TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22

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.

Von Neumann parameterization of self-adjoint extensions

Statement

Assume the Axiom of Choice. Let $T$ be a densely defined closed symmetric operator on a complex Hilbert space H, with first-variable-linear inner product, with deficiency subspaces $K_{\pm}$ (Deficiency subspaces and deficiency indices), and let $V : K_{+} \to K_{-}$ be a unitary operator. Then $D (T_{V}) = D (T) \oplus {u + V u : u \in K_{+}}, T_{V} (x + u + V u) = T x + i u - i V u$ defines a self-adjoint extension $T_{V}$ of $T$ ; the sum is direct and $D (T_{V})$ is dense. The map $V \mapsto T_{V}$ is a bijection from the set of unitary operators $K_{+} \to K_{-}$ onto the set of self-adjoint extensions of $T$ .

Facts & Assumptions

[A1]

For the given closed densely defined symmetric T, $K_{+} = \ker (T^{*} - i)$ and $K_{-} = \ker (T^{*} + i)$ are closed, and $H = ran (T + i) \oplus K_{+} = ran (T - i) \oplus K_{-}$ orthogonally. The linear map $C_{T} ((T + i) x) = (T - i) x$ is an isometric isomorphism between these ranges, and $(I - C_{T}) (T + i) x = 2 i x$ . Full AC licenses this deficiency-space interface, including its Hilbert-dimension convention. Deficiency subspaces and deficiency indices

[A2]

Under Countable Choice the Cayley correspondence sends a unitary U with $\ker (I - U) = {0}$ to the self-adjoint operator $S (I - U) y = i (I + U) y$ , with domain $ran (I - U)$ , and recovers $C_{S} = U$ . For self-adjoint S its Cayley transform satisfies $C_{S} (S + i) x = (S - i) x$ on D(S). Cayley correspondence between self-adjoint operators and unitaries Cayley transform of a self-adjoint operator

[A3]

D(T) is norm dense in H. Orthogonal decompositions have zero intersection and their squared norms add. A vector orthogonal to a dense subspace is zero: continuity of pairings follows from Cauchy-Schwarz. Densely defined, closed and closable operators, and cores Orthogonality and the orthogonal complement Cauchy–Schwarz: $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥ ∥ y ∥$ , with equality exactly for dependent pairs

[A4]

Full AC is assumed to use [A1]. It implies the Countable Choice required by [A2] directly: AC supplies a choice function for the range family of any given sequence of nonempty sets, and composing that choice function with the sequence gives the required indexed choices. No additional family of choices is made in the construction from the supplied unitary V. The Axiom of Choice The Axiom of Countable Choice ( ${A C}_{ω}$ )

Proof

technique · direct

Given: T as in the statement and a unitary $V : K_{+} \to K_{-}$ . Put $M_{+} = ran (T + i)$ and $M_{-} = ran (T - i)$ .

1.1

Define $U (m + u) = C_{T} m - V u$ for $m \in M_{+}$ and $u \in K_{+}$ . The orthogonal decomposition in [A1] makes this a uniquely defined linear map on H. Its two output terms lie in the orthogonal subspaces $M_{-}$ and $K_{-}$ , so $∥ U (m + u) ∥^{2} = ∥ C_{T} m ∥^{2} + ∥ V u ∥^{2} = ∥ m ∥^{2} + ∥ u ∥^{2} = ∥ m + u ∥^{2}$ . Since both component maps are onto their corresponding summands, U is onto H. Thus U is unitary and extends C_T; the minus sign on K_+ is necessary for the displayed plus sign in u+Vu.

A1A3A4

2.1

For $x \in D (T)$ , $(I - U) (T + i) x = 2 i x$ , hence $ran (I - U)$ contains D(T) and is dense. If Uz=z, then for every y in H, $⟨ z, (I - U) y ⟩ = ⟨ z, y ⟩ - ⟨ U z, U y ⟩ = 0$ , since a unitary preserves inner products. Consequently z is orthogonal to the dense D(T), so z=0 by [A3]. This proves $\ker (I - U) = {0}$ .

A1A3step 1.1

3.1

Apply [A2], licensed by [A4], to get the self-adjoint operator $S (I - U) y = i (I + U) y$ on $ran (I - U)$ , with $C_{S} = U$ . Because $(I - U) ((T + i) x + u) = 2 i x + u + V u$ , that domain equals $D (T) + {u + V u : u \in K_{+}}$ ; scalar multiplication by 2i maps D(T) onto itself. To prove the sum direct, suppose $x = u + V u \in D (T)$ . Then $(I - U) (T + i) x = 2 i x = (I - U) (2 i u)$ , and injectivity from step 2.1 gives $(T + i) x = 2 i u$ . The two sides lie in M_+ and K_+, whose intersection is zero. Hence u=0 and x=0. Also u+Vu=(I-U)u shows the parametrization of the second summand is injective. Thus every vector has a unique representation x+u+Vu, and the domain contains the dense D(T).

A1A2A3A4step 1.1step 2.1

4.1

On D(T), $S (2 i x) = i (I + U) (T + i) x = 2 i T x$ , so Sx=Tx. On the second summand, $S (u + V u) = S (I - U) u = i (I + U) u = i u - i V u$ . Linearity yields $S (x + u + V u) = T x + i u - i V u$ . Thus S is exactly the well-defined operator T_V in the statement and is a self-adjoint extension of T.

A1step 1.1step 3.1

5.1

Let R be any self-adjoint extension of T and put W=C_R. For x in D(T), $(R + i) x = (T + i) x$ , so $W (T + i) x = (R - i) x = (T - i) x$ . Thus W agrees with C_T on M_+ and maps M_+ onto M_-. For u in K_+ and m in M_+, $⟨ W u, W m ⟩ = ⟨ u, m ⟩ = 0$ , so Wu belongs to K_-. Conversely, for v in K_-, take the unique y with Wy=v. For every m in M_+, $⟨ y, m ⟩ = ⟨ v, W m ⟩ = 0$ , hence y belongs to K_+. This proves W(K_+)=K_-, without treating W* as the Cayley transform of R. Consequently V=-W restricted to K_+ is unitary from K_+ onto K_-, and the construction of step 1.1 returns U=W. The inverse correspondence [A2] then gives T_V=R.

A1A2A3step 1.1step 4.1

6.1

If T_V=T_{V'}, their Cayley transforms agree by [A2]. Step 3.1 identifies these transforms with the constructed U and U', whose restrictions to K_+ are -V and -V'. Hence V=V'. Along with steps 4.1 and 5.1, this proves the bijection. This includes empty parameter sets: if no such unitary exists, step 5.1 rules out every self-adjoint extension. If K_+=K_-={0}, the unique unitary of the zero spaces gives D(T_V)=D(T) and T_V=T, so T is already self-adjoint. If H={0}, every displayed map is its unique zero-space map and the same conclusion holds directly. No finite-dimensional or separability assumption is used. AC is used only through [A4].

A1A2A4step 1.1step 3.1step 4.1step 5.1∎

Depends on

Used by

Dependency tree · two levels

41 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Gerald Teschl, Mathematical Methods in Quantum Mechanics, second edition (standard reference, not scraped)
Theo Buehler and Dietmar A. Salamon, Functional Analysis (standard reference, not scraped)