What is the solution for two electrons on a sphere?

[bob012345]

TL;DR Summary

Two Particles on a 3D Sphere.

The Quantum Mechanical solution for a particle on a sphere is well known. I'm looking for a treatment of two particles on a sphere where both particles are electrons. I assume it's analytically solvable. Of course, I am not expecting someone to actually solve it from scratch (unless you want to) but please point me to an existing solution or help me set up the problem. Later, I'll want to address and compare it to two particles in a spherical box. Thanks.

 

[Vanadium 50]

I assume it's analytically solvable

Why?

 

[bob012345]

Why? Well, because it's a constrained two body problem and not an unconstrained three body problem. At least I think so, otherwise please let me know. Thanks.

 

[bob012345]

It took awhile to find these references. They all look very complicated dashing my hopes it was an easily or even exactly solvable problem. I'm not sure why it isn't. Any insight would be helpful.

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.79.062517
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.97.235431
https://journals.aps.org/prb/abstract/10.1103/PhysRevB.66.235118

 

[bob012345]

These references were disappointing as they went through much unnecessary complictions overgeneralizing and formalizing the problems. I'm still looking for a straightforward treatment. Perhaps from a graduate textbook if anyone has seen one. Thanks.

 

[TeethWhitener]

help me set up the problem. Later, I'll want to address and compare it to two particles in a spherical box.

Have you tried to write down the Hamiltonian for either of the problems? Where are you getting stuck?

 

[bob012345]

Have you tried to write down the Hamiltonian for either of the problems? Where are you getting stuck?

That's one of my problems, setting up the Hamiltonian from scratch and to see if it's separable. That's why I'd like to see a textbook reference. The paper by Loos and Gill starts off ok but then quickly gets into Hartree-Fock methods which ruin it in terms of simplicity (for me that is...).

 

[TeethWhitener]

Can you write down the Hamiltonian for two non-interacting particles on a sphere?

 

[bob012345]

Can you write down the Hamiltonian for two non-interacting particles on a sphere?

I think so, it would be the Helium Hamiltonian without Coulomb terms but with constant potential energy terms and no radial radial wavefunction, except a constant. But I want them to interact at least with the Coulomb term which would be a function of relative angle as in the Loos paper. That sound correct?

 

[TeethWhitener]

I think so

So write it out and show us where you get stuck.

The reason I’m doing this is because it isn’t clear from your descriptions what, exactly, you’re having trouble with. If you can write out the Schrödinger equation for the problem but are having trouble understanding which terms make an analytical solution problematic, we can try to address that. If you can’t write down the Schrödinger equation for the problem, we can try to address that. Etc etc.

 

[bob012345]

So write it out and show us where you get stuck.

The reason I’m doing this is because it isn’t clear from your descriptions what, exactly, you’re having trouble with. If you can write out the Schrödinger equation for the problem but are having trouble understanding which terms make an analytical solution problematic, we can try to address that. If you can’t write down the Schrödinger equation for the problem, we can try to address that. Etc etc.

I can't do LaTeX from my iPad (yet). But the Hamiltonian you requested for two non-interacting particles on a sphere should be just sum of two Hamiltonians for one particle say for particles called 1 and 2. The Hamiltonian is separable and the total wavefunction is the product of each particle wavefunction. Adding the Coulomb interaction would be the next step.

 

[Vanadium 50]

Adding the Coulomb interaction would be the next step.

Which is not trivial.

 

[bob012345]

Which is not trivial.

Absolutely not trivial! But I'll give it a try.

 

[PeterDonis]

the total wavefunction is the product of each particle wavefunction

Not if the particles are bosons or fermions. The wave function has to have the correct symmetry properties under particle exchange. You said electrons in the OP, which would be fermions, so the wave function would have to be unchanged except for a sign flip under particle exchange. A simple product of two one-particle wave functions will not have this property.

 

[Vanadium 50]

A simple product of two one-particle wave functions will not have this property.

True.

Getting a little deeper, the product of two one-particle wavefunctions will not capture the physics of the situation either. What is the wavefunction of one particle? It's uniform in solid angle, by symmetry. `And the other? Same thing. And the relation of one given the other? There isn't one in this case, and that's wrong - for spin-statistics reasons (as you said) and because of the interaction Hamiltonian.

 

[Vanadium 50]

I've thought about this some more.

1. There are only a half-dozen or so analytically solvable systems in QM. Even the helium atom is not one of these.
2. Consider the variables you would be using: φ and θ (or maybe cosθ ) for each electron, and an interaction term that's a whole mess of trig. Solvable? I am not sure it's even separable!
3. You want to do two things when writing down the Hamiltonian, and they may be at odds with each other. One is to write it in such a way as to ensure that the wavefunction is antisymmetric under electron exchange as well as to reflect the global φ symmetry, and the other is to write it so the whole mess of trig is as easy to work with as possible.

That said, is there anything we can say about the ground state?

Classically, the lowest energy state has the two electrons at antipodes, and the electrons moving as slowly as possible. Is this the same solution quantum mechanically? I'd like to think so, but I am beginning to suspect that it's not. If r is the 2nd electron's distance on the sphere from the first antipode, does r commute with the total Hamiltonian? Maybe...but probably not.

 

[HomogenousCow]

Well the Hamiltonian has a symmetry under the full rotation operator, so why not do perturbation theory off of the eigenstates of the total angular momentum operator. I really doubt there are any analytical solutions for this system, the potential looks terrible.

 

[bob012345]

True.

Getting a little deeper, the product of two one-particle wavefunctions will not capture the physics of the situation either. What is the wavefunction of one particle? It's uniform in solid angle, by symmetry. `And the other? Same thing. And the relation of one given the other? There isn't one in this case, and that's wrong - for spin-statistics reasons (as you said) and because of the interaction Hamiltonian.

I thought of mentioning exchange symmetry but didn't since I felt the very first step was literally non-interacting particles which I take to include exchange interactions also. I equate exchange interactions with spin (including spin 0). I'm assuming for the problem that spin of zero is different than non existence of spin. Also, only the ground state solution for one particle on a sphere is uniform. Excited states are the normalized Spherical Harmonics. If we include spin (and still no Coulomb) and they are bosons I can make the total wavefunction symmetric and for the original assumption of electrons, anti-symmetric. This includes both the spatial and spin parts. I believe that for the ground state in the two electron on a sphere case, since the spatial anti-symmetric wavefunction is zero, there is only the symmetric spatial part times the anti-symmetric spin function so it's a Singlet state. Again, this is including exchange symmetry but not Coulomb interaction so it's just for learning purposes.

 

[PeterDonis]

the very first step was literally non-interacting particles which I take to include exchange interactions also

This is not correct. Non-interacting particles are still either fermions or bosons.

I equate exchange interactions with spin (including spin 0).

That is not correct. Exchange interactions are not spin and don't work the same as spin.

I'm assuming for the problem that spin of zero is different than non existence of spin.

"Non existence of spin" is meaningless; there is no QM mathematical model that has that property.

for the ground state in the two electron on a sphere case, since the spatial anti-symmetric wavefunction is zero

Why do you think the spatial anti-symmetric wave function is zero?

 

[bob012345]

I've thought about this some more.

1. There are only a half-dozen or so analytically solvable systems in QM. Even the helium atom is not one of these.
2. Consider the variables you would be using: φ and θ (or maybe cosθ ) for each electron, and an interaction term that's a whole mess of trig. Solvable? I am not sure it's even separable!
3. You want to do two things when writing down the Hamiltonian, and they may be at odds with each other. One is to write it in such a way as to ensure that the wavefunction is antisymmetric under electron exchange as well as to reflect the global φ symmetry, and the other is to write it so the whole mess of trig is as easy to work with as possible.

That said, is there anything we can say about the ground state?

Classically, the lowest energy state has the two electrons at antipodes, and the electrons moving as slowly as possible. Is this the same solution quantum mechanically? I'd like to think so, but I am beginning to suspect that it's not. If r is the 2nd electron's distance on the sphere from the first antipode, does r commute with the total Hamiltonian? Maybe...but probably not.

Regarding your point #2, I was thinking it was solvable because it boils down to a restricted two body problem but now I'm not sure. It may not be solvable unless the problem can be fully cast in terms of the relative separation on the sphere.

 

[bob012345]

This is not correct. Non-interacting particles are still either fermions or bosons.
That is not correct. Exchange interactions are not spin and don't work the same as spin.
"Non existence of spin" is meaningless; there is no QM mathematical model that has that property.
Why do you think the spatial anti-symmetric wave function is zero?

Thanks for correcting my loose views regarding exchange symmetry and spin. I see exchange symmetries regards the broader concept of identical particles and is a larger and different concept than just spin. Perhaps I should have said non-treatment of spin for teaching purposes such as when they treated the Hydrogen atom solution in my undergrad QM class. I can see now I would construct the anti-symmetric wavefunction for the two particles on a sphere even if not considering spin but there's no need to not consider spin since what is an electron without spin? I think the spatial anti-symmetric ground state of the two electron particles on a sphere is zero because it's the difference of two identical constants.

 

[PeterDonis]

I think the spatial anti-symmetric ground state of the two electron particles on a sphere is zero because it's the difference of two identical constants.

What constants? Please write down the actual math.

 

[bob012345]

What constants? Please write down the actual math.

The normalized ground state for one particle on sphere is the lowest Spherical Harmonic function 1/SQ[4 Pi] which is a constant. If I have two particles and I construct an anti-symmetric wavefunction from the individual ground states for the spatial part (no Coulomb), don't I get zero? How would you then define the spatial part of the ground state? Thanks. Here is my resource;

https://www.kau.edu.sa/GetFile.aspx?id=162289&fn=13 Sphere and H atom.pdf

 

[PeterDonis]

If I have two particles and I construct an anti-symmetric wavefunction from the individual ground states for the spatial part (no Coulomb), don't I get zero?

No, because the term "spatial" is misleading here. The spatial wave function of a two-particle system on a sphere is not two functions each on a 2-sphere; it is one function on a 4-dimensional space. An antisymmetric function on a 4-dimensional space is not the same as the difference of two identical functions on a 2-sphere.

This is basic quantum mechanics, and in an "A" level thread it should already be part of your background knowledge.

 

[hilbert2]

This is quite an interesting problem as a simplified model of a multielectron atom... Anyone who's more familiar with this, does it make a significant difference in the qualitative features of the energy eigenfunctions, whether I define the electron-electron distance $r$ in the $e^2 /4\pi\epsilon_0 r$ term as the shortest path that stays on the sphere, or as the euclidean distance in the 3D space that the sphere is embedded in?

 

[HomogenousCow]

This is quite an interesting problem as a simplified model of a multielectron atom... Anyone who's more familiar with this, does it make a significant difference in the qualitative features of the energy eigenfunctions, whether I define the electron-electron distance $r$ in the $e^2 /4\pi\epsilon_0 r$ term as the shortest path that stays on the sphere, or as the euclidean distance in the 3D space that the sphere is embedded in?

That's an interesting thought. If we used the geodesic distance instead that could significantly simplify calculations, still not enough to make it exactly solvable but I think it would make perturbation terms very easy to calculate. You could probably exploit the rotational symmetry of the problem and "put" one of the spherical harmonics at the north pole, then the geodesic distance to the other harmonic is simply some multiple of the polar angle. That seems very do-able to arbitray orders in perturbation theory. However the functions you integrate over do need to be eigenstates of the total angular momentum operator so I don't know if that spoils it hm..

 

[bob012345]

No, because the term "spatial" is misleading here. The spatial wave function of a two-particle system on a sphere is not two functions each on a 2-sphere; it is one function on a 4-dimensional space. An antisymmetric function on a 4-dimensional space is not the same as the difference of two identical functions on a 2-sphere.

This is basic quantum mechanics, and in an "A" level thread it should already be part of your background knowledge.

I disagree it's misleading as I'm getting that from the treatment of the Helium ground state in my old QM text. I used the "A" level because I think that's the level of the problem not because it describes my abilities as such but I'm doing my best. :)

 

[PeterDonis]

I disagree it's misleading as I'm getting that from the treatment of the Helium ground state in my old QM text.

Then please give a specific reference and show your work. So far you have not written a single bit of math in this thread. That's not a good sign for an "A" level discussion.

I used the "A" level because I think that's the level of the problem not because it describes my abilities as such

I agree it's the level of the problem, but if you don't have the requisite background then perhaps you should take the time to develop it before trying to tackle this problem.

 

[bob012345]

Then please give a specific reference and show your work. So far you have not written a single bit of math in this thread. That's not a good sign for an "A" level discussion.
I agree it's the level of the problem, but if you don't have the requisite background then perhaps you should take the time to develop it before trying to tackle this problem.

We are so far discussing how to set up the problem and I've also been spending many hours in my books preparing my answers but I don't know how to do LaTeX yet. If I have to be an expert on solving this problem before I can ask about it here, I might not need to come here but I came here because I do need help. I sense your advice is just to become that expert first. But as I'm interacting here, I do feel I'm gaining confidence and learning. The ground state of two electrons on the sphere is discussed in the Loos paper but it's not crystal clear how they got there to me. You will have to give me time to put up my math if you insist that be done before I can get more help.

 

[TeethWhitener]

https://arxiv.org/abs/1004.3641If anyone’s interested in the state of the problem.

 

[PeterDonis]

If I have to be an expert on solving this problem before I can ask about it here,

Not an expert in solving this specific problem, but it's not clear that you have a general "A" level background in quantum mechanics. Without that you might not be able to get far on this specific problem no matter how much help we give you; it would amount to us just solving the problem and telling you the solution.

 

[PeterDonis]

You will have to give me time to put up my math

That's fine, nobody is under time pressure here. But if you are going to make claims like "the antisymmetric part of the two-particle spatial wave function is zero", you need to back them up with math.

 

[bob012345]

Not an expert in solving this specific problem, but it's not clear that you have a general "A" level background in quantum mechanics. Without that you might not be able to get far on this specific problem no matter how much help we give you; it would amount to us just solving the problem and telling you the solution.

Please define exactly what an "A" level is. I've had a lot of QM at the University of Illinois.

 

[PeterDonis]

lease define exactly what an "A" level is.

Graduate level knowledge of the subject matter. See here:

https://www.physicsforums.com/help/threadprefixes/

 

[HomogenousCow]

https://arxiv.org/abs/1004.3641If anyone’s interested in the state of the problem.

So as one would expect, the groundstate does correspond to the classical antipodal configuration.