I wanted to ask this question in physics section but I think it is most philosophical question than physical. Question is about Bell's theory that proves there are no hidden variables. My question is that if hidden variables exist and system condition is determinrd before experiment or system condition is not determined now and in the moment of observation it will determine, statistically is there any difference between the two? When we throw a dice: is there any difference if dice from the beginning know a particular number (for example 3) that will come or dice does'n know the number at begining but when it stops its particular number be specified? In both situations the probability is 1/6. So the bell's theory is philosophically wrong.

  • 1
    what exactly does "philosophically wrong" mean?
    – virmaior
    May 11, 2016 at 8:54
  • I mean it is logically or reasonably wrong no matter how you do the experiments.
    – Alberto
    May 11, 2016 at 9:13
  • As John Bell discussed superdeterminism in a BBC interview:"There is no need for a faster than light signal to tell particle A what measurement has been carried out on particle B, because the universe, including particle A, already "knows" what that measurement, and its outcome, will be." in my opinion it is the definition of hidden variables. Isn't it?
    – Alberto
    May 11, 2016 at 14:18
  • No, it is a known variable of a collection of particles, rather than a hidden variable on the particle.
    – user9166
    May 11, 2016 at 16:47
  • I can't get the point. It is known but not for us so it is hidden. If a collection of particles has a variable so each particle inherite the variable (or a function of it). So whats the diffrent? My understanding of all of these is that Copenhagen interpretation is wrong.
    – Alberto
    May 11, 2016 at 16:58

3 Answers 3


So first of all : for a macroscopic object like a dice, it's possible to know the result. If I'm not mistaken, this question is for a quantum particle with a parameter (like spin, or a charge). The Copenhagen interpretation says that the state of the particle is only determined when you perform a measure on it.

Bell's inequality ensures that the information between two intricate particles won't go faster than the speed of light and provides causality. I don't see any philosophical issue with that.

Bell's inequality are always violated (see also Alain Aspect and is experiment in 1981), and one as to find a way to live with that result, in a quantum relativistic world. At this point, several ideas can be discussed. Here is a link where peoples explain those points way better than me : https://en.wikipedia.org/wiki/Bell%27s_theorem

Something important is that when you do the measure on a particle, the state of the other one will change, but if you want to use this information, you have to travel next to this particle and you can't do this faster than the speed of light, thus the relativity won't be broken.

  • I know that we are talking about particles. And dice treat differently. I mean maybe Einstein was right and there is hidden variables that we don't know about them and therefor when we do a measurement the state of particle is determined before that not at the moment of measurement. And I want to say no experiment can distinguish between Copenhagen interpretation or hidden variables theories.
    – Alberto
    May 11, 2016 at 14:02
  • Dice are a collection of particles -- physics doesn't suddenly change at some scale. The statistical nature of the results just disappears into the Law of Large Numbers. The standard deviations become so narrow, we just ignore them. But the effect is just hidden, not absent.
    – user9166
    May 11, 2016 at 16:45
  • In a dice there is a really large number of particles and you have to take into account quantum decoherence. I think it's a totally different problem if you have macroscopic object. The laws to describe them can be very different. I guess we are starting a Schrodinger's cat discussion here.
    – JSFDude
    May 11, 2016 at 20:57

Bell's inequality doesn't say that there can't be any hidden variables. What it says is that any theory consistent with quantum mechanics can't have hidden variables that are both causal and local.

David Bohm's hidden variable theory, for example, is consistent with quantum mechanics but is non-local.

As to your point about pre-determined systems. The key point is that one can set up a quantum system whereby a random event in one part of that system can affect the state in another part faster than message could be sent. Sure you can argue that the original system "knew" about the state of the random event generator before it started but this very swiftly gets into the realms of magic.

  • "...original system "knew" about the state of the random event generator before it started..." I don't understand. Isn't this the definition of a local hidden variable?
    – Alberto
    May 11, 2016 at 16:51
  • @Alberto "Isn't this the definition of a local hidden variable?" No. Most attempts at a hidden variable theory would not require you to keep adding variables to ensure you get the outcome. Such a theory would have no predictive power and would require quantum mechanics as an input (to determine the outcome you need to match) so would be pointless.
    – Alex
    May 11, 2016 at 17:22
  • I don't said we should put QM away. QM can predict observations very well and a hidden variable theory is pointless as you said but my problem is about the interpretations of QM that refuse realism and locality. If we consider hidden variable or superdeterminism problems with QM will solve without requiring a new theory.
    – Alberto
    May 11, 2016 at 19:07
  • @Alberto To be clear, I didn't say HV theories are pointless. I said a local (and causal) one was because, to our current level of knowledge, any HV theory must match QM. Such a theory must be either non-local or non-causal. Also, your question was about Bell's inequality and its "wrongness". This comment seems to be about a lack of comfort with QM interpretations. You may want to think about asking another question on that topic.
    – Alex
    May 12, 2016 at 10:05

Bell's theorem shows that certain quantum results are incompatible with both local realist models where the results are predetermined, and local realist models where the particle doesn't "decide" what result to give until the experimenter measures it. Here's a simple analogy I wrote up a while ago to illustrate:

Suppose we have a machine that generates pairs of scratch lotto cards, each of which has three boxes that, when scratched, can reveal either a cherry or a lemon. We give one card to Alice and one to Bob, and each scratches only one of the three boxes. When we repeat this many times, we find that whenever they both pick the same box to scratch, they always get the same result--if Bob scratches box A and finds a cherry, and Alice scratches box A on her card, she's guaranteed to find a cherry too.

Classically, we might explain this by supposing that there is definitely either a cherry or a lemon in each box, even though we don't reveal it until we scratch it, and that the machine prints pairs of cards in such a way that the "hidden" fruit in a given box of one card always matches the hidden fruit in the same box of the other card. If we represent cherries as + and lemons as -, so that a B+ card would represent one where box B's hidden fruit is a cherry, then the classical assumption is that each card's +'s and -'s are the same as the other--if the first card was created with hidden fruits A+,B+,C-, then the other card must also have been created with the hidden fruits A+,B+,C-.

The problem is that if this were true, it would force you to the conclusion that if Alice and Bob are picking randomly which box to scratch on each trial (with a 1/3 chance of A, B, or C each time), then if they do this a large number of times, we should expect that in the subset of trials where Alice and Bob happened to pick different boxes to scratch, they should find the same fruit at least 1/3 of the time. For example, if we imagine Bob and Alice's cards each have the hidden fruits A+,B-,C+, then we can look at each possible way that Alice and Bob can randomly choose different boxes to scratch, and what the results would be:

Bob picks A, Alice picks B: opposite results (Bob gets a cherry, Alice gets a lemon)

Bob picks A, Alice picks C: same results (Bob gets a cherry, Alice gets a cherry)

Bob picks B, Alice picks A: opposite (Bob gets a lemon, Alice gets a cherry)

Bob picks B, Alice picks C: opposite results (Bob gets a lemon, Alice gets a cherry)

Bob picks C, Alice picks A: same results (Bob gets a cherry, Alice gets a cherry)

Bob picks C, Alice picks picks B: opposite results (Bob gets a cherry, Alice gets a lemon)

In this case, you can see that that if they are equally likely to pick each combination of boxes, then 2 times out of 6 when they choose different boxes, they will get the same fruit (i.e. a 1/3 chance of the same result). You'd get the same answer if you assumed any other preexisting state where there are two fruits of one type and one of the other, like A+,B+,C- or A+,B-,C-. On the other hand, if you assume a state where each card has the same fruit behind all three boxes, so either they're both getting A+,B+,C+ or they're both getting A-,B-,C-, then of course even if Alice and Bob pick different boxes to scratch they're guaranteed to get the same fruits with probability 1. So if you imagine that when multiple pairs of cards are generated by the machine, some fraction of pairs are created in inhomogoneous preexisting states like A+,B-,C- while other pairs are created in homogoneous preexisting states like A+,B+,C+, then the probability of getting the same fruits when you scratch different boxes should be somewhere between 1/3 and 1. 1/3 is the lower bound, though--even if 100% of all the pairs were created in inhomogoneous preexisting states, it wouldn't make sense for you to get the same answers in less than 1/3 of trials where you scratch different boxes, provided you assume that each card has such a preexisting state with "hidden fruits" in each box.

But now suppose Alice and Bob look at all the trials where they picked different boxes, and found that they only got the same fruits 1/4 of the time! That would be the violation of Bell's inequality, and something equivalent actually can happen when you measure the spin of entangled photons along one of three different possible axes. So in this example, it seems we can't resolve the mystery by just assuming the machine creates two cards with definite "hidden fruits" behind each box, such that the two cards always have the same fruits in a given box.

Something mathematically analogous is predicted in certain experiments where entangled photons are passed through polarizers at different angles, in which the probability that both photons will have the same result (both pass through their respective polarizers, or both are blocked) is cos^2(theta), where theta is the angle between the two polarizers. Suppose the experimenters decide in advance that on each trial, they will choose randomly between one of three angles for their polarizer: 0 degrees from vertical, 60 degrees, or 120 degrees. On a given trial, if both experimenters choose the same angle, then theta=0 so they are guaranteed to get the same result with probability 1 (both photons pass or both are blocked), but if they choose different settings the probability of getting the same result would be cos^2(±60) or cos^2(±120), which in both cases gives a probability of 1/4. But the Bell inequality whose derivation I sketched says that if you want to explain the perfect match when both experimenters make the same choice using local hidden variables, the probability of a match when they make different choices should be no lower than 1/3.

Note that if the lotto cards didn't decide what fruit to show until the experimenters picked which box to start scratching, then if Alice and Bob made their decisions at points in spacetime such, given the assumption of locality, no information about Alice's choice would be available at the position and time Bob made his own choice (and vice versa), then there would be no way to get the result that whenever they scratch the same box, they are guaranteed to get the same fruit. And the argument above shows that if you try to solve this by assuming the fruits were predetermined when the two cards were printed at a common location, before they were sent on their separate ways towards Alice and Bob, then you can't reproduce the fact that if they choose different boxes, the chance they'll get the same fruit would be 1/4.

  • at 60 degree if experiments says that probability is 1/4 , so in (1/3 -1/4=1/12) case that we expect to see entangled photons there is no entanglement?
    – Alberto
    May 12, 2016 at 23:44
  • I don't understand your question. In the formalism of QM, 100% of the photon pairs are entangled, and that entanglement determines the probability of 1/4. What is subtracting 1/4 from 1/3 supposed to represent physically?
    – Hypnosifl
    May 13, 2016 at 0:11

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