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Knill–Laflamme Recovery and Subsystem Codes

1 · Prerequisites

2 · Summary

This page proves the finite-dimensional Knill–Laflamme recovery criterion, constructs a trace-preserving syndrome recovery, and extends the criterion to a protected logical factor with a gauge subsystem. The proofs fix the library’s linear-first inner-product convention and preserve the tensor-factor order of the code isometry. A distance corollary distinguishes unknown errors, whose supports may form a union, from known erasure supports. The companion page checks the matrix conventions and recovery maps in small finite-dimensional examples.

3 · Logical flowchart

4 · Definitions, theorems and proofs

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Agreement on pure states implies complete agreement

Statement

Let $H, K, R$ be finite-dimensional complex inner-product spaces, with the library convention that the inner product is linear in its first argument. For $x \in H$ , put $P_{x} : H \to H$ , $P_{x} (z) = ⟨ z, x ⟩ x$ (rank one when $x \neq 0$ ). Write $B (H) = {End}_{C} (H)$ . If complex-linear maps $Φ, Ψ : B (H) \to B (K)$ agree on $P_{x}$ for every unit $x \in H$ , then $Φ = Ψ$ on $B (H)$ and $Φ \otimes {id}_{B (R)} = Ψ \otimes {id}_{B (R)}$ on $B (H \otimes R)$ , under the canonical matrix-unit identification.

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces $H, K, R$ , and complex-linear maps $Φ, Ψ$ satisfying the stated rank-one equalities.

[F1]

Real and complex inner product spaces, with the inner product linear in the first argument: the inner product is linear in the first argument and conjugate-linear in the second.

[F2]

Every finite-dimensional real or complex inner product space has an orthonormal basis: $H$ and $R$ have orthonormal bases, including the empty basis in dimension zero.

[F3]

The elementary tensors of two bases form the product basis of the tensor product: elementary tensors of basis vectors form a basis of the tensor product.

[F4]

Linear map between vector spaces over the same field: a complex-linear map preserves sums and complex scalar multiples.

Proof

1.1givenF1F4algebra

Put $Δ = Φ - Ψ$ . By [F4], $Δ$ is complex-linear. For every nonzero $x \in H$ , $P_{x} = ∥ x ∥^{2} P_{x / ∥ x ∥}$ , and $P_{0} = 0$ , so linearity gives $Δ (P_{x}) = 0$ for every $x \in H$ .

2.1step 1.1F1F4algebra

With $∣ u ⟩ ⟨ v ∣$ denoting the map $z \mapsto ⟨ z, v ⟩ u$ , the linear-first convention gives $P_{u + v} - P_{u - v} = 2 (∣ u ⟩ ⟨ v ∣ + ∣ v ⟩ ⟨ u ∣)$ and $P_{u + i v} - P_{u - i v} = 2 i (∣ v ⟩ ⟨ u ∣ - ∣ u ⟩ ⟨ v ∣) .$ If $S = P_{u + v} - P_{u - v}$ and $T = P_{u + i v} - P_{u - i v}$ , then $∣ u ⟩ ⟨ v ∣ = (S - T / i) / 4$ and $∣ v ⟩ ⟨ u ∣ = (S + T / i) / 4$ . Hence $Δ (∣ u ⟩ ⟨ v ∣) = 0$ for all $u, v$ .

3.1step 2.1F2F4algebrachoose

Choose an orthonormal basis $(e_{j})$ of $H$ using [F2]. Every $T \in B (H)$ has the expansion $T = \sum_{j, k} ⟨ T e_{k}, e_{j} ⟩ ∣ e_{j} ⟩ ⟨ e_{k} ∣$ , so these matrix units form a basis of $B (H)$ . Step 2.1 makes $Δ$ zero on that basis, hence on all of $B (H)$ ; thus $Φ = Ψ$ .

4.1step 2.1step 3.1F3F4algebrachoose∎

Choose an orthonormal basis $(r_{a})$ of $R$ . By [F3], $(e_{j} \otimes r_{a})$ is a basis of $H \otimes R$ and the operators $∣ e_{j} \otimes r_{a} ⟩ ⟨ e_{k} \otimes r_{b} ∣$ form a basis of $B (H \otimes R)$ . Each equals $(∣ e_{j} ⟩ ⟨ e_{k} ∣) \otimes (∣ r_{a} ⟩ ⟨ r_{b} ∣)$ , on which $(Δ \otimes {id}_{B (R)})$ is zero by step 2.1. Therefore the two amplified maps agree on the entire operator space.

LemmaStatement: AI-adaptedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Product operator matrix units form a basis

Statement

Let $H_{G}, H_{Q}$ be finite-dimensional complex inner-product spaces with fixed ordered orthonormal bases $(g_{s})_{s < m}$ and $(q_{α})_{α < n}$ ; either index set may be empty. Put $B (H) = {End}_{C} (H)$ . For each $s, t < m$ and $α, β < n$ , define $M_{s α, t β} x = ⟨ x, g_{t} \otimes q_{β} ⟩ (g_{s} \otimes q_{α}) .$ Then the family $(M_{s α, t β})_{s, t, α, β}$ is a basis of $B (H_{G} \otimes H_{Q})$ . If either factor is zero-dimensional, this is the empty basis of the zero endomorphism space.

Facts & Assumptions

Given: The finite-dimensional spaces and their ordered orthonormal bases in the Statement.

[F1]

Real and complex inner product spaces, with the inner product linear in the first argument: the complex inner product is linear in its first argument, so $x \mapsto ⟨ x, v ⟩$ is linear for fixed $v$ .

[F2]

The product basis is orthonormal and tensor dimensions multiply: the elementary product tensors $g_{s} \otimes q_{α}$ form an orthonormal basis of $H_{G} \otimes H_{Q}$ , including the empty and zero-dimensional cases.

[F3]

Linear map between vector spaces over the same field: a linear map is determined by its values on a spanning set.

[F4]

A finite list $v : n \to V$ is an ordered basis if and only if every $x \in V$ equals $\sum_{i < n} λ_{i} v_{i}$ for exactly one $λ : n \to F$ ; those scalars are the coordinates of $x$ in that ordered basis: each vector has a unique finite coordinate expansion in an ordered basis.

Proof

1.1F1F2algebra

By [F1], each $M_{s α, t β}$ is linear. Applying it to a product basis vector gives $M_{s α, t β} (g_{t^{'}} \otimes q_{β^{'}}) = g_{s} \otimes q_{α}$ when $(t^{'}, β^{'}) = (t, β)$ and $0$ otherwise, by [F2]. Thus these operators have the individual matrix-unit columns of the product basis.

2.1givenF2F3F4step 1.1algebra∎

If either basis is empty, [F2] makes $H_{G} \otimes H_{Q} = {0}$ ; its endomorphism space is the zero vector space, whose empty family is a basis, as asserted. Otherwise, for any $T \in B (H_{G} \otimes H_{Q})$ and each input basis vector $g_{t} \otimes q_{β}$ , write $T (g_{t} \otimes q_{β}) = \sum_{s, α} c_{s α, t β} (g_{s} \otimes q_{α})$ using [F2, F4]. The finite sum $\sum_{s, α, t, β} c_{s α, t β} M_{s α, t β}$ has the same value as $T$ on every basis vector by step 1.1, hence equals $T$ by [F3]. If a linear combination of the $M_{s α, t β}$ is zero, applying it to each input basis vector shows each coefficient is zero by [F2]. Therefore this family spans and is linearly independent.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Partial trace over the first tensor factor

Definition

Let $H_{G}, H_{Q}$ be finite-dimensional complex inner-product spaces, and fix the factor order $H_{G} \otimes H_{Q}$ . For ordered orthonormal bases $(g_{s})$ and $(q_{α})$ , define ${Tr}_{G} : B (H_{G} \otimes H_{Q}) \to B (H_{Q}), ({Tr}_{G} X)_{α β} = \sum_{s} X_{(s, α), (s, β)},$ where $X_{(s, α), (t, β)} = ⟨ X (g_{t} \otimes q_{β}), g_{s} \otimes q_{α} ⟩$ under the linear-first convention. Equivalently, this is the partial trace over the second factor after swapping the two wires. The partial-trace characterization and wire-permutation unitarity in The partial trace is characterized by the trace pairing and is basis-independent and Finite wire permutations are unitary basis permutations show that the map is independent of the chosen orthonormal bases. In particular, the same definition applies when $H_{Q} = H_{L} \otimes H_{R}$ , with that residual factor order left unchanged. For an operator on $(H_{G} \otimes H_{L}) \otimes H_{R}$ , use the canonical associator of the ordered tensor-register construction to view it on $H_{G} \otimes (H_{L} \otimes H_{R})$ before taking this partial trace.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

The Knill–Laflamme coefficient matrix is positive

Statement

Let $H_{L}, H_{P}, H_{O}$ be finite-dimensional complex inner-product spaces, $V : H_{L} \to H_{P}$ an isometric encoding of nonzero logical space, and $P = V V^{*}$ . For a finite family of linear error operators $E_{1}, \dots, E_{m} : H_{P} \to H_{O}$ , suppose scalars $C_{a b}$ satisfy $P E_{a}^{*} E_{b} P = C_{a b} P$ for all $a, b$ . Then the matrix $C = (C_{a b})$ is Hermitian positive semidefinite.

Facts & Assumptions

Given: The finite-dimensional encoding and error family in the Statement, and the displayed Knill–Laflamme scalar-compression identities.

[F2]

An isometric quantum encoding gives an orthogonal code projector: $V^{*} V = I_{H_{L}}$ , $P^{*} = P = P^{2}$ , and $P V = V$ .

[F3]

Real and complex inner product spaces, with the inner product linear in the first argument: the complex inner product is linear in its first argument and conjugate-linear in its second.

[F4]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ and Every linear map between finite-dimensional inner product spaces has a unique adjoint: finite-dimensional maps have unique adjoints satisfying $⟨ T u, v ⟩ = ⟨ u, T^{*} v ⟩$ .

Proof

1.1givenF2F4choosealgebra

By [F2], $P V = V$ , $V^{*} P = V^{*}$ and $V^{*} V = I_{H_{L}}$ . Multiplying $P E_{a}^{*} E_{b} P = C_{a b} P$ by $V^{*}$ and $V$ therefore gives $V^{*} E_{a}^{*} E_{b} V = C_{a b} I_{H_{L}}$ . Choose a unit vector $ψ \in H_{L}$ , possible because $H_{L} \neq {0}$ . The adjoint identity [F4] yields $C_{a b} = ⟨ E_{a}^{*} E_{b} V ψ, V ψ ⟩ = ⟨ E_{b} V ψ, E_{a} V ψ ⟩$ .

2.1step 1.1F3algebra∎

For any $z = (z_{1}, \dots, z_{m}) \in C^{m}$ , the first-argument linearity and conjugate symmetry in [F3] give $z^{*} C z = \sum_{a, b} \overline{z_{a}} z_{b} ⟨ E_{b} V ψ, E_{a} V ψ ⟩ = {∥ \sum_{b} z_{b} E_{b} V ψ ∥}^{2} \geq 0.$ Also $C_{b a} = \overline{C_{a b}}$ , so $C$ is Hermitian. The nonnegative quadratic form is exactly positive semidefiniteness.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

A unitary error-basis change gives orthogonal error ranges

Statement

Let $V : H_{L} \to H_{P}$ be an isometry with $H_{L} \neq {0}$ and $P = V V^{*}$ . Let $E_{1}, \dots, E_{m} : H_{P} \to H_{O}$ be a finite family satisfying $P E_{a}^{*} E_{b} P = C_{a b} P$ , and let $C$ be the resulting positive matrix. There is a unitary matrix $U$ such that for $F_{i} = \sum_{a} E_{a} U_{a i}$ one has $\sum_{i} F_{i} X F_{i}^{*} = \sum_{a} E_{a} X E_{a}^{*}$ for every $X$ , and $V^{*} F_{i}^{*} F_{j} V = δ_{i j} λ_{i} I_{H_{L}}$ with $λ_{i} \geq 0$ . If $λ_{i} = 0$ , then $F_{i} V = 0$ ; if $λ_{i} > 0$ , then $W_{i} = F_{i} V / \sqrt{λ_{i}}$ is an isometry and the ranges of the $W_{i}$ are pairwise orthogonal.

Facts & Assumptions

Given: The finite-dimensional isometry and error family in the Statement.

[F2]

An isometric quantum encoding gives an orthogonal code projector: $V^{*} V = I_{H_{L}}$ and $P^{*} = P = P^{2}$ , so $P V = V$ and $V^{*} P = V^{*}$ .

[F3]

The Knill–Laflamme coefficient matrix is positive: the scalar matrix $C$ in $P E_{a}^{*} E_{b} P = C_{a b} P$ is Hermitian positive semidefinite.

[F4]

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely: a Hermitian endomorphism of a finite-dimensional complex inner-product space has an orthonormal eigenbasis.

[F5]

Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ : each nonnegative real number has a unique nonnegative square root.

[F6]

Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : $(A B)^{*} = B^{*} A^{*}$ and $(λ A)^{*} = \overline{λ} A^{*}$ for compatible finite-dimensional maps.

[F7]

Real and complex inner product spaces, with the inner product linear in the first argument: the inner product is linear in its first argument and defines the norm by $∥ x ∥^{2} = ⟨ x, x ⟩$ .

Proof

1.1F3F4F5algebra

By [F3], $C$ is Hermitian positive semidefinite. Apply [F4] to obtain a unitary $U$ with $U^{*} C U = diag (λ_{1}, \dots, λ_{m})$ . For each eigenvector $u_{i}$ , positivity gives $λ_{i} = u_{i}^{*} C u_{i} \geq 0$ , and [F5] supplies $\sqrt{λ_{i}}$ .

1.2givenF6algebra

Expand the transformed operator sum. Unitarity and [F6] give $\sum_{i} F_{i} X F_{i}^{*} = \sum_{a, b} (\sum_{i} U_{a i} \overline{U_{b i}}) E_{a} X E_{b}^{*} = \sum_{a} E_{a} X E_{a}^{*} .$ Thus mixing preserves the represented map, including every zero-on-code error.

2.1givenF2step 1.1algebra

Compress the transformed products by $V$ . The scalar-compression identities and [F2] give $V^{*} F_{i}^{*} F_{j} V = (U^{*} C U)_{i j} I_{H_{L}} = δ_{i j} λ_{i} I_{H_{L}}$ .

3.1step 2.1F5F7algebra∎

If $λ_{i} = 0$ , then for every $x \in H_{L}$ , $∥ F_{i} V x ∥^{2} = λ_{i} ∥ x ∥^{2} = 0$ , so $F_{i} V = 0$ . If $λ_{i} > 0$ , then [F5] and step 2.1 give $W_{i}^{*} W_{j} = δ_{i j} I$ for positive eigenvalues. Hence each $W_{i}$ is isometric, and for $i \neq j$ its range is orthogonal to the range of $W_{j}$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Exact recovery of an encoded quantum code

Definition

Let $H_{L}, H_{P}, H_{O}$ be nonzero finite-dimensional complex inner-product spaces. An encoding isometry $V : H_{L} \to H_{P}$ has code projector $P = V V^{*}$ ; An isometric quantum encoding gives an orthogonal code projector gives $V^{*} V = I_{H_{L}}$ , so ${Ad}_{V} (X) = V X V^{*}$ is trace-preserving. Let $N : B (H_{P}) \to B (H_{O})$ be a channel. The code is exactly correctable for $N$ if there is a decoded recovery channel $R : B (H_{O}) \to B (H_{L})$ for which

$R \circ N \circ {Ad}_{V} = {id}_{B (H_{L})}$

as linear maps, where ${Ad}_{V} (X) = V X V^{*}$ . Equivalently, the physical recovery ${Ad}_{V} \circ R$ satisfies $({Ad}_{V} \circ R \circ N \circ {Ad}_{V}) (ρ) = V ρ V^{*}$ for every positive semidefinite trace-one logical operator $ρ$ .

It suffices to check equality on the rank-one projectors $P_{ψ}$ from Agreement on pure states implies complete agreement for unit vectors $ψ$ : by that lemma, agreement on those projectors implies agreement on every operator and after tensoring with the identity on any finite reference system.

For any chosen finite Kraus representation $N (X) = \sum_{a} E_{a} X E_{a}^{*}$ , trace preservation gives $\sum_{a} E_{a}^{*} E_{a} = I_{H_{P}}$ by Trace preservation is Kraus normalization.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Kraus operators in an identity-channel representation are scalar

Statement

Let $H$ be a nonzero finite-dimensional complex inner-product space with the library convention linear in the first argument. Let $K_{1}, \dots, K_{m} : H \to H$ be a finite family whose Kraus map is the identity channel, so $\sum_{j} K_{j} X K_{j}^{*} = X$ for every $X \in End (H)$ . Then for each $j$ there is a scalar $c_{j} \in C$ such that $K_{j} = c_{j} I_{H}$ .

Facts & Assumptions

Given: The finite Kraus family and identity-map equality in the Statement.

[F1]

Real and complex inner product spaces, with the inner product linear in the first argument fixes linearity in the first argument and conjugate-linearity in the second.

[F2]

Finite Kraus representation: the displayed finite operator sum represents the map on every endomorphism.

[F3]

Finite-dimensional quantum channel: a channel is completely positive and trace-preserving on the full endomorphism space.

[F4]

Every finite-dimensional real or complex inner product space has an orthonormal basis: $H$ has a finite orthonormal basis.

[F5]

For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ : a finite-dimensional inner-product space is the orthogonal direct sum of a subspace and its orthogonal complement.

[F6]

Every linear map between finite-dimensional inner product spaces has a unique adjoint: each finite linear operator has an adjoint satisfying the defining pairing identity.

Proof

1.1givenF1F2F3F5F6algebra

For a unit vector $ψ$ , define $P_{ψ} z = ⟨ z, ψ ⟩ ψ$ . The adjoint identity [F6] rewrites $K_{j} P_{ψ} K_{j}^{*} z = ⟨ z, K_{j} ψ ⟩ K_{j} ψ$ , so the assumed channel identity gives $\sum_{j} K_{j} P_{ψ} K_{j}^{*} = P_{ψ}$ . If $u ⊥ ψ$ , then $⟨ P_{ψ} u, u ⟩ = 0$ and $⟨ K_{j} P_{ψ} K_{j}^{*} u, u ⟩ = ∣ ⟨ u, K_{j} ψ ⟩ ∣^{2}$ . Thus $\sum_{j} ∣ ⟨ u, K_{j} ψ ⟩ ∣^{2} = 0$ , so each inner product vanishes. For a fixed $j$ , write $K_{j} ψ = c ψ + v$ with $v ⊥ ψ$ using [F5]. The vanishing inner product with $u = v$ gives $0 = ⟨ v, K_{j} ψ ⟩ = ∥ v ∥^{2}$ , so $v = 0$ and $K_{j} ψ \in span {ψ}$ .

2.1step 1.1F4algebrachoose

If $\dim H = 1$ , each linear endomorphism is scalar, so the conclusion holds. If $\dim H \geq 2$ , choose an orthonormal basis $(e_{1}, \dots, e_{n})$ by [F4]. Step 1.1 gives $K_{j} e_{r} = c_{j r} e_{r}$ . For $r \neq s$ , applying step 1.1 to $(e_{r} + e_{s}) / \sqrt{2}$ shows $c_{j r} e_{r} + c_{j s} e_{s} \in span {e_{r} + e_{s}}$ , hence $c_{j r} = c_{j s}$ . Thus $c_{j r}$ is independent of $r$ for each $j$ .

3.1step 2.1algebra∎

In either dimension case, each $K_{j}$ acts by one scalar on a basis of $H$ , and therefore $K_{j} = c_{j} I_{H}$ on all of $H$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Orthogonal error restrictions determine syndrome projections

Statement

Let $H_{L}, H_{O}$ be finite-dimensional complex inner-product spaces, and let $W_{i} : H_{L} \to H_{O}$ be finitely many isometries with pairwise orthogonal ranges. Then $P_{i} = W_{i} W_{i}^{*}$ are pairwise orthogonal projections, their sum $S = \sum_{i} P_{i}$ is an orthogonal projection satisfying $0 \leq S \leq I_{H_{O}}$ , and $Q = I_{H_{O}} - S$ is the orthogonal projection onto the residual complement. The empty family is allowed, with $S = 0$ and $Q = I_{H_{O}}$ .

Facts & Assumptions

Given: Finite-dimensional complex inner-product spaces $H_{L}, H_{O}$ and a finite family of isometries $W_{i} : H_{L} \to H_{O}$ with pairwise orthogonal ranges.

[F2]

An endomorphism is an orthogonal projection exactly when it is idempotent and self-adjoint: a self-adjoint idempotent on a finite-dimensional inner-product space is an orthogonal projection.

[F3]

The orthogonal projection $P_{W} v$ is the $W$ -component in $V = W \oplus W^{⊥}$ : an orthogonal projection selects the component in its range along the orthogonal complement.

[F4]

Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : $(A B)^{*} = B^{*} A^{*}$ and $T^{* *} = T$ for compatible finite-dimensional maps.

[F5]

Non-negative and positive operators: $0 \leq T$ means $T$ is self-adjoint and $⟨ T x, x ⟩ \geq 0$ for every $x$ .

[F6]

Real and complex inner product spaces, with the inner product linear in the first argument: $⟨ x, x ⟩ = ∥ x ∥^{2}$ and the inner product is positive definite.

[F7]

An isometric quantum encoding gives an orthogonal code projector: every finite-dimensional linear isometry $W$ satisfies $W^{*} W = I$ .

[F8]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ and Every linear map between finite-dimensional inner product spaces has a unique adjoint: adjoints exist and satisfy $⟨ W x, y ⟩ = ⟨ x, W^{*} y ⟩$ .

Proof

1.1givenF2F4F7F8algebra

By [F7, F8], each $W_{i}$ has an adjoint and $W_{i}^{*} W_{i} = I$ . For each $i$ , [F4] gives $P_{i}^{*} = (W_{i} W_{i}^{*})^{*} = W_{i} W_{i}^{*}$ , and $P_{i}^{2} = W_{i} (W_{i}^{*} W_{i}) W_{i}^{*} = W_{i} W_{i}^{*} = P_{i}$ by [F7]. The image of $P_{i}$ is contained in $ran W_{i}$ , while $P_{i} W_{i} x = W_{i} x$ for every $x$ ; hence its image equals $ran W_{i}$ . By [F2], $P_{i}$ is the orthogonal projection onto that range.

2.1givenstep 1.1F2F4F6F8algebra

If $i \neq j$ , orthogonality of the ranges gives $⟨ x, W_{i}^{*} W_{j} y ⟩ = ⟨ W_{i} x, W_{j} y ⟩ = 0$ for all $x, y \in H_{L}$ by [F8]. Taking $x = W_{i}^{*} W_{j} y$ shows $W_{i}^{*} W_{j} = 0$ by [F6]. Hence $P_{i} P_{j} = W_{i} (W_{i}^{*} W_{j}) W_{j}^{*} = 0$ . Thus $S^{*} = S$ and $S^{2} = \sum_{i} P_{i}^{2} + \sum_{i \neq j} P_{i} P_{j} = S$ ; [F2] makes $S$ an orthogonal projection. The same calculation gives $(I - S)^{*} = I - S = (I - S)^{2}$ , so [F2] makes $Q = I - S$ an orthogonal projection.

3.1step 1.1step 2.1F2F3F5F6algebra∎

Each $P_{i}$ fixes its own range and kills the other ranges, so $ran S$ is exactly the orthogonal sum of the $ran W_{i}$ . Since $Q = I - S$ is an orthogonal projection, its range is $\ker S$ , the orthogonal complement of the range of $S$ by [F3]. For every $x \in H_{O}$ , $⟨ S x, x ⟩ = ∥ S x ∥^{2} \geq 0$ and $⟨ Q x, x ⟩ = ∥ Q x ∥^{2} \geq 0$ , so [F5] gives $0 \leq S \leq I$ . For the empty family the sums give $S = 0, Q = I$ directly.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Preserving the logical marginal forces a gauge-only channel

Statement

Let $H_{G}, H_{L}$ be nonzero finite-dimensional complex inner-product spaces in the fixed order $H_{G} \otimes H_{L}$ . Let $Φ : B (H_{G} \otimes H_{L}) \to B (H_{G} \otimes H_{L})$ be a channel with a finite Kraus family $K_{1}, \dots, K_{m}$ . If ${Tr}_{G} (Φ (ρ)) = {Tr}_{G} (ρ)$ for every positive semidefinite trace-one operator $ρ$ on $H_{G} \otimes H_{L}$ , then there are operators $A_{j} : H_{G} \to H_{G}$ such that $K_{j} = A_{j} \otimes I_{H_{L}}$ for every $j$ and $\sum_{j} A_{j}^{*} A_{j} = I_{H_{G}}$ . Consequently $Φ = T_{G} \otimes {id}_{B (H_{L})}$ for the channel $T_{G} (X) = \sum_{j} A_{j} X A_{j}^{*}$ , and the same marginal identity holds after tensoring with the identity on any finite reference space. Conversely, any finite family $A_{j}$ with $\sum_{j} A_{j}^{*} A_{j} = I_{H_{G}}$ defines a channel with this marginal-preserving property.

Also let $J : H_{G} \otimes H_{L} \to H_{P}$ be an isometric encoding and let $D : B (H_{O}) \to B (H_{G} \otimes H_{L})$ be a channel. For every unit $g_{0} \in H_{G}$ , the gauge-reset map $X ⟼ J (P_{g_{0}} \otimes {Tr}_{G} (D (X))) J^{*},$ where $P_{g_{0}} x = ⟨ x, g_{0} ⟩ g_{0}$ , is a channel into the encoded subspace $im J$ .

Facts & Assumptions

Given: The finite-dimensional spaces and channel in the first part of the Statement; and, for its final assertion, the isometry $J$ and decoded channel $D$ there specified.

[F1]

Partial trace over the first tensor factor: in the order $H_{G} \otimes H_{Q}$ , the matrix entries are $({Tr}_{G} X)_{α β} = \sum_{s} X_{(s, α), (s, β)}$ , and the same formula applies with $H_{Q} = H_{L} \otimes H_{R}$ .

[F2]

Real and complex inner product spaces, with the inner product linear in the first argument and The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ : the pairing is linear in its first argument and $⟨ T x, y ⟩ = ⟨ x, T^{*} y ⟩$ .

[F3]

Every finite-dimensional real or complex inner product space has an orthonormal basis: finite-dimensional inner-product spaces have orthonormal bases. For a subspace $W$ of a finite-dimensional inner product space, $V = W \oplus W^{⊥}$ : a vector can be extended to an orthonormal basis by adjoining an orthonormal basis of its orthogonal complement.

[F4]

Finite Kraus representation and Finite-dimensional quantum channel: a finite Kraus family represents its channel on the full operator space, and a channel is completely positive and trace-preserving.

[F5]

Trace preservation is Kraus normalization: a finite Kraus family is trace-preserving exactly when its adjoint-square sum is the identity.

[F6]

An isometric quantum encoding gives an orthogonal code projector: $J^{*} J = I$ for an isometric encoding.

[F7]

Tensor products of finitely many linear maps and Tensor maps preserve composition, identities, and adjoints: tensor maps act on elementary tensors and $(A \otimes B)^{*} = A^{*} \otimes B^{*}$ .

[F8]

The product basis is orthonormal and tensor dimensions multiply: elementary product tensors form a basis and span the finite tensor product.

[F9]

A finite Kraus sum is completely positive: every finite Kraus sum is completely positive.

[F10]

Kraus and Choi characterizations of complete positivity: every finite-dimensional completely positive map has a finite Kraus representation.

[F11]

Product operator matrix units form a basis: for finite-dimensional factors with ordered bases, tensor products of operator matrix units form a basis of the endomorphisms of their product space. Finite iterated tensor products represent multilinear maps independently of parenthesization supplies the canonical reassociation for the ordered three-factor register.

Proof

1.1givenF1F2F3F4choosealgebra

Fix unit vectors $g \in H_{G}$ and $ℓ \in H_{L}$ , and write $P_{ℓ} x = ⟨ x, ℓ ⟩ ℓ$ . Extend $ℓ$ to an orthonormal basis $(e_{0}, e_{1}, \dots, e_{q - 1})$ of $H_{L}$ with $e_{0} = ℓ$ , using [F3]. For each $j$ , expand $K_{j} (g \otimes ℓ) = \sum_{k < q} y_{j k} \otimes e_{k}$ . Apply the marginal hypothesis to the rank-one density operator $P_{g \otimes ℓ}$ . By [F4] the output is $\sum_{j} P_{K_{j} (g \otimes ℓ)}$ . In each diagonal entry $k, k$ , the partial-trace formula [F1] sums the squared coordinates of $y_{j k}$ over an orthonormal basis of $H_{G}$ , so that entry is $\sum_{j} ∥ y_{j k} ∥^{2}$ . The required marginal is $P_{ℓ}$ , whose $k$ th diagonal entry is zero for every $k > 0$ . Since each squared norm is nonnegative, $y_{j k} = 0$ for all $j$ and $k > 0$ . Hence $K_{j} (g \otimes ℓ) = η_{j, g, ℓ} \otimes ℓ$ for some $η_{j, g, ℓ} \in H_{G}$ .

2.1step 1.1algebra

Fix a unit $g$ . If unit vectors $ℓ_{1}, ℓ_{2}$ are linearly independent, then $ℓ_{1} + ℓ_{2} \neq 0$ ; normalize their sum and apply step 1.1 to $ℓ_{1}, ℓ_{2}$ and $(ℓ_{1} + ℓ_{2}) / ∥ ℓ_{1} + ℓ_{2} ∥$ . Linearity of $K_{j}$ gives $η_{j, g, ℓ_{1}} \otimes ℓ_{1} + η_{j, g, ℓ_{2}} \otimes ℓ_{2} = η_{j, g, ℓ_{+}} \otimes (ℓ_{1} + ℓ_{2})$ . Extend $ℓ_{1}, ℓ_{2}$ to a basis and apply its two corresponding coordinate functionals to the second tensor factor; this gives $η_{j, g, ℓ_{1}} = η_{j, g, ℓ_{+}} = η_{j, g, ℓ_{2}}$ . If $ℓ_{2} = c ℓ_{1}$ for a unit scalar $c$ , linearity and $c \neq 0$ give the same equality directly. Thus $η_{j, g, ℓ}$ is independent of the unit vector $ℓ$ .

3.1step 1.1step 2.1F8constructchoose

Fix a unit $ℓ_{0} \in H_{L}$ . For unit $g$ , step 1.1 shows $K_{j} (g \otimes ℓ_{0})$ lies in $H_{G} \otimes span {ℓ_{0}}$ . Scaling gives the same for every $g$ . Define $A_{j} g$ by $K_{j} (g \otimes ℓ_{0}) = A_{j} g \otimes ℓ_{0}$ ; it is unique because $u \mapsto u \otimes ℓ_{0}$ is injective. The linearity of $K_{j}$ makes $A_{j}$ linear. By step 2.1, $K_{j} (g \otimes ℓ) = A_{j} g \otimes ℓ$ for all simple tensors. Since simple tensors span by [F8], $K_{j} = A_{j} \otimes I_{H_{L}}$ .

4.1F5F7step 3.1algebrachoose

Trace preservation [F5] and step 3.1 give $(\sum_{j} A_{j}^{*} A_{j}) \otimes I_{H_{L}} = I_{H_{G}} \otimes I_{H_{L}}$ , using the tensor-adjoint rule [F7]. Apply both sides to $g \otimes ℓ_{0}$ for arbitrary $g$ and a fixed unit $ℓ_{0}$ . Injectivity of $u \mapsto u \otimes ℓ_{0}$ yields $\sum_{j} A_{j}^{*} A_{j} = I_{H_{G}}$ .

5.1step 3.1step 4.1F5F9F11algebra

Define $T_{G} (X) = \sum_{j} A_{j} X A_{j}^{*}$ . Its finite Kraus form makes it completely positive by [F9], and the normalization from step 4.1 makes it trace-preserving by [F5]. Thus it is a channel. On a simple operator tensor $X_{G} \otimes X_{L}$ , step 3.1 gives $Φ (X_{G} \otimes X_{L}) = \sum_{j} (A_{j} X_{G} A_{j}^{*}) \otimes X_{L} = T_{G} (X_{G}) \otimes X_{L}$ . By [F11], tensor products of operator matrix units form a basis of the full operator space, so linearity gives $Φ = T_{G} \otimes {id}_{B (H_{L})}$ .

5.2F1F3F8F11step 4.1algebrachoose

For a finite reference space $H_{R}$ , choose orthonormal bases of $H_{G}, H_{L}, H_{R}$ by [F3]; [F8] gives the product basis of $H_{L} \otimes H_{R}$ . Use the canonical reassociation in [F11] and the operator matrix-unit basis there to write any operator $ρ$ on $(H_{G} \otimes H_{L}) \otimes H_{R}$ as $ρ = \sum_{s, t} ∣ g_{s} ⟩ ⟨ g_{t} ∣ \otimes ρ_{s t}$ , with $ρ_{s t} \in B (H_{L} \otimes H_{R})$ . Write $A_{j} g_{s} = \sum_{r} a_{r s}^{(j)} g_{r}$ . By the first-factor partial-trace formula [F1], the coefficient of $ρ_{s t}$ after applying $Φ \otimes {id}_{B (H_{R})}$ and tracing out $G$ is $\sum_{j, r} a_{r s}^{(j)} \overline{a_{r t}^{(j)}} = \sum_{j} (A_{j}^{*} A_{j})_{t s} = δ_{t s}$ . Therefore the output marginal is $\sum_{s} ρ_{s s} = ({Tr}_{G} \otimes {id}_{B (H_{R})}) (ρ)$ . The calculation applies to every operator, hence to every density operator and every finite reference.

6.1givenF1F2F5F6F7F8F9F10constructchoose∎

For the gauge-reset assertion, choose an orthonormal basis $(g_{s})$ of $H_{G}$ and define $C_{s} : H_{G} \otimes H_{L} \to H_{L}$ by $C_{s} (g \otimes ℓ) = ⟨ g, g_{s} ⟩ ℓ$ . Its adjoint is $C_{s}^{*} ℓ = g_{s} \otimes ℓ$ , so $\sum_{s} C_{s}^{*} C_{s} (g \otimes ℓ) = \sum_{s} ⟨ g, g_{s} ⟩ g_{s} \otimes ℓ = g \otimes ℓ$ ; simple tensors span, hence $\sum_{s} C_{s}^{*} C_{s} = I_{H_{G} \otimes H_{L}}$ . On the product basis, the matrix entries of $\sum_{s} C_{s} Y C_{s}^{*}$ are exactly the first-factor contraction entries in [F1], hence ${Tr}_{G} (Y) = \sum_{s} C_{s} Y C_{s}^{*}$ . Let $J_{g_{0}} : H_{L} \to H_{P}$ be $J_{g_{0}} ℓ = J (g_{0} \otimes ℓ)$ . It is an isometry by [F6]. Choose finite Kraus operators $D_{r}$ for $D$ using [F10]. Since $J_{g_{0}} Y J_{g_{0}}^{*} = J (P_{g_{0}} \otimes Y) J^{*}$ , the gauge-reset map equals $X \mapsto \sum_{s, r} J_{g_{0}} C_{s} D_{r} X D_{r}^{*} C_{s}^{*} J_{g_{0}}^{*}$ , so it has Kraus operators $J_{g_{0}} C_{s} D_{r}$ , and $\sum_{s, r} (J_{g_{0}} C_{s} D_{r})^{*} (J_{g_{0}} C_{s} D_{r}) = \sum_{r} D_{r}^{*} (\sum_{s} C_{s}^{*} J_{g_{0}}^{*} J_{g_{0}} C_{s}) D_{r} = \sum_{r} D_{r}^{*} D_{r} = I_{H_{O}}$ . Thus the map is completely positive by [F9] and trace-preserving by [F5], so it is a channel. Its range is contained in $im J$ by construction.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

An explicit syndrome recovery corrects a Knill–Laflamme error family

Statement

Let $V : H_{L} \to H_{P}$ be an isometry with $H_{L} \neq {0}$ and $P = V V^{*}$ . Let $N (X) = \sum_{a = 1}^{m} E_{a} X E_{a}^{*}$ be a finite Kraus representation of a channel from $B (H_{P})$ to $B (H_{O})$ . If $P E_{a}^{*} E_{b} P = C_{a b} P$ for all $a, b$ , then there is a finite Kraus recovery channel $R : B (H_{O}) \to B (H_{L})$ satisfying $R \circ N \circ {Ad}_{V} = {id}_{B (H_{L})}$ . It is obtained by diagonalizing $C$ , decoding each nonzero orthogonal error range, and sending the orthogonal complement to a fixed logical state. For this recovery and every $X \in B (H_{L})$ , the off-diagonal error action is $R (E_{a} V X V^{*} E_{b}^{*}) = C_{b a} X$ .

Facts & Assumptions

Given: The finite-dimensional code isometry, the finite Kraus channel, and the displayed scalar-compression condition.

[F1]

Quantum code as an isometric encoding and its code projector: $V$ is an isometric encoding and $P = V V^{*}$ .

[F2]

An isometric quantum encoding gives an orthogonal code projector: $V^{*} V = I_{H_{L}}$ .

[F3]

Finite-dimensional quantum channel and Finite Kraus representation: channels are completely positive and trace-preserving, and a finite Kraus family represents the map on the full operator space.

[F4]

Trace preservation is Kraus normalization: a finite Kraus representation is trace-preserving exactly when its adjoint-square sum is the identity.

[F5]

A finite Kraus sum is completely positive: every finite Kraus sum is completely positive.

[F6]

A unitary error-basis change gives orthogonal error ranges: after unitary error mixing, each nonzero restriction is $\sqrt{λ_{i}} W_{i}$ with $W_{i}^{*} W_{j} = δ_{i j} I$ ; zero-eigenvalue restrictions vanish.

[F7]

Orthogonal error restrictions determine syndrome projections: the ranges of the $W_{i}$ have orthogonal projections $P_{i} = W_{i} W_{i}^{*}$ and $Q = I - \sum_{i} P_{i}$ is the projection onto their orthogonal complement.

[F8]

Every finite-dimensional real or complex inner product space has an orthonormal basis: every finite-dimensional inner-product space has an orthonormal basis, including its finite-dimensional subspaces.

[F9]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Every linear map between finite-dimensional inner product spaces has a unique adjoint, and Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : adjoints exist uniquely in finite dimension, $T^{* *} = T$ , and $(S T)^{*} = T^{*} S^{*}$ .

[F10]

Agreement on pure states implies complete agreement: if two complex-linear maps agree on all rank-one projectors, they agree on the whole operator space and after tensoring with the identity on any finite reference.

Proof

1.1givenF1F6

Apply [F6] to diagonalize the coefficient matrix and replace the Kraus operators by the unitarily mixed family $F_{i}$ . The operator-sum map remains $N$ , the positive eigenvalues are $λ_{i}$ , and $F_{i} V = \sqrt{λ_{i}} W_{i}$ for $λ_{i} > 0$ , while $F_{i} V = 0$ when $λ_{i} = 0$ .

1.2F7F8constructchoose

For each $λ_{i} > 0$ , define $R_{i} = W_{i}^{*} : H_{O} \to H_{L}$ . Choose a unit vector $ℓ_{0} \in H_{L}$ and an orthonormal basis $(q_{s})$ of $Q H_{O}$ using [F8]; for each basis vector define $T_{s} x = ⟨ x, q_{s} ⟩ ℓ_{0}$ . These are finite Kraus operators.

2.1givenF2F4step 1.1algebra

Trace preservation [F4] and $V^{*} V = I$ from [F2] give $V^{*} (\sum_{i} F_{i}^{*} F_{i}) V = I_{H_{L}}$ . Using step 1.1, the left side is $(\sum_{i} λ_{i}) I_{H_{L}}$ . Since $H_{L}$ is nonzero, $\sum_{i} λ_{i} = 1$ .

2.2F3F4F5F7F9step 1.2algebra

By [F9], $R_{i}^{*} R_{i} = W_{i} W_{i}^{*}$ . From $T_{s} x = ⟨ x, q_{s} ⟩ ℓ_{0}$ and the adjoint definition, $T_{s}^{*} y = ⟨ y, ℓ_{0} ⟩ q_{s}$ ; hence $T_{s}^{*} T_{s} x = ⟨ x, q_{s} ⟩ q_{s}$ . Summing over the orthonormal basis of $Q H_{O}$ gives $\sum_{s} T_{s}^{*} T_{s} = Q$ . Therefore $\sum_{λ_{i} > 0} R_{i}^{*} R_{i} + \sum_{s} T_{s}^{*} T_{s} = \sum_{λ_{i} > 0} W_{i} W_{i}^{*} + Q = I_{H_{O}}$ by [F7]. The finite Kraus map is completely positive by [F5] and trace-preserving by [F4], so it is a recovery channel.

3.1step 1.1step 2.1step 1.2F6F7F9algebra

For positive $λ_{j}$ , $R_{i} F_{j} V = \sqrt{λ_{j}} W_{i}^{*} W_{j} = δ_{i j} \sqrt{λ_{j}} I_{H_{L}}$ ; if $λ_{j} = 0$ then $F_{j} V = 0$ . Also $T_{s} F_{j} V = 0$ because $Q W_{j} = 0$ . Hence for every $X \in B (H_{L})$ , $(R \circ N \circ {Ad}_{V}) (X) = \sum_{j} λ_{j} X = X$ by step 2.1.

4.1step 3.1F10

The equality in step 3.1 is equality of linear maps on all operators. Tensoring that equality with the identity on any finite reference gives the same equality there by [F10].

5.1step 1.1step 1.2step 2.2step 3.1F6F7F9algebra∎

The same recovery Kraus actions in step 3.1 show $R (F_{i} V X V^{*} F_{j}^{*}) = δ_{i j} λ_{i} X$ for every $X \in B (H_{L})$ ; zero-eigenvalue terms vanish because $F_{i} V = 0$ . Since $F_{i} = \sum_{a} E_{a} U_{a i}$ , unitarity gives $E_{a} = \sum_{i} F_{i} \overline{U_{a i}}$ . Therefore $R (E_{a} V X V^{*} E_{b}^{*}) = \sum_{i} \overline{U_{a i}} U_{b i} λ_{i} X = C_{b a} X$ , using $C = U diag (λ_{i}) U^{*}$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Exact recovery implies the Knill–Laflamme condition

Statement

Let $N (X) = \sum_{a = 1}^{m} E_{a} X E_{a}^{*}$ be a finite Kraus representation of a channel from $B (H_{P})$ to $B (H_{O})$ , and let $V : H_{L} \to H_{P}$ be an isometry with $H_{L} \neq {0}$ and $P = V V^{*}$ . If the code is exactly correctable in the sense of Exact recovery of an encoded quantum code, then there are scalars $C_{a b}$ such that $P E_{a}^{*} E_{b} P = C_{a b} P$ for all $a, b$ .

Facts & Assumptions

Given: The finite-dimensional encoding, Kraus channel, and an exact decoded recovery channel $R : B (H_{O}) \to B (H_{L})$ .

[A1]

Exact recovery of an encoded quantum code: exactness means $R \circ N \circ {Ad}_{V}$ is the identity channel on $B (H_{L})$ .

[A2]

Finite-dimensional quantum channel and Finite Kraus representation specify the CPTP channel and finite operator-sum conventions.

[A3]

Kraus and Choi characterizations of complete positivity: every finite-dimensional completely positive map has a finite Kraus representation.

[A4]

Trace preservation is Kraus normalization: a trace-preserving finite Kraus map satisfies $\sum_{j} R_{j}^{*} R_{j} = I$ .

[F1]

Kraus operators in an identity-channel representation are scalar: every operator in a finite Kraus representation of the identity channel is a scalar multiple of the identity.

[F2]

The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ and Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : adjoints satisfy the pairing identity and reverse compatible compositions.

Proof

1.1givenA1A2A3algebrachoose

By [A3], choose a finite Kraus representation $R (X) = \sum_{j} R_{j} X R_{j}^{*}$ . Composing its Kraus operators with the errors $E_{a}$ and the encoding $V$ gives the Kraus family $R_{j} E_{a} V : H_{L} \to H_{L}$ for $R \circ N \circ {Ad}_{V}$ . By exactness [A1], this family represents the identity channel.

2.1step 1.1F1

Apply [F1] to the finite Kraus family in step 1.1. For each $j, a$ there is a scalar $c_{j a}$ with $R_{j} E_{a} V = c_{j a} I_{H_{L}}$ .

3.1A4F2step 2.1algebra

Trace preservation of $R$ gives $\sum_{j} R_{j}^{*} R_{j} = I_{H_{O}}$ by [A4]. Therefore, for every $a, b$ , $V^{*} E_{a}^{*} E_{b} V = V^{*} E_{a}^{*} (\sum_{j} R_{j}^{*} R_{j}) E_{b} V = \sum_{j} (R_{j} E_{a} V)^{*} (R_{j} E_{b} V) = (\sum_{j} \overline{c_{j a}} c_{j b}) I_{H_{L}} .$

4.1givenstep 3.1algebra∎

Put $C_{a b} = \sum_{j} \overline{c_{j a}} c_{j b}$ . Since $P = V V^{*}$ , $P E_{a}^{*} E_{b} P = V (V^{*} E_{a}^{*} E_{b} V) V^{*} = C_{a b} P$ by step 3.1. This is the Knill–Laflamme condition for every pair of errors.

CorollaryStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

Distance corrects unlocated errors and known erasures

Statement

Let $V : H_{L} \to H_{P}$ be an isometric encoding into an ordered register of $n$ finite factors, as in Quantum code as an isometric encoding and its code projector, with code projector $P = V V^{*}$ and quantum distance $d$ as in Detectable operators and distance of a finite-register quantum code. Let $N (X) = \sum_{a} E_{a} X E_{a}^{*}$ be a channel on $H_{P}$ with a finite Kraus family, and let $t \geq 0$ be an integer. Support and weight have the factor-order convention of Support and weight of an operator on a finite register. If every Kraus error $E_{a}$ has weight at most $t$ and $2 t < d$ , then the code is exactly correctable. If a fixed, known erased set $S \subseteq [n]$ supports every Kraus error and $∣ S ∣ < d$ , then the code is exactly correctable. The first claim permits different unknown supports; the second uses the one known support.

Facts & Assumptions

Given: The finite code isometry and channel in the Statement, together with the stated support hypotheses.

[F1]

Support and weight of an operator on a finite register: $E$ is supported in $S$ when the declared factor permutation makes it $E_{S} \otimes I_{S^{c}}$ , and $wt (E)$ is the minimum support size.

[F2]

Finite registers and their ordered computational bases, Ordered tensor registers, their inner product, and wire permutations, Tensor products of finitely many linear maps, The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ , Tensor maps preserve composition, identities, and adjoints, and Finite wire permutations are unitary basis permutations: the register factors have the declared order; tensor maps act factorwise, preserve composition and adjoints, and wire permutations are bijective isometries of the ordered product spaces.

[F3]

Detectable operators and distance of a finite-register quantum code and Every operator below the quantum code distance is detectable: an operator of weight strictly below $d$ compresses to a scalar multiple of $P$ .

[F4]

Finite Kraus representation and Finite-dimensional quantum channel: the finite Kraus family represents a CPTP map.

[F5]

An explicit syndrome recovery corrects a Knill–Laflamme error family: if all products $E_{a}^{*} E_{b}$ compress to scalar multiples of the code projector, a finite trace-preserving recovery corrects the channel.

Proof

1.1givenF1F2choosealgebra

Fix Kraus errors $E_{a}, E_{b}$ and choose supports $S_{a}, S_{b}$ with $∣ S_{a} ∣, ∣ S_{b} ∣ \leq t$ , as allowed by [F1]. Put $U = S_{a} \cup S_{b}$ . Since $S_{a}, S_{b} \subseteq U$ , extend each local operator by identity on the additional factors of $U$ and use the wire permutations in [F2] to put both operators in the same factor order $H_{U} \otimes H_{U^{c}}$ . On every product basis vector in that order, each conjugated error acts on its own $S_{a}$ or $S_{b}$ coordinates and leaves every other coordinate fixed. The tensor-map definition in [F2] therefore gives forms ${\tilde{E}}_{a} \otimes I_{U^{c}}$ and ${\tilde{E}}_{b} \otimes I_{U^{c}}$ . The adjoint and composition identities in [F2] give $E_{a}^{*} E_{b} = ({\tilde{E}}_{a}^{*} {\tilde{E}}_{b}) \otimes I_{U^{c}}$ in that order. Thus $E_{a}^{*} E_{b}$ is supported in $U$ and has weight at most $∣ U ∣ \leq ∣ S_{a} ∣ + ∣ S_{b} ∣ \leq 2 t < d$ .

2.1step 1.1F3F4F5

By [F3], each product in step 1.1 is detectable, so there are scalars $C_{a b}$ with $P E_{a}^{*} E_{b} P = C_{a b} P$ . Applying [F5] gives one recovery for the entire Kraus family.

3.1givenF1F2F3F4F5algebra∎

If the known affected set is a fixed $S$ , every $E_{a}$ acts as the identity outside $S$ , and so does every product $E_{a}^{*} E_{b}$ by [F1,F2]. Its weight is at most $∣ S ∣ < d$ . Applying [F3] and then [F5] gives exact recovery. This uses the one known support $S$ rather than a union of two unknown supports.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6-luna)Open item page →

Exact correction of a protected subsystem

Definition

Let $H_{G}, H_{L}, H_{P}, H_{O}$ be nonzero finite-dimensional complex inner-product spaces with the fixed factor order $H_{G} \otimes H_{L}$ . Let $J : H_{G} \otimes H_{L} \to H_{P}$ be an isometric encoding and set $P = J J^{*}$ . The factors $H_{G}$ and $H_{L}$ are the gauge and protected logical subsystems. Let $N : B (H_{P}) \to B (H_{O})$ be a channel.

For any finite-dimensional residual space $H_{Q}$ , use the first-factor partial trace ${Tr}_{G}$ of Partial trace over the first tensor factor on $H_{G} \otimes H_{Q}$ . In particular, for a reference $H_{R}$ , take $H_{Q} = H_{L} \otimes H_{R}$ and leave that factor order unchanged.

A decoded recovery is a channel $R : B (H_{O}) \to B (H_{G} \otimes H_{L})$ . The subsystem $H_{L}$ is exactly correctable for $N$ under $J$ when some decoded recovery satisfies, for every finite-dimensional reference $H_{R}$ and every positive semidefinite trace-one operator $ρ$ on $(H_{G} \otimes H_{L}) \otimes H_{R}$ ,

$(({Tr}_{G} \otimes {id}_{B (H_{R}})) \circ ((R \circ N \circ {Ad}_{J}) \otimes {id}_{B (H_{R})})) (ρ) = ({Tr}_{G} \otimes {id}_{B (H_{R})}) (ρ),$

where ${Ad}_{J} (X) = J X J^{*}$ . The recovery may change the gauge state. The one-operator Kraus map ${Ad}_{J}$ is completely positive by A finite Kraus sum is completely positive and trace-preserving by $J^{*} J = I$ and Trace preservation is Kraus normalization. Thus $Φ = R \circ N \circ {Ad}_{J}$ is a channel by Finite-dimensional channels compose and tensor, and it has a finite Kraus representation by Kraus and Choi characterizations of complete positivity. Explicitly, if $(R_{b})$ and $(E_{a})$ represent $R$ and $N$ , then $(R_{b} E_{a} J)_{b, a}$ represents $Φ$ . Taking a one-dimensional reference in the displayed recovery condition gives precisely the marginal hypothesis of Preserving the logical marginal forces a gauge-only channel. By that lemma and its converse with arbitrary finite references, this is equivalently a corrected decoded channel of the form $T_{G} \otimes {id}_{B (H_{L})}$ for a channel $T_{G}$ on the gauge factor. For a unit vector $g_{0} \in H_{G}$ , resetting the decoded gauge and re-encoding uses $X \mapsto J (P_{g_{0}} \otimes {Tr}_{G} (R (X))) J^{*}$ , where $P_{g_{0}} (x) = ⟨ x, g_{0} ⟩ g_{0}$ ; the same supplier proves this is a channel. If $\dim H_{G} = 1$ , the marginal condition is ordinary exact recovery of the code subspace in the sense of Exact recovery of an encoded quantum code.

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

A degenerate code with a non-diagonal Knill–Laflamme matrix

Example

Take $H_{L} = C^{2}$ , $H_{P} = C^{2} \otimes C^{2}$ , with standard basis columns $e_{0} = (1, 0)^{T}$ and $e_{1} = (0, 1)^{T}$ , and define $V ψ = e_{0} \otimes ψ$ and $W ψ = e_{1} \otimes ψ$ . Put $P_{0} = e_{0} e_{0}^{*} = (\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array})$ and $P_{1} = e_{1} e_{1}^{*} = (\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array})$ . On the first factor use $X = (\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array})$ and $Z = (\begin{array}{cc} 1 & 0 \\ 0 & - 1 \end{array})$ . Define a channel with Kraus operators

$E_{1} = \frac{I_{4}}{\sqrt{3}}, E_{2} = \frac{X \otimes I_{2}}{\sqrt{3}}, E_{3} = \frac{Z \otimes I_{2}}{\sqrt{3}} .$

The code stores the logical qubit in the second factor. The errors $E_{1}$ and $E_{3}$ have the same restriction to the code, while $E_{2}$ moves the code to the orthogonal first-factor subspace.

Facts & Assumptions

Given: The finite-dimensional encoding and the three explicitly displayed error operators in the Example.

[F1]

Quantum code as an isometric encoding and its code projector supplies the isometric encoding and code-projector convention.

[F2]

Finite-dimensional quantum channel, Finite Kraus representation, and Trace preservation is Kraus normalization specify finite Kraus channels and their trace-preserving normalization.

[F3]

An explicit syndrome recovery corrects a Knill–Laflamme error family: the displayed recovery construction corrects every finite family satisfying the scalar-compression condition.

[F4]

A finite Kraus sum is completely positive: a finite Kraus sum is completely positive.

[F5]

Trace preservation is Kraus normalization: the adjoint-square sum of a trace-preserving Kraus family is the identity.

Verification

1.1givenF2F4F5algebra

Each of $I_{4}$ , $X \otimes I_{2}$ , and $Z \otimes I_{2}$ is unitary. Hence $\sum_{a = 1}^{3} E_{a}^{*} E_{a} = \frac{1}{3} (I_{4} + I_{4} + I_{4}) = I_{4}$ , so the displayed Kraus map is trace-preserving; it is completely positive because it is a finite Kraus sum.

2.1givenF1step 1.1algebra

With $P = V V^{*}$ , direct compression gives $V^{*} E_{a}^{*} E_{b} V = C_{a b} I_{L}, C = \frac{1}{3} (\begin{array}{ccc} 1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1 \end{array}) .$ Since $P = V V^{*}$ , this is equivalently $P E_{a}^{*} E_{b} P = C_{a b} P$ for every $a, b$ . In particular, $C$ is non-diagonal and has rank $2$ .

3.1step 2.1algebra

The orthonormal eigenvectors $u_{+} = (1, 0, 1) / \sqrt{2}$ , $u_{2} = (0, 1, 0)$ , and $u_{0} = (1, 0, - 1) / \sqrt{2}$ have eigenvalues $2 / 3, 1 / 3, 0$ , respectively. The corresponding mixed Kraus operators are $F_{+} = (I_{4} + Z \otimes I_{2}) / \sqrt{6} = \sqrt{2 / 3} (P_{0} \otimes I_{2})$ , $F_{2} = E_{2}$ , and $F_{0} = (I_{4} - Z \otimes I_{2}) / \sqrt{6} = \sqrt{2 / 3} (P_{1} \otimes I_{2})$ . The columns $u_{+}, u_{2}, u_{0}$ form a unitary matrix, so expanding the Kraus sum gives $\sum_{i} F_{i} Y F_{i}^{*} = \sum_{a} E_{a} Y E_{a}^{*}$ for every operator $Y$ .

4.1step 3.1algebra

By the definition of $W$ , one has $F_{+} V = \sqrt{2 / 3} V$ , $F_{2} V = \sqrt{1 / 3} W$ , and $F_{0} V = 0$ . The restrictions $V, W$ are isometries with orthogonal ranges, so the positive-eigenvalue error basis has exactly two syndromes; the zero eigenvalue error annihilates the code.

5.1step 4.1algebra

Define recovery Kraus operators $R_{+} = V^{*}$ and $R_{2} = W^{*}$ . Their adjoint products are $V V^{*} = P_{0} \otimes I_{2}$ and $W W^{*} = P_{1} \otimes I_{2}$ , so $R_{+}^{*} R_{+} + R_{2}^{*} R_{2} = I_{4}$ and the recovery is trace-preserving.

6.1step 2.1step 3.1step 5.1F1F3F5algebra∎

For the mixed errors, $R_{+} F_{+} V = \sqrt{2 / 3} I_{L}$ and $R_{2} F_{2} V = \sqrt{1 / 3} I_{L}$ ; all other products $R_{i} F_{j} V$ vanish. Thus for every $X \in B (H_{L})$ the corrected channel returns $(2 / 3 + 1 / 3) X = X$ . The unitary Kraus change preserves the original channel, so this recovery corrects the displayed three-error channel, including its degeneracy; this is the recovery construction in [F3].

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6-luna)Open item page →

The Knill–Laflamme criterion for a protected subsystem

Statement

Let $J : H_{G} \otimes H_{L} \to H_{P}$ be an isometric encoding between nonzero finite-dimensional complex inner-product spaces, let $P = J J^{*}$ , and let $N (X) = \sum_{a = 1}^{m} E_{a} X E_{a}^{*}$ be a finite Kraus representation of a channel from $B (H_{P})$ to $B (H_{O})$ . The protected subsystem $H_{L}$ is exactly correctable under $J$ , in the sense of Exact correction of a protected subsystem, if and only if there are operators $M_{a b} : H_{G} \to H_{G}$ such that $J^{*} E_{a}^{*} E_{b} J = M_{a b} \otimes I_{H_{L}} for every a, b .$

Equivalently, $P E_{a}^{*} E_{b} P = J (M_{a b} \otimes I_{H_{L}}) J^{*} .$ For any chosen orthonormal basis $(g_{k})$ of $H_{G}$ , put $Q_{k ℓ} = ∣ g_{k} ⟩ ⟨ g_{ℓ} ∣ \otimes I_{H_{L}}$ and $P_{k ℓ} = J Q_{k ℓ} J^{*}$ , where $∣ u ⟩ ⟨ v ∣$ means $x \mapsto ⟨ x, v ⟩ u$ under the library's linear-first convention. The same criterion is equivalent to the existence of scalars $λ_{a b k ℓ}$ such that $P_{k k} E_{a}^{*} E_{b} P_{ℓ ℓ} = λ_{a b k ℓ} P_{k ℓ} for every a, b, k, ℓ .$ These are equalities of physical-space operators using the embedded gauge matrix units $P_{k ℓ}$ .

Facts & Assumptions

Given: The finite-dimensional encoding $J$ , the finite Kraus channel $N$ , and the exact-correctability definition in Exact correction of a protected subsystem.

[F1]

Exact correction of a protected subsystem: a decoded channel $R$ witnesses exact correction when the recovered logical marginal equals the input logical marginal for every density operator, including every finite reference.

[F2]

Quantum code as an isometric encoding and its code projector and An isometric quantum encoding gives an orthogonal code projector: $J$ is isometric, $P = J J^{*}$ , and $J^{*} J = I_{H_{G} \otimes H_{L}}$ .

[F3]

Finite-dimensional quantum channel, Finite Kraus representation, Trace preservation is Kraus normalization, A finite Kraus sum is completely positive, and Kraus and Choi characterizations of complete positivity: channels are completely positive and trace-preserving, finite Kraus sums are completely positive, trace preservation is Kraus normalization, and finite-dimensional completely positive maps have finite Kraus representations.

[F4]

Preserving the logical marginal forces a gauge-only channel: a channel preserving the logical marginal has Kraus operators $A_{j} \otimes I_{H_{L}}$ , and the marginal identity holds with every finite reference; it also verifies the gauge-reset channel.

[F5]

An explicit syndrome recovery corrects a Knill–Laflamme error family: for its constructed ordinary recovery, if $C_{a b}$ is defined by $V^{*} E_{a}^{*} E_{b} V = C_{a b} I$ , then $R (E_{a} V X V^{*} E_{b}^{*}) = C_{b a} X$ for every operator $X$ .

[F6]

Exact recovery implies the Knill–Laflamme condition: exact recovery of an ordinary code implies the scalar-compression condition; An explicit syndrome recovery corrects a Knill–Laflamme error family proves its sufficiency.

[F11]

Exact correction of a protected subsystem and Exact recovery of an encoded quantum code: when the gauge factor is one-dimensional, the marginal recovery definition is ordinary encoded recovery.

[F7]

Every finite-dimensional real or complex inner product space has an orthonormal basis: every finite-dimensional inner-product space has an orthonormal basis.

[F8]

Real and complex inner product spaces, with the inner product linear in the first argument, The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ , Every linear map between finite-dimensional inner product spaces has a unique adjoint, and Adjoints satisfy $(S + T)^{*} = S^{*} + T^{*}$ , $(λ T)^{*} = \overline{λ} T^{*}$ , $(S T)^{*} = T^{*} S^{*}$ , and $T^{* *} = T$ : the pairing is linear in its first argument, finite-dimensional adjoints exist, and $(S T)^{*} = T^{*} S^{*}$ .

[F9]

Agreement on pure states implies complete agreement: equality of complex-linear maps on all operators gives equality after tensoring with the identity on any finite reference.

[F10]

Finite registers and their ordered computational bases, Ordered tensor registers, their inner product, and wire permutations, Tensor products of finitely many linear maps, Tensor maps preserve composition, identities, and adjoints, and The product basis is orthonormal and tensor dimensions multiply: the gauge/logical factor order is fixed, the maps $ι_{k} x = g_{k} \otimes x$ are isometries, and simple tensors span.

[F12]

Product operator matrix units form a basis: tensor products of operator matrix units form a basis of the endomorphisms of a finite-dimensional ordered product space.

Proof

1.1givenF1F2F3F4algebrachoose

For necessity, let $R : B (H_{O}) \to B (H_{G} \otimes H_{L})$ be an exact decoded recovery. By [F3], choose a finite Kraus family $(S_{r})$ for $R$ . The corrected map $Φ = R \circ N \circ {Ad}_{J}$ has Kraus operators $K_{r a} = S_{r} E_{a} J$ . It is completely positive by [F3], and $\sum_{r, a} K_{r a}^{*} K_{r a} = J^{*} \sum_{a} E_{a}^{*} (\sum_{r} S_{r}^{*} S_{r}) E_{a} J = J^{*} (\sum_{a} E_{a}^{*} E_{a}) J = J^{*} J = I_{H_{G} \otimes H_{L}}$ by [F2] and [F3]. Thus $Φ$ is a channel. By exactness [F1], it preserves the $H_{L}$ marginal; [F4] gives $S_{r} E_{a} J = A_{r a} \otimes I_{H_{L}}$ for some $A_{r a} : H_{G} \to H_{G}$ .

1.2givenF2F3F7F10constructchoose

For sufficiency, assume $J^{*} E_{a}^{*} E_{b} J = M_{a b} \otimes I_{H_{L}}$ . Let $d_{G} = \dim H_{G} > 0$ and choose an orthonormal basis $(g_{k})_{k < d_{G}}$ by [F7]. Define $ι_{k} : H_{L} \to H_{G} \otimes H_{L}$ by $ι_{k} x = g_{k} \otimes x$ , and put $F_{a k} = E_{a} J ι_{k}$ and ${\tilde{F}}_{a k} = F_{a k} / \sqrt{d_{G}}$ . By trace preservation [F3] and $J^{*} J = I$ from [F2], $\sum_{a, k} F_{a k}^{*} F_{a k} = \sum_{k} ι_{k}^{*} J^{*} (\sum_{a} E_{a}^{*} E_{a}) J ι_{k} = d_{G} I_{H_{L}} .$ Thus $({\tilde{F}}_{a k})$ is a finite Kraus representation of a channel $\tilde{N} : B (H_{L}) \to B (H_{O})$ .

2.1F2F3F4step 1.1algebraconstruct

Trace preservation of $R$ and $N$ gives $\sum_{r} S_{r}^{*} S_{r} = I_{H_{O}}$ and $\sum_{a} E_{a}^{*} E_{a} = I_{H_{P}}$ by [F3]. Hence $J^{*} E_{a}^{*} E_{b} J = J^{*} E_{a}^{*} (\sum_{r} S_{r}^{*} S_{r}) E_{b} J = \sum_{r} (S_{r} E_{a} J)^{*} (S_{r} E_{b} J) = (\sum_{r} A_{r a}^{*} A_{r b}) \otimes I_{H_{L}} .$ Set $M_{a b} = \sum_{r} A_{r a}^{*} A_{r b}$ . This proves the operator-valued condition.

2.2F3F5F8step 1.2algebra

For all $a, b, k, ℓ$ , the assumed condition gives ${\tilde{F}}_{a k}^{*} {\tilde{F}}_{b ℓ} = \frac{⟨ M_{a b} g_{ℓ}, g_{k} ⟩}{d_{G}} I_{H_{L}} .$ Apply [F5] to $\tilde{N}$ with the identity encoding of $H_{L}$ . It supplies a channel $S : B (H_{O}) \to B (H_{L})$ and a finite Kraus family $(S_{r})$ for which $S ({\tilde{F}}_{a k} X {\tilde{F}}_{b ℓ}^{*}) = \frac{⟨ M_{b a} g_{k}, g_{ℓ} ⟩}{d_{G}} X$ for every $X \in B (H_{L})$ .

3.1step 2.1F2F7F8algebrachoose

Fix an orthonormal gauge basis $(g_{k})$ using [F7]. The matrix entry of $M_{a b}$ in row $k$ , column $ℓ$ is $⟨ M_{a b} g_{ℓ}, g_{k} ⟩$ by [F8]. Therefore $P_{k k} E_{a}^{*} E_{b} P_{ℓ ℓ} = J (Q_{k k} (M_{a b} \otimes I_{H_{L}}) Q_{ℓ ℓ}) J^{*} = ⟨ M_{a b} g_{ℓ}, g_{k} ⟩ P_{k ℓ} .$ This proves the matrix-unit statement in the necessary direction.

3.2step 1.2step 2.2F2F3F7F9F10F12algebrachoose

Trace preservation and $J^{*} J = I$ imply $\sum_{a} M_{a a} \otimes I_{H_{L}} = J^{*} (\sum_{a} E_{a}^{*} E_{a}) J = I_{H_{G} \otimes H_{L}},$ so $\sum_{a} M_{a a} = I_{H_{G}}$ : applying the equality to each $g \otimes ℓ_{0}$ for fixed unit $ℓ_{0}$ gives $(\sum_{a} M_{a a} g - g) \otimes ℓ_{0} = 0$ , and tensoring with nonzero $ℓ_{0}$ is injective. Define the decoded recovery $T (X) = P_{g_{0}} \otimes S (X)$ for a fixed unit $g_{0} \in H_{G}$ . For a gauge matrix unit $∣ g_{k} ⟩ ⟨ g_{ℓ} ∣ \otimes X$ , the encoded channel output is $\sum_{a} F_{a k} X F_{a ℓ}^{*}$ ; applying $S$ and scaling the cross-term formula in step 2.2 from ${\tilde{F}}_{a k}$ to $F_{a k} = \sqrt{d_{G}} {\tilde{F}}_{a k}$ gives logical output $\sum_{a} S (F_{a k} X F_{a ℓ}^{*}) = \sum_{a} ⟨ M_{a a} g_{k}, g_{ℓ} ⟩ X = ⟨ g_{k}, g_{ℓ} ⟩ X = δ_{k ℓ} X .$ Here the scaling is $S (F_{a k} X F_{b ℓ}^{*}) = d_{G} S ({\tilde{F}}_{a k} X {\tilde{F}}_{b ℓ}^{*})$ , so the factor $d_{G}$ cancels the denominator in step 2.2. The input logical marginal of that matrix unit is also $δ_{k ℓ} X$ . By linearity and the product matrix-unit basis in [F12], the decoded map $T$ with output gauge reset preserves the logical marginal on every operator; [F9] extends the equality to every finite reference.

3.3F2step 2.1algebra

Since $P = J J^{*}$ , the operator-valued condition gives $P E_{a}^{*} E_{b} P = J (M_{a b} \otimes I_{H_{L}}) J^{*}$ . Conversely, multiplying this compressed equality by $J^{*}$ and $J$ and using $J^{*} J = I$ recovers the operator-valued condition.

4.1step 3.2F1F3F4F9algebraconstruct

The Kraus operators of $T$ are $T_{r} y = g_{0} \otimes S_{r} y$ . Since $g_{0}$ is unit, $\sum_{r} T_{r}^{*} T_{r} = \sum_{r} S_{r}^{*} S_{r} = I_{H_{O}}$ ; hence $T$ is a channel by [F3]. Step 3.2 proves it preserves the logical marginal on every operator, and [F9] extends that identity to every finite reference. The physical reset-and-encode map is a channel by [F4], so the protected subsystem is exactly correctable.

4.2F2F7step 3.3algebra

For a fixed orthonormal gauge basis, if the operator-valued condition holds, then $Q_{k k} (J^{*} E_{a}^{*} E_{b} J) Q_{ℓ ℓ} = ⟨ M_{a b} g_{ℓ}, g_{k} ⟩ Q_{k ℓ}$ , which gives the embedded matrix-unit condition after multiplying by $J$ and $J^{*}$ . Conversely, sum the embedded matrix-unit identities over $k, ℓ$ . Since $\sum_{k} P_{k k} = P$ and $J^{*} P_{k ℓ} J = Q_{k ℓ}$ , the resulting operator is $J^{*} E_{a}^{*} E_{b} J = (\sum_{k, ℓ} λ_{a b k ℓ} ∣ g_{k} ⟩ ⟨ g_{ℓ} ∣) \otimes I_{H_{L}}$ . This establishes equivalence for the fixed basis; because the operator-valued condition is basis-independent, it holds for every orthonormal gauge basis.

5.1F2F6F7F11algebra∎

If $\dim H_{G} = 1$ , every $M_{a b}$ is a scalar, so the criterion becomes $P E_{a}^{*} E_{b} P = C_{a b} P$ . Under the unitary identification $H_{G} \otimes H_{L} ≅ H_{L}$ , the first-factor trace leaves the logical operator unchanged, so the subsystem recovery definition [F11] is ordinary exact recovery. The necessity and sufficiency directions are [F6].

5 · Examples, counterexamples and false statements

None yet.

Knill–Laflamme Recovery and Subsystem Codes

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Agreement on pure states implies complete agreement

Statement

Facts & Assumptions

Proof

Product operator matrix units form a basis

Statement

Facts & Assumptions

Proof

Partial trace over the first tensor factor

Definition

The Knill–Laflamme coefficient matrix is positive

Statement

Facts & Assumptions

Proof

A unitary error-basis change gives orthogonal error ranges

Statement

Facts & Assumptions

Proof

Exact recovery of an encoded quantum code

Definition

Kraus operators in an identity-channel representation are scalar

Statement

Facts & Assumptions

Proof

Orthogonal error restrictions determine syndrome projections

Statement

Facts & Assumptions

Proof

Preserving the logical marginal forces a gauge-only channel

Statement

Facts & Assumptions

Proof

An explicit syndrome recovery corrects a Knill–Laflamme error family

Statement

Facts & Assumptions

Proof

Exact recovery implies the Knill–Laflamme condition

Statement

Facts & Assumptions

Proof

Distance corrects unlocated errors and known erasures

Statement

Facts & Assumptions

Proof

Exact correction of a protected subsystem

Definition

A degenerate code with a non-diagonal Knill–Laflamme matrix

Example

Facts & Assumptions

Verification

The Knill–Laflamme criterion for a protected subsystem

Statement

Facts & Assumptions

Proof

5 · Examples, counterexamples and false statements

Sources