Orlanthai

5 ms·

1.
by Orlanthai 6y ago
Continuing from the longer post below. The device has to have its contextual observable algebra develop a non-trivial center, not just be entangled as is mentioned in section 5 of the paper I linked. It's well known entanglement alone
2.
by Orlanthai 6y ago
> The state's evolution will be completely equivalent to (a linear superposition of) the evolution of |x>|x> and |y>|y>. That's a physically observable fact that's independent of your choice of basis (it's
3.
by Orlanthai 6y ago
See my reply above. You're just declaring we only experience Schmidt bases for no particular reason. Where are you getting this "clear connection" between experience and the decomposition in one particular basis. Do you have
4.
by Orlanthai 6y ago
Going from your last post. > 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 ther
5.
by Orlanthai 6y ago
I have. They still don't make Quantum Theory and Chaos similar. Rather for some systems QM can motivate ergodicity as well as classical chaos can. However that doesn't mean Quantum Probability and Classical Probability are alike s
6.
by Orlanthai 6y ago
I think that's different. Chaos still uses classical probability and the randomness is just ignorance of underlying initial conditions. This is very different from QM.
7.
by Orlanthai 6y ago
The lack of noncontextual hidden variables is the main implication of the Kochen-Specker theorem from which the inability to condition follows. It's not "backward" from the conclusion of no noncontextual hidden state, it&#x27
8.
by Orlanthai 6y ago
There are other alternatives such as the derivation of quantum theory within the GPT framework and many other axiomatic derivations. I've never seen the Born rule derived from unitary evolution and axioms for how subjective experience
9.
by Orlanthai 6y ago
Those are the same states so I'm not sure what you mean. The point is that there is no reason to select out any particular basis over another. You can't just retreat into "well this is the only basis I can experience" be
10.
by Orlanthai 6y ago
Measurement is not just the same as entanglement. If you try that you get paradoxes and you don't match actual observations. If you treat a measurement as entanglement then by the Kochen-Specker theorem you can't condition on the
11.
by Orlanthai 6y ago
Derivations like that don't work, you've just declared it by fiat but there's no such proof that is known to work.
12.
by Orlanthai 6y ago
It depends on what one calls the measurement problem. This solves the "consistency/small problem", i.e. treating the macroscopic apparatus as boolean is justified. It doesn't resolve the "outcome problem", i.e.
13.
by Orlanthai 6y ago
There's a continuous infinity of alternate basis expansions. There's no reason to think you'd experience along that particular basis.
14.
by Orlanthai 6y ago
Well you don't need to have nonlinear evolution to get what alpineidyll3 is saying. It's sufficient for the observable algebra of macroobservables to be commutative. This allows the evolution to be linear and have no interferences