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The finite-temperature trace entropy is continuous at occupation endpoints

Statement

For ϕ(x)=xlog⁡x+(1−x)log⁡(1−x) with 0log⁡0=0, the map D↦tr⁡ϕ(D) is continuous on the full domain Dr, including matrices with eigenvalues 0 or 1.

Facts & Assumptions

Given: The continuous scalar extension ϕ on [0,1] and Hermitian matrices D∈Dr.

[F1]

The entropy integrand and endpoint convention are as in the finite-temperature model (Fixed-particle finite-temperature density-matrix free energy).

[F2]

Ordered eigenvalues of complex Hermitian matrices change by at most the operator norm of the matrix perturbation (Ordered eigenvalues of Hermitian matrices are operator-norm Lipschitz).

[F3]

The natural logarithm satisfies log⁡′(x)=1/x and log⁡x=∫1xds/s for x>0 (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[F4]

For real or complex matrices, the induced operator norm is bounded by the Frobenius norm, ∥A∥2≤∥A∥F (Spectral and Frobenius norms are unitarily invariant, are given by singular values, and satisfy the sharp rank comparison).

Proof

1.1F1F3algebra

For 0<x<1, put t=1/x. By [F3], 0≤log⁡t=∫1tds/s≤∫1tds/s=2(t−1), so ∣xlog⁡x∣=(log⁡t)/t≤2/t→0. The other term (1−x)log⁡(1−x) tends to 0 as x↓0 by continuity of y↦ylog⁡y at y=1. Since ϕ(1−x)=ϕ(x), the same limits hold at 1. Thus the stipulated endpoint values make ϕ continuous on [0,1].

1.2F1F2F4given

By [F2], if Dk→D in Frobenius norm, then every ordered eigenvalue dk,j tends to the corresponding dj; indeed [F4] gives ∥Dk−D∥2≤∥Dk−D∥F→0. All these eigenvalues lie in [0,1] by the domain definition.

2.1F1step 1.1step 1.2algebra∎

Spectral functional calculus gives tr⁡ϕ(Dk)=∑j=1nϕ(dk,j) and likewise for D. Scalar continuity from step 1.1 makes each of the finitely many summands converge, so tr⁡ϕ(Dk)→tr⁡ϕ(D).

Depends on

Used by

Cited to discharge well-definedness by Fixed-particle finite-temperature density-matrix free energy.

Dependency tree · two levels

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Sources