CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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Cook existence is not completeness

Statement refuted

Every isometric wave operator has range equal to the full absolutely continuous subspace of the interacting Hamiltonian.

Statement references

Wave, scattering and channel operators

The Borel sigma-algebra of a topological space

Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$

A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included

$Q$ is countably infinite

Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable

If $μ$ is sigma-finite and $A$ is countably generated, then $L^{p} (μ)$ is separable for $1 \leq p < \infty$

Facts & Assumptions

Given: $H_{0} = L^{2} ((0, 1), B ((0, 1)), β; C)$ , $B ((0, 1))$ is the Borel sigma-algebra and $β (B) = λ_{1} (B)$ for every Borel $B \subseteq (0, 1)$ , $⟨ f, g ⟩ = \int_{0}^{1} f (x) \overline{g (x)} d β (x)$ is linear in its first variable, $H_{0} f (x) = x f (x)$ on $D (H_{0}) = H_{0}$ , $H = H_{0} \oplus H_{0}$ , $H = H_{0} \oplus H_{0}$ on $D (H) = H$ , and $J f = (f, 0)$ . Use $U_{0} (t) f = e^{i t x} f$ and $U (t) (f, g) = (e^{i t x} f, e^{i t x} g)$ .

[A1]

Assume the Axiom of Choice (The Axiom of Choice).

[A2]

Assume Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Completeness means $ran Ω = H_{a c} (H)$ , not merely that $Ω$ is an isometry (Wave, scattering and channel operators).

[F2]

Self-adjointness means equality with the adjoint, including its domain (Symmetric, self-adjoint and essentially self-adjoint operators).

[F3]

The spectral PVM of $H_{0} = M_{x}$ is $E_{0} (B) = M_{1_{B \cap (0, 1)}}$ ; its scalar measures are $B \mapsto \int_{B \cap (0, 1)} ∣ f (x) ∣^{2} d x$ . On the direct sum, $E (B) = E_{0} (B) \oplus E_{0} (B)$ is a PVM whose spectral integral is $H_{0} \oplus H_{0}$ , so uniqueness in the spectral theorem identifies it as the PVM of $H$ (Spectral theorem for unbounded self-adjoint operators (PVM form), Unbounded Borel functional calculus: domains, products, spectral mapping, Projection valued measure, Integral of a measurable function against a projection-valued measure).

[F4]

For self-adjoint $T$ , $e^{i t T}$ is a strongly continuous unitary group with generator $i T$ (A self-adjoint operator generates a strongly continuous unitary group).

[F5]

The wave-operator definition inserts $P_{a c} (H_{0})$ in its approximants and defines completeness by equality with the interacting absolutely continuous subspace (Wave, scattering and channel operators).

[F6]

The Borel sigma-algebra of $(0, 1)$ is generated by the countable basis of open intervals with rational endpoints intersected with $(0, 1)$ ; the rationals are countable and dense (The Borel sigma-algebra of a topological space, $Q$ is countably infinite, Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

[F7]

The Borel Lebesgue measure $β$ of $(0, 1)$ is finite, with $β ((0, 1)) = 1$ (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ , A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[F8]

A sigma-finite measure on a countably generated sigma-algebra has separable $L^{2}$ under Countable Choice (If $μ$ is sigma-finite and $A$ is countably generated, then $L^{p} (μ)$ is separable for $1 \leq p < \infty$ ).

Counterexample

Given: [A1], [A2], [F6]–[F8], the Hilbert spaces, operators, identification $J$ , and unitary groups specified above.

1.1A1A2F2F3F4F6F7F8givenalgebra

By [F6], the Borel sigma-algebra on $(0, 1)$ is countably generated, and by [F7] its measure is finite, hence sigma-finite. [F8] gives separability of $H_{0}$ ; a direct sum of two separable spaces is separable because pairs from the product of countable dense subsets are countable and dense, so $H$ is separable. Since $0 < x < 1$ , $∥ H_{0} f ∥^{2} = \int_{0}^{1} x^{2} ∣ f ∣^{2} d β \leq ∥ f ∥^{2}$ , and $⟨ H_{0} f, h ⟩ = ⟨ f, H_{0} h ⟩$ for every $f, h \in H_{0}$ . Thus every $h$ belongs to $D (H_{0}^{*})$ , with $H_{0}^{*} h = H_{0} h$ , proving self-adjointness by [F2]. The same componentwise pairing on the full domain $H$ proves $H^{*} = H$ . Its PVM is the multiplication family in [F3], so bounded Borel functional calculus gives $e^{i t H_{0}} = M_{e^{i t x}} = U_{0} (t)$ and $e^{i t H} = U (t)$ , consistently with [F4].

1.2F2F3F4F5givenalgebra

By [F3], every scalar spectral measure of $H_{0}$ is absolutely continuous, so $P_{a c} (H_{0}) = I$ . Thus the projection in [F5] drops out. Using $J f = (f, 0)$ , the two-space wave-operator approximants are exactly $U (t) J U_{0} (- t) f = (e^{i t x} e^{- i t x} f, 0) = (f, 0)$ for every $f \in H_{0}$ and every real $t$ . Since $H J = J H_{0}$ and both generators are bounded, direct differentiation gives $\frac{d}{d t} (e^{i t H} J e^{- i t H_{0}}) = i e^{i t H} (H J - J H_{0}) e^{- i t H_{0}} = 0$ . Thus the approximants are constant and both strong limits exist with $Ω_{\pm} f = (f, 0)$ . Their norm equals $∥ f ∥$ , so they are isometries.

1.3F3givenalgebra

By [F3], every vector $f \oplus g \in H$ has scalar spectral measure $B ⟼ \int_{B \cap (0, 1)} (∣ f (x) ∣^{2} + ∣ g (x) ∣^{2}) d x$ , which is absolutely continuous with respect to Lebesgue measure. Therefore $H_{a c} (H) = H$ .

2.1F1step 1.2step 1.3∎

The wave-operator range is $H_{0} \oplus {0}$ , which is a proper subspace of $H$ because the constant function $1 \in L^{2} ((0, 1))$ has norm one, so $0 \oplus 1$ is a nonzero vector outside the range. Hence the wave operators are isometric but not complete, refuting the statement.

Depends on

Dependency tree · 2 levels

Sources

D. R. Yafaev, Lectures on Scattering Theory, Lecture 1, Definitions 1.1–1.2