Oai280-C Operational Certification of Logical-Algebra Recovery Access-restricted decoders, connected perfect-tensor benchmarks, and simultaneous reference-preserving confidence bounds

We give a finite-dimensional certification framework for a specified decoder acting on a declared accessible interface. The target may be a logical code, a subsystem, or a finite direct-sum algebra. Error is half-diamond distance, so the claim includes preservation of correlations with independent references. We prove the required task separations, the full-code constant-complement criterion, a programme-to-channel theorem with explicit resource ledgers, and dimension-correct Choi-to-channel bounds. A five-qubit stabilizer tensor supplies a fully specified connected two-cell encoding. Its normalization, subsystem decoders, hybrid-algebra decoder, and several optimal access errors are proved locally. For a three-share regression task we also prove decoder uniqueness, a minimal coherent environment dimension, an exact streaming peak of two qubits, and a four-message lower bound in a specified one-way model. General mixed-target tomography and a qubit Bell-correlation procedure give simultaneous finite-confidence intervals for complete implemented channels. Calibration, access assumptions, transfer allowances, and all-decoder obstructions are kept separate. Explicit noise examples and shot-count calculations identify both useful certificates and certificate-specific limitations. No gravitational dictionary, continuum geometry, or device-independent certification is asserted.

An operational recovery statement has more content than an operator identity or an existence theorem. It specifies what information is to be returned, what interface is accessible, which decoder is implemented, what happens on every branch, and how the resulting error is bounded.

Operator-algebra quantum error correction already provides the appropriate mathematical language for quantum, classical, and hybrid targets BenyKempfKribs,BenyKempfKribsPRA. Approximate recovery has corresponding worst-case entanglement-fidelity formulations and constructive recovery results BenyOreshkov. Perfect-tensor networks provide finite bulk-to-boundary code models PastawskiEtAl. The present contribution is an explicit certification interface connecting a supplied finite code to a specified apparatus. It does not replace those theories.

The internal source notes supplied for this manuscript attribute related interfaces to N. Cheung's manuscripts En02-B/C, En05-C, Oai265-B, Oai290-A, Em05-B, XVII-A, and Oai280-A/B CheungEn02B,CheungEn02C,CheungEn05C,CheungOai265B, CheungOai290A,CheungEm05B,CheungXVIIA,CheungOai280AB. Their full texts and publication metadata were not supplied. Accordingly, these citations identify provenance and intended comparisons, not independently checked published theorems. Every mathematical result used below is proved here or attributed to an identified external article. Appendix app:provenance records this distinction.

outcome. For finitely many predeclared tasks we construct intervals

such that, under the printed access, preparation, measurement, and sampling hypotheses,

Here ${\Lambda }_{\star ,j}$ is the channel actually claimed, ${\mathcal{T}}_j$ is the declared target, ${\alpha }_{\mathrm{s}}$ is the validation failure budget, and $\beta$ covers separately justified calibration or transfer assumptions. This is a statement about complete channels, not selected successful branches.

All Hilbert spaces are finite dimensional. A channel is a completely positive trace-preserving linear map between full matrix algebras. We use Schrödinger-picture channels and Hilbert–Schmidt adjoints.

For channels with the same input and output spaces define

For channel differences, the diamond norm can be evaluated on states with an ancillary reference of dimension equal to the input dimension. Thus $d_{\diamond }$ includes arbitrary input-reference correlations. Its range is $[0,1]$. We use the normalized Choi convention throughout; see Section sec:choi. These conventions agree with the channel-distance discussion in WallmanFlammia.

Let $H$ be the input space.

1. A named effect $0\le F\le I_H$ defines the binary readout channel

   $${\mathcal{M}}_F(\rho )=\mathrm{Tr}(F\rho )|1⟩⟨1|+\mathrm{Tr}((I_H-F)\rho )|0⟩⟨0|.$$

2. Full-code recovery has target ${\mathrm{id}}_H$.

3. If $H=C_L\otimes S$, subsystem recovery has target

   $${\mathcal{T}}_L={\mathrm{Tr}}_S:\mathcal{B}(C_L\otimes S)⟶\mathcal{B}(C_L).$$

4. If

   $$H=\underset{\alpha }{⨁}(C_{\alpha }\otimes S_{\alpha }),\mathcal{M}=\underset{\alpha }{⨁}(\mathcal{B}(C_{\alpha })\otimes I_{S_{\alpha }}),$$

   set $K={⨁}_{\alpha }{C}_{\alpha }$. The algebra-restriction target is

   $${\mathcal{R}}_{\mathcal{M}}(\rho )=\underset{\alpha }{⨁}{\mathrm{Tr}}_{S_{\alpha }}(P_{\alpha }\rho P_{\alpha }).$$

The direct-sum output in (eq:algebra-target) is embedded in $\mathcal{B}(K)$. It contains the sector label and the quantum state within each sector. It does not retain cross-sector coherences.

The map ${\mathcal{R}}_{\mathcal{M}}$ is CPTP. Its adjoint, restricted to block-diagonal output operators, is

Consequently $\mathrm{ran}{\mathcal{R}}_{\mathcal{M}}^{†}=\mathcal{M}$.

Proof. Fix orthonormal bases $\{|k{⟩}_{\alpha }\}$ of $S_{\alpha }$. Kraus operators are

where ${\iota }_{\alpha }:C_{\alpha }\to K$ is the output inclusion. Their adjoint products sum to $I_H$. The adjoint formula follows by taking traces. Off-diagonal output blocks have zero adjoint image.

A claim consists of the following frozen data.

1. Input space $H$, output space $K$, and target $\mathcal{T}$.

2. An encoding and noise model, together with an accessible channel ${\mathcal{N}}_A:\mathcal{B}(H)\to \mathcal{B}(Q_A)$.

3. A decoder programme on $Q_A$ and fresh, initially uncorrelated ancillas only.

4. A complete tested channel ${\Lambda }_{0,A}$, including unsuccessful outcomes, abort replacements, and discarded flags.

5. A claimed channel ${\Lambda }_{\star ,A}$, and, if different, a justified allowance

   $$d_{\diamond }({\Lambda }_{\star ,A},{\Lambda }_{0,A})\le b_A.$$

6. A register-access audit, gate/instrument library, arrival schedule, live-register budget, storage ledger, and classical communication convention.

7. Validation preparations, measurements, shot counts, statistical allocations, calibration allowances, and the threshold ${\epsilon }_A$.

8. Required negative controls: effect versus coherent recovery, fixed versus arbitrary spectator, successful branch versus complete channel, and unrestricted versus resource-restricted decoding.

The external validation reference is never a decoder input.

A useful physical form is

All maps in this expression belong to the same declared experiment. An ideal code identity alone does not establish (eq:tested-channel) for an apparatus.

Suppose the only input carrying information to the decoder is a finite classical record, and all other decoder inputs are fixed and independent of the logical system and its reference. Then the resulting logical channel is measure-and-prepare:

It is entanglement-breaking. For a $d$-dimensional identity target, $d\ge 2$,

Proof. The record probabilities form a POVM $\{F_m\}$; conditional decoder outputs are states ${\sigma }_m$. For any reference $R$,

which is separable.

Let ${\Phi }_d$ be the maximally entangled state. A product state $\sigma \otimes \tau$ has overlap $\mathrm{Tr}(\sigma {\tau }^{\mathsf{T}})/d\le 1/d$ with ${\Phi }_d$. By convexity the same bound holds for separable states. Measuring $\{{\Phi }_d,I-{\Phi }_d\}$ therefore gives trace distance at least $1-1/d$ from ${\Phi }_d$. This is a permissible diamond-norm input.

Let

The effect $F=|0⟩⟨0|$ is preserved exactly, while

Proof. ${{\Delta }_Z}^{†}(F)=F$. The upper bound follows from ${\Delta }_Z-{\mathrm{id}}_2=({\mathrm{Ad}}_Z-{\mathrm{id}}_2)/2$ and the fact that two channels have half-diamond distance at most one. On $|+⟩⟨+|$, the output is $I_2/2$, at trace distance $1/2$ from $|+⟩⟨+|$.

These are the locally proved hard gates corresponding to the classical-record and effect-target separations described in the supplied Cheung source notes CheungIEn06A,CheungOai290A,CheungXVIIA.

Let $\mathcal{N}:\mathcal{B}(H)\to \mathcal{B}(Q)$ be CPTP, with Kraus operators $E_i:H\to Q$. The following are equivalent:

1. A CPTP map $D:\mathcal{B}(Q)\to \mathcal{B}(H)$ satisfies $D\mathcal{N}={\mathrm{id}}_H$.

2. Every product ${E_j}^{†}{E}_i$ is a scalar multiple of $I_H$.

3. A complementary channel of $\mathcal{N}$ has the form

   $$\hat{\mathcal{N}}(\rho )=\mathrm{Tr}(\rho )\omega$$

   for a fixed state $\omega$.

Proof. The standard Kraus complementary channel has entries

Hence (ii) and (iii) are equivalent, with the scalar matrix positive and of trace one.

For (i)$\Rightarrow$(ii), let $K_a$ be Kraus operators of $D$. The Kraus operators $K_a{E}_i$ represent the identity channel. Its Choi matrix has rank one, so every $K_a{E}_i=c_{ai}{I}_H$. Trace preservation of $D$ gives

For (ii)$\Rightarrow$(i), change the Kraus basis to diagonalize the positive scalar matrix. Remove zero Kraus operators and write

The ranges of the isometries $W_i$ are orthogonal. With $\Pi ={\sum }_i{W}_i{W}_i^{†}$, define

for an arbitrary fixed state $\tau$ on $H$. This map is CPTP and $D\mathcal{N}={\mathrm{id}}_H$.

Let $\mathcal{N}={\mathrm{Tr}}_S$ on $C_L\otimes S$. The identity decoder on $C_L$ recovers the subsystem exactly. A complementary channel returns ${\mathrm{Tr}}_{C_L}\rho$, which is not constant when $\dim S>1$. Thus subsystem recovery does not require a left inverse on the whole bulk.

This elementary example prevents an inappropriate full-code premise from entering an algebra or subsystem claim. It is consistent with the subsystem and algebra scope of BenyKempfKribs,BenyKempfKribsPRA and the full-code interface reported in CheungVIIG,CheungEm05B.

For CPTP maps $\mathcal{N}:\mathcal{B}(H)\to \mathcal{B}(Q)$ and $\mathcal{T}:\mathcal{B}(H)\to \mathcal{B}(K)$, the following are equivalent:

1. For all states $\rho ,\sigma$, $\mathcal{N}(\rho )=\mathcal{N}(\sigma )$ implies $\mathcal{T}(\rho )=\mathcal{T}(\sigma )$.

2. $\ker \mathcal{N}\subseteq \ker \mathcal{T}$ as complex linear subspaces.

3. $\mathrm{ran}{\mathcal{T}}^{†}\subseteq \mathrm{ran}{\mathcal{N}}^{†}$.

4. There is a linear map $L$ on $\mathrm{ran}\mathcal{N}$ such that $\mathcal{T}=L\mathcal{N}$.

No positivity or complete positivity of $L$ follows.

Proof. Let $X=X^{†}\in \ker \mathcal{N}$. Trace preservation gives $\mathrm{Tr}X=0$. For sufficiently small $t>0$, both $I_H/\dim H+tX$ and $I_H/\dim H-tX$ are states. Condition (i) implies $\mathcal{T}(X)=0$. Decomposing a general kernel element into Hermitian and anti-Hermitian parts proves (ii). The converse is immediate. Finite-dimensional orthogonal-complement duality proves (ii)$\iff$(iii). For (ii)$\Rightarrow$(iv), set $L(\mathcal{N}(X))=\mathcal{T}(X)$; the kernel inclusion makes this well defined. The remaining implication is immediate.

For $0<t<1$, the qubit depolarizing channel

is invertible as a linear map. It therefore determines every input snapshot. Nevertheless, no CPTP left inverse exists: the trace distance between the two computational-basis outputs is $t$, whereas their input trace distance is one, and channels cannot increase trace distance.

Theorem thm:snapshot is the local finite-dimensional comparison suggested by the En05-C source notes CheungEn05C. It is deliberately not promoted to a CPTP-decoder theorem.

The five-qubit code is a standard exact quantum error-correcting code LaflammeEtAl. We specify its phases and normalization rather than importing a decoder convention.

Number physical qubits $1,\dots ,5$ and set

Tensor-product symbols are omitted in Pauli strings. Let $G=⟨g_1,g_2,g_3,g_4⟩$ and

Define

The generators commute and are independent. The projector $P$ has rank two. Up to signs, the fifteen nonidentity elements of $G$ are the cyclic rotations of

In particular every nonidentity element has weight four. The vectors in (eq:encoder) are orthonormal.

Proof. Two Pauli strings commute when the number of positions at which their nonidentity factors anticommute is even. This verifies pairwise commutation. Multiplication of the four generators gives the fifteen strings listed by the three cyclic orbits; they are distinct and none is the identity. Thus the group has sixteen elements and the generators are independent. Since every nonidentity Pauli is traceless, $\mathrm{Tr}P=32/16=2$.

None of the fifteen strings contains only $I$ and $Z$. Hence $⟨00000|P|00000⟩=1/16$, proving the normalization of $|\overline{0}⟩$. Both $\overline{X}$ and $\overline{Z}$ commute with $G$, and they anticommute with each other. The vector $|\overline{0}⟩$ has $\overline{Z}$ eigenvalue $+1$; $|\overline{1}⟩$ has eigenvalue $-1$ and the same norm.

For clarity, the syndrome of a Pauli is the four-bit string recording its anticommutation with $(g_1,g_2,g_3,g_4)$.

For every set $E$ of at most two physical qubits,

This identity holds on all operators, not only states.

Proof. The syndrome table contains each nonzero four-bit string exactly once. A nonidentity Pauli of weight one has nonzero syndrome. A weight-two Pauli has the sum of two distinct one-site syndromes, again nonzero. Such a Pauli $Q$ anticommutes with a stabilizer generator, implying $V^{†}QV=0$. Expanding operators on $E$ in its Pauli basis proves (eq:privacy).

Let $B$ be a physical triple, $E=\bar{B}$, and fix computational bases and ascending register order. Define the square matrix

Thus every entry of $U_B$ is fixed by (eq:encoder).

The matrix $U_B$ is unitary and

Consequently

recovers the qubit exactly, including every external reference.

For an arbitrary accessible set $A$,

The infimum is over all CPTP decoders on the declared accessible set.

Proof. Privacy on $E$ gives

The eight columns in (eq:UB) are therefore orthonormal. Summing the four environment components proves (eq:block-channel), first on matrix units and hence on all operators. Tensoring this equality with a reference proves the reference statement. For larger access, discard to a triple.

For $|A|\le 2$, every decoder output is a replacement state $\tau$. On a Bell input, $\tau \otimes I_2/2$ has Bell overlap $1/4$. The Bell-projector measurement gives error at least $3/4$. The replacement $I_2/2$ attains this value because

so its half-diamond distance from ${\mathrm{id}}_2$ is at most $3/4$.

For a triple $B$, every CPTP map satisfying $D{{\mathcal{N}}_{\mathcal{B}}}^V={\mathrm{id}}_2$ equals $D_B$ on the whole matrix algebra $\mathcal{B}(({\mathbb{C}}^2{)}^{\otimes 3})$. Its Choi rank is four. Every pure-environment Stinespring implementation therefore requires environment dimension at least four, and a coherent output-plus-environment realization needs at least three qubits in total.

Proof. Let $K_a$ be Kraus operators of $D$. Kraus operators of $D{{\mathcal{N}}_{\mathcal{B}}}^V$ are

Because the identity channel has Choi rank one, each is scalar times $I_2$. Thus

for suitable vectors $v_a$. Trace preservation gives ${\sum }_a|v_a⟩⟨v_a|=I_4$, whence $D={\mathrm{Tr}}_{{\mathbb{C}}^4}\circ {\mathrm{Ad}}_{{U_B}^{†}}$.

The four Kraus operators $(I_2\otimes ⟨e|){U_B}^{†}$ are linearly independent, so the Choi rank is four. A Stinespring environment dimension bounds the number of linearly independent Kraus operators. The lower bound is attained by applying ${U_B}^{†}$ and retaining the two-qubit environment.

The coherent bound is not a lower bound on all measurement-assisted live-register implementations. That distinction will matter below.

Define

Every reduction of $|T⟩⟨T|$ to at most three of its six qubits is maximally mixed. Thus every balanced three-input/three-output reshaping of $\sqrt{8}T$ is unitary.

Proof. A triple containing $R$ contains at most two physical qubits. Equation (eq:privacy) applied to the Bell construction makes this reduction maximally mixed. A triple containing only physical qubits is a reduction of $V{V}^{†}/2=P/2$. By Lemma lem:stabilizer, no nonidentity stabilizer is supported on three qubits, so this reduction is $I_8/8$. Smaller reductions follow by tracing. The balanced reshaping statement is precisely the matrix form of a maximally mixed three-qubit reduction.

This is the finite perfect-tensor ingredient used in PastawskiEtAl. It is not an experimental certification result.

Use two copies of the same tensor. Each cell has one bulk input, four exterior physical legs, and one joining leg. Denote the left bulk by $L$, the right bulk by $S$, the exterior legs by

and the joining legs by $e,e^{\prime }$.

Reshape the block encoder into

Use the same convention for the right cell. Define

In components,

Both $F$ and $W$ are isometries. If

then $W=2C$ and $C^{†}C=I_{LS}/4$. Equation (eq:W) is a deterministic isometric implementation: prepare one Bell pair and apply the two isometries $F$. It does not postselect a Bell outcome.

Proof. Single-leg privacy gives

so $F^{†}F=I_4$. Equation (eq:W) now immediately gives $W^{†}W=I_4$. Direct contraction with the normalized Bell bra gives $C_{a,c;l,s}=2^{-1/2}{\sum }_e{V}_{a,e;l}{V}_{c,e;s}$, proving $W=2C$ and the claimed success operator.

The factor $2\sqrt{2}$ in (eq:W-components) is essential: contracting normalized perfect states without auditing the scale would not produce the stated isometry.

For a left exterior triple $K\subseteq \{a_1,\dots ,a_4\}$, let $b$ be the remaining exterior leg. Define

The order in the coefficient is the declared exterior order, with $b=h$ and the joining leg last.

The matrix $U_K$ is unitary and

for every operator $\rho$ on $L\otimes e$. Equivalently,

Proof. Equation (eq:UK) is the balanced reshaping $\sqrt{8}T$ from Proposition prop:perfect, hence is unitary. The second formula follows directly from $F_{a_K,b;l,e}=\sqrt{2}{V}_{a_K,b,e;l}$. Tracing $b$ proves the channel identity.

Let

If $A$ contains a left exterior triple $K$, then

satisfies

The equality holds for arbitrary $LS$ inputs, including spectator correlations and arbitrary external references.

If $A$ contains a triple on each side, applying both inverse unitaries and discarding their joining and auxiliary registers recovers the entire two-qubit bulk exactly.

Proof. Trace the entire right exterior output in (eq:W). Because $F_R$ is isometric, this is equivalent to tracing $S,e^{\prime }$. The left joining register is then $I_e/2$, independent of the remaining state on $L$ and its reference. Applying Lemma lem:patchidentity gives

This is an operator identity and proves (eq:subsystemexact).

For two triples, the two inverse unitaries return

Tracing the indicated registers returns ${\rho }_{LS}$. All steps commute with adjoining an untouched reference.

For all four exterior legs of a cell one may instead use the complete decoder

The second term completes the channel outside the isometry range.

On $L\otimes S$ choose sectors according to the computational basis of $S$:

Each sector contains one logical qubit. The restriction target is

Identify its output with $L\otimes S_{\mathrm{cl}}$.

A triple on each side supports exact recovery of ${\mathcal{T}}_{\mathrm{hyb}}$: use the full-bulk decoder of Theorem thm:patch and then dephase the returned $S$ register. The target preserves sector labels and within-sector quantum information, but not coherence between $s=0$ and $s=1$.

Proof. The full-bulk decoder returns $\rho$. Dephasing $S$ produces (eq:hyb).

If $|A|\le 2$ among the eight exterior legs, then

If only left exterior legs are accessible, their channel is

Proof. The second identity follows from the right trace used in the proof of Theorem thm:patch and

For two accessible legs on one side, block privacy proves the result.

For one leg on each side, first observe that $F^{†}QF=0$ for every nonidentity one-site exterior Pauli $Q$. Indeed, the reduction of $|T⟩⟨T|$ to its bulk leg, joining leg, and that exterior leg is maximally mixed; expanding its matrix elements gives this identity. Expectations of nonidentity Pauli operators supported on the selected two legs therefore vanish in (eq:W). Their Pauli expansion proves the stated replacement channel.

For the ideal encoding $W$, the optimal unrestricted half-diamond errors for the following declared targets and access classes are:

No value is asserted here for an unlisted mixed access pattern.

Proof. For $|A|\le 2$, the channel is constant. The subsystem optimum is $3/4$ by the qubit replacement argument; the upper bound is unchanged by composing with ${\mathrm{Tr}}_S$. For full two-qubit recovery, a maximally entangled four-dimensional input has overlap $1/16$ with every replacement Choi state. The replacement $I_4/4$ attains error $15/16$, by the sixteen-term two-qubit Pauli expansion of ${\mathrm{Dep}}_4$.

For the hybrid target, identify the output with $L\otimes S$ and write ${\mathcal{T}}_{\mathrm{hyb}}={\mathrm{id}}_L\otimes {\Delta }_{Z,S}$. Its normalized Choi state is supported on

It has support probability one. A replacement Choi state $\tau \otimes I_4/4$ has support probability $1/8$, because ${\mathrm{Tr}}_{L^{\prime }{S}^{\prime }}{P}_{\mathrm{hyb}}=I_{LS}/2$. Thus the error is at least $7/8$. For the upper bound, ${\mathcal{T}}_{\mathrm{hyb}}$ is the equal mixture of ${\mathrm{id}}_{LS}$ and ${\mathrm{Ad}}_{I_L\otimes Z_S}$, whereas ${\mathrm{Dep}}_4$ is the equal mixture of all sixteen Pauli conjugations. Their difference is $7/8$ times a difference of two channels, so its half-diamond norm is at most $7/8$.

For left-only triple or four-leg access, $L$ is returned exactly and no information about $S$ is available. Replacing $S$ by $I_2/2$ gives full-bulk error $3/4$; a Bell input on $S$ gives the matching lower bound. For the hybrid target, returning a fair sector label gives error $1/2$. The lower bound follows by using the two orthogonal sector inputs with the same logical state: they have identical accessible outputs, so a single decoder cannot be closer than $1/2$ to both orthogonal target outputs. The two-triple case is Theorem thm:patch.

A programme is a finite acyclic decision tree. Each node applies a declared operation to named live registers:

1. a unitary from a frozen finite library;

2. preparation of a fixed fresh ancilla, initially independent of the input, source, and validation reference;

3. a finite-outcome CP instrument whose outcome sum is TP;

4. explicit discard, classical update, or an allowed input arrival.

Every terminal branch returns a state on the same declared output space. An abort has an explicit replacement output. No operation acts on omitted source registers or the external validation reference. No primitive hides an undeclared quantum memory.

For this manuscript a sufficient frozen library contains: the explicitly defined ${U_B}^{†}$ and ${U_K}^{†}$ matrices, unitary extensions of $F$, single-qubit Clifford operations, Bell preparation, the two-qubit measurement used below, ordinary projective measurements, and explicit discards. A unitary extension of $F$ is fixed by completing its four columns with Gram–Schmidt in computational-basis order. Thus the finite library is determined by printed matrices, not by an unspecified synthesis oracle.

The decoder live-register peak is the maximum, over branches and times, of the number of decoder-held qubits. It includes fresh ancillas and returned output registers. Unarrived source registers are listed separately in a source-storage ledger; they cannot be operated on before arrival.

The storage ledger records every register's holding interval and location. The classical ledger records outcome alphabets and messages. A total experiment-storage claim must include both decoder storage and parked source storage. A decoder peak is not, by itself, a bound on total physical storage.

Every legal finite programme induces a CPTP map on its declared accessible input. Its action on reference-entangled inputs is the same map tensored with the reference identity. Register counting along every branch bounds its declared peak and storage costs.

Proof. Associate a CP map with each branch prefix by composing its operation maps, tensoring fresh states, and tracing explicit discards. For an instrument, summing the children yields the parent's trace-preserving continuation. Induction from the leaves to the root shows that the sum of complete branch maps is CP and TP. Replacing an abort by a fixed state is a TP continuation of that abort branch.

Every operation has support only on declared registers. Tensoring with an untouched reference commutes with all compositions and branch sums, proving the reference statement. The live-set update at each node is specified by arrivals, preparations, operations, and discards. Its cardinality and holding intervals therefore give the stated pathwise bounds.

This proves the programme interface described in CheungEn02B,CheungEn02C; no apparatus-level conformity is deduced from it.

We now restrict access to the regression triple $B=\{1,2,3\}$ of the five-qubit encoding. This is a precisely scoped resource theorem, not a claim about every access set of the connected encoding.

Multiplying the full logical Paulis by stabilizers gives

They satisfy $X_{B}V=VX$ and $Z_{B}V=VZ$. Consequently

Set

These are commuting independent Pauli operators. Their common eigenprojectors are

Each projector has rank one. Choose a one-qubit unitary $R_{pq}$ by

These two anticommuting axes specify a unitary up to global phase.

Consider streaming programmes with access exactly to $\{1,2,3\}$, arbitrary arrival order, no additional source-correlated quantum resource, and no undeclared quantum memory. The following programme recovers exactly with decoder peak two:

1. hold qubits $1,2$ and measure $\{{\Pi }_{pq}\}$;

2. retain $(p,q)$ classically and discard qubits $1,2$;

3. receive qubit $3$ and apply $R_{pq}$.

No legal exact programme in this streaming model has peak one.

Proof. The adjoint of the printed programme sends

It sends $Y=iXZ$ to the corresponding logical $Y$ representative, since ${R_{pq}}^{†}Y{R}_{pq}=pqX$. Together with the identity these operators span the output algebra. The programme therefore equals $D_B$, proving exact recovery. Its live peak is two.

For the lower bound, suppose an exact peak-one programme exists. Before another input qubit can arrive, the first input qubit must have been discarded, leaving only a finite classical record. Operations on that first qubit define a POVM $\{E_m\}$.

By Theorem thm:unique, the complete decoder is $D_B$. Every Kraus operator of $D_B$ has the form $(I_2\otimes ⟨v|){U_B}^{†}$. For a refinement according to the first record $m$, the corresponding effect is therefore in the commutant of $X_B$ and $Z_B$. But that effect is $E_m\otimes I$ on the three inputs. At every individual site the local factors of $X_B$ and $Z_B$ anticommute. Hence $E_m$ commutes with two independent Pauli axes and must equal $p_m{I}_2$.

After discarding the first qubit, record $m$ and the other two qubits are thus described by $p_m$ times the two-share reduced channel. Lemma lem:privacy makes that channel constant. No continuation can recover the logical qubit, a contradiction.

Suppose a sender holds qubits $1,2$, a receiver holds qubit $3$, and only a finite classical message is sent from sender to receiver. There is no extra shared resource. Exact recovery requires at least four possible messages, hence at least two bits in a fixed-length message model. Four messages suffice.

Proof. The preceding programme proves sufficiency. For necessity, let the sender's POVM be $\{E_m\}$ and the receiver's conditional channel be $T_m$. Exactness and Theorem thm:unique imply, on all three-qubit operators,

where $|{\psi }_{pq}⟩⟨{\psi }_{pq}|={\Pi }_{pq}$. Therefore

A unitary channel has rank-one Choi matrix, so a convex decomposition of it into channels contains only that same channel with positive weight. Any message having positive probability for sector $(p,q)$ must therefore implement ${\mathrm{Ad}}_{R_{pq}}$. The four conjugation channels are distinct by (eq:R). Their message supports are disjoint, so there must be at least four messages.

Theorems thm:unique–thm:messages concern different resource classes: a coherent Stinespring realization, a specified streaming triple, and a specified one-way architecture. They do not assert a universal lower bound for arbitrary larger access sets, pre-shared entanglement, hidden quantum storage, or a different source-arrival convention.

Measuring the three single-site factors of $Z_B=-Y_1{Z}_2{Y}_3$ sequentially and returning their signed parity gives exact logical $Z$ readout at peak one. Similarly, $X_B=-Z_1{X}_2{Z}_3$ gives exact logical $X$ readout.

These are quantum-to-classical channels. They do not constitute the coherent decoder in Theorem thm:stream. Proposition prop:EB makes the distinction quantitative.

Concatenate the five-qubit encoder to depth $h$ and retain three designated children at every node. Within the module class in which each child is fully decoded before being consumed by its parent, the depth-first measurement-assisted schedule has peak $h+1$, for $h\ge 1$. This is optimal in that module class.

The explicitly non-adaptive schedule that receives and holds all $3^h$ selected leaves before any decoding has peak $3^h$.

Proof. At height one, Theorem thm:stream gives peak two. At a parent, decode the first child and retain its one-qubit output. Decode the second child while holding that output, measure the two outputs using the parent two-qubit measurement, and discard them. Decode the third child and apply the parent feedforward. If the child peak is $p_{h-1}$, the parent peak is $p_{h-1}+1$. Thus $p_h=h+1$.

In the declared module class, two completed child outputs are needed for the parent measurement. One must be held while a second child is decoded. This gives the same recurrence as a lower bound. The non-adaptive peak follows from its stipulated initial holding of all selected leaves; later decoding need not exceed that count.

This is an architectural comparison, not a global scheduling theorem over every possible instrument decomposition.

Consider the complete qubit channel

A success flag may identify the identity branch with conditional Bell fidelity one. If the returned output discards that flag, then

Proof. The complete channel difference is $(1-\eta )({\mathrm{Dep}}_2-{\mathrm{id}}_2)$. Theorem thm:block gives the stated norm.

If the flag is retained, the target must specify its ideal output law and the channel distance must be recomputed on that larger output space. Conditional success normalization can also be nonlinear when success depends on the input; it is not then a channel replacement for the complete instrument.

Let ${\mathcal{N}}_{\mathrm{mod}}$ and ${\mathcal{N}}_{\mathrm{act}}$ have the same accessible output space, and let $D$ and $\tilde{D}$ have the same input and output spaces. If

then

Proof. Insert $D{\mathcal{N}}_{\mathrm{mod}}$ and $\tilde{D}{\mathcal{N}}_{\mathrm{mod}}$ and use the triangle inequality. Pre- and post-composition by channels contract diamond distance.

This is the local version of the approximation accounting described in CheungEn02B,CheungEn02C. A transfer allowance is a different statement: it compares the tested complete channel with the channel claimed elsewhere. It does not calibrate state preparation or measurement.

Let $d=\dim H$, $k=\dim K$, and use

For a linear map $\Lambda :\mathcal{B}(H)\to \mathcal{B}(K)$ define

For channels $\Lambda ,\mathcal{T}$ with input dimension $d$,

The output dimension need not equal the input dimension.

Proof. The lower bound uses ${\Phi }_d$ as a permissible input. For the upper bound, every pure state on $H\otimes H$ can be written as $(I_H\otimes M)|{\Phi }_d⟩$ with $\mathrm{Tr}(M^{†}M)=d$, and hence ${\Vert M\Vert }_{\infty }\le \sqrt{d}$. For $\Delta =\Lambda -\mathcal{T}$ its output is

The trace norm is at most $d{\Vert J(\Delta )\Vert }_1$. Convexity handles mixed inputs; Schmidt decomposition reduces a larger reference to dimension $d$. Taking the channel-discrimination supremum gives the result.

These inequalities are consistent with the normalized-Choi convention in Lemma 7 of WallmanFlammia; completely bounded norm formulations are also discussed in Watrous.

If $\Lambda :\mathcal{B}({\mathbb{C}}^d)\to \mathcal{B}({\mathbb{C}}^d)$ and $f=\mathrm{Tr}({\Phi }_{d}J(\Lambda ))$, then

Proof. The Bell-projector measurement proves the lower bound. For completeness, the pure-state trace-distance inequality

follows by purifying $\sigma$ with overlap $\sqrt{⟨\psi |\sigma |\psi ⟩}$ with a purification of $|\psi ⟩⟨\psi |$, evaluating the trace distance of the two pure states, and tracing the purifier. Apply this inequality to $J(\Lambda )$ and use Lemma lem:choi.

For $H=C_L\otimes S$, with dimensions $\ell ,s$ and $d=\ell s$,

The reference includes both $L^{\prime }$ and $S^{\prime }$. Testing only $L^{\prime }$ with a fixed spectator prepares a different, weaker task.

For the algebra target, with $c_{\alpha }=\dim C_{\alpha }$ and $s_{\alpha }=\dim S_{\alpha }$,

where each block pairs its output sector with the matching input reference sector. Equations (eq:subchoi) and (eq:algchoi) follow by expanding the normalized maximally entangled input in matrix units.

A single overlap with these mixed states does not permit the pure-target proof of Corollary cor:pure. The general trace-norm certificate below avoids that mistake.

If an independently established structural identity gives

then

Without that identity, tests with one fixed spectator do not establish the equality.

Proof. Composition by ${\mathrm{Tr}}_S$ gives the upper bound. For the lower bound, embed an arbitrary $C_L$ input with a fixed state of $S$, retaining its reference.

The following assumptions are part of the claim, not consequences of favorable frequencies.

Conditionally on frozen pilot and calibration information:

1. The decoder and target are fixed before held-out validation.

2. Each setting is applied to independent repetitions of one fixed complete channel ${\Lambda }_0$.

3. Validation inputs are independent of the apparatus' fresh ancillas and hidden systems. The reference is untouched by the decoder.

4. Each measurement has a declared bounded outcome range and a calibrated relation between its mean and the ideal Choi expectation.

5. Shot counts and confidence allocations are fixed; no unaccounted optional stopping or data-dependent setting removal is used.

6. Every branch belongs to the tested channel. Abort and loss trials are not silently deleted.

7. An access audit supports the declared interface.

8. Calibration and transfer statements hold jointly except on an event of probability at most $\beta$.

A drifted, memory-dependent, or preparation-dependent apparatus is not covered unless a separate theorem supplies an appropriate sampling model.

Put $D=kd$, the dimension of the output-reference space. Choose an orthonormal Hermitian basis $\{G_{\nu }{\}}_{\nu =1}^{D^2}$ satisfying ${\Vert G_{\nu }\Vert }_{\infty }\le 1$. For example, use the diagonal matrix units and, for $i<j$,

Measure the spectral observable $G_{\nu }$ on fresh Choi outputs. Let ${\hat{\theta }}_{\nu }$ be the mean of $n_{\nu }$ outcomes in $[-1,1]$. Allocate ${\delta }_{\nu }>0$ and set

Suppose calibration supplies

on the good calibration event. Define

No positivity projection of $\hat{J}$ is required.

If $d_{\diamond }({\Lambda }_{\star },{\Lambda }_0)\le b$ on the good calibration and transfer event, set

Then

The result applies to subsystem and algebra targets (eq:subchoi)–(eq:algchoi).

Proof. For independent outcomes in $[-1,1]$, Hoeffding's inequality Hoeffding gives