CounterexampleConstruction: AI-adaptedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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A centered exact band can be empty

Statement

There are two-level additive data and a canonical mean for which every zero-width band centered at that mean is empty. Consequently a normalized uniform band density is not defined for that choice.

Facts & Assumptions

Given: The finite additive ensemble convention in Additive spectral Hamiltonian and normalized ensembles.

[F1]

Additive spectral Hamiltonian and normalized ensembles: the second canonical probability is $p_{2} = e^{- β e_{2}} / (e^{- β e_{1}} + e^{- β e_{2}})$ and the $N$ -site energy label with $m$ occurrences of level $e_{2}$ has energy $m e_{2} + (N - m) e_{1}$ .

Proof

1.1givenF1algebra

Take $q = 2$ , $e_{1} = 0$ , $e_{2} = ϵ > 0$ , and choose finite $β > 0$ with $β ϵ = \log \sqrt{2}$ . By [F1], $p_{2} = \frac{e^{- β ϵ}}{1 + e^{- β ϵ}} = \frac{1}{1 + \sqrt{2}} = \sqrt{2} - 1.$ This number is irrational: if $\sqrt{2}$ were rational, writing it as a reduced fraction $a / b$ gives $a^{2} = 2 b^{2}$ , so $a$ is even, then $b$ is even, contradicting reduction. Hence $p_{2}$ is irrational.

2.1F1step 1.1algebra∎

For any integer $N \geq 1$ , an $N$ -site label has energy $m ϵ$ for some integer $m$ , whereas the canonical mean is $N E = N ϵ p_{2}$ . The zero-width band condition forces $m ϵ = N ϵ p_{2}$ , hence $m = N p_{2}$ . Since $p_{2}$ is irrational, $N p_{2}$ is not an integer. Thus no label satisfies the condition, $Γ_{N} = \emptyset$ , and $Π_{N} / ∣ Γ_{N} ∣$ is undefined. This shows why a nonemptiness or positive width hypothesis cannot be omitted.

Depends on

Additive spectral Hamiltonian and normalized ensembles

A centered exact band can be empty

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources