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Relative modular representation of finite quantum relative entropy

Statement

Let A,B be positive-definite operators on a nonzero finite-dimensional complex inner-product space H. On L(H), with Hilbert-Schmidt inner product ⟨X,Y⟩HS:=tr⁡(XY∗), define ΔB,A(X):=BXA−1. Then ΔB,A is positive definite and

D(A∥B)=⟨(−ln⁡ΔB,A)(A1/2),A1/2⟩HS.

The logarithm of ΔB,A is defined by its positive eigenvalues.

Facts & Assumptions

Given: H,A,B as in the Statement.

[F1]

Finite-dimensional quantum relative entropy: for positive-definite A,B, D(A∥B)=tr⁡(Aln⁡A)−tr⁡(Aln⁡B).

[F3]

A non-negative operator has a unique non-negative square root: A has a positive-definite square root A1/2.

[F5]

Real and complex inner-product spaces and their induced length: the inner product is linear in its first argument and conjugate-linear in its second, so Ejix=⟨x,ui⟩vj is linear.

[F6]

Every linear map between finite-dimensional inner product spaces has a unique adjoint: every linear map between finite-dimensional inner-product spaces has a unique linear adjoint, including zero-dimensional spaces.

[F7]

The adjoint T∗:W→V is characterised by ⟨Tv,w⟩W=⟨v,T∗w⟩V: the adjoint is characterized by ⟨Tx,y⟩=⟨x,T∗y⟩.

[F8]

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm: for positive real x,y, ln⁡(x/y)=ln⁡x−ln⁡y.

Proof

1.1givenF6F7F8F2F5algebrachoose

All adjoints of the finite-dimensional linear maps used below exist uniquely by [F6]. Choose orthonormal eigenbases (ui) and (vj) for A and B, with Aui=aiui, Bvj=bjvj and ai,bj>0, using [F2]. For each i,j define Ejix:=⟨x,ui⟩vj. These linear matrix units form an orthonormal basis of L(H) for ⟨⋅,⋅⟩HS by the adjoint pairing [F7] and direct trace evaluation. They satisfy BEji=bjEji and EjiA−1=ai−1Eji, so ΔB,AEji=(bj/ai)Eji. The quotient law [F8] gives −ln⁡(bj/ai)=−ln⁡bj+ln⁡ai, so spectral calculus on this basis shows that ΔB,A is positive definite and gives (−ln⁡ΔB,A)(X)=−(ln⁡B)X+X(ln⁡A).

2.1F1F3F4step 1.1algebra∎

Applying the formula in step 1.1 to A1/2 gives ⟨(−ln⁡ΔB,A)(A1/2),A1/2⟩HS=tr⁡(−Aln⁡B+A1/2(ln⁡A)A1/2). By trace cyclicity [F4], the second trace is tr⁡(Aln⁡A). The result equals D(A∥B) by [F1], proving the identity.

Depends on

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Dependency tree · two levels

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Sources