2,324 reputation
1122
bio website staff.science.uu.nl/~hooft101
location Utrecht, Netherlands
age
visits member for 1 year, 11 months
seen Feb 4 at 15:15

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.


Sep
5
comment Why do people categorically dismiss some simple quantum models?
For many physicists, this is all that matters, also in the 1960s. I presume Einstein did think of something like hidden variables. 2: WITH Hidden variables, Bell claims that the hidden variables are non-local, but what he really means is that the Ansatz equation that he starts with cannot be satisfied. My claim is that in a superdeterministic scenario that equation cannot hold, even if the evolution laws of a CA are local.
Sep
5
comment Why do people categorically dismiss some simple quantum models?
@user 7348: 1: if you don't care about hidden variables, quantum mechanics as it is, or more precisely quantum field theory, is entirely local. Locality means that if we have, in the Heisenberg notation, two field operators depending on space-time: $Op_1(x_1,t_1)$ and $Op_2(x_2,t_2)$, then they must commute if $(x_1,t_1)$ and $(x_2,t_2)$ are completely space-like separated. This holds for QFT and even (in spite of claims to the contrary) for string theory (if two points are spacelike separated in target space, they also are so in the world sheet - assuming we may ignore certain projections).
Aug
30
comment Why do people categorically dismiss some simple quantum models?
OK let me correct that last statement. Some pseudo-non-locality can enter in two ways: 1: the description of the vacuum state as a superposition of CA states. The vacuum makes discussion of Bell's inequalities very difficult in CA. 2: There are good reasons for imagining information loss to occur in the CA. You can still map that onto a quantum system (with full CPT invariance), but that mapping leads to holography and apparent non-locality.
Aug
30
comment Why do people categorically dismiss some simple quantum models?
They even don't violate locality, since the QFT has commutators that always vanish outside the light cone. It could be that there is some rudimentary non-locality in the mapping, but I have not really encountered that yet.
Aug
30
comment Why do people categorically dismiss some simple quantum models?
Superdeterminism is obviously there, if you care to give it some thought. Now, when people talk of "conspiracy" they really mean that they don't understand the result. But you can understand it this way: the CA can be treated as a fully grown quantum system, as a QFT. This QFT leads to correlations that look like conspiracy. Hence this apparent conspiracy is there. Don't be afraid of spooks. Mathematically, there's nothing wrong with them, just a bit difficult.
Aug
29
comment Why do people categorically dismiss some simple quantum models?
@user7348, Do you mean those pilot waves? I think these are ugly concoctions, but I do agree that they show that there is a possibility in principle. I think that elegance and plausibility will be important assets of a healthy theory. I can't make working field theories using Bohm. I am talking about much more fundamental principles. And most importantly: my theory IS quantum mechanics, not an "alternative".
Aug
28
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
It's essential to add interactions, making the CA non-trivial. Only then my arguments make sense. Without that, they also work but it's just formalities and semantics. Interactions turn your CA into a universal computer. You can still "solve" the CA equations but you can't speed them up using, say, renormalization group (RG) techniques, to go to larger time and distance cales. Therefore you would need a computer with Planckian dimensions. That does not exist today.
Aug
28
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
I'd be happy to continue in "chat" but I am not going to waste my time trying to figure out how to do this.
Aug
27
revised In 't Hooft beable models, do measurements keep states classical?
added 698 characters in body
Aug
27
revised In 't Hooft beable models, do measurements keep states classical?
added 698 characters in body
Aug
27
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
That these correlations are "absurd" or "ridiculous" is not a good enough argument to me. The mouse gut, the brain, a flipper machine, they all obey conservation laws such as energy and angular momentum, and the laws of thermodynamics. So they also obey unitarity. There's nothing absurd or ridiculous about that. Just do the CA - quantum mapping.
Aug
27
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
@ Mitchel: The CA is a universal computer, as such non integrable. I think my CA - quantum mapping just proves that the CA generates "miraculous finetuning". It does not exactly randomise. It computes more accurately than a mouse so, yes, it controls the mouse's gut. Now remember that the QFT generated by the CA still obeys quantum causality (i.e. commutators vanish outside the light cone), so the finetuning is still causal, which makes it a bit less "miraculous". We can never short circuit this universal computer, but we can compute its correlations, using the mapping on QM.
Aug
26
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
@ Mitchell: Do you agree with the following? Your "arbitrarily complicated macroscopic processes" do have a past; to change the setting of polarization filter $a$, you have to change that past. By resetting the past, the "local vacuum data", both at $a$ and at $b$, will have changed completely. My point is that, whether a photon at $a$ or at $b$ gets through, is difficult or impossible to compute in the deterministic theory. Only the chances can be calculated by using QM as a tool. Acceptable or not?
Aug
24
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
The sad thing is that I can't get it into people's heads that counterfactual measurements are not meaningful, even in CA theories. Operators that are not diagonal in the CA basis cannot be measured. The point that seems to escape all the time is that a CA can be so complex that no counterfactual measurements are needed to describe the outcome of a macroscopic observation. This is why non-diagonal operators may nevertheless be extremely useful to describe intermediate results of a quantum calculation.
Aug
24
comment Can superdeterminism resolve contextuality, entanglement and Shor's algorithm in quantum mechanics?
@ Mitchell: even if the logical chain does not pass through the mouse's gut, it does reset the data for the EPR experiment. So the photons are affected by the past data.
Aug
24
comment Does any particle ever reach any singularity inside the black hole?
The metric contains terms $(1-{2M\over r})$, so that if $r<0$ one must replace $r=-r',\ M=-M'$ to describe the extended region there. In Kerr, positive and negative r are smoothly connected. The singularity is only along the equator. We do have another pathology in that region: there is a region with closed timelike curves. Just look at the complete metric expression carefully. Now where is this wrong wrong wrong?
Aug
24
comment Does any particle ever reach any singularity inside the black hole?
I said that the SECOND horizon is unstable. It really means that objects falling in long after the observer himself, will get in his way, with infinite blue shift. There is also an issue about infalling objects in the new universe where your observer left the bh. But I thought we agreed there. What you did wrong was that you did not notice that, in the Kerr case, ignoring the problems with the second horizon, the observer can get out in two directions: $r\rightarrow\pm\infty$. The negative r direction corresponds with negative mass BH. Just look at the equs for metric, and Penrose diagram.
Aug
23
comment In 't Hooft beable models, do measurements keep states classical?
It's not a question of "thinking that ..." or "intuition", I am talking of simple mathematical facts. As explained in step 3 above, all that needs to be computed in a CA, with states $|Q_i(t)\rangle$ is the amplitudes $\langle Q_i(t_{\,1})|Q_j(t_{\,2})\rangle$. They are 0 or 1 (mostly 0). I can compute this in any basis I like. These are unitary transformations. There's nothing more to it than that. But I am done. I terminate this discussion.
Aug
22
comment Does any particle ever reach any singularity inside the black hole?
@Anixx: The answer by Gordon is correct; this is the paradoxical thing about Hawking radiation. It is not observed by the observer falling in.
Aug
22
comment Does any particle ever reach any singularity inside the black hole?
@John, you are right, but only in a formal sense: the infalling particle will close in to the horizon so rapidly, and the evaporation is so slow, that the external observer will not be able to distinguish the signals he receives from Hawking radiation. The signals from the particle will red-shift way too fast.