Timeline for Physical meaning of a complete set of compatible observables
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 7 at 23:36 | comment | added | FlatterMann | From an experimentalist's point of view the problem isn't even well posed. If I may throw my favorite ball rolling down a hill on an infinite plane into the ring: what does an observable mean in this case? Once the ball has rolled off the hill and is on the plane it's performing a free 2d motion. The z-axis is no longer an observable. Sounds stupid? Maybe, but isn't that exactly the problem of future cosmologists who won't be able to observe the big bang any longer? How does theory get around the dynamic selection of observables by nature in a rational way? | |
| Nov 7 at 18:18 | answer | added | mmfrgmpds | timeline score: 0 | |
| Oct 27 at 7:23 | history | became hot network question | |||
| Oct 27 at 7:08 | answer | added | Valter Moretti | timeline score: 9 | |
| S Oct 26 at 20:05 | history | suggested | Lee Mosher | CC BY-SA 4.0 | spelling and formatting |
| Oct 26 at 19:38 | review | Suggested edits | |||
| S Oct 26 at 20:05 | |||||
| Oct 26 at 14:23 | answer | added | ZeroTheHero | timeline score: 6 | |
| Oct 26 at 11:20 | answer | added | BioPhysicist | timeline score: 2 | |
| Oct 26 at 5:38 | comment | added | naturallyInconsistent | I have absolutely no idea what you mean by maximal, not least because Tobias above mentioned that it should be a minimal set. But consider the H atom that has the LRLP vector as an extra symmetry. Then you could have too many symmetries and their conserved quantities, i.e. too many labels, and that would make it annoying. In that case we usually just ignore some extra labels and make do with the minimal set, which is then nicer to work with. | |
| Oct 25 at 19:23 | comment | added | Tobias Fünke | There is no real "physical meaning" a priori. It is just that a CSCO can yield a nice way to label an orthonormal basis. Think of the hydrogen atom for example.--And, depending on the definition one employs, one calls a CSCO a set which is minimal; that is, if one removes one operator from that set, the remaining set does not yield a unique orthonormal basis, cf. Galindo and Pascual, QM book Vol I. | |
| Oct 25 at 18:54 | history | edited | Qmechanic♦ | CC BY-SA 4.0 | edited tags |
| Oct 25 at 18:51 | history | asked | Steppenwolf | CC BY-SA 4.0 |