If a stationary observer, 'A', observes the collapse of a wavefunction, does an observer, 'B', traveling at relativistic speed observe a different collapse of the same wavefunction? What do all the interpretations says about this topic?
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Marco Fabbri
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4Everyone will agree about the measurement outcome. In most interpretations, there is no "physical collapse". And it has never been seen experimentally... – Tobias Fünke Feb 02 '24 at 14:21
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1You can calculate the wavefunction before and after the collapse in each reference frame. There is no ambiguity. – Mauricio Feb 02 '24 at 14:23
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1Note also that in relativity we prefer to use fields and not wavefunctions. – Mauricio Feb 02 '24 at 14:26
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1@Mauricio I wouldn't say so without further caveats (and one has to be very precise here). There are problems with the description of relativistic measurements, even in (relativistic) QFT. See e.g. Quantum information and relativity theory. Peres and Treno. III B. Or Quantum Information in Relativity: The Challenge of QFT Measurements. Anastopoulos and Savvidou. Section 3.1 and the references therein. – Tobias Fünke Feb 02 '24 at 14:28
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1This may help - Does the collapse of the wave function happen immediately everywhere? – mmesser314 Feb 02 '24 at 14:39
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1https://physics.stackexchange.com/questions/749475/is-collapse-of-wavefunction-actually-relative/749500#749500 – alanf Feb 02 '24 at 14:42
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1@TobiasFünke the only problem I can see with that is when you have entangled states, but that is not the question here... – Mauricio Feb 02 '24 at 14:44
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1This could also help: https://physics.stackexchange.com/questions/588420/lorentz-transformation-of-wavefunction#:~:text=Consider%20two%20frames%20S%20and,%CF%88%E2%80%B2(x%E2%80%B2). – Mauricio Feb 02 '24 at 14:45
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Thank you everyone – Marco Fabbri Feb 02 '24 at 15:51
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Can the collapse be "observed"? – md2perpe Feb 02 '24 at 17:04