12
$\begingroup$

According to the special theory of relativity, distant simultaneity depends on the observer's reference frame.

And, according to the quantum theory, in the case of two entangled particles, a measure on one of the particles simultaneously affects the second one. Under which reference frame is this simultaneous?

Improve this question
$\endgroup$
1
  • 5
    $\begingroup$ Entanglement produces correlation, not causation. There is nothing in special relativity that forbids correlations without causality. Ontologically, of course, special relativity is simply not a complete theory and, more importantly, it's not a holy cow. Even if quantum mechanics would violate special relativity, the world wouldn't end. Quite the contrary, we would simply have cast SR aside as a poor description of nature and replaced it with something better. $\endgroup$
    – CuriousOne
    Commented Dec 23, 2014 at 12:58

2 Answers 2

Reset to default
14
$\begingroup$

It doesn't really matter, because the phrase "simultaneously affects the other particle" is misleading.

Let's suppose you have a pair of totally anticorrelated photons. You measure one of them, then you'll know the outcome of the other one. The phrase "the measurement simultaneously affects the other particle" is not physical, because until you actually measure the other particle, you can't even notice anything different. There is no "effect". The only thing we can meaningfully talk about is the two measurements of the two particles. Now, depending on the reference frame, one will come before the other (or they are simultaneous) and whatever we measure, one result will imply the other.

This is why I think that the term "the particle simultaneously affects the other particle" is not very good, because it implies something like an active link - but depending on the reference frame particle A would affect particle B or the other way round. There is no "one particle affecting the other". Only if you are in a specified reference frame, it looks like there is an immediate influence of one particle on another.

Improve this answer
$\endgroup$
8
  • $\begingroup$ This is a clear message. The uncertainty is in our knowledge about which particle is in which state. the particles are entangled since they are produced together. Perhaps somebody learned it by an other way, but the result of both thoughts are the same. So we not use the easier one? $\endgroup$
    – HolgerFiedler
    Commented Dec 23, 2014 at 13:25
  • 2
    $\begingroup$ @Martiin But when you make a measurement at point A, doesn't the mathematical description of what is happening at point B also change? State vector reduction doesn't just have a local influence, does it? $\endgroup$
    – neuronet
    Commented Dec 23, 2014 at 14:34
  • 1
    $\begingroup$ I think you could put this in a different way and say that this isn't true temporal simultaneity, but a kind of logical simultaneity. So for example, you might say that the instant you accept that the Peano axioms are true, you simultaneously accept that all the theorems they prove are also true. But that's not something that actually happens in time; it's a logical, not a physical consequence. $\endgroup$
    – senderle
    Commented Dec 23, 2014 at 15:52
  • $\begingroup$ "Only if you are in a specified reference frame, it looks like there is an immediate influence of one particle on another." yes, and this was the question...what reference frame(s)? I think the answer, as it stands, needs work (for one, within the formalism of nonrelativistic QM, with state vector reduction there is change at two locations at hte same time). Is there an expert on relativistic quantum theory here? $\endgroup$
    – neuronet
    Commented Dec 23, 2014 at 22:25
  • $\begingroup$ @neuronet: The problem you are having boils down to this: Is the wave-function epistemic or ontic. If it is epistemic, then it only represents our knowledge. As such, your problem vanishes, since there is no problem at having different knowledge in different reference frames (then, every reference frame would have a state according to its knowledge). Since the measurement outcomes are the same, nothing changes. $\endgroup$
    – Martin
    Commented Dec 23, 2014 at 23:44
-2
$\begingroup$

Measuring one particle does not affect the other at all. Bell's theorem explains that if you try to simulate an entangled quantum system by modelling a quantum system with a classical stochastic variable the result has to be non-local. However, quantum systems are described by Heisenberg picture observables, which are represented by Hermitian operators, not classical stochastic variables. The particles each exist in multiple versions that can interact with one another in interference experiments, which is why they can't be described by classical stochastic variables. Each particle's observables describe quantum information about the relations between the different versions of each particle, but this information can't be revealed by measurements on either particle alone:

http://arxiv.org/abs/quant-ph/9906007

http://arxiv.org/abs/1109.6223

http://arxiv.org/abs/quant-ph/0104033.

In each measurement, both of the outcomes happen and the correlations are established when the results are compared, not when the measurement is done on each particle.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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