CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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A zero-frequency Bose mode destroys trace normalization

Statement refuted

Assume the Axiom of Choice and Countable Choice. For a finite bosonic mode list with $ω_{1} = 0$ , $ω_{k} > 0$ for $2 \leq k \leq r$ , $ℏ > 0$ , and $β > 0$ , the full occupation partition sum for energies $E (n) : = \sum_{k = 2}^{r} ℏ ω_{k} (n_{k} + \frac{1}{2}), n \in N_{0}^{r},$ is finite and equals $\prod_{k = 2}^{r} {[2 \sinh ⁣ (\frac{β ℏ ω_{k}}{2})]}^{- 1},$ where the empty product is $1$ and $\sinh x : = (e^{x} - e^{- x}) / 2$ .

Statement references

Certified positive finite quadratic Bose Hamiltonian is normalizable

Explicit diagonal Fock domains and occupation trace

The Axiom of Choice

The Axiom of Countable Choice ( ${A C}_{ω}$ )

Facts & Assumptions

Given: An integer $r \geq 1$ , $ℏ > 0$ in joule-seconds, $β > 0$ in inverse joules, and finite positive angular frequencies $ω_{2}, \dots, ω_{r}$ in inverse seconds. The first mode has energy zero for every occupation.

[A1]

The Axiom of Choice and The Axiom of Countable Choice ( ${A C}_{ω}$ ): assume both Choice principles as stated; Countable Choice is the hypothesis of the diagonal occupation trace supplier.

[F1]

Certified positive finite quadratic Bose Hamiltonian is normalizable: the finite diagonalization partition formula assumes every normal frequency is strictly positive; it does not include a zero-frequency mode.

[F2]

Explicit diagonal Fock domains and occupation trace: finite bosonic mode occupations form the diagonal basis, and real diagonal energy multipliers are self-adjoint on their maximal weighted $ℓ^{2}$ domains under Countable Choice. Its finite geometric trace formula requires $q_{j} = e^{- β ϵ_{j}} < 1$ for every mode.

Counterexample

For $β > 0$ , define the occupation partition sum $Z_{β} : = \sum_{n \in N_{0}^{r}} e^{- β E (n)} \in [0, + \infty] .$ The first mode contributes no energy.

1.1A1F2givenalgebra

Apply [F2] to the reduced finite mode set ${2, \dots, r}$ with $ϵ_{k} = ℏ ω_{k}$ . Its factors satisfy $q_{k} = e^{- β ϵ_{k}} \in (0, 1)$ , and the set is finite, so the hypotheses $q_{*} < 1$ and $\sum_{k = 2}^{r} q_{k} < \infty$ hold. Its occupation partition sum is $\prod_{k = 2}^{r} (1 - q_{k})^{- 1}$ ; adding each oscillator's zero-point energy $ℏ ω_{k} / 2$ multiplies this by $e^{- β \sum_{k = 2}^{r} ℏ ω_{k} / 2}$ , giving $\prod_{k = 2}^{r} [2 \sinh (β ℏ ω_{k} / 2)]^{- 1}$ . If $r = 1$ , the reduced set is empty and its sum and product are both $1$ .

2.1A1givenF1F2step 1.1algebra

For each $k \geq 2$ , put $x_{k} = β ℏ ω_{k} > 0$ . Its one-mode occupation sum is $\sum_{n_{k} = 0}^{\infty} e^{- x_{k} (n_{k} + 1 / 2)} = \frac{e^{- x_{k} / 2}}{1 - e^{- x_{k}}} = {[2 \sinh (x_{k} / 2)]}^{- 1},$ which is positive and finite. If the first occupation is restricted to $0 \leq n_{1} \leq N$ and the remaining modes are summed, the resulting partial sum is $(N + 1) \prod_{k = 2}^{r} {[2 \sinh (x_{k} / 2)]}^{- 1} .$ This tends to $+ \infty$ as $N \to \infty$ ; when $r = 1$ the product is empty and the same partial sum is $N + 1$ . Hence $Z_{β} = + \infty$ .

3.1F1step 2.1algebra∎

Removing mode $1$ changes the occupation Hilbert space and leaves only the finite product over $k \geq 2$ . The full system has an additional countably infinite zero-energy occupation degeneracy, so its positive Gibbs weights cannot be normalized. The finite positive-frequency theorem does not apply at $ω_{1} = 0$ , and the omitted-mode product is the partition sum of a different model.

Remarks

The counterexample is finite in mode count; the divergence comes from the unbounded occupation of one zero-energy bosonic mode, not from an infinite-mode limit.

A zero-frequency Bose mode destroys trace normalization

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Remarks

Depends on

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