Oai280-B Finite-Confidence Certification of Boundary Reconstruction in Five-Qubit Perfect-Tensor Codes

We connect an explicit five-qubit encoding, a declared boundary access interface, and a complete deterministic decoder to a simultaneous finite-data quantum-channel certificate. The encoding is printed entrywise and realized by a fixed Clifford circuit. A syndrome calculation proves privacy of every boundary set of at most two qubits. Every set of at least three qubits admits an explicitly constructed accessible decoder; the optimal half-diamond recovery error for the smaller sets is exactly $3/4$. The associated six-qubit state is proved perfect.

For implemented channels, all failed attempts and decoder branches are retained. Conditional fixed-sample validation using three Bell correlations gives simultaneous reference-preserving error bounds, with calibration allowances in the fidelity interval and channel-transfer allowances outside the square root. We derive the certificate-specific sample requirements, an exact binomial alternative, and a linear conversion for independently established Pauli channels, including a physically randomized task.

The paper also constructs a two-pentagon tensor network with an explicit normalization and accessible subsystem decoders. A finite-spectator extension of the three-correlation certificate bounds recovery of one logical factor on arbitrary referenced bulk inputs. No experimental dataset, executed simulation, successful compilation, gravitational dictionary, metric reconstruction, Page curve, or inherited characteristic quantum coefficient is claimed.

. Quantum error correction; perfect tensor; boundary access; deterministic decoder; diamond norm; finite-confidence certificate; logical subsystem; complete failure accounting.

The five-qubit perfect code is established quantum error-correction theory Laflamme,KnillLaflamme. Perfect tensors furnish the building blocks of the holographic tensor-network codes of Pastawski, Yoshida, Harlow, and Preskill HaPPY. Experimental five-qubit encoding and decoding are also established Gong. We claim neither a new code nor a first implementation.

The contribution here is a connected operational construction:

The finite statistical certificate concerns the resulting end-to-end logical channel, not a region name or a sector tag.

Cheung's supplied En02-B record already identifies the ideal five-qubit access threshold and distinguishes resource-limited execution from information-theoretic recovery CheungEn02B. Oai280-A supplies the simultaneous three-correlation certificate Oai280A; Oai265-B and Oai290-A supply related finite-confidence and decision architectures CheungOai265B,CheungOai290A. The results needed below are proved locally with their actual finite hypotheses. No supplied companion manuscript is a theorem premise.

The paper supplies:

1. an entrywise $32\times 2$ encoding matrix and a gate sequence;

2. exact access results for all $32$ boundary subsets;

3. matrix-defined accessible decoders for every authorized set, and a gate-level two-slot streaming decoder for a named triple;

4. a complete channel and register ledger;

5. simultaneous finite-data certification and its acceptance guarantee;

6. certificate-specific feasibility calculations, exact binomial inversion, and justified structure-assisted conversion;

7. an analytically solved flagged-erasure benchmark;

8. a normalized two-pentagon network and a finite-factor subsystem certificate.

No validation data were supplied. Consequently no measured reconstruction map, numerical acceptance study, or hardware result can honestly be reported. The decision table in Section sec:reporting is a reporting interface, not an invented dataset.

The following distinctions are maintained throughout.

1. Encoding identification and decoder performance differ. A successful end-to-end certificate does not independently identify the physical encoder with the ideal tensor.

2. Access is an interface, not a caption. A decoder receiving an omitted syndrome or simulator side information does not implement the declared boundary task.

3. A branch is not a complete channel. Successful trials cannot be renormalized into the deterministic channel being certified.

4. An implemented-decoder exclusion is not an optimal-decoder exclusion. The latter requires a bound uniform over all permitted decoders.

5. The ideal threshold is cardinality. For one pentagon, connectedness of a subset in a drawing does not change its ideal recovery threshold.

6. Clifford circuitry does not imply Pauli noise. A structure-assisted bound requires an independently established channel structure or a separately specified randomization task.

7. Forty shots is a procedure-specific limitation. It is not a universal lower bound for every laboratory protocol.

8. A fixed spectator is not an arbitrary spectator. The stronger subsystem task must be compared with a partial-trace channel on all referenced inputs.

9. The unrelated analytic programme remains unrelated. Finite channel certification does not construct a characteristic distributional product or admit an inherited gravitational coefficient.

All Hilbert spaces in the proved operational results are finite dimensional. Channels are completely positive and trace preserving (CPTP) on full matrix spaces. Set

The diamond norm includes arbitrary independent external references. Its standard finite-dimensional conventions are those of Watrous Watrous.

Let

Boundary legs are ordered $1,2,3,4,5$, and computational basis strings are ordered lexicographically with leg $1$ first. Addition in binary exponents is modulo two.

For $x=(x_1,\dots ,x_5)\in \{0,1{\}}^5$, define

The matrix $V:C\to H_{\partial }$ is defined, for every row $x$ and column $i\in \{0,1\}$, by

Write $|0_L⟩=V|0⟩$, $|1_L⟩=V|1⟩$, and $\mathcal{V}(M)=VM{V}^{†}$.

Equation (eq:Vmatrix) specifies all $64$ matrix entries, including zeros and phases; it is not a definition only up to logical basis or local Clifford equivalence.

Let $H$ be the Hadamard gate and $S=\mathrm{diag}(1,i)$. Define

Products act on vectors from right to left.

The gate sequence

satisfies

It uses one $Z$, four Hadamards, four CNOTs, and five controlled $Z$ gates, with the displayed leg ordering.

Proof. After $Z_1$ and the four Hadamards, the state is

The four CNOTs change the first bit to $x_1=i+y_2+y_3+y_4+y_5$, so the final string has parity $i$. The five controlled $Z$ gates multiply its amplitude by $(-1{)}^{q(x)}$. This gives (eq:Vmatrix).

Each column has exactly $16$ entries of magnitude $1/4$, hence norm one. The columns have disjoint parity supports, hence are orthogonal. This also tests $V^{†}V=I_2$ directly.

The gate count is a count for the declared abstract library, not a nearest-neighbor compilation, a hardware depth, or an optimality claim.

Use the Pauli matrices $X,Y,Z$, with $Y=iXZ$, and set

\begin{split}
 g_1&=XZZXI,& g_2&=IXZZX,\\
 g_3&=XIXZZ,& g_4&=ZXIXZ,
 \end{split}
 \qquad
 \overline X=XXXXX,\quad \overline Z=ZZZZZ.
 \label{eq:stabilizers}

The range of $V$ is the simultaneous $+1$ eigenspace of $g_1,\dots ,g_4$. Moreover,

Proof. Let

with cyclic indices. Conjugating $X_j$ by $U_G$ gives $K_j$, so the $K_j$ stabilize $|G⟩$. They are independent: a product of a nonempty subset has a nonzero binary $X$-support on precisely that subset.

Direct multiplication gives

The four corresponding binary vectors are independent. Their common $+1$ eigenspace therefore has dimension $2$; equivalently the stabilizer projector has trace $32/16=2$.

Writing $Z^{\otimes 5}=Z_{\mathrm{all}}$, the entrywise formula gives

Every $g_j$ commutes with $Z_{\mathrm{all}}$, and hence stabilizes both columns. Thus their span is the whole two-dimensional stabilizer space.

The parity rule gives $\overline{Z}V=VZ$. Complementing every bit preserves parity up to one and satisfies

modulo two. Comparing the two columns in (eq:Vmatrix) therefore gives $\overline{X}V=VX$.

For $A\subseteq \{1,\dots ,5\}$, let $\bar{A}$ be its complement, and order both subsets increasingly. Let

be the corresponding tensor permutation. Define

These are two reductions of the same encoding.

For a Pauli operator $P$, its syndrome is the four-bit vector $s(P)$ specified by

The single-qubit syndromes are:

For every subset $B$ with $|B|\le 2$ and every matrix $M\in \mathcal{B}(C)$,

The identity holds on every matrix unit and after amplification by every independent reference.

Proof. The printed syndrome table follows by counting local Pauli anticommutations in (eq:stabilizers). Its $15$ entries are exactly the $15$ nonzero four-bit vectors.

A nonidentity Pauli of weight one has nonzero syndrome. A weight-two Pauli is a product of single-qubit Paulis on distinct legs. Its syndrome is the sum of two distinct entries and is therefore nonzero.

If $P$ anticommutes with some $g_j$, then

so $V^{†}PV=0$. Expanding an arbitrary observable on $B$ in its Pauli basis shows that the reduced channel has only its identity coefficient. Trace preservation fixes that coefficient as the right-hand side of (eq:privacy).

This is an operator identity for all $M$, not just a statement about diagonal input states. Tensoring the channel identity with a reference proves the last assertion.

This proves locally the ideal privacy assertion attributed to En02-B :privacy CheungEn02B.

Every reduction of $|T⟩$ to three of its six qubit legs is $I_8/8$. Consequently every reduction to at most three legs is maximally mixed.

Proof. A triple containing $Q$ consists of $Q$ and two boundary legs $B$. Lemma lem:privacy, applied to the Bell matrix-unit expansion, gives

A triple not containing $Q$ has a complementary triple containing $Q$. Since $|T⟩$ is pure and both triple spaces have dimension eight, their reduced states have the same eigenvalues. The first calculation therefore makes the second triple maximally mixed too. Smaller reductions follow by partial trace.

This is the finite perfect-tensor property used here. It supplies a HaPPY building block HaPPY, not an extended gravitational geometry.

We restate the needed exact recovery theorem, fixing its decoder domain and its off-support completion. The principle is standard exact quantum error correction KnillLaflamme; related complete-output formulations occur in Oai280A,CheungHawk04B,CheungEn07A,CheungEn07B.

Let $W:C_d\to R\otimes B$ be a finite-dimensional isometry and let $\mathcal{N},\mathcal{C}$ be its two reductions. The following are equivalent:

1. a CPTP decoder $D:\mathcal{B}(R)\to \mathcal{B}(C_d)$ satisfies $D\mathcal{N}={\mathrm{id}}_{C_d}$;

2. $\mathcal{C}(M)=\mathrm{Tr}(M){\tau }_B$ for one state ${\tau }_B$;

3. $\mathcal{C}(E_{ij})={\delta }_{ij}{\tau }_B$ for every matrix unit.

If $r=\mathrm{rank}{\tau }_B$, then $\dim R\ge dr$. The recovered channel preserves every independent external reference.

Proof. The equivalence of the last two statements is linearity. For the forward direction, dilate the decoder by an isometry $L:R\to C_d\otimes G$. Then

dilates the identity. Purity of its decoded marginal on a basis input gives $Z|i⟩=|i⟩|{\chi }_i⟩$. Recovery of $E_{ij}$ gives $⟨{\chi }_j,{\chi }_i⟩=1$. All normalized ${\chi }_i$ coincide. Tracing $C_{d}G$ proves constancy of the complement.

Conversely, diagonalize ${\tau }_B={{\sum }_{a=1}}^r{\lambda }_a|a⟩⟨a|$, with ${\lambda }_a>0$. Write

The complementary matrix-unit identities give

Thus $dr$ vectors in $R$ are orthonormal. Define the isometry $T:C_d\otimes {\mathbb{C}}^r\to R$ by $T|i,a⟩=|u_{ia}⟩$, and let ${\tau }_K={\sum }_a{\lambda }_a|a⟩⟨a|$. Then

For any fixed state $\omega$ on $C_d$,

is CP and trace preserving on all of $R$, and recovers every matrix unit. Equality of channels remains equality after reference amplification.

For $|A|=m\ge 3$, put $k=5-m$ and $K_A=({\mathbb{C}}^2{)}^{\otimes k}$. Define vectors and a matrix using the printed $V$:

For the encoding (eq:Vmatrix):

1. If $m\ge 3$, then

   $${\mathcal{C}}_A(M)=\mathrm{Tr}(M)\frac{I_{\bar{A}}}{2^k},{T_A}^{†}{T}_A=I_{C\otimes K_A},$$

   and

   $$\begin{array}{rl}{\mathcal{N}}_A(M)& =T_A(M\otimes I_{K_A}/2^k){T_A}^{†},\\ D_A(Y)& ={\mathrm{Tr}}_{K_A}({T_A}^{†}Y{T}_A)+\mathrm{Tr}((I_A-T_A{T}_A^{†})Y)\frac{I_2}{2}\end{array}$$

   satisfies $D_A{\mathcal{N}}_A={\mathrm{id}}_2$.

2. If $m\le 2$, then

   $${\mathcal{N}}_A(M)=\mathrm{Tr}(M)\frac{I_A}{2^m},$$

   and

   $$\underset{D:\mathcal{B}(H_A)\to \mathcal{B}(C)\mathrm{CPTP}}{\inf }{d}_{\diamond }(D{\mathcal{N}}_A,{\mathrm{id}}_2)=\frac{3}{4}.$$

3. In the second case, the complement is nonconstant and is itself exactly recoverable. More explicitly,

   $${\mathcal{C}}_A(M)=T_{\bar{A}}(M\otimes I_A/2^m){T_{\bar{A}}}^{†}.$$

Every assertion concerns the full logical qubit, including its reference entanglement.

Proof. For $m\ge 3$, the omitted set has size $k\le 2$, so Lemma lem:privacy gives the complementary matrix-unit identities. Comparing their entries gives

Expanding $V|i⟩$ in the omitted computational basis proves the first line of (eq:DA). The second line is the complete decoder in Theorem thm:exact; alternatively its traces add to $\mathrm{Tr}Y$, and direct substitution returns $M$.

For $m\le 2$, privacy makes the accessible channel replacement. Every subsequent decoder is therefore replacement by some state $\omega$. On a Bell input its output is $I_Q/2\otimes \omega$, whose Bell overlap is $1/4$. The Bell-projector test gives error at least $3/4$.

The decoder returning $I_2/2$ attains the bound. Indeed the completely depolarizing channel is

The triangle inequality gives $d_{\diamond }(\mathcal{T},\mathrm{id})\le 3/4$, matching the Bell lower bound.

Finally $|\bar{A}|\ge 3$, so the first part, with the roles interchanged, gives the displayed complement formula and its exact recovery. It cannot be constant because it recovers two distinct logical inputs.

Strings such as $124$ denote the set $\{1,2,4\}$. All subsets are covered below.

Here “recoverable” means the constructive channel (eq:DA); it is not just an existence label.

For each triple the decoder matrix is completely specified by (eq:TA). The corresponding omissions are:

For every triple, $T_A$ is an $8\times 8$ unitary, and $D_A={\mathrm{Tr}}_{K_A}\circ {\mathrm{Ad}}_{{T_A}^{†}}$.

Equation (eq:TA) is also an algorithmic circuit specification; it uses no quantum access to $\bar{A}$.

For $m\ge 3$, complete the orthonormal family $\{{u^A}_{i,b}\}$ by applying Gram–Schmidt to the computational basis in lexicographic order, omitting zero residuals. Set $t=m-k-1\ge 0$. Define a unitary $U_A$ that sends

and sends the added basis vectors, in their construction order, to the unused computational basis vectors in lexicographic order.

The decoding program is:

1. apply $U_A$ on the $m$ accessible qubits;

2. read the last $t$ qubits in the computational basis;

3. if their string is $0^t$, return the first qubit and discard the $k$ syndrome qubits;

4. otherwise discard all remaining data and return $I_2/2$.

For $t=0$, the third step is unconditional.

This program implements exactly (eq:DA). The successful projector is $T_A{T}_A^{†}$; coherences between it and its orthogonal complement are removed by the complete measurement.

For a fully finite gate description, one may decompose the specified matrix $U_A$ into two-level unitaries by column elimination. When eliminating entries $a,b$, use, if $r=(|a{|}^2+|b{|}^2{)}^{1/2}>0$,

Its product with $(a,b{)}^{\mathsf{T}}$ is $(r,0{)}^{\mathsf{T}}$. After finitely many eliminations the remaining diagonal phases complete the decomposition. Reversing the factors gives $U_A$. Here arbitrary two-level gates on the declared accessible register are admitted. No gate acts on an omitted register. This is not an efficiency or hardware-native compilation claim.

For triples the simpler specification is

whose entries are explicitly the conjugate transpose of (eq:TA) evaluated using (eq:Vmatrix).

For this triple define

For $\rho =(I+r_{x}X+r_{y}Y+r_{z}Z)/2$,

The representatives satisfy

Proof. For $R_Z$, multiplication by $\overline{Z}$ gives $R_Z\overline{Z}=g_3$, so its action on the code is logical $Z$. For $R_X$, $R_X\overline{X}=YIYXX$. The latter is the cyclic shift by two positions of $g_2{g}_3$, and is a stabilizer. Cyclic shifts preserve the code because (eq:Vmatrix) is cyclically invariant. Thus $R_X$ acts as logical $X$. Direct Pauli multiplication then gives the $R_Y$ assertions.

For completeness, the syndrome map on the six binary Pauli coordinates of $123$ has rank four. For example the syndromes of $X_1,X_2,Z_2,Z_3$ are independent. Its kernel has dimension two, and its three nonidentity elements are the unsigned versions of $R_X,R_Y,R_Z$. Every other nonidentity Pauli on the triple compresses to zero. Also ${\mathcal{N}}_{123}(I/2)=I_8/8$ by Proposition prop:perfect. Pauli expansion now gives (eq:N123).

Set

Receive carriers $1,2,3$ in that order. Retain carrier $1$ until carrier $2$ arrives, apply $B_{12}$, destructively read carriers $1,2$ as $t,s$, then receive carrier $3$, apply $C_{ts}{R_3}^{\mathrm{rot}}$, and return it. This is a complete CPTP decoder and satisfies

Its peak is two live quantum carriers, excluding the independent validation reference and the incoming carriers before admission to the decoder apparatus.

Proof. Conjugation gives

and hence

The third-qubit rotation maps $Z\mapsto X$, $Y\mapsto Z$, $X\mapsto Y$.

Thus, before the classical correction, the branch $t,s$ has probability $1/4$ and normalized output

Conjugation by $C_{ts}$ removes all three signs. Each of the four unnormalized branch maps is therefore $\rho \mapsto \rho /4$ on the encoded input. Summing the branches returns $\rho$. Linearity proves the equality on every matrix and every referenced input.

On arbitrary physical triple inputs, the program consists of unitaries, a complete two-qubit measurement, conditional unitaries, and quantum return. It is therefore CPTP without postselection. At most carriers $1,2$ are simultaneously live; after their consumption only carrier $3$ is live.

This locally verifies the decoding architecture discussed in En02-B and En02-E CheungEn02B,CheungEn02E. Cyclic relabeling gives such a program for the five cyclic consecutive triples. The matrix decoder above covers all ten triples without asserting the same optimized resource cost.

Carriers arrive in order $1,2,3$. A carrier must enter the apparatus before being operated on. At most one quantum system of qubit capacity is live in the apparatus at a time. No quantum scratch space, intermediate quantum output, preshared entangled ancilla, or access to the validation reference is available. Arbitrary classical memory and arbitrary permitted single-qubit instruments are allowed. A consumed carrier leaves only its classical history.

Under Assumption ass:stream,

In particular exact recovery in this interface requires peak at least two, and Proposition prop:streamdecoder attains it.

Proof. The Bell-input state is, by (eq:N123),

Before carrier $2$ enters a one-slot apparatus, carrier $1$ must be consumed. Each complete classical history from that carrier has a POVM effect

For a positive-probability outcome, the conditional state is

where

The independent commuting operators $A,B$ have rank-one joint projectors

Consequently

Every first factor is a state because its Bloch vector has length at most one. The conditional state is separable across $Q\mid 23$. Subsequent decoder operations, including classical conditioning, preserve this separability.

A separable two-qubit state has Bell overlap at most $1/2$: the claim holds for pure product states by Cauchy–Schwarz and extends by convexity. The Bell-projector test therefore gives half-diamond error at least $1/2$.

For attainment, destructively measure $Z_1,X_2,Z_3$ sequentially, report the negative product of their eigenvalues, and prepare the corresponding logical $X$-eigenstate. This realizes

Its Bell overlap is $1/2$, and the displayed mixture gives half-diamond error at most $1/2$. The program uses one slot, including its final state preparation.

The theorem reproduces, with a local proof, the specifically scoped streaming separation recorded in En02-C CheungEn02C. It is not unrestricted decoder optimality under an unspecified memory model.

On the authorized triple $123$, logical $Z$-readout can be implemented by sequentially measuring $Y_1,Z_2,Y_3$ and reporting their negative parity. Measuring and then preparing the reported logical $Z$-eigenstate realizes the logical rank-one Lüders instrument. Its unconditional channel is

Thus exact logical $Z$-statistics, and even the corresponding rank-one selective instrument, do not certify full-code recovery.

Proof. The measured product is $R_Z$, so Lemma lem:triple gives the logical projective effects. For a rank-one projector $P_t=|t⟩⟨t|$, the measure-and-prepare map is

including its action on a reference. Summing gives dephasing. Its Pauli mixture gives the upper error $1/2$, while the Bell overlap $1/2$ gives the matching lower bound.

This is the finite distinction emphasized in CheungEn02B,Oai280A,CheungOai290A,CheungXVIIA. Destructive readout retains its destructive meaning CheungI; it is not renamed as coherent decoding.

After the access cut, the decoder receives exactly its declared accessible interface. It may use:

1. quantum registers in $A$;

2. fresh local ancillas in fixed states independent of the logical input, omitted registers, and external reference;

3. records generated from operations on its received interface;

4. deterministic classical feedforward.

It receives no omitted register, omitted-register syndrome, hidden encoding environment, input label, validation reference, or privileged simulator state. Every terminal branch has a specified logical qubit output.

A joint inverse on all five legs is forbidden when it uses inaccessible legs. For $A=12345$, all five legs are accessible, and a joint inverse is not an access violation. This qualification repairs an overly broad version of the prohibition.

A finite decoder program satisfying Definition def:accesscontract defines a CPTP map $D_A:\mathcal{B}(H_A)\to \mathcal{B}(C)$. Its extension to a complete boundary input is

Equal accessible restrictions therefore give equal returned states, including when an independent reference is retained.

Conversely, for a CPTP map $\Psi :\mathcal{B}(H_A\otimes H_{\bar{A}})\to \mathcal{B}(C)$, the following are equivalent:

1. $\Psi =D{\mathrm{Tr}}_{\bar{A}}$ for a CPTP map $D$;

2. equal $A$-marginals always give equal returned states;

3. ${\Psi }^{\ast }(\mathcal{B}(C))\subseteq \mathcal{B}(H_A)\otimes I_{\bar{A}}$.

Proof. Fresh preparation, accessible gates, complete instruments, classical control, and complete branch summation compose to a CPTP map on the received interface. Induction over the program cannot introduce dependence on an unreceived input.

Factorization gives the adjoint formula and snapshot determination. Conversely, let $h=h^{\ast }$ be an adjoint output observable under statement two. If a traceless Hermitian $M$ satisfies ${\mathrm{Tr}}_{\bar{A}}M=0$, then for sufficiently small positive $t$, the states $I/\dim H\pm tM$ have equal $A$-marginals. Thus $\mathrm{Tr}(hM)=0$. The Hilbert–Schmidt orthogonal complement of $\ker {\mathrm{Tr}}_{\bar{A}}$ is $\mathcal{B}(H_A)\otimes I_{\bar{A}}$, proving statement three. Complex linearity handles all output observables.

Under containment write ${\Psi }^{\ast }(O)=D^{\ast }(O)\otimes I$. The identity inclusion reflects matrix positivity, so $D^{\ast }$ is unital and completely positive. Its trace dual is CPTP and gives statement one. Reference amplification follows from complete positivity and the channel identity.

This is the finite-factor specialization of the snapshot comparison in En05-C and Oai280-A CheungEn05C,Oai280A. No claim about a conformal net or a continuum observable limit follows. In particular, commuting payload channels alone are not the observable-limit hypotheses of En06-A CheungEn06A.

Suppose $\{{\mathcal{M}}_f{\}}_f$ is a complete instrument on the accessible interface and ${\mathcal{R}}_f$ is a CPTP return map for outcome $f$. The complete decoder is

For a failure branch, one may choose ${\mathcal{R}}_f(Y)=\mathrm{Tr}(Y)I_2/2$. The sum is CPTP because the instrument is complete.

Loss or leakage may require an enlarged input interface. For example, use ${\hat{H}}_j={\mathbb{C}}^2\oplus {H_j}^{\mathrm{flag}}$. Then the partial trace and decoder in the following equations are understood on those declared enlarged spaces. An unmodeled leakage subspace cannot be silently treated as a qubit input. Every flag branch must still be completed.

This is the finite complete-instrument distinction discussed in CheungVIIg,CheungEm05B,CheungEn02E. It does not require asserting results from another physical model.

Let

be the implemented encoder channel, and let ${\mathcal{E}}_{\partial }$ be the declared complete boundary process. Define

Tensor permutations are understood as in (eq:accesschannels). Thus:

The independent reference $Q$ bypasses this entire channel and is used only in final validation.

For an after-loss process ${\mathcal{E}}_A$, the experiment is instead

This is a different ordering. Equality requires its own theorem and is not presumed.

A live-name ledger records allocation, receipt, unitary support, measurement, consumption, classical storage, and quantum return on every branch. It can certify syntactic access and resource use for a trusted implementation description; it does not prove that uncharacterized hardware follows that description.

Suppose the channels have common declared interfaces and

Then

For an authorized ideal region, $D_A{\mathcal{N}}_A={\mathrm{id}}_2$.

Proof. Insert one intermediate composite for each component replacement. The triangle inequality and contraction of the diamond norm under CPTP pre- and postcomposition bound each replacement by its displayed allowance.

These are separately supplied component bounds. Favorable decoder validation does not establish them.

For deployment or another target channel with the same logical interfaces, independently justify

Direct certification uses ${\Lambda }_{\star ,A}={\Lambda }_{0,A}$ and $b_A=0$. A fitted simulator, a few matching marginal laws, or an ideal tensor identity is not by itself a diamond-distance transfer bound.

For channels $\mathcal{N},\tilde{\mathcal{N}}$ with common input and accessible output interfaces, define

Then

In particular, if $|A|\le 2$ and $d_{\diamond }({\mathcal{N}}_{0,A},{\mathcal{N}}_A)\le {\eta }_A$, every deterministic decoder obeys

Proof. For every fixed $D$, contraction and the reverse triangle inequality bound the difference of the two errors by the channel distance. Taking the infimum gives one inequality; interchanging the channels gives the other.

The predecoder distance in this proposition is not supplied by a certificate for one already decoded channel.

For a qubit channel $\Lambda$, set

The Pauli expansion is

Define

For a channel with input and output dimension $d$, normalized Bell Choi state $J_{\Lambda }$, and Bell overlap $F$,

More generally, for channels with common input dimension $D$,

Proof. The Bell input and Bell-projector effect give the lower bound. For a state $\rho$ and pure projector $P$,

This follows from the pure-state distance formula, convex decomposition of $\rho$, and concavity of the square root; it is also the pure-target case of FuchsVanDeGraaf.

To prove the Choi conversion, let $\Delta =\Phi -\Psi$. Unit referenced vectors can be written

where $A,B$ map the $D$-dimensional reference into the actual reference space and have Hilbert–Schmidt norm $\sqrt{D}$. Then

has trace norm at most $D{\Vert J_{\Delta }\Vert }_1$. A singular-value decomposition extends this bound to every trace-norm-one referenced operator. Taking the supremum and dividing by two proves (eq:choiconversion). The pure-target state bound gives the identity-channel result.

Let $\mathcal{A}$ be a finite nonempty predeclared collection of region–decoder tasks, and let $K=|\mathcal{A}|$. If multiple decoders or noise configurations for the same region are tested and selected using the validation results, they count as separate tasks in $K$.

For every $A\in \mathcal{A}$, the validation procedure:

1. prepares an independent Bell pair $QC$;

2. implements exactly the declared channel ${\Lambda }_{0,A}$ in (eq:implemented);

3. measures $XX,YY,ZZ$ on $Q,C_{\mathrm{out}}$ in separate predeclared settings;

4. assigns a complete bounded score to every attempted trial.

Decoder training precedes held-out validation. The same logical channel must be tested in all three settings.

Conditional on a declared training, selection, and calibration record $\mathcal{Z}$:

1. $n_{Aj}\ge 1$ is fixed before validation scores are observed;

2. scores $Z_{Aji}\in [-1,1]$ within each pair $A,j$ are independent and have a common mean $m_{Aj}$;

3. all three settings test the same CPTP logical channel ${\Lambda }_{0,A}$;

4. on a joint event $G$, with $\mathrm{Pr}(G^c)\le \beta$,

   $$\left|F_{\mathrm{obs},A}-F({\Lambda }_{0,A})\right|\le b_{\mathrm{cal},A},F_{\mathrm{obs},A}=\frac{1+m_{Ax}-m_{Ay}+m_{Az}}{4},$$

   and

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

All allowances are nonnegative, and ${\alpha }_{\mathrm{s}}+\beta <1$.

Independence between settings or tasks is not required for the coverage union bound. Adaptive stopping, validation-driven decoder tuning, and unaccounted channel drift are not covered by this fixed-sample contract.

Suppose the common actual tested state ${\rho }_A$ satisfies

and the complete score observable $M_{Aj}$ differs from ${\sigma }_j\otimes {\sigma }_j$ by operator norm at most ${\eta }_{Aj}$. Then a sufficient allowance is

Proof. Let $B_A=(I+M_{Ax}-M_{Ay}+M_{Az})/4$. Then $\Vert B_A-P_2\Vert \le {\sum }_j{\eta }_{Aj}/4$. The Bell projector is an effect, so its expectation changes by at most ${\delta }_A$ under the state comparison. Applying both bounds gives the claim.

The existence of a common ${\rho }_A$ matters. Setting-dependent preparation changes require their own calibration comparison. Assigning zero to a failed analyzer record does not automatically make that record an ideal Pauli measurement.

Define