Chapter 4 quoted Chandrasekhar's 1944 trial function for

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

Chapter 4 quoted Chandrasekhar's 1944 trial function for H−\mathrm{H}^-. Its spatial part is ψ(r1,r2)=a(r1)b(r2)+b(r1)a(r2)\psi(\mathbf r_1,\mathbf r_2)=a(r_1)b(r_2)+b(r_1)a(r_2), with a(r)=e−ζ1ra(r)=e^{-\zeta_1r} and b(r)=e−ζ2rb(r)=e^{-\zeta_2r} (rr in units of a0a_0), multiplied by the singlet spin state 12[α(1)β(2)−β(1)α(2)]\tfrac1{\sqrt2}[\alpha(1)\beta(2)-\beta(1)\alpha(2)]. With its best exponents, about 1.041.04 and 0.280.28, it gives −0.5133 Eh-0.5133\,E_{\rm h}. Hartree-Fock gives −0.4879 Eh-0.4879\,E_{\rm h} and the exact non-relativistic energy is −0.527751 Eh-0.527751\,E_{\rm h}. ((a)) Show that for ζ1=ζ2\zeta_1=\zeta_2 the state is a single Slater determinant with both electrons in one spatial orbital, and that for ζ1≠ζ2\zeta_1\ne\zeta_2 it is the sum of the two determinants built from the spin-orbitals aα,bβa\alpha,b\beta and from bα,aβb\alpha,a\beta. ((b)) Find the fraction of the correlation energy, measured from Hartree-Fock, that this function recovers, and the binding energy in eV it predicts against a hydrogen atom plus a free electron at rest, −0.5 Eh-0.5\,E_{\rm h}. ((c)) For any product f(r1)g(r2)f(r_1)g(r_2) the values at nucleus-centred distances obey ψ(0,R) ψ(R,0)=ψ(0,0) ψ(R,R)\psi(0,R)\,\psi(R,0)=\psi(0,0)\,\psi(R,R). Evaluate both sides for Chandrasekhar's ψ\psi, show that the left side is the larger whenever ζ1≠ζ2\zeta_1\ne\zeta_2, and say what this means for where one electron is found when the other is close to the nucleus.

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