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?

  • 3
    $\begingroup$ 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 May 25 '17 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. $\endgroup$ – WAH May 25 '17 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 May 25 '17 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. $\endgroup$ – WAH May 25 '17 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 May 25 '17 at 22:22

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. $\endgroup$ – WAH May 25 '17 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 May 25 '17 at 21:44
  • $\begingroup$ @WAH: Forgot to address by name earlier. $\endgroup$ – kpv May 25 '17 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.