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The minimal derivative has deficiency indices (1,1) and many self-adjoint extensions

Statement refuted

Assume the Axiom of Choice (and hence Countable Choice and Dependent Choice). Let $T$ be the minimal operator of A symmetric closed operator that is not self-adjoint, that is $T = - i d / d x$ on $D (T) = {f \in A C [0, 1] : f^{'} \in L^{2} (0, 1), f (0) = f (1) = 0}$ in $H = L^{2} (0, 1)$ . Then $T$ is a closed symmetric operator with $d_{+} (T) = d_{-} (T) = 1$ : $K_{+} = \ker (T^{*} - i) = C e^{- x}$ and $K_{-} = \ker (T^{*} + i) = C e^{x}$ . Consequently $T$ is not self-adjoint but has infinitely many self-adjoint extensions, and these are exactly the operators $T_{μ} f = - i f^{'}, D (T_{μ}) = {f \in A C [0, 1] : f^{'} \in L^{2} (0, 1), f (1) = μ f (0)}, ∣ μ ∣ = 1.$

Facts & Assumptions

[A1]

The minimal operator $T$ is densely defined, closed and symmetric but not self-adjoint, and $D (T^{*}) = {g \in A C [0, 1] : g^{'} \in L^{2} (0, 1)}$ with $T^{*} g = - i g^{'}$ (A symmetric closed operator that is not self-adjoint).

[A2]

Self-adjoint extensions of a closed symmetric operator correspond bijectively to unitary operators $K_{+} \to K_{-}$ , with domain $D (T) \oplus {u + V u : u \in K_{+}}$ and action $T_{V} (x + u + V u) = T x + i u - i V u$ (Von Neumann parameterization of self-adjoint extensions, Deficiency subspaces and deficiency indices).

[A3]

A closed symmetric operator has a self-adjoint extension if and only if its deficiency indices agree; its extensions are indexed by the unitaries between the deficiency subspaces (Existence of self-adjoint extensions is equality of deficiency indices).

Counterexample

technique · direct

Given: The minimal operator $T$ on $H = L^{2} (0, 1)$ .

1.1A1

By A symmetric closed operator that is not self-adjoint, $T$ is densely defined, closed, symmetric and not self-adjoint, and $D (T^{*}) = {g \in A C [0, 1] : g^{'} \in L^{2} (0, 1)}$ with $T^{*} g = - i g^{'}$ .

1.2A1

$\ker (T^{*} - i) = C e^{- x}$ and $\ker (T^{*} + i) = C e^{x}$ : the equations $T^{*} g = i g$ and $T^{*} g = - i g$ read $g^{'} = - g$ and $g^{'} = g$ .

2.1A2step 1.2

Hence $d_{+} = d_{-} = 1$ and $K_{+} \cap K_{-} = {0}$ ; by the von Neumann parameterization the self-adjoint extensions of $T$ correspond bijectively to the unitaries $V : C e^{- x} \to C e^{x}$ , that is, to the numbers $λ$ with $V (e^{- x}) = λ e^{x}$ and $∣ λ ∣ = ∥ e^{- x} ∥ / ∥ e^{x} ∥ = 1 / e$ .

2.2A2step 1.2

Domains: by the parameterization, $D (T_{V}) = D (T) \oplus {u + V u : u \in K_{+}}$ . For $u = c e^{- x}$ the element is $x_{T} + c (e^{- x} + λ e^{x})$ with $x_{T} (0) = x_{T} (1) = 0$ , so its endpoint values are $c (1 + λ)$ at $0$ and $c (e^{- 1} + λ e)$ at $1$ ; hence $D (T_{V})$ consists exactly of the absolutely continuous $f$ with $f^{'} \in L^{2}$ and $f (1) = μ f (0)$ , where $μ = (e^{- 1} + λ e) / (1 + λ)$ .

3.1A2A3step 2.1

In particular there are infinitely many self-adjoint extensions, so $T$ is not self-adjoint; the case of $λ$ equal to a suitable value reproduces the periodic operator of Periodic derivative and its unitary translation group.

3.2step 2.2

The map $λ \mapsto μ$ is a bijection from ${λ : ∣ λ ∣ = 1 / e}$ onto the unit circle: for $∣ λ ∣ = 1 / e$ one computes $∣ μ ∣ = 1$ , and for $∣ μ ∣ = 1$ the formula $λ = (μ - e^{- 1}) / (e - μ)$ inverts it and satisfies $∣ λ ∣ = 1 / e$ .

4.1A2step 3.1step 3.2∎

Therefore the self-adjoint extensions of $T$ are exactly the operators $T_{μ}$ of the statement, one for each $μ$ on the unit circle; $T$ itself is not among them because it is not self-adjoint.

Depends on

Used by

Self-adjoint extensions and deficiency indices: agreement pointer Remark

Dependency tree · two levels

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Sources

Dana P. Williams, Lecture Notes on the Spectral Theorem (standard reference, not scraped)
Gerald Teschl, Mathematical Methods in Quantum Mechanics, second edition (standard reference, not scraped)