Timeline for Triangular Tight Binding Model
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 2, 2023 at 16:58 | comment | added | Viking | Well, honestly I find it a bit surprising that a simple and quite symmetric Hamiltonian has such eigenvectors with no apparent symmetry. But it is what it is. | |
| Oct 2, 2023 at 15:16 | comment | added | Sid | I did check it with Mathematica. Here is the link. The first eigenvector is the same as I found it. But look at the second and the third. Do you think that is normal? | |
| Oct 2, 2023 at 14:14 | comment | added | Viking | Sometimes the results are weird and there is nothing you can do about it :) You can check (for example) with Mathematica that you have found the eigenvectors correctly. It can also help you decompose the initial state into these weird eigenvectors to figure out the time evolution. | |
| Oct 2, 2023 at 11:18 | comment | added | Sid | Yes, that is what I was asking. The eigenvector corresponding to $\lambda_1 = 2cos(\phi)$ turns out to be $\begin{bmatrix} 1\\ 1\\ 1 \end{bmatrix}$ Now, for $\lambda_2, \lambda_3$, the calculation is becoming dirty with some weird result. Is there any trick involved here for these two eigenvalues? | |
| Oct 2, 2023 at 3:08 | comment | added | Viking | The Hamiltonian that you wrote in matrix form matches the one in the basis decomposition that you wrote in the first equation of your question, if that's what you are asking. | |
| Oct 1, 2023 at 18:53 | comment | added | Sid | Oh, yes! My bad. I did check that $\lambda_2, \lambda_3$ that you wrote were in fact the eigenvalues but I am struggling to find the eigenvectors for those two eigenvalues. Now, I am doubting if I even calculated the Hamiltonian correctly. Could you have a look? | |
| Oct 1, 2023 at 18:52 | history | edited | Viking | CC BY-SA 4.0 |
added 5 characters in body
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| Oct 1, 2023 at 4:03 | history | answered | Viking | CC BY-SA 4.0 |