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Quantum spectral energy concentration at a differentiable pressure point

Statement

Use the translation-invariant finite-range spin model and pressure $p_{Φ}$ of Bounded quantum lattice pressure on cubes and van Hove sets. Let $β > 0$ and suppose $p_{Φ}$ is differentiable at $β$ . Put $e_{Φ} (β) : = - p_{Φ}^{'} (β)$ , which has units joules per site when $β$ is measured in inverse joules. For each nonempty finite region $Λ$ , let $ρ_{Λ, β} = e^{- β H_{Λ}} / Z_{Λ} (β)$ and define its energy spectral probability measure by $μ_{Λ, β} (C) = {Tr}_{Λ} (ρ_{Λ, β} E_{H_{Λ}} (C))$ , where $E_{H_{Λ}} (C)$ is the sum of the orthogonal eigenspace projections of $H_{Λ}$ with eigenvalues in the Borel set $C \subseteq R$ . Its finite-volume mean energy density is $m_{Λ, β} : = ∣ Λ ∣^{- 1} {Tr}_{Λ} (ρ_{Λ, β} H_{Λ})$ .

For every van Hove sequence $(Λ_{n})$ and every $ϵ > 0$ , $m_{Λ_{n}, β} \to e_{Φ} (β)$ and there are constants $c_{ϵ} > 0$ and $n_{0}$ such that

μ_{Λ_{n}, β} ⁣ ({E : ∣ E / ∣ Λ_{n} ∣ - e_{Φ} (β) ∣ \geq ϵ}) \leq e^{- c_{ϵ} ∣ Λ_{n} ∣} (n \geq n_{0}) .

This is a statement about the finite-volume Gibbs spectral distribution under the declared mathematical model; it supplies no empirical observation.

Facts & Assumptions

Given: $β > 0$ , the model of Translation-invariant finite-range spin interaction, and differentiability of its limiting pressure at $β$ .

[F1]

Bounded quantum lattice pressure on cubes and van Hove sets: for each $γ > 0$ , $p_{Λ_{n}} (γ) = ∣ Λ_{n} ∣^{- 1} \log Z_{Λ_{n}} (γ)$ converges to the finite $p_{Φ} (γ)$ along every van Hove sequence.

[F2]

Duhamel derivative and finite Kubo–Mori response: along a finite-dimensional Hermitian matrix path $M (s)$ , $\frac{d}{d s} \log Tr (e^{M (s)}) = Tr (ρ_{s} M^{'} (s))$ with $ρ_{s} = e^{M (s)} / Tr (e^{M (s)})$ . Only this first-derivative identity is used here.

[F3]

Complex spectral theorem: a normal endomorphism of a finite-dimensional complex inner product space has an orthonormal eigenbasis, and conversely: every finite-dimensional self-adjoint matrix has an orthonormal eigenbasis with real eigenvalues.

Proof

technique · convex secants and exponential Markov bounds for the finite spectral law

Given: [F1]–[F3], $β > 0$ , differentiability of $p_{Φ}$ at $β$ , and $ϵ > 0$ .

1.1F2F3algebra

For nonempty finite $Λ$ , diagonalize $H_{Λ}$ using [F3], with eigenvalues $E_{j}$ and Gibbs weights $w_{j} = e^{- β E_{j}} / Z_{Λ} (β)$ . Then $p_{Λ} (β) = ∣ Λ ∣^{- 1} \log \sum_{j} e^{- β E_{j}}$ is convex because $p_{Λ}^{''} (β) = ∣ Λ ∣^{- 1} (\sum_{j} w_{j} E_{j}^{2} - (\sum_{j} w_{j} E_{j})^{2}) = ∣ Λ ∣^{- 1} \sum_{j} w_{j} (E_{j} - \sum_{k} w_{k} E_{k})^{2} \geq 0$ . Applying [F2] to $M (β) = - β H_{Λ}$ also gives $p_{Λ}^{'} (β) = - Tr (ρ_{Λ, β} H_{Λ}) / ∣ Λ ∣$ .

1.2givenF3algebra

For nonempty finite $Λ$ , the measure $μ_{Λ, β}$ assigns weight $w_{j}$ to each eigenvalue, combining weights when eigenvalues coincide, and is a probability measure because $\sum_{j} w_{j} = 1$ . For $t \in (0, β)$ and $e = e_{Φ} (β)$ , each eigenvalue $E \geq ∣ Λ ∣ (e + ϵ)$ satisfies $e^{t E} \geq e^{t ∣ Λ ∣ (e + ϵ)}$ . Summing the corresponding weights gives $μ_{Λ, β} {E / ∣ Λ ∣ \geq e + ϵ} \leq \exp ⁣ (∣ Λ ∣ [p_{Λ} (β - t) - p_{Λ} (β) - t (e + ϵ)]) .$ The identity $\sum_{j} w_{j} e^{t E_{j}} = Z_{Λ} (β - t) / Z_{Λ} (β)$ gives the displayed exponent.

2.1F1step 1.1algebra

Fix $h \in (0, β)$ . Convexity in step 1.1 gives $(p_{Λ} (β) - p_{Λ} (β - h)) / h \leq p_{Λ}^{'} (β) \leq (p_{Λ} (β + h) - p_{Λ} (β)) / h$ . Along $Λ_{n}$ , [F1] makes the two secants converge for fixed $h$ to the corresponding secants of $p_{Φ}$ . Letting $h ↓ 0$ and using differentiability at $β$ squeezes $p_{Λ_{n}}^{'} (β) \to p_{Φ}^{'} (β)$ ; step 1.1 therefore proves $Tr (ρ_{Λ_{n}, β} H_{Λ_{n}}) / ∣ Λ_{n} ∣ \to e_{Φ} (β)$ .

2.2F1step 1.2algebra

Differentiability and $e = - p_{Φ}^{'} (β)$ give $p_{Φ} (β - t) - p_{Φ} (β) = t e + o (t)$ as $t ↓ 0$ . Choose $t \in (0, β)$ so the remainder has magnitude at most $t ϵ / 2$ . For all sufficiently large $n$ , convergence in [F1] at $β - t$ and $β$ makes the exponent bracket in step 1.2 at most $- t ϵ / 4$ . Thus the upper-tail probability is at most $e^{- t ϵ ∣ Λ_{n} ∣ / 4}$ .

3.1F1F3step 2.2algebra∎

For the lower tail, if $E \leq ∣ Λ ∣ (e - ϵ)$ then $e^{- t E} \geq e^{- t ∣ Λ ∣ (e - ϵ)}$ . The same finite spectral sum gives $μ_{Λ, β} {E / ∣ Λ ∣ \leq e - ϵ} \leq \exp ⁣ (∣ Λ ∣ [p_{Λ} (β + t) - p_{Λ} (β) + t (e - ϵ)]) .$ Since $p_{Φ} (β + t) - p_{Φ} (β) = - t e + o (t)$ , choose a possibly smaller fixed $t > 0$ so the limiting bracket is at most $- t ϵ / 2$ ; for large $n$ the bracket is at most $- t ϵ / 4$ . Let $t_{+}$ be the choice in step 2.2, let $t_{-}$ be this lower-tail choice, and set $c_{0} = \min (t_{+}, t_{-}) ϵ / 4 > 0$ . Once $∣ Λ_{n} ∣ \geq 4 \log 2 / c_{0}$ , the sum of the two tail bounds $2 e^{- c_{0} ∣ Λ_{n} ∣}$ is at most $e^{- c_{0} ∣ Λ_{n} ∣ / 2}$ ; hence $c_{ϵ} = c_{0} / 2$ works after increasing $n_{0}$ .

Quantum spectral energy concentration at a differentiable pressure point

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources