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Dirichlet momentum failure

Example

Assume the Axiom of Choice and Countable Choice (The Axiom of Choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $0 < L < \infty$ and define on $L^{2} (0, L; C)$ the operator $T : = - i D, D (T) : = {u \in H^{1} (0, L) : u^{*} (0) = u^{*} (L) = 0},$ with $H^{1}$ and traces as in One-dimensional Sobolev spaces on intervals. Then:

(a) $T$ is densely defined and symmetric, but not self-adjoint; its adjoint is $T^{*} = - i D, D (T^{*}) = H^{1} (0, L),$ with no boundary condition;

(b) the deficiency vectors are explicit: $\ker (T^{*} - i) = C e^{- x}, \ker (T^{*} + i) = C e^{x},$ both one-dimensional, so the deficiency indices of $T$ are $(1, 1)$ ;

(c) the self-adjoint extensions of $T$ are exactly the twisted-periodic momentum operators $P_{θ}$ , $θ \in [0, 2 π)$ , of The twisted-periodic momentum operator on an interval: $P_{θ} = - i D, D (P_{θ}) = {u \in H^{1} (0, L) : u^{*} (L) = e^{i θ} u^{*} (0)};$ in particular no self-adjoint extension keeps the Dirichlet condition $u^{*} (0) = u^{*} (L) = 0$ .

Verification

Given: $0 < L < \infty$ ; the operator $T = - i D$ with $D (T) = {u \in H^{1} (0, L) : u^{*} (0) = u^{*} (L) = 0}$ ; the twisted-periodic operators $P_{θ}$ of The twisted-periodic momentum operator on an interval.

Proof technique: direct computation of the adjoint and of the boundary form.

[F1] The normalized sine system $(e_{n})_{n \geq 1}$ is complete in $L^{2} (0, L)$ (Completeness of the Dirichlet sine system on an interval). Each sine has zero endpoint traces, so $e_{n} \in H_{0}^{1} (0, L)$ by Integration by parts and the endpoint-vanishing space on intervals clause (b); by the definition $H_{0}^{1} (0, L) = {\overline{C_{c}^{\infty} (0, L)}}^{∥ \cdot ∥_{H^{1}}}$ (One-dimensional Sobolev spaces on intervals), every finite sine sum is an $L^{2}$ limit of test functions because $∥ \cdot ∥_{2} \leq ∥ \cdot ∥_{H^{1}}$ . Thus $C_{c}^{\infty} (0, L)$ is dense in $L^{2}$ and contained in $D (T)$ . For $u, v \in H^{1} (0, L)$ , $\int_{0}^{L} u \overline{v^{'}} = u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)} - \int_{0}^{L} u^{'} \overline{v}$ (Integration by parts and the endpoint-vanishing space on intervals).

[F2] A vector $v \in L^{2}$ lies in $D (T^{*})$ exactly when $u \mapsto ⟨ T u, v ⟩$ is bounded on $D (T)$ ; then $⟨ T u, v ⟩ = ⟨ u, T^{*} v ⟩$ for all $u \in D (T)$ ; a self-adjoint operator has no proper symmetric extension (Adjoint of a densely defined operator, Symmetric, self-adjoint and essentially self-adjoint operators).

[F3] The equation $D u = c u$ on $(0, L)$ has the solutions $u (x) = A e^{c x}$ (The complex exponential by its power series and uniqueness of solutions of a first-order linear ODE with constant coefficient).

1.1F1

[Density and symmetry] By [F1], $C_{c}^{\infty} (0, L) \subseteq D (T)$ and $C_{c}^{\infty} (0, L)$ is dense, so $T$ is densely defined. For $u, v \in D (T)$ , [F1] gives $⟨ T u, v ⟩ - ⟨ u, T v ⟩ = - i (u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)}) = 0,$ since both traces of both functions vanish. Hence $T$ is symmetric.

2.1step 1.1F1F2algebra

[The adjoint] Let $v \in H^{1} (0, L)$ . For every $u \in D (T)$ , [F1] and the vanishing of $u^{*}$ at the endpoints give $⟨ T u, v ⟩ = \int_{0}^{L} (- i D u) \overline{v} = - i [u^{*} \overline{v^{*}}]_{0}^{L} + \int_{0}^{L} u \overline{(- i D v)} = ⟨ u, - i D v ⟩ .$ Hence $u \mapsto ⟨ T u, v ⟩$ is the restriction to $D (T)$ of the bounded functional $⟨ \cdot, - i D v ⟩$ on $L^{2}$ , so $v \in D (T^{*})$ and $T^{*} v = - i D v$ by [F2]. Conversely, if $v \in D (T^{*})$ , testing the identity $⟨ T u, v ⟩ = ⟨ u, T^{*} v ⟩$ against $u \in C_{c}^{\infty} (0, L) \subseteq D (T)$ shows that $- i D v$ (as a distribution) equals the $L^{2}$ class $T^{*} v$ ; hence $D v \in L^{2}$ and $v \in H^{1} (0, L)$ . Thus $D (T^{*}) = H^{1} (0, L)$ and $T^{*} = - i D$ . Since $H^{1} (0, L) ⊋ D (T)$ (for instance the constant function $1$ lies in $H^{1}$ but $1^{*} (0) = 1 \neq 0$ ), $T$ is not self-adjoint. This is (a).

3.1step 2.1F3

[Deficiency vectors] By step 2.1, $v \in \ker (T^{*} - i)$ means $v \in H^{1} (0, L)$ and $- i D v = i v$ , i.e. $D v = - v$ ; by [F3], $v (x) = A e^{- x}$ . Hence $\ker (T^{*} - i) = C e^{- x}$ , one-dimensional; similarly $D v = v$ gives $\ker (T^{*} + i) = C e^{x}$ . This is (b).

4.1step 2.1step 3.1F2algebra∎

[All self-adjoint extensions] Fix $θ \in [0, 2 π)$ . By The twisted-periodic momentum operator on an interval, $P_{θ}$ is self-adjoint with $D (P_{θ}) = {u \in H^{1} : u^{*} (L) = e^{i θ} u^{*} (0)}$ , and $P_{θ} u = - i D u = T^{*} u$ for $u \in D (P_{θ})$ ; since $D (T) \subseteq D (P_{θ})$ ( $0 = e^{i θ} \cdot 0$ ), $P_{θ}$ is a self-adjoint extension of $T$ . Conversely, let $S$ be a self-adjoint extension of $T$ , so $T \subseteq S$ and $S \subset T^{*}$ (a self-adjoint operator has no proper symmetric extension and $T^{*}$ is the maximal domain on which the action is $- i D$ by step 2.1); write $D_{1} : = D (S)$ , a linear subspace with $D (T) \subseteq D_{1} \subseteq H^{1} (0, L)$ . Since $S$ is symmetric, for all $u, v \in D_{1}$ , $0 = ⟨ S u, v ⟩ - ⟨ u, S v ⟩ = - i (u^{*} (L) \overline{v^{*} (L)} - u^{*} (0) \overline{v^{*} (0)}) .$ If every $u \in D_{1}$ had $u^{*} (0) = 0$ , then the identity would give $u^{*} (L) \overline{v^{*} (L)} = 0$ for all $u, v \in D_{1}$ , hence $u^{*} (L) = 0$ for all $u$ , so $D_{1} \subseteq D (T)$ and $S = T$ , contradicting that $T$ is not self-adjoint. Choose $v \in D_{1}$ with $v^{*} (0) \neq 0$ ; the identity for $u \in D_{1}$ and this fixed $v$ gives $u^{*} (L) = λ u^{*} (0)$ with $λ : = \overline{v^{*} (0)} / \overline{v^{*} (L)}$ , and testing the identity with $u = v$ gives $∣ v^{*} (L) ∣ = ∣ v^{*} (0) ∣$ , so $∣ λ ∣ = 1$ and $λ = e^{i θ}$ for some $θ \in [0, 2 π)$ . Hence $D_{1} \subseteq D (P_{θ})$ and $S \subseteq P_{θ}$ . Both operators are self-adjoint, so by [F2] the inclusion cannot be proper: $S = P_{θ}$ . This is (c).

Dirichlet momentum failure

Example

Verification

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