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(Sorry if I'm spamming your board with Popular Science speculations, but I just thought this could be an interesting thought experiment).

There are quite a few videos on youtube about Bell's Experiment. This experiment supposedly disproves the hypothesis of local hidden variables. We split entangled photons, measure them separately and while our results look individually random, they are somehow align if we compare them afterwards. This supposed to prove that photon "decided" its orientation at the time of measurement.

This heavily relies on our ability to "surprise" the photon with a random direction we're going to measure it. But what if that photon knew all along in which direction it was going to be measured. Since photons travel at the speed of light, they don't experience time, hence they should know the future, and there shouldn't be a way for us to "surprise" it at all.

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    $\begingroup$ Bell tests don’t have to use photons. $\endgroup$
    – Joe
    Sep 30 at 12:16
  • $\begingroup$ You're right, I'm sure. I've excused myself as being a non-expert. For whatever reason I've decided that it does, since youtube.com/watch?v=zcqZHYo7ONs this video talks about both about photons and Bells Theorem. Thank you for your correction. $\endgroup$
    – avloss
    Sep 30 at 13:21
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    $\begingroup$ I was with you up to the last sentence. If information from the future can reach the photon that's currently bouncing off your head, then information from the future can reach your head. $\endgroup$
    – WillO
    Sep 30 at 13:57
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  1. Bell's inequalities hold more generally. You can verify the existence of nonclassical correlations in a variety of platforms, including those that have nothing to do with photons or light in general.

  2. "This heavily relies on our ability to "surprise" the photon with a random direction we're going to measure it. But what if that photon knew all along in which direction it was going to be measured":

    I don't think this means anything. Bell's inequalities are a way to certify the existence of a specific type of correlations between measurement outcomes. It's not about the photon being "surprised" or not. It's about the way you interact with the photon during measurement. In other words, it's about the types of "questions" you "ask" the two photons.

    Any kind of classical correlation between the two parties isn't sufficient to violate Bell's inequalities. If you assume that the systems "know the direction along which they will be measured", then yes, you can violate the inequalities. This is often referred to as the superdeterminism loophole. However, it's worth remarking that any probability distribution can be obtained with such an assumption. In other words, you get a model which is not falsifiable, as it can "explain" any observation. Furthermore, you'd have to come up with a physical mechanism to explain how the information about the measurement choices somehow leaks to the detectors. In many situations, such a mechanism would be really weird and far-fetched.

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  • $\begingroup$ Thank you, makes sense. I'm only learning about "superdeterminism" here after asking this question. physical mechanism to explain how the information - I guess that was my whole question. I'm speculating (perhaps very naiively) that the physical mechanism is "time travel" - photon (or any other elementary particle travelling at the speed of light) knows the future. By "surprised" i meant exactly that, photon already "knows" what questions it will be asked. It's like going to a test, after hacking into the teachers computer and knowing all the questions in advance. $\endgroup$
    – avloss
    Sep 30 at 13:29
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    $\begingroup$ @avloss yes, but the sentence "the photon knows the future" doesn't really mean anything. The whole point of a physical description of a phenomenon is to describe how things now "explain" how things will be later. A description of current state based on the future state would be useless, how could you ever use it? Also, the "photons don't experience time" thing is a common misconception. It simply isn't meaningful to consider the reference frame of a photon in special relativity. $\endgroup$
    – glS
    Sep 30 at 14:02
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I do not think your argument "they don't experience time, hence they should know the future" has a very strong foundations...

But you can make particle to know the future pretty naturally if

  1. Physical laws are fully deterministic so that every experiment we make is predetermined by initial data
  2. The relevant particles have access to relevant initial data to make their predictions

In fact, both points are satisfied in universe governed by Newtonian physics, so they are not something too crazy to contemplate. It seems to me, that the overoptimistic claim that Bell's theorem disputes local hidden variable theories uses its own conclusion as an assumption. If we can model whole universe by deterministic local hidden variable theory, then Bell's theorem is not applicable.

See https://en.wikipedia.org/wiki/Superdeterminism

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  • $\begingroup$ Yeah, thank you, this is exactly what I was thinking. I'm surprised this is not mentioned in the popular science youtube videos, since it seems to be rather relevant. Also, a very entertaining idea. $\endgroup$
    – avloss
    Sep 30 at 13:16
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A)

by hidden variables I suppose it's meant a parameter that could change the results. Suppose the singlet state and a measurement only on one side with the operator $A\otimes 1$ and $A=diag(1,-1)$. We could start by saying that the final states in B were $\{|\phi_n\rangle=\pmatrix{\cos\phi_n\\\sin\phi_n}\}_{n=1}^N$. We take as starting point that the sum of the probabilities of measurement of A are 1 : $$p(+)+p(-)=1=\sum_{n=1}^N |\langle +\phi_n |\psi\rangle|^2+|\langle -\phi_n |\psi\rangle|^2$$

From this we deduce $N=2$ as unique condition and for example $p(+)=\frac{\sin^2\phi_1+\sin^2\phi_2}{2}$.

Thus only by knowing we have an intricated singlet state and measuring only one particle induce a possible variation in the probabilities. Measuring the probabilities would indicate the values of the phi's. Note that there could be more end Ector in B if a probability for each were given mimicking a mixed endstate.

B)

Bell experiments rely on Bell theorem. A version of it were : quantum correlation cannot be modeled with local hidden variables. This model is : if $A(a,v)=sign(\cos(a-v))$ then $C(a,b)=-\frac{1}{2\pi}\int_0^{2\pi}A(a,v)A(b,v)dv=-1+\frac{2}{\pi}(b-a)$

Now this correlation can only depend on the relative angle, or if you want the angles a and b are relative to a third choosable direction that we can always set to a, this gives $C(b-a)=-\frac{1}{2\pi}\int A(0,v)A(b-a,v)dv=-1+2(b-a)/\pi$ where b-a is now the relative angle smaller than pi.

If we now want to obtain the quantum cosine correlation, we perform a change of variable such that $-1+2(b-a)/\pi=-\cos (b'-a')$ Hence $b-a=\pi/2(1-\cos( b'-a'))$.

Putting back in the function A we get the model giving the cosine relationship with $A(b'-a',v)=sign(\cos(\frac{\pi}{2}(1-\cos(b'-a'))-v))$.

We can always set $a'$ to 0 and we get the desired function.

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