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 | |
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| Sep 6, 2020 at 19:56 | comment | added | J. Murray | @Noumeno If you are trying to be mathematically rigorous, I find the second convention preferable as well. However, at the standard physicists level of rigor I find it difficult to argue that bra-ket notation is not vastly more convenient, especially when doing things like perturbation theory. | |
| Sep 6, 2020 at 19:51 | vote | accept | Noumeno | ||
| Sep 6, 2020 at 19:51 | comment | added | Noumeno | I am trying to argue that the second convention is the better one, anyway I get your point. | |
| Sep 6, 2020 at 17:48 | comment | added | J. Murray | @Noumeno Either you're working in bra-ket notation or you aren't. If you are, then $\phi(x)$ should be considered to be the component of your vector along the basis vector $|x\rangle$. If you are not working in bra-ket notation, then the vector is the function $\phi$. It seems to me that you're trying to mix both conventions, which is not a good idea. | |
| Sep 6, 2020 at 17:40 | comment | added | Noumeno | Yes, I understood that, and I think that you are completely right. But I also think that we can think of function as complex vectors in infinite dimension, in your answer you seem to imply that we can only think this about ket vectors, also I find that $f(x)$ is not a number as you state, because it can be a number if $x$ is fixed, but it can also represent a function if $x$ is not fixed. So your answer has some problematic parts, do you agree or am I missing something regarding this? But the main message of the answer is correct in my opinion of course. | |
| Sep 6, 2020 at 17:30 | comment | added | J. Murray | @Noumeno I think you are misunderstanding my answer. You are free to work exclusively in the position representation if you wish, but if you want to use bra-ket notation, then you need to understand that there's a difference between an abstract operator (which acts on abstract kets) and the position-space representation of the operator (which acts on wavefunctions). | |
| Sep 6, 2020 at 17:16 | comment | added | Noumeno | I think that your answer may create a lot of confusion. Don't misunderstand me, I think your answer is completely correct but I have a problem with it. $\phi(x)$ is a number if $x$ is fixed, however we often interprete $\phi(x)$ not as a number but as a function, functions indeed form a complex vector space and can be thought as a vector with infinite entries. All you answer is based on the notion that $\phi(x)$ is a number in my question, but I intended to say $\phi(x)$ as a function. I think that this would be better to use instead of $|\phi\rangle$. Do you get my point? If not let me know. | |
| S Sep 4, 2020 at 22:50 | history | suggested | Sebastiano | CC BY-SA 4.0 |
Improved mathcode \iint
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| Sep 4, 2020 at 22:46 | review | Suggested edits | |||
| S Sep 4, 2020 at 22:50 | |||||
| Sep 4, 2020 at 21:12 | history | answered | J. Murray | CC BY-SA 4.0 |