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by lmm
2107 days ago
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What would you expect "experiencing along those other bases" to look like? If you expand along a different basis you just get something like: half a chance of experiencing (1/sqrt(3)|x> + 2/sqrt(3)|y>), and half a chance of experiencing (1/sqrt(3)|x> + 2/sqrt(3)|y>), so it amounts to the same thing. |
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> the structure of the wavefunction is that it divides cleanly into those two branches, and that's true in any basis.
It's not. It only has this Schmidt decomposition in one basis. In other bases there will be cross terms among the basis elements. What you're doing is privileging Schmidt bases as ones that give experiences. In another basis with states w,z say the state will be: |w>|w> + |z>|z> + |w>|z> + |z>|w>
So you won't be able to give this clean "experience" reading unless you posit we can't experience in things like the w,z basis here and only in Schmidt bases, but then you run into the problem that for real macroscopic systems they won't admit a Schmidt basis.
This seems like the kind of a "vague" Many Worlds where one doesn't look any deeper than pretending a macro-device is a qubit (e.g. no thermal states etc) and looking at one basis. There's a reason properly developed MWI is nothing like this such as the Spacetime State realism of Wallace and Timpson.
Why one would believe in quantum state realism at all is a separate question.
>Of course you can
No you can't, it's a direct consequence of the Kochen-Specker theorem. If the device is treated quantum mechanically and it enters an entangled state of the form you gave then you cannot perform conditioning as the Kochen-Specker theorem, via the non-uniqueness of Hilbert space orthogonal decompositions, prevents an unambiguous formulation of Bayes's law. I can link to papers proving this if you wish.
The fact that we do experiments where we can condition is, in light of this theorem, a demonstration that our measurement devices do not enter into the kind of CHSH states you're giving.