Fresh post! Bound for more discussion and more comments. Three electron three well matrix formulation

*(Don't stare on the picture too long!)

On the left of the picture above you can see the basis I used to create the 20x20 matrix much like the two electron two well system.This also is the paper I derived the matrix ( I have checked it multiple times) with the squares being the overlap integral and the squares with a asterisk are the negative ones. As you can see it's hermitian just as expected from the hamiltonian.

A hard part of this task was to determine when to put a minus sign in front of a transition. I have no idea why we put the minus sign before but I came up with a pattern and it seems like it works. I had a positive contribution I when \( \uparrow,.. \rightarrow ..,\uparrow\) and \( ..,\downarrow \rightarrow \downarrow,.. \) . Negative contribution -I when \( ..,\uparrow \rightarrow \uparrow,.. \) and \( \downarrow,.. \rightarrow ..,\downarrow \) . Also \( \uparrow\downarrow,.. \rightarrow \downarrow,\uparrow \) is negative I and \( \uparrow\downarrow,.. \rightarrow  \uparrow,\downarrow\) is positive I . I know this seems weird but I needed to come up with a pattern that applies in all cases i encountered and that incorporates the pattern used for the 2 electron 2well system .

Since its a huge matrix, I decided to plug in U=4ev and I= 1ev to get numerical values as solutions.(Note:since its a 20x20 matrix, it has 20 eigenvectors and eigenvalues.) Bellow you can see our beautiful hamiltonian H :

0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0
0  0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  1  0  0  0
0  0  0  0  0  0  0  0  0  0  1  0 -1  0  1  0 -1  0  0  0
0  0  0  0  0  0  0  0  0  0 -1  0  0  0 -1  0  0  0  0  0
0  0  0  0  0  0  0  0  0  0  0  1  0  0  0  1  0  0  0  0
0  0  0  0  0  0  0  0  0  0  0 -1  0  1  0 -1  0  1  0  0
0  0  0  0  0  0  0  0  0  0  0  0  0 -1  0  0  0 -1  0  0
0  0  0  0  0  0  0  0  4  0 -1  0 -1  0  0  0  0  0  0  0
0  0  0  0  0  0  0  0  0  4  0 -1  0  1  0  0  0  0  0  0
0  0  0  1 -1  0  0  0 -1  0  4  0  0  0  0  0  0  0  0  0
0  0  0  0  0  1 -1  0  0 -1  0  4  0  0  0  0  0  0  0  0
0  0  1 -1  0  0  0  0 -1  0  0  0  4  0  0  0  0  0  0  0
0  0  0  0  0  0  1 -1  0  1  0  0  0  4  0  0  0  0  0  0
0  0  0  1 -1  0  0  0  0  0  0  0  0  0  4  0  0  0  1  0
0  0  0  0  0  1 -1  0  0  0  0  0  0  0  0  4  0  0  0 -1
0  0  1 -1  0  0  0  0  0  0  0  0  0  0  0  0  4  0  1  0
0  0  0  0  0  0  1 -1  0  0  0  0  0  0  0  0  0  4  0 -1
0  0  0  0  0  0  0  0  0  0  0  0  0  0  1  0  1  0  4  0
0  0  0  0  0  0  0  0  0  0  0  0  0  0  0 -1  0 -1  0  4

( to try out your own I and U just use the form above and replace the values accordingly)

Eigenvalue/Eigenvector solutions and comparison with two electron two well system
Just like the 2 electron 2 well case we get two 0 eigenvalues for the eigenvectors:
(1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)=\( \downarrow, \downarrow, \downarrow \)
  and
(0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0) =\( \uparrow, \uparrow, \uparrow \)

Now lets start from the highest and lowest energy eigenvectors/eigenvalues

The highest energy,  E=5.819 eV I get the state:

(0,0,0,0,0,-0.122,0.244,-0.122,0,0.559,0,-0.509,0,0.509,0,-0.201,0,0.201,0,0)
Eigenvector,\( \mathbf{e}=-0.122 \cdot \uparrow, \downarrow, \downarrow +0.244 \downarrow, \uparrow, \downarrow-0.122\downarrow, \downarrow, \uparrow\)
\( + 0.559 \uparrow \downarrow, \downarrow, .. -0.509 \uparrow \downarrow, .., \downarrow+0.509 \downarrow, \uparrow \downarrow, ... -0.201 ..,\uparrow\downarrow, \downarrow + 0.201 \downarrow,..,\uparrow\downarrow \)

For comparison, in my older post with the derivation of the two electron two well system (also with I=1ev, U=4eV) I got a highest energy eigenvalue E=5eV.
with eigenvector,\(\mathbf{e}= 0.957\uparrow\downarrow,..+0.116 \uparrow, \downarrow -0.116 \downarrow, \uparrow+0.957 ..,\uparrow\downarrow\)

The lowest energy, E=-1.2 eV we get the eigenstate:

(0,0,0,0,0,0.365,-0.730,0.365,0,-0.088,0,-0.228,0,-0.211,0,2.11,0,0)
Eigenvector \( \mathbf{e}= 0.365 \uparrow, \downarrow, \downarrow - 0.730 \downarrow, \uparrow, \downarrow +0.365 \downarrow,\downarrow,\uparrow \)
\(-0.88 \uparrow \downarrow, \downarrow, .. - 0.228 \uparrow \downarrow, .. , \downarrow+0.228 \downarrow, \uparrow \downarrow, .. -0.211 ..,\uparrow\downarrow,\downarrow +0.211 \downarrow,..,\uparrow\downarrow \)

In my old post I also calculated the minimum eigenvalue,eigenvector with an eigenvalue of E=-1eV.
with eigenvector:\(\mathbf{e}=-1/9\uparrow\downarrow,.. +0.5\uparrow, \downarrow -0.5 \downarrow, \uparrow -1/9..,\uparrow\downarrow\)

As you can see that both lowest energy eigenstates do not completely avoid states with coulomb repulsion but still have the lowest energy. However we mentioned in class that for the 2\(e^-) 2well system the energy eigenstate has net spin 0 .  The 3\(e^-) 3well system does not have net spin 0, raising the following question: Is our previous assertion that there is a relationship between states we have obtained for the 2 electron system and anti-ferromagnetism well founded?
*Not sure if important:*
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 As a sanity check of my results, I wanted to mention that I did have a state of E=U=4eV for the three electron three well system that has the same energy as a eigenstate of the two electron system.

For 3e 3w system:
E=4eV,
\(\mathbf{e}= -1/2 \uparrow\downarrow,\downarrow,.. + 1/2 \uparrow\downarrow,..,\downarrow\)
\(+1/2 \downarrow, \uparrow\downarrow,.. - 1/2 ..,\uparrow\downarrow,\downarrow\)

For 2e 2w system:
E=4eV  with eigenvector:
=\(\mathbf{e}= \frac{1}{\sqrt{2}}[-(\uparrow\downarrow,..)+(..,\uparrow\downarrow)] \)
=\(.707[-(\uparrow\downarrow,..)+(..,\uparrow\downarrow)]\)
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If you are interested in more states other than the ones I already provided, you can read the next part of the post or explore them yourself using the basis and the results of the eigenvalues/eigenvector calculator in the end of the post.

Layout and Comparison of 4 lowest energy eigenstates

E=-1.2 eV

\(\mathbf{e}= 0.365 \uparrow, \downarrow, \downarrow - 0.730 \downarrow, \uparrow, \downarrow +0.365 \downarrow,\downarrow,\uparrow \)
\(-0.088 \uparrow \downarrow, \downarrow, .. - 0.228 \uparrow \downarrow, .. , \downarrow
+0.228 \downarrow, \uparrow \downarrow, .. -0.211 ..,\uparrow\downarrow,\downarrow
+0.211 \downarrow,..,\uparrow\downarrow \)

First note the high amplitude on the state without coulomb repulsion \(-0.730\downarrow, \uparrow, \downarrow\) which is connected via one hop to four coulomb repulsion states \(- 0.228 \uparrow \downarrow, .. , \downarrow\),\(
+0.228 \downarrow, \uparrow \downarrow, ..\),\( -0.211 ..,\uparrow\downarrow,\downarrow\) and
\(+0.211 \downarrow,..,\uparrow\downarrow\). Now note that 0.365 \( \uparrow, \downarrow, \downarrow\) and \(0.365 \downarrow,\downarrow,\uparrow\) have also a considerable amplitude and they do have their corresponding one hop coulomb repulsion pairs \(-0.228 \uparrow \downarrow, .. , \downarrow\),\(-0.211 ..,\uparrow\downarrow,\downarrow\)   and \(0.228 \downarrow,\uparrow\downarrow,..\), \( 0.211\downarrow,..,\uparrow\downarrow \) respectively. The above observations can yield the conclusion that the state \(\downarrow, \uparrow, \downarrow\) is special and has the highest amplitude because of it's ability to tunnel to four states. The overall form of the eigenvector allows tunneling between many states, even ones with coulomb repulsion, since the decrease in kinetic energy via reduction of confinement counteracts the coulomb repulsion. Ultimately note the very small amplitude state \(-0.088 \uparrow \downarrow, \downarrow,..\) that is only connected via one hop to \( \downarrow, \uparrow \downarrow,.. \). This last state will be discussed further under the next lowest state.

E=-1.162eV

\(\mathbf{e}= -0.369 \uparrow,\uparrow,\downarrow + 0.738 \uparrow, \downarrow, \uparrow -0.369 \downarrow, \uparrow, \uparrow\)
\( - 0.214 \uparrow\downarrow,..,\uparrow 0.214 \uparrow,\uparrow\downarrow,.. - 0.214 .., \uparrow\downarrow, \uparrow+ 0.214 \uparrow,..,\uparrow\downarrow \)

Note that this eigenvector is also composed of a high amplitude on the state without coulomb repulsion \( 0.738 \uparrow, \downarrow, \uparrow\) that is connected by one hop to the four states \(- 0.214 \uparrow\downarrow,..,\uparrow, 0.214 \uparrow,\uparrow\downarrow,.. ,- 0.214 .., \uparrow\downarrow, \uparrow\) and \(+ 0.214 \uparrow,..,\uparrow\downarrow\) . The eigenvector also has \(-0.369 \uparrow,\uparrow,\downarrow\) and \(-0.369 \downarrow, \uparrow, \uparrow \) with their one hop double tunneling partners \(0.214 \uparrow,\uparrow\downarrow,..\), \(0.214 \uparrow,..,\uparrow\downarrow\) and  \(- 0.214 \uparrow\downarrow,..,\uparrow\), \(- 0.214 .., \uparrow\downarrow\) respectively. This is strikingly similar to the lowest energy eigenstate but with up and down reversed. These tunneling relationships seem to be the key of what makes these two first states have so much lower energy!(Note:It turns out that including a state that has access to four electron states greately decreases the energy)

Another important note is that the lowest energy state had a small \(-0.088 \uparrow \downarrow, \downarrow,..\) state which seems to be the only striking difference between the two lowest energy states (everything else is the same if you reverse up and down directions). It seems to be the only explanation why the previous state has lower energy than this one. It allows the tunneling to a further state that must decrease the degree of confinement and hence the kinetic energy of the system.

E=-0.494eV

\(\mathbf{e}= 0.664 \uparrow,\uparrow,\downarrow -0.664 \downarrow,\uparrow,\uparrow\)
\(- 0.073\uparrow\downarrow,\uparrow,..-0.164 \uparrow\downarrow,.., \uparrow-0.164 \uparrow,\uparrow\downarrow,..-0.164..,\uparrow\downarrow, \uparrow- 0.164 \uparrow,..,\uparrow\downarrow+0.073 ..,\uparrow,\uparrow\downarrow \)

First note that this energy is higher from the previous one by 0.668 eV and that the two eigenstates discussed earlier have a difference of only 0.038 eV (only around 6% of 0.668eV). The first question to ask is why? So lets see the differences.

In contrast to the states presented previously, it does not include a high amplitude state that can tunnel to four different states via one hop. Also, it looks a lot like the previous eigenvector if one removed the contribution of the high amplitude \( \uparrow, \downarrow, \uparrow \) state and then renormalized accordingly. It can then be inferred that excluding the state that has access to four electron states increases the energy.

 More specifically, the eigenvector includes two high amplitude states, \( 0.664 \uparrow,\uparrow,\downarrow -0.664 \downarrow,\uparrow,\uparrow\) which can tunnel with one step only to two states each. More specifically, \(0.164\uparrow,..,\uparrow\downarrow \),\(0.164 \uparrow,\uparrow\downarrow,..\) and \(-0.164 \uparrow\downarrow,.., \uparrow\),\(-0.164..,\uparrow\downarrow, \uparrow \).

Finally the tiny amplitude contributions \(0.073 ..,\uparrow,\uparrow\downarrow\) and \(- 0.073\uparrow\downarrow,\uparrow,..\) seem to be similar to the small amplitude state in the lowest energy eigenvector . They reduce the energy allowing further tunneling although they include coulomb repulsion!

E=-0.472eV

\( \mathbf{e}=0.669 \uparrow,\downarrow,\downarrow -0.669 \downarrow,\downarrow,\uparrow \)
\(-0.150 \uparrow\downarrow,..,\downarrow -0.150 \downarrow, \uparrow\downarrow,.. -0.166 .., \uparrow\downarrow,\downarrow -0.166 \downarrow,..,\uparrow\downarrow-0.074 .., \downarrow, \uparrow\downarrow \)

The eigenstate above is related to the previous one just like the lowest with the next lowest energy eigenstate. It would be the same as the previous one if you switched the orientation of down and up (also noting that amplitudes are similar) however it differs in the small amplitude states.  This state has only one small amplitude state that includes coulomb repulsion \(-0.074 .., \downarrow, \uparrow\downarrow\), in contrast with the two states \(0.073 ..,\uparrow,\uparrow\downarrow\) and \(- 0.073\uparrow\downarrow,\uparrow,..\) we had before. I think that the 0.22 eV change in energy is solely because the two small amplitude states lower the kinetic energy and avoid coulomb repulsion more efficiently than the single amplitude state.

The two last eigenstates seem to have low energies using the same principle as the first two, tunneling. I think that they do a lousy job compered to the first two eigenstates but still get to have energy lower than \(E_0=0eV\).
Available info and solutions:
This is the basis used