Last year I posted some stuff about the two-electron potential well. It ought to be a pretty standard problem but it turns out you won't find it hardly anywhere. It's true that there is a standard textbook problem with two particles in a potential well, but they almost always stipulate that they are two "non-interacting" particles, so you simply get the product combinations of the standard single-electron solutions. I wanted to see what happens with actual electrons that repel each other. I was pretty surprised to find out that the shape of the solution depends on the size of the box. I delve more deeply into this when I take up the subject of the isoelectonic series of helium. But for the two-electron box, it turns out that for the very
small box, the solution tends towards the simple product function. For the very large box, the two electrons tend to stay away from each other at opposite ends of the box. And for the medium box, which is to say, for a box on the scale of typical atomic dimensions, it's a compromise, as shown for a typical case in the drawing below:
You construct this drawing (or I constructed it) by assuming A and B are each the familar sine wave solutions for the box, and just distort them a bit so they tend to keep away from each other. Then you symmetrize the wave function by reversing A and B, so that your final solution doesn't distinguish between the two electons. Because, of course, the electrons are supposed to be indistinguishable. It's basically what I show in the sketch below:
Like I said, I was pretty happy when I figure this out. What has happened recently is I've come to realize that this solution is wrong; or, at least, it's incomplete. You can tell it's wrong because the wave function for two electrons must be anti-symmetric. And if I reverse the roles of A and B in the function I've sketched above, I get exactly the same function back. It's symmetric, and that has to be wrong. It's easy to correct this and make it anti-symmetric. You just put a minus sign in the middle instead of a plus sign. But that's not right either. Yes, you've made the function anti-symmetric. But now the energy of the modified (antisymmetric) system turns out to be higher than that of the original symmetric case.
Why is the energy higher? Why indeed is the energy different at all? Remember how we constructed these cases. We took an electron in a box, and added a second electron. We then pushed the two electrons just slightly away from each other, to slightly reduce their coulombic interaction without distorting the wave function with too many "higher harmonics" with their inherently greater momentum content. The old compromise of kinetic and potential energy. It's a straightforward optimisation. Then, after we've optimised the energy, almost as an afterthough, we consider the symmetrization. Symmetrical or anti-symmetrical? The horrifying fact is,
regardless of which one we choose, the energy after symmetrization is
different from what we calculated when we optimized the individual electron wavefunctions. In fact, the symmetrical form (almost?) always has the lower energy in these situations. But from my point of view, the truly disturbing aspect is that we can't simply optimize first, symmetrize later: we have to do them both at the same time. It makes the calculation very difficult to visualize. And that's not even the reason I took up this question today. I said already that I came up with this solution last year and recently realized it was incomplete. And now I want to show what has to be done to fix it up. The problem is that I haven't taken into account the spin of the electrons. If I attempt to track the spin, and redraw my sketch of the function in terms of the sum of product functions, it ought to look like this:
You see I've got electron A with spin up, and B with spin down. But now let's consider the symmetry of what I've written. If you reverse the roles of A and B, you don't get the same function back. You don't get it back with the same sign, and you don't get it back with the opposite sign. You get a different function altogether. You now have to symmetrize
again. But this time I'm going to choose the antisymmetric option. I've written it out in full below:
That is what the solution has to look like for two particles in a box. It looks awfully complicated but there is really no simpler way to indicate it so far as I can tell. By the way, if you're familiar with the traditional way of writing two electrons in the "singlet" state vs the "triplet" state, you may be able to figure out that I've got them in a singlet state here. Which is good. But right now it all seems very complicated.
You construct this drawing (or I constructed it) by assuming A and B are each the familar sine wave solutions for the box, and just distort them a bit so they tend to keep away from each other. Then you symmetrize the wave function by reversing A and B, so that your final solution doesn't distinguish between the two electons. Because, of course, the electrons are supposed to be indistinguishable. It's basically what I show in the sketch below: 
Like I said, I was pretty happy when I figure this out. What has happened recently is I've come to realize that this solution is wrong; or, at least, it's incomplete. You can tell it's wrong because the wave function for two electrons must be anti-symmetric. And if I reverse the roles of A and B in the function I've sketched above, I get exactly the same function back. It's symmetric, and that has to be wrong. It's easy to correct this and make it anti-symmetric. You just put a minus sign in the middle instead of a plus sign. But that's not right either. Yes, you've made the function anti-symmetric. But now the energy of the modified (antisymmetric) system turns out to be higher than that of the original symmetric case. 
Why is the energy higher? Why indeed is the energy different at all? Remember how we constructed these cases. We took an electron in a box, and added a second electron. We then pushed the two electrons just slightly away from each other, to slightly reduce their coulombic interaction without distorting the wave function with too many "higher harmonics" with their inherently greater momentum content. The old compromise of kinetic and potential energy. It's a straightforward optimisation. Then, after we've optimised the energy, almost as an afterthough, we consider the symmetrization. Symmetrical or anti-symmetrical? The horrifying fact is, regardless of which one we choose, the energy after symmetrization is different from what we calculated when we optimized the individual electron wavefunctions. In fact, the symmetrical form (almost?) always has the lower energy in these situations. But from my point of view, the truly disturbing aspect is that we can't simply optimize first, symmetrize later: we have to do them both at the same time. It makes the calculation very difficult to visualize. And that's not even the reason I took up this question today. I said already that I came up with this solution last year and recently realized it was incomplete. And now I want to show what has to be done to fix it up. The problem is that I haven't taken into account the spin of the electrons. If I attempt to track the spin, and redraw my sketch of the function in terms of the sum of product functions, it ought to look like this: 
You see I've got electron A with spin up, and B with spin down. But now let's consider the symmetry of what I've written. If you reverse the roles of A and B, you don't get the same function back. You don't get it back with the same sign, and you don't get it back with the opposite sign. You get a different function altogether. You now have to symmetrize again. But this time I'm going to choose the antisymmetric option. I've written it out in full below: 
That is what the solution has to look like for two particles in a box. It looks awfully complicated but there is really no simpler way to indicate it so far as I can tell. By the way, if you're familiar with the traditional way of writing two electrons in the "singlet" state vs the "triplet" state, you may be able to figure out that I've got them in a singlet state here. Which is good. But right now it all seems very complicated. 
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