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Logical Kraus operators of the completed transpose decoder

Statement

Let d≥1 be an integer, let HP,HO be nonzero finite-dimensional complex inner-product spaces, and let V:Cd→HP be a linear isometry. Set P=VV†. Let N:L(HP)→L(HO) be a quantum channel with finite Kraus representation N(X)=∑bEbXEb†, where Eb:HP→HO. Here L(H) denotes the complex-linear endomorphisms of H.

Set A=N(P) and let S be the orthogonal projector onto supp⁡A=(ker⁡A)⊥. The operator A−1/2 is the inverse square root on this support and zero on ker⁡A. For any code state σ, meaning σ≥0, tr⁡σ=1 and PσP=σ, define Ra=PEa†A−1/2 and the completed transpose recovery T(Y)=∑aRaYRa†+tr⁡((IHO−S)Y)σ.

The logical decoder D:L(HO)→L(Cd) given by D(Y)=V†T(Y)V is completely positive and trace-preserving (CPTP). On encoded noisy inputs its composition has Kraus operators Kab=V†Ea†A−1/2EbV, so DN(VXV†)=∑a,bKabXKab† for every X∈L(Cd).

Facts & Assumptions

Given: The spaces, maps and operators in the Statement.

Goal: Prove the asserted identities and conclusions.

[F1]

The completed transpose map is CPTP, code-supported, and agrees with its support-normalized part on every noisy code-supported operator. Support-normalized transpose recovery and completion

[F2]

An isometric encoding is X↦VXV†, with V†V=I and VV†=P. Compressing any code-supported quantum channel T by V†(⋅)V gives a CPTP logical decoder, as justified by ancilla conjugation and composition. Code restriction and decoded channel types

[F3]

A finite Kraus representation expresses a complex-linear map as a finite sum of conjugations, with the identity valid for every input operator. Finite Kraus representation

Proof

Given: The spaces, maps and operators in the Statement.

Goal: Prove the asserted identities and conclusions.

Proof technique: direct.

1.1givenF1F2

Compression Z↦V†ZV is completely positive. Every output Z of T satisfies PZP=Z, so tr⁡(V†ZV)=tr⁡(PZ)=tr⁡Z; the trace equality follows by expanding both traces in orthonormal bases adapted to the code. Thus D is trace preserving and completely positive.

2.1F1F2F3step 1.1∎

For VXV†=P(VXV†)P, the completion term vanishes after the noise. Substitute Ra=PEa†A−1/2 into the composition, and use V†P=V† and PV=V. This gives exactly the displayed Kab and channel formula.

Depends on

Used by · 2

Dependency tree · 2 levels

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