Timeline for Why doesn't Gaussian wavepacket broadening in position mean there will be a shortening in momentum?
Current License: CC BY-SA 4.0
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| Oct 19, 2020 at 15:17 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Sep 18, 2020 at 12:30 | comment | added | Alex Gower | Yeah sorry I shouldnt have used the word uncertainty, just more a general 'spread' word if you agree | |
| Sep 18, 2020 at 12:24 | comment | added | BioPhysicist | @AlexGower The wave function and the probability distribution don't have uncertainties. It's the measurements themselves. I agree with the rest though. | |
| Sep 18, 2020 at 12:11 | comment | added | Alex Gower | Hi, sorry I think we are on the same page, but I was just asking about why you were considering putting 'the central peak narrows' for the momentum wavefunction in your answer since (the way I'm reading it) that statement is sort of suggesting that there is some heuristic reduction in the wavefunction uncertainty, when we're both agreeing now that there need not be and there isnt. Why not just leave the answer as 'yes the momentum wavefunction CHANGES as it must since its a FT, but neither the momentum wavefunction nor the momentum probability distribution has any uncertainty changes' | |
| Sep 18, 2020 at 11:21 | comment | added | BioPhysicist | @AlexGower Just so that we are on the same page here, wave functions don't have uncertainty. The momentum uncertainty (I term I really do not like) refers to the standard deviation of momentum measurements if you prepared multiple "copies" of the state and then performed momentum measurements on all of them. As for what you are saying, I am not sure what you are getting at. Yes, the momentum wave function is changing in that manner and the uncertainty isn't changing, but I've already shown that. | |
| Sep 18, 2020 at 9:36 | comment | added | Alex Gower | Would it not just be fine to say that the momentum wavefunction doesn't have any narrowing of uncertainty/peaks? The fact that it changes at all is reassuring to me as it means it is losing no information from the position wavefunction. From what I've read on the other answers, wouldn't it just be fine if its uncertainty didn't decrease at all, its shape just changed within the same envelope. (I'm talking about the momentum wavefunction here not the probability distribution) | |
| Sep 18, 2020 at 8:49 | comment | added | Alex Gower | What would the significance be of just the central peak narrowing? | |
| Sep 17, 2020 at 17:47 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Sep 17, 2020 at 16:27 | comment | added | BioPhysicist | @Andrea Yes, that is a good and correct point to make. Maybe it would be better to say the central peak narrows? | |
| Sep 17, 2020 at 16:19 | comment | added | Andrea | This is a fantastic answer, but there is a little mistake. You say that '$\Phi$ narrows', but I don't think it does. While it does get higher and higher frequency (because $\Psi$ gains support in larger positions), it is still always contained in the enveloppe $\sqrt{|\Phi|^2}$, which, as you noted, does not change. | |
| Sep 17, 2020 at 15:40 | vote | accept | Alex Gower | ||
| Sep 17, 2020 at 15:36 | comment | added | BioPhysicist | @AlexGower I have edited the answer accordingly. | |
| Sep 17, 2020 at 15:36 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Sep 17, 2020 at 13:47 | comment | added | BioPhysicist | @AlexGower $\Delta x$ and $\Delta p$ do not have to be inversely proportional. The only time you would know anything for sure (before just actually doing the calculations) is if you knew $\Delta x\Delta p=\hbar/2\pi$ and then you made a new state with a smaller $\Delta x$ or smaller $\Delta p$. Then you would know that the other value had to increase. But even then, it wouldn't have to necessarily be an inversely proportional increase. As for the Fourier Transform stuff, I will edit my answer accordingly when I have time. | |
| Sep 17, 2020 at 9:14 | comment | added | Alex Gower | And I guess will the 'bulk area' idea, I was wondering how exactly the fourier transform of the position wavefunction (i.e. giving wavefunction in momentum space) is changed after this wavepacket broadening. Obviously its standard deviation hasn't changed, but I would expect there to be some difference since I though the FT doesn't lose any information from its input and its input has definitely changed. Otherwise you wouldn't be able to get back to this 'broadened' position wavefunction from the momentum wavefunction using an inverse FT. | |
| Sep 17, 2020 at 9:10 | comment | added | Alex Gower | Ah okay I thought it could be something like this. So am I correct in saying the Uncerainty Principle does not mean that the uncertainties of x and p are inversely proportional, but their bounds? It's just confusing since some sources online make it seem like an increase in uncertainty of one must definitely cause a decrease in uncertainty of the other, but I guess this isn't the case then? | |
| Sep 17, 2020 at 2:48 | history | edited | BioPhysicist | CC BY-SA 4.0 |
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| Sep 17, 2020 at 1:34 | history | answered | BioPhysicist | CC BY-SA 4.0 |