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LemmaStatement: AI-generatedProof: AI-generated
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Recovery optimization on orthogonal flagged branches

Statement

Let H=Cd with d≥1, and let K=⨁j=1mKj be a finite orthogonal direct sum of nonzero finite-dimensional complex inner-product spaces. Concretely, K is the space of tuples v=(v1,…,vm) with vj∈Kj, with coordinatewise vector operations and inner product ⟨v,w⟩K=∑j=1m⟨vj,wj⟩Kj, linear in the first argument. Write ιj:Kj→K for the map putting its input in coordinate j and zero in every other coordinate, and Pj=ιjιj∗ for the orthogonal projector onto its image; the star denotes the adjoint for these inner products. Let pj≥0, with ∑jpj=1, and let each Nj:L(H)→L(Kj) be a quantum channel. Define N(X)=∑jpjιjNj(X)ιj∗. Then, for entanglement fidelity at the maximally mixed input,

max⁡R:L(K)→L(H) channelFe(R∘N)=∑jpjmax⁡Rj:L(Kj)→L(H) channelFe(Rj∘Nj).

A decoder attaining the left side can be constructed by projecting onto the orthogonal blocks and applying a maximizing decoder for the corresponding block. The hypothesis that the weights are fixed independently of the input is part of this statement; input-dependent branch weights are not covered.

Facts & Assumptions

[F1]

A quantum channel is a complex-linear completely positive trace-preserving map; a decoder is a channel with the indicated input and logical output. Complete positivity means positivity after tensoring with every finite-dimensional identity map. Code restriction and decoded channel types

[F2]

For ∣Φd⟩=d−1/2∑a=0d−1∣a⟩⊗∣a⟩, entanglement fidelity is the overlap of ∣Φd⟩ with the output of id⁡⊗E on ∣Φd⟩⟨Φd∣. Entanglement fidelity on a finite logical space

[F3]

For a linear map T:L(A)→L(B), its input-first Choi matrix is J(T)=∑a,b∣a⟩⟨b∣⊗T(∣a⟩⟨b∣) in an ordered orthonormal basis of A. Such a map is a channel exactly when J(T) is positive semidefinite and its partial trace over B is IA. Choi matrix with input factor first Finite channel, Kraus, Choi, and Stinespring equivalence

[F5]

A complex inner product is linear in the first argument, conjugate symmetric, and positive definite: ⟨x,x⟩≥0, with equality exactly when x=0. Real and complex inner-product spaces and their induced length

[F6]

An adjoint T∗ of a linear map T:A→B satisfies ⟨Tx,y⟩B=⟨x,T∗y⟩A for all x∈A,y∈B. The adjoint T^*:W→ V is characterised by ⟨ Tv,w⟩_W=⟨ v,T^*w⟩_V

[F7]

For compatible rectangular matrices, tr⁡(AB)=tr⁡(BA). For A∈ M_m× n(F) and B∈ M_n× m(F), tr(AB)=tr(BA)

[F8]

Vectors are orthogonal when their inner product is zero, and two subspaces are orthogonal when every vector in one is orthogonal to every vector in the other. An orthonormal basis is an ordered basis whose vectors are pairwise orthogonal and each have induced norm one. Orthogonal vectors and subspaces, orthogonal and orthonormal sets, and orthonormal bases

[F9]

For a subspace W of a finite-dimensional inner-product space V, the orthogonal projection sends each v∈V to the unique w∈W for which v−w is orthogonal to every vector in W. When treated as an endomorphism of V, this map is followed by the inclusion W↪V. The orthogonal projection P_Wv is the W-component in V=W⊕ W^⊥

Proof

Given: The finite spaces, fixed probabilities and channels in the statement.

Goal: Establish the displayed equality and construct a maximizing decoder.

1.1givenF5construct

The coordinatewise operations make the tuple space a complex vector space. The pairing in the Statement is linear in its first argument and conjugate symmetric by [F5]. Its value on (v,v) is a finite sum of nonnegative terms, vanishing exactly when every vj=0, so it is positive definite and hence an inner product. Extending a basis of each Kj by zero in the other coordinates gives a finite basis of K: these vectors span coordinatewise, and a linear relation vanishes in each coordinate only when all coefficients vanish. Thus K is a finite-dimensional complex inner-product space.

1.2F1F2F3F4F8

Each branch supremum is attained. Indeed, in fixed bases, [F3] identifies branch decoders with positive semidefinite Choi matrices J whose output partial trace is IKj. This set is nonempty: the replacement channel Y↦tr⁡(Y)IH/d is completely positive and trace-preserving. The set is closed by its matrix inequalities and linear partial-trace equations. It is bounded because tr⁡J=dim⁡Kj, its diagonal entries are nonnegative, and positivity of each 2×2 principal submatrix gives ∣Jab∣2≤JaaJbb. The matrix coordinates form a finite-dimensional real Euclidean space, so [F4] gives compactness. The Choi formula determines a map by its values on matrix units, and the overlap in [F2] is a linear function of those values; hence the branch objective is continuous. By [F4] it attains a maximum.

2.1givenF5F6F8F9step 1.1algebra

Define the coordinate map qj:K→Kj by qj(v)=vj. For x∈Kj and v∈K, the pairing gives ⟨ιjx,v⟩K=⟨x,vj⟩Kj, so qj is the adjoint ιj∗ by [F6]; uniqueness follows from [F5], since a vector orthogonal to every vector, including itself, is zero. Also ⟨ιjx,ιjy⟩K=⟨x,y⟩Kj, so each inclusion is an isometry. Coordinate evaluation yields ιj∗ιj=IKj and ιk∗ιj=0:Kj→Kk when k≠j. The map Pj=ιjqj retains exactly coordinate j; it is the identity on im⁡ιj and zero on the other coordinate subspaces, which are orthogonal to that image by the summed pairing. For every v∈K, the vector Pjv lies in im⁡ιj and the residual v−Pjv has zero coordinate j, so it is orthogonal to every vector in that image by [F8]. Thus [F9], viewed as an endomorphism of K, identifies Pj as the orthogonal projector onto im⁡ιj, and ∑jPj=IK because the sum restores every coordinate of every tuple.

3.1givenF1F2F7step 2.1

For any decoder R on K, define Rj(Y)=R(ιjYιj∗). Conjugation by ιj is completely positive, since every finite amplification is conjugation by an inclusion, and it preserves trace because ιj∗ιj=IKj. Composition with R therefore makes each Rj a channel. By linearity of the channel action and of the overlap in [F2], Fe(RN)=∑jpjFe(RjNj). Thus every left-side value is at most the sum of the branchwise suprema.

4.1givenF1F7step 2.1step 3.1step 1.2∎

Choose a maximizing channel Rj for each of the finitely many branches and define R(Y)=∑jRj(ιj∗Yιj). Every compression is completely positive, and a finite sum of completely positive maps is completely positive. Also tr⁡R(Y)=∑jtr⁡(ιj∗Yιj)=tr⁡Y, since ∑jPj=IK; hence R is a decoder. The coordinate identities in step 2.1 give ιj∗ιj=IKj and ιk∗ιj=0 for k≠j, so R(ιjYιj∗)=Rj(Y) for every Y∈L(Kj). The identity in step 3.1 therefore gives the sum of the branch maxima. This achieves the upper bound from step 3.1, proving the equality and the asserted construction.

Depends on

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