Qm Math Coulomb — Examples

[Normalization] By polar coordinates and [F2], $\mathrm{\parallel }{\psi }_{100}{\mathrm{\parallel }}_2^2=N^{2}4\pi {\int }_0^{\mathrm{\infty }}{r}^2{e}^{-2\alpha r}dr=({\gamma }^3\mathrm{/}2)2\mathrm{/}{\gamma }^3=1$.

[Free maximal realization without arbitrary choice] For the real multiplier $t(\xi )=4{\pi }^2\mathrm{\mid }\xi {\mathrm{\mid }}^2$, truncate any $g\in L^2$ to $g_n={\mathbf{1}}_{\{t\le n\}}g$ to prove density of $D(M_t)$. It is symmetric because $t$ is real. If $g\in D(M_t^{\ast })$ with adjoint value $h$, testing against every $L^2$ function supported in $\{t\le n\}$ gives $h=tg$ there. These sets increase to all space, so $tg=h\in L^2$ and $D(M_t^{\ast })=D(M_t)$. Thus $M_t$ and its unitary conjugate $T$ are self-adjoint, with $D(T)=H^2$ by [F3]. For $z\mathrm{\notin }\mathbb{R}$, $R_0(z)={\mathcal{F}}_2^{\ast }{M}_{(t-z{)}^{-1}}{\mathcal{F}}_2$ is the actual two-sided inverse, since $t\mathrm{/}(t-z)$ is bounded and $(T-z)R_0(z)=I$, $R_0(z)(T-z)=I$ on $D(T)$.

[Hardy relative bound and Neumann realization] Put $Bu=-\gamma u\mathrm{/}r$ for $u\in H^2$. By [F3] and Fourier Cauchy–Schwarz, $\mathrm{\parallel }Bu{\mathrm{\parallel }}_2\le 2\gamma \mathrm{\parallel }\mathrm{\nabla }u{\mathrm{\parallel }}_2\le 2\gamma \mathrm{\parallel }Tu{\mathrm{\parallel }}_2^{1\mathrm{/}2}\mathrm{\parallel }u{\mathrm{\parallel }}_2^{1\mathrm{/}2}\le \epsilon \mathrm{\parallel }Tu{\mathrm{\parallel }}_2+{\gamma }^2{\epsilon }^{-1}\mathrm{\parallel }u{\mathrm{\parallel }}_2$ for every $\epsilon >0$. Choose $Y>16{\gamma }^2$ and $\epsilon =1\mathrm{/}4$. The exact free multipliers give $\mathrm{\parallel }T{R}_0(\pm iY)\mathrm{\parallel }\le 1$, $\mathrm{\parallel }{R}_0(\pm iY)\mathrm{\parallel }\le 1\mathrm{/}Y$, and hence $\mathrm{\parallel }B{R}_0(\pm iY)\mathrm{\parallel }<1\mathrm{/}2$. The operator series ${\sum }_{j\ge 0}(-B{R}_0(\pm iY){)}^j$ converges in norm, and $R_0(\pm iY)(I+B{R}_0(\pm iY){)}^{-1}$ maps into $D(T)$ and is the two-sided inverse of $C\mp iY$ for $C=T+B$ on $H^2$. This proves surjectivity of both nonreal signs with the exact factor order.

[Radial law and moments] By [F1], $p(r)=r^2{\int }_{S^2}\mathrm{\mid }{\psi }_{100}(r\mathrm{\Omega }){\mathrm{\mid }}^{2}d\sigma =4a_0^{-3}{r}^2{e}^{-2r\mathrm{/}{a}_0}$. By [F2], its integral is $4a_0^{-3}2\mathrm{/}(2\mathrm{/}{a}_0{)}^3=1$ and its first moment is $4a_0^{-3}6\mathrm{/}(2\mathrm{/}{a}_0{)}^4=3a_0\mathrm{/}2$.

[Exact adjoint domain] $C$ is densely defined and symmetric because both $T$ and the real multiplication $B$ are symmetric on $H^2$. Let $y\in D(C^{\ast })$, fix $z=iY$, and let $x=(C-z{)}^{-1}(C^{\ast }-z)y\in D(C)$. Then $(C^{\ast }-z)(y-x)=0$. Since $\mathrm{ran}(C-\bar{z})=L^2$, adjoint pairing shows $\ker (C^{\ast }-z)=\{0\}$; consequently $y=x\in D(C)$. Thus $D(C^{\ast })=D(C)=H^2$ and $C=C^{\ast }$, not merely a formally symmetric differential expression. All constructions are explicit sequences/truncations or single-vector existential arguments under CC.

[Weak $H^2$ domain of the actual exponential] On $r>0$ let $f=e^{-\alpha r}$ and ${\mathrm{\Omega }}_i=x_i\mathrm{/}r$. Its classical derivatives are ${\mathrm{\partial }}_{i}f=-\alpha f{\mathrm{\Omega }}_i$ and ${\mathrm{\partial }}_j{\mathrm{\partial }}_{i}f={\alpha }^{2}f{\mathrm{\Omega }}_i{\mathrm{\Omega }}_j-\alpha f({\delta }_{ij}-{\mathrm{\Omega }}_i{\mathrm{\Omega }}_j)\mathrm{/}r$. Each is in $L^2({\mathbb{R}}^3)$: the first derivatives are bounded times an exponential and second derivatives are bounded by $2{\alpha }^{2}f+2\alpha f\mathrm{/}r$, with ${\int }_0^1{r}^2\mathrm{/}{r}^{2}dr<\mathrm{\infty }$ and exponential tails. These are also the weak derivatives on all space. Indeed, integrate against a compactly supported smooth test function over $r>\delta$ and let $\delta \downarrow 0$; each inner-sphere boundary term has magnitude bounded by a constant times its area $4\pi {\delta }^2$, because $f$ and its first derivatives are bounded. It tends to zero, and the volume integrals converge since the displayed derivatives are locally integrable. Thus no origin-supported delta term is present and $f\in H^2$, even though it is not classically differentiable at zero. Summing the weak second derivatives gives $-\mathrm{\Delta }f=(-{\alpha }^2+2\alpha \mathrm{/}r)f$ almost everywhere; hence $Cf=-{\alpha }^{2}f$ because $\gamma =2\alpha$. Multiplication by $N$ preserves the domain and gives the claimed eigenvalue $-{\gamma }^2\mathrm{/}4$.

[The level is the ground level, locally] For a compactly supported smooth $v$, integrate $\mathrm{div}\mathrm{\Omega }=2\mathrm{/}r$ on punctured balls; the inner flux of $\mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{\Omega }$ is $O({\delta }^2)$ and vanishes. Expanding the square gives $\mathrm{\parallel }\mathrm{\nabla }v+\alpha \mathrm{\Omega }v{\mathrm{\parallel }}_2^2=\mathrm{\parallel }\mathrm{\nabla }v{\mathrm{\parallel }}_2^2+{\alpha }^2\mathrm{\parallel }v{\mathrm{\parallel }}_2^2-2\alpha \int \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}r$. By $H^1$ smooth density and the Hardy inequality, all these terms extend continuously to $H^1$: for example $\int \mathrm{\mid }v{\mathrm{\mid }}^2\mathrm{/}r\le \mathrm{\parallel }v\mathrm{/}r{\mathrm{\parallel }}_2\mathrm{\parallel }v{\mathrm{\parallel }}_2$. For $v\in D(C)=H^2$, Fourier pairing then gives $⟨Cv,v⟩+{\alpha }^2\mathrm{\parallel }v{\mathrm{\parallel }}_2^2=\mathrm{\parallel }\mathrm{\nabla }v+\alpha \mathrm{\Omega }v{\mathrm{\parallel }}_2^2\ge 0$. Thus $C\ge -{\alpha }^2$. For any real $\lambda <-{\alpha }^2$, Cauchy–Schwarz gives $\mathrm{\parallel }(C-\lambda )v{\mathrm{\parallel }}_2\ge (-{\alpha }^2-\lambda )\mathrm{\parallel }v{\mathrm{\parallel }}_2$; self-adjointness makes its range closed, and the orthogonal complement of the range is $\ker (C-\lambda )=\{0\}$, so its range is all $L^2$ and its inverse is bounded. Such $\lambda$ is outside the spectrum. Step 4.1 supplies an eigenvector at the lower bound, so this is a ground vector. No negative-spectrum theorem or AC spectral calculus is needed.

[Degree-zero Laguerre identity without Kummer dependence] The degree-zero generalized Laguerre polynomial is $L_0^{(a)}(s)=1$: equivalently the degree-zero normalized solution of $s{v}^{\mathrm{\prime }\mathrm{\prime }}+(a+1-s)v^{\mathrm{\prime }}=0$ is $v\equiv 1$. In the reduced $l=0$ equation $-u^{\mathrm{\prime }\mathrm{\prime }}-\gamma u\mathrm{/}r=-{\alpha }^{2}u$, substitution $u=r{e}^{-\alpha r}v(2\alpha r)$ produces $s{v}^{\mathrm{\prime }\mathrm{\prime }}+(2-s)v^{\mathrm{\prime }}=0$ since $\gamma =2\alpha$. Its degree-zero normalized solution $L_0^{(1)}=1$ therefore gives exactly $u=r{e}^{-\alpha r}$. Normalization supplies $u_{10}=2a_0^{-3\mathrm{/}2}r{e}^{-r\mathrm{/}{a}_0}$ and ${\psi }_{100}=u_{10}{Y}_{00}\mathrm{/}r$ with the same constant as step 1.1. This is the claimed special polynomial-exponential case; it does not invoke a general radial completeness theorem.

[Conclusion] Steps 1.1, 4.1, 2.2 and 5.2 prove (a)–(d), and steps 1.2, 2.1 and 3.1 prove its exact self-adjoint realization, step 4.1 proves the weak ground-vector domain, and step 5.1 proves the ground lower bound under CC alone.

[The vector and its coefficients] Since $g$ is compactly supported in $(0,\mathrm{\infty })$ and $Y_{00}$ is bounded, $\psi$ is a finite-block element in $L^2$; by [F1] its polar transform is supported in the $(0,0)$-block with entry $g$, so $\mathrm{\parallel }\psi {\mathrm{\parallel }}_2=\mathrm{\parallel }g{\mathrm{\parallel }}_2<\mathrm{\infty }$ and $\psi \mathrm{\ne }0$. This is (a).

[Positive continuous-spectrum component] The function $h(E):={\int }_0^{\mathrm{\infty }}{\phi }_0(\sqrt{E},r)g(r)dr$ is continuous by [F4] and satisfies $h(1)=1$ by the choice of $g$; hence there is $\delta >0$ with $\mathrm{\mid }h(E)\mathrm{\mid }\ge 1\mathrm{/}2$ on $(1-\delta ,1+\delta )$. By [F2] and the block identity [F3], $⟨E_C((0,\mathrm{\infty }))\psi ,\psi ⟩=⟨E_{A_0}((0,\mathrm{\infty }))g,g⟩={\int }_0^{\mathrm{\infty }}{\rho }_0(E)\mathrm{\mid }h(E){\mathrm{\mid }}^{2}dE\ge {\int }_{1-\delta }^{1+\delta }{\rho }_0(E)\mathrm{\mid }h(E){\mathrm{\mid }}^{2}dE\ge \frac{1}{4}{\int }_{1-\delta }^{1+\delta }{\rho }_0(E)dE>0$, because ${\rho }_0>0$ and the interval has positive length. This is (b).

[The bound states do not span $L^2$] The negative spectral subspace $E_C((-\mathrm{\infty },0))L^2$ is exactly the closed span of the bound states by [F3]; since $(0,\mathrm{\infty })$ and $(-\mathrm{\infty },0)$ are disjoint Borel sets, the spectral projections are orthogonal, so the component $E_C((0,\mathrm{\infty }))\psi \mathrm{\ne }0$ of step 2.1 is orthogonal to the whole negative spectral subspace. Hence $\psi$ is not in the closed span of the bound states, which is therefore a proper subspace of $L^2({\mathbb{R}}^3;\mathbb{C})$. This is (c).

[Conclusion] Parts (a), (b), (c) are steps 1.1, 2.1 and 3.1.

[The product vector is not an eigenvector] By [F2], $(T_R\otimes I)\mathrm{\Psi }=(T_R\phi )\otimes u_1$ and $(I\otimes C_{\mu })\mathrm{\Psi }=\phi \otimes (C_{\mu }{u}_1)=\phi \otimes E_1{u}_1=E_1\mathrm{\Psi }$ by [F1]; hence $(T_R\otimes I+I\otimes C_{\mu })\mathrm{\Psi }=E_1\mathrm{\Psi }+(T_R\phi )\otimes u_1$, and the difference from $E_1\mathrm{\Psi }$ is $(T_R\phi )\otimes u_1$. By [F3] $T_R\phi \mathrm{\ne }0$ and $u_1\mathrm{\ne }0$, so this difference is nonzero: $\mathrm{\Psi }$ is not an eigenvector at $E_1$. The full no-eigenvalue result Two-body reduced mass (d) also rules out every other energy, so $\mathrm{\Psi }$ is not an eigenvector at any energy. This is (a).

[Energy strictly above the threshold] Taking pairings with $\mathrm{\Psi }$ and using [F2] and [F1], $⟨(T_R\otimes I+I\otimes C_{\mu })\mathrm{\Psi },\mathrm{\Psi }⟩=\mathrm{\parallel }{u}_1{\mathrm{\parallel }}^2⟨T_R\phi ,\phi ⟩+\mathrm{\parallel }{u}_1{\mathrm{\parallel }}^2{E}_1\mathrm{\parallel }\phi {\mathrm{\parallel }}^2=\mathrm{\parallel }{u}_1{\mathrm{\parallel }}^2(E_1\mathrm{\parallel }\phi {\mathrm{\parallel }}^2+⟨T_R\phi ,\phi ⟩)$, and $⟨T_R\phi ,\phi ⟩=\frac{4{\pi }^2{\mathrm{\hslash }}^2}{2M}\int \mathrm{\mid }\xi {\mathrm{\mid }}^2\mathrm{\mid }\widehat{\phi }(\xi ){\mathrm{\mid }}^{2}d\xi >0$ by [F3] applied to the nonzero function $\phi$ (the integrand is nonnegative and not a.e. zero). Hence the energy of $\mathrm{\Psi }$ is strictly larger than $E_1\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}^2$, and since $\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}^2=\mathrm{\parallel }\phi {\mathrm{\parallel }}^2\mathrm{\parallel }{u}_1{\mathrm{\parallel }}^2$, $\mathrm{\Psi }\mathrm{/}\mathrm{\parallel }\mathrm{\Psi }\mathrm{\parallel }$ is a unit vector with energy $E_1+⟨T_R\phi ,\phi ⟩\mathrm{/}\mathrm{\parallel }\phi {\mathrm{\parallel }}^2>E_1$, this vector lies strictly above the bottom $E_1$ of the spectrum; by the no-eigenvalue theorem of Two-body reduced mass (d) no nonzero vector at all attains $E_1$. This is (b).

[Relative eigenvalue versus total spectrum] The relative vector $u_1$ is an eigenvector of $C_{\mu }$ at $E_1$ by [F1]; but by the actual no-eigenvalue theorem of Two-body reduced mass (d), no nonzero vector at all satisfies the two-body eigenvalue equation, and $E_1$ is the bottom of the continuous spectrum of the total operator. This is (c).

[Conclusion] Parts (a), (b), (c) are steps 1.1, 2.1 and 3.1.