Rank-one attractive perturbation has at most one negative eigenvalue

By [F2], $H_0$ is self-adjoint. If $E<0$, multiplication by $(\mathrm{\mid }p{\mathrm{\mid }}^2-E{)}^{-1}$ is a bounded inverse of $H_0-EI$: it is bounded by $\mathrm{\mid }E{\mathrm{\mid }}^{-1}$, and $\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}(\mathrm{\mid }p{\mathrm{\mid }}^2-E)\le 1$ puts its range in $D(H_0)$. Therefore $\sigma (H_0)\subseteq [0,\mathrm{\infty })$. For $B_n=\{p:\mathrm{\mid }p\mathrm{\mid }<1\mathrm{/}n\}$, [F3] gives $0<\mathrm{\mid }{B}_n\mathrm{\mid }<\mathrm{\infty }$; the unit vectors $f_n=\mathrm{\mid }{B}_n{\mathrm{\mid }}^{-1\mathrm{/}2}{\mathbf{1}}_{B_n}$ lie in $D(H_0)$ and satisfy $\mathrm{\parallel }{H}_0{f}_n\mathrm{\parallel }\le n^{-2}$. If $0\in \rho (H_0)$, then $1=\mathrm{\parallel }{f}_n\mathrm{\parallel }\le \mathrm{\parallel }{H}_0^{-1}\mathrm{\parallel }\mathrm{\parallel }{H}_0{f}_n\mathrm{\parallel }\to 0$, a contradiction. Thus $\inf \sigma (H_0)=0$.

By [F3], $\mathrm{\parallel }g{\mathrm{\parallel }}_2^2=4\pi \mathrm{\mid }C{\mathrm{\mid }}^2{\int }_0^{\mathrm{\infty }}{r}^2{e}^{-r^2}dr<\mathrm{\infty }\mathrm{.}$ For $E=-{\kappa }^2<0$, the rank-one scalar in [F1] is $m(\kappa )=⟨(H_0+{\kappa }^2{)}^{-1}g,g⟩={\int }_{{\mathbb{R}}^3}\frac{\mathrm{\mid }C{\mathrm{\mid }}^2{e}^{-\mathrm{\mid }p{\mathrm{\mid }}^2}}{\mathrm{\mid }p{\mathrm{\mid }}^2+{\kappa }^2}dp\mathrm{.}$ By [F1] and step 1.1, its zeros in $1+\lambda m(\kappa )$ correspond exactly to negative eigenvalues, each with multiplicity one.

If $0<{\kappa }_1<{\kappa }_2$, then $(\mathrm{\mid }p{\mathrm{\mid }}^2+{\kappa }_1^2{)}^{-1}>(\mathrm{\mid }p{\mathrm{\mid }}^2+{\kappa }_2^2{)}^{-1}$ everywhere; since $\mathrm{\mid }g(p){\mathrm{\mid }}^2>0$, integration gives $m({\kappa }_1)>m({\kappa }_2)$. On any compact interval $[a,b]\subset (0,\mathrm{\infty })$, the integrands are bounded by $\mathrm{\mid }g(p){\mathrm{\mid }}^2\mathrm{/}{a}^2\in L^1$, so dominated convergence [F4] gives continuity of $m$ there. Also $0<m(\kappa )\le \mathrm{\parallel }g{\mathrm{\parallel }}_2^2\mathrm{/}{\kappa }^2$, hence $m(\kappa )\to 0$ as $\kappa \to \mathrm{\infty }$.

The integrands for $m(1\mathrm{/}n)$ increase pointwise to $\mathrm{\mid }g(p){\mathrm{\mid }}^2\mathrm{/}\mathrm{\mid }p{\mathrm{\mid }}^2$ as $n\to \mathrm{\infty }$. Monotone convergence [F4] gives ${\lim }_{n}m(1\mathrm{/}n)=I_0$. Since step 3.1 makes $m$ strictly decreasing in $\kappa$, its limit as $\kappa \downarrow 0$ equals this supremum. By [F3], $I_0=4\pi \mathrm{\mid }C{\mathrm{\mid }}^2{\int }_0^{\mathrm{\infty }}{e}^{-r^2}dr\in (0,\mathrm{\infty })\mathrm{.}$

The function $1+\lambda m(\kappa )$ is continuous and strictly increasing on $(0,\mathrm{\infty })$ by step 3.1, with endpoint limits $1+\lambda I_0$ and $1$ by steps 3.1 and 4.1. When $1+\lambda I_0<0$, the endpoint limits give finite $0<a<b$ with $1+\lambda m(a)<0<1+\lambda m(b)$; [F5] on $[a,b]$ gives a positive zero, unique by strict increase. If $1+\lambda I_0\ge 0$, then $m(\kappa )<I_0$ for every $\kappa >0$ by step 3.1 and the endpoint limit in step 4.1; because $\lambda <0$, this gives $1+\lambda m(\kappa )>1+\lambda I_0\ge 0$, so no positive zero exists. Therefore a positive zero exists exactly when $1+\lambda I_0<0$. By step 2.1 and [F1], these zeros correspond exactly to negative eigenvalues.