# 2-electron system under compression

• A
Gold Member

## Summary:

Two electrons are put in a potential well of decreasing width. What happens to the mean distance of electrons?

## Main Question or Discussion Point

I did some calculations for the ground state energy and wave function of a system of two electrons put in a finite-depth 2D potential well. Regardless of the shape of the potential well (square or circular), the expectation value of the electron-electron distance ##\langle r_{12}\rangle = \langle\psi_0 \left.\right|r_{12}\left|\right.\psi_0\rangle## first becomes smaller when the diameter of the well is decreased. But at some point it increases again. Is this because the two electrons "give up" trying to fit in the potential well when it's too narrow, or is it more likely that I haven't set up my diffusion monte carlo parameters properly?

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anuttarasammyak
Gold Member
It seems the two electrons are in the same ground state with opposite z spins. Is it so?

Gold Member
I think the two electrons are in the same ground state with opposite z spins. Is that so?
They probably have opposite z-spins but the ground state wavefunction is not a product of two one-electron wave functions when the Coulomb interaction is included.

• anuttarasammyak
anuttarasammyak
Gold Member
Thanks. So as well as you I think high energy of close Coulomb interaction in narrow well would not allow the electrons to occupy the bound states.

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Gold Member
Thanks. So as well as you I think high energy of close Coulomb interaction in narrow well would not allow the electrons to occupy the binding-states.
Yes, when the potential well is made less wide, the expected kinetic energy of the confined electrons increases and at some point the potential barrier isn't high enough to confine them with the Coulomb repulsion and all that.

anuttarasammyak
Gold Member
In a simple 1d well case effective non dimensional potential be
$$- \frac{2mVw^2}{\hbar^2}$$
where V>0 is potential wall height and w is its width. Such ##Vw^2## evaluation could work as for well energy part in this case also.

Non dimensional Coulomb repulsion energy is similarly
$$\alpha \frac{2m\ e^2\ w}{4\pi\epsilon_0\ \hbar^2\ }$$
where non dimension ##\alpha > 1 ## approximately.

So I suppose the sum should be negative at least in order electrons are in bound states. The condition is
$$\frac{\alpha e^2}{4\pi\epsilon_0} < Vw$$
deep and/or wide enough well.

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PeterDonis
Mentor
2019 Award
Is this because the two electrons "give up" trying to fit in the potential well when it's too narrow, or is it more likely that I haven't set up my diffusion monte carlo parameters properly?
I don't see how this question is answerable if we aren't seeing your actual model.

They probably have opposite z-spins
There shouldn't be any "probably" about it. Since there's nothing to distinguish one electron from the other, the wave function has to be antisymmetric under particle exchange. That means either the spatial part is symmetric and the spin part is antisymmetric (opposite z spins), or the spatial part is antisymmetric and the spin part is symmetric (same z spins). These two cases will not have the same energy, so only one of them can be the ground state. So the spin relationship (opposite or same) should be definitely known in the ground state, not "probably" anything.

the ground state wavefunction is not a product of two one-electron wave functions when the Coulomb interaction is included.
This is true, but it doesn't change what I said above.

Gold Member
Got this problem solved myself. Thanks for input.

• kent davidge
Would you mind showing us some graphs? Sounds very interesting.

• vanhees71
Gold Member
Would you mind showing us some graphs? Sounds very interesting.
For example, one calculation had 2 electrons inside a potential well the shape of a circular disk, and potential energy jumping from 0 to 20000 Hartree when crossing the circular boundary. A parameter ##r## is a multiplier of all linear dimensions of the system, with the area of the circle being ##3.14## Bohr##^2## when ##r=1.0## and ##(3.14## Bohr##^2 )r^2## when ##r## has some other value. The ground state energy ##E_0## of the system behaves like this as a function of ##r##: The solid line is a case where the electrons have been made non-interacting and the dashed line describes one with the Coulomb force included. A function of form ##E_0 \propto r^{-2}## fits well in the non-interacting case (and it fits perfectly if the depth of the well, ##V_0##, is infinite).

The equivalent result for ##\langle r_{12} \rangle## as a function of ##r## is increasing for ##r>1.5## but decreases when going from ##r=1.0## to ##r=1.5##.

I'm not telling too much detail as I'm possibly trying to publish the results somewhere at some point. I already got myself convinced that ##\langle r_{12}## doesn't increase monotonically as a function of ##r## when the well depth ##V_0## is not infinite.

• kent davidge
Thanks. I don't see the interacting ground energy increasing in that graph, does that occur beyond 4 Bohrs?

Gold Member
Thanks. I don't see the interacting ground energy increasing in that graph, does that occur beyond 4 Bohrs?
The ##E_0 (r)## should not increase on any interval, unlike the electron-electron distance expectation value. When you make the potential well larger, the two particles can have a smaller kinetic energy and also more effectively minimize the Coulomb potential energy.

Edit: A plot of the data points for ##\langle r_{12}\rangle## as a function of ##r## looks like this for a well depth of 20000 Hartree. The quantities on the axes have dimensions of Bohr and Hartree, but can equally well be seen as dimensionless. Last edited:
I see, curious to see how the expectation value behaves for small trap radii.