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TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Exact recovery implies the Knill–Laflamme condition

Statement

Let N(X)=∑a=1mEaXEa∗ be a finite Kraus representation of a channel from B(HP) to B(HO), and let V:HL→HP be an isometry with HL≠{0} and P=VV∗. If the code is exactly correctable in the sense of Exact recovery of an encoded quantum code, then there are scalars Cab such that PEa∗EbP=CabP for all a,b.

Facts & Assumptions

Given: The finite-dimensional encoding, Kraus channel, and an exact decoded recovery channel R:B(HO)→B(HL).

[A1]

Exact recovery of an encoded quantum code: exactness means R∘N∘Ad⁡V is the identity channel on B(HL).

[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 ∑jRj∗Rj=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.

Proof

1.1givenA1A2A3algebrachoose

By [A3], choose a finite Kraus representation R(X)=∑jRjXRj∗. Composing its Kraus operators with the errors Ea and the encoding V gives the Kraus family RjEaV:HL→HL for R∘N∘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 cja with RjEaV=cjaIHL.

3.1A4F2step 2.1algebra

Trace preservation of R gives ∑jRj∗Rj=IHO by [A4]. Therefore, for every a,b, V∗Ea∗EbV=V∗Ea∗(∑jRj∗Rj)EbV=∑j(RjEaV)∗(RjEbV)=(∑jcja‾cjb)IHL.

4.1givenstep 3.1algebra∎

Put Cab=∑jcja‾cjb. Since P=VV∗, PEa∗EbP=V(V∗Ea∗EbV)V∗=CabP by step 3.1. This is the Knill–Laflamme condition for every pair of errors.

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