# Helium

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 $\psi_1$.

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

• $H = -\frac{1}{2}\Delta+W+2V_2$

We compute stationary points by a gradient method, and the total energy as $2TE(\Psi_1)$. Computing in cylindrical coordinates without angular variation around the cylinder axis and a far-field condition $\Psi_1 =0$, 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 showing good agreement with observed table value):