Timeline for Why doesn't velocity enter into Heisenberg's Uncertainty Principle?
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 28, 2017 at 21:27 | comment | added | Zhe | It doesn't matter how close these points are. You are trying to use your macroscopic intuition of interpolation. That doesn't work at the quantum level. You cannot tell what path was taken between these two points. | |
| Oct 28, 2017 at 17:37 | comment | added | a-cyclohexane-molecule | @TheLostGuardian0, how will you obtain these points? Each measurement will affect the future path of the electron, so your points will not be an accurate reflection of the electron's original trajectory. | |
| Oct 28, 2017 at 14:19 | vote | accept | TheLostGuardian0 | ||
| Oct 28, 2017 at 14:15 | comment | added | TheLostGuardian0 | @Zhe But what if we have millions of such points? Wouldn't that solve the problem, as the path would be revealed by the various possible places it was in? | |
| Oct 27, 2017 at 21:23 | comment | added | Zhe | Here's a good question: if you have something at point A and something at point B at a later time, what's the path between these two points? Turns out the more certainty you have about those two positions, the less you can tell about which path was taken. | |
| Oct 27, 2017 at 20:00 | comment | added | Ivan Neretin | There is no microscope that would determine the path of the electrons. Also, Bohr-Sommerfield model fails when it comes to 2 or more electrons. So it goes. | |
| Oct 27, 2017 at 18:22 | comment | added | TheLostGuardian0 | @a-cyclohexane-molecule my bad I should have worded my question better. If we didn't need momentum to find the trajectory, then why go for quantum mechanical model instead of sticking to Bohr-Sommerfield model? (We are able to determine the path of the electrons by using a microscope so this is similar to Bohr's model) | |
| Oct 27, 2017 at 18:14 | comment | added | a-cyclohexane-molecule | @TheLostGuardian0, I'm not sure what you mean by needing momentum. You don't need to measure (or care about) momentum at all if you're only interested in position; the Heisenberg principle just says that you can't measure both simultaneously accurately. | |
| Oct 27, 2017 at 18:10 | comment | added | TheLostGuardian0 | @a-cyclohexane-molecule I appreciate all your help, but I'm still not clear as to why we need momentum? I mean isn't position enough if we have large samples? Momentum is only required if we have 1 sample right? | |
| Oct 27, 2017 at 18:05 | comment | added | a-cyclohexane-molecule | @TheLostGuardian0, do note that you'll need to tag Ivan (as I've tagged you here) for him to get a notification about your comment. With regards to your question, you wouldn't get exact agreement, but it will be a better and better approximation as you increase the number of measurements. | |
| Oct 27, 2017 at 17:59 | comment | added | TheLostGuardian0 | Okay, so let's say they aren't 100%identical. Now by using our electron microscope we know where the electron "has been" but we can't predict its future because of the energy of the photon. Now even if the hydrogen atoms aren't 100% identical wouldn't my method directly(if the 100pictures are overlapped) give us a picture that looks exactly like the probability density graph for 1s? | |
| Oct 27, 2017 at 17:44 | comment | added | Ivan Neretin | No-cloning theorem is an overkill. Let's put it another way: just how are you going to make sure that these 100 atoms are 100% identical? | |
| Oct 27, 2017 at 17:38 | history | answered | a-cyclohexane-molecule | CC BY-SA 3.0 |