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Schwartz/distribution spaces and Hilbert Fock conventions — Examples

1 · Prerequisites

2 · Summary

The companion example compares one-mode bosonic CCR and fermionic CAR with the exact domains of creation, annihilation and the number operator. Its square-summable coefficient sequence belongs to both creation and annihilation domains but lies outside the number-operator domain; the displayed convergence and divergence arguments establish this distinction locally.

Use the Fock pairing, sector conventions and operator domains from qft-mf01. The example is a mathematical boundary witness for unbounded domains and does not assign particle statistics, model a measurement or report an empirical result. Its cited source context and source-verification limitations are retained in the item and batch coverage record. Draft placement does not grant independent proof acceptance.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-15 (gpt-6-luna, gpt-6.1-sol, and delegated Codex review/repair agents; see completed run evidence)Open item page →

One-mode CCR/CAR and a vector outside the number-operator domain

Example

Let $H = C e$ with $∥ e ∥ = 1$ , and write $∣ n ⟩ = e^{\otimes n}$ for its normalized bosonic $n$ -particle vector, including $∣ 0 ⟩ = Ω$ . On the one-mode fermionic Fock space write $∣ 1 ⟩ = c^{†} (e) Ω$ . The one-mode bosonic operators satisfy the CCR on the common invariant finite-particle domain $D_{f i n}$ ; the fermionic operators satisfy the CAR on the whole two-dimensional fermionic Fock space. The bosonic number operator from Second-quantized multipliers, joint spectral measure and exact domains is $N ∣ n ⟩ = n ∣ n ⟩$ with $D (N) = {(c_{n})_{n \geq 0} : \sum_{n \geq 0} n^{2} ∣ c_{n} ∣^{2} < \infty} .$ The Fock vector $Ψ = \sum_{n = 0}^{\infty} (n + 1)^{- 3 / 2} ∣ n ⟩$ belongs to $D (a (e)) \cap D (a^{†} (e))$ but not to $D (N)$ .

Facts & Assumptions

Given: Countable Choice The Axiom of Countable Choice ( ${A C}_{ω}$ ), a unit vector $e$ , the one-mode bosonic and fermionic Fock constructions, and the maximal bosonic creation/annihilation domains.

[A1]

For the number-operator calculation take $X = {*}$ , $A = {\emptyset, {*}}$ , $μ ({*}) = 1$ , and $ω : X \to R$ with $ω (*) = 1$ . Then $L^{2} (X, μ; C)$ is one-dimensional, with unit vector $e = 1_{{*}}$ , and its one-particle multiplier is exactly $1$ .

[F1]

On one-mode bosonic sectors, $a (e) ∣ n ⟩ = \sqrt{n} ∣ n - 1 ⟩$ for $n \geq 1$ , $a (e) Ω = 0$ , and $a^{†} (e) ∣ n ⟩ = \sqrt{n + 1} ∣ n + 1 ⟩$ . The fermionic one-mode maps are bounded and satisfy the CAR. Bosonic and fermionic Fock operators with exact domains

[F2]

The maximal domains are $D (a (e)) = {(c_{n}) : \sum_{n \geq 1} n ∣ c_{n} ∣^{2} < \infty}$ and $D (a^{†} (e)) = {(c_{n}) : \sum_{n \geq 0} (n + 1) ∣ c_{n} ∣^{2} < \infty}$ . Bosonic and fermionic Fock operators with exact domains

[F3]

For the one-point measure space specified in [A1], the second-quantized multiplier is the number operator, acts by $n$ on sector $n$ , and has the exact square-summability domain stated above. Second-quantized multipliers, joint spectral measure and exact domains

Verification

1.1F1givenalgebra

For bosons, the ladder formulas [F1] give $a (e) a^{†} (e) ∣ n ⟩ = (n + 1) ∣ n ⟩$ and $a^{†} (e) a (e) ∣ n ⟩ = n ∣ n ⟩$ for every $n \geq 0$ , where the second identity at $n = 0$ uses $a (e) Ω = 0$ . Their difference is $∣ n ⟩$ ; linear extension proves $[a (e), a^{†} (e)] = I$ on the finite-particle core. For fermions, $c (e) Ω = 0$ , $c^{†} (e) Ω = ∣ 1 ⟩$ , $c (e) ∣ 1 ⟩ = Ω$ , and $c^{†} (e) ∣ 1 ⟩ = 0$ . Hence ${c (e), c^{†} (e)}$ fixes both basis vectors, while $c (e)^{2} = (c^{†} (e))^{2} = 0$ ; these identities hold on the whole two-dimensional fermionic Fock space.

1.2F3givenalgebra

The squared norm of $Ψ$ is $\sum_{n \geq 0} (n + 1)^{- 3}$ , which is finite: on each block $2^{k} \leq n + 1 < 2^{k + 1}$ there are $2^{k}$ terms, each at most $2^{- 3 k}$ , so the block sum is at most $2^{- 2 k}$ and the geometric series of these bounds converges. By [F3], $Ψ \in D (N)$ would require $\sum_{n \geq 1} n^{2} / (n + 1)^{3} < \infty$ . Since $n + 1 \leq 2 n$ for $n \geq 1$ , $n^{2} / (n + 1)^{3} \geq 1 / (8 n)$ . The harmonic series diverges because each block $2^{k} \leq n < 2^{k + 1}$ has $2^{k}$ terms at least $2^{- (k + 1)}$ , so every block sum is at least $1 / 2$ . Therefore $Ψ \notin D (N)$ .

2.1F2step 1.2algebra∎

By [F2], the annihilation-domain sum for $Ψ$ is $\sum_{n \geq 1} n / (n + 1)^{3} \leq \sum_{n \geq 1} n^{- 2}$ , and the creation-domain sum is $\sum_{n \geq 0} (n + 1)^{- 2}$ . Each is finite: the $k$ th dyadic block of $\sum_{n \geq 1} n^{- 2}$ has at most $2^{k}$ terms each at most $2^{- 2 k}$ , so its block sum is at most $2^{- k}$ ; the shifted series has the same convergence. Thus $Ψ \in D (a (e)) \cap D (a^{†} (e))$ while step 1.2 proves $Ψ \notin D (N)$ . This is a domain distinction for the stated operators, not a physical or empirical claim.