LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pass

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.

The position-momentum commutator identity on the product domain

Statement

Assume the Axiom of Countable Choice and the Axiom of Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ ), The Axiom of Choice). Fix integers $d \geq 1$ , $r \geq 1$ and $ℏ > 0$ , let $H^{(r)} = L^{2} (R^{d}; C^{r})$ be the component-valued space with the componentwise position and momentum operators $Q_{j}, P_{k}$ of Componentwise position and momentum operators on the component-valued wave-function space clauses (a) and (b), with domains $D (Q_{j}) = {f = (f_{l}) : x_{j} f_{l} \in L^{2} for all l}, D (P_{k}) = {f = (f_{l}) : D_{k} f_{l} \in L^{2} for all l},$ and let $D_{k}$ denote the weak partial derivative (Weak derivative of a locally integrable function). Then:

(a) Product rule. If $f \in H^{(r)}$ has $x_{j} f \in L^{2} (R^{d}; C^{r})$ and $x_{j} D_{j} f \in L^{2} (R^{d}; C^{r})$ (the latter read componentwise), then $x_{j} f$ lies in $D (P_{j})$ and $D_{j} (x_{j} f) = f + x_{j} D_{j} f;$

(b) The product domain. $D (Q_{j} P_{j}) = {f \in D (P_{j}) : x_{j} D_{j} f \in L^{2} (R^{d}; C^{r})}$ and $D (P_{j} Q_{j}) = {f \in D (Q_{j}) : D_{j} (x_{j} f) \in L^{2} (R^{d}; C^{r})}$ , and their intersection is $D (Q_{j} P_{j}) \cap D (P_{j} Q_{j}) = {f \in H^{(r)} : D_{j} f, x_{j} f, x_{j} D_{j} f \in L^{2} (R^{d}; C^{r})};$

(c) Commutator identity. For every $f \in D (Q_{j} P_{j}) \cap D (P_{j} Q_{j})$ , $(Q_{j} P_{j} - P_{j} Q_{j}) f = i ℏ f;$

(d) Core. The componentwise Schwartz class $S (R^{d}; C^{r})$ (Schwartz space and its seminorms) is contained in $D (Q_{j} P_{j}) \cap D (P_{j} Q_{j})$ , and (c) holds on it.

No identity of unbounded products is asserted outside $D (Q_{j} P_{j}) \cap D (P_{j} Q_{j})$ ; in particular no identity is asserted on $D (P_{j})$ alone.

Facts & Assumptions

Given: The Axiom of Countable Choice and the Axiom of Choice; integers $d, r \geq 1$ ; $ℏ > 0$ ; the componentwise operators $Q_{j}, P_{k}$ with the displayed domains; a class $f \in H^{(r)}$ ; a coordinate index $j$ .

[F1]

Domains and actions. $Q_{j}$ and $P_{k}$ are self-adjoint with the maximal componentwise domains displayed in the Statement, $(Q_{j} f)_{l} = x_{j} f_{l}$ and $(P_{k} f)_{l} = - i ℏ D_{k} f_{l}$ componentwise, and $D (P_{k}) = {f : ξ_{k} F_{2}^{(r)} f \in L^{2}}$ (Componentwise position and momentum operators on the component-valued wave-function space clauses (a), (b)). Hence, for a class $f$ , the products $Q_{j} P_{j} f$ and $P_{j} Q_{j} f$ are defined exactly by the memberships $P_{j} f \in D (Q_{j})$ and $Q_{j} f \in D (P_{j})$ respectively, that is $x_{j} D_{j} f \in L^{2}$ and $D_{j} (x_{j} f) \in L^{2}$ in the componentwise reading; no other hypothesis enters the definitions of the product domains.

[F2]

Weak Leibniz rule with a smooth factor. Let $Ω \subseteq R^{n}$ be open, $η \in C^{\infty} (Ω; C)$ , and $u \in L_{l o c}^{1} (Ω; C)$ ; suppose $u$ has weak derivatives $D^{γ} u \in L_{l o c}^{1} (Ω; C)$ for all $γ \leq α$ . Then $η u$ has a weak $α$ -derivative, and its almost-everywhere class is $D^{α} (η u) = \sum_{β \leq α} (\binom{α}{β}) (D^{β} η) D^{α - β} u$ (Weak Leibniz rule with a smooth factor). With $α = e_{j}$ and $η (x) = x_{j}$ this gives $D_{j} (x_{j} u) = u + x_{j} D_{j} u$ in $L_{l o c}^{1}$ for every $u \in L_{l o c}^{1}$ whose weak $j$ -derivative exists in $L_{l o c}^{1}$ .

[F3]

Componentwise reading and Schwartz core. Weak differentiation, multiplication by $x_{j}$ and the $L^{2}$ condition are read componentwise; a class lies in $S (R^{d}; C^{r})$ exactly when all of its components lie in the scalar Schwartz class (Schwartz space and its seminorms, The space $L^{p} (μ)$ as the quotient by null functions); the Schwartz class is contained in $D (Q_{j}) \cap D (P_{k})$ , is mapped into itself by $Q_{j}$ and $P_{k}$ , and satisfies the canonical commutation relation $(Q_{j} P_{j} - P_{j} Q_{j}) f = i ℏ f$ there (Canonical commutation relations on the component-valued core and commutation with the internal factor clauses (a) and (d)).

[F4]

Local square-integrability. $L^{2} (R^{d}; C^{r}) \subseteq L_{l o c}^{1} (R^{d}; C^{r})$ componentwise, by the Cauchy-Schwarz bound on bounded measurable sets (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors, The space $L^{p} (μ)$ as the quotient by null functions).

Proof

technique · apply the weak Leibniz rule with the smooth factor $x_{j}$ componentwise, read the product domains off the definitions, and assemble the commutator with the two actions

Given: The data of the statement and [F1]-[F4].

1.1givenF2F4algebra

[The product rule] Let $f \in H^{(r)}$ with $x_{j} f \in L^{2}$ and $x_{j} D_{j} f \in L^{2}$ . Componentwise, $f_{l} \in L^{2} \subseteq L_{l o c}^{1}$ by [F4]. The occurrence of $D_{j} f_{l}$ in the hypothesis uses the stated locally integrable weak derivative; no global $L^{2}$ membership of $D_{j} f_{l}$ is needed for clause (a). Thus [F2] with $α = e_{j}$ and $η = x_{j}$ applies and gives $D_{j} (x_{j} f_{l}) = f_{l} + x_{j} D_{j} f_{l}$ in $L_{l o c}^{1}$ for every $l$ . The right-hand side is a component of an $L^{2}$ class by hypothesis, so $D_{j} (x_{j} f) \in L^{2}$ componentwise and $x_{j} f \in D (P_{j})$ with the displayed identity. This is (a).

2.1step 1.1F1algebra

[The two product domains] By [F1], $D (Q_{j} P_{j}) = {f : P_{j} f \in D (Q_{j})} = {f \in D (P_{j}) : x_{j} D_{j} f \in L^{2}}$ and $D (P_{j} Q_{j}) = {f : Q_{j} f \in D (P_{j})} = {f \in D (Q_{j}) : D_{j} (x_{j} f) \in L^{2}}$ , which are the first two displays of (b). For the intersection, the inclusion $\subseteq$ is immediate from these displays; conversely, if $f \in H^{(r)}$ has $D_{j} f$ , $x_{j} f$ and $x_{j} D_{j} f$ in $L^{2}$ , then $f \in D (P_{j}) \cap D (Q_{j})$ and step 1.1 gives $D_{j} (x_{j} f) = f + x_{j} D_{j} f \in L^{2}$ , so $f$ lies in both product domains. This is (b).

3.1step 1.1step 2.1F1algebra

[The commutator] Let $f \in D (Q_{j} P_{j}) \cap D (P_{j} Q_{j})$ . By [F1] and step 2.1, $(Q_{j} P_{j} f)_{l} = x_{j} (- i ℏ D_{j} f_{l}) = - i ℏ x_{j} D_{j} f_{l}$ componentwise, while $(P_{j} Q_{j} f)_{l} = - i ℏ D_{j} (x_{j} f_{l}) = - i ℏ (f_{l} + x_{j} D_{j} f_{l})$ by step 1.1. Subtracting componentwise, $((Q_{j} P_{j} - P_{j} Q_{j}) f)_{l} = - i ℏ x_{j} D_{j} f_{l} + i ℏ f_{l} + i ℏ x_{j} D_{j} f_{l} = i ℏ f_{l}$ , so $(Q_{j} P_{j} - P_{j} Q_{j}) f = i ℏ f$ . This is (c).

4.1step 2.1step 3.1F3∎

[The core] Every component of a Schwartz class lies in the scalar Schwartz class, which is closed under multiplication by $x_{j}$ and under $D_{j}$ and is contained in $L^{2}$ ; hence $D_{j} f$ , $x_{j} f$ and $x_{j} D_{j} f$ are in $L^{2}$ , and step 2.1 places $S (R^{d}; C^{r})$ in the intersection. The identity of (c), together with [F3], gives (d).

The position-momentum commutator identity on the product domain

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources