Coulomb Hamiltonian

The three-dimensional attractive Coulomb potential is singular, but it does not force a new operator domain or an energy unbounded below. The standard Hamiltonian acts on $H^2(\mathbb R^3)$, and a completed-square identity gives its exact lower bound. The operator construction is proved in Domains of Hamiltonians; here we develop the quantitative model estimates and their dependence on the chosen realization.

Required background. Domains of Hamiltonians provides the Coulomb relative-bound estimate and Kato–Rellich theorem. Quadratic Forms explains how lower bounds extend from a form core.

The Coulomb singularity preserves the full-space domain

Let

$$H_\kappa=-\alpha\Delta-\frac{\kappa}{|x|},\qquad \alpha>0,\quad\kappa>0,$$

on $L^2(\mathbb R^3)$. The Hardy and interpolation estimates in the required background show that multiplication by $1/|x|$ has relative operator bound zero with respect to $-\Delta$. Therefore

$$D(H_\kappa)=H^2(\mathbb R^3),\qquad Q(H_\kappa)=H^1(\mathbb R^3).$$

The operator is self-adjoint, and $C_c^\infty(\mathbb R^3)$ is an operator core. No upper bound on this nonrelativistic coupling $\kappa$ is needed for these statements. The form is

$$q_\kappa[\psi]=\alpha\|\nabla\psi\|^2 -\kappa\int_{\mathbb R^3}\frac{|\psi(x)|^2}{|x|}\,d^3x.$$

These hypotheses are specific to the three-dimensional Laplacian and Coulomb singularity. A different kinetic operator or a more singular potential requires a new estimate.

A completed square gives the sharp lower bound

Put $\widehat r=x/|x|$ away from zero. For a smooth compactly supported function and $c>0$, integration by parts gives

$$\begin{aligned} \|\nabla\psi+c\widehat r\psi\|^2 &=\|\nabla\psi\|^2+c^2\|\psi\|^2 +c\int\widehat r\cdot\nabla|\psi|^2\,d^3x\\ &=\|\nabla\psi\|^2+c^2\|\psi\|^2 -2c\int\frac{|\psi|^2}{|x|}\,d^3x. \end{aligned}$$

Here $\nabla\cdot\widehat r=2/|x|$ as a distribution in three dimensions. There is no delta term: the flux through a small sphere is proportional to its area and tends to zero. The Coulomb form is continuous in the $H^1$ norm, so density extends the identity to every $\psi\in H^1$.

Choose $c=\kappa/(2\alpha)$. Then

$$q_\kappa[\psi]= \alpha\|\nabla\psi+c\widehat r\psi\|^2 -\frac{\kappa^2}{4\alpha}\|\psi\|^2.$$

It follows that $H_\kappa\geq-\kappa^2/(4\alpha)$. The bound is sharp because the square vanishes for $\psi(x)=Ce^{-c|x|}$, which belongs to the form and operator domains. The full sequence of hydrogen energies and its radial interpretation belong to Hydrogen Atom as a Schrödinger Operator.

The calculation proves more than a variational upper estimate: it gives a lower bound for every state in the form domain and exhibits a state attaining it.

Removing the origin changes the operator seed

The core statement uses smooth functions on all of $\mathbb R^3$, including those nonzero at the origin. Replacing that seed by $C_c^\infty(\mathbb R^3\setminus\{0\})$ gives a different symmetric operator.

The punctured seed is not essentially self-adjoint. In the s-wave sector the radial endpoint is limit-circle, and additional self-adjoint realizations are possible. The full-space $H^2$ domain selects the ordinary Coulomb realization; it does not leave a free contact parameter.

The punctured smooth functions are nevertheless a form core for the ordinary Coulomb form. A small radial cutoff removes the origin with an $H^1$ error tending to zero, and the Coulomb term is form continuous. Thus its Friedrichs extension recovers the standard Hamiltonian even though its operator closure does not. Operator-core and form-core statements must not be interchanged.

Finitely many centres remain controlled

For fixed points $a_j\in\mathbb R^3$ and finitely many real strengths $g_j$, consider

$$H=-\alpha\Delta+\sum_{j=1}^N\frac{g_j}{|x-a_j|}.$$

Translation invariance of the relative-bound estimate and finiteness of the sum give self-adjointness on $H^2$. Define $K=\sum_j\max\{-g_j,0\}$. If $K>0$, assign kinetic coefficients $\alpha_j=\alpha\max\{-g_j,0\}/K$ to the attractive terms. Applying the single-centre lower bound to each such form gives

$$H\geq-\sum_{g_j<0}\frac{g_j^2}{4\alpha_j} =-\frac{K^2}{4\alpha}.$$

Repulsive terms are nonnegative and can be dropped for this estimate. For $K=0$, $H\geq0$. The bound is uniform in the positions of finitely many fixed centres, including coincident centres. It is not a many-particle stability theorem with a constant independent of particle number or total charge.

Exercise: localization does not produce Coulomb collapse

For normalized smooth $\psi$, put $\psi_\lambda(x)=\lambda^{3/2}\psi(\lambda x)$, $\lambda>0$. Determine its energy and minimize over $\lambda$.

Solution

Writing $T=\alpha\|\nabla\psi\|^2>0$ and $C=\kappa\int|\psi|^2/|x|>0$, a change of variables gives $q_\kappa[\psi_\lambda]=\lambda^2T-\lambda C$. Its minimum is $-C^2/(4T)$ at $\lambda=C/(2T)$, and its energy grows to positive infinity as $\lambda\to\infty$. Kinetic localization costs one more power of $\lambda$ than the Coulomb attraction gains. The sharp form inequality above additionally bounds this minimum by $-\kappa^2/(4\alpha)$, uniformly over the initial shape.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — Schrödinger operators and the hydrogen ground state.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapters 6, 10, and 11 — Coulomb self-adjointness, radial realizations, and atomic Hamiltonians.