Timeline for Equivalence between wavefunction and Dirac ket notation
Current License: CC BY-SA 3.0
14 events
when toggle format | what | by | license | comment | |
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Apr 26, 2020 at 8:41 | history | edited | Qmechanic♦ |
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Aug 8, 2016 at 7:13 | history | tweeted | twitter.com/StackPhysics/status/762547343656939520 | ||
Aug 7, 2016 at 23:42 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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Aug 7, 2016 at 23:34 | answer | added | Emilio Pisanty | timeline score: 3 | |
Aug 7, 2016 at 23:33 | history | edited | knzhou | CC BY-SA 3.0 |
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Aug 7, 2016 at 23:32 | answer | added | knzhou | timeline score: 4 | |
Aug 7, 2016 at 23:29 | comment | added | user108787 | What's wrong with using it the way Dirac designed it, as shorthand? Just use it all the way up till you have to do the integration. Unless you want to get deep into math country, that's what I would suggest, FWIW. | |
Aug 7, 2016 at 23:28 | history | edited | user126566 | CC BY-SA 3.0 |
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Aug 7, 2016 at 23:24 | comment | added | user108787 | The history, interchangeability and notation used in kets and bras is pretty well covered here en.m.wikipedia.org/wiki/Bra%E2%80%93ket_notation | |
Aug 7, 2016 at 23:23 | comment | added | CuriousOne | You can always write a function as an integral of a function and an integral of functions is... a function. | |
Aug 7, 2016 at 23:21 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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Aug 7, 2016 at 23:15 | comment | added | DanielSank | The equation you wrote, $\lvert \Psi \rangle = \int_n d^n \mathbf{r}\Psi(\mathbf{r})$ is not correct, or perhaps it's not clear what you mean by $\mathbf{r}$. I think you mean to write $\lvert \Psi \rangle = \int d^n \mathbf{r} \Psi(\mathbf{r}) \lvert \mathbf{r} \rangle$. | |
Aug 7, 2016 at 23:14 | review | First posts | |||
Aug 8, 2016 at 0:53 | |||||
Aug 7, 2016 at 23:12 | history | asked | user126566 | CC BY-SA 3.0 |