The unique negative eigenvalue of the attractive delta interface operator

[Form of a negative-energy eigenfunction] Let $E<0$, put $E=-{\kappa }^2$ with $\kappa >0$, and let $0\mathrm{\ne }u\in D(A_{\alpha })$ satisfy $A_{\alpha }u=Eu$. On $(0,\mathrm{\infty })$ the function satisfies $u^{\mathrm{\prime }\mathrm{\prime }}={\kappa }^{2}u$ by [F1]. Put $g=u^{\mathrm{\prime }}-\kappa u$; then $g^{\mathrm{\prime }}=-\kappa g$, so [F2] gives $(e^{\kappa x}g{)}^{\mathrm{\prime }}=0$ and $g=C{e}^{-\kappa x}$. Integrating $(e^{-\kappa x}u{)}^{\mathrm{\prime }}=C{e}^{-2\kappa x}$ shows $u(x)=c_{+}{e}^{\kappa x}+d_{+}{e}^{-\kappa x}$. Since $u\in L^2(0,\mathrm{\infty })$, necessarily $c_{+}=0$. The same calculation on $(-\mathrm{\infty },0)$ gives $u(x)=c_{-}{e}^{\kappa x}+d_{-}{e}^{-\kappa x}$ there, and square integrability forces $d_{-}=0$. Continuity of the global representative at $0$ gives $d_{+}=c_{-}=:a$, so $u(x)=a{e}^{-\kappa \mathrm{\mid }x\mathrm{\mid }}$. Since $u\mathrm{\ne }0$, $a\mathrm{\ne }0$.

[The eigenvalue at zero] Suppose $A_{\alpha }u=0$ with $u\mathrm{\ne }0$. By [F1], $u^{\mathrm{\prime }\mathrm{\prime }}=0$ classically on each half-line, so $u$ is affine on each half-line; an affine function in $L^2(0,\mathrm{\infty })$ or $L^2(-\mathrm{\infty },0)$ vanishes, so $u=0$, a contradiction. Hence $0$ is not an eigenvalue.

[Jump condition forces $\alpha <0$] For $u(x)=a{e}^{-\kappa \mathrm{\mid }x\mathrm{\mid }}$ the one-sided traces of the derivative are $u^{\mathrm{\prime }\ast }(0+)=-\kappa a$ and $u^{\mathrm{\prime }\ast }(0-)=+\kappa a$, so the interface condition of [F1] reads $-2\kappa a=\alpha a$, i.e. $\alpha =-2\kappa <0$ and $E=-{\kappa }^2=-{\alpha }^2\mathrm{/}4$. Hence a negative eigenvalue can exist only for $\alpha <0$, and then it equals $-{\alpha }^2\mathrm{/}4$; every such eigenfunction is a multiple of $x\mapsto e^{\alpha \mathrm{\mid }x\mathrm{\mid }\mathrm{/}2}$ (because $\kappa =-\alpha \mathrm{/}2$). This proves the nonexistence part of (a) and the uniqueness statement of (b) up to normalisation.

[Existence and normalisation] Let $\alpha <0$, $\kappa :=-\alpha \mathrm{/}2>0$, $E_{\alpha }=-{\kappa }^2=-{\alpha }^2\mathrm{/}4$, and ${\psi }_{\alpha }(x):=\sqrt{\kappa }{e}^{-\kappa \mathrm{\mid }x\mathrm{\mid }}$. Then $\mathrm{\parallel }{\psi }_{\alpha }{\mathrm{\parallel }}^2=\kappa {\int }_{\mathbb{R}}{e}^{-2\kappa \mathrm{\mid }x\mathrm{\mid }}dx=\kappa \cdot \frac{1}{\kappa }=1,$ using ${\int }_0^{\mathrm{\infty }}{e}^{-2\kappa x}dx=(2\kappa {)}^{-1}$ and symmetry. By [F2], ${\psi }_{\alpha }$ is continuous, $H^1$ on each half-line with derivative $-\kappa \mathrm{sgn}(x){\psi }_{\alpha }$ a.e., and $H^2$ on each half-line with $D^2{\psi }_{\alpha }={\kappa }^2{\psi }_{\alpha }$; its derivative jump is ${\psi }_{\alpha }^{\mathrm{\prime }\ast }(0+)-{\psi }_{\alpha }^{\mathrm{\prime }\ast }(0-)=-2\kappa {\psi }_{\alpha }(0)=\alpha {\psi }_{\alpha }(0)$, so ${\psi }_{\alpha }\in D(A_{\alpha })$ by [F1], and $A_{\alpha }{\psi }_{\alpha }=-D^2{\psi }_{\alpha }=-{\kappa }^2{\psi }_{\alpha }=E_{\alpha }{\psi }_{\alpha }$. Hence $E_{\alpha }$ is a negative eigenvalue for every $\alpha <0$, and $\ker (A_{\alpha }-E_{\alpha })=\mathbb{C}{\psi }_{\alpha }$ by step 2.1. This completes (a) and (b).

[Scaling] For any $\alpha \in \mathbb{R}$ and $\mathrm{\hslash },m>0$, let $H_{\alpha }:=({\mathrm{\hslash }}^2\mathrm{/}2m)A_{\alpha }$ with the same domain. Since the scalar is real and positive, the adjoint identity gives $H_{\alpha }^{\ast }=({\mathrm{\hslash }}^2\mathrm{/}2m)A_{\alpha }^{\ast }=H_{\alpha }$, so $H_{\alpha }$ is self-adjoint. If $\alpha <0$, part (b) gives $H_{\alpha }{\psi }_{\alpha }=({\mathrm{\hslash }}^2\mathrm{/}2m)E_{\alpha }{\psi }_{\alpha }=-{\mathrm{\hslash }}^2{\alpha }^2\mathrm{/}(8m){\psi }_{\alpha }$. With $g:={\mathrm{\hslash }}^2\alpha \mathrm{/}(2m)$, one has $\alpha =2mg\mathrm{/}{\mathrm{\hslash }}^2$, hence $-{\mathrm{\hslash }}^2{\alpha }^2\mathrm{/}(8m)=-m{g}^2\mathrm{/}(2{\mathrm{\hslash }}^2)$. No eigenvalue claim for $\alpha \ge 0$ is made in this scaling clause. This proves (c).