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An explicit syndrome recovery corrects a Knill–Laflamme error family

Statement

Let V:HL→HP be an isometry with HL≠{0} and P=VV∗. Let N(X)=∑a=1mEaXEa∗ be a finite Kraus representation of a channel from B(HP) to B(HO). If PEa∗EbP=CabP for all a,b, then there is a finite Kraus recovery channel R:B(HO)→B(HL) satisfying R∘N∘Ad⁡V=id⁡B(HL). 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∈B(HL), the off-diagonal error action is R(EaVXV∗Eb∗)=CbaX.

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=VV∗.

[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 λiWi with Wi∗Wj=δijI; zero-eigenvalue restrictions vanish.

[F7]

Orthogonal error restrictions determine syndrome projections: the ranges of the Wi have orthogonal projections Pi=WiWi∗ and Q=I−∑iPi 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.

[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 Fi. The operator-sum map remains N, the positive eigenvalues are λi, and FiV=λiWi for λi>0, while FiV=0 when λi=0.

1.2F7F8constructchoose

For each λi>0, define Ri=Wi∗:HO→HL. Choose a unit vector ℓ0∈HL and an orthonormal basis (qs) of QHO using [F8]; for each basis vector define Tsx=⟨x,qs⟩ℓ0. These are finite Kraus operators.

2.1givenF2F4step 1.1algebra

Trace preservation [F4] and V∗V=I from [F2] give V∗(∑iFi∗Fi)V=IHL. Using step 1.1, the left side is (∑iλi)IHL. Since HL is nonzero, ∑iλi=1.

2.2F3F4F5F7F9step 1.2algebra

By [F9], Ri∗Ri=WiWi∗. From Tsx=⟨x,qs⟩ℓ0 and the adjoint definition, Ts∗y=⟨y,ℓ0⟩qs; hence Ts∗Tsx=⟨x,qs⟩qs. Summing over the orthonormal basis of QHO gives ∑sTs∗Ts=Q. Therefore ∑λi>0Ri∗Ri+∑sTs∗Ts=∑λi>0WiWi∗+Q=IHO 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 λj, RiFjV=λjWi∗Wj=δijλjIHL; if λj=0 then FjV=0. Also TsFjV=0 because QWj=0. Hence for every X∈B(HL), (R∘N∘Ad⁡V)(X)=∑jλjX=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(FiVXV∗Fj∗)=δijλiX for every X∈B(HL); zero-eigenvalue terms vanish because FiV=0. Since Fi=∑aEaUai, unitarity gives Ea=∑iFiUai‾. Therefore R(EaVXV∗Eb∗)=∑iUai‾UbiλiX=CbaX, using C=Udiag⁡(λi)U∗.

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