Why H Forms Molecule H2, while He Does not form He2

Why does the Hydrogen atom H form a H2 molecule, while the Helium atom He does not form a He2 molecule?

Standard Quantum Mechanics StdQM offers a vague explanation of physics:

• The two electrons of H2 combine to form a bond.
• Out of the four electrons of He2, two electrons combine to form a bond and the other two electrons form an anti-bond, altogether a no-bond. 

StdQM computation (without explanation of physics) agrees with observation:  

• H2 exists as a molecule with minimal energy -1.17 Hartree at kernel distance 1.4 atomic units au, compared to the energy -1 of separated H atoms. The dissociation energy of H2 is thus 0.17 Hartree.
• He2 does not form a molecule since the energy of He2 for kernel distances smaller than 5 atomic units is not lower than that of separated He atoms.  

StdQM computations for He2 is viewed to be very difficult because of the 12 spatial dimensions involved for 4 electrons and so results are very scarce if any.  

Let us now show that RealQM as a new alternative to StdQM delivers (i) results in agreement with StdQM computation and (ii) explanation of physics more convincing than StdQM. 

RealQM computations with this code on a uniform 3d grid with mesh size 0.1 au gives the following energies E depending on kernel distance D normalised to zero for maximal distance D=4 au:

•  D = 4       E = 0.00
•  D = 3.2    E = -0.04
•  D = 2.4    E = 0.003
•  D = 1.6    E = 0.20

We see essentially no energy drop from 4 to 1.6 au giving the message that two He atoms do not form a chemical bond into He2. The total energy of two He atoms is -5.806 and so the drop of 0.04 for D=3.2 represents a relative drop less than 0.01 on a 0.1 au grid in accordance with second order convergence,

RealQM offers the following explanation of the physics of bond for H2 and no-bond for He2:

The secret of the bond for H2 is hidden in the electron configuration of RealQM as consisting of non-overlapping electron densities. As two H atoms approach the two electron densities meet at a free boundary which is a midway normal plane to the line between the kernels with equal non-zero density together with vanishing normal derivative on both sides of the plane. This means that electron density can accumulate between the kernels thus decreasing kernel potential energy by profiting from both kernels, without increase of kinetic energy, thus forming a bond. This explanation is different from that given by StdQM based on bonding orbitals, and is based on the new electron configuration of RealQM. 

You can follow the physics by running this code. Very educating.

With the explanation for H2 in mind we now turn to He and meet the question why apparently the same explanation does not work for He. We recall that the RealQM electron configuration of a He atom consists of two half-spherical distributions meeting at a plane through the kernel as free boundary. Let us now imagine two He atoms approaching with the free boundary planes normal to the line between the kernels, which thus two half-spherical electron distributions meeting at a free boundary between the kernels and the other two outside. Again there will be an accumulation of charge density between the kernels, which  will now create a jump in electron density between inner and our electrons, which will push the inter-atom free boundary outwards and so act to decrease the inter-kernel electron accumulation and disable bonding. Computation shows that this effect is real. 

In short: The new atom/molecule model of RealQM offers an explanation based on physics of both the bond of H2 and the no-bond of He2.

RealQM for He2 at kernel distance 2.4 au with energy 0.003 Hartree above two He atoms indicating no-bond. Notice shift of free boundary beyond kernel to reduce accumulation of charge between kernels allowing yellow curves to take on same value at inter-atom free boundary.