According to the Uncertainty Principle we cannot calculate the precise position or momentum of a quantum particle at the same time, trying to calculate one of them would make the other uncertain. One of the examples is the Single Slit Experiment in which a photon or any other quantum particle passed through an extremely narrow slit is seen landing at multiple locations on the detecting screen on the other side of the slit, since we are certain about the position of particle passing through the slit we get an uncertainty in its momentum, the result of which is seen on the screen. My question is what if there is no scope given for the quantum particle to be uncertain of its position or momentum by constraining its motion in only one dimension i.e. instead of passing the particle through a single slit, what if we pass it through an infinitely long continuation of slits which are approximately of the size of quantum particle? In this way the particle would have no scope to obtain an uncertainty in momentum since it would have no place to land upon at multiple positions and hence the Uncertainty Principle would be violated?
3 Answers
According to the Uncertainty Principle we cannot calculate the precise position or momentum of a quantum particle at the same time, trying to calculate one of them would make the other uncertain.
This is a misunderstanding: not calculate, but measure.
The position and momentum of a particle cannot be simultaneously measured with arbitrarily high precision. There is a minimum for the product of the uncertainties of these two measurements. There is likewise a minimum for the product of the uncertainties of the energy and time.
Since you are talking of single slits, here is what the interference pattern looks like:
The width of the slit is such as to conform with the energy of the electron so that the interference patterns would be visible.
The distribution on the right is the probability distribution of where a single electron will end up after passing the slit. You can see that maybe a fourth of the electrons will miss going through a second slit you want to set up. And that is a geometric progression. If you make the aperture smaller, less electrons will be going to the central peak.
My question is what if there is no scope given for the quantum particle to be uncertain of its position or momentum by constraining its motion in only one dimension i.e. instead of passing the particle through a single slit, what if we pass it through an infinitely long continuation of slits which are approximately of the size of quantum particle?
The probability of a particle passing through even 20 slits will be effectively zero.
The size of the quantum particle is its de Broglie wavelength, another rule of thumb that describes the probabilistic solutions of quantum mechanical entities.
The Heisenberg uncertainty is a rule of thumb way of talking of the way quantum mechanics constrains probabilities in interactions, in this case "electron hitting single slit".
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Your response seems to imply that the HUP can be violated as long as the probability of violation is low. – user1247 Mar 26 '19 at 17:34
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@user1247 can you point out the sentence when this can be interpreted as you say? The Hup is not violated afaik ,ever. – anna v Mar 26 '19 at 18:04
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You make the statement "The probability of a particle passing through even 20 slits will be effectively zero," without otherwise providing a reason why the questioner's logic fails. This seems to imply that you are saying that the logic is sound, but that violations of the uncertainty principle are just very rare. If it helps, the logic still goes through for only 2 slits in sequence. – user1247 Mar 26 '19 at 18:07
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Once an electron has manifested at an outer ring,for the first scatter, the second scatter can either bring it in the central peak , but there is a high probability that it will be again at an outer location, the probability of getting out of the concesutive distributions is cumulative and high.. You cannot constrain the direction of the electron . – anna v Mar 26 '19 at 19:17
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We know you can't constrain the direction of the electron... nonetheless every once in a while it will go through two slits in succession and be detected. That's the point of the thought experiment... – user1247 Mar 26 '19 at 19:22
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@user1247 Yes, it has a diminishing probability of going through many slits, as a large percentage gets out of the beam each time. For two slits in a row if you start with 20 electrons you may be left with 15, for twenty slits with zero is all I am saying. Thus you cannot defeat the HUP, because it is the HUP that takes them out of the beam line. If you mean that because the electron is somewhere in the middle circle you know its position versus its momentum with accuracies smaller than HUP , if you put in the numbers, you will disillusioned.the uncertainty in the momentum where – anna v Mar 27 '19 at 05:18
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where the electron hit was measured and the uncertainty in the position of the hit will fulfill the HUT if you put in the numbers. If you make the slit very narrow, the situation gets worse http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1 – anna v Mar 27 '19 at 05:24
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You just implied that you can defeat the HUP 15/20 times, since 15/20 times you have both position and angle data to arbitrary accuracy. – user1247 Mar 27 '19 at 18:36
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@user1247 You have completely misunderstood. All these locations have a delta(x) and all the momenta . a delta(p), (the wave packet of the particle) Δ(x) Δ(p)>h/2π . – anna v Mar 27 '19 at 18:54
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No, you have completely understood: the momenta's delta(p) can be made arbitrarily small by increasing the distance between the sequential slits. That's the entire point: getting the momentum from the angle between the slits. I'm not saying that HUP is wrong, but just that you haven't addressed the question! – user1247 Mar 27 '19 at 19:24
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@user1247 you do not know what you are talking about. The momenta come with a width from the original beam, a quantum mechanical width for each electron. Look what happens to diffraction when the slits are made smaller http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslitd.html#c1 – anna v Mar 27 '19 at 20:01
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I understand what happens when the slits are made smaller. I'm afraid you are simply responding hastily without taking the time to actually read what is written. Diffraction happens. It exists. Momentum becomes uncertain when the slit width is made small. We know these things. This is not what the question is about. The question is about the inference of the momentum direction (before the diffraction at the second slit) from the angle between the two slits. – user1247 Mar 28 '19 at 00:00
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I think my question was misinterpreted, when I mean passing the electron through an infinitely long continuum of slits, I intended to state the the continuum of slits is in the direction of the Detection Screen. Imagine it being like a tunnel of slits that guides the electron in one dimension until it is detected on the screen. This way both its momentum and position can be measured without any certainty. – Ajinkya Naik Dec 26 '19 at 14:51
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I gave the argument for two slits, for a tunnel of slits the electron would never go thourgh the end. And how do you propose to measure the momentum? The slit experiment is not good for measuring electron momenta,one can use the wavevelegth but with a tunnel of slits there is no possibility of measuring the wavelength. – anna v Dec 26 '19 at 16:45
Your proposed configuration limits further the position of the particle, but the more you constrain the position, the wider the approximation in momentum!
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My statement is simply the Heisenberg's uncertainty principle. If you disagree, you disagree with the fundamentals of quantum mechanics. – Michele Grosso Mar 27 '19 at 13:57
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You seem to think that the uncertainty principle requires the uncertainty in position times the uncertainty in momentum to be equal to some amount. But that is not what the uncertainty principle says. It says that there is a constant such that the product of the uncertainty must be at least that constant. It does not violate the uncertainty principle for the product of the uncertainties to be more than that constant.
You also seem to be working without a precise definition of "uncertainty". Uncertainty, in this context, is simply another term for the standard deviation of the distribution. Even if there are an infinite number of places where the particle can be, that doesn't mean that there is infinite uncertainty. If the probabilities of locations far from the center go to zero fast enough, the total uncertainty will be finite. The normal distribution, for instance, has infinite spread, but finite standard deviation.
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