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Yes, the many-worlds interpretation is supposed to be time symmetric. Consider this toy example with a particle than can be in either of the states $\left|1\right\rangle$ or $\left|2\right\rangle$. Additionally, we denote the no particle state as $\left|0\right\rangle$. The particle gets emitted by our source $S$ that can be in its ground state $\left|S_0\...


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This experiment has been done, first by Birgit Dopfer in 1998, then later by Dr. John Cramer of the University Of WA. In Dopfer's experiment, there was a "coincidence detector" which is basically an AND gate to filter out only the entangled pairs. By moving the detector in the beam of photons not going to the double-slit, the information about the photon's ...


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There is no derivation of the probabilistic nature of the wavefunction: it is an interpretation (postulate), the only one that makes the theory consistent. From Sakurai, Modern Quantum Mechanics: Schrödinger published his famous wave equation in February 1926 in the famous paper Quantisierung als Eigenwertproblem (Quantization as an Eigenvalue Problem),...


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Question 1 is based on the false premise that B measurement does not always obey a 50:50 probability. In fact persons A and B both always have a 50:50 probability of getting a spin up. That particles A and B are entangled means that there is a correlation between the measurements of A and those of B: when A is up, B is down; when B is down, A is up. But this ...


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Your question 1 has absolutely nothing to do with entanglement, and for that matter absolutely nothing to do with quantum mechanics. You can formulate exactly the same question this way: You and I each have in our pockets a tennis ball that is either red or green. We somehow each know for certain that they are of the same color (and that either color is ...


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What kind of information are we talking about in superluminal information transport? It might be more intuitive to call it superluminal communication. For the prohibited kind of superluminal information transport, you need to be able to achieve the following. At the start of the protocol, A holds a classical bit which either has value 0 or 1, and which is ...


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There is absolutely no "freedom for interpretations" here. Both books should – and all other books that are not completely wrong do – agree with these formulae and agree that $a_n$ never indicates an eigenvalue. In both books and all of science, $a_n$ is the complex probability amplitude such that $|a_n|^2$ represents the probability that the system has ...



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