The Helium atom as two electron charge densities restricted to half-spaces meeting at plane through the kernel as free boundary under a Bernoulli condition = homogeneous Neumann  + continuity.  The basic “two-valuedness” in stdQM expressed by Pauli’s Exclusion Principle is here replaced by a the two-valuedness of the half-space subdivision, not asking for any ad hoc introduction of the notion of spin.

As first crucial test case beyond Hydrogen, we now consider realQM for Helium with N=2 seeking a minimiser among wave functions \psi =\psi_1+\psi_2 arguing that by symmetry we may assume that \psi_1 is supported by the half-space \Omega_1=\{x:x_1\ge 0\} and \psi_2 by \Omega_2=\{x:x_1\le 0\} with the plane \Gamma_1=\{x:x_1=0\} as free boundary. More precisely, we assume by symmetry that \psi_2(x_1,x_2,x_3)=\psi_1(-x_1,x_2,x_3) for x_1\le 0, which allows restriction of attention to \psi_1. As shown below, realQM passes this test successfully!

The ground state \Psi_1 then appears as the minimiser over H^1(\Omega_1) of the total energy

  • TE(\psi_1)=\frac{1}{2}\int_{\Omega_1}\vert\nabla \psi_1\vert^2dx+\int_{\Omega_1}W\psi_1^2dx+\int_{\Omega_1}V_2\psi_1^2dx

under the normalisation

  • \int_{\Omega_1}\psi_1^2(x)dx=1

where the electron potential V_2(x) for x\in\Omega_1 is given by

  • V_2(x)=\int_{\Omega_2}\frac{\psi_1^2(-y_1,y_2,y_3)}{2\vert x-y\vert}dy.

We observe that V_2 is the solution of the equation -\Delta V_2=2\pi f in \Re^3 with f(x)=\psi_1^2(-x_1,x_2,x_3) for x_1<0 and f(x)=0 else, which gives an alternative way of computing V_2 other then direct integration as above.

The condition of stationarity with respect to variations of \Psi_1 takes the form of the eigenvalue problem

  • H\Psi_1=E_1\Psi_1\quad\mbox{in }\Omega_1,\quad\frac{\partial\Psi_1}{\partial n}=0\quad\mbox{on } \Gamma_1,

for the Hamiltonian

  • H = -\frac{1}{2}\Delta+W+2V_2.

Computation with realQM in spherical coordinates with azimuthal symmetry on a mesh with 200 points in radial direction and 100 in polar angle (on my iPad), gives the value -2.904 (Hartree) for the ground state energy of Helium in good agreement the observed value and benchmark computations with stdQM:

  • Pekeris (1959): -2.903724376    (best stdQM)
  • Koki (2009): -2.9042 (in better agreement with observation).


Computing in cylindrical coordinates without angular variation around the cylinder axis, we obtain on a 40\times 40 mesh (observing in particular the homogeneous Neumann condition on the plane x_1=0 as the free boundary again showing good agreement with observed table value: