It's been a while, but I finally got to the bottom of my problem with the helium atoms. You may remember that I had two isolated protons and I tried to solve the Schroedinger equation by sharing two electrons between them so each atom looked like a miniature helium. Now I know what I did wrong.
You can account for the energy of the system by adding up five terms. They are:
(1) the kinetic energy of electron A
(2) the kinetic energy of electron B
(3) the potential energy of electron A
(4) the potential energy of electron B
(5) the repulsion energy of electron A versus electron B
If you have a solution to the Shroedinger equation, and you make a new wave function where all these terms are exact multiples of your old solution, then the new wave function will also be a solution. That's what I was trying to do.
I took the helium atom solution and spread it out in space so it was twice as wide. Then I cloned it and put one replica at proton A and one replica at proton B. Looking at the five components of system energy, it appeared to me that each one was exactly one quarter of the original, giving me a valid solution. That was my mistake.
The kinetic energy of electron A is indeed one quarter of the original, and so is the kinetic energy of electron B. It works because the del-squared operator automatically gives you one-quarter the result when you double the linear dimension.
The potential energy of electron A is also one quarter of the original, as is the potential energy of electron B. It works because at each atom you have one-eighth the energy: half the nuclear charge, half the electron charge, and twice the distance. At first glance you might think there ought to be extra terms in the potential energy on account of the attraction of proton A for electron B and vice versa, but I can reduce these terms arbitrarily close to zero by putting the atoms far apart. No, the potential energy works out OK. It is the repulsion energy which is messed up.
The repuslion energy of the two electrons appears at first glance to work out exactly the same as the potential energy. At each atom you have half an electron repelling half an electron at twice the distance: one-eighth the energy. Double it for the second atom and you are back to one quarter, so everything seems proportional. But it isn't.
I am not a fan of the probability density interpretation of the wave function but in this instance I don't have a better explanation. The interpretation that works is not that you have half an electron repelling half an electron. It is that you have a 50% probability of a whole electron repelling a whole electron. This gives you twice the energy as what I calculated, so this term goes out of whack with the other four terms.
It has to work this way because otherwise, you could apply this technique in the opposite direction and solve the doubly ionized beryllium atom (Be++) as a squeezed-down replica of the helium atom. All the energy levels would be exactly four times as big. In fact you do just this when going from the hydrogen atom to the He+ ion. It works in that case because with only one electron there is no repulsion term. The isoelectronic series of hydrogen consists scaled copies of the identical wave function. But the isoelectronic series of helium doesn't work that way.
So I can't create mini-helium by sharing two electrons between two isolated protons. But that doesn't mean my problem doesn't have a solution. It just means that the wave function I chose does not minimize the energy of the system. There is a solution, and it is the one suggested by Spectracat: the hydrogen negative ion. It means you can take the wave function of H- and clone it so each proton gets a copy. Then you share the two electrons between the two protons. It looks strange but it's a solution of the Schroedinger equation. Each proton has two "half-electrons" bound to it. This solution comes in two versions: symmetric and antisymmetric. If you take sums and differences you get back the traditional solutions where two electrons are at A and a bare proton at B, and vice versa.