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Suppose you have a particle in deep space, and two observers, such that none of these particles interact with the other two unless either observer decide to. Then if one observer makes a measurement for the particle and the wave function collapses for that value, is that measurement also now determined for the other observer or is it only determined once the second observer either interacts with the particle himself or the other observer.

edit: I think I am asking if a measurement locks in a pure state for only those who have actually interacted with the system or if the system is in a locked in state as far as the entire universe is concerned, whether they have actually interacted with it or not. For example if you have a particle in a box in front of you right now, and you measure some observable, collapsing the wave function. does it make sense to say that the wave function has also collapsed for me, I just don't know it yet, or does it only make sense to say that the wavefunction has collapsed for you, until a time that I measure it or you tell me, otherwise I still say its wavefunction is "un-collapsed". (the more i think about this question the more it seems to be fundamentally unanswerable)

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    $\begingroup$ What you're proposing is basically the Wigner's Friend thought experiment, "in space". $\endgroup$ – Emilio Pisanty Nov 28 '20 at 12:27
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    $\begingroup$ Are you asking if a measurement "locks in" a pure state for a system ? $\endgroup$ – StephenG Nov 28 '20 at 12:56
  • $\begingroup$ i have edited my question for clarification of what i at least think i am asking $\endgroup$ – Matthew Hill Nov 28 '20 at 16:36
  • $\begingroup$ The two particles are in a joint (and entangled) superposition for you. $\endgroup$ – A_P Nov 28 '20 at 17:17
  • $\begingroup$ Also note that in real life, observers tend to be large objects, so decoherence becomes relevant. All this means is that the superposition is so messy that you have no hope of detecting that it is one, and hence are free to think of there being a definite (if unknown) result for all practical purposes. $\endgroup$ – A_P Nov 28 '20 at 17:19
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Might as well turn my comments into an answer.

The two particles are in a joint (and entangled) superposition for you. In one branch, the particle has some state (A) and the observer "sees" that state, and likewise for the other state (B).

Note that in real life, observers tend to be large objects, so decoherence becomes relevant. All this means is that the superposition is so messy that you have no hope of detecting or exploiting it, and hence are free to think of there being a definite (if unknown) result for all practical purposes.

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