Timeline for Eigenfunctions of compatible observables that are not shared
Current License: CC BY-SA 4.0
9 events
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| May 17, 2020 at 1:08 | comment | added | user87745 | @Mew It's not as edgy a case as you might imagine ;) In fact, we rarely care about compatible observables if both of them are non-degenerate--because we'd automatically know about the value of the other by simply measuring one. Think about it, how many examples can you recall where we use two compatible observables with both having non-degenerate spectra? About your other point, yes, but it would only work in one direction, if you measure $m=0$ first, that wouldn't tell you anything about $l$. Also, getting lucky means that you happen to land in a non-degenerate eigen-subspace of an operator. | |
| May 17, 2020 at 0:55 | comment | added | Mew | That's actually how I imagined it when I asked the question, so it turns out that train of thought is an edge case after all. To add to what you said, I think that could also take place with degenerate spectra, but you'd need to get lucky: e.g., for $Y^m_l$, having $l$ measure to be $0$ would mean the system must be, phrased as generally as possible, in a "superposition of $l=0$ states", which can of course only be $Y^0_0$, an eigenstate of both observables $l$ and $m$. | |
| May 17, 2020 at 0:29 | comment | added | user87745 | +1: Excellently thorough answer! @Mew Notice that the fact that measuring one operator does not bring the state to an eigenstate of the second operator has everything to do with the degeneracy of the spectra of the operators. If you have two compatible operators and none of their spectra is degenerate then measuring the one would necessarily bring the state to an eigenstate of both the operators. | |
| Apr 26, 2020 at 16:45 | comment | added | Mew | Right. Key take-away, based on your closing sentence: compatibility of two observables does not mean that measuring the first brings the system into an eigenfunction of the second and vice versa. Instead, it means that measuring the second after measuring the first (and vice versa) doesn't invalidate the first of both measurements. | |
| Apr 26, 2020 at 16:43 | vote | accept | Mew | ||
| Apr 26, 2020 at 16:15 | comment | added | Jahan Claes | @Mew Edited, maybe it answers both your questions now? | |
| Apr 26, 2020 at 16:15 | history | edited | Jahan Claes | CC BY-SA 4.0 |
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| Apr 26, 2020 at 15:48 | comment | added | Mew | Thanks for the input! I'm not accepting the answer just yet, as I want my second question to be answered as well. Also, although I'm only taking my first course in QM, I have to say I don't really dig Griffiths all that much. He's kind of all-over-the-place and doesn't refer to rigorous mathematics as often as he should (e.g. Sturm-Liouville theory), and he tends to use problems and examples as theorems. It also takes him just short of 100 pages to get to observables. | |
| Apr 25, 2020 at 17:36 | history | answered | Jahan Claes | CC BY-SA 4.0 |