Timeline for Another derivation of canonical position-momentum commutator relation
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2016 at 5:13 | comment | added | knzhou | With a little practice, though, you'll be able to do this as fast as you can write. It looks like there are a lot of rules, but I already used almost every rule there is in my answer! There's not too much to learn beyond that. | |
| Sep 5, 2016 at 5:13 | comment | added | knzhou | @VladimirVargas Yeah, usually people start with flippiefanus's picture. (Note, however, that his expression $\hat{p} = -i\hbar \partial_x$ only works if all functions are position-space coefficients, which can be confusing if you want to, say, switch between position space and momentum space.) | |
| Sep 5, 2016 at 5:08 | comment | added | Vladimir Vargas | Oh, I see! Maybe practice will give me that third eye to see things more quickly. Although now I'm not sure that the notes I'm reading are the best for a quantum mechanics beginner. Thank you very much. | |
| Sep 5, 2016 at 5:00 | history | edited | knzhou | CC BY-SA 3.0 |
added 1081 characters in body
|
| Sep 5, 2016 at 4:50 | comment | added | knzhou | @VladimirVargas Yeah, I made a jump there. I'll edit in a justification. | |
| Sep 5, 2016 at 4:47 | vote | accept | Vladimir Vargas | ||
| Sep 5, 2016 at 4:47 | comment | added | Vladimir Vargas | Hi, it is not very clear to me why $\hat{p}$ enters the integral and acts only on the coefficient of $|x'\rangle$. Thank you for your excellent answer. | |
| Sep 5, 2016 at 4:37 | history | answered | knzhou | CC BY-SA 3.0 |