Timeline for Is there a "position operator" for the "particle on a ring" quantum mechanics model?
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| Aug 6, 2023 at 8:38 | comment | added | Tobias Fünke | Hi, thanks for the reply. You are right, I agree here: If you use the periodic extension, it makes sense, because the product of two periodic functions (with the same period) is periodic again. Now everything is well-defined. This is also used in one of the answers you've linked. But then I find the notation a bit strange, which however was my misinterpretation. | |
| Aug 6, 2023 at 4:52 | comment | added | J. Murray | Hello @TobiasFünke - are you perhaps unconsciously imposing continuity on these periodic functions? For example, the function $\psi(x) = x$ on $[0,2\pi)$ can be extended to a periodic function by copying and pasting it over the whole real line. It won't be continuous (hence the subtlety mentioned in my comment above) but that's not a general requirement of the Hilbert space. | |
| Aug 5, 2023 at 22:06 | comment | added | Tobias Fünke | Hi J. Murray, but if your Hilbert space is $L^2(S^1)$, then you operate on equivalence classes of periodic functions -so the operator must map such an equivalence class to another one. But this is not possible in general, due to the periodicity requirement. As far as I see, this has nothing to do with the interval you choose. Do I miss something obvious here? | |
| May 4, 2021 at 2:13 | comment | added | J. Murray | @d_b Well, by considering $\mathbb R\bmod 2\pi$ we ensure that $\phi \in [0,2\pi)$ so nothing is multivalued. That being said, unless $\psi(0)=0$, the action of $\hat \Phi$ does introduce a discontinuity in the wavefunction. This isn't necessarily a problem, but it does make the position/momentum uncertainty relation more subtle than it is on the line because the domain of the operator $\hat P_\Phi \circ \hat \Phi$ is not the same as the domain of the operator $\hat \Phi \circ \hat P_\Phi$, c.f. this nice answer by ACuriousMind. | |
| May 4, 2021 at 1:56 | comment | added | d_b | Isn't there some issue with this operator being multivalued, making it better to work with $e^{i \hat{\Phi}}$ instead? | |
| May 4, 2021 at 1:30 | history | answered | J. Murray | CC BY-SA 4.0 |