1
$\begingroup$

In this scenario let us take a particle that can have either spin up or spin down. Let's say that observer 1 observed the particle's spin to be down. But observer 2 hasn't observed it yet. So, then would the wave-function have collapsed from his/her perspective? Would the particle be in a superposition of an observed/unobserved state (if there is such a thing)? Also, how would a similar scenario play out when there is a pair of entangled particles?

$\endgroup$
  • 4
    $\begingroup$ Related: the Wigner's friend thought experiment. $\endgroup$ – knzhou Oct 2 '17 at 13:58
  • 1
    $\begingroup$ .... or, more accurately, the story of how Wigner's friend, a.k.a. your Observer 1, got him(/her)self entangled with the experiment. $\endgroup$ – Emilio Pisanty Oct 2 '17 at 14:01
  • $\begingroup$ "Also, how would a similar scenario play out when there is a pair of entangled particles?" Please ask one question per post. $\endgroup$ – DanielSank Oct 2 '17 at 18:05
  • $\begingroup$ Wavefunction collapse isn't a general feature of quantum mechanics. It only exists in the Copenhagen interpretation. Therefore questions like this don't have objective answers that can be verified by experiment. $\endgroup$ – Ben Crowell Oct 2 '17 at 18:38
1
$\begingroup$

This question has no answer (or rather it has multiple speculative answers), and all comes down to quantum interpretation.

I don't think it's controversial to say that the majority preference (if one makes such assumptions) is to believe that no, the particle will not still be in superposition to one observer once observed by another, as its macroscopic interactions (e.g. measurement) have caused the particle to decohere absolutely.

There are however interpretations such as qbism (quantum Bayesian) that through taking a Bayesian approach assume all states are knowledge based, so for instance one can imagine that by observing the spin of a particle, to an external non-observer you are now in a superposition along with the particle---this secondary person still has no further knowledge of the system.

There are also interpretations where this question isn't even considered relevant or it's just unclear what assumption one should form.

Sorry if this isn't satisfactory; that's just how it is. If you'd like any more information on a specific interpretation (or any other follow-up), let me know and I'll try to expand this answer accordingly.

$\endgroup$

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.