ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)

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One-charge soft photon count example

Physical scope

An exact numerical specialization of the prescribed-current positive free coherent-state model and finite-cutoff photon-count prediction. It describes a prepared model state, not an observed or simulated emission event.

Example

Use the one-charge prescribed current of Regulated soft coherent photon-count prediction with final velocity $v = \frac{1}{2}$ along $\hat{z}$ . Choose any upper photon-energy cutoff $Λ > 0$ in joules and set $λ = Λ / e$ , where $e$ is Euler's number. The angular coefficient is $C = \frac{α}{π} (2 \log 3 - 2) > 0.$ The exact cutoff coherent state therefore has mean photon count $C$ , vacuum probability $e^{- C}$ , and mean photon energy $C Λ (1 - e^{- 1})$ joules. Each halving of $λ$ increases the mean count by $C \log 2$ ; the mean energy remains bounded and tends to $C Λ$ as $λ ↓ 0$ . These are model predictions for the prepared free state.

Facts & Assumptions

Given: Countable Choice, the coupling normalization, and the conditional count law and energy formula of Regulated soft coherent photon-count prediction.

[A1]

Countable Choice ${A C}_{ω}$ is carried by The Axiom of Countable Choice ( ${A C}_{ω}$ ) and by the conditional theorem used here.

[F1]

QFT example units, metric, gauge, and pairing conventions: $g_{e m}^{2} = 4 π α$ with $g_{e m} > 0$ ; $α$ is a positive dimensionless coupling, distinct from the electron gyromagnetic factor.

[F2]

Regulated soft coherent photon-count prediction: for $0 < v < 1$ the prescribed one-charge current has $C (v) = \frac{α}{π} (v^{- 1} \log ((1 + v) / (1 - v)) - 2) > 0$ ; at cutoffs $0 < λ < Λ$ , $p_{n} = e^{- C \log (Λ / λ)} (C \log (Λ / λ))^{n} / n!$ and the mean photon energy is $C (Λ - λ)$ joules. The claim concerns the constructed free state, not full interacting-QED emission.

Verification

Given: [A1], [F1]–[F2], $v = \frac{1}{2}$ , $Λ > 0$ , and $λ = Λ / e$ .

1.1A1F1F2algebra

Substituting $v = \frac{1}{2}$ into [F2] gives $C = (α / π) (2 \log 3 - 2) > 0$ , and $\log (Λ / λ) = \log (e) = 1$ . Thus the mean count is $C$ , $p_{0} = e^{- C}$ , and the general count formula specializes to $p_{n} = e^{- C} C^{n} / n!$ .

2.1F1F2algebra∎

Since $λ = Λ / e$ , the energy formula in [F2] gives $C (Λ - λ) = C Λ (1 - e^{- 1})$ joules. After $j$ successive halvings, $λ_{j} = Λ e^{- 1} 2^{- j}$ , so $\log (Λ / λ_{j}) = 1 + j \log 2$ and the mean count is $C (1 + j \log 2)$ . The mean energy is $C (Λ - λ_{j}) \to C Λ$ and [F2] gives no Fock-vector or projective-ray limit when $C > 0$ . These remain predictions for the prescribed current and prepared free state.

One-charge soft photon count example

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels