in this section we introduce a first application of the Hartree-Fock methos for the helium atom. In order to carry out the calculation we shall use the electronic Hamiltonian within the Born-Oppenheimer approximation.
The helium atom has two electrons, and we will label their coordinates and , wich are combined position and spin coordinates
, where the spin can assume two values
.
The Born-Oppenheimer Hamiltonian for the helium atom reads:
(66) |
Since electrons are fermions, the wave function should be antisymmetric under an exchange of coordinates. We use the following antisymmetric trial wave function for the ground state:
(67) |
We shall now replace this into the Schrödinger equation. Since the Hamiltonian does not act on the spin degree of freedom, we are left with:
(68) |
We now have a single-particle Hamiltonian for the wave-function of a single electron. However, the Hamiltonian deppends on the wave-function we are looking for. This is a self-consistent problem: is the solution to the Schrödinger equation, but the Hamiltonian deppends on itself. To solve this kind of problem, we start searching for a solution with some trial wave-function , which is used for constructing the potential. SOlving the Schrödinger equation for this potential one finds a new ground state , which is in turn used to build a new potential. The procedure is repeated until the ground state and the corresponding energy at step do not deviate appreciably from those in the previous step.
The wave-function we have used is called uncorrelated because of the fact that the probability
for finding a particle at position and another one at is uncorrelated, i.e. can be written as the product of two one-electron probabilities:
(70) |
The Coulomb term in (70) is called Hartree potential. We will see that in more complex many-body problems as second term arises, due to the fermionic exchange statistics.