When send in a small opening particle acquire the spread of impulse p. E.g. there can be sometimes a huge value of p, which the particle doesn't have initially. So is with Energy too. The question is where do these surplus energy and momentum come from? I never encountered that in books. I think there is no other place where it can come from except the atoms of the opening. But in that case follows the question what if the opening is cooled down to near -273? Take into account that one can have very slow particles so they don't deliver much energy but must take away. So they will cool down the opening and the whole barrier but not under -273,16.
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$\begingroup$ Have you heard of the term 'zero point energy'? $\endgroup$– Jun Seo-HeCommented Aug 30, 2021 at 19:36
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$\begingroup$ This might help physics.stackexchange.com/questions/460009/… $\endgroup$– MauricioCommented Aug 30, 2021 at 23:40
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$\begingroup$ @JunSeo-He Do you mean the lowest E level of a quantum oscillator? What does it help? $\endgroup$– MercuryCommented Aug 31, 2021 at 6:12
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1$\begingroup$ It means that even in absolute 0 in which classical particles stop moving , due to the zero point energy Heisenberg's uncertainty principle still holds $\endgroup$– Jun Seo-HeCommented Aug 31, 2021 at 11:57
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$\begingroup$ Should your comment mean that the passing tru the slit particle takes away energy from the zero point energy of the atoms in the slit? I don't think zero energy can be taken away from them. Logically if you take it they will be left without zero energy and and it wouldn't be appropriate to call it zero E. $\endgroup$– MercuryCommented Aug 31, 2021 at 15:56
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2 Answers
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Subatomic particles are (or are associated with) wave packets (of finite size). Fourier analysis tells us that if the packet is very short, the frequency of the wave cannot be accurately determined. I believe this is the source of the uncertainty principle. I am not aware of any situation where a particle can gain energy by passing through a small opening. It gains energy by interacting with other particles or fields. A particle confined to the volume of a nucleus (by the strong nuclear force) would have at any instant a very uncertain energy.
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$\begingroup$ What you believe that is the source of HUP is the mathematical reason. What happens in the nucleus stays behind the nucleus and is not so obvious. Nevertheless this could be some question. $\endgroup$– MercuryCommented Sep 1, 2021 at 14:56
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The Heisenberg principle is related to uncertainties of measurements. The particle is in a QM-state, described by a wave function. If we make some measurement that tells us that it is confined in a very small location, there is a limitation in my max. accuracy of knowledge about its momentum.
For example, it is necessary a electronic microscope to be sure about where things are, for small distances. But that means that electrons collide on whatever is being observed, and the device manages to form an image on the screen from the scattered electrons. Electrons are used instead of photons exactly because they have smaller wave lengths, what also means bigger momentum.
So, we can say that the source of the momentum of the observed particles is the momentum of the incoming ones, that must be greater and greater for smaller accuracy of the location.
answered Aug 30, 2021 at 21:25
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$\begingroup$ Do you mean that after passing a slit (opening) the particle does not have greater impulse but it is introduced just after observing it by a electron microscope? $\endgroup$– MercuryCommented Aug 31, 2021 at 6:13
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$\begingroup$ The point is: how do you know that the particle are there in the slit opening without some type of detector? $\endgroup$ Commented Aug 31, 2021 at 12:30
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$\begingroup$ After the slit there is a detector of the particle (passive registrator of the event) which is moved along y (perpendicular axis through the slit and source). It just registers the y coordinate which is proportional to the p in y direction. Do you say that it can change p(y) and tgis does not happen at the slit? $\endgroup$– MercuryCommented Aug 31, 2021 at 15:39
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$\begingroup$ And if such event is registered it is obvious that the particle went through the slit, because there are not any other paths. $\endgroup$– MercuryCommented Aug 31, 2021 at 15:47