Timeline for How to get the position operator in the momentum representation from knowing the momentum operator in the position representation?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 22, 2018 at 19:56 | answer | added | Voulkos | timeline score: 2 | |
| Nov 24, 2013 at 6:44 | answer | added | Art Brown | timeline score: 7 | |
| Nov 23, 2013 at 23:04 | answer | added | Qmechanic♦ | timeline score: 6 | |
| Nov 18, 2013 at 7:59 | answer | added | David Z | timeline score: 7 | |
| Nov 17, 2013 at 20:25 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
added 52 characters in body; edited title
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| Nov 17, 2013 at 20:17 | answer | added | user26143 | timeline score: 21 | |
| Nov 17, 2013 at 19:21 | answer | added | joshphysics | timeline score: 26 | |
| Nov 17, 2013 at 10:03 | comment | added | Qmechanic♦ | One cannot derive $\hat{x}=ih \frac{\partial}{\partial p}$ just from the formula $\hat{p}=-ih \frac{\partial}{\partial x}$ alone. For instance $\hat{x}=\hat{c}+ih \frac{\partial}{\partial p}$ is also consistent, where $\hat{c}$ is a Casimir operator (e.g. proportional to the identity operator ${\bf 1}$). One needs to make further (conventional) assumptions in order to derive eq. (2). See e.g. this, this, and possibly this Phys.SE post. | |
| Nov 17, 2013 at 5:22 | review | Close votes | |||
| Nov 18, 2013 at 4:23 | |||||
| Nov 17, 2013 at 3:52 | review | First posts | |||
| Nov 17, 2013 at 5:03 | |||||
| Nov 17, 2013 at 3:33 | history | asked | user34032 | CC BY-SA 3.0 |