Timeline for Decomposing $|\Psi \rangle$ into the $|L=1 \, m_z \rangle$ basis
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 10, 2020 at 8:40 | comment | added | Davide Morgante | @TheoreticalMinimum Yep, you could. In the beginning I was thinking of an hydrogen-like problem so that the kets I've used made the most sense. But clearly it's just a matter of notation! | |
| Jun 9, 2020 at 23:14 | comment | added | user224659 | @DavideMorgante I think this notation is a bit misleading.. one should at least put $|n,l\rangle$ in the front braket. | |
| Jun 9, 2020 at 21:41 | comment | added | Davide Morgante | I'm happy it helped! | |
| Jun 9, 2020 at 21:38 | comment | added | ChemiCalChems | Ok, seems nice! Thank you so much! | |
| Jun 9, 2020 at 21:37 | comment | added | Davide Morgante | Ohw ok, then it's perfectly fine. But still you cannot identify a ket $|n\rangle$ with a wavefunction. Since you're only interested in the radial part, and so on the angular momenta, you can just call it $|n\rangle$ and leave it at that since you don't have the specifics of the radial part. It'll just count as a placeholder for it. | |
| Jun 9, 2020 at 21:34 | comment | added | ChemiCalChems | It's not defined to be an electron in a hydrogen-like atom, just the wavefunction of a spinless particle. | |
| Jun 9, 2020 at 21:31 | comment | added | Davide Morgante | If I'm not mistaken there could be an error in the definition of the $\Psi$ wavefunction since, on the top of my head, the only radial solution proportional to $e^{-r}$ should be $R_{10}$ which is in contrast with the $l=1$ of the angular part. But I could be wrong here | |
| Jun 9, 2020 at 21:28 | comment | added | Davide Morgante | Well that's not good enough. You should check the radial wavefunction solutions to the hydrogen atom $R_{nl}(r)$ which will depend on the $n$ quantum number. When you find the right $R_{nl}$ then you'll have the $n$ to write down the associated ket $|n\rangle$ | |
| Jun 9, 2020 at 21:23 | vote | accept | ChemiCalChems | ||
| Jun 9, 2020 at 21:23 | comment | added | ChemiCalChems | I've been knocking myself in the head for far too long, this makes sense. Just one question, I assume I'd be defining $|n\rangle = e^{-r}$ myself, right? | |
| Jun 9, 2020 at 21:22 | history | answered | Davide Morgante | CC BY-SA 4.0 |