Timeline for Operator norm of creation and annihilation operators
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2013 at 9:54 | vote | accept | Stephan Kulla | ||
| Jul 7, 2013 at 9:33 | answer | added | Vibert | timeline score: 10 | |
| Jul 6, 2013 at 18:34 | comment | added | user10851 | @Vibert I second the call for that to be an answer. There's really nothing left to say. | |
| Jul 6, 2013 at 17:07 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 4 characters in body; edited tags
|
| Jul 6, 2013 at 16:54 | history | edited | dmckee --- ex-moderator kitten | CC BY-SA 3.0 |
latex improvement
|
| Jul 6, 2013 at 16:42 | review | First posts | |||
| Jul 6, 2013 at 18:46 | |||||
| Jul 6, 2013 at 16:30 | comment | added | Stephan Kulla | @Vibert: Thanks for the quick answer. If you want, you can write an quick answer, so I can accept it. | |
| Jul 6, 2013 at 16:28 | comment | added | Vibert | No. Consider the one-particle sector, with states $|n\rangle$ having occupation number $n$. Then $a |n\rangle = \sqrt{n} |n-1\rangle,$ so $\|a |n\rangle \| = \sqrt n.$ This means that $a$ is not bounded. | |
| Jul 6, 2013 at 16:24 | history | asked | Stephan Kulla | CC BY-SA 3.0 |