Timeline for Why is the wave function of a particle with definite momentum $p$ given as $e^{ipx/\hbar}$?
Current License: CC BY-SA 4.0
12 events
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| Feb 2, 2023 at 8:16 | comment | added | Tobias Fünke | To expand my previous comment: With $X$ as a multiplication and $P$ as a derivative (up to constants) operator. Essentially, this boils down to the fact that states with definite momentum are eigenfunctions of the momentum operator (by definition). | |
| Feb 2, 2023 at 8:06 | comment | added | Tobias Fünke | Well, if you accept that $[X,P]=i \hbar I$, then you can derive that $\langle x|p\rangle =: \psi_p(x) =\ldots$... It is an easy exercise | |
| Feb 2, 2023 at 8:03 | comment | added | CBBAM | @TobiasFünke I haven't, my curiosity was more towards where this equation came from in the first place and if it had any physical or mathematical motivation. | |
| Feb 2, 2023 at 7:57 | comment | added | Tobias Fünke | Have you tried to solve the corresponding SE? | |
| Feb 2, 2023 at 7:55 | vote | accept | CBBAM | ||
| Feb 2, 2023 at 7:55 | comment | added | CBBAM | @TobiasFünke Yes, non-relativistic. | |
| Feb 2, 2023 at 7:45 | comment | added | Tobias Fünke | In which context? In (non-relativistic) QM? | |
| Feb 2, 2023 at 7:37 | answer | added | John Rennie | timeline score: 1 | |
| Feb 2, 2023 at 7:00 | comment | added | Rhino | Also, it is the solution to Schrodinger's equation and, actually, also other more general relativistic wave equations. | |
| Feb 2, 2023 at 6:30 | history | edited | Qmechanic♦ |
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| Feb 2, 2023 at 6:23 | comment | added | march | Well, $\lambda = h/p= 2\pi\hbar/p$ is de-Broglie. After that, you're just writing down the general expression for a plane wave with that wavelength. So the reason that we write that wave function down is that it's essentially the same as de-Broglie's hypothesis, and in the end (after a tortuous path) it matches experimental realities. | |
| Feb 2, 2023 at 6:20 | history | asked | CBBAM | CC BY-SA 4.0 |