Qm Math Intervals And Contacts — Examples

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

[The adjoint] Let $v\in H^1(0,L)$. For every $u\in D(T)$, [F1] and the vanishing of $u^{\ast }$ at the endpoints give $⟨Tu,v⟩={\int }_0^L(-iDu)\bar{v}=-i[u^{\ast }\overline{v^{\ast }}{]}_0^L+{\int }_0^{L}u\overline{(-iDv)}=⟨u,-iDv⟩\mathrm{.}$ Hence $u\mapsto ⟨Tu,v⟩$ is the restriction to $D(T)$ of the bounded functional $⟨\cdot ,-iDv⟩$ on $L^2$, so $v\in D(T^{\ast })$ and $T^{\ast }v=-iDv$ by [F2]. Conversely, if $v\in D(T^{\ast })$, testing the identity $⟨Tu,v⟩=⟨u,T^{\ast }v⟩$ against $u\in C_c^{\mathrm{\infty }}(0,L)\subseteq D(T)$ shows that $-iDv$ (as a distribution) equals the $L^2$ class $T^{\ast }v$; hence $Dv\in L^2$ and $v\in H^1(0,L)$. Thus $D(T^{\ast })=H^1(0,L)$ and $T^{\ast }=-iD$. Since $H^1(0,L)⊋D(T)$ (for instance the constant function $1$ lies in $H^1$ but $1^{\ast }(0)=1\mathrm{\ne }0$), $T$ is not self-adjoint. This is (a).

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

[All self-adjoint extensions] Fix $\theta \in [0,2\pi )$. By The twisted-periodic momentum operator on an interval, $P_{\theta }$ is self-adjoint with $D(P_{\theta })=\{u\in H^1:u^{\ast }(L)=e^{i\theta }{u}^{\ast }(0)\}$, and $P_{\theta }u=-iDu=T^{\ast }u$ for $u\in D(P_{\theta })$; since $D(T)\subseteq D(P_{\theta })$ ($0=e^{i\theta }\cdot 0$), $P_{\theta }$ is a self-adjoint extension of $T$. Conversely, let $S$ be a self-adjoint extension of $T$, so $T\subseteq S$ and $S\subset T^{\ast }$ (a self-adjoint operator has no proper symmetric extension and $T^{\ast }$ is the maximal domain on which the action is $-iD$ 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 $u,v\in D_1$, $0=⟨Su,v⟩-⟨u,Sv⟩=-i(u^{\ast }(L)\overline{v^{\ast }(L)}-u^{\ast }(0)\overline{v^{\ast }(0)})\mathrm{.}$ If every $u\in D_1$ had $u^{\ast }(0)=0$, then the identity would give $u^{\ast }(L)\overline{v^{\ast }(L)}=0$ for all $u,v\in D_1$, hence $u^{\ast }(L)=0$ 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 $v^{\ast }(0)\mathrm{\ne }0$; the identity for $u\in D_1$ and this fixed $v$ gives $u^{\ast }(L)=\lambda u^{\ast }(0)$ with $\lambda :=\overline{v^{\ast }(0)}\mathrm{/}\overline{v^{\ast }(L)}$, and testing the identity with $u=v$ gives $\mathrm{\mid }{v}^{\ast }(L)\mathrm{\mid }=\mathrm{\mid }{v}^{\ast }(0)\mathrm{\mid }$, so $\mathrm{\mid }\lambda \mathrm{\mid }=1$ and $\lambda =e^{i\theta }$ for some $\theta \in [0,2\pi )$. Hence $D_1\subseteq D(P_{\theta })$ and $S\subseteq P_{\theta }$. Both operators are self-adjoint, so by [F2] the inclusion cannot be proper: $S=P_{\theta }$. This is (c).

[Orthonormality and completeness] For $n,m,n^{\mathrm{\prime }},m^{\mathrm{\prime }}$, $⟨e_{nm},e_{n^{\mathrm{\prime }}{m}^{\mathrm{\prime }}}⟩=\left({\int }_0^L{e}_n\stackrel{‾}{e_{n^{\mathrm{\prime }}}}\right)\left({\int }_0^L{e}_m\stackrel{‾}{e_{m^{\mathrm{\prime }}}}\right)={\delta }_{n{n}^{\mathrm{\prime }}}{\delta }_{m{m}^{\mathrm{\prime }}}$ by [F1] and Fubini. If $g\in L^2((0,L{)}^2)$ is orthogonal to all $e_{nm}$, then for every $n$ the function $h_n(y):={\int }_0^{L}g(x,y)e_n(x)dx$ (defined as zero on exceptional sections) is measurable and satisfies $\mathrm{\mid }{h}_n(y){\mathrm{\mid }}^2\le {\int }_0^L\mathrm{\mid }g(x,y){\mathrm{\mid }}^{2}dx$ a.e. by Cauchy-Schwarz, hence $h_n\in L^2(0,L)$ by Tonelli; moreover $g(x,y)e_n(x)e_m(y)$ is absolutely integrable by Cauchy-Schwarz on the square, so ${\int }_0^L{h}_n\overline{e_m}=0$ for all $m$ by Fubini, so $h_n=0$ a.e. by completeness of $(e_m)$. Fubini also gives a full-measure set $Y_0$ on which $g(\cdot ,y)\in L^2(0,L)$. For each $n$, choose a full-measure set $Y_n$ on which $h_n(y)=0$; Countable Choice permits choosing these sets simultaneously, and countable subadditivity makes $Y_0\cap {\bigcap }_n{Y}_n$ full measure. For every $y$ in this intersection, all sine coefficients of $g(\cdot ,y)$ vanish, so completeness of $(e_n)$ gives $g(\cdot ,y)=0$ a.e. in $x$. Fubini then yields $g=0$ a.e. on the square. This is (a).

[Diagonal operator and eigenspace] By [F2] applied to the complete orthonormal family $(e_{nm})$ and the real numbers $({\mu }_{nm})$ (indexed by ${\mathbb{N}}^2$, which is countable), define $D({\mathrm{\Delta }}_D):=\{g\in H:{\sum }_{n,m\ge 1}{\mu }_{nm}^2\mathrm{\mid }⟨g,e_{nm}⟩{\mathrm{\mid }}^2<\mathrm{\infty }\}$ and ${\mathrm{\Delta }}_{D}g:={\sum }_{n,m\ge 1}{\mu }_{nm}⟨g,e_{nm}⟩e_{nm}$. To match the supplier's indexing by $\mathbb{N}=\{0,1,\dots \}$, enumerate $(n,m)$ by increasing $n+m\ge 2$, and within each sum by increasing $n$; this lists every positive pair once. The supplier proves convergence on this domain and self-adjointness, with ${\mathrm{\Delta }}_D{e}_{nm}={\mu }_{nm}{e}_{nm}$; a vector $x={\sum }_{n,m}{c}_{nm}{e}_{nm}\in D({\mathrm{\Delta }}_D)$ lies in $\ker ({\mathrm{\Delta }}_D-\mu )$ exactly when ${\sum }_{n,m}({\mu }_{nm}-\mu )c_{nm}{e}_{nm}=0$, which by uniqueness of coefficients forces $c_{nm}=0$ whenever ${\mu }_{nm}\mathrm{\ne }\mu$. The equation $n^2+m^2=5$ with $n,m\ge 1$ has exactly the solutions $(1,2)$ and $(2,1)$, so $\ker ({\mathrm{\Delta }}_D-\mu )=\mathbb{C}{e}_{12}\oplus \mathbb{C}{e}_{21}$ is two-dimensional. This proves (b) and the displayed eigenspace assertion in (c).

[One-dimensional comparison] The operator $A$ in the comparison has precisely the domain and action required by [F3]; the hypotheses $L>0$, $\alpha ,\beta \in \mathbb{R}$, $V\in C([0,L];\mathbb{R})$ and both choice assumptions are explicit. Thus its spectrum consists of eigenvalues with one-dimensional eigenspaces. By step 2.1, the square operator instead has a two-dimensional eigenspace at $5{\pi }^2\mathrm{/}{L}^2$. This establishes the stated contrast and completes (c).

[Normalisation] ${\int }_{\mathbb{R}}\mathrm{\mid }\psi {\mathrm{\mid }}^2=\frac{1}{2}{\int }_{\mathbb{R}}{e}^{-\mathrm{\mid }x\mathrm{\mid }}dx={\int }_0^{\mathrm{\infty }}{e}^{-x}dx=1$.

[Membership and eigenvalue] By [F2], $\psi$ is $H^1$ with $H^2$ half-line restrictions, and its derivative jump is ${\psi }^{\mathrm{\prime }\ast }(0+)-{\psi }^{\mathrm{\prime }\ast }(0-)=-{\psi }^{\ast }(0)$, so $\psi \in D(A_{-1})$ by [F1] with $\alpha =-1$. On each half-line, $A_{-1}\psi =-D^2\psi =-\frac{1}{4}\psi$, so $A_{-1}\psi =-\frac{1}{4}\psi$. This is the verification of (a) for the explicit function; uniqueness of the negative eigenvalue and of its eigenspace is part (b) of The unique negative eigenvalue of the attractive delta interface operator, specialised to $\alpha =-1$: $-{\alpha }^2\mathrm{/}4=-1\mathrm{/}4$ and $\sqrt{-\alpha \mathrm{/}2}=\sqrt{1\mathrm{/}2}$.

[The general and scaled statements] Part (b) is exactly The unique negative eigenvalue of the attractive delta interface operator (b), and part (c) is its (c): for $\alpha <0$, $E_H=-{\mathrm{\hslash }}^2{\alpha }^2\mathrm{/}(8m)=-m{g}^2\mathrm{/}(2{\mathrm{\hslash }}^2)$ with $g={\mathrm{\hslash }}^2\alpha \mathrm{/}(2m)$, and at $\alpha =-1$ this equals $-{\mathrm{\hslash }}^2\mathrm{/}(8m)$.