So a few months ago a research team did the following experiment: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.118.060401

They shot lasers at certain detectors trying to detect the spin of the photons and to prove that there is an underlying hidden variable that influences all of these properties, for which quantum entanglement exists. The measurement of the spin depends on the orientation of the detector. To each detector is a telescope attached and pointing to a 600ly distant star. And based on the wavelengths of the photons that hit the telescope, the detector is reoriented.

So now my question is, is this really random? I mean yes, the photons from the sun were created even before the laser were set up. But since we live all in the same universe, nothing is really COMPLETELY random, just to a certain degree. Isn't bells theorem even a proof that causality doesn't even exist at a certain degree or that true randomness doesn't exist?

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    $\begingroup$ No, it is not a proof of that. QM postulates that the outcome of a measurement is random. If randomness comes from hidden variables (thus most likely not 100% random), or is inherent to our universe, is something that nobody knows. $\endgroup$
    – user126422
    Commented May 25, 2017 at 16:21
  • $\begingroup$ Not quite. We can show experimentally that Bell's Inequality is violated; therefore there are no local hidden variable theories. Which is to say that if special relativity holds, then we know that results of experimental measurements aren't deterministic. $\endgroup$ Commented May 25, 2017 at 21:17
  • $\begingroup$ But if quantum entanglement, special relativity and no hidden variables hold at the same time, how can you then describe all of these events by a single theory? I mean, the only way that comes to my mind, is that information can be send in the past, to influence the past, so that all of these axioms can be true. $\endgroup$
    – Maxim
    Commented May 25, 2017 at 21:31
  • $\begingroup$ It's precisely the entanglement that creates the correlation between the measurements, which then violates Bell's Inequality. The postulate of wavefunction collapse in quantum mechanics means that the state of the system can change instantaneously over infinite distance. This instantaneous change over infinite distance doesn't violate the speed limit from the speed of light because no information is transmitted between observers. $\endgroup$ Commented May 25, 2017 at 21:56
  • $\begingroup$ You might take a look at devices designed to generate actual random (not psuedorandom) number. en.wikipedia.org/wiki/… $\endgroup$
    – David Elm
    Commented May 25, 2017 at 22:22

2 Answers 2



In order to understand the motivation behind this experiment a good understanding of the assumptions on which Bell's theorem stands is required. There are two such assumptions:

  1. Locality (there is no instant communication/influence between the particles'source - S and each detector - D1, D2). In other words S does not instantly influence D1 or D2, D1 does not instantly influence D2 or S and D2 does not instantly influence D1 or S. A theory that denies this assumption is a non-local theory. The de Broglie–Bohm interpretation of QM is an example of such theory.

  2. Independence (the state of D1 and D2, or D1 and S or D2 and S are not correlated as a result of a past common cause). A theory that denies this assumption is a so-called superdeterministic theory. No such theory is currently available in complete form, but 't Hooft's cellular automaton interpretation of QM is a step in this direction.

The experiment you mention tries to put some limits on superdeterministic theories by pushing the hypothetical common cause 600 years into the past. This is not about randomness per se (S, D1 and D2 could each evolve according to a deterministic process) but about the absence of correlations between the states of these devices.

In my opinion the relevance of such an experiment is minimal. We have experimental proof that such common-cause type correlations apply to the gravitational interaction in the case of galaxies for example (the stars in a galaxy do not move independently - they more or less orbit the galactic center) and there is little reason to assume that the electromagnetic interaction which represents the domain of Bell theorem experiments is more limited in this respect.


There is no scientific experimental way to prove absence of causality. Because that would result into a paradox.

For example, causality is - you do this (in given conditions) and so and so will happen.

Any experimental proof, by definition has to be repeatable. If causality did not exist, we could not rely on any experiment including the experiment that disproves causality. Because how do we know that the results are caused by what we did, or by randomness. Science itself would become vague in that case.

There is causality.

Randomness refers to "our inability" to compute the exactness, which becomes impossible at certain levels, specially at quantum levels.

Even in classical case, we can compute how long it will take for a pool of water to evaporate, but we can not compute which day/time, a specific water molecule will evaporate. The process is too random that it is beyond our computational capabilities.

Then, when we go too close to computations, we need to measure things too closely and that process could change our computations. For example, poking an entangled particles, ends its entanglement.

  • $\begingroup$ I don't believe you're using words correctly. Causality means a result does not violate special relativity. I think you mean reproducibility when you write causality. $\endgroup$ Commented May 25, 2017 at 21:15
  • $\begingroup$ I looked up Wikipedia - 1. the relationship between cause and effect. 2. the principle that everything has a cause. $\endgroup$
    – kpv
    Commented May 25, 2017 at 21:44
  • $\begingroup$ @WAH: Forgot to address by name earlier. $\endgroup$
    – kpv
    Commented May 25, 2017 at 23:32
  • $\begingroup$ I think WAH is right - the word deterministic is more suitable here $\endgroup$ Commented Jul 19, 2019 at 9:07

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