Variational Method - Helium Atom Ground State

Helium Atom Ground State

The helium atom consists of two electrons with mass m and electric charge −e, around an essentially fixed nucleus of mass M ≫ m and charge +2e. The Hamiltonian for it, neglecting the fine structure, is:

where ħ is the reduced Planck constant, ε₀ is the vacuum permittivity, r_i (for i = 1, 2) is the distance of the i-th electron from the nucleus, and |r₁ − r₂| is the distance between the two electrons.

If the term V_ee = e2/(4πε₀|r₁ − r₂|), representing the repulsion between the two electrons, were excluded, the Hamiltonian would become the sum of two hydrogen-like atom Hamiltonians with nuclear charge +2e. The ground state energy would then be 8E₁ = −109 eV, where E₁ is the Rydberg constant, and its ground state wavefunction would be the product of two wavefunctions for the ground state of hydrogen-like atoms:

where a₀ is the Bohr radius and Z = 2, helium's nuclear charge. The expectation value of the total Hamiltonian H (including the term V_ee) in the state described by ψ₀ will be an upper bound for its ground state energy. <V_ee> is −5E₁/2 = 34 eV, so is 8E₁ − 5E₁/2 = −75 eV.

A tighter upper bound can be found by using a better trial wavefunction with 'tunable' parameters. Each electron can be thought to see the nuclear charge partially "shielded" by the other electron, so we can use a trial wavefunction equal with an "effective" nuclear charge Z < 2: The expectation value of H in this state is:

This is minimal for Z = 27/16; Shielding reduces the effective charge to ~1.69. Substituting this value of Z into the expression for H yields 729E₁/128 = −77.5 eV, within 2% of the experimental value, −78.975 eV.

Even closer estimations of this energy have been found using more complicated trial wave functions with more parameters. This is done in physical chemistry via Variational Monte Carlo