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Finite quantum mutual information identities and the Holevo identity

Statement

For a finite-dimensional bipartite state ρAB,

I(A:B)ρ=S(ρA)+S(ρB)−S(ρAB)=D(ρAB∥ρA⊗ρB).

For a finite classical-quantum state ρXB=∑xpxΠx⊗ρx, its mutual information equals its Holevo information:

I(X:B)ρ=S ⁣(∑xpxρx)−∑x:px>0pxS(ρx)=χ({px,ρx}).

Facts & Assumptions

Given: The finite-dimensional states in the Statement, with the root linear-first convention.

[F1]

Finite quantum mutual, conditional, and conditional mutual information: I(A:B)ρ=S(ρA)+S(ρB)−S(ρAB) and I(X:B)ρ=S(ρX)+S(ρB)−S(ρXB).

[F2]

Finite-dimensional quantum relative entropy: on support inclusion, D(A∥B)=tr⁡(Aln⁡A)−tr⁡(Aln⁡B) with support-restricted spectral logarithms.

[F3]

Joint state support is contained in the tensor product of its marginal supports: supp⁡(ρAB)⊆supp⁡(ρA⊗ρB), so the QRE in the Statement is finite.

[F4]

Finite classical-quantum states and Holevo information: ρXB=∑xpxΠx⊗ρx, ρB=∑xpxρx, and χ=S(ρB)−∑px>0pxS(ρx).

[F5]

The partial trace is characterized by the trace pairing and is basis-independent: tr⁡((Tr⁡BR)T)=tr⁡(R(T⊗IB)).

[F6]

Partial trace over the first tensor factor: Tr⁡A is the first-factor partial trace and has entries (Tr⁡AR)jj′=∑iR(i,j),(i,j′).

[F7]

Partial trace over a finite tensor factor: the second-factor partial trace has entries (Tr⁡BR)ii′=∑jR(i,j),(i′,j).

[F8]

Ordered tensor registers, their inner product, and wire permutations: tensor product bases retain the displayed factor order and use the product inner product.

[F10]

Finite-dimensional von Neumann entropy: entropy is −tr⁡(Rln⁡R), with zero-eigenvalue terms omitted.

Proof

1.1givenF2F3

By [F3], the support condition in [F2] holds for ρAB and ρA⊗ρB, so the relative entropy is evaluated by its finite trace formula.

2.1givenF2F8F9F11step 1.1algebrachoose

Choose orthonormal eigenbases of the marginals using [F9], with positive eigenvalues ai for ρA on its support and bj for ρB on its support. Their product vectors form an orthonormal eigenbasis of ρA⊗ρB on its support by [F8], with eigenvalues aibj>0. Since [F11] gives ln⁡(aibj)=ln⁡ai+ln⁡bj, spectral calculus gives ln⁡(ρA⊗ρB)=(ln⁡ρA)⊗IB+IA⊗(ln⁡ρB) on that support.

3.1F1F2F3F5F6step 2.1algebra

Extend each marginal logarithm by zero on its kernel, which does not affect the support-restricted traces by [F2] and [F3]. Taking T=ln⁡ρA in [F5] gives tr⁡(ρAB((ln⁡ρA)⊗IB))=tr⁡(ρAln⁡ρA). For the second marginal, expanding the finite product-basis trace and using the entries in [F6] gives tr⁡(ρAB(IA⊗ln⁡ρB))=tr⁡(ρBln⁡ρB). Combining these identities with [F2] and step 2.1 yields D(ρAB∥ρA⊗ρB)=S(ρA)+S(ρB)−S(ρAB)=I(A:B)ρ.

4.1givenF1F4F7F9F10F11step 2.1algebra∎

In the orthogonal blocks of ρXB, the eigenvalues are pxλxj, where λxj are the eigenvalues of ρx; zero-probability blocks contribute only zeros. By [F7] and tr⁡ρx=1, ρX=∑xpxΠx; its eigenvalues are (px), and the B-marginal is ρˉB=∑xpxρx by [F4]. The block eigenvalues follow from [F9]. For px>0 and λxj>0, applying [F11] and expanding −(pxλxj)ln⁡(pxλxj) and using ∑jλxj=1 gives S(ρXB)=S(ρX)+∑x:px>0pxS(ρx) by [F10]. Substitution into the definition [F1] cancels S(ρX) and proves I(X:B)ρ=χ({px,ρx}).

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