The one-centre Coulomb model: bound energies, n^2-degeneracy, completeness of the bound states in the negative spectral subspace and the continuum

[Eigenpairs and degeneracy] By [F3] clause (a), ${\psi }_{nl\bar{m}}\in D(C_{\gamma })$ and $C_{\gamma }{\psi }_{nl\bar{m}}=-\frac{{\gamma }^2}{4n^2}{\psi }_{nl\bar{m}}$; by [F5] the eigenpair correspondence gives ${\psi }_{nl\bar{m}}\in D(H_{\kappa })$ and $H_{\kappa }{\psi }_{nl\bar{m}}=c_0(-\frac{{\gamma }^2}{4n^2}){\psi }_{nl\bar{m}}$. With $c_0={\mathrm{\hslash }}^2\mathrm{/}2m$ and $\gamma =2m\kappa \mathrm{/}{\mathrm{\hslash }}^2$ one computes $c_0\frac{{\gamma }^2}{4}=\frac{{\mathrm{\hslash }}^2}{2m}\cdot \frac{4m^2{\kappa }^2}{4{\mathrm{\hslash }}^4}=\frac{m{\kappa }^2}{2{\mathrm{\hslash }}^2}$, so $H_{\kappa }{\psi }_{nl\bar{m}}=E_n{\psi }_{nl\bar{m}}$ with $E_n=-\frac{m{\kappa }^2}{2{\mathrm{\hslash }}^2{n}^2}$; the identity $E_n=-\frac{{\mathrm{\hslash }}^2}{2m{a}_{\kappa }^2{n}^2}$ follows from $a_{\kappa }={\mathrm{\hslash }}^2\mathrm{/}(m\kappa )$. Normalising preserves the eigenpair relation and orthogonality is inherited from [F3] clause (a); the eigenspace of $C_{\gamma }$ at $-{\gamma }^2\mathrm{/}(4n^2)$ is exactly the span of the ${\psi }_{nl\bar{m}}$ with the stated labels, of dimension $n^2$, and by [F5] the eigenspaces of $H_{\kappa }$ at $E_n$ are the same, so the ${\mathrm{\Psi }}_{nl\bar{m}}$ of fixed $n$ span exactly that eigenspace. This proves (a) and the eigenspace claim of (b).

[Completeness in the negative subspace] By [F3] clause (b) the family $({\mathrm{\Psi }}_{nl\bar{m}})$ is a complete orthonormal family of $E_{C_{\gamma }}((-\mathrm{\infty },0))L^2$; by [F5] the spectral projections satisfy $E_{H_{\kappa }}((-\mathrm{\infty },0))=E_{C_{\gamma }}(c_0^{-1}(-\mathrm{\infty },0))=E_{C_{\gamma }}((-\mathrm{\infty },0))$, so the same family is a complete orthonormal family of $E_{H_{\kappa }}((-\mathrm{\infty },0))L^2$. This is (c).

[No nonnegative eigenvalue] Suppose $H_{\kappa }\psi =\lambda \psi$ with $\psi \mathrm{\ne }0$ and $\lambda \ge 0$; by [F5] this is equivalent to $C_{\gamma }\psi =(\lambda \mathrm{/}{c}_0)\psi$ with $\lambda \mathrm{/}{c}_0\ge 0$, and $\psi \in D(C_{\gamma })$. Write $U\psi =(u_{l\bar{m}})$; by the block identity [F4] one has $U\psi \in D({⨁}_{(l,\bar{m})}{A}_l)$ and $(A_l{u}_{l\bar{m}}{)}_{l,\bar{m}}=U{C}_{\gamma }{U}^{\ast }U\psi =(\lambda \mathrm{/}{c}_0)U\psi$, so $A_l{u}_{l\bar{m}}=(\lambda \mathrm{/}{c}_0)u_{l\bar{m}}$ for every $(l,\bar{m})$. Since $U$ is unitary and $\psi \mathrm{\ne }0$, some component $u_{l_0{\bar{m}}_0}$ is nonzero, and $\lambda \mathrm{/}{c}_0\ge 0$ is an eigenvalue of $A_{l_0}$, contradicting [F2] (no eigenvalue $k^2>0$, and no eigenvalue $0$). Hence every eigenvalue of $H_{\kappa }$ is negative; by [F2] clauses (b) and (d) together with [F5] the negative eigenvalues of $H_{\kappa }$ are exactly the numbers $c_0(-\frac{{\gamma }^2}{4n^2})=-\frac{{\gamma }^2{c}_0}{4n^2}$, $n\ge 1$, with the eigenspaces of step 1.1. This completes (b).

[Non-exhaustion and the continuum] By [F3] clause (c) the nonnegative spectral subspace of $C_{\gamma }$ is nonzero, indeed infinite dimensional; by [F5] at $B=[0,+\mathrm{\infty })$, $E_{H_{\kappa }}([0,+\mathrm{\infty }))=E_{C_{\gamma }}([0,+\mathrm{\infty }))$, so the nonnegative spectral subspace of $H_{\kappa }$ is nonzero and infinite dimensional, and by step 1.2 the closed span of the bound states, namely the negative spectral subspace, is a proper subspace of $L^2({\mathbb{R}}^3;\mathbb{C})$; this is the non-completeness in (d). For a unit vector $\psi$ in the nonnegative subspace the outcome law obeys ${\mu }_{\psi }^{H_{\kappa }}(B)=\mathrm{\parallel }{E}_{H_{\kappa }}(B)\psi {\mathrm{\parallel }}^2$ by [F6], and for Borel $B\subseteq (-\mathrm{\infty },0)$ the multiplicativity in [F5] with $B\cap [0,+\mathrm{\infty })=\mathrm{\varnothing }$ gives $E_{H_{\kappa }}(B)\psi =E_{H_{\kappa }}(B)E_{H_{\kappa }}([0,+\mathrm{\infty }))\psi =E_{H_{\kappa }}(\mathrm{\varnothing })\psi =0$, so strictly negative energy outcomes have probability zero.

[Energy-law reading] For a unit $\psi$, [F6] gives the sigma-additive law ${\mu }_{\psi }^{H_{\kappa }}$ with ${\mu }_{\psi }^{H_{\kappa }}(B)=\mathrm{\parallel }{E}_{H_{\kappa }}(B)\psi {\mathrm{\parallel }}^2$; put $N:=\{E_n:n\ge 1\}$ and $D:=(-\mathrm{\infty },0)\setminus N$. By step 1.1 and [F6], each ${\mathrm{\Psi }}_{nl\bar{m}}$ lies in the range of $E_{H_{\kappa }}(\{E_n\})$; multiplicativity [F5] therefore gives $E_{H_{\kappa }}(D){\mathrm{\Psi }}_{nl\bar{m}}=0$. Step 1.2 says these vectors have dense linear span in the negative spectral subspace, so boundedness of the projection makes it vanish on that entire subspace. Since $D\subseteq (-\mathrm{\infty },0)$, [F5] also gives $E_{H_{\kappa }}(D)=E_{H_{\kappa }}(D)E_{H_{\kappa }}((-\mathrm{\infty },0))=0$. Thus the restricted law is carried by $N$, and [F6]'s atom rule gives ${\mu }_{\psi }^{H_{\kappa }}(\{E_n\})=\mathrm{\parallel }{E}_{H_{\kappa }}(\{E_n\})\psi {\mathrm{\parallel }}^2=w_n$, so additivity gives ${\sum }_n{w}_n={\mu }_{\psi }^{H_{\kappa }}((-\mathrm{\infty },0))=\mathrm{\parallel }{E}_{H_{\kappa }}((-\mathrm{\infty },0))\psi {\mathrm{\parallel }}^2\le 1$. For $\psi ={\mathrm{\Psi }}_{nl\bar{m}}$ step 1.1 gives ${\mu }_{\psi }^{H_{\kappa }}={\delta }_{E_n}$. This is (e).

[Assembly and scope] Steps 1.1 and 2.1 give (a) and (b), step 1.2 gives (c), step 2.2 gives (d) and step 2.3 gives (e); only the declared model, P1, P2, P4, P5 and the cited mathematics were used, so no measured line, transition rate, photon coupling or empirical result is asserted, and the absolute continuity of the positive spectral subspace is not claimed for the scaled model. This completes the proof with the recorded scope (f).