# Two electrons in a potential well

• conway

#### conway

This question came up in another thread I started about the wave functions of helium vs hydrogen. It's gone off on something of a tangent so I hope I'm not out of line in starting a fresh thread with this topic.

I've basically sketched what I would think are the only two realistic candidate solutions for two electrons in a potential well. I basically looked for the simplest solution where the electrons made some kind of effort to stay away from each other and wrote it out as a simple product function. Since the product function wasn't symmetric, I reversed the two electrons and added the two functions together. Standard symmetrization. Then I noticed that the anti-symmetric combination looked like it might have lower energy, in this case, than the symmetric combination. So I've got these two wave functions and I can't see any obvious way to tell which would be the correct ground state.

I also haven't figured out if the spin states have a significant effect on the energy. Of course the symmetric combination would have the singlet spin state and the other combination would have the triplet state. I just don't know if I have to worry about that yet.

I posted my sketches in the other thread. I don't know how to make the thumbnails appear below so I'm just going to put these links in and see if they work. I hope the meaning of the diagrams is obvious:

SYMMETRIC FUNCTION:
https://www.physicsforums.com/attachment.php?attachmentid=24045&d=1267583181

ANTISYMMETRIC FUNCTION:
https://www.physicsforums.com/attachment.php?attachmentid=24046&d=1267585014

So just to check, the problem is a single potential well, centred at x=0 (eg, a harmonic well). Put two electrons in, what do they do (label a and b)?

If the electron repulsion was turned off completely:
The lowest energy state would be to put the electrons both in the groundstate of the potential well, and also in a spin singlet. (Antisymmetric spin, and symmetric spatial)
A:$$\phi_1(x_a)\phi_1(x_a)|\rm singlet\rangle$$
The spatial part sketched, would look like a circular blob centred on xa=0,xb=0.

The next lowest state, would be to have one electron in the first excited state of the single particle system, and the other in the groundstate. Obvioulsy this needs to be symmetrised, so the options are
the three options:
B:$$[\phi_1(x_a)\phi_2(x_b)-\phi_1(x_b)\phi_2(x_a)]|\rm triplet\rangle$$
or
C:$$[\phi_1(x_a)\phi_2(x_b)+\phi_1(x_b)\phi_2(x_a)]|\rm singlet\rangle$$

Now, if you consider turning on weak repulsive interactions:
State A will still be the lowest energy, but the circular blob will distort so that it begins to look like what you have called 'SYMMETRIC FUNCTION'. It must be symmetric, because the spin part is antisymmetric.

The 3 State Bs will be the next energy level. B is lower than C because the repulsive energy of interaction between the electrons is automatically smaller in the antisymmetrised spatial wave function state than the symmetrised spatial wave function state.

State B will begin to look like you 'ANTI-SYMMETRISED' state.

If the repulsive interaction is sufficiently strong, the levels might begin to cross and it won't be easy to make statements like the above.

• Najah
Awesome analysis, Peter. Thanks. I'm still thinking about it but I know I'll have some more questions. I have to admit I just now realized that the 1-dimensional well isn't going to be all that useful in developing real physical insights because the 1/r potential really depends on being in three dimensions. So for the 1-dimensional well it's not even clear what I should be using for a potential. Still, some of the ideas you've explained should be helpful in thinking about the 3-d case.

Okay, I'm going to try and interpret what this means in the three-dimensional case. At first, I would have never believed that the shape of the wave function could change based on the strength of the interaction. But as I think about it, maybe it does. Just imagine that there might be a crossover point where the triplet state actually becomes lower in energy than the singlet!

Of course we don't have to vary the strength of the interaction: we can just use real electrons and vary the size of the box. At first I thought that with a very large box, the interaction would be insignificant and both electrons would occupy the symmetric singlet state. But then I thought about it: as you reduce the size of the box, the kinetic energy goes as 1/L^2, while the potential goes as 1/L. So its the kinetic that has to dominate for a very small box. In the limit, the two electrons occupy the identical singlet state.

What I think happens as the box gets bigger is the state starts to deform as Peter pointed out, the "lobes" heading northwest and southeast on the one-dimensional diagram. In the 3-d case of course you get 3 degenerate states, one for each axis, but the idea is the same. At first the spherical blob is just distorted, but then peaks form and the valley deepens.

Now, when the lobes are far apart and the valley is deepened almost to zero, it takes very little to flip one side negative, and voila you have the triplet state. So the singlet and triplet become degenerate for a very large box. They don't actually cross over.
Although there is no critical size of crossover, there are interesting points along the way such as when the line between the two lobes first gets a "kink". I wonder if this has ever been calculated.

A couple of points:

Peter, I notice that you ask if I could confirm that I am talking about a potential well, eg. harmonic, centered on the origin. Actually I am thinking about the simple box, or infinite well; but descriptively, the solutions are going to be similar. Also, as a small point, in my sketches I've actually placed the left border of the box at the origin. Again, doesn't really matter...

In the meantime I've thought about the physical interpretation some more. I wonder if we can use the language of atomic orbitals to describe the states we are talking about; your ground state A would be the 1s^2 state, the difference state B would be labeled (triplet)1s2p and C would be (singlet)1s2p. My conclusion from my previous post is that in the limit of the very large box, the 1s^2 and the triplet(1s2p) states become degenerate.

I think I can explain it a little better now. For the very small box, the interaction is zero and the three states have their simple representation as the sums and differences of product states. For the square well centered at the origin, your functions phi1 and phi2 are just

phi1 = coskx

phi2 = sin2kx

(with k chosen appropriately so it just fits the box). I'm going to drop the "k" from here on to simplify the formulas a bit.

As the box gets bigger, phi1=cos(x) gets distorted by mixing in a little bit of sin(2x). In the 1-d case, this has the effect of distorting the blob in my diagram. As the second harmonic gets stronger and starts to dominate, the blob separates. At the same time, the 1s2p state phi2 = sin(2x) gets distorted by mixing in some of the ground state. In the limit of the very large box, the mixing is complete:

ph1 = phi2 = cosx +/- sin2x.

Note that the 1s^2 state and the (singlet) 1s2p state have converged into the identical state, which is the new ground state! (Neglecting spin, the (triplet) 1s2p is also degenerate.)

Of course, an interesting thing you can do when you have degenerate states is you can recombine them with sums and differences to get new basis states. In this instance, you just recover the simple product states:

phi1 + phi2 = {cos(x1)) + sin(2x1)}*{cos(x2) -sin(2x2)}

phi1 - phi2 = {cos(x1)) - sin(2x1)}*{cos(x2) + sin(2x2)}

The physical meaning of these states is that electron A sits at the left of the box and electron B at the right. The other state is the same except the two electrons are reversed.

The 3d case is of course harder to visualize as the box grows and transforms, but when you get to the limit of the very large box, I believe the physics is the same: the 1s^2 and the 1s2p states have converged into a hybrid which pushes the two electrons to opposite sides of the box.

Physically, this is correct! The reason the 1s state is lower than the 2p state for the square well is only because of the kinetic energy. And for the very large box, the kinetic energy of the two states becomes so low that the differences is negligible compared to the repulsive force. So it makes sense for the electrons to push each other as far apart as possible, since the "penalty" for confinement is in this case negligible.

Last edited:
A couple of points:

Peter, I notice that you ask if I could confirm that I am talking about a potential well, eg. harmonic, centered on the origin. Actually I am thinking about the simple box, or infinite well; but descriptively, the solutions are going to be similar. Also, as a small point, in my sketches I've actually placed the left border of the box at the origin. Again, doesn't really matter...

In the meantime I've thought about the physical interpretation some more. I wonder if we can use the language of atomic orbitals to describe the states we are talking about; your ground state A would be the 1s^2 state, the difference state B would be labeled (triplet)1s2p and C would be (singlet)1s2p. My conclusion from my previous post is that in the limit of the very large box, the 1s^2 and the triplet(1s2p) states become degenerate.

I think I can explain it a little better now. For the very small box, the interaction is zero and the three states have their simple representation as the sums and differences of product states. For the square well centered at the origin, your functions phi1 and phi2 are just

phi1 = coskx

phi2 = sin2kx

(with k chosen appropriately so it just fits the box). I'm going to drop the "k" from here on to simplify the formulas a bit.

As the box gets bigger, phi1=cos(x) gets distorted by mixing in a little bit of sin(2x). In the 1-d case, this has the effect of distorting the blob in my diagram. As the second harmonic gets stronger and starts to dominate, the blob separates. At the same time, the 1s2p state phi2 = sin(2x) gets distorted by mixing in some of the ground state. In the limit of the very large box, the mixing is complete:

ph1 = phi2 = cosx +/- sin2x.

Note that the 1s^2 state and the (singlet) 1s2p state have converged into the identical state, which is the new ground state! (Neglecting spin, the (triplet) 1s2p is also degenerate.)

Of course, an interesting thing you can do when you have degenerate states is you can recombine them with sums and differences to get new basis states. In this instance, you just recover the simple product states:

phi1 + phi2 = {cos(x1)) + sin(2x1)}*{cos(x2) -sin(2x2)}

phi1 - phi2 = {cos(x1)) - sin(2x1)}*{cos(x2) + sin(2x2)}

The physical meaning of these states is that electron A sits at the left of the box and electron B at the right. The other state is the same except the two electrons are reversed.

The 3d case is of course harder to visualize as the box grows and transforms, but when you get to the limit of the very large box, I believe the physics is the same: the 1s^2 and the 1s2p states have converged into a hybrid which pushes the two electrons to opposite sides of the box.

Physically, this is correct! The reason the 1s state is lower than the 2p state for the square well is only because of the kinetic energy. And for the very large box, the kinetic energy of the two states becomes so low that the differences is negligible compared to the repulsive force. So it makes sense for the electrons to push each other as far apart as possible, since the "penalty" for confinement is in this case negligible.

I don't have time for a detailed commentary right now, but this is all rather muddled, and you mix and match concepts that are not compatible ... it seems like you may be on the right track, but you need to be a bit more explicit about what you mean.

1) You say "particle in infinite box" ... that means that the potential for a single particle in the box is zero, period. You are certainly free to put more than one electron in it and consider what happens when you vary the electron-electron interaction, however, changing the size of the box will have no intrinsic effect on the potential energy (in this post and others you indicate that you think that changing the size of the box will change the *potential energy* in the 1-electron case ... it does not).

2) In your analysis, you say "For the very small box, the interaction is zero ..." and later "As the box gets bigger, phi1=cos(x) gets distorted by mixing in a little bit of sin(2x)". Those statements cannot be applied consistently to the same system. If there is no interaction, there will be no "distortion" or "mixing" of the wavefunctions when the box size is changed. In the case of zero interaction, the wavefunctions will stay basically the same when the box size is changed .. only the k-constant will change.

I don't have time for a detailed commentary right now, but this is all rather muddled, and you mix and match concepts that are not compatible ... it seems like you may be on the right track, but you need to be a bit more explicit about what you mean.

1) You say "particle in infinite box" ... that means that the potential for a single particle in the box is zero, period. You are certainly free to put more than one electron in it and consider what happens when you vary the electron-electron interaction, however, changing the size of the box will have no intrinsic effect on the potential energy (in this post and others you indicate that you think that changing the size of the box will change the *potential energy* in the 1-electron case ... it does not).

Spectracat, I appreciate your contribution. It was peteratcam who made the very useful and insightful suggestion that we analyze the two-electron case by starting with zero interaction and gradually "turning on" the repulsion. What I did was to observe that this was mathematically equivalent to varying the scale, since the kinetic and potential energies vary differently as you make the box bigger or smaller. The "no-repulsion" case turns out, somewhat counter-intuitively, to correspond to the limit of the very small box. And as the box becomes very large in size, the potential, or repulsive force, dominates the kinetic energy term.

2) In your analysis, you say "For the very small box, the interaction is zero ..." and later "As the box gets bigger, phi1=cos(x) gets distorted by mixing in a little bit of sin(2x)". Those statements cannot be applied consistently to the same system. If there is no interaction, there will be no "distortion" or "mixing" of the wavefunctions when the box size is changed. In the case of zero interaction, the wavefunctions will stay basically the same when the box size is changed .. only the k-constant will change.

Again, the interaction goes to zero for the very small box because the kinetic term, with its 1/r-squared dependence, dominates the potential which goes as 1/r. As the box gets bigger, it is the influence of the potential term which cause the states to distort. Of course the easiest way to analyze the distortion of the ground state is to perturb it by mixing in a bit of the next highest state, and this is physically reasonable in this case. It is significant that when we take the limit of the state being an equal mixture of the 1s^2 and 1s2p waveshapes, we get a physically reasonable picture for the large box: the two electrons stay more or less on opposite halves of the box.

I haven't yet quite figured out how to construct a spherically symmetric (or cubically symmetric at least) solution for the cubical box, but I'm guessing there ought to be one; possibly just a superposition cases I've already identified along the x,y, and z axes. The physical interpretation would be that the two electrons are always at opposite ends of the box but there is no preferred orientation. I think this is similar to the wave function for helium.

Spectracat, I appreciate your contribution. It was peteratcam who made the very useful and insightful suggestion that we analyze the two-electron case by starting with zero interaction and gradually "turning on" the repulsion. What I did was to observe that this was mathematically equivalent to varying the scale, since the kinetic and potential energies vary differently as you make the box bigger or smaller. The "no-repulsion" case turns out, somewhat counter-intuitively, to correspond to the limit of the very small box. And as the box becomes very large in size, the potential, or repulsive force, dominates the kinetic energy term.

Again, the interaction goes to zero for the very small box because the kinetic term, with its 1/r-squared dependence, dominates the potential which goes as 1/r.

No, this is wrong ... the kinetic energy is proportional to L-2, which is the size of the box ... the potential interaction is proportional to r12-1, which is the separation between the particles ... this term needs to be included in the Hamiltonian, and is non-trivial to solve for. In any case, the interaction most assuredly does not vanish in the limit of a very small box.

As the box gets bigger, it is the influence of the potential term which cause the states to distort. Of course the easiest way to analyze the distortion of the ground state is to perturb it by mixing in a bit of the next highest state, and this is physically reasonable in this case. It is significant that when we take the limit of the state being an equal mixture of the 1s^2 and 1s2p waveshapes, we get a physically reasonable picture for the large box: the two electrons stay more or less on opposite halves of the box.

This is at least qualitatively correct, but you are still pretty "loose" in your terminology. I think what you mean is summarized by the following:

1) The one-electron solutions form a complete set of basis states
2) To a zero-order approximation, we can consider the two electrons to be occupying two different states, and ignore the interaction between them.
3) In the limit of very weak interactions, you can solve for the corrections to the zero-order wavefunctions using second-order perturbation theory, with the 1/r12 interaction is the perturbing Hamiltonian.
4) As the strength of the interaction increases, higher order perturbative terms start to contribute, and it makes sense to solve the problem variationally. We can do this by diagonalizing the variational matrix, where the off-diagonal matrix elements express the coupling between the states due to the 1/r12 interaction term.

I haven't yet quite figured out how to construct a spherically symmetric (or cubically symmetric at least) solution for the cubical box, but I'm guessing there ought to be one; possibly just a superposition cases I've already identified along the x,y, and z axes. The physical interpretation would be that the two electrons are always at opposite ends of the box but there is no preferred orientation. I think this is similar to the wave function for helium.

You can solve for this using the well-known particle in a spherical box (PISB) wavefunctions for the one-electron zero-order states. These are expressed a sine function for the radial part (indexed by quantum number n), multiplied by the spherical harmonics for the angular part (indexed by quantum numbers l and m). The solution will share some characteristics of the helium wavefunction, but of course it will be fundamentally different since the Hamiltonian lacks the Coulomb potential from the nucleus.

I think what you mean is summarized by the following:

1) The one-electron solutions form a complete set of basis states
2) To a zero-order approximation, we can consider the two electrons to be occupying two different states, and ignore the interaction between them.
3) In the limit of very weak interactions, you can solve for the corrections to the zero-order wavefunctions using second-order perturbation theory, with the 1/r12 interaction is the perturbing Hamiltonian.
4) As the strength of the interaction increases, higher order perturbative terms start to contribute, and it makes sense to solve the problem variationally. We can do this by diagonalizing the variational matrix, where the off-diagonal matrix elements express the coupling between the states due to the 1/r12 interaction term.

I appreciate that you are trying to be polite by saying "I think this is what you mean...", because if I meant what I actually said, then I would be talking nonsense. The sad fact is I probably did mean almost exactly what I said; I certainly didn't mean to say the things you've listed here. The things you list are very conventional prescriptions for solving equations, but they say nothing at all about what the actual wave function looks like in a specific circumstance. I on the other hand, have said that here is a box and this is what the wave functions look like: when the box is very small, the behavior approaches this limit, and when the box is large, the behavior approaches a different limit. Furthermore, I've said that you can do this not by applying perturbation theory or diagonalizing a matrix, but by drawing pictures and making physical arguments about what is going on.

Whether I'm right or wrong, I think that what I said is pretty much exactly what I meant to say. (And I'm pretty sure I'm right about the limit of the small box.)

Furthermore, I've said that you can do this not by applying perturbation theory or diagonalizing a matrix, but by drawing pictures and making physical arguments about what is going on.

And that can be a very effective tool for understanding problems, but you have to make sure that your wavefunctions, and your arguments, are correct ones. The solutions you sketched are reasonable enough guesses for the lowest two one-electron wavefunctions in this problem, and peteratcam's approach to building up the two-electron solutions was also reasonable. However, you have (at least) twice made statements that indicate you have a misconception about an important part of this problem: In post #4 you wrote:
But then I thought about it: as you reduce the size of the box, the kinetic energy goes as 1/L^2, while the potential goes as 1/L. So its the kinetic that has to dominate for a very small box ...
and in post #7
Again, the interaction goes to zero for the very small box because the kinetic term, with its 1/r-squared dependence, dominates the potential which goes as 1/r.

Those statements are both incorrect, for the reasons I stated above. The potential *inside* the box does not depend at all on the size of the box for the one-electron solutions. The potential inside the box is zero if we are neglecting the electron-electron repulsion. The *unperturbed* state energies (i.e. kinetic energies of zero-potential PIB states) do indeed scale as the inverse square of the box dimension. The electron-electron repulsion scales as 1/r12, the inverse of the distance between the electrons, which is a variable in this system, and thus can become arbitrarily small. Basically, no matter how small you make L, by definition, r12 can always be smaller, since the electrons are inside the box. It would make sense to talk about the expectation value of 1/r12, once the wavefunctions are known, as representing a measure of how significant a contribution the electron-electron repulsion makes to the total energy of the state.

Whether I'm right or wrong, I think that what I said is pretty much exactly what I meant to say. (And I'm pretty sure I'm right about the limit of the small box.)

I guess it depends what you mean by small ... at any size scale relevant to models of electrons in atoms, your statement is incorrect; you certainly cannot assume that the electron-electron repulsion goes to zero when the box approaches atomic size. If you continued to shrink the box down toward the Planck scale .. then I have no idea what would happen .. but it is my suspicion that you *still* would not be able to ignore the electron electron interactions in this case.

Spectracat, I hope you will not take this as any kind of put-down but you might not want to be too hasty in dismissing my arguments about the role of scale. No, I am not talking about the Planck length: the crossover point almost certainly occurs around the typical atomic dimensions. So I am saying that a pair of electrons confined to a cubical box of say, a picometer, would basically have the same solution as your hypothetical "non-interacting fermions". You might want to consider the possibility that I am correct, and for exactly the reasons I keep emphasizing: that the kinetic energy dominates the "potential" (by which I mean the energy associated with the electron-electron repulsion).

I am also surprised that you reject my application of 1/L as a reasonable indicator of the strength of the repulsive interaction. You reject my 1/L on the grounds that it is not equal to the expectation value of 1/r12. How different could it be, apart from a small scale factor? Yes, the ratio changes somewhat over the entire range, but it obviously must approach a constant at the small scale limit.

When I first opened up this problem, it never occurred to me that the shape of the solutions might depend on the size of the box. Peteratcam's suggestion opened my eyes to this possibility, which quickly became a certainty. It is truly surprising that the "weak-interaction limit" corresponds to the tiny box, not the huge box: but that's the way it goes.

I'm going to see if I can't make a verifiable prediction based on my theory. You know how there is a series of hydrogen-like ions: H, He+, Li++ etc. where the energies are 1,4, 9 Rydbergs etc. We might naively expect that there is also a corresponding series of helium-like ions: He, Li+, B++, etc. characterized by two electrons around a single nucleus. My theory says that unlike the hydrogen series, the helium series will not be characterized by a simple progression based on geometric similarity. In particular, as you get to very heavy nuclei with their small confinement diameters, the two-electron ion should have very nearly twice the energy of the single electron ion: where He+ is -54ev and He is -79 (very different from -104 (edit: should be -108!)) I am suggesting that U90+ should much more nearly double the energy of U91+. So, if the one-electron ion is 91^2 Rydbergs (approximately -8300 eV) the two-electron U90+ ion should be close to -15000 or -16000 eV: definitely not the -12400 or so you would expect from a mathematical proportionality to helium.

This would be an example of where my approach would apply to cases on the atomic scale. I wonder if there is verifiable data one way or the other on this point?

(Edit: This isn't really an edit but more of a correction: where I said 8300 eV I obviously meant 8300 Rydbergs, which is close to 110,000 eV. So just read "Rydbergs" where I've said "eV" in the last section.)

Last edited:
I figured out that I can tighten up my estimate for the energy levels of Uranium. First of all, the uranium atom with one electron and 92 protons should have 92^2 Rydbergs of binding energy based on similarity to the hydrogen atom: I make this out to be 115,000 eV. For the helium atom, the naive model of no-interaction would give us a value of -54 eV for the ion and -108 eV for the neutral atom. The actual atom has an energy of -79 eV. We can say the interaction between the electrons has added 29 eV to the system.

For the uranium atom, the interaction energy is greater by a geometric factor of 92: not 92-squared! so we should add something like 3000 eV to the system. So where the naive model gives us -230,000 eV the corrected calculation for the interaction energy should give close to -227,000 eV. It's a very small correction compared to the pronounced effect in helium.

This calculation assumes the shape of the electron cloud doesn't change between helium and uranium. Actually it becomes relatively more compact, which probably adds a couple thousand more eV to the interaction energy.

Spectracat, I hope you will not take this as any kind of put-down but you might not want to be too hasty in dismissing my arguments about the role of scale. No, I am not talking about the Planck length: the crossover point almost certainly occurs around the typical atomic dimensions. So I am saying that a pair of electrons confined to a cubical box of say, a picometer, would basically have the same solution as your hypothetical "non-interacting fermions". You might want to consider the possibility that I am correct, and for exactly the reasons I keep emphasizing: that the kinetic energy dominates the "potential" (by which I mean the energy associated with the electron-electron repulsion).

I am also surprised that you reject my application of 1/L as a reasonable indicator of the strength of the repulsive interaction. You reject my 1/L on the grounds that it is not equal to the expectation value of 1/r12. How different could it be, apart from a small scale factor? Yes, the ratio changes somewhat over the entire range, but it obviously must approach a constant at the small scale limit.

When I first opened up this problem, it never occurred to me that the shape of the solutions might depend on the size of the box. Peteratcam's suggestion opened my eyes to this possibility, which quickly became a certainty. It is truly surprising that the "weak-interaction limit" corresponds to the tiny box, not the huge box: but that's the way it goes.

Your statements about the "shapes" of the wavefunctions changing with the size of the box is a bit vague. On the one hand, of course they will change, they will get bigger. But I guess you mean that you think their aspect ratio will also change (i.e. shapes with respect to the box size), and that is what you find surprising. I would make the following comments about that:

1) it is clearly not correct in the limit of non-interacting particles. In that case, you just have the one-electron solutions, which only change their size (not shape) with the size of the box
2) for a fixed, non-zero interaction between the electrons (i.e. Vrep/r12), it is not completely clear to me that the shapes will change ... it certainly seems logicial on the face of it, but in order to find the correct steady state solutions, you need to correctly account for exchange and correlation over the entire range of values of r12, and those effects can be hard to guess.

Peteratcam's post did not say anything about changing the size of the box, it talked about altering the strength of the repulsive electron-electron interaction, these are not the same thing.

It is not "surprising" that the weak-interaction limit corresponds to a small box on the atomic scale, because I have said, that *supposition* is incorrect. Your ionic examples show this ... your argument actually shows that the interaction energy is higher for a smaller box .. it does not approach zero, which is what the "weak-interaction" limit implies to me. Yes, the kinetic contribution increases faster, but that is a separate argument. I don't think it is clear that a higher kinetic energy should wash out the electron-electron interaction effects .. after all, they are confined to a smaller box, so they are less able to avoid each other. So, that argument does not seem physically sensible.

Finally, your ionic energy progression argument is interesting, but you are not comparing apples with apples. In the case of a uranium ion, I can perhaps accept that as being in the particle in a box limit, where the confining coulomb potential from the nucleus can be considered so steep as to be approximated by an infinitely deep well. However, as you decrease the nuclear charge, that approximation gets worse and worse, and certainly in the case of helium, the infinite potential approximation is quite poor. In the uranium case, the potential pushes the electrons together and forces them to interact ... even if that interaction distorts the wavefunction, the *relative* energetic penalties are negligible compared to moving some of the electron density away from the center of the box. In the helium case, the curvature of the potential is noticable on the scale of the electron-electron repulsion, so the box size cannot be taken to be constant.

Peteratcam's post did not say anything about changing the size of the box, it talked about altering the strength of the repulsive electron-electron interaction, these are not the same thing.

Yes they are. That's my point.

Your ionic examples show this ... your argument actually shows that the interaction energy is higher for a smaller box .. it does not approach zero, which is what the "weak-interaction" limit implies to me.

That's like saying you don't agree that (x+1)/(x-squared) approaches zero as x approaches infinity "because x keeps getting bigger and bigger".

Yes, the kinetic contribution increases faster, but that is a separate argument. I don't think it is clear that a higher kinetic energy should wash out the electron-electron interaction effects ..

For the helium atom, the non-interacting model gives you an error of almost 30% on the energy of the system. For uranium, if my calculation is correct, the error is less than 2%. I would say the interaction effect is pretty much getting "washed out" by that point.

after all, they are confined to a smaller box, so they are less able to avoid each other. So, that argument does not seem physically sensible

That's why I call it "surprising". But it is correct.

Anyhow, there's no need to quibble over the details: I've actually put my money where my mouth is and posted a calculation based on my "theory" . I've gone and calculated ionization energies for heavy nucleii. This data should exist somewhere. If I'm wrong, show me the numbers.

Yes they are. That's my point.

You have an astounding level of confidence for someone making qualitative predictions based on your own deductive reasoning. That is fine, but it makes it an awful lot harder for you to appreciate your own mistakes, since you don't think they exist. Show me a proof from *anywhere* that supports your statement.

That's like saying you don't agree that (x+1)/(x-squared) approaches zero as x approaches infinity "because x keeps getting bigger and bigger".

No, you missed my point. The interaction energy is not a relative term, it is an absolute term. For your system, it gets bigger when the box size decreases, period. My statement is equivalent to saying that x+1 gets bigger when x gets bigger. Whether or not other terms in the energy increase fast enough so that the increasing potential can be neglected is, as I said, a separate argument, which needs to account for the other factors I mentioned.

For the helium atom, the non-interacting model gives you an error of almost 30% on the energy of the system. For uranium, if my calculation is correct, the error is less than 2%. I would say the interaction effect is pretty much getting "washed out" by that point.

But helium and uranium are not equivalently good models for a particle in a box ... you are saying "helium is like a big box, and uranium is like a small box", and this is not true .. as I pointed out, while uranium might be a good model PIB, helium surely is not. Therefore the comparison you are making is not valid. In one case (helium" the wavefunction can "relax" into a larger part of the potential energy surface (i.e. bigger box) by paying a small energy penalty ... this is equivalent to nuclear screening, which is why there is a difference between the He ionization energies. In the other case, the potential is so steep that the wavefunction can never find it's way into a bigger box due to the repulsive forces.

That's why I call it "surprising". But it is correct. Again with the astounding confidence ... I don't suppose you have a proof of this ...

Anyhow, there's no need to quibble over the details: I've actually put my money where my mouth is and posted a calculation based on my "theory" . I've gone and calculated ionization energies for heavy nucleii. This data should exist somewhere. If I'm wrong, show me the numbers.

Look them up yourself if you like, but in any case, you have yet to establish that they are relevant to the discussion at hand. Even if they show the trend that you predict, I don't think it will be for the reasons you have hypothesized, because you have not taken the different shapes of the nuclear coulomb potentials into account properly in the two cases.

No, you missed my point. The interaction energy is not a relative term, it is an absolute term.

"Relative" and "absolute" are relative terms. The energy is either relative or absolute depending on how you choose to think of it. My point is that it is more useful in this case to treat it as a relative term.

But helium and uranium are not equivalently good models for a particle in a box ... you are saying "helium is like a big box, and uranium is like a small box", and this is not true .. as I pointed out, while uranium might be a good model PIB, helium surely is not.

How can one be a good model and not the other? They are both simple 1/r potentials.

...Show me a proof from *anywhere* that supports your statement. ...Again with the astounding confidence ... I don't suppose you have a proof of this ...

Look them up yourself if you like, but in any case, you have yet to establish...

Physics is not so much a game of rigorous proofs as it is of putting together convincing evidence; but to the extent that "proofs" are part of the game, it seems only fair to me that at some point the onus of proof shifts from "prove that you're right" to "prove that I'm wrong". I feel we passed that point when I posted my calculations about uranium. I've given everyone a golden opportunity to prove that I don't know what I'm talking about, and I've taken quite a gamble in doing so.

... that they are relevant to the discussion at hand. Even if they show the trend that you predict, I don't think it will be for the reasons you have hypothesized, because you have not taken the different shapes of the nuclear coulomb potentials into account properly in the two cases.

I don't like those "even if you're right you're wrong" arguments. If my numbers turn out to be right, then I think my theory is entitled to a small amount of deference on that account; and you have to recognize that if you still want to dispute my explanations, you will be arguing from a somewhat weakened position.

My numbers look pretty good. The first four two-electron ions are H-, He, Li+, and Be++.
Using the data for ionization energies I get a constant of 16.05 eV per electron per nucleon to account for the repulsion energy. The error is 2.3 eV for helium, and the explanation is interesting.

The most naive model for the helium atom basically puts two non-interacting electrons in the same hydrogen-type ground state. The energy would then be -8 Rydbergs, or 108.8 eV. The actual energy is -79eV. So we look for a correction in terms of the electron-electron interaction.

The most straightforward calculation would be to treat the two electrons as simple cloud charges and do a classical energy calculation. The result of this calculation is quite close to the true value, but it is a little high, by about 2 eV. It is off because the electrons are not in a simple product state, and in fact it is the same error given by my semi-empirical formula ( -108.8 + 2 x 16.05 = 76.7). For larger ions the error disappears rather quickly, and by the time I get to beryllium it is around 1%. In this way we can see that for beryllium and beyond, the two-electron cloud is basically a product state of the familiar hydrogen ion solution, properly scaled for charge.

OK, I made a mistake and I have to fix it:

My numbers look pretty good. The first four two-electron ions are H-, He, Li+, and Be++.
Using the data for ionization energies I get a constant of 16.05 eV per electron per nucleon to account for the repulsion energy. The error is 2.3 eV for helium, and the explanation is interesting.

The most naive model for the helium atom basically puts two non-interacting electrons in the same hydrogen-type ground state. The energy would then be -8 Rydbergs, or 108.8 eV. The actual energy is -79eV. So we look for a correction in terms of the electron-electron interaction.

I was doing good up to here. The mistake is coming up:

The most straightforward calculation would be to treat the two electrons as simple cloud charges and do a classical energy calculation.

Maybe that's the most straightforwad thing to do, but it's not what'd generally done. I just did it myself to check and I got an energy for the helium atom of -91.6 eV. No one gets that number, because they do something different. Yes, they do the classical cloud-vs-cloud energy calculation, but they also compensate for the screening by using an adjustable number less than 2 for the effective central potential. In other words, a two-parameter optimization.

The result of this calculation is quite close to the true value, but it is a little high, by about 2 eV. It is off because the electrons are not in a simple product state, and in fact it is the same error given by my semi-empirical formula ( -108.8 + 2 x 16.05 = 76.7).

Well, not quite. It's also off a little more because the wave function is still only approximate. But I'm almost right because most of the error remains no matter how many parameters you use to optimize, because this method essentially assumes the two electrons are in a true product state. That's where most of the error comes from.

For larger ions the error disappears rather quickly, and by the time I get to beryllium it is around 1%. In this way we can see that for beryllium and beyond, the two-electron cloud is basically a product state of the familiar hydrogen ion solution, properly scaled for charge.

Yes, for beyond beryllium the wave function approaches a true product state, but it's not as I first thought simply the product state of the familiar hydrogen waveforms. It's a modified waveform that you'd essentially get for the helium atom if you kept refining the optimization described above.

I think I've got it straight now.

No, I don't have it straight yet. Going back to the source of all knowledge (Wikipedia) I get the following information:

Calculation 1: No interaction
Energy level: -108.8 eV

Calculation 2: Z=2, include classical cloud-vs-cloud energy
Energy level: -75 eVf

(Yes, I got confused when I did this wrong, missing a factor of 2 and getting repulsion energy of 17 instead of 34.)

Calculation 2: Modify the wave functions as though Z were variable and optimize:
Minimum energy level: -77.5 for Z = 1.69

ACTUAL ENERGY LEVEL: -78.95

What all these calculations have in common is that they treat the electron-electron interaction as simple problem in classical electrostatics. I'm thinking this would be right if the wave function was a simple product state (which I believe it becomes in the limit of heavy nucleii) but I wonder if it doesn't limit your ability to get the right answer for helium. I'm going to try and check it out...

So just to check, the problem is a single potential well, centred at x=0 (eg, a harmonic well). Put two electrons in, what do they do (label a and b)?

If the electron repulsion was turned off completely:
The lowest energy state would be to put the electrons both in the groundstate of the potential well, and also in a spin singlet. (Antisymmetric spin, and symmetric spatial)
A:$$\phi_1(x_a)\phi_1(x_a)|\rm singlet\rangle$$
The spatial part sketched, would look like a circular blob centred on xa=0,xb=0.

The next lowest state, would be to have one electron in the first excited state of the single particle system, and the other in the groundstate. Obvioulsy this needs to be symmetrised, so the options are
the three options:
B:$$[\phi_1(x_a)\phi_2(x_b)-\phi_1(x_b)\phi_2(x_a)]|\rm triplet\rangle$$
or
C:$$[\phi_1(x_a)\phi_2(x_b)+\phi_1(x_b)\phi_2(x_a)]|\rm singlet\rangle$$

Now, if you consider turning on weak repulsive interactions:
State A will still be the lowest energy, but the circular blob will distort so that it begins to look like what you have called 'SYMMETRIC FUNCTION'. It must be symmetric, because the spin part is antisymmetric.

The 3 State Bs will be the next energy level. B is lower than C because the repulsive energy of interaction between the electrons is automatically smaller in the antisymmetrised spatial wave function state than the symmetrised spatial wave function state.

State B will begin to look like you 'ANTI-SYMMETRISED' state.

If the repulsive interaction is sufficiently strong, the levels might begin to cross and it won't be easy to make statements like the above.

Still working this out. I've drawn pictures of your three states (hope it shows well in the thumbnail!) and noticed something odd. I can't make state B decay into state A (triplet to ground state). If I take a superposition of the two wave functions, I get a blob that oscillates from upper left to lower right. But in the actual well, there is no net motion of charge. It's basically electrons A and B swapping positions left and right. No radiation.

On the other hand, if I mix states A and C, it radiates nicely. Now both electrons are towards the left of the box, then they move towards the right, back and forth. Energy is given off, and state C decays into state A.

Can anyone explain why the mixture of A and B doesn't radiate?

#### Attachments

• 2E_WELL.JPG
20 KB · Views: 374