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Existence of self-adjoint extensions is equality of deficiency indices

Statement

Assume the Axiom of Choice. Let $T$ be a densely defined closed symmetric operator on $H$ , with deficiency indices $d_{\pm} (T)$ (Deficiency subspaces and deficiency indices). Then $T$ has a self-adjoint extension if and only if $d_{+} (T) = d_{-} (T)$ . Moreover $T$ is self-adjoint if and only if $d_{+} (T) = d_{-} (T) = 0$ , and a densely defined symmetric (not necessarily closed) operator $S$ is essentially self-adjoint if and only if $d_{\pm} (\overline{S}) = 0$ , equivalently $\ker (S^{*} \mp i) = {0}$ ; if $d_{+} (\overline{S}) = d_{-} (\overline{S})$ , then $S$ has self-adjoint extensions.

Facts & Assumptions

[A1]

Unitary operators $K_{+} \to K_{-}$ correspond bijectively to self-adjoint extensions of $T$ ; every such unitary is onto by definition, and the Cayley transform $U_{S}$ of a self-adjoint extension restricts to a unitary $K_{+} \to K_{-}$ (Von Neumann parameterization of self-adjoint extensions).

[A2]

Every Hilbert space has a complete orthonormal family, and two Hilbert spaces are unitarily isomorphic exactly when their orthonormal bases have the same cardinality; a unitary $K_{+} \to K_{-}$ exists exactly when $\dim K_{+} = \dim K_{-}$ (Existence of a maximal orthonormal family, and maximality as completeness, A Hilbert space with a given orthonormal basis is $ℓ^{2}$ of the index set, Orthonormal families, complete orthonormal systems and Hilbert bases).

[A3]

A closed symmetric operator is self-adjoint if and only if $\ker (T^{*} \mp i) = {0}$ , equivalently $ran (T \pm i) = H$ (Range criterion for self-adjointness).

[A4]

For a densely defined symmetric $S$ one has $\overline{S} = S^{* *}$ and $S^{*} = (\overline{S})^{*}$ ; $S$ is essentially self-adjoint exactly when $\overline{S}$ is self-adjoint (Closability is equivalent to density of the adjoint domain, Symmetric, self-adjoint and essentially self-adjoint operators).

Proof

technique · direct

Given: A densely defined closed symmetric operator $T$ , and a densely defined symmetric operator $S$ .

1.1A1A2

If $d_{+} (T) = d_{-} (T)$ , then by [A2] there is a unitary $V : K_{+} \to K_{-}$ (equal Hilbert dimensions), and [A1] produces a self-adjoint extension $T_{V}$ of $T$ . Conversely, if $T$ has a self-adjoint extension $S$ , then by [A1] its Cayley transform restricts to a unitary $K_{+} \to K_{-}$ , so $\dim K_{+} = \dim K_{-}$ by [A2].

1.2A3

$T$ is self-adjoint if and only if $d_{+} = d_{-} = 0$ : if both deficiency subspaces are zero then $\ker (T^{*} \mp i) = {0}$ and [A3] applies; conversely a self-adjoint $T$ has $T^{*} = T$ and $T \mp i$ injective by the estimate $∥ (T \mp i) x ∥ \geq ∥ x ∥$ , so both kernels vanish.

2.1A4step 1.1step 1.2

For $S$ symmetric, $\overline{S}$ is closed and symmetric by [A4], and $\ker (S^{*} \mp i) = \ker ((\overline{S})^{*} \mp i)$ because $S^{*} = (\overline{S})^{*}$ ; applying steps 1.1-1.2 to $\overline{S}$ gives: $S$ is essentially self-adjoint, meaning $\overline{S}$ self-adjoint, if and only if $d_{\pm} (\overline{S}) = 0$ ; if $d_{+} (\overline{S}) = d_{-} (\overline{S})$ then $\overline{S}$ has a self-adjoint extension, which is also an extension of $S$ .

3.1

All the stated equivalences are steps 1.1, 1.2 and 2.1. ∎

Depends on

Used by

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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)