Timeline for Heisenberg uncertainty and Lorentz contraction
Current License: CC BY-SA 3.0
20 events
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| Jun 25, 2019 at 9:07 | comment | added | user87745 | @WillO Your last comment is really important I guess. The procedure laid out in your answer does exactly what you said--it makes a change of units. However, how things will be seen in a frame in relative motion wrt the lab frame will not be captured by this. Because, the relevant transformation is not $x\to ax$ but rather $x\to ax+bt$. So, we will inevitably have to include the dependence on time because there is no shared instant of time by the two observers. And then, things will get very ugly IMO because Schrodinger equation is not Lorentz invariant. | |
| Sep 16, 2016 at 5:33 | comment | added | WillO | @knzhou --- just to re-emphasize (though I realize it's probably unnecessary): The various problems with SR vs nonrelativistic QM seem to me to be red herrings, because we're not really invoking anything about SR here --- just two observers, one of whom has decided to measure distance in inches and the other in meters --- and those observers will disagree about the values of $\Delta x$ and $\Delta p$ (though not about their product). | |
| Sep 16, 2016 at 5:31 | comment | added | WillO | @knzhou: All I'm saying is that we have one observer who assigns the coordinate $ax$ to the point where the other observer assigns the coordinate $x$, where $a$ is a Lorentz factor. That means that if the observers agree on the value of the wave function at each point, and one says that function is $f(x)$, the other must say it's (up to a scale factor) $f(ax)$ --- and then the scale factor has to be $\sqrt{a}$ to keep the square of the integral equal to $1$. SR has almost nothing to do with it--- all that matters is that one observer multiplies the other's coordinates by a fixed factor. | |
| Sep 16, 2016 at 5:27 | comment | added | knzhou | It's not that I'm worried about how you're applying special relativity; I think special relativity can't be applied at all. The theory of nonrelativistic quantum mechanics is, well, nonrelativistic. It allows wavefunctions to go faster than the speed of light. It has the energy-momentum relationship $E = p^2/2m$ instead of $E = mc^2$. So I don't see how something like the Lorentz boost can be used. | |
| Sep 16, 2016 at 5:22 | comment | added | WillO | @knzhou: I am comparing the frames of two observers, in the same place at the same time, in relative motion with respect to each other --- and I am assuming a wave function that does not vary with time. Does that help? I do agree that if the wave function is changing over time, things are probably more complicated. | |
| Sep 16, 2016 at 5:20 | comment | added | knzhou | The math looks right, but I'm not convinced that this is the right relationship between the lab and boosted wavefunctions. The Schrodinger equation is not relativistic. | |
| Sep 16, 2016 at 5:19 | comment | added | WillO | @knzhou: I understand your concern --- but do you see a problem with the math? | |
| Sep 16, 2016 at 5:18 | history | edited | WillO | CC BY-SA 3.0 |
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| Sep 16, 2016 at 5:17 | comment | added | knzhou | This is very strange. The momentum distribution after a boost should be a shifted version of the original momentum distribution. Under your reasoning, a boost would change the momentum of a plane wave by a multiplicative factor, rather than simply adding a constant to it. | |
| Sep 16, 2016 at 5:12 | history | undeleted | WillO | ||
| Sep 16, 2016 at 5:12 | history | edited | WillO | CC BY-SA 3.0 |
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| Sep 16, 2016 at 4:44 | history | edited | WillO | CC BY-SA 3.0 |
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| Sep 16, 2016 at 4:09 | history | deleted | WillO | via Vote | |
| Sep 16, 2016 at 4:09 | history | undeleted | WillO | ||
| Sep 16, 2016 at 4:08 | history | edited | WillO | CC BY-SA 3.0 |
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| Sep 16, 2016 at 3:48 | history | edited | WillO | CC BY-SA 3.0 |
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| Sep 16, 2016 at 3:42 | history | deleted | WillO | via Vote | |
| Sep 16, 2016 at 1:46 | comment | added | Oti | Why is $/Delta x$ larger in the lab frame? Shouldn't it be smaller due to Lorentz contraction? The uncertainty relation should be preserved in the lab frame, i.e. $/Delta x' /Delta p' = /Delta x /Delta p /ge 1/2$. | |
| Sep 15, 2016 at 20:55 | history | edited | WillO | CC BY-SA 3.0 |
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| Sep 15, 2016 at 20:47 | history | answered | WillO | CC BY-SA 3.0 |