Timeline for Why is an operator the quantum mechanical analogue of an observable?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 2, 2019 at 12:00 | history | tweeted | twitter.com/StackPhysics/status/1123920277224685573 | ||
| Apr 24, 2019 at 22:12 | comment | added | march | I feel like the title and the text don't match. The title is a kind of deep question that requires us to work backward from experiments to the mathematical structure that allows us to predict the outcomes of those experiments. The text is more about understanding the structure itself rather than understanding where it comes from. | |
| Apr 24, 2019 at 18:04 | answer | added | eranreches | timeline score: 0 | |
| Apr 24, 2019 at 14:11 | history | became hot network question | |||
| Apr 24, 2019 at 13:40 | comment | added | FGSUZ | But all functions we deal with can be written as a linear combination of plane waves. | |
| Apr 24, 2019 at 13:17 | answer | added | yuggib | timeline score: 11 | |
| S Apr 24, 2019 at 12:28 | history | suggested | ohneVal | CC BY-SA 4.0 |
Title and text improvements
|
| Apr 24, 2019 at 12:24 | answer | added | Andrew Steane | timeline score: 2 | |
| Apr 24, 2019 at 12:21 | comment | added | gented | Essentially because experimentally one can see that energy levels (or angular momentum or the like) are discretised and happen to be exactly spaced as eigenvalues of some matrices. This implies that more or less the theory must be a linear operator theory on a Hilbert space. | |
| Apr 24, 2019 at 12:07 | review | Suggested edits | |||
| S Apr 24, 2019 at 12:28 | |||||
| Apr 24, 2019 at 11:31 | history | asked | sangstar | CC BY-SA 4.0 |