at most $1/2+\zeta$. Arbitrary processing of $B$ cannot increase these bounds.

Proof. Apply the diamond comparison to ${\rho }_{QL}$. Use the triangle inequality through ${\tau }_B$ for the second claim. For equally likely states, the best success probability is $(1+d_{\mathrm{tr}})/2$, giving the third claim. CPTP contraction proves the last assertion.

This is privacy against the named complete complement. It is not a cryptographic security theorem against an adversary also granted unmodelled receiver data or public transcripts. Those systems would have to be included in the adversary interface.

The qubit isometry

preserves the computational-basis beacon exactly at $R$, but both marginal channels dephase. Their nonconstant complements exclude exact whole-qubit recovery.

Let

The Choi overlap of ${\mathrm{Ad}}_W$ with identity is

which approaches one as $d$ grows. Nevertheless

Indeed, $W$ maps $(|0⟩+|1⟩)/\sqrt{2}$ to an orthogonal state. Every computational-basis beacon is unchanged. A decoder applying $W^{†}$ repairs the channel, but the uncorrected receiver has maximal worst-case error.

The distinction is between a specified decoder, an optimal decoder, a selected test, and a full-code claim CheungOai290280.

A deterministic a posteriori route to full-code certification is a complete Choi comparison. If a calibrated reconstructed Choi matrix $\hat{J}$ satisfies

then

This follows from the Choi conversion already proved. Obtaining the calibration allowance $t$ is an additional experimental-statistical task, not the observation of one beacon.

Fix a pure payload input $|\psi ⟩$ independently of $U$. After tracing $E$, the vault state is

This is generally a mixed random-projector ensemble. The entropy below is that of late radiation $P_k$, not automatically that of the full receiver $P_{k}E$.

At a cut $D=rb$,

Moreover,

Proof. Twirl ${\mathsf{S}}_{P{P}^{\prime }}\otimes I_{B{B}^{\prime }}$. Its identity and full-swap coefficients are $r(b^2-1)/(D^2-1)$ and $b(r^2-1)/(D^2-1)$. The vault purity is $1/h$, giving (eq:purity). The global $PBE$ state is pure, so the $P$ marginal has rank at most $\min (r,bh)$. Finally $S(\rho )\ge -{\log }_2\mathrm{Tr}{\rho }^2$; Jensen's inequality gives the mean lower bound.

The coherent rank envelope balances $r$ against $bh$, not against a classical record alphabet. The pure Page–Sen ensemble is the specialization $h=1$, not the general $h>1$ preparation CheungHawk05ABC,CheungHawk07A.

Let

Choose $\zeta >0$ and

Declare

A set of Haar devices of measure at least $1/(4d)$ satisfies:

1. $z_k(U)\le \zeta$ for every $k_{\ast }\le k\le K$;

2. one fixed decoder core has half-diamond error at most $\min \{a_d,\sqrt{2\zeta }\}$ at every retained cut;

3. for the fixed pure input,

   $$S({P_{k_{\mathrm{peak}}}}^U)>{\log }_2(h\sqrt{d}/2)\ge \max \{S({P_{k_{\ast }}}^U),S({P_{k_{+}}}^U),S({P_K}^U)\};$$

4. $B_K$ is one physical qubit.

The ensemble mean has the same strict peak relative to $k_{\ast }$ and $k_{+}$.

Proof. Write $t=\sqrt{d}=2^s$. The first three cuts have dimensions

The cutoff has $b_K=2$.

At the first cut, (eq:momentbound) gives

Therefore

At the peak, put $\mu =ht/2$. Equation (eq:purity) yields

Hence

Markov's inequality implies

On this event, $S({P_{\mathrm{peak}}}^U)>{\log }_2\mu$. The rank bound at $k_{\ast }$ and $k_{+}$ is $\mu$; at $K$ it is $2h\le \mu$. Subtracting the possible leakage-failure set leaves measure greater than $1/(4d)$.

Nested monotonicity propagates leakage, and Proposition prop:decoder supplies the common receiver policy. Also $p_{\mathrm{q}}<1/\mu$, so the mean entropy lower bound is strictly above ${\log }_2\mu$ at the peak.

For $d=4^s$, $s\ge 2$, and $0<\epsilon <1$, it suffices to choose

Then the family in Theorem thm:turnover has full-diamond error at most $\epsilon$ on all retained cuts. For $d=16$, its measure allowance is $1/64$. For larger $d$, the proved allowance is $1/(4d)$, not a dimension-independent $1/64$.

This new theorem does not assert turnover for every payload dimension or every Haar sample. Arbitrary qubit dimensions have the separate recovery theorem of Section sec:scaling.

For destructive readout in a prescribed basis,

The measured name is consumed; $T$ survives.

For every fixed process with an append-only record,

This is the classical chain rule, including adaptive readout. It concerns the unedited record of one process, not a selected subsequence or a reset memory.

Thus coherent entropy may rise and fall while the album cannot. Neither curve determines recovery by itself CheungHawk05ABC. In the nested model, leakage and optimal recovery error remain nonincreasing through the coherent peak.

A finite convoy is a new model with payloads $L_1,\dots ,L_N$, independent resource factors $H_i{E}_i{F}_i$, specified unitary devices, declared release interfaces, and fresh decoder workspaces. The payloads themselves need not be independent: they may be mutually entangled and entangled with one external reference.

For session $i$, suppose the complete preparation is independent of all payloads and other session resources, and its decoded channel satisfies

Suppose the actual independent-session output channel is ${⨂}_i{\Lambda }_i$. Then

The bound preserves every joint payload–reference correlation.

If session $i$'s device event has probability at least $1-p_i$, all session certificates hold jointly with probability at least $1-{\sum }_i{p}_i$, when this lower bound is positive. Device-event independence is unnecessary for this assertion.

Proof. Replace the tensor factors by identity one at a time. Each difference has diamond norm at most $2e_i$, since the remaining tensor factors are channels of diamond norm one. The triangle inequality proves (eq:convoy). Its diamond formulation already includes mutually entangled inputs. The probability statement is the union bound.

The tensor-product hypothesis is supplied by the newly declared independent resources and noninteracting session operations. It is not inferred from good marginal tests.

Fix $N$, payload dimension $d=2^{\ell }$, total full-diamond allowance $0<\epsilon <1$, and $0<p<1$. Use $N$ fresh copies of the same resource and port design. Select one device matrix $U$ certified to per-session full-diamond error $\epsilon /N$. Then the same decoder core can be used separately in all sessions, and the convoy's joint full-diamond error is at most $\epsilon$.

A sufficient per-session released dimension is

At first retained residual dimension $b_{\ast }$, $f=1$, it suffices to take

rounded upward to a power of two. The device-selection probability is at least $1-p$, not a fresh input-outcome probability for each ship.

Proof. Apply Theorem thm:threshold with per-session target $\epsilon /N$, and Theorem thm:convoy with $e_i=\epsilon /(2N)$. The same device property applies to every identically prepared copy; it need not be resampled.

This construction consumes $N{\log }_{2}h$ Bell pairs in total. It does not show that one Bell resource or one unreset vault serves $N$ payloads. Its $N^2$ allowance is a sufficient finite joint-error budget, not an asymptotic capacity or throughput optimum.

Laboratory occupancy depends on whether the sessions run in parallel or sequentially. The total stock of resource halves, retained outputs, supplier holdings, and workspaces must remain in the separate storage ledger.

Memory channels require a stronger contract than an isolated single-use channel KretschmannWernerMemory.

Let $M$ be a reusable device memory, and fix its specified reset state ${\omega }_M$. For session $i$, let

be the complete returned session map, including its receiver and reset action. Suppose, on the entire joint input space,

Previous returned payloads are isolated from later device operations. Then, for any initial memory and any joint payload–reference state, the final returned payloads and memory differ from ideal identity return with fixed ${\omega }_M$ by half-diamond error at most

Proof. Insert untouched identity wires for prior outputs and future inputs. Replace each actual session by ${\mathrm{id}}_{L_i}\otimes {\mathcal{T}}_{{\omega }_M}$, one at a time. All maps before and after a replacement are channels, so diamond contraction bounds the replacement by $u_i$. The ideal sequence returns each payload unchanged and resets the memory. Sum the bounds.

Condition (eq:memorycontract) includes memory correlations with all untouched systems. It is stronger than testing the device only when $M={\omega }_M$.

One explicit realization is to discard the old processing memory into an accounted-for environment, prepare a fresh independent resource, perform a certified session, and prepare the designated output memory state. That is resource replacement, not a free cyclic reset. No physical thermal cost follows until an implementation is specified CheungEn07A.

Consider an exact first-session interface

The logical complement is replacement and has zero leakage. Nevertheless $M$ is correlated with $S$.

If $S$ is discarded after returning $A$, then the accessible resource ${\Phi }_{SM}$ is gone. The remaining marginal $I_M/h$ does not restore it. A proposed second teleportation session cannot replace a missing receiver half by the fact that this marginal is unchanged.

More generally, a Bell state on $QM$ and the product state $I_Q/h\otimes I_M/h$ have identical $M$ marginals but trace distance $1-h^{-2}$. Marginal equality does not establish a reference-preserving reset.

If a single payload is sent through successive specified decoded links ${\Lambda }_j:L\to L$, then

The proof is the same telescoping argument. Every link still needs its own resource, access, and delay ledger. This licenses a declared quantum network, not a geometric shortcut.

If the entire returned interface is classical, every deterministic decoder has

The full-diamond floor is $2(1-1/d)$.

Proof. A decoder of a classical channel is measure-and-prepare and hence entanglement breaking EntanglementBreaking. The Bell-input output is separable. A product vector has Bell squared overlap at most $1/d$; convexity gives the same bound for a separable state. Testing the Bell projector proves the claim.

This is an obstruction for an album-only interface. It cannot be applied to a record plus a surviving quantum receiver without checking that combined channel CheungHawk05ABC,CheungEm03AF.

For $d=2^{\ell }$, prepare $\ell$ Bell pairs $HE$ before the unknown payload. A Bell transform on $LH$, followed by Enter, implements