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

[conway]

We know that Helium can have basically the same solution as the Hydrogen atom if there is only one electron. You get the same equation with a different factor on the potential and the basic solution is the same as the Hydrogen atom. But I'm trying to figure out why it can't work the other way around.

If you have two hydrogen atoms side by side and one electron, then treating them as two potential wells you get the well-known pair of ground states, symmetric and antisymmetric, with the symmetric state just a little bit lower in energy than the antisymmetric. You can presumably fill this state with one electron, in which case each hydrogen atom should have half an electron.

Now add a second electron. Shouldn't each of the two hydrogen atoms now be able to incorporate half of the second electron just the way the helium atom does? I can't see how the equations should be any different. So the two hydrogen atoms would end up sharing two half-electrons each.

It's true that this solution is not the lowest energy state for the two hydrogen atoms. There is a lower energy state where each of them has exactly one whole electron, which is the familiar solution. (There is an even lower state where the two atoms come together and form a molecule, but in this situation we're sort of assuming for sake of argument that the atoms are kept apart.)

So my quasi-helium states are not the lowest energy state, but they still seem like they should be a valid state that satisfies the relevant equations. It doesn't seem like this happens in practise because the corresponding spectral lines are not present as far as I know. But I can't see why not.

 

[torquil]

Now add a second electron. Shouldn't each of the two hydrogen atoms now be able to incorporate half of the second electron just the way the helium atom does? I can't see how the equations should be any different. So the two hydrogen atoms would end up sharing two half-electrons each.

It's true that this solution is not the lowest energy state for the two hydrogen atoms. There is a lower energy state where each of them has exactly one whole electron, which is the familiar solution. (There is an even lower state where the two atoms come together and form a molecule, but in this situation we're sort of assuming for sake of argument that the atoms are kept apart.)

So my quasi-helium states are not the lowest energy state, but they still seem like they should be a valid state that satisfies the relevant equations. It doesn't seem like this happens in practise because the corresponding spectral lines are not present as far as I know. But I can't see why not.

Such a "mixed" state for two hydrogen atoms be pretty unlikely (high energy), since the nuclei are separated, I would think. If the two hydrogen atoms are far apart, the electron state needs to have an energy close to a free electron, in order to travel between the atoms with a significant probbility.

When you put the atoms closer to each other, then the energy of the mixed state becomes lower, and when the atoms are close enough together it will be responsible for the H_2 molecular bond, right?

Anyway, surely the equations describing 2 hydrogen atoms are very different from one helium atom, since only in one case can the positive charge be considered to be at one point.

Torquil

 

[conway]

Maybe I should back up a step. Forget about the helium atom for the time being; let's just establish that a single electron can be shared between two protons that are far apart.

When you write the wave function for a single hydrogen atom, you get a certain solution for the ground state. This is similar to the finite potential well. There is nothing to stop us from writing down two potential wells, separated by an arbitrary distance, and solving the equation. The solution looks exactly the same, only the shape of the wave function is duplicated in each location. You can put one electron into the wave function, and it is shared between the two potential wells. The energy is exactly the same as if you had only one potential well.

There is one twist to the situation: the wave function in the second well can be the same or it can have the opposite sign of the first well. These are the symmetric and antisymmetric solutions. The symmetric solution has slightly lower energy and hence lower frequency.

For a solution of the equation, you can put the electron in either the symmetric or the antisymmetric state. Or, you can put it in a superposition of the two. Because the states have slightly different frequency, the wave will slowly go in and out of phase with each other, causing the probability to shift back and forth from one well to the other.

Whenever the probability is all in one side, it looks like an ordinary hydrogen atom with a solitary proton some distance away. The energy of the proton is zero, and the energy of the hydrogen atom is -13.6 eV. This energy is a constant whether the electron is at one atom or the other, or whether it is shared between both.

So when the electron is shared, the energy of each atom is the same, and it is -6.8 eV.

 

[torquil]

Thanks, I understand a bit better now what you were saying.

I looked up the H2 molecule in my molecular quantum mechanics book (Atkins). He writes the following about both calculational methods that are discussed (VB=valence bond method & MO=molecular orbital method):

1) Both methods emphasize the importance of the displacement of charge into the internuclear region. The lowest energy wavefunction in each case is the one where contructive interference occurs in the bonding region.

2) The lowest energy is achieved with both electrons in the same symmetric spatial function and so the spin function has to be the antisymmetric singlet: electrons pair on bond formation.

So I guess this is something like what you were after?

Torquil

 

[SpectraCat]

We know that Helium can have basically the same solution as the Hydrogen atom if there is only one electron. You get the same equation with a different factor on the potential and the basic solution is the same as the Hydrogen atom. But I'm trying to figure out why it can't work the other way around.

I am really not clear what you mean by this

If you have two hydrogen atoms side by side and one electron, then treating them as two potential wells you get the well-known pair of ground states, symmetric and antisymmetric, with the symmetric state just a little bit lower in energy than the antisymmetric. You can presumably fill this state with one electron, in which case each hydrogen atom should have half an electron.

Ok .. I agree that there are two linear combinations of the atomic *orbitals* as you describe, and each linear combination can accommodate *two* electrons. I don't know what half an electron is however ... that is a nonsensical statement. I think you mean to say that the wavefunction, and therefore the probability density, is distributed equivalently between the two wells, giving an equal probability of finding the electron on either nucleus. You will never be able to measure half an electron

Now add a second electron. Shouldn't each of the two hydrogen atoms now be able to incorporate half of the second electron just the way the helium atom does? I can't see how the equations should be any different. So the two hydrogen atoms would end up sharing two half-electrons each.

Not sure what you are after here .. this situation is quite different that you describe. First of all, you are ignoring spin, which you cannot do. Are you talking about a triplet or a singlet state in your example? This matters because only eigenfunctions of the spin operator can qualify as valid wavefunctions for the system you describe.

Second if the distance between the two wells if sufficiently large that they don't interact (as I think you are trying to establish here), then the symmetric and anti-symmetric combinations of the 1s orbitals will be effectively degenerate.

It's true that this solution is not the lowest energy state for the two hydrogen atoms. There is a lower energy state where each of them has exactly one whole electron, which is the familiar solution. (There is an even lower state where the two atoms come together and form a molecule, but in this situation we're sort of assuming for sake of argument that the atoms are kept apart.)

Ok .. now I am lost ... what distinction are you trying to draw between two whole electrons and 4 half electrons distributed among the two wells ... the entire idea makes my head hurt.

So my quasi-helium states are not the lowest energy state, but they still seem like they should be a valid state that satisfies the relevant equations. It doesn't seem like this happens in practise because the corresponding spectral lines are not present as far as I know. But I can't see why not.

Ok, maybe I see what you are getting at. You are talking about the triplet state of two paired H-atoms at a large internuclear separation. The electrons are unpaired, with one in the symmetric and one in the anti-symmetric combination (if they were both in the same "orbital", their spins would have to be paired in the singlet state). This is actually the ground state configuration .. there is a small energy penalty for pairing the spins in the same orbital to form the singlet state.

I think you will find that you can recreate your "4 half-electron" combination simply by taking the inear combination of the two possible configurations of the two-electron state I described above ... i.e. the first configuration would have the "left" electron in the symmetric state and the "right" electron in the antisymmetric state, and they would be swapped in the second configuration.

 

[Bob S]

In the hydrogen molecule ion (2 protons plus 1 electron), the 2 protons are separated by 1.06 Angstroms (2 Bohr radii) and the dissociation (binding) energy is about 2.78 eV, according to Pauling and Wilson, Introduction to Quantum Mechanics, page 330. In the helium atom, the 2 protons are 100,000 times closer together.

Bob S

 

[SpectraCat]

In the hydrogen molecule ion (2 protons plus 1 electron), the 2 protons are separated by 1.06 Angstroms (2 Bohr radii) and the dissociation (binding) energy is about 2.78 eV, according to Pauling and Wilson, Introduction to Quantum Mechanics, page 330. In the helium atom, the 2 protons are 100,000 times closer together.

Bob S

Sure, but he explicitly said he wasn't considering bound molecular states, but rather the states of separated atoms. That said, I am not totally clear on the connection the the helium atom either ... I am waiting for clarification on that.

 

[conway]

I'm trying to solve the simplest possible problem with two electrons. I've chosen the problem of two isolated hydrogen atoms. Yes, we could solve them one at a time, which everyone remembers having done in 3rd year or whenever. But I'd like to see how it looks if you solve it as a two-electron problem. Because you should obviously get the same answer.

So you solve for the wave functions, and you get the symmetric and the antisymmetric solutions. Now it gets a little funny. You can put both electrons in the lowest state, but only if you make their spins opposite. That seems a little arbitrary if the atoms are far apart. But never mind.

What really gets me is if you take the six-dimensional wave function of the helium atom, spread it out in space by a factor of two on account of the unit charge of hydrogen vs the double charge of helium, and apply it to this problem, it appears to me that it should be a valid solution of the Schroedinger equation for two isolated hydrogen atoms. Each atom has the same 1/r potential, two half-electrons vying for the same territory with the same mutual interaction energy (scaled back by the same factor)...how is this not a correct solution? That's what I'm trying to figure out.

 

[Bob S]

What really gets me is if you take the six-dimensional wave function of the helium atom, spread it out in space by a factor of two on account of the unit charge of hydrogen vs the double charge of helium, and apply it to this problem, it appears to me that it should be a valid solution of the Schroedinger equation for two isolated hydrogen atoms. Each atom has the same 1/r potential, two half-electrons vying for the same territory with the same mutual interaction energy (scaled back by the same factor)...how is this not a correct solution? That's what I'm trying to figure out.

The two protons will not share the two electrons, or even one, unless they are so close that they will form a hydrogen molecule ion (with one electron), or a neutral molecule (with two electrons), because these two solutions are a lower energy state than two free hydrogen atoms sharing one or two electrons..

Bob S

 

[conway]

I don't think there's anything controversial about the two free hydrogen atoms sharing one electron. It seems like very standard potential-well stuff. It only really gets weird when I try to apply similar reasoning to the second electron.

 

[ytuab]

There are the H2 molecule model using the Bohr orbit.

See Bohr's 1913 molecular model revisited article (PNAS 2005).

It is generally believed that the old quantum theory, as presented by Niels Bohr in 1913, fails when applied to few electron systems, such as the H2 molecule. Here, we find previously undescribed solutions within the Bohr theory that describe the potential energy curve for the lowest singlet and triplet states of H2 about as well as the early wave mechanical treatment of Heitler and London. We also develop an interpolation scheme that substantially improves the agreement with the exact ground-state potential curve of H2 and provides a good description of more complicated molecules such as LiH, Li2, BeH, and He2.

In conclusion, we find a simple extension of the Bohr molecular model that gives a clear physical picture of how electrons create chemical bonding. At the same time, the description is surprisingly accurate, providing good potential energy curves for relatively complex many body systems..

But I think this molecule model is too simple and incomplete.
Now isn't 1910's - 1920's.
So this paper should have used more rigorous methods in calculating the Coulomb force and so on, and found better orbits, I think.

But as an approximation, this result shows a good tendency.

 

[SpectraCat]

I'm trying to solve the simplest possible problem with two electrons. I've chosen the problem of two isolated hydrogen atoms. Yes, we could solve them one at a time, which everyone remembers having done in 3rd year or whenever. But I'd like to see how it looks if you solve it as a two-electron problem. Because you should obviously get the same answer.

So you solve for the wave functions, and you get the symmetric and the antisymmetric solutions. Now it gets a little funny. You can put both electrons in the lowest state, but only if you make their spins opposite. That seems a little arbitrary if the atoms are far apart. But never mind.

It's not arbitrary, it's quantum mechanics. You have stipulated in your example that the two H-atoms are to be considered as a single system, with one wavefunction describing the electrons. That means that the symmetric (S) and anti-symmetric (A) combinations you describe are common orbitals and are governed by the Pauli exclusion principle. So you have to pair the spins if you want both electrons in the same state .. as I said, this costs energy, even at arbitrarily large distances, so the triplet state is lower in energy.

What really gets me is if you take the six-dimensional wave function of the helium atom, spread it out in space by a factor of two on account of the unit charge of hydrogen vs the double charge of helium, and apply it to this problem, it appears to me that it should be a valid solution of the Schroedinger equation for two isolated hydrogen atoms. Each atom has the same 1/r potential, two half-electrons vying for the same territory with the same mutual interaction energy (scaled back by the same factor)...how is this not a correct solution? That's what I'm trying to figure out.

Ok .. I really have no idea what you are talking about. Your intended reference system of the helium atom is not comparable to the case of two separated H-atoms in any meaningful way. There is a HUGE difference between an electron interacting with ONE nucleus of charge 2, and an electron interacting with TWO nuclei, each of charge one. The first is the +1 helium ion, a simple extension of the one-electron solutions to the H-atom, but with Z=2. The second is the hydrogen molecular ion, a 3-body problem that is formally not possible to solve without an approximation. If you make the usual Born-Oppenheimer approx., then you can solve it using elliptical coordinates as is done in standard QM texts.

And you keep talking about half-electrons ... as I told you before ... that is nonsensical. There is no such thing as a half electron.

 

[conway]

And you keep talking about half-electrons ... as I told you before ... that is nonsensical. There is no such thing as a half electron.

I can accept if you disagree with where I think the calculation would go, but I don't get it when you say you don't understand what I am trying to do.

I thought you agreed that for two isolated protons and one electron, the mathematical solution gives you a wave function which is equally shared between the two protons. For the sake of argument, I've been saying that each atom gets half an electron. If this terminology is not the most accurate, it seems to me that it is at least descriptive. If you have a better way to describe what the electron is doing, just apply it to the following paragraph:

Using my terminology, I would describe what I am trying to solve with the two-electron case as follows: I have two isolated protons and four half-electrons. Each proton gets two half-electrons. The analogy with the helium atom seem pretty obvious; just divide the charge in two everywhere. And I can't see why it wouldn't lead to an exact mathematical solution.

 

[SpectraCat]

I can accept if you disagree with where I think the calculation would go, but I don't get it when you say you don't understand what I am trying to do.

I thought you agreed that for two isolated protons and one electron, the mathematical solution gives you a wave function which is equally shared between the two protons. For the sake of argument, I've been saying that each atom gets half an electron. If this terminology is not the most accurate, it seems to me that it is at least descriptive. If you have a better way to describe what the electron is doing, just apply it to the following paragraph:

Using my terminology, I would describe what I am trying to solve with the two-electron case as follows: I have two isolated protons and four half-electrons. Each proton gets two half-electrons. The analogy with the helium atom seem pretty obvious; just divide the charge in two everywhere. And I can't see why it wouldn't lead to an exact mathematical solution.

You are using (fuzzy) logic to rationalize something instead of writing down the math. I understand that you think characterizing equal probability density of a 1-electron wavefunction on two separated nuclei as each atom having "half an electron" is ok ... I am telling you that I think it is not, because it leads to confusion, as you have already noticed.

However, I think I finally understand what you are saying .. you are not comparing two H-atoms to 1 He atom ... you are comparing two H-atoms to two "half-helium" atoms, which I again think is non-sensical and has led you to your rather odd conclusion.

In any case, I can perhaps help ease your confusion here. The issue is the 1/|r1-r2| electron-electron interaction term in the potential. As you know, the wavefunctions in this case look like super-positions of ground state 1-electron H-atom wavefunctions (1s orbitals), with essentially no probability density in the space between the atoms. So, the electron-electron interaction term will be *drastically* higher for the case where both electrons are around the same center. In fact, those cases *do* correspond to a helium-atom like configuration, although it is better to call it an H- ion-like configuration, because the nuclear charges is just Z=1.

In terms of electronic structure calculations, this interaction gives rise to the J (Coulomb) and K (exchange) two-electron integrals in the Hartree-Fock treatment of atomic and molecular energy calculations.

So anyway, does this answer your question? The states you are talking about do exist in principle, but they correspond to the superposition of the two "separated H- ion plus proton" states, and they are much higher in energy.

 

[conway]

Spectracat, please forgive the apparent sarcasm in what follows. But I'm not quite sure where this ended up.

However, I think I finally understand what you are saying .. you are not comparing two H-atoms to 1 He atom ... you are comparing two H-atoms to two "half-helium" atoms, which I again think is non-sensical and has led you to your rather odd conclusion.

In any case, I can perhaps help ease your confusion here...

So I'm wrong?

So anyway, does this answer your question? The states you are talking about do exist in principle, but they correspond to the superposition of the two "separated H- ion plus proton" states, and they are much higher in energy.

So I'm right?

 

[SpectraCat]

Spectracat, please forgive the apparent sarcasm in what follows. But I'm not quite sure where this ended up.
So I'm wrong?

EDIT: I didn't say you were wrong, I said you were confused. I think that is accurate.

So I'm right?

I guess you are free to draw whatever conclusions you want ... I put a fair amount of time and effort into analyzing your case an putting it into the *proper* quantum mechanical context. If you already knew the answer, then why did you waste my time?

EDIT2: FWIW, you are also "wrong" in the sense that your analysis using a helium atom doesn't give the right result for the energies of the superposition states I mentioned in my last post. Applying the Hamiltonian to those states yields half the ground state energy of the H- ion, not half the ground state energy of the helium atom.

 

[conway]

...The states you are talking about do exist in principle, but they correspond to the superposition of the two "separated H- ion plus proton" states, and they are much higher in energy.

No, those aren't the states I'm talking about. When you solve the H- ion you get something essentially different from the helium atom because the forces don't scale equally. You have the same interaction force between the electrons but only half the central attraction of the nucleus. You definitely don't "recover" an equivalent to the helium atom or anything else by taking the superposition of this system with a bare proton. That's not how superposition works.

What I've done is identify a system where all the forces scale exactly the same way, so the solution should be a scaled replica of the helium atom. I don't know why I'm having difficulty making myself clear on this. It is possible that I am incorrect in the calculation, but as far as the physical system which I am analyzing, I don't know how I could describe it any better than I already have.

 

[ytuab]

In both the helium atom and the hydrogen molecule, the two electrons' orbitals have the symmetrical structure.
But there are some differences between them, I think.

The helium has only one nucleus, so both the two electrons of the helium are moving around all over the helium atom.
If we pay our attention to the "Coulomb force" in the hydrogen molecule, in the space which is closer to the nucleus 1, the Coulomb force from the nucleus 1 is always stronger than the Coulomb force from the nucleus 2. (The electron-electron interaction is smaller than the electron-nucleus interaction, because the two electrons are mainly at the opposite positions of the nucleus.)

So one electron of the hydrogen molecule mainly belongs to one of the two nuclei rather than always moving around all over the hydrogen molecule (H2).

Actually, the bonding energy of H2 is only 4.7467 eV, which value is much smaller than the absolute value of the helium ground state energy (-79.005 eV).
And the internuclear distance of H2 is 0.7414 A(angstrom), which is longer than the Bohr radius. (In the hydrogen molecule ion, it is much longer, as Bob S says in #6.)

There is a difference in the nuclear movement,too. This motion of H2 is probably more complex than the helium.
But if we try to consider this effect correctly and compare these energies, even in the helium atom, it is difficult to calculate the energy including the nuclear movement.(See this thread.)

Because the Schroedinger equation doesn't have the clear orbitals, so it can't judge whether the nucleus is stopping or not. (For example, when the nucleus is just at the center of the two electrons, the nucleus stops, and this effect vanish only at this point.)

 

[conway]

In both the helium atom and the hydrogen molecule, the two electrons' orbitals have the symmetrical structure.
But there are some differences between them, I think.

The helium has only one nucleus...

But my question wasn't about the helium atom or the hydrogen molecule. It was about two separated hydrogen atoms.

 

[torquil]

But my question wasn't about the helium atom or the hydrogen molecule. It was about two separated hydrogen atoms.

Was my quote from the Atkins book in post #4 relevant for your problem? You didn't comment on it, so I don't know if you missed it or not

Torquil

 

[SpectraCat]

No, those aren't the states I'm talking about. When you solve the H- ion you get something essentially different from the helium atom because the forces don't scale equally. You have the same interaction force between the electrons but only half the central attraction of the nucleus. You definitely don't "recover" an equivalent to the helium atom or anything else by taking the superposition of this system with a bare proton. That's not how superposition works.

What I've done is identify a system where all the forces scale exactly the same way, so the solution should be a scaled replica of the helium atom. I don't know why I'm having difficulty making myself clear on this. It is possible that I am incorrect in the calculation, but as far as the physical system which I am analyzing, I don't know how I could describe it any better than I already have.

Ok, you kind of missed my point, so perhaps I didn't explain it very well. I'll try again.

In the case of two non-interacting H-atoms, as we are considering here, there are a total of six ways of placing the electrons into the ground state atomic orbitals (i.e. no electrons in 2s or higher). There are four degenerate cases corresponding to the cases where each nucleus has 1 electron: up-up, up-down, down-up and down-down, where up and down refer to the orientation of the electron spins on some space fixed axis. The energy for each of these states is -27.2 eV (2 x -13.6 eV, neglecting fine structure). The other two cases are the ion-pair states I described previously, where both electrons are around a single nucleus. The energy for those cases is just the energy of the free hydride ion, or about -14.36 eV (estimating from electron affinity of H-atom of 73 kJ/mol).

So you see, the ion pair states are *way* higher in energy. In the current model, those are the only states that have an electron-electron repulsion term in them. So, since helium has electron-electron repulsion, you will have to build those states into whatever linear combination you might take to get your model.

Anyway, this is the concept I was trying to get across before. If we are going to think about your example in terms of concepts we recognize from simpler systems, the we are pretty much forced to break it down in these terms. I suppose that you could build an ansatz around the helium atom instead of hydrogen, but it is easy to see that it will have problems describing some common systems. Take for example an isolated H-atom ... how could your model cope with that? Even if we assume it makes sense to talk about half-electrons, it is not consistent with any experimental measurements to think of hydrogen consisting of two half-electrons orbiting a proton. For one thing, it implies that there is a non-zero contribution from electron-electron repulsion in a one-electron system! So, even if you were able to make the math work out somehow, I don't see your model as being at all intuitive, which means it would have little predictive or even pedagogical value.

But perhaps I am wrong .. I would be interested to see if there is some way you can cook this all together to get the right energy for your two H-atom test case.

 

[conway]

Torquil, I'm terribly sorry I didn't respond to your original post but honestly it deals with a different problem than the one I was asking about. SpectraCat was closest to dealing with my actual question so I got involved in a discussion with him. He's just come back with some interesting points which I'm going to try to respond to now.

Ok, you kind of missed my point, so perhaps I didn't explain it very well. I'll try again.

In the case of two non-interacting H-atoms, as we are considering here, there are a total of six ways of placing the electrons into the ground state atomic orbitals (i.e. no electrons in 2s or higher). There are four degenerate cases corresponding to the cases where each nucleus has 1 electron: up-up, up-down, down-up and down-down, where up and down refer to the orientation of the electron spins on some space fixed axis. The energy for each of these states is -27.2 eV (2 x -13.6 eV, neglecting fine structure). The other two cases are the ion-pair states I described previously, where both electrons are around a single nucleus. The energy for those cases is just the energy of the free hydride ion, or about -14.36 eV (estimating from electron affinity of H-atom of 73 kJ/mol).

For starters, I have to commend you for coming up with the correct value for the energy of the H- ion, apparently by reasoning from chemical thermodynamics. I found the same value last week in this paper by A. R. P. Rau:
www.ias.ac.in/jarch/jaa/17/113-145.pdf

The other thing I'd like to point out is that when you solve the system of two isolated hydrogen atoms with two protons, you also get novel states that don't occur when you simply solve the hydrogen atoms one at a time. So in that sense we are on the same page.
How can you be sure that there are not more unexpected solutions?

So you see, the ion pair states are *way* higher in energy. In the current model, those are the only states that have an electron-electron repulsion term in them. So, since helium has electron-electron repulsion, you will have to build those states into whatever linear combination you might take to get your model.

Here is where you seem to lay down an unjustified condition: namely, that if I am proposing helium-like states then I must create them as superpositions of the six states you have allowed. This seems pretty arbitrary. In particular, the linear combination which you suggest of

1/sqrt(2)|H- over here and H+ over there> + 1/sqrt(2)|H+over here and H- over there>

does not give the same solution as I am proposing: for one thing, the energy of -14.3eV is different from the -19.5 eV which I get for a combination of mini-heliums (based on the helium ground state energy of -78eV).

You can get some insight as to why these species (hypothetical or not) are different if you look at Rau's paper. In H-, the two electrons really don't occupy the same orbital: the second one is farther from the nucleus. Of course you then symmetrize the function so the electrons are indistinguishable, but there really are in a sense two different wave functions. If I'm undertanding the implications correctly, it means you can symmetrize them or you can choose to antisymmetrize them and presumably this means for the antisymmetric combination, you are allowed to put the electrons in the triplet state. This is very different from helium: one electron does exactly what the other does and you can't antisymmetrize the spatial wave functions because then everything would go to zero. So the total wave function is symmetric and the electrons have to go in the singlet state.

Anyway, this is the concept I was trying to get across before. If we are going to think about your example in terms of concepts we recognize from simpler systems, the we are pretty much forced to break it down in these terms. I suppose that you could build an ansatz around the helium atom instead of hydrogen, but it is easy to see that it will have problems describing some common systems. Take for example an isolated H-atom ... how could your model cope with that?

Well, wasn't that my question to begin with: why do I seem to get new solutions for the four-body system that I didn't get when I analyze it as a pair of isolated two-bodies?

Even if we assume it makes sense to talk about half-electrons, it is not consistent with any experimental measurements to think of hydrogen consisting of two half-electrons orbiting a proton. For one thing, it implies that there is a non-zero contribution from electron-electron repulsion in a one-electron system!

You make a great deal of my phrase "half-electrons". True, quantum mechanics does not recognize half-electrons, but it places no restrictions on the wave function of an electron being distributed in two widely separated locations. This is the only sense in which I have ever used the phrase "half-electrons".

So, even if you were able to make the math work out somehow, I don't see your model as being at all intuitive, which means it would have little predictive or even pedagogical value.

But does the math work or doesn't it? If it works, it seems to me there should be consequences.

 

[conway]

Let's talk about Spectracat's superposition of states, quoting my description from the previous post:

1/sqrt(2)|H- over here and H+ over there> + 1/sqrt(2)|H+ over here and H- over there>

I hope this is clear enough. Let's recap: I suggested solving two isolated protons as a four-body system. The expected hydrogen atom solutions come up, and also apparently this variation on the hydrogen negative ion. SpectraCat is doubtful that my "mini-heliums" are a valid solution.

Looking at the wave function above, we can see that you can just as well put a minus sign and get an antisymmetric version. Then, using the familiar procedure, you can combine the symmetric and antisymmetric to return the traditional, location based representation: a negative ion here and a bare proton there, and vice versa.

What I'm going to suggest is that the mixed function above is awfully close to the first excited state of Helium. In other words, an exact replica of the 1s2p state spread out in space. The energy of the hydrogen negative ion comes out quite close to the expected one quarter the value for helium (-14.5eV based on helium, -14.36 eV for the actual H- ion).

So the question is: if this procedure works, why can't we get an even lower value (more stable) for the H- ion by using my ground-state helium model? I think the answer may be that parity screw you up. Notice that for the excited state of helium, phi(r1,r2) is a different state from phi(r2,r1). So you can take sums and differences and make a new basis. But for the ground state, swapping r2 and r1 returns exactly the same function. So you can't do the same trick with sums and differences.

 

[SpectraCat]

Torquil, I'm terribly sorry I didn't respond to your original post but honestly it deals with a different problem than the one I was asking about. SpectraCat was closest to dealing with my actual question so I got involved in a discussion with him. He's just come back with some interesting points which I'm going to try to respond to now.



For starters, I have to commend you for coming up with the correct value for the energy of the H- ion, apparently by reasoning from chemical thermodynamics. I found the same value last week in this paper by A. R. P. Rau:
www.ias.ac.in/jarch/jaa/17/113-145.pdf

Well, it's just a cheap parlor trick (definition of electron affinity), but thanks. It's hard to find tabulations of gas phase ion energies, but electron affinities are easy to find.

The other thing I'd like to point out is that when you solve the system of two isolated hydrogen atoms with two protons, you also get novel states that don't occur when you simply solve the hydrogen atoms one at a time. So in that sense we are on the same page.
How can you be sure that there are not more unexpected solutions?

What "novel" states are you talking about? Unless I misunderstand the question, you only get the six cases I mentioned above (more on that below), if we neglect electronically excited states.

Here is where you seem to lay down an unjustified condition: namely, that if I am proposing helium-like states then I must create them as superpositions of the six states you have allowed. This seems pretty arbitrary.

It's really not arbitrary at all. The stipulation here is that the H-atoms are non-interacting .. therefore their energies are degenerate, and their individual wavefunctions are just the ground state H-atom wavefunctions (1s-orbitals). The added wrinkle of electron spin produces the four possibilities I listed above. The only other choices for putting 2 electrons around the nuclei are the ion pair states (again, neglecting electronically excited states). What else could there be? The H-atom orbitals are a complete set of solutions, so once you have exhausted the ways of populating the 1s orbital, there are no more possibilities without involving the n=2 (or higher) shell.

Now, once you start to bring the H-atoms closer, and they start interacting, those higher-lying orbitals do start to get mixed into the solutions (a la configuration interaction), along with the ion-pair states, and things get very messy. You can approximate this variationally, or with perturbation theory, but it isn't a lot of fun ... better to let a computer do it.

In particular, the linear combination which you suggest of

1/sqrt(2)|H- over here and H+ over there> + 1/sqrt(2)|H+over here and H- over there>

does not give the same solution as I am proposing: for one thing, the energy of -14.3eV is different from the -19.5 eV which I get for a combination of mini-heliums (based on the helium ground state energy of -78eV).

Yeah, this is just one of those manifestations of the problems I have been indicating you will get with this model. Basically, a He-atom is a crappy model for an H-atom, even given the stipulations you have introduced. It has built in electron-electron repulsion, which may not be appropriate in all situations (such as the one we are considering). Also, even though you are correct that the forces scale properly, what about other considerations, such as the average radius, and the nuclear screening? Those are going to be wildly different from the He-atom case. So, I don't think it is at all clear that you can simply take 25% of the He-atom energy as the correct energy for this system. You would need to solve the Coulomb and exchange integrals for this specific case, and come up with some way to deal with the electron correlation. It's going to be messy, and worse, that is what one ends up doing for molecules *anyway*, even when we use computational techniques based on expansion in the (simpler) basis of 1-electron atomic orbitals that is typically chosen. So it is hard to see what advantages derive from your treatment ...

You can get some insight as to why these species (hypothetical or not) are different if you look at Rau's paper. In H-, the two electrons really don't occupy the same orbital: the second one is farther from the nucleus.

Do you think the situation is any different for the He atom? If you designate a "primary" electron, then the "secondary" electron in He will experience a reduced nuclear charge due to screening effects, so you end up in the same place. The effect is more dramatic in the H- case to be sure, but it is just a question of degree ... the phenomenology is the same in both cases.

Of course you then symmetrize the function so the electrons are indistinguishable, but there really are in a sense two different wave functions. If I'm undertanding the implications correctly, it means you can symmetrize them or you can choose to antisymmetrize them and presumably this means for the antisymmetric combination, you are allowed to put the electrons in the triplet state. This is very different from helium: one electron does exactly what the other does and you can't antisymmetrize the spatial wave functions because then everything would go to zero. So the total wave function is symmetric and the electrons have to go in the singlet state.

No, this is not correct. The electrons are paired in a singlet state in both cases. The triplet configurations correspond to excited states ... again, they are closer lying in the H- case, but the logic is the same.

Well, wasn't that my question to begin with: why do I seem to get new solutions for the four-body system that I didn't get when I analyze it as a pair of isolated two-bodies?

I think it is because you have introduced a bunch of extra terms (e.g. electron-electron interaction) that are not present in the physical system under consideration.

You make a great deal of my phrase "half-electrons". True, quantum mechanics does not recognize half-electrons, but it places no restrictions on the wave function of an electron being distributed in two widely separated locations. This is the only sense in which I have ever used the phrase "half-electrons".

Fair enough .. it seemed like you were using it in a different context. Just remember thought that it is just as nonsensical to talk about the energy (or any other observable) for "half the wavefunction". Those quantities are defined in terms of integrals over the *entire* wavefunction. That is why I think the "half-electron" idea is a bad one, even in the context you describe above.

But does the math work or doesn't it? If it works, it seems to me there should be consequences.

I think that unfortunately it doesn't, for the reasons I have outlined above.

 

[SpectraCat]

Let's talk about Spectracat's superposition of states, quoting my description from the previous post:

1/sqrt(2)|H- over here and H+ over there> + 1/sqrt(2)|H+ over here and H- over there>

I hope this is clear enough. Let's recap: I suggested solving two isolated protons as a four-body system. The expected hydrogen atom solutions come up, and also apparently this variation on the hydrogen negative ion. SpectraCat is doubtful that my "mini-heliums" are a valid solution.

Looking at the wave function above, we can see that you can just as well put a minus sign and get an antisymmetric version. Then, using the familiar procedure, you can combine the symmetric and antisymmetric to return the traditional, location based representation: a negative ion here and a bare proton there, and vice versa.

Yup. That is the second ion-pair configuration I mentioned.

What I'm going to suggest is that the mixed function above is awfully close to the first excited state of Helium. In other words, an exact replica of the 1s2p state spread out in space. The energy of the hydrogen negative ion comes out quite close to the expected one quarter the value for helium (-14.5eV based on helium, -14.36 eV for the actual H- ion).

No, I really don't think so. First of all, the first excited state of helium is 1s¹2s¹. Second, the H- ion states are singlets, and the first excited state of helium is a triplet, so the symmetry is wrong. Also, not sure where you got that 14.5 eV number .. and it is simply wrong to take 25% of a helium result as the correct result for this system, for any state, for the reasons I laid out in the previous post.

So the question is: if this procedure works, why can't we get an even lower value (more stable) for the H- ion by using my ground-state helium model? I think the answer may be that parity screw you up. Notice that for the excited state of helium, phi(r1,r2) is a different state from phi(r2,r1). So you can take sums and differences and make a new basis. But for the ground state, swapping r2 and r1 returns exactly the same function. So you can't do the same trick with sums and differences.

No, this isn't correct. If the wavefunctions are properly antisymmetrized (which they must be in the end), you can never get a different wavefunction back from swapping two elecrons ... you get the same wf back, multiplied by -1.

 

[conway]

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.

 

[conway]

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.

 

[conway]

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.

 

[peteratcam]

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 \psi(x_1,x_2) 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 into \psi_1(x_1)\psi_2(x_2) (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.

 

[conway]

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 let's 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...

 

[conway]

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.

 

[conway]

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.

 

[conway]

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?

 

[conway]

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.