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TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)
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Boundary energy bounds thermal mutual information

Statement

Assume the Axiom of Choice. Let HA,HB be nonzero finite-dimensional complex inner-product spaces. Let HA=HA∗ act on HA, let HB=HB∗ act on HB, and let H∂=H∂∗ act on HA⊗HB. Set

H:=HA⊗IB+IA⊗HB+H∂.

For β>0, define Zβ:=Tr⁡(e−βH) and the finite Gibbs density ρβ:=e−βH/Zβ. Then

I(A:B)ρβ≤2β∥H∂∥op,

where I(A:B) uses dimensionless entropy in nats and ∥K∥op:=sup⁡∥v∥=1∥Kv∥ is the operator norm.

For the optional lattice-family conclusion, fix m≥1. For each member s of a specified family, take a finite site set Λs⊆Zm, a partition Λs=As⊔Bs, and nonzero finite-dimensional complex inner-product spaces Ks,x for x∈Λs. Use the lexicographic order on Zm in Hs,X:=⨂x∈XKs,x, with the empty tensor product defined as C, and identify Hs,Λs with Hs,As⊗Hs,Bs in the displayed factor order. Let Js be a finite set of interaction labels with nonempty supports Xs,j⊆Λs and self-adjoint terms hs,j on Hs,Xs,j. For Y∈{Λs,As,Bs} containing Xs,j, write h^s,jY for the extension of hs,j to Hs,Y by the identity on every factor in Y∖Xs,j. The crossing-label set is

J∂,s:={j∈Js:Xs,j∩As≠∅ and Xs,j∩Bs≠∅},∣∂As∣:=∣J∂,s∣.

Let JA,s:={j∈Js:Xs,j⊆As} and JB,s:={j∈Js:Xs,j⊆Bs}. Set HAs:=∑j∈JA,sh^s,jAs, HBs:=∑j∈JB,sh^s,jBs, and H∂,s:=∑j∈J∂,sh^s,jΛs. Under the stated identification these give the displayed Hamiltonian decomposition. For each s, assume βs>0 and define ρβs by the Gibbs formula above. If one specified C≥0 satisfies ∥H∂,s∥op≤C∣∂As∣ for every family member, the bound gives I(As:Bs)ρβs≤2βsC∣∂As∣ memberwise. No finite-range condition or additional geometric scaling assumption is used. When H∂,s and each interaction term are measured in joules, βs has units J−1 and C has units J, so the bound is dimensionless.

Facts & Assumptions

Given: The spaces and Hermitian matrices in the Statement, β>0, natural logarithms, and a fixed kB>0 with units J/K when physical units are assigned. Put T=(kBβ)−1.

[A1]

The Axiom of Choice: every family of nonempty sets has a choice function. This is inherited through the Gibbs variational supplier in [F1] and its [F2] relative-entropy-domain dependency Quantum spectral entropy and relative-entropy domains; the finite matrix calculation in this item makes no infinite-dimensional selection.

[F1]

Finite Gibbs free-energy and entropy optimization: for a finite Hermitian Hamiltonian, its Gibbs density minimizes Tr⁡(Hω)−TSphys(ω) over all density matrices, including rank-deficient competitors, with Sphys=kBS and T=(kBβ)−1. Its relative-entropy-domain dependency, recorded as [F2] in that supplier, carries Choice through the declared supplier chain; its own [A1] records energy and units.

[F2]

Partial traces of a finite bipartite state are states: both marginals of a finite bipartite density operator are density operators.

[F3]

Finite partial trace, Schmidt decomposition and finite no-signalling: for every finite bipartite endomorphism R and local endomorphism A, Tr⁡((Tr⁡BR)A)=Tr⁡(R(A⊗IB)), with the corresponding identity for tracing the first factor.

[F4]

Dimensionless entropy, mutual and conditional information: S(ρ)=−Tr⁡(ρlog⁡ρ) and I(A:B)ρ=S(ρA)+S(ρB)−S(ρAB).

[F5]

Finite matrix spectral and state calculus: finite Hermitian matrices have orthonormal eigenbases with real eigenvalues; positive matrices have nonnegative spectral weights, and the finite trace is cyclic.

[F6]

Finite mixing, product entropy, and bipartite subadditivity: for finite-dimensional density matrices, the physical entropy of a tensor product is the sum of the physical entropies of its factors, including rank-deficient factors. Its declared supplier chain carries the same Choice hypothesis in [A1].

Proof

1.1givenF1F2F5algebra

By [F5], H has finitely many real eigenvalues, so e−βH is positive definite and Zβ>0. Thus ρβ is a finite Gibbs density to which [F1] applies. Its marginals ρA:=Tr⁡Bρβ and ρB:=Tr⁡Aρβ are density operators by [F2].

2.1givenF4F5F6step 1.1algebra

Set τ:=ρA⊗ρB. Tensor products of orthonormal eigenbases show that τ is positive semidefinite with eigenvalues piqj, whose sum is (∑ipi)(∑jqj)=1; hence τ is a density operator. By [F6], Sphys(τ)=Sphys(ρA)+Sphys(ρB). Since kB>0 and Sphys=kBS, division by kB gives S(τ)=S(ρA)+S(ρB). Therefore [F4] gives S(τ)−S(ρβ)=I(A:B)ρβ.

3.1A1givenF1step 2.1algebra

Define Fβ(ω):=Tr⁡(Hω)−β−1S(ω) for finite density matrices. Since TSphys=TkBS=β−1S, [F1] applies to this functional and yields Fβ(ρβ)≤Fβ(τ). Rearranging and using step 2.1 gives β−1I(A:B)ρβ≤Tr⁡(H(τ−ρβ)).

4.1step 1.1step 2.1step 3.1F3F5algebra

The coordinate formula in [F3] gives Tr⁡B(ρA⊗ρB)=ρA and Tr⁡A(ρA⊗ρB)=ρB, since each factor has trace one. The defining trace-pairing identities in [F3], with cyclicity from [F5], then show that the expectations of HA⊗IB in τ and ρβ agree, and likewise for IA⊗HB. Thus both local terms cancel in the trace on the right of step 3.1, leaving Tr⁡(H(τ−ρβ))=Tr⁡(H∂(τ−ρβ)).

5.1givenF5step 3.1step 4.1algebra∎

Diagonalize H∂=∑jλjPj by [F5]. For any density matrix ω, the numbers wj:=Tr⁡(Pjω) are nonnegative (each is a sum of nonnegative diagonal expectations of ω on an orthonormal basis of ran⁡Pj) and sum to one. Thus ∣Tr⁡(H∂ω)∣=∣∑jλjwj∣≤∑j∣λj∣wj≤max⁡j∣λj∣=∥H∂∥op. Applying this absolute bound to τ and ρβ gives Tr⁡(H∂(τ−ρβ))≤2∥H∂∥op. Combine this with steps 3.1 and 4.1 and multiply by β>0 to obtain the claimed inequality. Substituting the additional family bound ∥H∂∥op≤C∣∂A∣ yields its stated conditional area scaling.

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