Potential well with two electrons

In summary, the conversation is discussing the best way to ask a question about the energy levels of electrons in a potential well. The question is whether to ask about spinless, non-interacting electrons or to consider the unwanted complications of spin and repulsion. The answer is that the wavefunction for two non-interacting electrons can be found using the Slater determinant, and the charge density for two electrons is additive.
  • #1
110
0
I've been meaning to ask this question for a while but I thought I'd google it first. I found out that this question was put to this forum a few years ago by peter038, without a definite resolution. So I thought maybe we could try again. Part of the problem is to decide exactly what is the best question to ask.

Where do we start? We want to ask the simplest possible question and yet we don't want to make it TOO simple. We know that electrons have spin 1/2 (so two of them can occupy the same state) and we know that they repel each other. Both of these are unwanted complications; so the temptation is to ask the question for spinless, non-interacting electrons. Is this a good question to ask? My guess is that any number of such particles would then be able to fit into the ground state. This partially defeats the purpose of my question, since what I am really interested in is how the energy levels fill up one after another. But even if that's the case, I still have to start by asking the question: what does the wave function look like for two spinless, non-repelling electrons in a potential well, if one of them is in the ground state and the other is in the first excited state?
 
Physics news on Phys.org
  • #2
If you neglect interactions between the two electrons, then the wavefunction of the combined system is found by computing the slater determinant, and is a pretty easy task. These wavefunctions also form a complete set for the interacting system, and you can do perturbation ordinary theory to find the interacting result.
 
  • #3
Ok. I looked up the Slater determinant and basically for the two-particle case it tells me to take the product of the two single-particle wave functions as the independent probability of the particles being at x1 and x2...then, reverse the polarity for the case of the particles being swapped and add the two amplitudes. So for the two particles in a 1-d box occupying the ground state and the 1st excited state, I get something like
sinx sin2y - sin2x siny

How do I calculate where the charge is? I don't want to know "there is such-and-such probability of this electron here and that electron there...". I'd like a charge density just like the one I know how to calculate in the single-particle case. Does this exist in the present example?
 
  • #4
The charge density for two electrons is additive. So for a product wf [tex]\psi(r_1, r_2) = \phi_1(r_1) \phi_2(r_2)[/tex], the charge density is proportional to [tex]\left|\phi_1(r)|^2 +\left|\phi_2(r)|^2[/tex]
 

Suggested for: Potential well with two electrons

Replies
7
Views
689
Replies
15
Views
1K
Replies
5
Views
733
Replies
1
Views
551
Replies
7
Views
309
Replies
3
Views
583
Replies
2
Views
429
Replies
4
Views
234
Replies
5
Views
654
Replies
2
Views
473
Back
Top