2,755 reputation
1324
bio website staff.science.uu.nl/~hooft101
location Utrecht, Netherlands
age
visits member for 2 years, 9 months
seen Jan 20 at 23:07

Theoretical physicist, Utrecht University.

University Professor. Elementary particle physics, quantum gravity, black holes, quantum mechanics.

Won some prizes (such as the nobel prize 1999), but please don't hold that against me.


Aug
20
awarded  Nice Answer
Aug
19
comment Discreteness and Determinism in Superstrings?
@Ron: But you might consider staying on. Once you agree that the beable basis exists (or might exist), just continue doing QM there. Observe however, that you can do the same with any totally classical system such as the planets obeying Newton's laws. Their evolution law (at integer time steps) is also a permutator. You may pause at the question how the "Earth-Mars exchange operator" evolves with time, and conclude that you can understand the physics of the system without solving the problem, but you might also add that operator to your set of observables. It's the same planets you talk about.
Aug
19
comment Discreteness and Determinism in Superstrings?
@Ron, OK, maybe this is again just a question of semantics. I would say that Feynman diagrams represent a causal QFT (if done correctly), no matter which gauge choice or coordinate choice you use. Same for string theory.
Aug
19
revised Discreteness and Determinism in Superstrings?
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Aug
19
revised Why do people categorically dismiss some simple quantum models?
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Aug
19
revised Why do people categorically dismiss some simple quantum models?
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Aug
19
comment In 't Hooft beable models, do measurements keep states classical?
@Ron: Then let me clarify the statement. Call the states in the CA basis $|n\rangle$, where $n=1, ... N$. Whatever wave function $\psi$ I start with, I can call the "probability" that $n$ is realized $\rho_n=|\langle n|\psi\rangle|^2$. Because the evolution is a permutation, this $\rho$ evolves classically. The phase factors in $\langle n|\psi\rangle$ never play any role; they're unobservable. This is because the evolution operator won't let you generate superimposed states from CA states.
Aug
18
revised Why do people categorically dismiss some simple quantum models?
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Aug
18
revised Why do people categorically dismiss some simple quantum models?
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Aug
18
revised Why do people categorically dismiss some simple quantum models?
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Aug
18
answered In 't Hooft beable models, do measurements keep states classical?
Aug
18
comment In 't Hooft beable models, do measurements keep states classical?
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
18
awarded  Popular Question
Aug
17
awarded  Revival
Aug
17
answered Discreteness of Spacetime and Violation of Lorentz symmetry
Aug
17
awarded  Supporter
Aug
17
awarded  Commentator
Aug
17
comment Why do people categorically dismiss some simple quantum models?
@ Lubos Motl, you still didn't get it: the ammonia molecule is in the SM, and there superpositions are meaningful. But if you use the CA as a basis, superposition of two basis elements acts as a classical composition with classical probabilities. "Superdeterminism" here boils down to saying that 2 ammonia states that are not orthogonal only become "ontic" if you make them orthogonal by including other quantum states elsewhere, and making those orthogonal. It looks inelegant but I do think that this actually happens.
Aug
17
awarded  Self-Learner
Aug
17
answered Why do people categorically dismiss some simple quantum models?