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Zero-energy bosonic mode trace obstruction

Statement refuted

For a periodic bosonic box, deleting the zero wavevector only in the continuum mode integral is sufficient to leave the actual finite-box Gibbs trace finite even if the zero-energy mode remains in the discrete occupation space.

Facts & Assumptions

Given: Assume Countable Choice The Axiom of Countable Choice (ACω) for the Fock basis and Lebesgue-null singleton suppliers. Let I0=(Z3∖{0})×{1,2} be the enumerated base mode set of Actual photon box trace and Planck energy/entropy flux, with its stated max-norm shell and lexicographic ordering.

[A1]

Form the ordered disjoint union I′={0⋆}⊔I0, placing the new label 0⋆ first. Give 0⋆ energy ϵ0⋆=0 and each i=(n,s)∈I0 its energy from Actual photon box trace and Planck energy/entropy flux. Let ΩB′ be the set of finite-support maps m:I′→N0, let E′(m)=∑i∈I′ϵimi, and set H′=ℓ2(ΩB′,C) with its occupation basis. Define H′ by (H′a)m=E′(m)am on D(H′)={a∈H′:∑mE′(m)2∣am∣2<∞}, and define the bounded positive diagonal multiplier Gβ by Gβ∣m⟩=e−βE′(m)∣m⟩. Its diagonal partition sum is Zβ′:=∑m∈ΩB′e−βE′(m)∈[0,+∞]. This is the stipulated extension of the finite-box diagonal model by one zero-energy bosonic mode, without a zero-point term.

[F1]

Occupation bases and actual CCR/CAR domains, Statement: for a specified enumerated mode set, finite-support bosonic occupation configurations give an orthonormal basis of the bosonic Fock space. The current batch 45 evidence records this draft carrier as Step-1-ready; that is not independent review or proof acceptance, so the use remains conditional on its inspected interface.

[F2]

Trace class operator, Definition: a trace-class operator is compact and its singular values have finite sum. The compactness condition uses the exact definition in def-compact-linear-operator: the image of every bounded set must have compact closure; compactness of that closure is the finite-open-subcover property in def-metric-compactness.

[F3]

Actual photon box trace and Planck energy/entropy flux, Statement: the base model has nonzero wavevectors only and its corresponding diagonal Gibbs trace is finite under the explicitly stated summability conditions.

[F4]

The Axiom of Countable Choice (ACω): the Fock-basis supplier assumes ACω.

[F5]

Every affine hyperplane of Rn, and hence every proper linear subspace, is Lebesgue null, Statement: assuming ACω, every proper linear subspace of R3 is Lebesgue measurable and null; in particular, {0} is null.

Counterexample

Given: The explicit extended mode and occupation model in [A1].

1.1givenA1F1F3F4

For each integer r≥0, let m(r)∈ΩB′ have occupation r at 0⋆ and occupation zero at every i∈I0. These are distinct finite-support configurations, so [F1] makes ∣m(r)⟩ an orthonormal sequence in H′.

2.1givenA1F2F5step 1.1algebra∎

The diagonal multiplier satisfies E′(m(r))=rϵ0⋆=0, hence Gβ∣m(r)⟩=∣m(r)⟩. Let B={a∈H′:∥a∥≤1} and let K=Gβ(B)‾. If Gβ were compact, its image of the bounded set B would have compact closure K by the compact-operator definition recorded in [F2]. All vectors ∣m(r)⟩ belong to K. The sets K∩{x∈H′:∥x−y∥<1/2}, for y∈K, form an open cover of K. Distinct vectors ∣m(r)⟩ have distance 2, while every such ball has diameter at most 1; no finite subfamily covers all these vectors. This contradicts compactness, so Gβ is not compact and therefore is not trace class by [F2]. The diagonal partition sum Zβ′ contains ∑r=0∞e−βrϵ0⋆=∑r=0∞1=∞. The continuum zero-mode set {0}⊂R3 has Lebesgue measure zero by [F5], but that fact does not remove the infinitely many discrete zero-mode occupation states.

Depends on

Dependency tree · 2 levels