For a two-electron atom or ion of nuclear charge (fixed

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Atomic, Molecular & Optical Physics · Many-electron atoms and the periodic table

Deriving the Spectrum, Ch Self-consistent fields and correlation

Problem

For a two-electron atom or ion of nuclear charge ZZ (fixed, non-relativistic nucleus), a normalised single-exponential trial orbital ϕζ(r)∝e−ζr/a0\phi_\zeta(r)\propto e^{-\zeta r/a_0} used for both electrons gives the energy E(ζ)=ζ2−2Zζ+58ζ(in units of Eh, ζ dimensionless).E(\zeta) = \zeta^2 - 2Z\zeta + \tfrac58\zeta \qquad (\text{in units of } E_{\rm h},\ \zeta \text{ dimensionless}). For this one-parameter trial function the Hartree self-consistency condition (each electron sees the mean field produced by the other) is equivalent to the ordinary variational condition dE/dζ=0dE/d\zeta=0, because both orbitals are forced to be identical. For helium (Z=2Z=2) this iteration converges to ζ∗=1.6875\zeta^*=1.6875. ((a)) Apply the same self-consistency condition to Li+\mathrm{Li^+} (Z=3Z=3, two electrons). Find ζ∗\zeta^* and the corresponding ground-state energy in EhE_{\rm h} and in eV. ((b)) The exact non-relativistic ground-state energy of Li+\mathrm{Li^+} is −7.2799 Eh-7.2799\,E_{\rm h}. Find the missing energy in eV and compare its size with the analogous gap for helium, −77.49 eV-77.49\,\mathrm{eV} (single-exponential) against −79.01 eV-79.01\,\mathrm{eV} (exact).

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