Timeline for Canonical commutation relations-bounded vs. unbounded picture
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Aug 4, 2020 at 7:44 | comment | added | MaoWao | To reiterate what Christian Remling said: You have to define the operator $[Q,P]$. If $P$ and $Q$ are not everywhere defined, it is not obvious what their commutator is supposed to be. What is its domain? Do you want it to be closable and densely defined? | |
| Aug 3, 2020 at 17:35 | history | edited | truebaran | CC BY-SA 4.0 |
added 936 characters in body
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| Aug 3, 2020 at 14:10 | comment | added | Christian Remling | (1) has no immediate meaning for unbounded operators, so one has to give a precise interpretation, and one way to proceed would be (2) (until the last few lines of your post, I thought this is what you were doing). Of course, if this path is taken, then your question disappears. In any event, your question isn't precise until you clarify what (1) means for unbounded operators. | |
| Aug 3, 2020 at 13:26 | review | Close votes | |||
| Aug 6, 2020 at 17:26 | |||||
| Aug 3, 2020 at 13:06 | comment | added | Francois Ziegler | Does this answer your question? The Stone-von Neumann theorem without exponentials? | |
| Aug 3, 2020 at 12:53 | history | asked | truebaran | CC BY-SA 4.0 |