LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generated

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The fixed-particle density-matrix domain is nonempty and compact

Statement

For an integer $n \geq 2$ and a real $0 < r < n$ , put $D_{r} : = {D \in H_{n} : 0 ⪯ D ⪯ I, tr D = r} .$ This agrees with the domain of Fixed-particle finite-temperature density-matrix free energy when $r$ is an integer. Then $D_{r}$ is a nonempty compact subset of $H_{n}$ .

Facts & Assumptions

Given: An integer $n \geq 2$ , a real $0 < r < n$ , the displayed domain $D_{r}$ , and the Frobenius topology on the Hermitian matrices.

[F1]

For integer $1 \leq r < n$ , Fixed-particle finite-temperature density-matrix free energy defines $D_{r}$ by the constraints $D = D^{*}$ , $0 ⪯ D ⪯ I$ , and $tr D = r$ . For real $0 < r < n$ , the displayed formula in the Statement defines the same notation locally.

[F2]

A Hermitian matrix has an orthonormal eigenbasis (Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely).

[F3]

A subset of finite-dimensional real coordinate space is compact exactly when it is closed and bounded; no choice principle is used (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

Proof

1.1F1givenconstruct

The matrix $D_{*} = (r / n) I$ has eigenvalues $r / n \in (0, 1)$ and trace $r$ , so [F1] gives $D_{*} \in D_{r}$ . Thus the domain is nonempty.

1.2F1given

The conditions $D = D^{*}$ , $D ⪰ 0$ , $I - D ⪰ 0$ , and $tr D = r$ define a closed subset of the real coordinates of $H_{n}$ : the first and last are continuous equalities, while positive semidefiniteness is closed because it is the intersection of the closed inequalities $⟨ D x, x ⟩ \geq 0$ over all fixed vectors $x$ .

1.3F1F2givenalgebra

If $d_{1}, \dots, d_{n} \in [0, 1]$ are the eigenvalues of $D \in D_{r}$ , then $∥ D ∥_{F}^{2} = tr (D^{2}) = \sum_{j} d_{j}^{2} \leq \sum_{j} d_{j} = r$ . Thus $D_{r}$ is bounded.

2.1F3step 1.2step 1.3algebra∎

The real vector space $H_{n}$ has dimension $n^{2}$ and is isomorphic to $R^{n^{2}}$ by its real diagonal entries and real and imaginary off-diagonal coordinates. Closedness from step 1.2 and boundedness from step 1.3 give compactness by [F3].

Depends on

Used by

The finite-temperature free energy attains a minimum on the density domain Theorem

Dependency tree · two levels

50 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel Arovas, Thermodynamics and Statistical Mechanics, §7.4 Variational Density Matrix Method, UC San Diego / Physics LibreTexts (standard reference, not scraped)