I'd like to ask a very specific question about the entanglement nonlocality. I know that it is not possible to send a faster than light signal using this phenomenon, so that's not what I'm asking about. Still, the entanglement - let's think in particular about the GHZ experiment and about the example in this answer - seems to involve a sort of signal exchanged between the entangled spins. After reading the other posts and comments, the main explanation I've found is based on the following points:

  1. we can dismiss realism in favor of locality => that would imply there are no hidden variables and the results of the measurements connot be predefined
  2. the results can be explained in terms of classical correlation, without any causal effect (on those spacelike separated measurement processes)

While point 1 is clear enough to me (after the answer, comments and links in reply to my previous question), I fail to see - in math equations - where is point 2 coming from. Actually the classical concept of correlation is defined for 2 separated distributions of probability and by definition they cannot include reference to each other. In quantum mechanics the mathematical part boils down to either the collapse of a wavefunction (=> nonlocal, not a correlation) or a state vector that includes both indices (refers to both the spacelike separated components) and - afaik - it translates back to a classical concept of causal effect, not to what is intended by correlation.

Difference between correlation and dependence

Classically, correlation does not imply dependence and it is not expressed - by definition - as a relation of dependence. You simply have two separated series of events and their probability distributions and then you compute their correlations. There are several articles showing that the entanglement could violate local causality, for example look at the Experimental test of nonlocal causality.

Local causality is the combination of what we call causal parameter independence— there is no direct causal influence from the measurement setting Y (X) to the other party’s outcome A (B) — and causal outcome independence, stating that there is no direct causal influence from one outcome to the other.

more specifically, they write

Local causality captures the idea that there should be no causal influence from one side of the experiment to the spacelike separated other side. Formally, this is a constraint on the conditional probability distributions: p(a|b,x,y,λ) = p(a|x,λ) and p(b|a,x,y,λ) = p(b|y,λ). We would like to stress that local causality is not equivalent to signal locality, which follows from special relativity and imposes constraints on the observable probabilities only: p(a|x, y) = p(a|x) and p(b|x, y) = p(b|y). The natural generalization of signal locality to include the hidden variable is typically referred to as parameter independence or locality: p(a|x,y,λ) = p(a|x,λ) and p(b|x,y,λ) = p(b|y,λ) (36). Parameter independence, together with what is often referred to as outcome independence p(a|b,x,y,λ) = p(a|x,y,λ) and p(b|a,x,y,λ) = p(b|x,y,λ), then implies local causality.

From the same article, the conclusion is

quantum mechanics allows for correlations that violate this inequality, therefore witnessing its incompatibility with causal models that satisfy local causality and measurement independence.

In other words, the point is that Quantum Mechanics satisfies setting independence but violates outcome independence (and - afaik - there's no clear explanation why), so it also violates local causality (implied by both locality and outcome independence).

  • $\begingroup$ What does it mean for two distributions to be "separated"? $\endgroup$
    – WillO
    Jul 5, 2017 at 10:25
  • $\begingroup$ @WillO I mean that classically correlation does not imply dependence and it is not expressed - by definition - as a relation of dependence. You simply have two separated series of events and their probability distributions and then you compute their correlations. $\endgroup$
    – Xlsx2020
    Jul 5, 2017 at 10:42
  • $\begingroup$ I've edited and expanded my question to address some concerns raised by the only comment received till now. Can you show me why this question still deserves a -1? $\endgroup$
    – Xlsx2020
    Jul 5, 2017 at 13:17
  • $\begingroup$ Related: "For example, in Bell's Theorem, we notice that we cannot send superluminal signals so we assume there are no superluminal influences in reality, but this gets us into conflict with the experimentally observed predictions". They study the implications of Bell's Theorem and time symmetry. $\endgroup$
    – Xlsx2020
    Jul 5, 2017 at 17:50
  • 1
    $\begingroup$ "the results can be explained in terms of classical correlation" - who says that? The whole point of entanglement is that the results can't be explained in terms of classical correlation. $\endgroup$
    – tparker
    Jul 6, 2017 at 16:13

1 Answer 1


The main point to consider is that the non-signaling theorem cannot be reduced to a notion of an absolute

no influence can exist which is faster than the speed of light.

The non-signalling theorem does not prohibit the existence of instantaneous influences in the formation of nonlocal correlations: such influences are indeed superluminal.

In other words, outcome independence doesn't matter as locality and the non-signalling theorem was introduced, on such purpose, as a theorem for avoiding conflict with special relativity:

no superluminal influence can exist that can be controlled for signalling purposes.

A more in depth read on the subject can be found in this IOP article.


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