Coulomb essential spectrum

[Essential spectrum of the free operator] Fix $\lambda \ge 0$ and $\epsilon >0$ and put $S_{\epsilon }:=\{\xi \in {\mathbb{R}}^3:4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2\in (\lambda -\epsilon ,\lambda +\epsilon )\}$. If $\lambda >0$ choose ${\xi }_0$ with $4{\pi }^2\mathrm{\mid }{\xi }_0{\mathrm{\mid }}^2=\lambda$; continuity of $\xi \mapsto 4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$ gives a nonempty open ball around ${\xi }_0$ contained in $S_{\epsilon }$. If $\lambda =0$ take the ball $\{\mathrm{\mid }\xi \mathrm{\mid }<\sqrt{\epsilon }\mathrm{/}(2\pi )\}$. In either case $S_{\epsilon }$ contains a nonempty open ball, hence by [F6] contains infinitely many pairwise disjoint nonempty open balls and has infinite-dimensional $L^2$ space; by [F1] the projection $E_{H_0}((\lambda -\epsilon ,\lambda +\epsilon ))={\mathcal{F}}_2^{\ast }{M}_{{\mathbf{1}}_{S_{\epsilon }}}{\mathcal{F}}_2$ has range ${\mathcal{F}}_2^{\ast }{M}_{{\mathbf{1}}_{S_{\epsilon }}}{L}^2$, which is infinite dimensional. By [F4] applied to $H_0$ we get $\lambda \in {\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(H_0)$. Conversely, if $\lambda <0$ and $0<\epsilon <\mathrm{\mid }\lambda \mathrm{\mid }$, then $(\lambda -\epsilon ,\lambda +\epsilon )\cap [0,\mathrm{\infty })=\mathrm{\varnothing }$, so $S_{\epsilon }=\mathrm{\varnothing }$ and $E_{H_0}((\lambda -\epsilon ,\lambda +\epsilon ))=0$; since the projection vanishes on an open interval around $\lambda$, [F4] shows $\lambda \mathrm{\notin }\sigma (H_0)$. Hence ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(H_0)=[0,\mathrm{\infty })$.

[The resolvent difference is compact] Fix a nonreal $z$ and use the convention $R_A(z)=(z-A{)}^{-1}$ of Resolvent and spectrum of an unbounded operator. Both resolvents are bounded by [F2] and have range $D(C)=D(H_0)$. For $f\in L^2$ put $g=R_{H_0}(z)f$; then $(z-H_0)g=f$ and $(z-C)g=f+\gamma Mg$. Applying $R_C(z)$ gives $g=R_C(z)f+\gamma R_C(z)Mg$, hence $R_C(z)-R_{H_0}(z)=-\gamma R_C(z)M{R}_{H_0}(z)$. The factor $M{R}_{H_0}(z)$ is compact by [F2] (changing the sign of the free resolvent preserves compactness), and $R_C(z)$ is bounded; therefore this difference is compact by [F5].

[Weyl's theorem and the essential spectrum] By step 2.1 the resolvent difference of $C$ and $H_0$ is compact at every nonreal $z$, so [F3] gives ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(C)={\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(H_0)=[0,\mathrm{\infty })$ by step 1.1. The boundedness below of $C$ is [F2]. This is (a).

[Negative spectrum is discrete] The spectrum of the self-adjoint operator $C$ is the disjoint union of ${\sigma }_{\mathrm{d}}(C)$ and ${\sigma }_{\mathrm{e}\mathrm{s}\mathrm{s}}(C)=[0,\mathrm{\infty })$ by [F4]; hence every $\lambda \in \sigma (C)\cap (-\mathrm{\infty },0)$ lies in ${\sigma }_{\mathrm{d}}(C)$, which by definition means that $\lambda$ is an eigenvalue of $C$, is isolated in $\sigma (C)$ and has finite-dimensional eigenspace. Each such $\lambda$ admits a rational open interval $(p,q)$ containing $\lambda$ whose intersection with the spectrum is exactly $\{\lambda \}$. Fix an enumeration of $\{(p,q)\in {\mathbb{Q}}^2:p<q\}$ and assign to $\lambda$ its least eligible pair. Two different spectral points cannot receive the same pair, because that interval has exactly one spectral point. This is an injection into a countable set, so the negative eigenvalue set is at most countable; no arbitrary choices of overlapping right endpoints are used. This is (b).

[Completeness of the negative eigenspaces] Let $S$ be the closed linear span of the eigenspaces of $C$ for negative eigenvalues and let $P_{-}:=E((-\mathrm{\infty },0))$. Every eigenvector of $C$ with negative eigenvalue $\lambda$ satisfies $E(\{\lambda \})u=u$ by [F4], hence lies in $P_{-}{L}^2$ because $\{\lambda \}\subseteq (-\mathrm{\infty },0)$; therefore $S\subseteq P_{-}{L}^2$. For the reverse inclusion let $f\in P_{-}{L}^2$ be orthogonal to $S$ and put $\mu :=E_f$. Since $f=P_{-}f$ and $E_f(B)=\mathrm{\parallel }E(B)f{\mathrm{\parallel }}^2$, the measure $\mu$ is concentrated on $(-\mathrm{\infty },0)$, i.e. $\mu (\mathbb{R})=\mu ((-\mathrm{\infty },0))$, and by step 4.1 the set $\{{\lambda }_1,{\lambda }_2,\dots \}=\sigma (C)\cap (-\mathrm{\infty },0)$ is at most countable and consists of eigenvalues. By [F4] the support of $\mu$ is contained in $\sigma (C)$: if $B\cap \sigma (C)=\mathrm{\varnothing }$ then $E(B)=0$. Hence $\mu (\mathbb{R})=\mu (\{{\lambda }_n\})={\sum }_n\mu (\{{\lambda }_n\})={\sum }_n\mathrm{\parallel }E(\{{\lambda }_n\})f{\mathrm{\parallel }}^2$, and each term vanishes because $E(\{{\lambda }_n\})f$ is an eigenvector with negative eigenvalue and $f\perp S$. Therefore $\mu =0$, so $\mathrm{\parallel }f{\mathrm{\parallel }}^2=0$ and $f=0$; thus $P_{-}{L}^2\subseteq S$ and (c) holds.