Why doesn't Hydrogen have the same wave functions as Helium

  • Thread starter conway
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  • #26
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SpectraCat,

You've made some detailed comments on my last two posts, which I've read carefully. I'd like to respond to them individually but at this point in the discussion the "multiple quote/response" format starts to get unwieldy so I'll just try to go over the essentials. If there's something that I neglect to respond to please point it out.

What I did in my last posts was try to adapt my system of representation to include your case of the H- ion. The energy comes out wrong for my mini-heliums but I noticed that it was quite close for the first excited state. (Or the third excited state...whatever. BTW, the number -14.5 is based on the energy of -58 eV for the 1s2p state taken from the NIST spectral database, scaling down by a factor of 4. If I'd used the lowest excited state as you point out being 1s2s it's -59 eV, so it's still pretty close.) Anyhow, I needed a reason why the H- ion corresponds to the excited state of Helium and not the ground state, and I came up with an argument using parity whereby only the triplet states can be symmetrized in the way I need. The argument falls down not least because as you point out, the H- ion occurs in the singlet state, which I found by looking through on-line papers. I still don't know how you knew that.

Furthermore, if the symmetrization trick worked, I should also get excited states for the H- ion corresponding to all the Helium excited states, and it's been proven (again, as I found going through the papers) that no excited states exist for the negative ion.

On the point of when you're allowed to take sums and differences...I was technically wrong to argue that it works when you have two different electrons doing different things. Yes, both electrons must always be interchangeable; there is no such thing as this one being here and that one being there, once you've symmetrized the wave function. And yet there was a grain of truth to what I had put forward: the situations where you get the sums and differences are in fact when you have degenerate states, and these TEND to be cases where, for example the 1s2s Helium state, where we analyze from our "human" perspective that one electron fills the lowest orbital, and the second one goes to the next higher orbital; we indeed get a solution that satisfies the Schroedinger equation, but then almost as an afterthought we say "of course the electrons are indistinguishable" and we symmetrize the solution. Unlike the helium ground state, for example, where the two electrons are in the same state from the get-go.

In any case, I'm still baffled by a lot of questions. You had me 90% convinced that I should be able to get the correct solution for the hydrogen negative ion by solving the four-body problem for two electrons and two isolated protons, and then de-symmetrizing the solutions. It doesn't work, and I think the reason is that it's essentially a mis-application of the superposition principle. You can take the hydrogen negative ion and superpose the electron configuration with another one of two free electrons, but you can't "superpose" the ion with just a bare proton and get a neutral atom. It's just wrong, and it embarrasses me that I can't come up with a clear argument as to what exactly is wrong with it.

Is it still somehow wrong if you take a system of a bare proton and a negative ion, swith their locations, and then superpose the two systems to get two neutral atoms, each sharing two half-electrons? On the surface it's almost exactly the same thing which I just finished complaining about in the last paragraph...but this version is at least more convincing because the electrons are all accounted for. If you REVERSE this methodology...solve for the two neutral atoms, each sharing two "half-electrons", and then anti-symmetrize the system to get the familiar location basis...you get back the hydrogen ion plus the bare proton. It seems to me that you have been claiming that if I followed my program correctly, this is what I would end up doing, with all its attendant mathematical complications.

There is unfortunately quite a lot more I want to work out regarding these questions, but this is more than enough for one post.
 
  • #27
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This has been a very weird problem for me. I was so sure I had scaled everything correctly and come up with those mini-helium solutions to the Schroedinger equation. It seems I was wrong. Oh, I scaled correctly...up to a point. But I left something out, and I think I know what.

There is in fact a kind of mini-helium-type solution to the equation, but it's different. It is, despite my vehement protestations to the contrary, SpectraCat's negative hydrogen ion. Here is how you get it.

Since I am looking at two-electron solutions for the system (with the protons far apart), it's apparent that one such solution is to have a negative hydrogen ion at A and and a bare proton at B. But following my general principles, its ALWAYS more interesting to look at the symmetric solutions. You get these by reversing A and B, then taking sums and differences...you know the drill. So I end up with half an ion at A and half an ion at B. Notice, both "half-ions" consist of two half-electrons, so they are both neutral. Ions without a charge if you like. They are kind of like my mini-helium atoms except the electron configurations and of course the energies are different.

As an aside: yes, this is all standard QM so far. There's no need to nitpick over my use of the term "half-electron...clearly, I'm talking about a wave function which is localized at two different spots. Nothing illegal about that, just the way I choose my basis states.

But there's something very very disturbing about these little half-ions. The energy doesn't scale properly! They wave functions are exactly the wave function of the hydrogen negative ion, only half-filled. Let's add up the energy. There are five terms: two kinetic, two potential for each half electron, and one interaction term. When you go from the regular ion (plus the bare proton) to the symmetric pair of half ions, what happens to the energy terms?

The kinetic term for each location is given by the same wave-function, only half filled. So its half at A plus half at B which adds up. The potential term for each electron...same thing. But the interaction term is screwed up. At A, you have half an electron repelling half an electron over the identical geometry...it's one quarter the energy. The same at B. The interaction term gives you only half the energy.

This has been driving me crazy. It's the reason I originally thought that mini-helium was the correct solution for the equation: because all the energy terms scale correctly. But it turns out I made a mistake. No, there's nothing wrong with my scaling. It's more interesting than that, if I'm correct. I think I know how to make the energy add up.

I'll see if anyone else wants to bite on this before I post my solution.
 
  • #28
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I wasn't going to post my answer unless someone else seemed to be engaged in the problem, and no one else has posted in the last two weeks. Oddly enough, however, during that period the view counter for this thread has risen from 775 to 1003. It seems people are checking in but no one has commented. So I feel I ought to come clean: yes, I thought I had identified where the energy was, but I was wrong. I still can't make it add up.

I'm sure this surprising new development is of no concern to the majority of people who believe that the problem, whatever it is, exists only in my head. It's quite apparent to me that I did not succede in explaining to anyone the actual nature of my problem. And obviously there is no point in trying to carry on a discussion in these circumstances.

What I have done, in the meantime, is simplify the question to make it easier to deal with. Instead of hydrogen and helium wave function, I have gone back to the basics and re-formulated my problem in terms of the one-dimensional potential well. Two wells and two electrons, that is. In this form the problem becomes simple enough that I can actually draw out the wave functions. And the difficulty I had with the extra solutions, the quasi-heliums, doesn't disappear in the simplified version of the problem. It's still there.

If anyone is interested I will try to make some diagrams of the wave functions and post them.
 
  • #29
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What I have done, in the meantime, is simplify the question to make it easier to deal with. Instead of hydrogen and helium wave function, I have gone back to the basics and re-formulated my problem in terms of the one-dimensional potential well. Two wells and two electrons, that is. In this form the problem becomes simple enough that I can actually draw out the wave functions.
Perhaps this is where the trouble lies.
The wave function is a function on configuration space, not on real space.
*Only* for a single particle, does configuration space coincide with real space.
For two quantum particles, each moving in 1-dimension, the wave function is a function [tex]\psi(x_1,x_2)[/tex] and to draw it, you would need to draw it as a contour map in 2-D, or some fancy surface plot.

For non-interacting particles, the wavefunction will factorise in to [tex]\psi_1(x_1)\psi_2(x_2)[/tex] (perhaps with some symmetrisation).
For interacting particles it will not! That is why helium is hard, but hydrogen is easy.

If you want an example to study, the only nontrivial remotely solvable system I can think of would be to study two particles joined by a spring in a harmonic well.

I am interested, so you could post your diagrams if you want someone to have a look.
 
  • #30
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OK, thanks for the input. I'm going to start off by uploading a picture of a 2-electron well
2-electron well.jpg (haven't uploaded before so lets see if this works.) What I'm showing here is that the two electrons crowd each slightly towards opposite ends of the well. The story isn't that simple of course: I have to show how the function looks in a two-dimensional plot, and it has to be properly symmetrized with respect to swapping of the two particles. But let's try this first. Not I have to get back to watching American Idol...
 

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  • #31
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Well, my upload seemed to work. In my browser the little thumbnail view shows as much as I hoped to convey with the diagram. Now I've drawn what the function would look like if it was a simple product of electrons A and B. I hope the diagram is self-explanatory. You can see it can't be the correct wave function until it is symmetrized. That will be my next diagram.
 

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  • #32
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And here is the symmetrized wave function, as promised, for the one-dimensional well with two electrons. Of course I haven't done any actual mathematics but it isnt' hard to see that the ground state must be something along these lines. It's not a simple product function, but it's basically the sum of two product functions. You could of course also take the difference which would give you an antisymetric state, presumably with higher energy. Is it true that it has higher energy? I'm pretty sure it would but I haven't exactly thought about it yet.

I have to say I find the multi-dimensional wave functions hard to think about, even for the one-dimensional case. For the simple product function I originally posted, it is easy for me to see what the kinetic and potential energies are: electron B creates a potential which electron A must "live" in; and the kinetic energies are just given by the p-squared operator (second derivative) operating on the individual electrons. So I can work out the energy. Does the energy change when I symmetrize the function, giving the result shown in the attached image? I'm not quite sure yet.
 

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  • #33
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My early impressions of the energy problem is that the kinetic energies are correctly given by working them out in the simple product function representation; in other words, it doesn't change when you symmetrize them. But the potential energies are different. It's actually looking to me as thought the antisymmetric function might have a lower energy, because it pushes the electrons a little farther away from each other. (I'm neglecting the spin interaction here in case that matters...)

OK, this wasn't my original question, but can anyone explain which one, the symmetric or the antisymmetric case, looks more like the correct ground state for two electrons in a potential well?
 

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  • #34
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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.
 

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