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I have a question about decoherence. Assume there is a macroscopic black hole floating around and you have some macroscopic object with you with a huge number of internal degrees of freedom. Conventional decoherence theory predicts decoherence. If this macroscopic object is dumped into the black hole, will decoherence still happen? There are two environments here: the internal degrees of freedom of the macroscopic object which are coarse-grained over, and the exterior of the black hole. Which environmental degrees of freedom should you trace over? If black hole complementarity is right, the internal degrees of freedom and the external degrees of freedom don't commute, so clearly, unless some mathematical generalization of the partial trace can be invented, you can only trace over one or the other. A trace over the former leads to decoherence, but a trace over the latter can't for the simple reason that nothing can escape the black hole. Only thermal Hawking radiation can escape, but it is so scrambled that it carries no decoherence information.

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2 Answers 2

Grow up and be a no-nonsense physicist. Translation - become a positivist. Ask what are the empirical experimental observations that can be made here. Any more metaphysics is mere sophistry and illusion, and has to be committed to the flames.

Any experimenter jumping into the black hole to measure the macroscopic object can never deliver us any empirical data. No experiment performed inside a black hole can ever have any empirical results. Commit what is inside the black hole to the flames.

The only things which are empirically observable are the Hawking radiation. They do not decohere what is inside the hole.

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The question by the OP is misguided as it stands. Both choices of partial traces end up with decoherence. That is easy to see when the partial trace is over the internal degrees of freedom. It is also the case when it is over exterior of the black hole precisely because Hawking radiation is thermal and decoherent. Correct me if I am wrong, but what I guess the OP wanted to ask is whether the preferred pointer basis match up. They do not because a pointer inside the black hole can never be compared with a pointer outside the black hole.

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