Timeline for In 't Hooft beable models, do measurements keep states classical?
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| May 6, 2013 at 14:48 | history | edited | Řídící | CC BY-SA 3.0 |
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| Aug 30, 2012 at 19:20 | comment | added | Ron Maimon | @CuriousGeorge: It is logically possible, and this is why I wasn't sure if 't Hooft's stuff was right.or not for a long time. I tried to rearrange what he was doing into a normal probabilistic form, and failed. Then when he started getting the impossible results--- local failure of Bell's theorem, reproducing exact QM with no decoherence--- these things that just can't happen, I understood that the only way you can get these things is if the projections aren't equivalent. It is very hard to prove, because the measurement operators are complicated and macroscopic. | |
| Aug 30, 2012 at 7:13 | comment | added | Curious George | @Ron Is it possible that the question is "wrong"? Couldn't one take the opposite point of view? Couldn't the question be: is bayesian reduction in the CA states probability able to reproduce the projection to an eigenstate of an internal subsystem to some very good approximation? Isn't this the same thing as the decoherence interpetation of the "collapse of the wave-function" where the density matrix of the subsystem becomes diagonal because it gets entangled with a macroscopic apparatus that measures that observable? (just a thought, I didn't follow the details, I might be missing something) | |
| Aug 18, 2012 at 13:58 | answer | added | G. 't Hooft | timeline score: 20 | |
| Aug 18, 2012 at 13:21 | comment | added | G. 't Hooft | Let me add a question, for comparison: My favorite "classical" theory is the planetary system, assuming that planets move as point particles under Newton's laws. Yoy can actually introduce non-commuting operators there as well. The "Earth-Mars exchange operator" puts Mars where Earth is and Earth where Mars is (and some simple rules about their velocities and moons). The eigenvalues of this operator are $\pm 1$. We can calculate how it evolves. Is this an observable? | |
| Aug 15, 2012 at 17:16 | comment | added | Ron Maimon | @Christoph: That's a very nice argument for compatibility--- but it seems to be ultimately semiclassical, since you are relying on the classical equation of motions being exact. Assuming quantum mechanics gives you orthogonal structure, the symplectic structure is automatic in one of t'Hooft's formulations, in which he takes the formal classical system path integral (the Martin Siggia Rose formalism path integral for a Hamiltonian system) and phase rotates it to a different basis (as he always does in these things). The classical Hamiltonian gives you an additional symplectic structure. | |
| Aug 15, 2012 at 11:25 | comment | added | Christoph | @annav: my argument is valid for ordinary QM - no idea if there's a better one for t'Hooft's model... | |
| Aug 15, 2012 at 11:17 | comment | added | Christoph | @annav: the complex structure is related to the 2-out-of-3 property; the real and imaginary parts of the hermitian product are symmetric and anti-symmetric forms which induce a metric and symplectic structure on the projective Hilbert space; the metric structure provides the probabilities and the symplectic one the dynamics via Hamilton's equations; if we require metric and symplectic structure to be compatible, we get an almost-complex structure for free even if we consider the projective space as a real manifold | |
| Aug 15, 2012 at 11:04 | comment | added | anna v | @Physikslover yes, but does the algebraic structure emerge from this CA model or is it imposed by hand? | |
| Aug 15, 2012 at 10:20 | comment | added | Physiks lover | @annav complex numbers are ordered pairs of real numbers $(a,b)$ that satisfy the axioms of that algebraic structure. | |
| Aug 15, 2012 at 8:02 | comment | added | Ron Maimon | Also, this t'Hooft thing is so weird and new that nobody has the proper preparation to think about it, so don't feel intimidated. But it drives one nuts not to know the answer, and oscillate between "yes" and "no" for so long. | |
| Aug 15, 2012 at 7:58 | comment | added | Ron Maimon | I can't really give a full answer, because I don't claim to reproduce QM from CA models--- I admit I can't do it, I just don't see it as necessarily impossible, given holographic nonlocality. to see an example, consider a circle automaton where all sites move to the right. You can define perturbations to the uniform probability steady state which are complex waves which move to the right and to the left. These are formal waves, every real distribution is an equal sum of complex conjugate pairs, but I don't see any reason why formal complex waves can't end up describing some interior thing. | |
| Aug 15, 2012 at 7:53 | comment | added | Ron Maimon | @annav: the "i" can be considered as a two-by-two matrix, you don't have to use complex numbers. When you have complex eigenvalue pairs for a linear evolution (this happens generically when you have entropy preserving motion of a probability system that doesn't have a reversing measure), you can introduce complex vectors for convenience, they correspond to time-reflected waves as usual, so this is not a showstopper. It's a good question, but not this question, it's one of the things one should adress. t'Hooft introduces complex numbers by diagonalization--- eigenvectors of real Hamiltonians. | |
| Aug 15, 2012 at 6:49 | comment | added | anna v | Ron, this is way over my mathematical background but I am curious about something simpler:can complex numbers appear from real numbers without introducing the "i" by hand? | |
| Aug 14, 2012 at 14:53 | history | tweeted | twitter.com/#!/StackPhysics/status/235388604891926528 | ||
| Aug 14, 2012 at 14:24 | history | edited | Ron Maimon | CC BY-SA 3.0 |
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| Aug 14, 2012 at 14:18 | history | asked | Ron Maimon | CC BY-SA 3.0 |