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Being a layman in the sense that I enjoy reading about physics concepts, but lack the deeper mathematical understanding, I've been trying to wrap my mind around the concept of quantum entanglement and why it is so "spooky".

So, in my thought experiment, imagine:

  1. We have an electron with known quantifiable properties represented macroscopically by a container of 100 black beads.
  2. We have another electron represented by another container of 100 white beads.
  3. For the sake of the example, let us assume that the size of each bead isn't exact, but the density of each is exactly that of some fluid whose volume and density remains fairly constant despite external influences (I don't want to get caught up in some pedantic argument over this aspect of the experiment, but I want to keep it simple, so we'll call it water).
  4. We "entangle" the electrons. (Macroscopically, we dump the two sets of beads into a Galton Board with an even number of bins whose results we cannot see. We know that the statistical distribution of the beads follows a given "wave function" - in this case the binomial probability curve. But that is all we know.)
  5. We "separate" the electrons and move them to distant locations. (Macroscopically, we place the beads from even numbered bins into a container of water that is allowed to spill water out to maintain a constant volume. We do the same for the odd numbered bins. We seal and ship each container of water and beads to opposite coasts.)

The result is that I have two containers weighing exactly the same and externally look identical. But I do not know anything about their content collectively until I analyze the content of one of the containers.

I have a measuring device that can count the beads in a given container and tell me the number of black versus white beads, and another that can tell me the total weight of the beads in a given container. When I take these measurements (my "wave function collapses"), I know what results my counterpart will have measured in the other container whether he or she does so at the same instant or a year from now.

Macroscopically, the results of my example appear no different to me than the results of even the most convoluted of experiments to test quantum entanglement. There's no "spooky action at a distance" going on here. Entangling was the action and that didn't happen at a distance. Measuring is not an action - it's just observation. I can't change the results found in a container as I measure some property of its contents. I'm not violating the speed of light or causing some information paradox.

The point is, I'm just not seeing the mystery here. It just seems like intuitive common sense to me. So, I feel as though I am missing something significant about quantum entanglement. Is it possible to take my macroscopic example and modify it in some way to demonstrate the big mystery that I must be missing?

Mitselplik
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  • You didn't mention Bell's Theorem in your question. That is essential to understanding what you are missing. In modern parlance, the usual examples are constructed with polarization entangled photons. At various angle settings, there are no classical (macroscopic) examples that match the predictions of quantum mechanics without resorting to apparently nonlocal influences. I modestly suggest this page which I authored (I could not determine whether such self-reference is acceptable here, so I hope a moderator will correct me if not): https://drchinese.com/David/Bell_Theorem_Easy_Math.htm – DrChinese Jun 12 '23 at 17:31

1 Answers1

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Consider the four observables "weight of container A", "weight of container B", "number of beads in A" and "number of beads in B". You can treat these as classical random variables governed by a joint probability distribution (as you've told us explicitly in point 4.) So there is no entanglement here.

The fact that you can write down an example with no entanglement does not tell us anything about what happens in an example with entanglement.

As for the question in your final paragraph, suppose that you repeat this experiment many times and discover that a) whenever you and your faraway friend both measure number, you always get exactly complementary results (if one measures 98, the other measures 102). b) Whenever one of you measures number and the other measures weight, you always get exactly complementary results (if one gets, say, 89 beads then the other gets 111 grams). c) Whenever you both measure weight, you always get the same result (if one gets 101 grams, the other gets 101 grams). Now the observables are entangled --- you can confirm with a little computation that there is no joint probability distribution that fits these observations. Therefore they can't have been generated by anything like your Galton board.

[What happens if you both measure both weight AND number? Then the above is impossible. Therefore this will only work in a situation where you are each constrained to measure only one or the other --- as is the case in real-world entanglement.]

WillO
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  • You can't generate non-classical probabilistic behavior unless you measure non-commuting observables. This means that if you measure the weight, you change the number. But if what you mean by weight and number is what everybody else means by them, this doesn't happen. So, even if you are constrained to measure only one or the other, this analogy doesn't give any real intuition for quantum mechanics. – Peter Shor Jun 12 '23 at 12:41
  • Furthermore, your explanation of complementary bases is incorrect. In quantum mechanics, if one person measures weight, the other person gets a random number. – Peter Shor Jun 12 '23 at 12:47
  • @PeterShor: "If you measure the weight you change the number" seems to me to be a misleading way of putting it; in fact there is no well-defined number after you measure the weight (and usually there is no well-defined number before you measure the weight either). – WillO Jun 12 '23 at 12:47
  • But you said that if one person measures 89 beads, then the other gets 111 grams ... this contradicts your last comment — 111 is well-defined. – Peter Shor Jun 12 '23 at 12:48
  • @PeterShor : Re your second comment, I agree that this particular example cannot be realized in quantum mechanics. I continue to think it is a perfectly accurate illustration of what it means for four variables to have no joint probability distribution, which is key to understanding what is sometimes called "spooky" about the distributions that are possible in quantum mechanics. – WillO Jun 12 '23 at 12:49
  • @PeterShor: We have A) the setup described by the OP, B) things that can occur in QM, and C) the example I gave. I claim that all three are distinct. I claim that B) and C) both differ from A) in exactly the way that the OP is trying to understand. And I claim that B) is simpler than C), so it was a better illustration for the OP's purpose. He is not asking what happens in QM; he is asking how QM differs from his example. The answer is "in essentially the same way that C) differs from your example, and C might be easier to grasp quickly". – WillO Jun 12 '23 at 12:51
  • Re the "this contradicts your last comment". No. The "last comment" referred to said that there is no well-defined number in QM, which was directly responsive to your previous comment. There is a well-defined number in the example I gave. It is not contradictory for two different scenarios to have different properties. – WillO Jun 12 '23 at 12:56
  • Suppose you put 20 particles in a box. You close it and then weigh it. Now, you open the box again and count the particles. Are there still 20? If there aren't, doesn't that mean that weighing the box in some sense changes the number of particles? – Peter Shor Jun 12 '23 at 13:01
  • To me, "changes the number of particles" implies "changes the state from one eigenstate of the number observable to another", which, in the examples I believe you have in mind, is not what happens. The new state is not an eigenstate of the number observable at all. So I don't like this language, but in view of the fact that you are Peter Shor, I would start from the presumption that you mean the right thing, though you are expressing it in different language than I would choose. – WillO Jun 12 '23 at 13:20
  • TY Peter for responding. Examples I find to explain the concept differ very little from a "joint probability distribution" so how can my example be dismissed? Take electron spin being up or down for example. 50% chance either way - no different than if I randomly chose one of a pair of gloves and then observed which glove I have. In your response, a) makes sense, b) doesn't represent correlation since weight and amount are independent, but observations of both will correlate, and c) doesn't correspond with any example I've ever seen. Can you explain more? – Mitselplik Jun 12 '23 at 13:35
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    @Mitselpik: You need to consider four random variables --- say "spin of first electron when measuring apparatus is held at an angle of $A$ degrees", "spin of first electron when measuring apparatus is held at an angle of $B$ degrees", and the same for the second electron. With entanglement, there is no joint probability distribution for these four observables. That is the essence of entanglement. In the example I gave, there is also no joint probability distribution for the four observables. – WillO Jun 12 '23 at 13:46
  • @PeterShor is objecting to the fact that the statistics I posited in my example are not the same as the statistics you'd get in quantum mechanics. I wasn't trying to reproduce QM; I was giving the simplest example I could think of where you can check for yourself that there is no joint probability distribution. If you want to narrow your focus to just two observables that can be measured simultaneously (or in fact any number of observables that can be measured simultaneously) you won't be able to see this phenomenon. – WillO Jun 12 '23 at 13:48
  • I hope I understand you both at least a little better. Is the takeaway that it's not about exact measurements, but the probability of seeing a given result for some aspect of B when that aspect of A is measured. Is that closer to the mark? – Mitselplik Jun 12 '23 at 13:51
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    See my answer here: https://physics.stackexchange.com/a/330571/4993 – WillO Jun 12 '23 at 13:54