If you were to explain the Einstein-Podolsky-Rosen paradox to high school students (age 16, with no particular strength in math), what kind of intuitive example would you provide to make things understandable?
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1Are you asking for "real life application" as the title suggests (which would be hard I think... how can a paradox have real world applications?) or for intuitive examples to make things understandable as the body suggest (which is a good question, +1)? Please edit the title or the body consistently – glS Jan 09 '15 at 15:20
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There is a border case of quantum mechanics which can be explained classically in just one sentence: For two entangled photons in vacuum, both spacetime intervals are 0, everything is happening simultaneously. See the (still unanswered) question Local EPR-experiments with photons in vacuum? – Moonraker Jan 09 '15 at 15:44
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1I have just edited the title. Now it should be clearer. – Javier Arias Jan 09 '15 at 15:49
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Some people use "EPR paradox" interchangeably with Bell's theorem showing that violations of Bell inequalities rule out local hidden variable explanations for QM statistics--they're closely related, but Bell's theorem was more definitive in ruling out all possible local hidden variable explanations. If you're interested in a simple illustration of why Bell inequality violations can't be explained with local hidden variables that predetermine measurement results, see my lotto card example here. – Hypnosifl Jan 09 '15 at 17:04
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This paper on doing practical quantum key distribution with polarization entangled photons might be a candidate. The EPR paradox facilitates distribution of correlations (in the measurement outcomes that can be converted into a stream of bits) to two parties in a secure way. Any act of eavesdropping during the communication disturbs the quantum states, changing the measurement outcomes. This results in errors that warn the parties of the eavesdropping. – jayann Jan 09 '15 at 20:37
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Although to clarify/prevent confusion, I must add that EPR paradox (or Bell's theorem) is not essential for proving why quantum key distribution is secure. – jayann Jan 09 '15 at 20:40
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The Mermin-Peres magic square game? – Kevin Kostlan Jan 09 '15 at 21:24
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Here's a good paper explaining the basics of EPR and Bell's theorem: Is there a moon when nobody looks? Reality and the quantum theory" (pdf) by N. David Mermin. – Hypnosifl Jan 11 '15 at 00:12
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The EPR paradox is based on something that would seem obvious to any good high school student. If you have a uranium atom and wait for it to decay, then you get a thorium atom and an alpha particle. The alpha particle has a lot of energy, so it shoots off in one direction, and the thorium atom, by recoil, shoots off in the other direction. So if the uranium atom starts off in the middle of a football field, and you sit with a geiger counter at one end, then if you catch an alpha particle, the guy at the other end of the field will catch a thorium atom. It's not much of a paradox.
The sticky point is that according to the Copenhagen interpretation of quantum mechanics, until the moment you detected the alpha particle, it MIGHT have been anywhere: because up to that point, the alpha particle has something called a "wave function". The actual moment of detection is called the "collapse of the wave function" and it's a major philosophical issue in quantum mechanics. But that's not the point of the EPR paradox.
The real problem is that just as the alpha particle might have been anywhere, the thorium atom "might" have been anywhere too...because it also has a wave function. But once you find the alpha particle at one end of the field, suddenly there's no choice for the thorium atom...it has to be at the other end. The paradox is that something which happened at one end of the field instantaneously affected the possible outcomes that could take place a hundred yards (or a hundred light-years) away.
It sounds like an easy paradox to resolve: you just have to realize that at the moment the uranium atom decayed, the alpha particle went one way and the thorium atom went the other. That would solve the problem, but .... for reasons you can't explain to a high-school student, that would lead to enormous contradictions in the whole theory of quantum mechanics.
I'm not sure if anyone in 1936 thought that the EPR was a practical problem and not just a philosophical problem. The "thought experiment" proposed by EPR (Podalsky specifically, if I'm not mistaken), while not identical to the version I've posted here, was not something you could set up in practise. People talk a lot about how Bell changed everything when he proposed his famous inequality in 1964, but for my money it was Bohm who turned this into a practical question in 1950 when he proposed a version of EPR paradox involving spin states.
I talk about this in greater detail in my blogpost "Einstein => Bohm => Feynmann => Bell": http://marty-green.blogspot.ca/2011/11/einsteinbohmfeynmannbell.html
EDIT: I said the other day that you could "solve" the paradox by simply assuming that the thorium and alpha particles took off on specific pathways at the moment they were created, but this would create "enormous difficulties" which you "couldn't explain to a high school student". When someone (like me) says something like that, what he really means is that he doesn't understand it himself. So I thought about it and I'm going to try to explain what the contradictions are.
The problem is that both alpha "particles" and thorium "atoms" behave as waves. They interfere like waves. In particular, you could set up double-slit experiments at opposite ends of the football field and observe diffraction patterns...
Well actually, you can't do it with a single uranium atom in the middle. You could do it but there's no diffraction pattern with only one particle. You'd need a block of uranium, so it's constantly emitting particles in all directions. Then you could set up a double-slit at each end of the field (or a hundred light-years apart) and observe diffraction patterns.
But the thing about the two-slit experiment is if you watch the "particles" as they pass through the slit, you destroy the diffraction pattern. So what you do is set up your geiger counters at one end and catch the particles before they can go through the slits. They have to be really small geiger counters because the slits are really close together...but that's just a technical problem. The point is that assuming you can distinguish which slit they were going for, you now know which slit the other particles are heading for at the far end of the field...or a hundred light years away. So this should destroy the interference pattern.
But that means something you did at one end of the field (or the universe) instantly affected something that someone else could observe at the other end. That's a real problem.
And that's where the EPR paradox leads. I'm afraid I've actually overstated the case, because I appear to have designed a "practical" system that allows for faster-than-light communication. And that's not allowed in anyone's quantum mechanics. So I've done something wrong, but I can't exactly see what. Anyone care to weigh in?
Marty Green
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1So, is this EPR paradox basically the same thing as quantum entanglement? – Time4Tea Jan 09 '15 at 21:39
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Interesting to learn more about the history of these ideas, thanks for posting the link to your blog. One minor correction: the name is "Feynman", not "Feynmann". – Hypnosifl Jan 09 '15 at 21:50
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It is in fact BORIS PODOLSKY, not Podalsky. Never mind, thanks a lot for the insightful comment. Would it be philosopically fair to say that the thorium and the alpha particle are somehow bounded by a trace (say, a sub-index with the same feature or number which relates them as, say, two separate occurrence of a unique underlying phenomenon? – Javier Arias Jan 10 '15 at 12:21
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I can't comment on your question as to how the thorium atom is linked to the alpha particle. In standard QM, we say they do not each have their own independent wave function, but instead there is a single six-dimensional wave function which contains both of them. (That's the wave function that "collapses" when either one is detected.) – Marty Green Jan 10 '15 at 16:08
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1The answer to your question in the edit is the same as the ones I gave here and here to questions about why you can't use a delayed choice quantum eraser to create an FTL communicator--it turns out that if a particle is entangled in such a way that there's even the potential to learn which slit one went through by measuring the other in the right way, then there will be no double-slit interference pattern in the particle's total probability of landing at various points on the screen, – Hypnosifl Jan 10 '15 at 23:32
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(continued) although there can still be an interference pattern in the *coincidence count" between photons on one side and photons that landed at a particular spot on the other side. A diagram of just the type of setup you describe can actually be seen in fig. 2 of this paper, and the author writes "particle 2 can serve to find out which slit particle 1 passed and therefore no interference pattern arises." – Hypnosifl Jan 10 '15 at 23:41
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@Hypnosifl: "... there's even the potential to learn which slit one went through by measuring the other in the right way, then there will be no double-slit interference pattern ...". Magic! – bright magus Jan 11 '15 at 10:25
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@hypnosifl your answer is very helpful and I am still trying to work out the consequences; but I see now that the example I gave fails to create faste-than-light communication for a more prosaic reason. You DO get an interference pattern because when I substitued the block of uranium for the single uranium atom, I lost the perfect entanglement of the momenta of the thorium and alpha particles. The momenta were only entangled because there was nowhere else to go when the SINGLE uranium atom decayed. There is no entanglement with a block of uranium. – Marty Green Jan 11 '15 at 15:29
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@Marty Green - I don't understand what you mean--isn't a "block of uranium" just a large collection of uranium atoms? If each individual uranium atom has "perfect entanglement of the momenta of the thorium and alpha particles", and thus the predicted probability distribution for each individual thorium or alpha particle going through slits would lack a double-slit interference pattern, then a large collection won't show a double-slit interference pattern either. – Hypnosifl Jan 11 '15 at 15:36
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If you have only one uranium atom, then when it decays, the momentum of the thorium must be equal and opposite to the momentum of the alpha particle. If there's a block or uranium, no such restriction. The momentum is divided three ways, so measuring the alph particle tells you nothing about the momentum of the thorium atom. – Marty Green Jan 11 '15 at 15:53
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@Marty Green - But there is some spacing between the atoms, so immediately after the decay (if we imagine making regular measurements at the position of each atom at very short time intervals) a uranium atom would have disappeared and there would just be an entangled thorium and alpha particle in the same vicinity, only after enough time has passed for either of them to move to the position of some other uranium atom (or other particle in the block) and interact with it could they transfer any momentum to the rest of the block. – Hypnosifl Jan 11 '15 at 17:28
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(continued) So it's still true in this example that you there is the possibility in principle of determining the momentum of the alpha particle by measuring the thorium atom some very short time after the decay, before it's had time to reach some other particle and interact with it, and this is all that's required to ensure the alpha particle's probability distribution on the screen won't show any double-slit interference. – Hypnosifl Jan 11 '15 at 17:29
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@hypnosifl I'm not sure that the very problematical measurement system you propose actually meets the condition of being executable even "in principle". But if it is, you're saying that if you had a properly isolated block of uranium emitting a continuous flux of alpha particles...that you wouldn't be able to set up a double-slit system and measure a diffraction pattern? I don't think that's right. I think you would get a diffraction pattern. – Marty Green Jan 11 '15 at 17:41
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@Marty Green - Are you saying it's problematic from a practical point of view, or even from a theoretical point of view? In the theory of QM one often considers measurements of a precision that would be almost impossible to achieve in practice but doesn't violate any of the basic principles of the theory like the uncertainty principle. And yes, I'm saying that according to the theoretical principles I quoted in this answer, I think there should be no double-slit diffraction pattern in this case, though there is still single-slit diffraction. – Hypnosifl Jan 11 '15 at 18:10
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In general, it's a good idea to be suspicious if you find yourself going to extreme lengths to justify how something might be measurable. – Marty Green Jan 11 '15 at 19:08
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@Marty Green - But there are many things that would be very difficult to measure in practice, but physicists think they are likely to be true simply because they follow naturally as theoretical predictions from theories that have a lot of observational support for their other predictions (examples include Hawking radiation from black holes, and gravitational waves). If we just want to analyze what QM predicts would happen in this experiment theoretically, would you agree that all that matters to the argument is what it would theoretically be possible to measure? – Hypnosifl Jan 16 '15 at 16:19
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(continued) To put it another way, are you arguing that we should be suspicious that the theory of QM would actually predict no double-slit interference here if we did a detailed mathematical analysis (which would seem to imply that the physicists I quoted in the other answer were speaking sloppily when they suggested that any possibility of determining which-path information in principle from an entangled partner would destroy the double-slit interference) or are you arguing we should be suspicious that the theory of QM's predictions would be correct in this experiment? – Hypnosifl Jan 16 '15 at 16:25
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I appreciate your interest in my opinion, but I don't think I could have been more clear in my last statement: you should be suspicious of highly contrived measurement proposals. Surely you've read Feynmann's story of how people proposed different methods of measuring which slit the electron went through (he was drawing on the extended correspondence between Bohr and Einstein on this question) only to show that all such efforts were ultimately frustrated by the uncertainty principle. – Marty Green Jan 16 '15 at 16:44
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And yes, I think physicists are extremely sloppy in this kind of thing. Like the authors of that paper you referred me to who used exactly my setup of an example of producing entangled pairs. It only works for isolated atoms, not a block. They forgot to take into account the momentum absorbed by the uranium sample. – Marty Green Jan 16 '15 at 16:45
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@Marty Green - But an important element of Feynman's story was that there was always a good theoretical reason why attempts to violate the uncertainty principle would fail in principle, the reply was never just "this experiment would be too difficult to carry out in practice". If you missed that, I think you missed the point of Feynman's discussion! – Hypnosifl Jan 17 '15 at 22:54
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(cont.) Whereas if you do appreciate that point, then you should make clear whether your argument is that a detailed analysis would find theoretical reasons the experiment I mention can't be done in principle, regardless of practical considerations (if so, what do you think they are? If you don't know, what makes you so confident there would be such theoretical reasons, confident enough to say the professional physicists who wrote that paper must have "forgot" something?), or if you think there is reason to believe the theory itself is incorrect about what would happen in this experiment. – Hypnosifl Jan 17 '15 at 22:58
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But your professional physicists DID get it wrong. They show the particles shotting out of the block along opposite trajectories, towards distant detectors. You already pointed out that they should have had submicroscopic detectors in the interstitial spaces between the uranium atoms (!). That's not what they're showing in their picture. – Marty Green Jan 18 '15 at 02:23
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@Marty Green - Which specific paper that I linked to are you referring to? What type of particle is shown shooting out, might it be a type that has a high probability of traveling through a small piece of Uranium without being scattered or absorbed? In that case there'd be no need for an "interstitial" detector. Also, would you say that the same logic applies to crystals used in "parametric down-conversion", where a single photon goes in and if it interacts in the right way with the crystal, two entangled photons can emerge? – Hypnosifl Jan 18 '15 at 23:32
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(cont.) Finally, are you refusing to answer my question about whether your argument is that A) we should suspect that a detailed theoretical analysis would give a diff. prediction (just as purely theoretical analyses were able to show why the uncertainty principle wouldn't be violated in Feynman's story), or B) that we should suspect the theory is wrong? For someone who complained about the arguments of physicists being "extremely sloppy" before, it seems rather sloppy to say we should be "suspicious" of an argument but not distinguish these very different types of suspicions! – Hypnosifl Jan 18 '15 at 23:36
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You should know that I suffer from a low-level personality disorder which compels me to disagree with authority figures. Recently a Provincial Court judge has upheld the University of Winnipeg's decision to bar me from the campus because of my "single-minded determinedness to demonstrate that (my) point of view is the superior one". You can read the full decision here (quoted passage from paragraphs 46-50): http://www.canlii.org/en/mb/mbpc/doc/2014/2014mbpc42/2014mbpc42.html – Marty Green Jan 19 '15 at 02:23