ExampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generated

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Poisson number distribution

Example

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $(g_{n})_{n \in N}$ be the Hermite family of $L^{2} (R, d y)$ , $N = a^{†} a$ the number operator of Closed ladder domains, and for $z \in C$ let $ψ_{z} = e^{- ∣ z ∣^{2} / 2} \sum_{n} \frac{z^{n}}{\sqrt{n!}} g_{n}$ be the coherent vector of Coherent-state convergence. Define

$p_{n} : = ∣ ⟨ g_{n}, ψ_{z} ⟩ ∣^{2} = e^{- ∣ z ∣^{2}} \frac{∣ z ∣^{2 n}}{n!} (n \in N) .$

Then:

(a) $(p_{n})_{n \in N}$ is a probability distribution on $N$ : $p_{n} \geq 0$ and $\sum_{n} p_{n} = 1$ ;

(b) $⟨ N ⟩ : = ⟨ N ψ_{z}, ψ_{z} ⟩ = \sum_{n} n p_{n} = ∣ z ∣^{2}$ ;

(d) the distribution is the Poisson distribution with parameter $λ = ∣ z ∣^{2}$ , and it is nontrivial for every $z \neq 0$ : $p_{n} > 0$ for all $n$ .

Facts & Assumptions

Given: the coherent vector $ψ_{z}$ with coefficients $c_{n} = e^{- ∣ z ∣^{2} / 2} z^{n} / \sqrt{n!}$ in the orthonormal family $(g_{n})$ ; the numbers $p_{n} = ∣ c_{n} ∣^{2}$ .

[F1]

$⟨ g_{n}, g_{m} ⟩ = δ_{n m}$ and $c_{n} = ⟨ ψ_{z}, g_{n} ⟩$ with $ψ_{z} = e^{- ∣ z ∣^{2} / 2} \sum_{n} \frac{z^{n}}{\sqrt{n!}} g_{n}$ ; $⟨ N ψ_{z}, ψ_{z} ⟩ = ∣ z ∣^{2}$ ; $ψ_{z} \in D (N^{k})$ for every $k$ (Coherent-state convergence(a),(b),(e), Hermite orthogonality(a)).

[F2]

For $t = ∣ z ∣^{2} \geq 0$ , $\sum_{n \geq 0} \frac{t^{n}}{n!} = e^{t}$ and the series converges absolutely (The complex exponential by its power series, The complex exponential series converges absolutely for every complex argument).

[F3]

For an orthonormal family, $⟨ N ψ_{z}, ψ_{z} ⟩ = \sum_{n} n ∣ c_{n} ∣^{2}$ and, for $ψ_{z} \in D (N^{2})$ , $⟨ (N - ∣ z ∣^{2} I)^{2} ψ_{z}, ψ_{z} ⟩ = \sum_{n} (n - ∣ z ∣^{2})^{2} ∣ c_{n} ∣^{2}$ ; these are the coefficient formulas of the diagonal operator on the orthonormal family (Coherent-state convergence and The space $L^{p} (μ)$ as the quotient by null functions, Complex Lp classes and Euclidean test-function conventions).

Verification

1.1F1F2algebra

[(a) probability distribution] $p_{n} = ∣ c_{n} ∣^{2} = e^{- ∣ z ∣^{2}} ∣ z ∣^{2 n} / n! \geq 0$ , and $\sum_{n} p_{n} = e^{- ∣ z ∣^{2}} \sum_{n} \frac{(∣ z ∣^{2})^{n}}{n!} = e^{- ∣ z ∣^{2}} e^{∣ z ∣^{2}} = 1$ by [F2].

1.2F1F3

[(b) mean] By [F3] and [F1], $\sum_{n} n p_{n} = \sum_{n} n ∣ c_{n} ∣^{2} = ⟨ N ψ_{z}, ψ_{z} ⟩ = ∣ z ∣^{2}$ .

1.3F2F3algebra

[(c) variance] Expanding $(N - ∣ z ∣^{2} I)^{2} = N^{2} - 2 ∣ z ∣^{2} N + ∣ z ∣^{4} I$ and using [F3], $\sum_{n} (n - ∣ z ∣^{2})^{2} p_{n} = e^{- ∣ z ∣^{2}} \sum_{n} (n^{2} - 2 ∣ z ∣^{2} n + ∣ z ∣^{4}) \frac{∣ z ∣^{2 n}}{n!}$ . With $t = ∣ z ∣^{2}$ , the three series are $\sum_{n} n^{2} \frac{t^{n}}{n!} = (t^{2} + t) e^{t}$ , $\sum_{n} n \frac{t^{n}}{n!} = t e^{t}$ and $\sum_{n} \frac{t^{n}}{n!} = e^{t}$ , obtained from the absolutely convergent exponential series by reindexing ( $\sum_{n \geq 1} \frac{n t^{n}}{n!} = t \sum_{m \geq 0} \frac{t^{m}}{m!}$ and $\sum_{n \geq 2} \frac{n (n - 1) t^{n}}{n!} = t^{2} \sum_{m \geq 0} \frac{t^{m}}{m!}$ ); multiplying by $e^{- t}$ gives $(t^{2} + t) - 2 t^{2} + t^{2} = t = ∣ z ∣^{2}$ .

2.1step 1.1step 1.2step 1.3∎

[(d) Poisson form] By (a)-(c) the distribution has total mass $1$ , mean $∣ z ∣^{2} = λ$ and variance $∣ z ∣^{2} = λ$ , and its weights are $e^{- λ} λ^{n} / n!$ , which is the Poisson distribution with parameter $λ$ ; if $z \neq 0$ then $λ > 0$ and $p_{n} > 0$ for every $n$ , so no outcome is excluded.

Poisson number distribution

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources