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What would be measured if you measure two entangled qubits at exactly the same time?

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    $\begingroup$ Welcome to QCSE. Can you edit your question to provide more clarification on what you are asking? If Alice and Bob share two entangled qubits, e.g. $\frac{1}{\sqrt 2}(\vert 0_A0_B\rangle+\vert 1_A1_B\rangle)$, and Alice and Bob measure at the "same" time, it's the same as measuring at "different" times - they will either simultaneously get $0$ or $1$. $\endgroup$
    – Mark S
    Sep 13 '20 at 23:39
  • $\begingroup$ Sure, I think that I am trying to ask, would you get the same measurement or a different one? Because quantum entangled particles correlate to each other which means that you can deduce one thing from the other. So if you measure two entangled particles, at exactly the same time wouldn't the result be random (the two qubits would randomly choose between zero or one even if they are entangled meaning that it would be luck if you measure them the same)? $\endgroup$ Sep 15 '20 at 1:14
  • $\begingroup$ So if the quantum bit is separated by 1 light second. And you measure it is impossible to get the exact same answer, unless there is a communication travelling faster than the speed of light. $\endgroup$ Sep 15 '20 at 1:17
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The same thing as if you hadn't. The ordering of independent measurements never matters, entanglement or no entanglement.

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  • $\begingroup$ Would this actually happen because two quantum entangled particles deduce something about the other particle, so if you measure at exactly the same time wouldn't the result be random (either a zero or one)? $\endgroup$ Sep 15 '20 at 1:15
  • $\begingroup$ So if the quantum bit is separated by 1 light second. And you measure it at exactly the same time it is impossible to get the exact same answer, unless there is a communication travelling faster than the speed of light. $\endgroup$ Sep 15 '20 at 1:18
  • $\begingroup$ @BenoyDrenzla Sort of. There is no way to implement all the effects of entanglement, using classical mechanics, without faster than light communication. That's basically Bell's theorem. But even underlying classical mechanisms that used FTL communication would have to be insensitive to the ordering of the measurements because that's what quantum mechanics requires. In collapse based interpretations, where collapse moves at FTL speeds, you can check by calculation that it doesn't matter which one collapses first or if there is a simultaneous collapse of both. The same probabilities come out. $\endgroup$ Sep 15 '20 at 2:28
  • $\begingroup$ Thank you, it has helped me to understand quantum bits much more. $\endgroup$ Sep 15 '20 at 3:12

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