2,895 reputation
1325
bio website staff.science.uu.nl/~hooft101
location Utrecht, Netherlands
age
visits member for 2 years, 11 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
22
awarded  Nice Answer
Aug
21
awarded  Good Question
Aug
21
comment Why do people rule out local hidden variables?
The strange thing is that this point of resetting the experiment, both for Alice and for Bob, should be obvious from the rules everyone uses in QM; it is the reason why QM disobeys the Bell inequalities. All you have to say is that a hidden variable theory must have the same property, and this is possible, even if it is a local theory, but you have to accept its deterministic nature all the way. Don't dilute that with "free will".
Aug
21
comment Why do people rule out local hidden variables?
What this really means is that, strictly speaking, you can't make a "counterfactual" observation. You choose the operator you want to measure, and you can't change your mind anymore. The usual objections people make to this are based on ideas they claim to have about "free will". As you might infer from a previous answer, it has something to do with religion...
Aug
21
comment Why do people rule out local hidden variables?
This goes all the way to the Big Bang, and even distant quasars are causally connected (as most cosmologists will tell you, this is a consequence of inflation). All you need to realize is that, therefore, any such change of mind resets the experiment completely, and this takes away all contradictions. No "conspiracy" is needed, just a reset, like switching your computer off and on again. Note: Both Alice and Bob may measure something else ...
Aug
21
comment Why do people rule out local hidden variables?
I'm afraid that the correct answer to the question is that people do not understand superdeterminism. "Superdeterminism is silly", is what you often hear. However, denying it is far sillier. It is obvious that, in a Bell-like experiment, if either Alice or Bob or both want to "change their mind" (regardless whether it is on the basis of the number of mouse droppings, or the fluctuations of a distant quasar), then this change has its roots in the past.
Aug
21
awarded  Necromancer
Aug
21
awarded  Revival
Aug
20
answered Can every particle be regarded as being a combination of Black holes and White holes?
Aug
20
answered Does any particle ever reach any singularity inside the black hole?
Aug
20
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
"Superdeterminism" becomes a lot less ridiculous if you simply state the following: In a Bell-like experiment, neither Alice nor Bob can "change their minds" as to what to measure, without having one or more changes of the settings of the experiment in the distant past. If you do change any of these settings, not only the number of mouse droppings might or might not change, but the entire experiment is reset. This already removes Bell's contradiction. No non-local physical law is needed.
Aug
20
comment In 't Hooft beable models, do measurements keep states classical?
And well, before anybody else makes a nasty remark about it: discussing vacuum fluctuations this way requires second quantization. I have not yet been able to do second quantization of my formalism for (super) string theory; other little math obstacles will have to be put out of the way first. Will the second quantized string still act as a CA? I expect yes, but I don't know. I was hoping more experienced string guys would jump in.
Aug
20
comment In 't Hooft beable models, do measurements keep states classical?
Finally: in the CA models of QM, there's something else: the physical vacuum, $|\,\emptyset\,\rangle$, must be a superposition of all (or nearly all?) CA states $|n\rangle$. We can still choose the phase angles there and it is tempting to define these as being 0 (all inner products positive). That will uniquely define most other phases in the QM picture. May come out handy.
Aug
20
comment In 't Hooft beable models, do measurements keep states classical?
Rotations over arbitrary angles and Lorentz transformations are symmetry operators of such a kind. This may be the reason why we, humans, have become blind to the distinction between beables and changeables: we are mistaken in thinking that our ontological world has all these continuous symmetries!
Aug
20
comment In 't Hooft beable models, do measurements keep states classical?
Nothing stops me from using genuine QM amplitudes when defining how the Earth-Mars exchange operator works; it's just semantics, not physics. Same in a CA. All operators that are not diagonal on the CA basis are "physically useless", but they may be mathematically useful when turning to another basis. If you look at the math in my papers, you see an astonishing fact: the evolution operator works the same way on all beables, as it works on interchange operators (changeables)! This makes symmetry transformations between them possible!
Aug
20
comment Is it possible to reproduce the energy spectrum of quantum chaos using classical cellular automata?
Some say it's trivial, some say it's impossible, some say it's hardly impressive, some say it's wrong wrong wrong. For me, it's as if I hear Moslem extremists argue with Jehovah's witnesses. Isn't is about time that we consider the mathematical equations? The simpler the systems we are discussing, the more transparent things become. What we could try to agree about is that there's nothing mystical about QM. If a "universe" is very small, "ontology" doesn't mean so much anymore, but just consider such systems, and ask yourself what they could become when you replace them by more complex worlds.
Aug
20
comment Discreteness and Determinism in Superstrings?
@Ron: No, according to my rules, which are exactly as in QM, perturbations don't have to be small, not $\delta\rho$ but $\psi$ is the wave function. Its sign can be positive or negative, and its absolute square is the probability. Earth-Mars interchange acts on $\psi$, not $\rho$.
Aug
20
comment Why do people categorically dismiss some simple quantum models?
@Scary Monster: The claim is that any CA can be cast in the language of QM, although in most cases the QM models you get will be uninteresting; there will be states, and they will obey Schroedinger equations. Now many CA models are computationally universal, so certainly not integrable, and therefore the asociated QM theory is also expected to be non-trivial. But of course the math is much harder; it's much more instructive to search for cases where you can do (perturbative) calculations.
Aug
20
comment Why do people categorically dismiss some simple quantum models?
@ Motl: In ordinary applications of QM you can ignore this, since the templates are good enough, but not in questions of the interpretation of QM.
Aug
20
comment Why do people categorically dismiss some simple quantum models?
@ Motl: Apparently you axiomatize QM by basing it on "postulates". Clearly you won't understand my theory if you are not prepared to make any amendments, since your postulates are imprecise. You said that "experiment has shown that one can superimpose quantum states". Not true, you can only do this with the templates you are using, but not in the real world. When you consider superposition of two states, you ignore the environment of these two states, which are never the same, hence always orthogonal.