MADNESS 0.10.1
Hartree-Fock equations for the helium atom

The source is here.

Points of interest
  • application of the Coulomb and Helmholtz Green's functions
  • smoothing of the potential and initial guess
  • manual evaluation of the solution along a line
Background

The Hartree-Fock wave function is computed for the helium atom in three dimensions without using spherical symmetry.

The atomic orbital is an eigenfunction of the Fock operator

\begin{eqnarray*}
   \hat{F} \phi(r) &=& \epsilon \phi(r) \\
   \hat{F} &=& -\frac{1}{2} \nabla^2 - \frac{2}{r} + u(r) \\
   u(r) &=& \int \frac{\rho(s)}{| r - s |} d^3s \\
   \rho(r) &=& \phi(r)^2
\end{eqnarray*}

that depends upon the orbital via the Coulomb potential ( $ u(r) $) arising from the electronic density ( $ \rho(r) $).

Implementation

Per the usual MADNESS practice, the equation is rearranged into integral form

\[
    \phi = - 2 G_{\mu} * ( V \phi)
\]

where $ \mu = \sqrt{-2E} $ and $G_{\mu}$ is the Green's function for the Helmholtz equation

\[
    \left( - \nabla^2 + \mu^2  \right) G(r,r^{\prime}) = \delta(r,r^{\prime})
\]

The initial guess is $ \exp(-2r) $, which is normalized before use. Each iteration proceeds by

The kinetic energy operator is denoted by $T~=~-\frac{1}{2}
\nabla^2 $. Thus, one would expect to compute the kinetic energy with respect to a wave function $ \phi $ by $ <T \phi, \phi >
$. In this particular example, the wave functions goes to zero smoothly before the boundary so we apply the product rule for differentiation and the kinetic energy is equal to $ < \nabla \phi, \nabla \phi >
$.

The source is here.