9. Hydrogen and central potentials
Position in the course: Lesson 9 of 10. Complete the preceding derivation and use the explained exercises to check understanding.
Prerequisites: Angular momentum and radial differential equations
1. Separate a central-force problem
For a nonrelativistic electron and nucleus, eliminate center-of-mass motion and use the reduced mass \(\mu=m_eM/(m_e+M)\). A central potential depends only on \(r\), so write \(\psi=R_{nl}(r)Y_{lm}(\theta,\phi)\). Set \(u=rR\) to remove the first radial derivative. The resulting one-dimensional-looking equation is
The centrifugal term penalizes small radius for nonzero \(l\). The domain is the half-line, with admissible behavior at the origin and normalizable behavior at infinity. These conditions distinguish this equation from a full-line box problem.
2. Recover the Coulomb ground state
For hydrogen take \(V=-e^2/(4\pi\epsilon_0r)\). Try \(\psi=Ne^{-r/a}\) with \(l=0\). Since \(\nabla^2e^{-r/a}=(a^{-2}-2/(ar))e^{-r/a}\), substituting into \(H\psi=E\psi\) yields
Choose \(a=4\pi\epsilon_0\hbar^2/(\mu e^2)\) to cancel the radial term. The constant is the energy \(E_1=-\hbar^2/(2\mu a^2)\). Normalization uses \(4\pi\int r^2e^{-2r/a}dr=\pi a^3\), so \(N=(\pi a^3)^{-1/2}\). The derivation reveals why the exponential length scale is fixed rather than freely chosen.
3. Radial density is not point density
For the normalized ground state the radial probability density is \(P(r)=4r^2e^{-2r/a}/a^3\). Differentiating \(\ln P=\ln4+2\ln r-2r/a-3\ln a\) gives \(2/r-2/a=0\), so the most probable radius is \(a\). The point density \(|\psi(r)|^2\) instead peaks at the origin. Integrating \(rP(r)\) gives \(\langle r\rangle=3a/2\), different again. A maximum, a mean and a point density answer different questions.
4. General spectrum and its scope
Polynomial termination of the normalizable Coulomb radial solutions gives \(E_n=E_1/n^2\), with \(n=1,2,\ldots\) and \(l=0,\ldots,n-1\). The radial node count is \(n-l-1\); angular structure comes from \(Y_{lm}\). For a nuclear charge \(Z\), radius scales as \(a/Z\) and energy as \(Z^2\). These exact results apply to a one-electron Coulomb model. Electron-electron repulsion removes the simple hydrogenic degeneracy in many-electron atoms; relativistic, spin and finite-nucleus effects also alter precision spectra.
5. Worked virial and isotope checks
For the ground state \(\langle1/r\rangle=1/a\), so \(\langle V\rangle=-e^2/(4\pi\epsilon_0a)=2E_1\). Therefore \(\langle T\rangle=E_1-\langle V\rangle=-E_1\). The Coulomb virial relation \(2\langle T\rangle=-\langle V\rangle\) holds. Increasing nuclear mass increases \(\mu\) slightly, contracts \(a\) and increases the magnitude of the binding energy. This reduced-mass correction is not the whole isotope shift in a real atom.
6. Exercises and reasoning
How many radial nodes has a \(3p\) orbital? \(n=3,l=1\), so one. Its angular nodal structure is additional and should not be counted as radial nodes.
Does \(P(0)=0\) imply zero density at the nucleus? No. \(P\) includes the volume of a spherical shell, which vanishes at zero radius. The \(1s\) point density there is finite. This distinction matters when interpreting contact properties and orbital plots.
Analytical teaching schematic, not simulation data.
Further conceptual check
For a \(2p\) state, the radial density starts with an additional power of radius because the regular radial function behaves as \(r^l\) near the origin. The centrifugal barrier and regularity therefore explain suppression of nonzero-\(l\) density at the nucleus. This does not imply that all excited states have zero contact density: excited \(s\) states have different behavior.
7. References and study connections
The derivations and toy arithmetic are original teaching constructions. No molecular simulation is reported here.
8. Related theory and practice
Molecular methods · Quantum Monte Carlo · Gaussian · VASP