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Finite KMS condition exactly determines the Gibbs density

Statement

Let $H$ be a nonzero finite-dimensional complex Hilbert space, let $H = H^{*}$ on $H$ with energy units J, let $ℏ > 0$ have units J s, and let $β > 0$ have units J $^{- 1}$ . Write $Tr$ for the ordinary finite-dimensional trace. Assume the finite spectral-calculus and unital C $^{*}$ -algebra clauses [A1]–[A2] below. Define the finite-matrix Heisenberg group

$α_{t}^{H} (X) : = e^{i t H / ℏ} X e^{- i t H / ℏ} (X \in B (H), t \in R) .$

A state $ω$ on the unital C $^{*}$ -algebra $B (H)$ satisfies the $β$ -KMS strip condition of Physical-time KMS strip equilibrium condition for $α^{H}$ if and only if

$ω (X) = \frac{Tr (e^{- β H} X)}{Tr (e^{- β H})} (X \in B (H)) .$

Here $β$ has units J $^{- 1}$ , $H$ has units J, and the physical-time strip has width $β ℏ$ seconds. The result is finite-dimensional and makes no infinite-volume or relaxation claim.

Facts & Assumptions

Given: $H$ , $D = \dim H \geq 1$ , $H = H^{*}$ , $ℏ > 0$ , $β > 0$ , a state $ω$ on $B (H)$ , and the finite-matrix group $α_{t}^{H}$ .

[A1]

Finite matrix spectral and state calculus: every finite-dimensional Hermitian matrix has an orthonormal eigenbasis with real eigenvalues, and the finite matrix trace is cyclic. Its current proof correctly normalizes rational-coordinate approximants without claiming the normalized coordinates remain rational; its Step 1 readiness is current. This is authoring readiness, not independent approval.

[A2]

Finite matrix C-star algebra and density-state duality: for $D \geq 1$ , $B (H)$ with its ordinary operator norm is a nonzero unital complex C $^{*}$ -algebra. The current finite-matrix carrier is a draft with a complete local proof; no independent acceptance is claimed.

[F1]

Physical-time KMS strip equilibrium condition: a $β$ -KMS state has a bounded continuous function on the closed strip $S_{β ℏ}$ , holomorphic inside, with lower boundary $ω (A α_{t}^{H} (B))$ and upper boundary $ω (α_{t}^{H} (B) A)$ .

[F2]

Bounded KMS strip norm and zero-edge uniqueness: two bounded continuous functions holomorphic inside the same strip and with the same lower edge agree throughout, including the upper edge.

[F3]

Finite unitary dynamics and Gibbs KMS identity: for finite self-adjoint $H$ , the normalized Gibbs trace state $X \mapsto Z^{- 1} Tr (e^{- β H} X)$ satisfies the finite KMS boundary identity with the physical-time width $β ℏ$ .

Proof

technique · diagonalize $H$, use matrix units to derive the state weights from the strip edges, and invoke the finite Gibbs boundary calculation for the converse

Given: [A1]–[A2], [F1]–[F3], $H$ , $β$ , $ℏ$ , and $ω$ .

1.1A1A2algebra

By [A1]–[A2], choose an orthonormal basis $u_{1}, \dots, u_{D}$ with $H u_{i} = E_{i} u_{i}$ and $E_{i} \in R$ ; [A2] puts the finite matrix algebra in the C $^{*}$ -dynamical setting required by Physical-time KMS strip equilibrium condition. Let $E_{i j}$ be the matrix unit that maps $u_{j}$ to $u_{i}$ and all other basis vectors to zero. Direct multiplication gives $E_{a b} E_{c d} = δ_{b c} E_{a d}$ , and the exponential action gives $α_{z}^{H} (E_{i j}) = e^{i z (E_{i} - E_{j}) / ℏ} E_{i j}$ for $z \in C$ . This formula is used only for finite matrices, where the exponentials and scalar entries are entire.

2.1A2F1F2step 1.1algebra

Fix $i \neq j$ and apply [F1] to $A = E_{j j}$ , $B = E_{i j}$ . Since $E_{j j} E_{i j} = 0$ , the lower boundary of its KMS function is identically zero. By [F2] this KMS function vanishes on the whole strip. Its upper edge at $t = 0$ is $ω (E_{i j} E_{j j}) = ω (E_{i j})$ , so $ω (E_{i j}) = 0$ for every off-diagonal matrix unit.

2.2A2F1F2step 1.1algebra

For arbitrary $i, j$ , set $Δ = E_{i} - E_{j}$ and use [F1] with $A = E_{j i}$ , $B = E_{i j}$ . The lower edge is $e^{i t Δ / ℏ} ω (E_{j j})$ . The candidate $g (z) = e^{i z Δ / ℏ} ω (E_{j j})$ is entire and bounded on $0 \leq Im z \leq β ℏ$ , since $∣ e^{i (t + i y) Δ / ℏ} ∣ = e^{- y Δ / ℏ}$ and $y$ ranges over a compact interval. It has the same lower boundary as the KMS function, so [F2] identifies them. At the upper edge and $t = 0$ , the KMS identity is therefore $ω (E_{i i}) = e^{- β (E_{i} - E_{j})} ω (E_{j j})$ .

3.1givenstep 2.2A1algebra

The ratios in step 2.2 show that $e^{β E_{i}} ω (E_{i i})$ is a constant $c$ independent of $i$ . Since $\sum_{i} E_{i i} = I$ and $ω (I) = 1$ , $1 = c \sum_{i} e^{- β E_{i}}$ ; the sum is positive and finite, so $c = Z^{- 1}$ where $Z = Tr (e^{- β H}) > 0$ . Thus $ω (E_{i i}) = e^{- β E_{i}} / Z$ . This also covers $D = 1$ and degenerate eigenvalues.

4.1step 2.1step 3.1algebra

Every $X \in B (H)$ has a matrix-unit expansion $X = \sum_{i, j} X_{i j} E_{i j}$ . By linearity, step 2.1, and step 3.1, $ω (X) = \sum_{i} X_{i i} e^{- β E_{i}} / Z = Z^{- 1} Tr (e^{- β H} X)$ . Therefore every finite KMS state is the Gibbs trace state.

5.1A2F1F3∎

Conversely, [F3] supplies the entire bounded finite Gibbs correlation with the stated lower and upper edges, which is exactly [F1]'s finite-matrix KMS condition; [A2] ensures that this KMS definition applies to $B (H)$ . The normalized Gibbs trace state is therefore KMS, proving the reverse implication.

Remarks

The matrix-unit ratio proof does not assume that $ω$ is diagonal or invariant before the KMS condition is used. Equal-energy degeneracies are handled by the same ratio, and step 2.1 removes off-diagonal entries even within a degenerate eigenspace.

Finite KMS condition exactly determines the Gibbs density

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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