Timeline for Why do we represent states vectors with ket vectors?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 6, 2020 at 19:51 | vote | accept | Noumeno | ||
| Sep 4, 2020 at 21:12 | answer | added | J. Murray | timeline score: 4 | |
| Sep 4, 2020 at 17:46 | comment | added | Cosmas Zachos | "But there are some operations that make sense only if applied on functions and not on ket vectors. " Like what? Have you read up on Dirac's standard text? | |
| Sep 4, 2020 at 16:43 | comment | added | G. Smith | In friendly 3D linear algebra we almost always think of vectors in the context of a specific representation of them in some base. Didn’t you ever abstractly write a matrix as $\mathbf{M}$ and a vector as $\mathbf{v}$? | |
| Sep 4, 2020 at 16:20 | comment | added | DanielSank | I will say one thing though: this whole issue has absolutely nothing to do with quantum mechanics. This whole issue is just a question of whether we're working in a specific basis or with a basis-independent representation of the vectors. Notations such as $A \psi(x)$ are not consistent and shouldn't be used. I'll write a more comprehensive answer once we sort out your previous question. | |
| Sep 4, 2020 at 16:19 | comment | added | DanielSank | I would like to answer this, but I am reluctant to spend time on it because I wrote a pretty extensive answer on your other question that has not yet been resolved. I offered to follow up on your confusions in chat, and I would be happy to improve the answer to the point that it can be accepted. Please let me know if you would be interested in that. | |
| Sep 4, 2020 at 16:15 | comment | added | Vercassivelaunos | I disagree with your assertion that choosing a representation of a vector is easier. In fact, I try to avoid it as much as I can. The specific representation is superfluous information, and superfluous things are superfluous at best, confusing at worst. | |
| Sep 4, 2020 at 16:14 | comment | added | Jeanbaptiste Roux | Wavefunctions $\psi(x)$ represent the components of the vectors $|\psi \rangle$. When we are using on operator acting on $|\psi \rangle$, one can use the same operator but in a different representation on $\psi(x)$. For example the momentum operator $-i \hbar \frac{d}{dx}$ is acting on $\psi(x)$, but it is a special representation of the general operator $\hat{p}_x$ acting on $|\psi \rangle$. | |
| Sep 4, 2020 at 15:57 | history | edited | Qmechanic♦ |
edited tags; edited tags
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| Sep 4, 2020 at 15:54 | history | asked | Noumeno | CC BY-SA 4.0 |