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, ri (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 Vee = 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 Vee) in the state described by ψ0 will be an upper bound for its ground state energy. <Vee> is −5E1/2 = 34 eV, so
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
Read more about this topic: Variational Method
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