Timeline for When studying the hydrogen atom, why do we seek simultaneous eigenfunctions of $\hat{L}^2$, $\hat{L}_z$, and $\hat{H}$?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 1, 2020 at 13:13 | comment | added | Mr. Palomar | Thanks for the response, it makes sense to me now. | |
| Aug 1, 2020 at 11:43 | comment | added | Philip | @Mr.Palomar I'm not completely sure I understand, but yes to the first part. The question as to what determines a CSCO isn't (as far as I know) one that can be solved entirely mathematically. If we had "ignored" spin and thought that a system could be completely specified by the eigenvalues $| nlm\rangle$, then we'd be surprised by what happens when we place the atom in an external magnetic field: we would get more splittings in the spectral lines than expected! (Indeed this is why the "true" Zeeman effect is called "anomalous": it couldn't be explained without introducing Spin.) | |
| Aug 1, 2020 at 11:24 | comment | added | Mr. Palomar | Thanks for the response. The question whether a set of operators forms a CSCO isn't merely a mathematical one, correct? For instance, when adding spin, what we're doing is we enlarge the ambient Hilbert space (tensoring with $\mathbb{C}^2$, I guess) so that there's more 'space' for an additional operator. Had we not enlarged it in the first place, $\big\{\hat{H},\hat{L}^2,\hat{L}_z\big\}$ would've been a perfectly fine CSCO. Do I understand that correctly? | |
| Aug 1, 2020 at 10:10 | history | answered | Philip | CC BY-SA 4.0 |