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Dr. W. M. Stuckey shows that the origin of quantum entanglement is none other than Einstein's own Principle of Relativity (No Preferred Frame of Reference). Ei

Dr. W. M. Stuckey shows that the origin of quantum entanglement is none other than Einstein's own Principle of Relativity (No Preferred Frame of Reference). Einstein's famous 1905 paper on Relativity applied this principle to Translational Frames, showing it requires the Universal constant c (speed of light) to be the same in all such frames. Were it not, the frame in which c was highest would be the only frame at rest. But the same principle must require there to be no Preferred Orientation. This leads to the requirement that Planck's constant h be the same in all frames. If the Stern-Gerlach experiment could give results between +h and -h, then the orientation producing the maximum value would be a preferred frame. And because of that, when Alice and Bob measure entangled quantum particles, their combined results must violate Bell's inequality. But to my mind, the biggest take-away is that Einstein's Principle of Relativity absolutely requires that conservation can only be on average. All right, that's a ridiculously condensed summary. Enough to make your head spin :-) The title paper is for general audiences, and references the original paper at https://www.nature.com/articles/s41598-020-72817-7.pdf https://www.nature.com/articles/s41598-020-72817-7.pdf

jjbinx007 6y ago
I think I have a reasonable mental model of how certain QM processes work by visualising waves in transit coalescing as particles when measured. But I have absolutely no idea how to visualise entanglement. Any tips? Or do we just have to shut up and calculate?
- diegoperini 6y ago
  My understanding is, Looking at a single object with a fixed angle camera produces similar observations to an entagled pair when the pair is in a similar configuration and where each object in the pair is observed by their own camera except one of the cameras sees a negated result.
- GoblinSlayer 6y ago
  Often you can visualize entanglement as a superposition of independently evolving states, when those states are solutions of Schrodinger equation.
- lisper 6y ago
  The key to understanding entanglement for me was to understand that the wave function does not live in physical 3-D space, it lives in configuration space. A wave function that lives in physical 3-D space is a special case that applies only to a system that consists of a single unentangled particle. In that case, physical 3-D space and configuration space are the same. But in general, a wave function for N particles will live in a 3N-dimensional configuration space.
  - mlmonge 6y ago
    It seems to me that the configuration space of which you speak essentially explains the wave/particle duality. My B.S. Physics education from years ago never explained it as such as far as I can recall. I'm far from any expertise understanding, but this makes much sense. Any further pedagogical commentary would be most appreciated!
    - lisper 6y ago
      > configuration space ... explains the wave/particle duality I don't think so. You can have a wave function in physical 3-D space as a (very common) special case and you still have the wave-particle dichotomy. Why do you think configuration space explains WPD?
      - elcritch 6y ago
        Really WPD more fundamentally arrives from the “conversion” of the probability wave (living in configuration space as you put it) into properties in 3D physical space. Measuring the QM system is analogous to a Fourier transformation and has the same mathematical limitations arising from a Fourier transform from wave to discrete space. Depending on the question you ask (tied to the convolution function) you get either discrete particle answers or wavelike answers. The probability wave does live in 3D space AFAICT, but the QM properties like spin, charge, momentum etc are tied into that 3D space and form a combined configuration space. The really odd part to me is that at macro scales the probability waves collapse neatly into classic physics in 3D space, but still react in quantum fashion at small local atomic scales. As in the configuration spaces generally can only be determined for small subsystems but not a whole macro system without the “conversion” step.
- tim333 6y ago
  The way I see it is particles go all possible ways until you narrow things down with a measurement or something along those lines - an interaction that fixes where the thing is. (So in the two slit experiment with one particle it goes through both slits until its position is pinned down by hitting the screen.) In the entanglement experiment described the particles have angular momentum every which way until the angular momentum of one is pinned down by measurement whereupon the other one is also pinned down to the opposite by conservation of momentum. There is still a sort of spooky action at a distance when that happens or perhaps a splitting of the multiverse 'at a distance' into many worlds where the spins point different ways.
  - pault 6y ago
    I wish, for the love of god, that the early pop science about quantum physics hadn't used the phrase "when observed" when describing the wave collapse. If they had said "when interacted with by another force" (or anything along those lines), we wouldn't have loads of new age dummies talking about how the particle "knows" it is being observed by a conscious mind. No quantum woo, no Deepak Chopra.
    - hellotomyrars 6y ago
      I don't think you're giving the people who peddle bullshit enough credit. They'll just figure something else out.
    - strogonoff 6y ago
      > "when interacted with by another force" But that is not the same, right? I mean, if it interacted with a force and no observation was made, the wave function doesn’t collapse, does it? Honest question (to avoid any defensiveness, I should disclose that I don’t subscribe to panpsychism). > or anything along those lines Any suggestions?
      - pault 6y ago
        > if it interacted with a force and no observation was made, the wave function doesn’t collapse, does it? Any two systems interacting will cause the collapse. It doesn't matter if the system is attached to a scientist or not. > Any suggestions? No, I'm a software developer, not a quantum physicist. :)
- Waterfall 6y ago
  Have you heard of Bohemian or pilot wave theory for QM? It's very rarely discussed except in darker corners of the internet but it's surprising to me it's not used as visualization. https://en.m.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory https://en.m.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theo...
  - leephillips 6y ago
    I didn’t know I was writing for the “darker corners of the internet”! https://arstechnica.com/science/2017/07/a-brief-history-of-quantum-alternatives/ https://arstechnica.com/science/2017/07/a-brief-history-of-q...
    - dllthomas 6y ago
      Oh, yeah, Ars is pretty dark, but where it really comes up a lot is amongst child pornographers and terrorists. In fact, it's been shown that when a new pilot wave paper lands, productivity (such as it is) in those areas drops precipitously for a time. In fact it's a robust enough result that it forms a strong argument for increasing science funding. More seriously, I interpreted "darker corners of the internet" in the parent to be a bit tongue in cheek, but to generally be indicating fora where there's a higher ratio of layman to expert, and crackpot to serious practitioner. There was no claim that it isn't discussed outside of that setting, just that it occurs more frequently there (as a proportion of QM discussions in general). This squares with my (poorly informed) general impression.
  - throwaway_pdp09 6y ago
    "have you not heard of this rarely talked-about theory that exists in the dark corners where only whspers live? Well here's a link to the wikipedia page"
- deleted 6y ago
  [deleted]
- flubert 6y ago
  >Or do we just have to shut up and calculate? Edwin James had some interesting commentary on things like this: "From his reply to EPR, we find that Bohr's position was like this: 'You may decide of you own free will, which experiment to do. If you do experiment E1 you will get Result R1. If you do E2 you will get R2. Since it is fundamentally impossible to do both on the same system, and the present theory correctly predicts the results of either, how can you say that the theory is incomplete? What more can one ask of a theory?' While it is easy to understand and agree with this on the epistemological level, the answer that I and many others would give is that we expect a physical theory to do more than merely predict experimental results in the manner of an empirical equation; we want to come down to Einstein's ontological level and understand what is happening when an atom emits light, when a spin enters a Stern-Gerlach magnet, etc. The Copenhagen theory, having no answer to any question of the form: 'What is really happening when - - -?', forbids us to ask such questions and tries to persuade us that it is philosophically naive to want to know what is happening. But I do want to know, and I do not think this is naive; and so for me QM is not a physical theory at all, only and empty mathematical shell in which a future theory may, perhaps, be built." https://bayes.wustl.edu/etj/articles/cmystery.pdf https://bayes.wustl.edu/etj/articles/cmystery.pdf ...and which he goes on to makes some interesting observations about the Bell Inequalities. "Just as Bell revealed hidden assumptions in vonNeumann's argument,so we need to reveal the hidden assumptions in Bell's argument. There are at least two of them, both of which require the Jeffreys view point about probability to recognize..."
  - kgwgk 6y ago
    Typo: Jaynes, not James
    - flubert 6y ago
      Ack. I can't believe I messed that up. He wrote an awesome book: "Probability Theory: The Logic of Science". https://www.amazon.com/Probability-Theory-Science-T-Jaynes/dp/0521592712 https://www.amazon.com/Probability-Theory-Science-T-Jaynes/d... And there is a website with more information and a collection of his papers: https://bayes.wustl.edu/ https://bayes.wustl.edu/ https://bayes.wustl.edu/etj/node1.html https://bayes.wustl.edu/etj/node1.html
AnimalMuppet 6y ago
> But the same principle must require there to be no Preferred Orientation. This leads to the requirement that Planck's constant h be the same in all frames. If the Stern-Gerlach experiment could give results between +h and -h, then the orientation producing the maximum value would be a preferred frame. OK, I think I understand that. > And because of that, when Alice and Bob measure entangled quantum particles, their combined results must violate Bell's inequality. Could you be slightly less ridiculously condensed here? Give a one-or-two-paragraph, accessible-to-the-semi-layman explanation of why this means the result must violate Bell's Inequality? > But to my mind, the biggest take-away is that Einstein's Principle of Relativity absolutely requires that conservation can only be on average. And the same request here. Why does the principle of relativity require that?
- rssoconnor 6y ago
  I can give it a try. The linked nature article had a lot of details. > > But to my mind, the biggest take-away is that Einstein's Principle of Relativity absolutely requires that conservation can only be on average. > > And the same request here. Why does the principle of relativity require that? The way I read this comment was that "the principle of relativity cannot conserve angular momentum on a per-trial basis". In a Mermin Device a pair of entangled spin particles is set to two Stern-Gerlach experiments. The two particles has net (spin) angular momentum of 0 because that's was the net angular momentum of starting material. But if you measure the angular momentum of the two particles in two non-parallel directions, and if we also require that the only answers you are allowed to get are +hbar/2 or -hbar/2, then the sum of the angular momentum you get by adding +/-hbar/2 times one direction plus +/-hbar/2 times a different direction can never be 0. If angular momentum cannot be preserved on a per-trial basis, then I suppose it must be preserved on average, because, I suppose if it isn't preserved on average, then I don't think you can say that angular momentum is preserved at all. > And because of that, when Alice and Bob measure entangled quantum particles, their combined results must violate Bell's inequality. > > And because of that, when Alice and Bob measure entangled quantum particles, their combined results must violate Bell's inequality. > > Could you be slightly less ridiculously condensed here? Give a one-or-two-paragraph, accessible-to-the-semi-layman explanation of why this means the result must violate Bell's Inequality? The really short answer is that if we preserve the angular momentum on average then it entails that the correlations we observe from Mermin Device must match the correlations predicted by quantum mechanics, and therefore violate Bell's inequality for the same reason that predictions of quantum mechanics do. In more detail, if we take the results of a measurement where Alice measures angular momentum in the vertical direction and Bob measures the angular momentum off vertical by theta degrees where Alice gets a result of +hbar/2, then in order for angular momentum to be preserved, Bob's measurement would have to be -cos(theta)hbar/2. Of course Bob is only allowed to get hbar/2 or -hbar/2, so if we want angular momentum to be preserved on average then when we take an ensemble of trials, and filter out only those trials were Alice measures hbar/2, then the average of all of Bob's measurements for those trials should be -cos(theta)hbar/2. That requires that the probability Bob geting hbar/2 when Alice does is (1-cos(theta))/2 (= sin^2(theta/2)), which I believe is the value predicted by quantum mechanics. Once you have the predictions made by quantum mechanics, a violation of Bell's inequality follows by the usual arguments.
wnoise 6y ago
Invariance under velocity changes boosts the whole apparatus. And speeds less than C are observable, and do change w.r.t. an unboosted observer. Rotational invariance would rotate both the source and the detector, and there would be no surprise that the possible results and statistics over them are unchanged. The quantum surprise is that rotating the source relative to the detector leaves the possible results unchanged (though the statistics do change).

jjbinx007 6y ago

I think I have a reasonable mental model of how certain QM processes work by visualising waves in transit coalescing as particles when measured. But I have absolutely no idea how to visualise entanglement. Any tips? Or do we just have to shut up and calculate?

diegoperini 6y ago
My understanding is, Looking at a single object with a fixed angle camera produces similar observations to an entagled pair when the pair is in a similar configuration and where each object in the pair is observed by their own camera except one of the cameras sees a negated result.
GoblinSlayer 6y ago
Often you can visualize entanglement as a superposition of independently evolving states, when those states are solutions of Schrodinger equation.
lisper 6y ago
The key to understanding entanglement for me was to understand that the wave function does not live in physical 3-D space, it lives in configuration space. A wave function that lives in physical 3-D space is a special case that applies only to a system that consists of a single unentangled particle. In that case, physical 3-D space and configuration space are the same. But in general, a wave function for N particles will live in a 3N-dimensional configuration space.
- mlmonge 6y ago
  It seems to me that the configuration space of which you speak essentially explains the wave/particle duality. My B.S. Physics education from years ago never explained it as such as far as I can recall. I'm far from any expertise understanding, but this makes much sense. Any further pedagogical commentary would be most appreciated!
  - lisper 6y ago
    > configuration space ... explains the wave/particle duality I don't think so. You can have a wave function in physical 3-D space as a (very common) special case and you still have the wave-particle dichotomy. Why do you think configuration space explains WPD?
    - elcritch 6y ago
      Really WPD more fundamentally arrives from the “conversion” of the probability wave (living in configuration space as you put it) into properties in 3D physical space. Measuring the QM system is analogous to a Fourier transformation and has the same mathematical limitations arising from a Fourier transform from wave to discrete space. Depending on the question you ask (tied to the convolution function) you get either discrete particle answers or wavelike answers. The probability wave does live in 3D space AFAICT, but the QM properties like spin, charge, momentum etc are tied into that 3D space and form a combined configuration space. The really odd part to me is that at macro scales the probability waves collapse neatly into classic physics in 3D space, but still react in quantum fashion at small local atomic scales. As in the configuration spaces generally can only be determined for small subsystems but not a whole macro system without the “conversion” step.
tim333 6y ago
The way I see it is particles go all possible ways until you narrow things down with a measurement or something along those lines - an interaction that fixes where the thing is. (So in the two slit experiment with one particle it goes through both slits until its position is pinned down by hitting the screen.) In the entanglement experiment described the particles have angular momentum every which way until the angular momentum of one is pinned down by measurement whereupon the other one is also pinned down to the opposite by conservation of momentum. There is still a sort of spooky action at a distance when that happens or perhaps a splitting of the multiverse 'at a distance' into many worlds where the spins point different ways.
- pault 6y ago
  I wish, for the love of god, that the early pop science about quantum physics hadn't used the phrase "when observed" when describing the wave collapse. If they had said "when interacted with by another force" (or anything along those lines), we wouldn't have loads of new age dummies talking about how the particle "knows" it is being observed by a conscious mind. No quantum woo, no Deepak Chopra.
  - hellotomyrars 6y ago
    I don't think you're giving the people who peddle bullshit enough credit. They'll just figure something else out.
  - strogonoff 6y ago
    > "when interacted with by another force" But that is not the same, right? I mean, if it interacted with a force and no observation was made, the wave function doesn’t collapse, does it? Honest question (to avoid any defensiveness, I should disclose that I don’t subscribe to panpsychism). > or anything along those lines Any suggestions?
    - pault 6y ago
      > if it interacted with a force and no observation was made, the wave function doesn’t collapse, does it? Any two systems interacting will cause the collapse. It doesn't matter if the system is attached to a scientist or not. > Any suggestions? No, I'm a software developer, not a quantum physicist. :)
Waterfall 6y ago
Have you heard of Bohemian or pilot wave theory for QM? It's very rarely discussed except in darker corners of the internet but it's surprising to me it's not used as visualization. https://en.m.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theory https://en.m.wikipedia.org/wiki/De_Broglie%E2%80%93Bohm_theo...
- leephillips 6y ago
  I didn’t know I was writing for the “darker corners of the internet”! https://arstechnica.com/science/2017/07/a-brief-history-of-quantum-alternatives/ https://arstechnica.com/science/2017/07/a-brief-history-of-q...
  - dllthomas 6y ago
    Oh, yeah, Ars is pretty dark, but where it really comes up a lot is amongst child pornographers and terrorists. In fact, it's been shown that when a new pilot wave paper lands, productivity (such as it is) in those areas drops precipitously for a time. In fact it's a robust enough result that it forms a strong argument for increasing science funding. More seriously, I interpreted "darker corners of the internet" in the parent to be a bit tongue in cheek, but to generally be indicating fora where there's a higher ratio of layman to expert, and crackpot to serious practitioner. There was no claim that it isn't discussed outside of that setting, just that it occurs more frequently there (as a proportion of QM discussions in general). This squares with my (poorly informed) general impression.
- throwaway_pdp09 6y ago
  "have you not heard of this rarely talked-about theory that exists in the dark corners where only whspers live? Well here's a link to the wikipedia page"
deleted 6y ago
[deleted]
flubert 6y ago
>Or do we just have to shut up and calculate? Edwin James had some interesting commentary on things like this: "From his reply to EPR, we find that Bohr's position was like this: 'You may decide of you own free will, which experiment to do. If you do experiment E1 you will get Result R1. If you do E2 you will get R2. Since it is fundamentally impossible to do both on the same system, and the present theory correctly predicts the results of either, how can you say that the theory is incomplete? What more can one ask of a theory?' While it is easy to understand and agree with this on the epistemological level, the answer that I and many others would give is that we expect a physical theory to do more than merely predict experimental results in the manner of an empirical equation; we want to come down to Einstein's ontological level and understand what is happening when an atom emits light, when a spin enters a Stern-Gerlach magnet, etc. The Copenhagen theory, having no answer to any question of the form: 'What is really happening when - - -?', forbids us to ask such questions and tries to persuade us that it is philosophically naive to want to know what is happening. But I do want to know, and I do not think this is naive; and so for me QM is not a physical theory at all, only and empty mathematical shell in which a future theory may, perhaps, be built." https://bayes.wustl.edu/etj/articles/cmystery.pdf https://bayes.wustl.edu/etj/articles/cmystery.pdf ...and which he goes on to makes some interesting observations about the Bell Inequalities. "Just as Bell revealed hidden assumptions in vonNeumann's argument,so we need to reveal the hidden assumptions in Bell's argument. There are at least two of them, both of which require the Jeffreys view point about probability to recognize..."
- kgwgk 6y ago
  Typo: Jaynes, not James
  - flubert 6y ago
    Ack. I can't believe I messed that up. He wrote an awesome book: "Probability Theory: The Logic of Science". https://www.amazon.com/Probability-Theory-Science-T-Jaynes/dp/0521592712 https://www.amazon.com/Probability-Theory-Science-T-Jaynes/d... And there is a website with more information and a collection of his papers: https://bayes.wustl.edu/ https://bayes.wustl.edu/ https://bayes.wustl.edu/etj/node1.html https://bayes.wustl.edu/etj/node1.html

AnimalMuppet 6y ago

> But the same principle must require there to be no Preferred Orientation. This leads to the requirement that Planck's constant h be the same in all frames. If the Stern-Gerlach experiment could give results between +h and -h, then the orientation producing the maximum value would be a preferred frame. OK, I think I understand that. > And because of that, when Alice and Bob measure entangled quantum particles, their combined results must violate Bell's inequality. Could you be slightly less ridiculously condensed here? Give a one-or-two-paragraph, accessible-to-the-semi-layman explanation of why this means the result must violate Bell's Inequality? > But to my mind, the biggest take-away is that Einstein's Principle of Relativity absolutely requires that conservation can only be on average. And the same request here. Why does the principle of relativity require that?

rssoconnor 6y ago
I can give it a try. The linked nature article had a lot of details. > > But to my mind, the biggest take-away is that Einstein's Principle of Relativity absolutely requires that conservation can only be on average. > > And the same request here. Why does the principle of relativity require that? The way I read this comment was that "the principle of relativity cannot conserve angular momentum on a per-trial basis". In a Mermin Device a pair of entangled spin particles is set to two Stern-Gerlach experiments. The two particles has net (spin) angular momentum of 0 because that's was the net angular momentum of starting material. But if you measure the angular momentum of the two particles in two non-parallel directions, and if we also require that the only answers you are allowed to get are +hbar/2 or -hbar/2, then the sum of the angular momentum you get by adding +/-hbar/2 times one direction plus +/-hbar/2 times a different direction can never be 0. If angular momentum cannot be preserved on a per-trial basis, then I suppose it must be preserved on average, because, I suppose if it isn't preserved on average, then I don't think you can say that angular momentum is preserved at all. > And because of that, when Alice and Bob measure entangled quantum particles, their combined results must violate Bell's inequality. > > And because of that, when Alice and Bob measure entangled quantum particles, their combined results must violate Bell's inequality. > > Could you be slightly less ridiculously condensed here? Give a one-or-two-paragraph, accessible-to-the-semi-layman explanation of why this means the result must violate Bell's Inequality? The really short answer is that if we preserve the angular momentum on average then it entails that the correlations we observe from Mermin Device must match the correlations predicted by quantum mechanics, and therefore violate Bell's inequality for the same reason that predictions of quantum mechanics do. In more detail, if we take the results of a measurement where Alice measures angular momentum in the vertical direction and Bob measures the angular momentum off vertical by theta degrees where Alice gets a result of +hbar/2, then in order for angular momentum to be preserved, Bob's measurement would have to be -cos(theta)hbar/2. Of course Bob is only allowed to get hbar/2 or -hbar/2, so if we want angular momentum to be preserved on average then when we take an ensemble of trials, and filter out only those trials were Alice measures hbar/2, then the average of all of Bob's measurements for those trials should be -cos(theta)hbar/2. That requires that the probability Bob geting hbar/2 when Alice does is (1-cos(theta))/2 (= sin^2(theta/2)), which I believe is the value predicted by quantum mechanics. Once you have the predictions made by quantum mechanics, a violation of Bell's inequality follows by the usual arguments.

wnoise 6y ago

Invariance under velocity changes boosts the whole apparatus. And speeds less than C are observable, and do change w.r.t. an unboosted observer. Rotational invariance would rotate both the source and the detector, and there would be no surprise that the possible results and statistics over them are unchanged. The quantum surprise is that rotating the source relative to the detector leaves the possible results unchanged (though the statistics do change).