1
$\begingroup$

If two particles are entangled, with one moving near the speed of light relative to the other, would observing one tell us something about the other in the other's future or past?

$\endgroup$
  • $\begingroup$ In the classical EPR experiment we have two photons moving even at the speed of light relative to each other, in the opposite directions. So measuring the momentum of one tells us the momentum of the other. $\endgroup$ – Conifold Jun 21 '17 at 1:44
1
$\begingroup$

To determine that the particles were entangled, you need to measure both of them, and then compare the results of the measurements. You would then indeed potentially find that the measurements were superluminally correlated.

However, in order to compare the results, and realize they were correlated, you need to be in the forward light cone of both particles. Therefore no superluminal communication is possible. This is true independently of the particles' relative speeds.

$\endgroup$
  • 1
    $\begingroup$ We find that the measurements are correlated, we do not find that they are superluminally correlated. That can only be inferred from violations of Bell inequalities, if one assumes that they were supposed to hold in the first place. But that assumption is only justified under classical "local realism" about quantum particles, so the inference is invalid. $\endgroup$ – Conifold Jun 21 '17 at 1:40
0
$\begingroup$

Let us take a simple example, the decay of a pi0 to two photons. These photons are entangled because the pi0 has spin 0 whereas a photon has spin one, so one must have spin projection parallel to its motion and the other antiparallel.

One of the two gammas goes off to the moon, the other is detected in the lab and its spin measured. If it is +1 we immediately know that the one that went to the moon had spin -1. Time does not enter the problem because the lab is at rest.

Take a Z0 to μ+μ- . The same spin argument holds and again time does not affect the outcome, measuring the charge of the one in the lab, tells us what the charge of the one that went to the moon is, in addition to the knowledge of spin orientation, because it should add up to 1, the spin of the Z0.

$\endgroup$

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.