TheoremStatement: 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.

Position-momentum uncertainty on the maximal domains of the line

Statement

Assume the Axiom of Choice and the Axiom of Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix $ℏ > 0$ and let $Q$ and $P$ be the position and momentum operators of Position and momentum operators on Euclidean L2 on $L^{2} (R; C)$ ; by Self-adjointness, Fourier representation and translation identities of position and momentum both are self-adjoint, $D (P) = W^{1, 2} (R; C)$ and $D (Q) = {f : x f \in L^{2}}$ . Let $f \in D (Q) \cap D (P)$ have $∥ f ∥ = 1$ , and put $a : = ⟨ Q f, f ⟩ = \int_{R} x ∣ f (x) ∣^{2} d x, b : = ⟨ P f, f ⟩,$ $x_{Q} : = (Q - a) f, y_{P} : = (P - b) f, Δ Q : = ∥ x_{Q} ∥, Δ P : = ∥ y_{P} ∥ .$ Then:

(a) $a, b \in R$ , and with $m_{1} : = \int_{R} f^{'} \overline{f} d x$ and $m_{2} : = \int_{R} x f^{'} \overline{f} d x$ (absolutely convergent complex integrals, $f^{'}$ the weak derivative) one has $Re m_{1} = 0, Re m_{2} = - \frac{1}{2}, b = ℏ Im m_{1};$

(b) $⟨ x_{Q}, y_{P} ⟩ = - \frac{i ℏ}{2} + ℏ Im m_{2} - a b$ ; consequently $Im ⟨ x_{Q}, y_{P} ⟩ = - \frac{ℏ}{2}$ and $Δ Q Δ P \geq \frac{ℏ}{2};$

(c) equality $Δ Q Δ P = \frac{ℏ}{2}$ holds if and only if $x_{Q}, y_{P}$ are linearly dependent and $Re ⟨ x_{Q}, y_{P} ⟩ = 0$ ;

(d) no assumption on the products $Q P$ or $P Q$ is used: the assertion holds on the full maximal domain of the pair, and the means $a, b$ , the centered vectors $x_{Q}, y_{P}$ and the variances are finite on it.

The statement concerns a preparation vector only; it asserts no apparatus error, jointly measured distribution or physical interpretation.

Facts & Assumptions

Given: a unit vector $f \in W^{1, 2} (R; C)$ with $x f \in L^{2}$ , its weak derivative $f^{'} \in L^{2}$ , the numbers $a, b$ , the vectors $x_{Q}, y_{P}$ and the variances.

[F1]

$f$ has a unique continuous locally absolutely continuous representative $f^{*}$ with $f^{*} (x) - f^{*} (y) = \int_{y}^{x} f^{'}$ ; moreover $∣ f^{*} (x) ∣ \leq C_{f} : = ∥ f ∥ + ∥ f^{'} ∥ < \infty$ for all $x$ and $f^{*} (x) \to 0$ as $∣ x ∣ \to \infty$ (One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives, Endpoint trace bound and decay of H1 functions clause (c)).

[F2]

There is $χ \in C_{c}^{\infty} (R)$ with $0 \leq χ \leq 1$ , $χ = 1$ on $[- 1, 1]$ , $χ = 0$ outside $[- 2, 2]$ ; for $R > 0$ , $χ_{R} (x) : = χ (x / R)$ satisfies $χ_{R} = 1$ on $[- R, R]$ , $χ_{R} = 0$ outside $[- 2 R, 2 R]$ , and $∣ x χ_{R}^{'} (x) ∣ \leq \sup_{t} ∣ t χ^{'} (t) ∣ = : C_{χ} < \infty$ (Explicit compactly supported smooth cutoffs).

[F3]

Complex conjugation preserves $W^{1, 2}$ and $D (\overline{f}) = \overline{D f}$ : this follows from the weak-derivative definition applied to the conjugate test functions, since $\int \overline{f} φ^{'} = \overline{\int f \overline{φ^{'}}} = - \overline{\int D f \overline{φ}} = - \int \overline{D f} φ$ for real-valued $φ$ and hence for all test functions by linearity.

[F4]

Smooth-factor Leibniz rule: if $η$ is smooth and $D^{β} η$ is bounded for $∣ β ∣ \leq 1$ and $u \in W^{1, 2}$ , then $η u \in W^{1, 2}$ with $D (η u) = (D η) u + η D u$ (Weak Leibniz rule with a smooth factor).

[F5]

Weak product rule for bounded Sobolev factors: if $u, v \in W^{1, 2} (R; C) \cap L^{\infty}$ , then $u v \in W^{1, 2}$ with $D (u v) = (D u) v + u (D v)$ (Weak product rule for bounded Sobolev functions).

[F6]

If $g \in W^{1, 1} (R; C)$ has compact support, then $\int_{R} D g = 0$ : the absolutely continuous representative of $g$ (for $p = 1$ ) vanishes outside a compact set, so its limiting difference and the difference formula give zero (One-dimensional $W^{1, p}$ functions have unique absolutely continuous representatives, applied with $p = 1$ ).

[F7]

Cauchy–Schwarz and its consequence: $∣ ⟨ v, w ⟩ ∣^{2} = (Re ⟨ v, w ⟩)^{2} + (Im ⟨ v, w ⟩)^{2}$ and $∥ v ∥ ∥ w ∥ \geq ∣ ⟨ v, w ⟩ ∣ \geq ∣ Im ⟨ v, w ⟩ ∣$ (Cauchy–Schwarz: $∣ ⟨ u, v ⟩ ∣ \leq ∥ u ∥ ∥ v ∥$ , with equality exactly for linearly dependent vectors).

[F8]

Dominated convergence: if $g_{R} \to g$ pointwise and $∣ g_{R} ∣ \leq G$ with $\int G < \infty$ , then $\int g_{R} \to \int g$ (Dominated convergence).

[F9]

Covariance-refined uncertainty of Covariance-refined preparation uncertainty without operator products: for symmetric densely defined operators $A, B$ and a unit vector $u \in D (A) \cap D (B)$ , with centered vectors $x = (A - a) u$ , $y = (B - b) u$ and $z : = ⟨ x, y ⟩$ , one has $Δ A Δ B \geq ∣ z ∣ \geq ∣ Im z ∣$ , with $∣ z ∣^{2} = (Re z)^{2} + (Im z)^{2}$ ; equality in $Δ A Δ B \geq ∣ z ∣$ holds exactly when $x$ and $y$ are linearly dependent (clauses (b) and (c) there, the equality clause being that of the Cauchy–Schwarz step).

Proof

1.1givenF7algebra

[Integrability of the two pairings] Since $f, x f, f^{'} \in L^{2} (R)$ , Cauchy–Schwarz gives $\int ∣ f^{'} \overline{f} ∣ \leq ∥ f^{'} ∥ ∥ f ∥ < \infty$ and $\int ∣ x f^{'} \overline{f} ∣ \leq ∥ x f ∥ ∥ f^{'} ∥ < \infty$ ; so $m_{1}, m_{2}$ are well-defined complex numbers.

1.2givenF2F3F4F5algebra

[Smooth cutoffs and products in W^{1,2}] For $R > 0$ put $η_{R} (x) : = x χ_{R} (x)$ and $u_{R} : = η_{R} f$ . By [F2], $η_{R}$ and $η_{R}^{'}$ are bounded (by $2 R$ and $1 + C_{χ}$ ), so [F4] gives $u_{R} \in W^{1, 2}$ with $D u_{R} = (χ_{R} + x χ_{R}^{'}) f + x χ_{R} f^{'}$ ; also $u_{R}$ is bounded and compactly supported. By [F3] and [F1], $\overline{f} \in W^{1, 2} \cap L^{\infty}$ with $D \overline{f} = \overline{f^{'}}$ . Hence [F5] applies to $u_{R}$ and $\overline{f}$ : $g_{R} : = u_{R} \overline{f} \in W^{1, 2} (R; C)$ , compactly supported, with $D g_{R} = (χ_{R} + x χ_{R}^{'}) ∣ f ∣^{2} + 2 x χ_{R} Re (f^{'} \overline{f}),$ using $f^{'} \overline{f} + \overline{f^{'}} f = 2 Re (f^{'} \overline{f})$ .

2.1givenstep 1.2F6algebra

[The integrated identity] By [F6] applied to the compactly supported $W^{1, 2}$ (hence $W^{1, 1}$ ) class $g_{R}$ , $0 = \int_{R} D g_{R} = \int_{R} χ_{R} ∣ f ∣^{2} + \int_{R} x χ_{R}^{'} ∣ f ∣^{2} + 2 Re \int_{R} x χ_{R} f^{'} \overline{f}$ .

3.1givenstep 1.1step 2.1F8algebra

[Limits as $R \to \infty$ ] As $R \to \infty$ : (i) $χ_{R} ∣ f ∣^{2} \to ∣ f ∣^{2}$ pointwise with $0 \leq χ_{R} ∣ f ∣^{2} \leq ∣ f ∣^{2} \in L^{1}$ , so $\int χ_{R} ∣ f ∣^{2} \to ∥ f ∥^{2} = 1$ by [F8]; (ii) $∣ x χ_{R}^{'} ∣$ is bounded by $C_{χ}$ and supported in ${∣ x ∣ \geq R}$ , so $∣ \int x χ_{R}^{'} ∣ f ∣^{2} ∣ \leq C_{χ} \int_{{∣ x ∣ \geq R}} ∣ f ∣^{2} \to 0$ by absolute continuity of the integral; (iii) $x χ_{R} f^{'} \overline{f} \to x f^{'} \overline{f}$ pointwise with $∣ x χ_{R} f^{'} \overline{f} ∣ \leq ∣ x f^{'} \overline{f} ∣ \in L^{1}$ by step 1.1, so $\int x χ_{R} f^{'} \overline{f} \to m_{2}$ by [F8]. Passing to the limit in step 2.1 gives $1 + 2 Re m_{2} = 0$ , that is $Re m_{2} = - \frac{1}{2}$ .

4.1givenstep 3.1F6F8algebra

[The second identity] Applying the same construction with $η_{R}$ replaced by $χ_{R}$ (so $g_{R}^{0} : = χ_{R} f \overline{f} \in W^{1, 2}$ , compactly supported, with $D g_{R}^{0} = χ_{R}^{'} ∣ f ∣^{2} + 2 χ_{R} Re (f^{'} \overline{f})$ ) and using [F6], $0 = \int χ_{R}^{'} ∣ f ∣^{2} + 2 Re \int χ_{R} f^{'} \overline{f}$ ; as $R \to \infty$ the first term tends to $0$ as in step 3.1(ii) and the second to $2 Re m_{1}$ by dominated convergence ( $∣ f^{'} \overline{f} ∣ \leq \frac{1}{2} (∣ f ∣^{2} + ∣ f^{'} ∣^{2}) \in L^{1}$ ). Hence $Re m_{1} = 0$ .

5.1givenstep 4.1F3algebra

[The means] $a$ is real because $Q$ is self-adjoint and $f \in D (Q)$ ; $b$ is real because $P$ is self-adjoint and $f \in D (P)$ . Moreover $b = ⟨ P f, f ⟩ = \int_{R} (- i ℏ f^{'}) \overline{f} d x = - i ℏ m_{1}$ , and $Re m_{1} = 0$ gives $- i ℏ m_{1} = ℏ Im m_{1}$ , real as required. This is (a).

6.1givenstep 3.1step 5.1algebra

[The pairing] Expand $⟨ x_{Q}, y_{P} ⟩ = ⟨ Q f, P f ⟩ - b ⟨ Q f, f ⟩ - a ⟨ f, P f ⟩ + a b ⟨ f, f ⟩$ . Here $⟨ Q f, f ⟩ = a$ and $⟨ f, P f ⟩ = \overline{⟨ P f, f ⟩} = b$ (both means real), so the last three terms sum to $- a b + a b - a b = - a b$ . And $⟨ Q f, P f ⟩ = \int x f \overline{(- i ℏ f^{'})} d x = i ℏ \int x f \overline{f^{'}} d x = i ℏ \overline{m_{2}} = i ℏ (- \frac{1}{2} - i Im m_{2}) = - \frac{i ℏ}{2} + ℏ Im m_{2}$ , using step 3.1. Hence $⟨ x_{Q}, y_{P} ⟩ = - \frac{i ℏ}{2} + ℏ Im m_{2} - a b$ , with imaginary part $- \frac{ℏ}{2}$ .

7.1givenstep 6.1F7F9algebra

[The uncertainty bound] By [F9] (or [F7] applied to $x_{Q}$ and $y_{P}$ ), $Δ Q Δ P = ∥ x_{Q} ∥ ∥ y_{P} ∥ \geq ∣ ⟨ x_{Q}, y_{P} ⟩ ∣ \geq ∣ Im ⟨ x_{Q}, y_{P} ⟩ ∣ = \frac{ℏ}{2}$ . This is (b).

8.1givenstep 6.1step 7.1F9algebra∎

[Equality and scope] By the equality clause of [F9], equality in $∥ x_{Q} ∥ ∥ y_{P} ∥ \geq ∣ ⟨ x_{Q}, y_{P} ⟩ ∣$ holds exactly when $x_{Q}, y_{P}$ are linearly dependent; given step 6.1, the further condition $∣ ⟨ x_{Q}, y_{P} ⟩ ∣ = \frac{ℏ}{2}$ is exactly $Re ⟨ x_{Q}, y_{P} ⟩ = 0$ by the identity $∣ ⟨ x_{Q}, y_{P} ⟩ ∣^{2} = (Re ⟨ x_{Q}, y_{P} ⟩)^{2} + (Im ⟨ x_{Q}, y_{P} ⟩)^{2}$ of [F9]. This is (c). No product $Q P$ or $P Q$ was used anywhere: the means, the centered vectors and the pairings were formed from $Q f$ and $P f$ only, so (d) holds.

Position-momentum uncertainty on the maximal domains of the line

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources