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Deterministic models. Clarification of the question:

The problem with these blogs is that people are inclined to start yelling at each other. (I admit, I got infected and it's difficult not to raise one's electronic voice.) I want to ask my question without an entourage of polemics.

My recent papers were greeted with scepticism. I've no problem with that. What disturbes me is the general reaction that they are "wrong". My question is summarised as follows:

Did any of these people actually read the work and can anyone tell me where a mistake was made?

Now the details. I can't help being disgusted by the "many world" interpretation, or the Bohm-de Broglie "pilot waves", and even the idea that the quantum world must be non-local is difficult to buy. I want to know what is really going on, and in order to try to get some ideas, I construct some models with various degrees of sophistication. These models are of course "wrong" in the sense that they do not describe the real world, they do not generate the Standard Model, but one can imagine starting from such simple models and adding more and more complicated details to make them look more realistic, in various stages.

Of course I know what the difficulties are when one tries to underpin QM with determinism. Simple probabilistic theories fail in an essential way. One or several of the usual assumptions made in such a deterministic theory will probably have to be abandoned; I am fully aware of that. On the other hand, our world seems to be extremely logical and natural.

Therefore, I decided to start my investigation at the other end. Make assumptions that later surely will have to be amended; make some simple models, compare these with what we know about the real world, and then modify the assumptions any way we like.

The no-go theorems tell us that a simple cellular automaton model is not likely to work. One way I tried to "amend" them, was to introduce information loss. At first sight this would carry me even further away from QM, but if you look a little more closely, you find that one still can introduce a Hilbert space, but it becomes much smaller and it may become holographic, which is something we may actually want. If you then realize that information loss makes any mapping from the deterministic model to QM states fundamentally non-local—while the physics itself stays local—then maybe the idea becomes more attractive.

Now the problem with this is that again one makes too big assumptions, and the math is quite complicated and unattractive. So I went back to a reversible, local, deterministic automaton and asked: To what extent does this resemble QM, and where does it go wrong? With the idea in mind that we will alter the assumptions, maybe add information loss, put in an expanding universe, but all that comes later; first I want to know what goes wrong.

And here is the surprise: In a sense, nothing goes wrong. All you have to assume is that we use quantum states, even if the evolution laws themselves are deterministic. So the probability distributions are given by quantum amplitudes. The point is that, when describing the mapping between the deterministic system and the quantum system, there is a lot of freedom. If you look at any one periodic mode of the deterministic system, you can define a common contribution to the energy for all states in this mode, and this introduces a large number of arbitrary constants, so we are given much freedom.

Using this freedom I end up with quite a few models that I happen to find interesting. Starting with deterministic systems I end up with quantum systems. I mean real quantum systems, not any of those ugly concoctions. On the other hand, they are still a long way off from the Standard Model, or even anything else that shows decent, interacting particles.

Except string theory. Is the model I constructed a counterexample, showing that what everyone tells me about fundamental QM being incompatible with determinism, is wrong? No, I don't believe that. The idea was that, somewhere, I will have to modify my assumptions, but maybe the usual assumptions made in the no-go theorems will have to be looked at as well.

I personally think people are too quick in rejecting "superdeterminism". I do reject "conspiracy", but that might not be the same thing. Superdeterminism simply states that you can't "change your mind" (about which component of a spin to measure), by "free will", without also having a modification of the deterministic modes of your world in the distant past. It's obviously true in a deterministic world, and maybe this is an essential fact that has to be taken into account. It does not imply "conspiracy".

Does someone have a good, or better, idea about this approach, without name-calling? Why are some of you so strongly opinionated that it is "wrong"? Am I stepping on someone's religeous feelings? I hope not.

References:

"Relating the quantum mechanics of discrete systems to standard canonical quantum mechanics", arXiv:1204.4926 [quant-ph];

"Duality between a deterministic cellular automaton and a bosonic quantum field theory in $1+1$ dimensions", arXiv:1205.4107 [quant-ph];

"Discreteness and Determinism in Superstrings", arXiv:1207.3612 [hep-th].


Further reactions on the answers given. (Writing this as "comment" failed, then writing this as "answer" generated objections. I'll try to erase the "answer" that I should not have put there...)

First: thank you for the elaborate answers.

I realise that my question raises philosophical issues; these are interesting and important, but not my main concern. I want to know why I find no technical problem while constructing my model. I am flattered by the impression that my theories were so "easy" to construct. Indeed, I made my presentation as transparent as possible, but it wasn't easy. There are many dead alleys, and not all models work equally well. For instance, the harmonic oscillator can be mapped onto a simple periodic automaton, but then one does hit upon technicalities: The hamiltonian of a periodic system seems to be unbounded above and below, while the harmonic oscillator has a ground state. The time-reversible cellular automaton (CA) that consists of two steps $A$ and $B$, where both $A$ and $B$ can be written as the exponent of physically reasonable Hamiltonians, itself is much more difficult to express as a Hamiltonian theory, because the BCH series does not converge. Also, explicit $3+1$ dimensional QFT models resisted my attempts to rewrite them as cellular automata. This is why I was surprised that the superstring works so nicely, it seems, but even here, to achieve this, quite a few tricks had to be invented.

@RonMaimon. I here repeat what I said in a comment, just because there the 600 character limit distorted my text too much. You gave a good exposition of the problem in earlier contributions: in a CA the "ontic" wave function of the universe can only be in specific modes of the CA. This means that the universe can only be in states $\psi_1,\ \psi_2,\ ...$ that have the property $\langle\psi_i\,|\,\psi_j\rangle=\delta_{ij}$, whereas the quantum world that we would like to describe, allows for many more states that are not at all orthonormal to each other. How could these states ever arise? I summarise, with apologies for the repetition:

  • We usually think that Hilbert space is separable, that is, inside every infinitesimal volume element of this world there is a Hilbert space, and the entire Hilbert space is the product of all these.
  • Normally, we assume that any of the states in this joint Hilbert space may represent an "ontic" state of the Universe.
  • I think this might not be true. The ontic states of the universe may form a much smaller class of states $\psi_i$; in terms of CA states, they must form an orthonormal set. In terms of "Standard Model" (SM) states, this orthonormal set is not separable, and this is why, locally, we think we have not only the basis elements but also all superpositions. The orthonormal set is then easy to map back onto the CA states.

I don't think we have to talk about a non-denumerable number of states, but the number of CA states is extremely large. In short: the mathematical system allows us to choose: take all CA states, then the orthonormal set is large enough to describe all possible universes, or choose the much smaller set of SM states, then you also need many superimposed states to describe the universe. The transition from one description to the other is natural and smooth in the mathematical sense.

I suspect that, this way, one can see how a description that is not quantum mechanical at the CA level (admitting only "classical" probabilities), can "gradually" force us into accepting quantum amplitudes when turning to larger distance scales, and limiting ourselves to much lower energy levels only. You see, in words, all of this might sound crooky and vague, but in my models I think I am forced to think this way, simply by looking at the expressions: In terms of the SM states, I could easily decide to accept all quantum amplitudes, but when turning to the CA basis, I discover that superpositions are superfluous; they can be replaced by classical probabilities without changing any of the physics, because in the CA, the phase factors in the superpositions will never become observable.

@Ron I understand that what you are trying to do is something else. It is not clear to me whether you want to interpret $\delta\rho$ as a wave function. (I am not worried about the absence of $\mathrm{i}$, as long as the minus sign is allowed.) My theory is much more direct; I use the original "quantum" description with only conventional wave functions and conventional probabilities.


(New since Sunday Aug. 20, 2012)

There is a problem with my argument. (I correct some statements I had put here earlier). I have to work with two kinds of states: 1: the template states, used whever you do quantum mechanics, these allow for any kinds of superposition; and 2: the ontic states, the set of states that form the basis of the CA. The ontic states $|n\rangle$ are all orthonormal: $\langle n|m\rangle=\delta_{nm}$, so no superpositions are allowed for them (unless you want to construct a template state of course). One can then ask the question: How can it be that we (think we) see superimposed states in experiments? Aren't experiments only seeing ontic states?

My answer has always been: Who cares about that problem? Just use the rules of QM. Use the templates to do any calculation you like, compute your state $|\psi\rangle$, and then note that the CA probabilities, $\rho_n=|\langle n|\psi\rangle|^2$, evolve exactly as probabilities are supposed to do.

That works, but it leaves the question unanswered, and for some reason, my friends on this discussion page get upset by that.

So I started thinking about it. I concluded that the template states can be used to describe the ontic states, but this means that, somewhere along the line, they have to be reduced to an orthonormal set. How does this happen? In particular, how can it be that experiments strongly suggest that superpositions play extremely important roles, while according to my theory, somehow, these are plutoed by saying that they aren't ontic?

Looking at the math expressions, I now tend to think that orthonormality is restored by "superdeterminism", combined with vacuum fluctuations. The thing we call vacuum state, $|\emptyset\rangle$, is not an ontological state, but a superposition of many, perhaps all, CA states. The phases can be chosen to be anything, but it makes sense to choose them to be $+1$ for the vacuum. This is actually a nice way to define phases: all other phases you might introduce for non-vacuum states now have a definite meaning.

The states we normally consider in an experiment are usually orthogonal to the vacuum. If we say that we can do experiments with two states, $A$ and $B$, that are not orthonormal to each other, this means that these are template states; it is easy to construct such states and to calculate how they evolve. However, it is safe to assume that, actually, the ontological states $|n\rangle$ with non-vanishing inner product with $A$, must be different from the states $|m\rangle$ that occur in $B$, so that, in spite of the template, $\langle A|B\rangle=0$. This is because the universe never repeats itself exactly. My physical interpretation of this is "superdeterminism": If, in an EPR or Bell experiment, Alice (or Bob) changes her (his) mind about what to measure, she (he) works with states $m$ which all differ from all states $n$ used previously. In the template states, all one has to do is assume at least one change in one of the physical states somewhere else in the universe. The contradiction then disappears.

The role of vacuum fluctuations is also unavoidable when considering the decay of an unstable particle.

I think there's no problem with the above arguments, but some people find it difficult to accept that the working of their minds may have any effect at all on vacuum fluctuations, or the converse, that vacuum fluctuations might affect their minds. The "free will" of an observer is at risk; people won't like that.

But most disturbingly, this argument would imply that what my friends have been teaching at Harvard and other places, for many decades as we are told, is actually incorrect. I want to stay modest; I find this disturbing.

A revised version of my latest paper was now sent to the arXiv (will probably be available from Monday or Tuesday). Thanks to you all. My conclusion did not change, but I now have more precise arguments concerning Bell's inequalities and what vacuum fluctuations can do to them.

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Dear @QuestionAnswers, you're misinterpreting what I say: I don't think that these questions are illegal to ask. They're legal to be asked, they were asked about 90 years ago and they were answered 85 years ago. It's silly, and not illegal, to ask them again in 2012 because physics has known the answer for quite some time. It's a pretty long time. 85 years after physicists determined that heliocentrism was more right than geocentrism, it was generally viewed as silly to question heliocentrism again. Learning in the modern age should be faster but it's apparently not. –  Luboš Motl Aug 16 '12 at 6:10
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@LubošMotl: Yes, of course you can't simulate quantum computation. The thing is, these types of quantum computing states are incredibly entangled, and very hard to realize without decoherence spoiling them, so much so that we haven't realized any such states experimentally. The question is whether you can simulate workaday QM, lots of decoherence, no quantum computer around, by a linearly scaling computer. You could say "do collapse", but that's harder than it looks computationally. In a discrete QM analog, you get automatic collapse, and you can always do monte-carlo. –  Ron Maimon Aug 16 '12 at 8:58
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@LubošMotl: You are repeating ridiculous propaganda as if it were fact, this is just deplorable fear-mongering. Zeilinger's "classes of models" is a very myopic class that doesn't include mine or any other reasonable nonlocal model. Locality and Lorentz invariance are unrelated, the naive arguments nonwithstanding. Here is a Lorentz invariant nonlocal action: $\int {\phi(x)\phi(y) \over ((x-y)^2 + 1)^{.73}} d^4x d^4y$, there are lots of others. Locality is absent in string theory, there are no local fields, and it is completely absent in AdS/CFT bulk, where the whole spacetime is emergent. –  Ron Maimon Aug 16 '12 at 17:09
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... even if it is physically wrong, if mathematically right, it is a prescription for reducing highly entangled states. Of course entanglement is the norm! But we usually call it "collapse", not entanglement, and it usually only goes one-way--- reducing a quantum system in complexity. The delicate cases of quantum computation require fine-tuning to make the entanglement go back and forth, not getting collapse, but nontrivial computation. Most cases, you can approximate entanglement as collapse. A "realist approximation" to quantum mechanics is useful for automatic collapse simulation. –  Ron Maimon Aug 16 '12 at 17:19
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The real issue I think has been exposed most clearly by Ron Maimon and some others in their earlier contributions: the problem is that in a CA the "ontic" wave function of the universe can only be in specific modes of the CA. This means that the universe can only be in states psi_1, psi_2, ... that have the preoperty ( psi_i | psi_j ) = delta_ij (apologies for this non-latex notation), whereas the quantum world that we would like to describe allow for many more states that are not at all orthonormal to each other. How could these states ever arise? –  G. 't Hooft Aug 16 '12 at 19:42

7 Answers 7

I can tell you why I don't believe in it. I think my reasons are different from most physicists' reasons, however.

Regular quantum mechanics implies the existence of quantum computation. If you believe in the difficulty of factoring (and a number of other classical problems), then a deterministic underpinning for quantum mechanics would seem to imply one of the following.

  • There is a classical polynomial-time algorithm for factoring and other problems which can be solved on a quantum computer.
  • The deterministic underpinnings of quantum mechanics require $2^n$ resources for a system of size $O(n)$.
  • Quantum computation doesn't actually work in practice.

None of these seem at all likely to me. For the first, it is quite conceivable that there is a polynomial-time algorithm for factoring, but quantum computation can solve lots of similar periodicity problems, and you can argue that there can't be a single algorithm that solves all of them on a classical computer, so you would have to have different classical algorithms for each classical problem that a quantum computer can solve by period finding.

For the second, deterministic underpinnings of quantum mechanics that require $2^n$ resources for a system of size $O(n)$ are really unsatisfactory (but maybe quite possible ... after all, the theory that the universe is a simulation on a classical computer falls in this class of theories, and while truly unsatisfactory, can't be ruled out by this argument).

For the third, I haven't seen any reasonable way to how you could make quantum computation impossible while still maintaining consistency with current experimental results.

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@Jim: A deterministic theory acting in a $2^n$-dimensional Hilbert space would indeed meet my criteria for a reasonable thing to think about. The problem is that it seems hard to reconcile a deterministic theory that increases with the size of the system with the criterion of locality. But this seems to be exactly what 't Hooft hopes to happen ... the equations for the deterministic theory itself aren't local, but they somehow translate into local physics. –  Peter Shor Aug 17 '12 at 17:07
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@Peter Shor: I have always opted for your 3rd possibility: the "error correcting codes" will eventually fail. The quantum computer will not work perfectly (It will be beaten by a classical computer, but only if the latter would be scaled to Planckian dimensions). This certainly has not yet been contradicted by experiment. –  G. 't Hooft Aug 17 '12 at 20:45
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A question: If you view things holographically, do you need to look at the entire universe to get a complete description of local dynamics? If it does, it's not very useful for doing calculations. And if it doesn't, it's hard for me to understand where the power of quantum computation is coming from. –  Peter Shor Aug 18 '12 at 16:26
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Finally a meaningful and valid answer, Prof @PeterShor. Ron: there don't exist any "approximately quantum" theories. Theories are either classical, quantum, or logically inconsistent. However, I agree with your comment that holography doesn't cause any problem for quantum computing because for the relevant states - which are far from forming black holes etc. - the laws for the qubits in the quantum computer are as local in the right variables as they've always been. Prof Shor, if you don't try to make the density of degrees of freedom too high (grav. collapse!), things work just like before. –  Luboš Motl Aug 20 '12 at 6:15
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@Lubos: unitarity is a very good condition, but saying that it is required for consistency requires a very strange definition of consistency. It does seem to be required to make thermodynamics work, but if you look at topological quantum error correction a la Kitaev, there are good reasons for believing that there could be laws of physics in which unitarity breaks down at the Planck scale but is preserved at macroscopic scales. –  Peter Shor Aug 21 '12 at 14:34

This could have been a comment, but as it actually anwers the question asked in the title, I'll post it as such:

As far as I can tell there's no rational reason to dismiss these models out of hand - it's just that quantum mechanics (QM) has set the bar awfully high: So far, there's no experimental evidence that QM is wrong, and no one has come up with a viable alternative.

Ultimately, your theory needs to reproduce all experimentally verified predictions of QM (or rather may only deviate within the experimental precision). However, there's of course no need to reproduce arbitrary predictions - in fact, if you did, you'd end up with a re-formulation - ie a new interpretation - of ordinary QM. If your model tells us large-scale quantum computation is impossible, then it's up to the experimentalists to prove you wrong.

Any objections beyond that are just psychology at work: It takes quite some effort for most people to convince themselves that QM is a valid description of the world we live in, and once such a believe is ingrained, it easily becomes dogma.

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[ text I had put here is moved to the original question, but I prefer not to erase the comments that were posted here. ]

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Dear Prof. 't Hooft, the most obvious reason for downvoting this answer is that it doesn't answer the original question. The second reason is that the question and the answer were posted by the same user which is... strange. The third reason is that the content is entirely invalid. Superpositions are never "superfluous" in any quantum mechanical theory. They must always be allowed, the coefficients must always be allowed to be complex, and the phases always matter. –  Luboš Motl Aug 17 '12 at 14:56
    
You would see that if you looked at least at one particular system, e.g. the ammonia molecule motls.blogspot.cz/2012/08/two-state-systems-masers.html?m=1 , and figured out what can be measured and in what state the molecule is actually finding itself at various times. The complex probability amplitudes always matter - after all, the rate of change of the phase as a function of one observable tells us about the value of the complementary observable, and so on. For an actual answer to your question, see e.g. motls.blogspot.cz/2012/08/… –  Luboš Motl Aug 17 '12 at 14:58
    
@ 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. –  G. 't Hooft Aug 17 '12 at 20:39
    
@G.'tHooft would you mind also transferring the relevant content of the comments to your question? It's much better for the site if this sort of information appears in questions or answers rather than (just) in comments. –  David Z Aug 22 '12 at 20:50

Foundational discussions are indeed somewhat like discussions about religious convictions, as one cannot prove or disprove assumptions and approaches at the foundational level.

Moreover, it is in the nature of discussions on the internet that one is likely to get responses mainly from those who either disagree strongly (the case here) or who can add something constructive (difficult to do in very recent research). I think this fully explains the responses that you get.

I myself read superficially through one of your articles on this and found it not promising enough to spend more time on the technical issues.

However, I agree that both many-worlds and pilot waves are unacceptable physical explanations of quantum physics, and I am working on an alternative interpretation.

In my view, particle nonlocality is explained by negating particles any ontological existence. Existent are quantum fields, and on the quantum field level, everything is local. Nonlocal features appear only when one is imposing on the fields a particle interpretation, which, while valid under the usual asssumptions of geometric optics, fails drastically art higher resolution. Thus nothing needs to be explained in the region of failure. Just as the local Maxwell equations for a classsical electromagnetic field explain single photon nonlocality (double slit experiments), and the stochastic Maxwell equations explain everything about single photons (see http://www.mat.univie.ac.at/~neum/ms/optslides.pdf), so local QFT explains general particle nonlocality.

My thermal interpretation of quantum mechanics (see the section http://www.mat.univie.ac.at/~neum/physfaq/topics/found0.html from my theoretical physics FAQ at http://www.mat.univie.ac.at/~neum/physfaq/physics-faq.html, and Chapter 10 of my book http://lanl.arxiv.org/abs/0810.1019) gives a view of physics consistent with actual experimental practice and without any of the strangeness introduced by the usual interpretations. I believe this interpretation to be satisfactory in all respects, though it requires more time and effort to analyse the standard conundrums along these lines, with a clear statistical mechanics derivation to support my so far mainly qualitative arguments.

In presenting my foundational views in online discussions, I had similar difficulties as you; see, e.g., the PhysicsForums thread ''What does the probabilistic interpretation of QM claim?'' http://www.physicsforums.com/showthread.php?t=480072

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The problem is that your interpretation is standard QM, and whether you take fields or particles as the basic variables, you still have the exponential explosion. The field wavefunction is just as horrendous as particle wavefunction. The model t'Hooft would like (and I would like too), would not require exponential resources for a large grid. You can't just rejigger the philosophy to make this happen, by changing the "ontic" status of objects in the theory (honestly, a positivist doesn't even care about the ontic status). –  Ron Maimon Aug 18 '12 at 2:32
    
@RonMaimon: Interpretations of quantum physics have nothing per se to do with computational complexity, but with making commong sense out of the standard quantum phenomena. –  Arnold Neumaier Aug 26 '12 at 9:54
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Yes, ordinarily this is true, and this is why this question is not ordinary. The point of 'tHooft's line of investigation is to formulate a new theory (not quantum mechanics) which is nevertheless approximately quantum, but which does not suffer from the exponential explosion of computatinal complexity that plagues ordinary QM, and therefore is experimentally different from ordinary QM for the case of a quantum computer. The goal of such a theory is to not be experimentally different in cases where you would disagree with present experiments. This is a tough thing to do. –  Ron Maimon Aug 26 '12 at 10:00
    
@RonMaimon: While this may be the case in his model, it is not his expressed goal. Instead, he wrote above: ''I can't help being disgusted by the "many world" interpretation, or the Bohm - deBroglie "pilot waves", and even the idea that the quantum world must be non-local is difficult to buy. I want to know what is really going on, and in order to try to get some ideas, I construct some models with various degrees of sophistication.'' –  Arnold Neumaier Aug 26 '12 at 10:04
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Yes, I know, the stated motivations invite a lot of philosophical answers, but this is really not the motivation. I know this from understanding the holographic principle, reading the papers, and from talking to him briefly once ten years ago. The motivation is to make small hidden variables, meaning a number of bits which makes classical computation of reasonable size given holographic constraints, and which reproduces QM approximately. It is not to reproduce QM exactly. But he makes a mistake in his papers, and gets too good a QM, so he thinks it reproduces it locally and almost exactly. –  Ron Maimon Aug 26 '12 at 10:09

There are two questions here: Why criticize your models? And are there better ideas? I will try to answer the second question in a separate answer. Here I only give some comments of a general nature to adress the first question.

I personally agree with you, and I think most people who care about this stuff do too, that it is disconcerting to have a theory in which the information produced by observations is not contained in the theory itself, but is produced out of thin-air by an act of measurement. The natural idea is that when we see a bit of information produced through an act of observation, then the value of this bit was somehow contained in the complete description of nature independent of the act of observation. This was Einstein's reality principle, and I agree that it is preferred for a theory to obey it.

When a theory doesn't obey the reality principle, one has to note that macroscopic reality does obey it, and find the bits in the macroscopic world by a philosophically contorted roundabout exercize in mysticism. But since physics is empirical, and positivism is fruitful, I take the point of view that any framework that explains the results of observations must ultimately be philosophically ok, even if it requires contortions, and even if it is not correct! So Newton's mechanics, even though it is wrong, is not necessarily empirically disproved given only observations of human beings and so on, so it must not be philosophically incompatible with free will. Similarly, quantum mechanics might be wrong, but we have no empirical data that shows it is wrong, so it should be philosopically consistent to say QM is all there is. This means QM should describe observers too, and if there is no mathematical contradiction with this view, there should not be a philosophical contradiction either, even if there is a contradiction with experiment. This is the many-worlds philosophy, and it's the self-consistent answer if quantum mechanics is correct. It might be annoying, but I don't think it is too annoying--- one should just learn to live with many-worlds as a fine philosophical position.

But it is wrong to just say "many-worlds" at this point, because the quantum description has not been tested in the realm where the many-worlds have a real logical-positivist manifestation--- most obviously when doing factoring of enormously large numbers using a quantum computer. Until we do this, it is definitely conceivable that nature is only very closely approximately quantum for small systems of a few particles, in the cases we tested the theory already, and is just not quantum for highly entangled many particle systems.

Even if the world turns out to be really quantum, and a quantum computer factors numbers all the time, finding a deterministic substructure is useful for giving a computationally tractable small truncation of quantum mechanics in cases which are not a quantum computer, and it is possible that this truncation can be useful for quantum simulations. This is so needed that I think finding a substructure to quantum mechanics is a central important problem, personally, regardless of whether it turns out to be right. For this reason, I devoted much time to understanding your approach.

The problem with your construction is that it works too well, it is too easy to transform a quantum system into a beable basis, so that the global wavefunction is evolving in a way that is deterministic using the global Hamiltonian. Since you introduce the Hilbert space early and use it to do the transformation of basis to the internal states of the automaton, there is no obvious barrier to transforming a quantum computer into a beable basis, nor is there any barrier to violating Bell's inequality locally. These do not suggest that the no-go theorems are flawed, rather they suggest that the transformation to a beable basis with a permutation Hamiltonian does not produce a true classical system.

The precise way in which I believe that this system fails to be classical is in state-preparation on the interior. The process of state-preparation involves a measurement, entangling some interior subsystem with a macroscopic subsystem, and then a reduction of the macroscopic system according to Born's rule, leaving a pure quantum state of the interior subsystem. In your paper on Born's rule, you suggested how reduction should happen in a CA system, but your precise models don't really respect this intuition, in that the measurement of intermediate states always produces one of the eigenstates of the observable in the interior, no matter how complicated the observable and how entangled its eigenstates. This is what allows you to reproduce quantum mechanics on the interior subsystems, I am somewhat certain that this does not keep the state unsuperposed in the beable basis. Because these internal reductions don't respect the probability structure, you are really doing quantum mechanics not CA, and this is the only reason that you have such an easy time sidestepping the no-gos.

The fact that you sidestep the no-gos with no difficulty suggests strongly that your construction is leaving the space of allowed classical probability distributions on the CA somehow. The only place where this can happen is during internal state-preparation, during measurements of internal operators. This is how you prepare Bell states or quantum computers, after all. These interior operations must be producing states (after projection) which cannot be interpreted as classical probability states of the automaton, although the Hamiltonian evolution never does this. This is not a proof, but I would bet lots of money (if I had any). I asked for a proof here: In t'Hooft beable models, do measurements keep states classical?

This is part I of the answer, I post it separately, so that people who agree with this part don't have to upvote the second part, which is devoted to a different approach to getting quantum mechanics out of automata, to answer the second question.

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This question tries to reproduce quantum mechanics from classical automata with a probabilistically unknown state.

Probability distributions on Automata states

Start with a classical CA and a probability distribution on the CA. To keep things general, I allow the CA to have some non-determinstic evolution, but only stochastic probability, no quantum evolution, and it's not necessary, you can always put the probability in the initial conditions, with no stochasticity at intermediate times, it's just an option.

The first point about these stochastic systems is detailed here: Consequences of the new theorem in QM? (in the section on duck feet). If the flow of probability is always between states where the probability is only infinitesimally different from te stationary distribution, then the classical flow conserves entropy, and is reversable, even if it is probabilistic and diffusive. This is the central motivation for the construction, and one should review how the particle diffusing the in the heat exchanger diffuser bounces back and forth reversibly from room to room, in a linear way described by an operator with a mostly complex eigenvalue, even though it is at all times only diffusing between different allowed regions.

Consider a classical probability distribution on a CA, $\rho(B)$ where B is the state of all the bits comprising the automaton, then

$$ I = - \sum_B \rho(B) \log(\rho(B)) $$

is the information contained in fully knowing the automaton state, above that provided by the distribution. If you make a perturbation to first order, changing $\rho$ to $\rho+\delta\rho$, you find

$$ I = - \sum_B (\rho(B)+\delta\rho(B)) \log(\rho(B) + \delta\rho(B)) $$

When $\rho$ is uniform, the first order correction vanishes since the sum of $\delta\rho$ is zero, and the second order correction gives a quadratic metric structure on $\delta\rho$.

$$ I = - \sum_B \delta\rho(B)^2$$

This is what I identify as a pre-quantum structure on the space of perturbations to the uniform distribution. The reason it is so symmetrical (like a sphere, not like a simplex) is because the perturbation is small. The reversibility is required by conservation of entropy, and conservation of entropy requires that all the transformations on $\delta\rho$ are orthogonal.

The picture to zeroeth order is that nearly every state is equally likely, but some states are slightly more likely than others, and the information revealed by experiments is only producing a slight bias for some states rather than others. These slight biases then are more symmetric than the underlying space of probabilities on automata states, because these distributions never deviate enough from uniformity to see the corners of the probability space simplex. The corners are the states where the automata bits are known for certain, and if you are always far from these, you can find a symmetric and reversible probabilistic dynamics.

Here is the central problem with this approach--- it is impossible for an information containing perturbation $\delta \rho$ to be everywhere small. The reason is that an everywhere small $\delta \rho$ necessarily produces a state which is almost indistinguishable from the uniform state, and which therefore makes a perturbation which corresponds to you having learned much less than even 1 bit of information. For example, if you have N bit automaton, and you make a distribution where the probability of every bit value is between ${1\over 2} - \epsilon$ and ${1\over 2} + \epsilon$, you get an information content bounded above by a small multiple of $\epsilon$ bits.

The reason is that learning even one bit of information about an automaton state roughly cuts down the number of states you can occupy by a factor of 2. This means that the true probability distribution must be significantly small on at least half the configurations, and it cannot be a small perturbation. This means that the information expansion breaks down, and this is where I was stuck for a long time

Locally small perturbations

The reason the notion of "small perturbation" is failing is because a small perturbation, as in the duck-feet example, is not globally small, it only has the property that the ratio of the probabilities between two nearby states is small. If the states are made by independently varying lots of bits, there are many states with the same ratio of probability.

The fix might as well be the following easy trick: just raise everything to the M-th power. If you have a system with states indexed by i an integer in the range 1,2,...,N, and a perturbation

$$(\rho_i + \delta\rho_i) $$

You can take the M-th tensor power of $\rho$, to produce a product distribution on the tensor space with M indices $i_1,i_2,...,i_M$. This product distribution is defined by the condition that changing every value i from one value to another produces the same ratio change in probability.

Now it is allowed for $\delta\rho$ to be small even when the information in $\delta\rho$ is not, because the M-th power isn't small at all. In fact, in this system, because it's a tensor product, if you know that the information content of $\rho+\delta\rho$ is I bits overall, then you learn that

$$ M \sum_B \delta\rho^2 = 1 $$

In other words, the finite information perturbations to the stationary distribution on a system with M-copies forms a (real, not complex) Hilbert space, ever more perfectly as M goes to infinity. If the dynamics is duck-feet, meaning that the entropy is conserved with the small perturbation, then the time evolution of $\delta\rho$ is necessarily an orthogonal transformation, no matter what the underlying stochastic or deterministic evolution law is.

The basic idea is that you can make a quantum mechanics emerge from stochastic evolution of systems with many identical copies, under the condition that the copies interact symmetrically with each other, so that you don't know which copy is which.

To see how the inner product comes out, you consider the mutual information, which tells you how independent two different distributions are. To lowest order, this is found by taking the information in $\delta\rho_1$ and $\delta\rho_2$ and subtracting the information in $\delta\rho_1$ and $\delta\rho_2$ separately. Since these are the norms, you find

$$ I_{12} = ||\delta\rho_1 + \delta\rho_2 ||^2 - || \delta\rho_1 ||^2 - ||\delta\rho_2||^2 = \langle\rho_1 , \rho_2 \rangle $$

So that if you have two distributions, they share states to the degree that their inner product is nonzero.

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Are you familiar with this paper: arxiv.org/pdf/1111.6597.pdf and if so, is it applicable to your kind of construction? –  user1247 Aug 19 '12 at 19:43
    
@user1247: That constraint is relatively weak, in that it assumes distant systems are described by independent ensembles (this is again locality creeping in). It was discussed here: physics.stackexchange.com/questions/17170/… . I read it and understood the argument, and found it interesting, but it doesn't apply to this type of thing, as you can explicitly see by constructing entangled states in the above: they always share the statistical ensemble, no matter how far apart they are, their description is not by concatenation of each individual one. –  Ron Maimon Aug 19 '12 at 20:30

There's no doubt it's possible to reproduce quantum integrable models rather efficiently and simply using classical systems. And of all integrable systems, harmonic oscillators are one of the simplest. The real challenge is to reproduce quantum nonintegrable systems. Can you reproduce quantum chaos? Can you reproduce quantum nonintegrable spin models over a 1d spatial lattice? Trying perturbation theory from an integrable models runs into the problem that the number of feynman diagrams grows exponentially with the number of loops.

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Its just the start in the future he may be able to. –  Asphir Dom Aug 18 '12 at 10:35
    
There is doubt! It is not trivial to reproduce simple quantum mechanics from cellular automata, and I don't think 'tHooft does it (although I think he came very close to doing so, and intuitively he is spot on). –  Ron Maimon Aug 19 '12 at 21:35
    
@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. –  G. 't Hooft Aug 20 '12 at 15:57

protected by David Z Aug 26 '12 at 1:06

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