I'm not a physicist (computer science background), but always was fascinated with the concept of entanglement. As many, the first intuition I had was - well, the two particles share some knowledge from the get go, so no matter how far you hold them, they still behave as one unit. I understand that this is called the hidden variables theory and that is generally refuted by Bell's theorem, and later in experiments.

Part 1 of the question

I understand that a recent "no loopholes" experiment in 2015 [1] "proved" Bell's theorem (e.g. that the hidden variable theory can't explain / predict all entanglement situations)

Is there a good non physicist resource that explains how that experiment disproves this?

Part 2 of the question

Here is my computer science laymen question, I'm sure there is a basic flaw in this thought experiment but I'm not sure what it is. Let's say I have 2 RSA tokens created together with the same seed value and same random number generator, until I press the button, I have no clue what value it will show, but if I press the buttons on both tokens, no matter how far they are, they will all produce the same number, this is not a surprising fact and no need to perform an experiment as it's well known (actually no need of two tokens, the fact that the server and the token have the same seed and generate the same number is enough)

I understand that a single quantum particle has no way of holding all the complexity of a chip, the best scientist are working hard just to get one stable qbit, and this is just a very ridiculous and theoretical thought experience, but let's say it was possible to minimize RSA tokens to the quantum level, ant man style, (sorry for causing some of you to cringe) - what in that 2015 experiment has that debunks a "hidden RSA token style variables" theory? What in Bell's theorem (I tried to read it, I don't have enough background in physics to even make it past the first paragraph) negates a hidden deterministic pseudo-random number generator like mechanism?

What are the holes in the "RSA token hidden variable" idea? I'm sure others came up with this analogy, can you help me find the source? And is there a good source that refutes it?

Part 3 of the question

Bell was quoted saying:

"There is a way to escape the inference of superluminal speeds and spooky action at a distance. But it involves absolute determinism in the universe, the complete absence of free will."

[2] about super-determinism is exactly that RSA token explanation?

If so, I don't get it, the fact we have deterministic RSA tokens doesn't imply we have no free will, it just implies that the RSA has a pseudo-random number generator, it doesn't negate the fact our brains might have "true" randomness, so can't there be a coesistence of determinism in quantum entanglement but no determinism in our own mind? why are they necessarily related?


[1] Hensen, B; Bernien, H; Dréau, AE; Reiserer, A; Kalb, N; Blok, MS; Ruitenberg, J; Vermeulen, RF; Schouten, RN; Abellán, C; Amaya, W; Pruneri, V; Mitchell, MW; Markham, M; Twitchen, DJ; Elkouss, D; Wehner, S; Taminiau, TH; Hanson, R. "Loophole-free Bell inequality violation using electron spins separated by 1.3 kilometres". Nature. 526: 682–686. arXiv:1508.05949 Freely accessible. Bibcode:2015Natur.526..682H. doi:10.1038/nature15759. PMID 26503041.

[2] The quotation is an adaptation from the edited transcript of the radio interview with John Bell of 1985. See The Ghost in the Atom: A Discussion of the Mysteries of Quantum Physics, by Paul C. W. Davies and Julian R. Brown, 1986/1993, pp. 45-46

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    $\begingroup$ I recommend the book 'Quantum reality' by Nick Herbert. ISBN13: 9780385235693 Bell's theorem and the implications of it being experimentally verified are discussed in chapter 11, p.199-211 (But don't skip, the exposition in the preceding chapters is helpful.) I found Herbert's discussion of Bell's theorem very transparent and very convincing. For a non-technical book this is the most in-depth and thorough one that I know of. $\endgroup$
    – Cleonis
    Aug 17, 2018 at 14:34
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    $\begingroup$ It would be better if you asked one question at a time. $\endgroup$ Aug 17, 2018 at 23:18
  • $\begingroup$ @PM2Ring Thanks for your insight, I appreciate it. I've deleted my comment. $\endgroup$
    – safesphere
    Aug 18, 2018 at 3:56

4 Answers 4


I'm going to update part your RSA token analogy to reflect the experimentally observed behavior of entangled particles. Hopefully this helps demonstrate why the hidden-variables theory doesn't work.

Let's say I have 2 RSA tokens created together with the same seed value and same random number generator, until I press the button, I have no clue what value it will show, but if I press the buttons on both tokens, no matter how far they are, they will all produce the same number

Let's start from the same point with two "entangled" RSA tokens that have the same seed value and same random number generator. But let's change our experiment so that we randomly select which token to press the button on first. Repeat this many times, recording the order pressed and number generated for each token.

running experiment... each row is a result where { order pressed, number generated }

Token A Token B
{ 0, 123 } { 1, 345 }
{ 1, 345 } { 0, 123 }
{ 1, 345 } { 0, 123 }
{ 0, 123 } { 1, 345 }
{ 0, 123 } { 1, 345 }
... ...

Looking at the results we see there's a clear correlation between being pressed first and generating the number 123, and being pressed second and generating the number 345. This suggests that the random number generator for this particular seed value generates 123 on the first press and 345 on the second press.

But what's spOoOoky is that this holds true regardless of which token is randomly selected to be pressed first. How could that be? How does the second token pressed know that it's second and therefore to generate the second "random number" for its given seed value?

One explanation is that somehow information is being shared between these two entangled RSA tokens so that they know when the other has been pressed. This is the "spooky action at a distance" explanation. Another explanation is that the order and numbers generated were predetermined.

But what isn't an explanation is that the tokens were generated with hidden state that could be used to algorithmically produce these results given the random sequencing of button presses.

Moving from analogy to reality, this behavior can be seen in experiments when two entangled particles are passed through polarized filters at randomly selected angles of 0°, 22.5°, or 45°. What's observed is that the behavior of the second particle passing through a 45° filter (whether it gets through or not) is statistically correlated to whether the first particle passed through the 0° or 22.5° filter. I.e., the behavior of the second particle is influenced by the randomly selected path of the first particle. Spooky!

Hopefully this helps. I have a computer science background myself and started with your same line of reasoning. I think its particularly hard for programmers to accept that there isn't a hidden state explanation for entanglement. My personal opinion - there's shared state in some universal memory that we haven't discovered yet, where the cosmic speed limit doesn't apply to information transfer.

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    $\begingroup$ Great answer. The missing part for me was that there is the "was measured first" thing, which they can't agree upon in advance since it's a result of an external action. It's an if statement that has an external factor (if(otherWasMeasured) return x else y), the otherWasMeasured is an external state vs internal state. I assume I can rename otherWasMeasured to hasOurJointWaveFunctionCollapsedAlready?. Which means the "propagation" of the collapse of a wave function, is immediate, either collapsed fully or not collapsed, regardless of space? $\endgroup$
    – Eran Medan
    Jul 22, 2022 at 23:44

I have token $A$. You have token $B$. Let's pretend there are only two possible values, $x$ and $y$, for each token.

Experiment One: I observe $A$ and you observe $B$. We do this repeatedly and find probabilities $p,q,r,s$ for the observations $(x,x)$, $(x,y)$, $(y,x)$ and $(y,y)$. These probabilities will necessarily add to one. No matter what the values of $p,q,r,s$ might be, there is no problem explaining this outcome with classical physics and classical probability. We simply posit that whatever is creating the tokens creates an $(x,x)$ pair with probability $p$, an $(x,y)$ pair with probability $q$, and so on. It is, in effect, a pseudo-random number generator, just as you've said.

Experiment Two: I observe $A$ but you don't observe $B$. Based on our explanation for Experiment One, what do you predict for the outcome of Experiment Two? You might well reason that I will observe an $x$ only when the pair is of the form $(x,x)$ or $(x,y)$, and the probability of that is $p+q$. Call that Eran's Theorem.

Now what if I perform Experiment Two and discover that Eran's Theorem is violated? I get $x$ with some probability much higher than $p+q$. How do you explain this? You might, in your desperation, suggest the possibility of superdeterminism --- I have no free will, and therefore am able to "choose" to perform Experiment Two not at random times, but only at times when an observation of $x$ is particularly likely.

In fact Eran's Theorem is never violated, so you don't have to worry about this.

But in just-slightly-more-complicated situations, with three or more possible observations instead of just two, we have Bell's Theorem, which tells us that, given the outcomes of one set of experiments, and given the expectation that those outcomes are driven by a pseudo-random number generator, the outcomes of some other experiment have to be such-and-such. This is the exact analogue of Eran's Theorem, in a slightly more complicated setting. And this time, the predictions really are violated.

If you want to understand why some people have been tempted to consider superdeterminism, you don't need to understand the details of Bell's Theorem (though in fact the details are quite easy to master and require only a little algebra; no physics at all). Just think about Eran's Theorem, if you prefer. If Eran's Theorem were violated, how would you attempt to explain that violation?

  • $\begingroup$ I'll do my best to understand the math, but I'm sure it's solid. But one assumption I'm not sure I am fully on board is the probabilities being known. I think that in an actual token, with a really complex PRNG that is also time based (is time a non local hidden variable?), it's almost impossible to know the probabilities, e.g. let's say you return x if the number generated is event and y if its odd, since it's time based, how would you calculate the expected value? by pressing every second? every hour? random times? how many times? This probably doesn't change the math, right? $\endgroup$
    – Eran Medan
    Aug 16, 2018 at 20:42
  • $\begingroup$ I wish I could understand this paper better: arxiv.org/pdf/quant-ph/0301059.pdf $\endgroup$
    – Eran Medan
    Aug 16, 2018 at 21:12
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    $\begingroup$ Re the probabilities being known: In the case of quantum mechanics, you know the probabilities because the theory tells you what they have to be, and you confirm this by repeating the experiments a great many times over and discovering that the results are consistent with the theory's predictions. $\endgroup$
    – WillO
    Aug 17, 2018 at 1:46
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    $\begingroup$ If you don't like all the variables, take this case: Whenever only I choose to look at my token, it says $x$. Whenever only you choose to look at your token, it says $x$. Whenever we both choose to look at our tokens, they both say $y$. Try explaining that with a pseudo-random number generator. $\endgroup$
    – WillO
    Aug 17, 2018 at 1:48
  • $\begingroup$ "If Eran's Theorem were violated, how would you attempt to explain that violation?" In some circumstances, the probability (x,x)=p and probability (x,y)=q were measured. But now you get x at probability considerably bigger than p+q. That is unlikely, but it will occasionally happen. But it happens more than can be explained by chance. That says that something has changed since you measured and got p and q. The old probabilities no longer fit the reality. That happens sometimes. $\endgroup$
    – J Thomas
    Aug 17, 2022 at 19:38

The answer to your 1st question.

In fact, Bell's inequality can be deduced using only elementary math.

Suppose you have three coin-flip generators, not necessarily fair, and may be connected. There are, in total, 8 different possible outcomes. After fiddling with various inequalities and equalities, you can prove that, assuming they are just ordinary coin-flip generators, a certain inequality holds. This is Bell's inequality.

Assuming a local hidden variable theory exists, i.e. there is some invisible variable causing all of the apparent randomness, and does not transport information that exceed the speed of light. Then one can construct a system which, the hidden variable theory predicts its behavior as the 3 coin-flippers described above. Therefore, the experiment outcome must agree with Bell's inequality.

However, the experiment results showed that this inequality is not true. Therefore such a theory cannot hold.


Here is my computer science laymen question, I'm sure there is a basic flaw in this thought experiment but I'm not sure what it is. Let's say I have 2 RSA tokens created together with the same seed value and same random number generator, until I press the button, I have no clue what value it will show, but if I press the buttons on both tokens [...] they will all produce the same number,

As I understand it, what you are describing here is two identical Rube Goldberg machines, identical to such a degree that they are guaranteed to produce the same output when triggered to produce output. Well, if the setup is guaranteed to produce the same output each time, why bother with the Rube Goldberg machine? You can just as well store that output plain and simple.

Sure, when you eyeball the Rube Goldberg machine you're not able so call what it will output, but that's not relevant. You set up another identical instance of that Rube Goldberg machine, you run it, and then you know the output that the other two identical instances will produce when it's their turn to be triggered.

By ensuring that every instance of the Rube Goldberg machine setup always produces the same output any randomness has been eliminated.

Acknowledging the criticism by Norbert Schuch in a comment I'm striking out the glove analogy part of my answer. I'm not sure the reasons that Norbert Schuch gives are definitive. But there are additional consideration to view it as unsuitable, so: strike.
To reproduce the quantum physics case the outcome must be inherently random.

I rather like the following visualisation. A rubber handglove can be turned inside out; as you turn it inside out a righ-handed glove is flipped to left-handed, and vice versa.
Thought demonstration:
A glove is flattened in such a way that when it is released it will either pop out to the left-handed form or the right-handed form. The other glove from the pair is prepared in the same way. Also, the pair of gloves is created in such a way that they are an entangled pair. The two gloves are send to different places. At a later point in time the two gloves are triggered to pop out. (Doesn't have to happen simultaneously.) Quantum physics says that whether they pop to left-handed or right-handed is truly random, but if they are entangled they will always pop out into the mirror opposite of each other.

  • $\begingroup$ one difference is that RSA tokens are time based, they don't always produce the same value, the value depends on time, can time be considered an external hidden variable? $\endgroup$
    – Eran Medan
    Aug 17, 2018 at 18:29
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    $\begingroup$ The glove analogy is horrible and utterly misleading (as it is entirely classical). Also, it is not true that randomness is needed to violate Bell's inequality (and randomness wouldn't fix the RSA example). There are deterministic distributions which violate it (namely p(x,y|a,b)=1 iff x+y = a*b mod 2, which can be deterministic, e.g. by having always x=1). $\endgroup$ Aug 17, 2018 at 23:20
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    $\begingroup$ @EranMedan I hadn't looked up how RSA tokens work, so I didn't know about the time dependency. Note: the measurements of the entangled entities do not have to happen simultaneously so an analogy that involves time dependency is a departure from the actual physics. In experiments the measurements are usually as simultaneously as possible, but that is only for the efficiency of the experiment. So that is key: whether the measurements of the entangled particles occur simultanously or not makes no difference. The correlation of the measurement results is independent of that. $\endgroup$
    – Cleonis
    Aug 18, 2018 at 3:54
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    $\begingroup$ The glove stuff isn't horrible, IMHO, but it doesn't really help. I assume that when a glove is flattened it's impossible to determine how it will pop out before you make it pop. That part is fine. The problem is that there's nothing in the glove system that forces a pair of gloves to be correlated, it just happens by magic, so it doesn't actually explain anything. I guess it's ok as an introduction to how a system that has entangled components appears to behave, but it's only a marginal improvement on the pair of socks in two boxes scenario, which is bad because it implies a hidden variable. $\endgroup$
    – PM 2Ring
    Aug 18, 2018 at 4:34

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