En04-A Under the Enter Philosophy—Readout, Recovery, and Residual Response: Magic, Complete Complements, and the Limits of a Geometric Interpretation

We formulate the proposed connection between quantum magic and matter-responsive geometry at a finite, already-prepared quantum carrier interface. The complete output is an isometry $V:C\to R\otimes B$, and residual response means that its complete complementary channel is not constant on the logical code. Exact recovery from $R$ is equivalent to absence of that response.

The proposed unconditional implication from non-Clifford resources to residual response is false. We prove explicit counterexamples: a Clifford copying circuit has a nonconstant complement, whereas non-Clifford logical evolution and even a fixed syndrome containing bipartite non-local magic can coexist with a constant complement and exact logical recovery. We therefore separate magic, residual response, and physical geometric backreaction.

For a specified controlled-rotation family, we construct a non-Clifford responding program and obtain the exact formulas

A general first-order residual-observable witness gives a quantitative sufficient condition for response under a declared deformation. A Pauli-compression criterion separately characterizes constant complements for Clifford encoders.

Destructive Born–Lüders readout consumes measured quantum names and returns a classical interface. Readout-and-drop of an unobserved residual is exactly partial trace and does not further change the radiation channel. The four-Majorana recovery obstruction, finite-channel certificates, access-preserving endpoint continuity, and finite-proxy recurrence are retained with their proper scope. No metric, field-to-memory receiver, admitted analytic quantum coefficient, or gravitational backreaction theorem is constructed.

{{0.93} The corrected logical structure. \[

non-Clifford resource \ \ residual response,
Clifford encoding \ \ constant complete complement,
constant complete complement \ \ exact whole-code recovery from R,
residual response \ \ physical geometric backreaction.

A responding non-Clifford model can be proved after its coupling, code, preparation, access partition, and response witness are fixed. It cannot be inferred merely from the word “magic.” }

The operational question addressed here begins after a preparation has produced a finite quantum carrier. It concerns a complete encoding or evolution map, an accessible output, an inaccessible residual, a quantum decoder, and a separate destructive observation.

The proposed slogan

contains several logically distinct statements. In the present channel formulation, the relevant operational condition is non-constancy of the complete residual channel. However, neither ordinary non-Cliffordness nor a generic notion of non-local magic is equivalent to this condition. Moreover, an inaccessible quantum memory is not automatically a geometric degree of freedom.

Exact quantum correction and no-hiding supply the recovery criterion KnillLaflamme,BraunsteinPati. Magic belongs to a resource theory relative to a specified stabilizer structure VeitchEtAl. These are different mathematical structures.

A related but more specialized literature studies nontrivial area operators and perturbative matter–geometry correlations in holographic or skewed subsystem codes Cao2024,CaoEtAl2026. Its hypotheses, subsystem decompositions, and magic quantities are not identical to a generic channel $C\to RB$. Section sec:geometry explains the separation.

Landmark I gives a Born-law and positive-branch residual characterization inside its declared strategy class CheungI (thm:main). En03-A supplies the minimal linear interface and ownership discipline CheungEn03A. We define the finite fragment locally and use the supplied characterization only at its stated documentary level.

Hawk04-A supplies complete preservation, the global restore map, the exact complementary-channel criterion, the precise no-hiding conclusion, and the four-Majorana control CheungHawk04A. Hawk04-B distinguishes quantum return, destructive observation, complete apparatus outputs, and operational discard CheungHawk04B. The statements needed below are proved locally.

Hawk02-E supplies the regulated microscopic architecture and the separation of source energy from residual independence CheungHawk02E. The complete-process and apparatus disciplines of En02-B, En02-C, En02-E, and Em03-C are maintained CheungEn02B,CheungEn02C,CheungEn02E,CheungEm03C. Availability of a decoder, existence of a channel, and executable implementation under a resource budget remain distinct.

The XIV/XV/XVI and OpenAI catalogue comparisons are documentary separations only. They do not supply a premise for any channel, magic, or response theorem in this paper.

1. Magic is not leakage. A non-Clifford gate can act entirely within the recoverable logical sector. Conversely, a Clifford interaction can copy a basis label to the residual.

2. The restore map does not move the diary into the residual. It recovers the diary from the complete output $RB$. Residual-only recovery follows when the radiation channel is constant, not merely from global preservation.

3. Readout-and-drop is not permanent physical deletion. It is a trace in the observer's channel semantics. Its complete unitary dilation retains omitted apparatus outputs.

4. Residual response is not yet geometric response. One must identify actual geometric observables and show that they distinguish the declared matter inputs in the same physical model.

The completed result is consequently a corrected operational bridge with explicit witnesses and counterexamples, not a proof of the original unconditional slogan.

Let $C\simeq {\mathbb{C}}^d$, $d\ge 2$, be a finite logical code. Output Hilbert spaces are separable. Channels act normally on trace-class operators and are completely positive and trace preserving, abbreviated CPTP.

For an isometry

set

The residual $B$ includes every inaccessible output: preparation purifiers, retained memories, uncollected modes, displaced carriers, and inaccessible apparatus factors.

Distances are

All channel statements retain arbitrary untouched references. For finite input, standard diamond-norm and Choi conventions are as in Watrous.

Define the replacement channel

and the quantities

The complete residual responds to the logical code if $\mathcal{C}$ is not a replacement channel. It is nonresponding if $\mathcal{C}={\mathcal{T}}_{\sigma }$ for one fixed state $\sigma$.

This concerns variation with the logical input at a fixed encoding or fixed time. Dependence of a fixed replacement state ${\sigma }_{\lambda }$ on an external coupling parameter $\lambda$ is not, by itself, logical-input response.

No “rigid geometry” is defined by Cliffordness. When convenient, we call a proved nonresponding interface an operationally rigid interface. This terminology carries no metric assertion.

For isometries into a common output space,

The same allowance holds after a common partial trace. CPTP pre- and postprocessing contract half-diamond distance.

On fixed output spaces, recovery error and distance to the replacement-channel set are one-Lipschitz:

Proof. For normalized vectors $u,v$,

Apply this to referenced isometric outputs, then use convexity and partial-trace contraction.

For each decoder, the triangle inequality changes its error by at most the channel comparison allowance. Taking infima in both directions proves recovery stability. The same argument applies to distance from the replacement-channel set.

Put $a_d=1-d^{-2}$. Then

Every channel from $C$ has recovery error at most $a_d$ and leakage distance from replacement channels at most $a_d$.

Proof. For any replacement state, the Bell-input output is $I_Q/d\otimes \sigma$. Its overlap with the target Bell projector is $d^{-2}$, giving the lower bound $a_d$.

The uniform twirl over the $d^2$ Weyl unitaries is ${\mathcal{T}}_{I_C/d}$. Their Bell outputs are mutually orthogonal. The triangle inequality and that Bell input give the matching distance $a_d$.

A decoder which ignores its input and prepares $I_C/d$ gives the recovery upper bound. For a channel $\Phi$,

Postprocessing contraction gives the leakage upper bound.

Let $J:C\to H_{\mathrm{in}}$ be an isometry, let $|\Omega ⟩$ be a fixed normalized ancilla, and let $U$ be unitary on the complete input space. Then

is an isometry. Its complete channel has the CPTP left inverse

For every code–reference trace-class operator $X$,

Proof. The preparation and evolution compose to an isometry. Both terms of (eq:restore) are completely positive; their traces sum to $\mathrm{Tr}Y$. The second term vanishes on the encoded range, and $\mathcal{L}(VX{V}^{†})=X$.

An isometric embedding preserves nonzero singular values, including after tensoring with a reference identity.

The map $\mathcal{L}$ returns $C$ from $RB$. It neither deletes the code nor asserts that $B$ alone carries a recoverable copy. This is the restore structure in CheungHawk04A (thm:global).

For $V:C\to R\otimes B$, the following are equivalent:

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

2. $\mathcal{C}={\mathcal{T}}_{\sigma }$ for one residual state $\sigma$;

3. for every code matrix unit $E_{ij}=|i⟩⟨j|$,

   $$\mathcal{C}(E_{ij})={\delta }_{ij}\sigma ;$$
4. $$J_{\mathrm{n}}(\mathcal{C})=\frac{I_Q}{d}\otimes \sigma ;$$

5. for every bounded residual observable $O$,

   $$V^{†}(I_R\otimes O)V\text{is a scalar multiple of }{I}_C.$$

Here

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

An exact complete decoder is

Proof. Linearity and expansion of the Bell projector give equivalence of (ii)–(iv). Duality gives

Thus (ii) implies (v). Conversely, in (v), evaluate at ${\tau }_C$ to identify the scalar as $\mathrm{Tr}(O\mathcal{C}({\tau }_C))$. Since bounded observables separate trace-class operators, $\mathcal{C}(X)=\mathrm{Tr}(X)\mathcal{C}({\tau }_C)$, proving (ii).

Assume (ii), and diagonalize

The support condition permits the expansion

Constancy on every matrix unit gives

Consequently $A(|i⟩|\alpha ⟩)=|r_{i\alpha }⟩$ is an isometry, and ${\sigma }_K={\sum }_{\alpha }{p}_{\alpha }|\alpha ⟩⟨\alpha |$ gives (eq:normalform). Formula (eq:decoder) is CPTP and recovers every matrix unit.

Conversely, dilate an exact decoder by an isometry $L:R\to C\otimes G$. Stinespring dilation is the standard channel structure theorem Watrous,KSW. The map $Z=(L\otimes I_B)V$ has

because the decoded marginal is pure. Exact recovery of $E_{ij}$ forces $⟨{\eta }_j,{\eta }_i⟩=1$. All ${\eta }_i$ coincide. Tracing the decoded register and $G$ therefore gives (iii).

These are the exact correction/no-hiding conditions KnillLaflamme,BraunsteinPati used in CheungHawk04A (thm:exact). The local proof additionally identifies the residual-observable test used below.

If $\mathcal{N}$ is constant, $B$ exactly recovers the code. If $\mathcal{C}$ is constant, $R$ exactly recovers it. Both complementary outputs cannot independently recover the entire unknown code when $d\ge 2$.

Proof. Interchange $R$ and $B$ in Theorem thm:exact. An exactly recoverable output has constant complement, which cannot itself recover two distinct inputs.

For a finite input, ${\delta }_{\mathrm{full}}=0$ if and only if $\mathcal{C}$ is constant. If two code states satisfy

then

If $O=O^{†}$, $\Vert O\Vert \le 1$, and

then

Proof. For every replacement state, the triangle inequality gives

Also $w\le {\Vert \mathcal{C}({\rho }_0-{\rho }_1)\Vert }_1=2x$.

If leakage is zero, choose ${\sigma }_n$ with $d_{\diamond }(\mathcal{C},{\mathcal{T}}_{{\sigma }_n})\to 0$. Evaluation at ${\tau }_C$ gives ${\sigma }_n\to \mathcal{C}({\tau }_C)$ in trace norm. Replacement-channel distance equals the corresponding state distance, so $\mathcal{C}={\mathcal{T}}_{\mathcal{C}({\tau }_C)}$.

A positive witness proves that the residual distinguishes some logical inputs. It does not prove residual-only recovery of the entire code. Joint recovery is guaranteed by the complete isometry; local recovery requires its own theorem.

Types are

with observable algebras

Classical registers are block diagonal. No conversion from a quantum register to a classical register is implicit.

A judgement $P:T\to T^{\prime }$ consumes the entire input interface and returns its declared output interface. Sequential and tensor composition require matching types. A live quantum name is not duplicated or silently omitted. This follows the minimal ownership reading of CheungEn03A, rather than an unprovided larger calculus.

The primitives are pure allocation, specified finite unitaries, destructive computational readout, classical blockwise control, and classical discard:

A specified unitary constant is a semantic object, not an assertion of exact finite-gate compilation, efficiency, or hardware realization.

For a measured subsystem $S$ and surviving quantum subsystem $T$, write

The destructive readout instrument is

The measured names are consumed. For a pure input,

No normalized residual is assigned to a zero-probability branch.

These are the destructive Born–Lüders residual formulas supplied by Landmark I within its stated strategy class CheungI (thm:main). We define the instrument directly; we do not reconstruct a measurement-uniqueness theorem from incomplete hypotheses.

Every finite acyclic legal program denotes a CPTP hybrid-output channel and has path maps $K_h$ satisfying

Reported outcomes group the unnormalized branch maps $X\mapsto K_{h}X{K}_h^{†}$. The same statement holds with arbitrary untouched references. Every branch terminates within the finite height of its instruction tree.

Proof. Allocation and unitary gates are isometric. Readout satisfies ${\sum }_x{M}_x^{†}{M}_x=I$. Classical discard sums blocks without conditional normalization.

Backward induction on the instruction tree proves completeness. Sequential composition composes the path maps; coarse-graining adds their CP maps. Tensoring each path map with a reference identity proves the complete-channel statement. A finite acyclic tree has finite maximum branch length.

This is the local version of the complete-history accounting in CheungEn02B (thm:B1) and CheungEn02C. A selected success branch is not substituted for a deterministic decoder.

After a carrier dictionary exists, the coherent body has type

and form

a := new; (C,a) := EncodeAndPrepare(C,a); (R,B) := CoherentBody(C,a); returnQ(R,B);

It contains no Enter.

A local decoder has type $P_{\mathcal{D}}:Q_R\to {Q_C}^{\mathrm{out}}$. A finite CPTP decoder can be implemented semantically by

w := new; (Cout,G) := DecoderUnitary(R,w); y := Enter(G); dropBit(y); returnQ(Cout);

The Kraus isometry

has a finite unitary extension after allocation. Readout and discard of $G$ give ${\sum }_j{D}_{j}X{D}_j^{†}$. The diary output remains quantum.

Applying this decoder locally to the coherent output yields a joint program returning (Cout,B). The observer controlling only Cout sees $\mathcal{DN}$. A radiation marginal is therefore a declared access restriction, not a rule silently deleting a live residual wire.

For non-power-of-two dimensions, all logical spaces are subspaces of padded qubit registers. Physical channels are completed on invalid padded inputs, for example by fixed-state preparation. Such padding does not establish a physical field-to-qubit receiver.

A prescribed terminal measurement may be appended after quantum return:

Cout := BasisUnitary(Cout); x := Enter(Cout); returnBit(x);

Its target is the measurement channel $\mathcal{M}$, not ${\mathrm{id}}_C$. By contraction,

The converse for one chosen measurement fails: computational dephasing preserves its computational record but not off-diagonal matrix units.

Enter is not a magic evaluator. A program may be mathematically analysed for nonstabilizerness, but the primitive semantics of $\mathsf{Enter}$ is the instrument (eq:enter).

Fix a qubit tensor decomposition and its Pauli frame. A Clifford unitary normalizes the Pauli group. A pure stabilizer state is a Clifford image of a computational basis state. The convex hull of such states is denoted $\mathsf{STAB}$. A state outside this convex hull has stabilizer magic in this declared convention VeitchEtAl.

The classification concerns operations and prepared states, not a promise that every arbitrary diary input is a stabilizer state. In particular, a Clifford program can process an unknown nonstabilizer diary.

Bosonic Gaussianity, fermionic Gaussianity, and qubit Cliffordness are not identified. An arbitrary qubit dictionary can preserve channel recovery while changing the Pauli-frame description. A magic claim must consequently fix its dictionary before evaluation.

For

with local output unitaries,

These unitaries may be non-Clifford.

Proof. The residual channel is ${\mathcal{C}}^{\prime }={\mathrm{Ad}}_{U_B}\mathcal{C}$. Replacement states are carried bijectively to their conjugates, and unitary channel distances are invariant. On radiation, a decoder may undo $U_R$, and conversely may prepend it when comparing the two tasks.

An output unitary mixing $R$ and $B$ is not covered: it changes the accessible task.

There is a Clifford coherent program with

There is a non-Clifford coherent program with

Proof. For the first example, allocate $B$ in $|0⟩$ and apply ${\mathrm{CNOT}}_{C\to B}$, returning the original carrier as $R$. Then

and both marginals are computational dephasing

The two residual basis outputs are orthogonal, giving ${\delta }_{\mathrm{full}}\ge 1/2$. The replacement $I_2/2$ gives the matching upper bound: for every referenced input the two diagonal reference blocks multiply output differences of trace norm one. Thus the half-diamond distance is at most $1/2$.

The inputs $|+⟩$ and $|-⟩$ have identical radiation outputs. A decoder of error $\epsilon$ therefore satisfies $1\le 2\epsilon$. The identity decoder has error $1/2$, proving the recovery formula.

For the second example, let

The complement is ${\mathcal{T}}_{|0⟩⟨0|}$, and $T^{†}$ recovers exactly. The gate is non-Clifford because $TX{T}^{†}=(X+Y)/\sqrt{2}$, which is not a Pauli. For input $|+⟩$, the output has Bloch components $(1/\sqrt{2},1/\sqrt{2},0)$, outside the convex hull of the six Pauli eigenstates.

The CNOT copies a basis label, not an arbitrary unknown quantum state. There is no violation of no-cloning.

The Choi state of $V_{\mathrm{copy}}$ is

a stabilizer state. Thus its recovery obstruction does not require magic in the encoding Choi state.

To make “non-local” precise for the next counterexample, define a faithful pure-state indicator across a fixed bipartition:

It is zero exactly when the state is locally unitary-equivalent to a pure stabilizer state. We use it only as an indicator: no monotonicity under arbitrary local operations is asserted. Related non-local magic constructions distinguish removable local magic from magic in correlations Cao2024,CaoEtAl2026.

Every reduced density matrix of a pure qubit stabilizer state is proportional to a projector. Its nonzero eigenvalues are equal, and its rank is a power of two.

Proof. For an $n$-qubit pure stabilizer state with stabilizer group $\mathcal{S}$,

Tracing the complementary subsystem leaves only Paulis supported on the retained subsystem $A$:

If ${\mathcal{S}}_A$ has $2^r$ elements, its group average $2^{-r}{\sum }_{P_A\in {\mathcal{S}}_A}{P}_A$ is a projector. Therefore ${\rho }_A$ has flat nonzero eigenvalues $2^{r-n_A}$ and rank $2^{n_A-r}$.

This standard stabilizer structure is also discussed in FattalEtAl; the needed statement is proved above.

Let $0<\kappa <\pi /4$, and prepare the fixed syndrome

Define

Then

so ${\delta }_{\mathrm{full}}={{\epsilon }_{\mathrm{rec}}}^R=0$.

Nevertheless,

and the complete encoding Choi state has

Proof. Tracing $R_0{R}_1$ gives the displayed replacement channel. Discarding $R_1$ and returning $R_0$ recovers the code exactly.

The two nonzero eigenvalues of the $B$-reduction are unequal. Local unitaries preserve them, whereas a pure stabilizer reduction must have a flat nonzero spectrum by Lemma lem:flat. Thus no local unitaries convert ${\chi }_{\kappa }$ to a stabilizer state. The local unitary groups are compact and the stabilizer set is finite, so the minimum distance is strictly positive.

The Choi state is $|{\Phi }_d{⟩}_{Q{R}_0}\otimes |{\chi }_{\kappa }{⟩}_{R_{1}B}$. For the Choi magic statement, take a qubit code, or more generally $d=2^k$, with its fixed qubit structure. Across $QR:B$, its $B$-reduction has the same two unequal eigenvalues. The same argument applies.

For the two-qubit syndrome, the indicator can be evaluated:

Indeed, the largest overlap with a product state is ${\cos }^2\kappa$, whereas the largest overlap with a maximally entangled two-qubit state is $(\cos \kappa +\sin \kappa {)}^2/2$. These are the two possible Schmidt spectra of a two-qubit pure stabilizer state. Both overlaps are strictly below one for $0<\kappa <\pi /4$.

This counterexample concerns bipartite non-local magic, including in a complete Choi state. It does not identify that quantity with the specialized perturbative tripartite quantity of CaoEtAl2026. The fixed syndrome carries no logical-input response.

Under the generic prepared-carrier assumptions, neither

nor

is valid. Conversely, ${\delta }_{\mathrm{full}}>0$ does not imply magic.

Proof. Use Proposition prop:independence and Theorem thm:NLcounterexample.

Let $C$ contain $k$ logical qubits and let $A$ contain $m$ allocated ancilla qubits in $|0^m⟩$. A Clifford unitary $U$ gives

with a declared output partition $RB$.

For each Hermitian residual Pauli $P_B$, write

The complete complement is constant if and only if every such Pauli satisfies one of:

1. $P_A$ has an $X$ or $Y$ factor, so $⟨0^m|P_A|0^m⟩=0$;

2. $P_C=I_C$.

Proof. Compression gives

The ancilla expectation is zero when an $X$ or $Y$ factor is present, and otherwise is one with the phase carried by $s$. Thus every compression is scalar precisely under the stated condition.

Residual Paulis form an operator basis. The criterion for all of them is therefore equivalent to the observable criterion in Theorem thm:exact.

Cliffordness makes this test algebraic; it does not predetermine its answer. The CNOT copying example fails the test because the residual $Z$ compresses to logical $Z$.

A nonresponding Clifford baseline is

A less trivial baseline prepares a Bell syndrome on $R_{1}B$ while keeping $R_0=C$. Then $B$ is fixed in $I_2/2$, even though it is entangled with a radiation syndrome. Constancy does not require a product output or a pure residual.

On one logical qubit and one allocated residual qubit, define

Initialize $B$ in $|0⟩$ and return the first qubit as $R$. Then

The typed body is

b := new; (R,B) := ControlledRy_theta(C,b); returnQ(R,B);

It is a finite semantic program with a specified unitary constant. No exact compilation into a fixed finite gate library is asserted.

For (eq:rotationisometry),

and

The exact quantities are

For $0<\theta <\pi /2$, the unitary $U_{\theta }$ is non-Clifford and its output on the stabilizer input $|+⟩|0⟩$ has positive bipartite non-local magic across $R:B$. At $\theta =\pi /2$, the unitary is Clifford but the complement is still nonconstant.

Proof. Partial traces of the two columns in (eq:rotationisometry) give the displayed channels. The residual pure states have overlap $\cos \theta$, hence trace distance $\sin \theta$. Lemma lem:witness gives the lower bound ${\delta }_{\mathrm{full}}\ge \sin \theta /2$.

Let

Each basis output has distance $\sin \theta /2$ from ${\sigma }_{\theta }$. For an arbitrary referenced input, the output difference is

where the diagonal reference blocks $X_i$ are positive and ${\sum }_i\mathrm{Tr}{X}_i=1$. The trace norm is therefore at most $\sin \theta$. This proves the matching half-diamond upper bound.

The radiation channel has

The identity decoder has error $(1-c)/2$, using an upper triangle bound and the orthogonal Bell outputs of $I$ and $Z$. For any decoder of error $\epsilon$, the ideal outputs of $|+⟩$ and $|-⟩$ have distance one, while their channel outputs have distance $c$. Contraction and the triangle inequality give $1\le 2\epsilon +c$. This proves the recovery formula.

For input $|+⟩$, the complete output is

Its radiation reduction has eigenvalues $(1+\cos \theta )/2$ and $(1-\cos \theta )/2$. For $0<\theta <\pi /2$, these are nonzero and unequal. Lemma lem:flat shows that the state is not locally equivalent to a stabilizer state. In particular, $U_{\theta }$ cannot be Clifford, because it maps a stabilizer input to a nonstabilizer output.

At $\theta =\pi /2$, $R_y(\pi )=XZ$, and

where $\mathrm{CZ}$ acts first. This is Clifford. On the allocated $|0⟩$ residual, its code isometry is the copying isometry of Proposition prop:independence.

This is the properly completed “responding non-Clifford” fragment: a specified non-Clifford coupling and a direct residual witness, rather than an inference from gate classification alone. The baseline $\theta =0$ is nonresponding, and the response is proved on the same code, preparation, and partition.

For small $\theta$,

The exact relation is

This relation belongs to this family, not to arbitrary encodings.

Magic on $|{\Psi }_{\theta }⟩$ vanishes at both endpoints, while leakage is zero only at the first endpoint. Thus even in this entangling family, leakage is not a faithful magic measure.

Let $V_0:C\to RB$ have constant complement, let $K=K^{†}$ be bounded, and define

Fix code states ${\rho }_0,{\rho }_1$, put $X={\rho }_0-{\rho }_1$, and choose a bounded Hermitian residual observable $O$ with $\Vert O\Vert \le 1$. Suppose

Then

In particular, the complement is nonconstant for

Proof. The derivative of the conjugated trace-class operator is $-\mathrm{i}[K,V_{0}X{V}_0^{†}]$. The double commutator has trace norm at most $4{\Vert K\Vert }^2{\Vert X\Vert }_1$. Taylor's integral remainder is therefore bounded by $2{\lambda }^2{\Vert K\Vert }^2{\Vert X\Vert }_1\le 4{\lambda }^2{\Vert K\Vert }^2$.

The baseline residual image of $X$ is zero by constancy. Hence

Apply (eq:observablewitness). The assumption $m\ne 0$ entails $\Vert K\Vert >0$.

The relevant additional premise is a nonzero residual matrix element of the coupling. A label such as “interacting,” “non-Gaussian,” or “non-Clifford” is not that premise. A vanishing first-order witness also does not rule out higher-order response.

For a finite encoded residual, the observer may explicitly add

with body

x := Enter(B); dropBit(x); returnQ(R);

This is an external operation, not a conclusion about autonomous evolution.

For every joint state, including an external reference,

Consequently,

with exactly the pre-existing radiation recovery error.

More generally, a trace-preserving local operation on $B$, with all its outputs ignored, leaves the untouched reference and $R$ unchanged.

Proof. Partial trace is the sum of the diagonal blocks in any orthonormal basis. For a general local channel, its Heisenberg adjoint is unital. It therefore leaves every observable $A\otimes I_B$ on the untouched factors unchanged.

Thus the phrase “empty the bin” is only operational forgetting at this interface. It does not prove annihilation of the physical residual or its apparatus.

For an infinite residual, partial trace remains well defined. The theorem does not provide a finite program that measures infinitely many modes.

For the instrument (eq:enter), define

This is an isometry. Tracing $E$ gives the observable destructive readout instrument. The complete output preserves the input globally.

Proof. Completeness of the $M_x$ gives ${W_{\mathrm{read}}}^{†}{W}_{\mathrm{read}}=I$. Tracing the orthogonal $E$-labels removes cross-outcome terms. Apply Theorem thm:global.

For a fully measured qubit,

The phase is absent from each classical marginal but retained jointly. The carrier $K$ in this coherent dilation is a quantum system; its observable $\mathsf{Bit}$ interface arises after $E$ is omitted.

This complete apparatus ledger is the distinction emphasized in CheungHawk04B,CheungEm03C. Operational discard may move a record carrier to an inaccessible environment in a unitary realization; it does not remove that environment from the complete preservation claim.

If the only retained output of a complete process is classical, every subsequent deterministic reconstruction of $C$ has error at least

The same bound holds for every measure-and-prepare channel followed by a decoder. Computational readout followed by basis-state preparation attains it.

Proof. A finite classical-output channel has the form

A decoder prepares one state for each outcome, giving a measure-and-prepare channel. Its Bell-input output is separable.

For a normalized product vector $u\otimes v$,

Convexity gives the same Bell-overlap upper bound for separable states. Testing the Bell projector gives error at least $1-1/d$.

Computational dephasing is

The $d$ Bell outputs are orthogonal, yielding distance $1-1/d$; the triangle inequality gives the matching upper bound.

The entanglement-breaking characterization is established in HorodeckiShorRuskai. The quantitative statement is proved locally. A surviving quantum memory or supplied quantum side information changes the hypothesis.

Residual measurement outcomes supplied to the observer may assist quantum recovery. For example,

has dephasing radiation. Measuring $B$ gives a uniform bit $x$; applying $Z^x$ to $R$ recovers exactly. Dropping the bit without using it leaves dephasing. The bit alone contains no diary. This is an access change, not a record-only quantum decoder.

Take four Majoranas with

and freeze

For $0<\mu <\Lambda <\infty$, let

The Hamiltonian is specified independently of Enter.

The field estimates

give arbitrarily small relative bound of the interaction with respect to $H_B$. Since $H_S$ and $A$ are bounded, Kato–Rellich gives self-adjointness on $D(H_B)$ Kato. The finite cutoffs make the displayed norms finite.

Prepare the source in $|11{⟩}_S$ and the bath in vacuum. Swap an even-parity two-mode diary into the source:

The displaced carrier is the fixed state $|11{⟩}_D$. The logical source states $s_0,s_1$ have zero source energy and coupling eigenvalues $a_0=-1,a_1=1$.

For a declared packet projection $P_{\mathfrak{r}}$, set

Radiation is the collected bath; the source, carrier, and uncollected bath form the complete residual.

The complete code vectors, up to a common phase, are

They give

and the exact formulas

In particular, ${{\epsilon }_{\mathrm{rec}}}^R\ge 1/2$ and ${\delta }_{\mathrm{full}}\ge 1/2$. The complete output is exactly reversible.

Proof. On source eigenvalue $a_i$, let $q_i=g{a}_i{h_B}^{-1}f$. The Weyl identity gives

where $\mathcal{W}(q)=e^{a^{†}(q)-a(q)}$. The vacuum therefore evolves to the coherent amplitude $(e^{-\mathrm{i}t{h}_B}-I)q_i={\eta }_i(t)$. The phase is common for $a_i=\pm 1$. Orthogonal packet decomposition factors the coherent state. Tracing the orthogonal source labels gives the radiation channel.

Its two pure output states have trace distance $z_R$. Their optimal equal-prior discrimination probability is $(1+z_R)/2$, obtained by the positive part of their difference. For any radiation decoder, its Bell overlap is

The two diagonal decoder effects constitute a discrimination POVM. The Bell-projector test gives error at least $(3-z_R)/4$.

Bath parity interchanges the two states, so an optimal discrimination POVM can be symmetrized to have equal conditional errors. Measurement followed by preparation of the inferred basis state gives

Its Pauli probabilities are

Orthogonal Pauli Bell outputs and the triangle inequality give half-diamond error $1-p_I$, proving the radiation formula.

Define the isometry

Then

Retracting the range of $L_t$ gives the dephasing recovery error $(1-c_R)/2$, by the same upper bound and $|+⟩$, $|-⟩$ lower bound used in Theorem thm:rotation.

Isometric embedding and CPTP retraction preserve leakage. For ${\Delta }_c$, Pauli covariance and convexity show that twirling any candidate replacement state to $I_2/2$ cannot increase its distance. Relative to that replacement, the Pauli coefficients, in order $I,Z,X,Y$, are

The half-diamond norm is half the sum of their absolute values: a Bell input gives the lower bound and the triangle inequality gives the upper bound. Hence

The complete columns are orthonormal, so global recovery follows from Theorem thm:global.

The radiation obstruction is CheungHawk04A (thm:four-majorana); the residual formulas are the refinements in CheungHawk04B. They are rederived here.

At $g=0$, or at $t=0$, collected radiation is vacuum and $c_R=1$. Therefore