LemmaStatement: AI-adaptedProof: 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.

Closed ladder domains

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $H$ be a complex Hilbert space (Hilbert space) with inner product linear in the first argument, let $(e_{n})_{n \in N}$ , $N = {0, 1, 2, \dots}$ , be a complete orthonormal family of $H$ (Orthonormal families, complete orthonormal systems and Hilbert bases), and for $x \in H$ write $c_{n} : = ⟨ x, e_{n} ⟩$ , so that $x = \sum_{n} c_{n} e_{n}$ and $\sum_{n} ∣ c_{n} ∣^{2} = ∥ x ∥^{2}$ . Define

$D (a) : = {x : \sum_{n} n ∣ c_{n} ∣^{2} < \infty}, a x : = \sum_{m \geq 0} \sqrt{m + 1} c_{m + 1} e_{m};$ $D (a^{†}) : = {x : \sum_{n} (n + 1) ∣ c_{n} ∣^{2} < \infty}, a^{†} x : = \sum_{m \geq 1} \sqrt{m} c_{m - 1} e_{m};$ $D (N) : = {x : \sum_{n} n^{2} ∣ c_{n} ∣^{2} < \infty}, N x : = \sum_{n} n c_{n} e_{n} .$

Then:

(a) $a, a^{†}, N$ are well-defined linear operators with the displayed domains; these domains contain the algebraic span $D_{0} = span {e_{n}}$ and are dense in $H$ ; and $∥ a x ∥^{2} = \sum_{n} n ∣ c_{n} ∣^{2}$ , $∥ a^{†} x ∥^{2} = \sum_{n} (n + 1) ∣ c_{n} ∣^{2}$ , $∥ N x ∥^{2} = \sum_{n} n^{2} ∣ c_{n} ∣^{2}$ ;

(b) $a$ and $a^{†}$ are closed, densely defined operators with $a^{*} = a^{†}$ and $(a^{†})^{*} = a$ ;

(d) $D (a a^{†}) = D (a^{†} a) = D (N)$ ; on $D (N)$ one has $a^{†} a = N$ and $a a^{†} = N + I$ , hence $[a, a^{†}] x = a a^{†} x - a^{†} a x = x$ for every $x \in D (N)$ ; and $D (N)$ is exactly the set of $x \in H$ for which both $a (a^{†} x)$ and $a^{†} (a x)$ are defined;

(e) $N$ is self-adjoint with $N e_{n} = n e_{n}$ and spectrum $σ (N) = N$ (Diagonal spectrum), and for $x \in D (N)$ the partial sums $x_{M} : = \sum_{n \leq M} c_{n} e_{n}$ satisfy $x_{M} \to x$ and $N x_{M} \to N x$ ;

(f) if $H = L^{2} (R, d y)$ with the Hermite orthonormal family $e_{n} = g_{n}$ of Hermite recurrence, then the operators $a_{P}, a_{P}^{†} : P \to P$ of that theorem are the restrictions of $a, a^{†}$ to $P$ , and $a$ and $a^{†}$ are exactly the closures of $a_{P}$ and $a_{P}^{†}$ .

Facts & Assumptions

Given: a complex Hilbert space $H$ , a complete orthonormal family $(e_{n})$ , and the operators $a, a^{†}, N$ displayed above.

[F1]

$⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ , $x = \sum_{n} ⟨ x, e_{n} ⟩ e_{n}$ with $\sum_{n} ∣ ⟨ x, e_{n} ⟩ ∣^{2} = ∥ x ∥^{2}$ for every $x \in H$ , and a series $\sum_{n} a_{n} e_{n}$ converges in $H$ exactly when $\sum_{n} ∣ a_{n} ∣^{2} < \infty$ , its sum then satisfying $∥ \sum_{n} a_{n} e_{n} ∥^{2} = \sum_{n} ∣ a_{n} ∣^{2}$ (Orthonormal families, complete orthonormal systems and Hilbert bases, Parseval equivalences for an orthonormal family).

[F2]

For a densely defined $T$ : $y \in D (T^{*})$ exactly when $x \mapsto ⟨ T x, y ⟩$ is bounded on $(D (T), ∥ \cdot ∥)$ , and then $T^{*} y$ is the unique $w$ with $⟨ T x, y ⟩ = ⟨ x, w ⟩$ for all $x \in D (T)$ ; $T^{*}$ is closed; $T$ is closed exactly when $T = (T^{*})^{*}$ is an adjoint (Adjoint of a densely defined operator, The adjoint is well defined, closed, and reverses inclusions).

[F3]

$∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ and $∣ \sum_{n \in F} a_{n} \overline{b_{n}} ∣ \leq (\sum_{n \in F} ∣ a_{n} ∣^{2})^{1 / 2} (\sum_{n \in F} ∣ b_{n} ∣^{2})^{1 / 2}$ for finite $F$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F4]

The self-adjoint diagonal operator of Diagonal adjoint and graph-core lemma with $λ_{n} = n$ is precisely $N$ on $D (N)$ , and its spectrum is the closure of ${n : n \in N} = N$ (Diagonal spectrum).

[F5]

In the concrete case $H = L^{2} (R, d y)$ with $e_{n} = g_{n}$ , the operators $a_{P}, a_{P}^{†}$ of Hermite recurrence act on the finite expansions by $a_{P} g_{n} = \sqrt{n} g_{n - 1}$ and $a_{P}^{†} g_{n} = \sqrt{n + 1} g_{n + 1}$ , and $P = span {g_{n}}$ (Hermite recurrence(e),(f)).

Proof

1.1givenF1F3algebra

[Well-definedness and densities] For $x \in D (a)$ the coefficients $\sqrt{m + 1} c_{m + 1}$ are square summable with $\sum_{m} (m + 1) ∣ c_{m + 1} ∣^{2} = \sum_{n \geq 1} n ∣ c_{n} ∣^{2} < \infty$ , so the series $a x = \sum_{m} \sqrt{m + 1} c_{m + 1} e_{m}$ converges in $H$ and $∥ a x ∥^{2} = \sum_{n \geq 1} n ∣ c_{n} ∣^{2}$ by [F1]; the same computation gives the formulas for $a^{†} x$ and $N x$ . Linearity is coefficientwise. Every finite combination of the $e_{n}$ lies in all three domains, and $D_{0}$ is dense in $H$ by [F1], so the three domains are dense.

1.2givenF1F2F3algebra

[ $a^{*} = a^{†}$ ] First let $y \in D (a^{*})$ and $w : = a^{*} y$ . Testing the defining identity $⟨ a x, y ⟩ = ⟨ x, w ⟩$ on $x = e_{m}$ ( $m \geq 1$ ) gives $\sqrt{m} \overline{⟨ y, e_{m - 1} ⟩} = \overline{⟨ w, e_{m} ⟩}$ , that is $\sqrt{m} ⟨ y, e_{m - 1} ⟩ = ⟨ w, e_{m} ⟩$ ; hence $\sum_{m \geq 1} m ∣ ⟨ y, e_{m - 1} ⟩ ∣^{2} = \sum_{m \geq 1} ∣ ⟨ w, e_{m} ⟩ ∣^{2} \leq ∥ w ∥^{2} < \infty$ ; testing $x = e_{0}$ gives $⟨ e_{0}, w ⟩ = 0$ , hence $⟨ w, e_{0} ⟩ = 0$ . Therefore $y \in D (a^{†})$ and $a^{†} y = \sum_{m \geq 1} \sqrt{m} ⟨ y, e_{m - 1} ⟩ e_{m} = \sum_{m \geq 1} ⟨ w, e_{m} ⟩ e_{m} = w$ by [F1]. Thus $a^{*} \subseteq a^{†}$ . Conversely let $y \in D (a^{†})$ and $x \in D (a)$ ; then $⟨ a x, y ⟩ = \sum_{m \geq 0} \sqrt{m + 1} c_{m + 1} \overline{d_{m}}$ with $d_{m} = ⟨ y, e_{m} ⟩$ , absolutely convergent by [F3] since $\sum_{m} (m + 1) ∣ c_{m + 1} ∣^{2} = ∥ a x ∥^{2} < \infty$ and $\sum_{m} ∣ d_{m} ∣^{2} = ∥ y ∥^{2}$ , and the same series equals $\sum_{n \geq 1} c_{n} \overline{\sqrt{n} d_{n - 1}} = ⟨ x, a^{†} y ⟩$ ; hence $∣ ⟨ a x, y ⟩ ∣ = ∣ ⟨ x, a^{†} y ⟩ ∣ \leq ∥ x ∥ ∥ a^{†} y ∥$ , so the functional is bounded on $(D (a), ∥ \cdot ∥)$ , so $y \in D (a^{*})$ and $a^{*} y = a^{†} y$ . Therefore $a^{*} = a^{†}$ .

1.3givenF1F2F3algebra

[ $(a^{†})^{*} = a$ and closedness] Let $y \in D ((a^{†})^{*})$ and $w : = (a^{†})^{*} y$ ; testing against $x = e_{m}$ with $m \geq 0$ gives $\sqrt{m + 1} ⟨ y, e_{m + 1} ⟩ = ⟨ w, e_{m} ⟩$ , so $\sum_{m \geq 0} (m + 1) ∣ ⟨ y, e_{m + 1} ⟩ ∣^{2} = \sum_{m} ∣ ⟨ w, e_{m} ⟩ ∣^{2} < \infty$ , that is $\sum_{n \geq 1} n ∣ ⟨ y, e_{n} ⟩ ∣^{2} < \infty$ ; thus $y \in D (a)$ and $a y = \sum_{n \geq 1} \sqrt{n} ⟨ y, e_{n} ⟩ e_{n - 1} = \sum_{m \geq 0} ⟨ w, e_{m} ⟩ e_{m} = w$ . Conversely for $y \in D (a)$ and $x \in D (a^{†})$ the same two absolutely convergent series show $⟨ a^{†} x, y ⟩ = ⟨ x, a y ⟩$ , so $y \in D ((a^{†})^{*})$ with $(a^{†})^{*} y = a y$ . Hence $(a^{†})^{*} = a$ . Since $a = (a^{†})^{*}$ and $a^{†} = a^{*}$ are adjoints of densely defined operators, [F2] makes both closed.

1.4givenF1algebra

[Kernels] If $a x = 0$ then $\sqrt{m + 1} c_{m + 1} = 0$ for all $m$ , so $c_{n} = 0$ for $n \geq 1$ and $x = c_{0} e_{0}$ ; conversely $a e_{0} = 0$ . If $a^{†} x = 0$ then $\sqrt{m} c_{m - 1} = 0$ for all $m \geq 1$ , so $c_{n} = 0$ for all $n$ and $x = 0$ .

1.5givenF1algebra

[Products, commutator and exact domain] For $x = \sum_{n} c_{n} e_{n}$ , membership in $D (a a^{†})$ means $\sum_{n} (n + 1) ∣ c_{n} ∣^{2} < \infty$ and $\sum_{m \geq 1} m ∣ \sqrt{m} c_{m - 1} ∣^{2} = \sum_{n} (n + 1)^{2} ∣ c_{n} ∣^{2} < \infty$ . The second condition implies the first, and, for $x \in H$ , is equivalent to $\sum_{n} n^{2} ∣ c_{n} ∣^{2} < \infty$ since $n^{2} \leq (n + 1)^{2} \leq 2 n^{2} + 2$ . On this domain the coefficient of $e_{n}$ in $a a^{†} x$ is $(n + 1) c_{n}$ . Likewise $x \in D (a^{†} a)$ means $\sum_{n} n ∣ c_{n} ∣^{2} < \infty$ and $\sum_{m \geq 0} (m + 1) ∣ \sqrt{m + 1} c_{m + 1} ∣^{2} = \sum_{n \geq 1} n^{2} ∣ c_{n} ∣^{2} < \infty$ ; the latter implies the former, and the coefficient of $e_{n}$ in $a^{†} a x$ is $n c_{n}$ , including zero at $n = 0$ . Hence both product domains equal $D (N)$ , $a^{†} a = N$ , $a a^{†} = N + I$ there, and $[a, a^{†}] x = x$ . Their common domain is exactly $D (N)$ .

2.1step 1.1F4

[(e) self-adjointness and spectrum] By [F4] the operator with domain $D (N)$ and action $N e_{n} = n e_{n}$ is self-adjoint with spectrum $N$ , and step 1.1 identifies it with the diagonal operator of [F4]. The partial-sum statement is the graph-core clause of Diagonal adjoint and graph-core lemma with $λ_{n} = n$ .

3.1step 1.1step 1.3F5∎

[(f) closure of the concrete ladder] In the concrete case $H = L^{2} (R, d y)$ , $e_{n} = g_{n}$ and $P = span {g_{n}}$ by [F5]; on the finite expansion $φ = \sum_{n \leq M} c_{n} g_{n}$ the operator $a$ of this lemma acts by $a φ = \sum_{m \leq M - 1} \sqrt{m + 1} c_{m + 1} g_{m} = a_{P} φ$ , and similarly $a^{†} φ = a_{P}^{†} φ$ ; hence $a_{P} \subseteq a$ and $a_{P}^{†} \subseteq a^{†}$ . Since $a$ is closed by step 1.3 and $a_{P} \subseteq a$ , the closure satisfies $\overline{a_{P}} \subseteq a$ . Conversely, for $x \in D (a)$ the partial sums $x_{M} = \sum_{n \leq M} c_{n} g_{n} \in P$ satisfy $x_{M} \to x$ and $a_{P} x_{M} = a x_{M} \to a x$ by step 1.1, so $(x, a x) \in \overline{Γ (a_{P})}$ and $a \subseteq \overline{a_{P}}$ ; hence $a = \overline{a_{P}}$ , and the argument for $a^{†}$ is identical.

Closed ladder domains

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources