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The helium atom

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 ${\bf x}_x$ and ${\bf x}_2$, wich are combined position and spin coordinates ${\bf x}_i = \{{\bf r}_i,s_i\}$, where the spin can assume two values $s_i = \uparrow,\downarrow$. The Born-Oppenheimer Hamiltonian for the helium atom reads:

\begin{displaymath}
H=-\frac{1}{2}\nabla^2_{r_1} - \frac{1}{2}\nabla^2_{r_2} - \frac{Z}{r_1} - \frac{Z}{r_2} + \frac{1}{r_{12}}.
\end{displaymath} (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:

\begin{displaymath}
\Psi({\bf x}_1,{\bf x}_2) = \phi({\bf r}_1)\phi({\bf r}_2) \...
...gle_2 - \vert\downarrow\rangle_1\vert\uparrow\rangle_2\right].
\end{displaymath} (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:

\begin{displaymath}
\left[-\frac{1}{2}\nabla^2_{r_1} - \frac{1}{2}\nabla^2_{r_2}...
...({\bf r}_1)\phi({\bf r}_2)
= E \phi({\bf r}_1)\phi({\bf r}_2).
\end{displaymath} (68)

We now multiply both sides from the left by $\phi({\bf r}^*_2)$ and we integrate over ${\bf r}_2$, arriving to:
\begin{displaymath}
\left[-\frac{1}{2}\nabla^2_{r_1} - \frac{Z}{r_1} +\int d^3r_...
...2-{\bf r}_2\vert} \right]\phi({\bf r}_1)
= E' \phi({\bf r}_1),
\end{displaymath} (69)

where several integrals we have removed the constant terms that do not deppend on ${\vec r}_1$ by absorving them into $E'$. The third term on the left hand side can be recognized as the Coulomb energy of particle 1 in the electric field generated by the charge density of particle 2.

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: $\phi$ is the solution to the Schrödinger equation, but the Hamiltonian deppends on $\phi$ itself. To solve this kind of problem, we start searching for a solution with some trial wave-function $\phi^{(0)}$, which is used for constructing the potential. SOlving the Schrödinger equation for this potential one finds a new ground state $\phi^{(1)}$, which is in turn used to build a new potential. The procedure is repeated until the ground state $\phi^{(i)}$ and the corresponding energy $E^{(i)}$ at step $i$ 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 $P({\bf r}_1,{\bf r}_2)$ for finding a particle at position ${\bf r}_1$ and another one at ${\bf r}_2$ is uncorrelated, i.e. can be written as the product of two one-electron probabilities:

\begin{displaymath}
P({\bf r}_1,{\bf r}_2)=P({\bf r}_1)P({\bf r}_2).
\end{displaymath} (70)

This does not mean that the electrons do not feel each other: in the depetrmination of the wave function, the term $1/\vert{\bf r}_1-{\bf r}_2\vert$ was taken into account. But this was done in an average way: it is not the actual position of particle 2 that determines the wave-function for electron 1, but the average charge distribution of electron 2. This is why this approach is usually referred-to as mean field.

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.


next up previous
Next: A program for the Up: The Hartree-Fock method Previous: The Born-Oppenheimer approximation
Adrian E. Feiguin 2009-11-04