Sorting out the helium atom

I said in my last post that I was going to generalize the helium atom so I could calculate the ground state energy for the whole isoelectonic series. The first step is to work out the helium atom.

The key to working out the energy is pretty simple (in retrospect! it was hard for me to do it the first time). There are three pieces: the kinetic energy, the potential energy, and the interaction energy (which comes from the electron-electron repulsion.) The starting point will be what I called the "naive model", where we take the wave functions to be hydrogen like, we drop both electrons into the ground state, and add the interaction energy as an afterthought. Here is how it goes.

With just one electron, helium scales from the hydrogen atom by a factor of two on linear dimension, which gives a factor of four on energy. So where hydrogen is -1 Ry, He+ (singly ionized) is -4 Ry. This is true theoretically and experimentally; it's pretty much an exact result.

We can look inside this quantity of energy and break it down further into kinetic and potential. This is the same thing you can do with a planet orbiting the sun. The kinetic energy is positive and the potential energy is negative. There is something called the virial theorem which tells you to expect these quantities to be in the ration of 2:1; it works for planets, and without getting into the reasons why, suffice to say it works for atoms too. The kinetic energy of He+ is +4 Ry and the potential is -8 Ry. Total: -4 Ry.

Now we add the second electron and assume it goes into exactly the same energy state as the first. Total energy of the atom is now -8 Ry. But that's neglecting the mutual repulsion of the electrons. As I mentioned in my last post, that's an exact integral that comes to 2.5 Ry, giving a grand total of -5.5 Ry. Let's write it out:

(eqn. 1) -16 Ry potential + 8 Ry kinetic + 2.5 Ry interactive = -5.5 Ry

That's what I've called the "naive model". Now we're ready to have some fun.

The first thing to do is to optimize the wave function to try and minimize the total energy. The calculation above assumes that the wave function is just the same as the hydrogen wave function, scaled by a factor of two on linear dimension. In fact, because the two electrons are crowding each other, we might expect the wave function to spread out a bit to try and compensate. The wave function is essentially of the form exp(-r), and the easiest way to spread it out is to put a variable parameter k into the exponential decay. The nice thing is that with this simple change, the three terms in our energy equation are altered in a very simple way.

(The traditional explanation is a little different, as you can see on the Wikipedia page. They talk about mutual screening and "effective nuclear charge Z". This is confusing for a couple of reasons, mostly because Z is normally reserved for nuclear charge number and later I'm going to need to vary just that parameter. So I stick with what I'm calling a relaxation parameter, k, which just relaxes the shape of the wave function without any particular justification.)

The question is, how does my relaxation factor change the energy? Looking at it term by term, it's pretty simple. The kinetic energy is quadratic in k and the other terms are linear. Why? The kinetic is quadratic because in quantum mechanics you differentiate twice to get KE, and each differentiation gives a factor of k. The potential is linear because of the 1/r term in the equation for potential energy. Change the average distance by a factor of k, and the energy goes by the same factor. So we can write, without any particular difficulty, the revised energy equation as follows:

(eqn. 2) -16k (potential) + 8k^2 (kinetic) + 2.5k (interactive) = Energy

You have got to check out Wikipedia to see how much work they went through to get from (eqn. 1) to (eqn. 2). I did it with virtually no work at all by using a simple scaling argument. Once you're here, it's a very simple first-year calculus exercise to idenfiy the minium energy as occuring when k= 27/32. (On Wikipedia they get 27/16, but they define their optimization parameter a little differently.) You get Energy = 729/128 Ry, or just about 77.5 eV. It's about 1.5 eV short of the true energy.

This is where they basically leave off, letting you think that if you keep tinkering with the shape of the wave function, you'll get closer and closer to the true energy. You won't. You have to do something really radical to improve on this calculation much beyond the point where we've already taken it. You have to come up with a wave function that includes the coordinates of both electrons independently. But that's a story for another day.

What we're going to do next is take the optimized function for helium and generalize it to the rest of the isoelectronic series. We're going to need to introduce a parameter Z for nuclear charge. And we're going to think about how the energy terms in (eqn. 2) scale with Z.

It's going to be pretty easy, but it's just a little different than when we scaled to k. This time, the potential and kinetic terms are both going to be quadratic, and only the interaction will be linear. It works this way because a factor of Z on nuclear charge gives you a factor of 1/Z on linear dimension. For kinetic energy, this gives you Z-squared because of the double differentiation. For potential energy, you get a factor of Z for dimension and another factor of Z for the increased charge: hence, Z-squared. For the interaction however, although you get a factor of Z on dimension, the charges are the same two electrons you started out with, so there's no additional factor. It's just linear in Z.

So we take eqn. 2 and just put Z or Z-squared in front of the terms as appropriate:

(eqn. 3) -16k(Z^2) (potential) + 8(k^2)(Z^2) (kinetic) + 2.5kZ (interactive)

Except that we have to fix it up so that when Z=2, we get back (eqn. 2) unaltered. So we just divide the quadratic terms by 4 and the linear terms by 2:

(eqn. 4) -4k(Z^2) (potential) + 2(k^2)(Z^2) (kinetic) + 1.25kZ (interactive) = Energy

And that's our great result. We're going to look at the implication of this in our next post.