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In the experiment where electrons are sent one by one through a slit on a screen behind which there is an electron detector, the electron is said to have a definite position at the time it crosses the slit (which can be made very narrow to make $\Delta x$ as small as we like) so it must have a large uncertainty in momentum which is why the electron must diffract. Now is it not possible to measure the time taken by the electron to reach the detector wall and thus compute its momentum? How does the diffraction process make determining its momentum harder? If I know the position of the point which lights up in the detector wall and the time taken by the electron to reach there from the slit I can easily calculate $p_x, p_y$. Does this not violate the uncertainty relation if my slit is arbitrarily small.?

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  • $\begingroup$ Isn't precisely that what wave-particle duality means? $\endgroup$ – rodrigo Dec 20 '15 at 12:00
  • $\begingroup$ Yes I'm aware of that. But does this not imply charge $e$ is spread across an area on the detector screen? As far as I know if you use electrons one by one in the double slit exp., it gradually builds up an interference pattern. There's a lumpiness to it. This lumpiness is destroyed in this scenario. $\endgroup$ – Weezy Dec 20 '15 at 12:03
  • $\begingroup$ The classic explanation is that the charge is not spread on the detector screen because the detection is an observation of the particle and that makes the wave function to collapse. $\endgroup$ – rodrigo Dec 20 '15 at 12:06
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    $\begingroup$ It feels contradictory? Well, that is expected. QM is unintuitive and it looks contradictory and weird. However it is the best available explanation to the observed phenomena. $\endgroup$ – rodrigo Dec 20 '15 at 13:29
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    $\begingroup$ @Weezy: You changed the question! Before it was about wave-particle duality, now it is about uncertainty principle, although of course they are related. I'm no expert, but I'd say that the key to your new question is that the UP says that `position and momentum cannot be known simultaneously. $\endgroup$ – rodrigo Dec 20 '15 at 18:17
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The problem discussed here is about the duality of wave/particle. The physics of an electron is quantum mechanics, and the duality wave-particle is crucial. In fact, an electron is neither a particle(sphere) neither a wave. In the double slit experiment, the electron can be considered as a particle when the detector perturb the system, or like a wave without the detector perturbing the system.

If the electron is behaving like a particle, it will NOT have an interference pattern, and passing like a sphere through only one in the holes. But when the electron is not perturbed with a measurement, it will behave like a wave PASSING through all holes. More here(https://en.wikipedia.org/wiki/Double-slit_experiment)

An interesting question is the scale of the measurement(perturbation). We expect that an infinitesimal perturbation to change a little bit the behavior of the system, and a smaller perturbation to change drastically the behavior of the system destroying the wave behavior. But in fact even a smaller perturbation destroy the wave nature of the particle.

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  • $\begingroup$ I've edited the question a bit to make my doubt more clearer. Please have a look. Thanks. $\endgroup$ – Weezy Dec 20 '15 at 18:15
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    $\begingroup$ The electron is an quantum mechanical system, in the PAth integral formulation of quantum mechanics, en.wikipedia.org/wiki/Path_integral_formulation, the electron will go through all possible holes of the panel, the direction of propagation(as a navigational map) cannot be known with respect to the velocity. Simply put:if you know when the electron will light the detector(final state), that doesn't mean that the electron propagates in straight line from the hole in the panel. Also, the electron diffract as a consequence that is behaving like a wave. $\endgroup$ – Mikey Mike Dec 20 '15 at 18:30
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Tennis balls. You have to aim at either slit. Direction in a straight line we all can understand. You cannot fire a single electon. No gun ever made can do that. Imagine that you have to fire one electron/photon through a vacuum at the speed of light at a detector. The single photon will not disappear when fired through a gap. It will hit the detector. Another seperate photon when fired through another gap (n)mm away (n)seconds later will hit the same detector screen at the angle between the gun and the second slit. The gun has to change its firing angle to fire the the single photon. Ergo, when the third electron/photon is fired at the wall between the gaps the photon will hit the wall. A detectoral dud. Fire a million electrons/photons in a broad field you will get a diffraction pattern like a wave on the detector. This is because light bounces off things. (air molecules in particular.) It is the spread of many photons (zillions) at once that causes the diffraction pattern. In reality you cannot expect to fire one electron and watch it go through two slits. In reality you cannot observe anything other than complex fractal moments of differentiation.

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Electron diffraction pattern is different than the diffraction of laser light as for the former you need many electrons to pass the double-slit one by one in order to observe the fringes, but in a classic light like a laser (photons are so many that it is considered as a classic EM field) as soon as the laser irradiates the double-slit the pattern is created. So in the electron diffraction when you are speaking of interference fringes you are basically considering the spatial probability of electrons' distribution.

Although for one electron passing the double-slit, the probability of electron hitting different parts differ based on quantum physics, still it hits one place on your detector behind the double-slit and one needs to wait for a while to see a pattern the same as light diffraction pattern. So still your charge is conserved in amount and you do not need to integrate over the area of your detector to measure the electron charge.

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  • $\begingroup$ Whoever down-voted me can elaborate why I am wrong? $\endgroup$ – Amin R. Dec 20 '15 at 17:28
  • $\begingroup$ I've edited the question a bit to make my doubt more clearer. Please have a look. Thanks. $\endgroup$ – Weezy Dec 20 '15 at 18:15

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