LemmaStatement: AI-generatedProof: AI-generated

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A Knill–Laflamme decoder retaining the code syndrome

Statement

Let $V : H_{L} \to H_{P}$ be a finite-dimensional isometry with nonzero logical space, $P = V V^{†}$ , and let $L_{1}, \dots, L_{m}$ be finite matrices. Assume the scalar Knill–Laflamme products for $E_{0} = I, E_{a} = L_{a}$ : $V^{†} E_{a}^{†} E_{b} V = c_{a b} I$ . Put $S = \sum_{a} L_{a}^{†} L_{a}$ and $r = \sum_{a} c_{a a} \geq 0$ . There is a CPTP decoder $R$ such that, writing $V (X) = V X V^{†}$ , $R V (X) = X$ , $R (\sum_{a} L_{a} V (X) L_{a}^{†}) = r X$ and $R (S V (X)) = R (V (X) S) = r X$ . Moreover $R (V X Z^{†}) = R (Z X V^{†}) = 0$ whenever $V^{†} Z = 0$ . These are linear-map identities and persist under any finite reference extension.

Facts & Assumptions

[F1]

Isometric encoding and a decoded recovery have the stated finite channel types. Code restriction and decoded channel types

[F2]

A finite scalar-product error family can be unitarily mixed into positive-weight orthogonal isometric error ranges, with zero-weight restrictions vanishing. A unitary error-basis change gives orthogonal error ranges

[F3]

For finite-dimensional complex inner-product spaces, a complex-linear map given by a finite Kraus sum is a quantum channel when the Kraus operators have adjoint-square sum equal to the input identity (condition 3 implies condition 1). Finite channel, Kraus, Choi, and Stinespring equivalence

[F4]

The tensor product of finite-dimensional linear maps is the unique linear map with the specified factorwise action on elementary tensors. Tensor products of finitely many linear maps

Proof

Given: The isometry and scalar products including the identity. Goal: Construct a decoder with the claimed branch and cross-term identities.

1.1givenF1F2

Write $ℓ_{a} = c_{0 a}$ and $F_{a} = L_{a} - ℓ_{a} I$ . Then $V^{†} F_{a} V = 0$ and $V^{†} F_{a}^{†} F_{b} V = (c_{a b} - \overline{ℓ_{a}} ℓ_{b}) I$ . Apply [F2] to this centered family: its nonzero mixed restrictions are $\sqrt{λ_{j}} W_{j}$ , where the $W_{j}$ are isometries with pairwise orthogonal ranges. Since each $F_{a} V$ is orthogonal to $im V$ , every $W_{j}$ range is also orthogonal to that code range. The scalar Gram matrix is positive because $∥ \sum_{a} z_{a} F_{a} V x ∥^{2}$ is its quadratic form times $∥ x ∥^{2}$ .

2.1F1F2F3step 1.1

Put $Q = I - V V^{†} - \sum_{j} W_{j} W_{j}^{†}$ . Choose a unit logical vector $ℓ_{*}$ and an orthonormal basis $q_{k}$ of $Q H_{P}$ . The decoder has Kraus operators $V^{†}$ , $W_{j}^{†}$ , and $∣ ℓ_{*} ⟩ ⟨ q_{k} ∣$ . Their adjoint-square sum is $V V^{†} + \sum_{j} W_{j} W_{j}^{†} + Q = I$ , so [F3], applied with input space $H_{P}$ and output space $H_{L}$ , makes their finite Kraus sum a completely positive trace-preserving map. On a code/complement cross term every Kraus congruence vanishes: either the code factor or its orthogonal factor is annihilated. This proves the cross-term identity and $R V = id$ .

3.1givenF1step 1.1step 2.1

Expand $L_{a} V$ in the mutually orthogonal ranges $V, W_{j}$ . Each coefficient is scalar, so the decoder sends $L_{a} V X V^{†} L_{a}^{†}$ to the sum of their squared magnitudes times $X$ , which is $c_{a a} X$ because $V^{†} L_{a}^{†} L_{a} V = c_{a a} I$ . Summing gives $r X$ . Finally $V^{†} S V = r I$ , hence $S V = r V + Z$ with $V^{†} Z = 0$ . The cross-term property in step 2.1 gives both anticommutator identities. These are equalities on the full logical matrix space.

4.1F4step 2.1step 3.1∎

Let $H_{A}$ be any finite-dimensional complex reference space, and write ${id}_{A} = {id}_{L (H_{A})}$ . For $H = H_{L}$ or $H_{P}$ , identify $L (H) \otimes L (H_{A})$ with $L (H \otimes H_{A})$ by sending $A \otimes B$ to the operator $x \otimes y \mapsto A x \otimes B y$ . The latter operator is supplied by [F4]; in finite ordered bases the tensors of matrix units act as the matrix units of the product basis, so this correspondence is a linear isomorphism. By [F4], the amplification of any linear map $T$ on operator spaces satisfies $(T \otimes {id}_{A}) (X \otimes B) = T (X) \otimes B$ . Apply the identities in steps 2.1 and 3.1 to each $X \otimes B$ ; linearity gives the identities for every operator on the reference-extended space, since these elementary operators span it. This includes the cross-term maps for each fixed $Z$ with $V^{†} Z = 0$ .

Depends on

Used by · 2

Dependency tree · 2 levels

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