Hydrogen Atom as a Schrödinger Operator

The hydrogen Hamiltonian has both bound and scattering sectors. Its familiar discrete energy levels describe only the bound-state spectral subspace. To state the model precisely, one must also specify the operator domain and the origin condition inherited by the radial problem. Hydrogen Atom develops the separation-of-variables solution and wavefunctions; this page explains their operator-theoretic setting.

Required background. Coulomb Hamiltonian proves the full-space self-adjointness and lower bound. Spectral Types distinguishes eigenvalues from continuous spectrum.

The full-space Hamiltonian fixes the radial realization

Consider the spinless, nonrelativistic relative-coordinate Hamiltonian for two point particles,

$$H=-\alpha\Delta-\frac{\kappa}{r},\qquad \alpha=\frac{\hbar^2}{2\mu}>0,\qquad \kappa>0,$$

on $\mathcal H=L^2(\mathbb R^3,d^3x)$. Here $\mu$ is the reduced mass. The operator is self-adjoint on $H^2(\mathbb R^3)$, with form domain $H^1(\mathbb R^3)$. These statements refer to the ordinary Coulomb model, without an additional contact interaction at the origin.

Using orthonormal spherical harmonics, write

$$\psi(r,\Omega)=\sum_{\ell=0}^{\infty} \sum_{m=-\ell}^{\ell}\frac{u_{\ell m}(r)}{r}Y_{\ell m}(\Omega).$$

The norm identity

$$\|\psi\|^2=\sum_{\ell,m}\int_0^\infty|u_{\ell m}(r)|^2\,dr$$

defines a unitary map onto the Hilbert direct sum of radial $L^2(0,\infty;dr)$ spaces. The factor $r$ has absorbed the original volume measure. Under this map, $H$ decomposes into copies of

$$h_\ell=-\alpha\frac{d^2}{dr^2} +\frac{\alpha\ell(\ell+1)}{r^2}-\frac{\kappa}{r}.$$

This equality includes domains. Each $h_\ell$ is the Friedrichs realization of the radial expression, and the direct-sum operator domain additionally requires $\sum_{\ell,m}\|h_\ell u_{\ell m}\|^2<\infty$. Choosing unrelated radial extensions would change the full-space Hamiltonian.

Square integrability at the origin is not the complete condition

For solutions of the radial eigenvalue equation, the two leading behaviors for $\ell\geq1$ are $r^{\ell+1}$ and $r^{-\ell}$. The latter is not square-integrable in $dr$ near zero. This is the limit-point case: the singular endpoint supplies no free self-adjoint boundary parameter in these channels.

For $\ell=0$, both independent local solutions are square-integrable in $dr$. One vanishes linearly; the other has a nonzero constant leading term with Coulomb-induced $r\log r$ corrections. This is the limit-circle case, so square integrability alone does not choose a self-adjoint realization.

The full-space energy form selects the vanishing trace,

$$u_{00}(0)=0.$$

For radial eigenfunctions it gives the usual regular behavior $u_{00}(r)\sim Cr$. A nonzero limiting value would give $\psi\sim C/r$, which lies locally in $L^2(\mathbb R^3)$ but not in $H^1$. The other radial extensions describe modified short-distance physics; they are not extra boundary choices left unspecified by the full-space $H^2$ Coulomb operator.

The ordinary radial boundary discussion is developed in Boundary Conditions for Radial Wavefunctions. The extension-theoretic distinction explains why normalizability by itself can admit unphysical solutions for this model.

The bound-state basis spans a spectral subspace

The exact Coulomb spectral theorem gives

$$E_n=-\frac{\mu\kappa^2}{2\hbar^2n^2},\qquad n=1,2,\ldots,$$

with multiplicity $n^2$ before spin is included. These isolated eigenvalues accumulate only at the continuum threshold zero. The continuous spectrum is $[0,\infty)$ and is absolutely continuous; zero is not an eigenvalue, and there is no singular-continuous spectrum. These are properties of the specified Coulomb realization, not consequences of writing down the negative-energy solutions alone.

Let $P_n$ project onto the eigenspace at $E_n$. The spectral decomposition gives the strong operator identity

$$I=P_{\mathrm b}+P_{\mathrm{ac}},\qquad P_{\mathrm b}=\sum_{n=1}^{\infty}P_n.$$

Thus the hydrogen orbitals form an orthonormal basis of $P_{\mathrm b}\mathcal H$, not of all $L^2(\mathbb R^3)$. A normalized state orthogonal to all bound eigenvectors is possible: it has unit spectral weight in the scattering sector.

Time evolution preserves these sectors. A generic packet is therefore a sum of bound-state evolution and an energy integral over the continuum. The generalized functions in the latter require a spectral normalization; they cannot be added to the bound basis as normalized Hilbert vectors. See Coulomb Continuum States for the physical continuum solutions.

Exercise: a normalizable radial function with the wrong origin behavior

Let $u(r)=Ce^{-r/a}$, with $a>0$ and $C$ chosen so $\|u\|_{L^2(dr)}=1$. Is $\psi(r,\Omega)=u(r)Y_{00}(\Omega)/r$ in the standard hydrogen form domain?

Solution

It is a normalized Hilbert vector by the radial norm identity. Near zero, however, $\psi$ behaves as $C Y_{00}/r$, so its radial derivative behaves as $-C Y_{00}/r^2$. Its kinetic form integral contains $\int_0^\varepsilon r^2r^{-4}dr$, which diverges. Hence $\psi\notin H^1(\mathbb R^3)$ and cannot belong to $D(H)$ either. It remains a legitimate initial Hilbert vector for the unitary group, but it is not an admissible finite-form-energy radial eigenfunction of this realization.

References

• Hall, Brian C., Quantum Theory for Mathematicians, Graduate Texts in Mathematics 267, Springer, 2013, doi:10.1007/978-1-4614-7116-5 — angular decomposition and the hydrogen spectrum.
• Teschl, Gerald, Mathematical Methods in Quantum Mechanics, 2nd ed., Graduate Studies in Mathematics 157, American Mathematical Society, 2014, author’s text, Chapters 9–10 — singular radial operators and the Coulomb spectral problem.