Timeline for Is the Momentum Operator a Postulate?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 11, 2016 at 0:32 | comment | added | Juan Perez | Yes. You need to start from somewhere. Any of those can be the relevant postulate, but one of them has to be. You can't just deduce them from the other postulates. | |
| Nov 5, 2016 at 16:11 | comment | added | OkThen | 2)Now is you should start from somewhere to obtain this. The people here seem to like symmetry arguments. But you could have started from the uncertainty principle, for example. Any physical thing you can think of that leads you to think about this algebra will suffice. | |
| Nov 5, 2016 at 16:11 | comment | added | OkThen | Two pieces answer. 1) Please, don't say you are sorry. Questions are encouraged. About your question: Well, you have to start from somewhere. By now you realize the important stuff is [x,p] = non-zero number independent of p and x. Any number is ok, because I can absorb it and redefine x and p to obtain the usual stuff (no one told you they should directly correspond to classical things). The non-trivial part is to motivate a relation $[x,p] \neq 0$. | |
| Nov 5, 2016 at 3:08 | comment | added | Juan Perez | Sorry to keep replying, but I'm a bit confused. I understand now that P = -iℏd/dx is not a postulate. But [x,p] = iℏ is one, right? (If it's not a postulate, but a choice, I could choose something else, like [x,p] = 0, or [x,p] = 78, and get equivalent results?) | |
| Nov 5, 2016 at 2:13 | comment | added | OkThen | Unfortunately, I don't think I made myself clear. There is no need for an additional postulate to obtain P = d/dx. Anything that satisfies [x,P]=i is valid. And you can write these representations much in the same way you obtain the Pauli Matrices just from the angular momentum algebra. All of these people answers show you that exact same thing. | |
| Nov 5, 2016 at 0:15 | comment | added | Juan Perez | You just said that you take the abstract algebra [x,p] = iℏ as a postulate. What I meant to say is that that postulate specifically gives the relationship between x and p, and that you couldn't obtain a working expression of the p operator (which, as you say does depend on what representation you choose) without it. | |
| Nov 4, 2016 at 2:56 | comment | added | OkThen | @JuanPerez No, why would you say that? It's a representation. The content of the theory are states and operarors where the norm gives probability. A representation is just a way to represent these things. It doesn't change the physical content! It is also not necessary, it's just convenient. | |
| Nov 4, 2016 at 2:45 | comment | added | Juan Perez | Another reply suggesting there is a postulate specifically for the form of p (with respect to x). Thanks. Guess the textbooks I read weren't that thorough. | |
| Nov 4, 2016 at 2:26 | history | answered | OkThen | CC BY-SA 3.0 |