Timeline for What is the physical states in Heisenberg picture?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 20, 2016 at 9:20 | vote | accept | Shing | ||
| May 20, 2016 at 9:20 | comment | added | Shing | I see it now... indeed, we can't take measurement as the physical states of the particles. Thanks for the elaboration :) | |
| May 19, 2016 at 17:26 | comment | added | OON | @Shing I think there's very simple idea that helps very much to get various "strange" stuff about quantum theory (like e.g. about identical particles). Classically we used to think about observables as the properties of the system. But when we do quantum theory the observables are the measurements we do. | |
| May 19, 2016 at 17:14 | comment | added | OON | @Shing For starters, I can use the same operators no matter what was the initial state of the system (and thus the corresponding Schrodinger state at that time) | |
| May 19, 2016 at 8:40 | comment | added | Shing | Thanks for answering, I am quite satisfied by your answer, but I do not understand how operators themselves can't be physical states? | |
| May 18, 2016 at 12:16 | history | answered | OON | CC BY-SA 3.0 |