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First principles modelling of physical phenomena has been very successful in physics. The largest limitation is perhaps the fact that many QM problems are NP hard so we would need really powerful computers if we want greater accuracy. But any QM model should be, in principle, still computable to any desired level of precision.

My question is: is this correct?

My problem with a positive answer would be that some local rule cellular automaton are Turing universal, which would imply that entanglement could be simulated by a model that uses a classical local rule. This seems wrong, doesn't it?

  • The answer to the first question is "yes, as far as we know". The answer to the second question is "no, it's not wrong." – DanielSank May 13 '15 at 18:16
  • @DanielSank but the last paragraph in the question imply that you can have a local theory of QM: the one simulated by the local CA. And it is agreed QM can not be modeled by a local hidden variables theory. –  May 13 '15 at 18:21
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    What makes you think in the first place that it can't??? Most better introductory QM textbooks contain all the necessary math to do the very computation that you are looking for. – CuriousOne May 13 '15 at 19:16
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    @curious one: did you really understand the question? the point is than then that the interpretation of Bell's theroem must be wrong –  May 13 '15 at 20:05
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    If the question that you wanted to ask is the question that you did ask, then yes. Quantum entanglement can be simulated on a digital computer to any degree that you like. This has nothing to do with Bell's theorem, whatsoever. I think you are simply mixing up a bunch of stuff that is completely independent. – CuriousOne May 13 '15 at 20:16
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    @CuriousOne I disagree with you, to me the connection is pretty clear, but perhaps I might have to put it as a separate question and make it more explicit –  May 13 '15 at 20:18
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    I think you really need to reformulate your question because this one has a trivial answer, which you don't seem to like (for whatever reason). The question of computability is not a physics question to begin with, of course. None but about a dozen trivial Hamiltonian systems can be computed the way you seem to want to compute them. In that sense classical mechanics or ANY physics doesn't agree with ANY computability requirement. Entanglement, of course, is one of the trivial cases that can be computed, so you picked the wrong one to get upset about nature being uncomputable. – CuriousOne May 13 '15 at 20:27
  • If Nature were to run numerical integrations of the Schrodinger equation as our computers do and then sample an outcome of an experiment from the corresponding calculated probability distribution, then yes in this sense a classical local theory can emulate any quantum system. But, I don't think these are the systems ruled out by Bell's theorem for we implement already quantum mechanical rules in the computer. – gatsu May 14 '15 at 09:14

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Here are some facts:

  1. As others have said, the evolution of a quantum state, including entanglement, can be simulated arbitrarily well classically with sufficient resources. That said, it does generally require exponential resources due to the exponential growth of Hilbert space.

  2. Of course, a (classical) computer can't deterministically predict the outcome of a particular measurement, only give correct probabilities. So it is very important to distinguish between simulating deterministic quantum state evolution classically (which is no problem) and actually replacing quantum mechanics with a classical model (which can't ever happen). The difference comes at the actual measurement.

  3. A computer can also simulate any number of impossible things. You can make a computer simulation where energy disappears, objects travel faster than light, etc etc.

  4. When you run a Bell test on your computer program, at some level what it will do is assign the outcome of one measurement, then communicate that to the other entangled particle so that they both have correllated outcomes in the right way. In other words, the whole program relies on the two "particles", however they are stored in the computer, being close enough to communicate with each other. As a result, a classical computer could never pass a loophole free Bell inequality test. Specifically, if you load the same program onto two computers and send them far apart, they will never be able to give measurements with the same outcomes as measurements on two entangled particles would.

  5. Notice once again that it's no problem for both the computers to know what state they're supposed to be in before you measure them. It's getting the two measurement outcomes to be properly correllated (in all measurement bases) that just isn't possible.

Rococo
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    So the last two points are saying classical computers can simulate quantum entanglement, but that doesn't mean your program can actually entangle the computer itself in any way, right? – Christopher King May 14 '15 at 02:06
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    Yes, that's basically the same statement. – Rococo May 14 '15 at 02:55
  • Looking back at this old answer, I see that I said (now corrected) "Actually, modelling the evolution of a quantum system is not even (believed to be) NP hard- if it was, a quantum computer could solve NP problems!" This is incorrect, I was mixing up NP hard and NP complete. Simulating a general quantum evolution is, I believe, generally expected to be NP hard but not NP complete. – Rococo Feb 10 '24 at 05:06
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Quite inefficiently compared to "direct implementation" in quantum hardware, but yes, entanglement defintely can be simulated to arbitrary degree of precision give adequate memory and time resources in a conventional, entirely classical digital computer! Just go and solve multi-particle Schrodinger equation.

And this fact has nothing to do with Bohmian pilot wave or any other correct interpretation of quantum mechanics.

Slaviks
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    If you can simulate it to any degree of precision then it means that you can have a theory that is in principle local (the rule 101 CA) but violates Bell's inequalities. –  May 13 '15 at 20:46
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    @brucesmitherson: One can generate random numbers to any required degree of randomness, which is all that is needed to satisfy your trivial requirement. You need to think this trough some more. – CuriousOne May 13 '15 at 20:50
  • @brucesmitherson How are you going to reproduce the Bell correlations between space-like separated events just because you have a computer in front of you that can predict the answer? You would need to be sending signals faster than the speed of light, making this simulation a non-local hidden variable theory. – Mark Mitchison May 14 '15 at 00:26
  • @CuriousOne There wouldn't be random number generation, since you just list all the possibilities, not sample them. – Christopher King May 14 '15 at 02:02
  • @PyRulez: In that case one probably doesn't even need a computer for many entanglement scenarios... it should be just a bit of linear algebra, right? – CuriousOne May 14 '15 at 02:18
  • @markmitchision Full computer is a full computer (Turing complete), I am not required to have the representation of state (Hilbert space vector/density matrix) to be in any way local on my simulation hardware – Slaviks May 14 '15 at 05:31
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    @Slaviks I agree with your answer. I was just pointing out that the existence of classical simulations which can reproduce the output of quantum measurements does not violate Bell's conclusions about the incompatibility of local realism with physics, which OP seems to be confused about. A classical computer cannot reproduce Bell correlations between space-like separated events without superluminal signalling. (Here I am talking about genuine, physical space-time events, not simulated events on a computer.) – Mark Mitchison May 14 '15 at 10:22
  • @MarkMitchison This is where I have problem understanding you:"A classical computer canno reproduce ...genuine, physical space-time events". What do you mean by "reproduce" if not simulation? An alternative universe? – Slaviks May 14 '15 at 13:30
  • @Slaviks You presumably agree that there cannot be a classical (ie. local and realistic) model underlying bell correlations? On the other hand, a classical computer system can be explained in terms of local realism, even though it may also simulate arbitrarily entangled states. The OP stated in a comment above that this latter fact implies that bell correlations are in fact locally realistic. – Mark Mitchison May 14 '15 at 15:39
  • I am trying (and evidently failing!) to explain to him/her why this is not the case. I think it is probably difficult for you to understand what I am trying to say since you did not share the OP's confusion in the first place! – Mark Mitchison May 14 '15 at 15:39
  • @markmitchinson To strengthen out the confusions is my intent. The (nim-existing) "classical rralistic & local models" is a class of imaginary universes that are ruled out by the Bell's theorem. My proposed simulatuon is obviously a very nonlocal mapping between the reality being simulates and the actual simulation code. – Slaviks May 14 '15 at 19:57
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    @Slaviks Agreed. And it may obviously be a non-local mapping to us, but clearly not to the OP based on their comments. That was all I was trying to clarify. – Mark Mitchison May 14 '15 at 23:40
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I will try my own answer, which is mostly thinking out loud, and am ready to delete it if it is too non-sensical.

For starter, we need to agree on what we mean by theory; in particular a theory of what and with which logical rules.

Let's roughly say that a theory is set of true statements about objects of interest for the theory with rules that enables one to relate some true statements to others and hence discover new true statements from existing ones.

  • A classical theory would be a theory whose logical structure follows propositional 2-valued (classical) logic

  • A quantum theory would be a theory whose logical structure follows instead quantum logic (and where for instance there is non-distributivity of the disjunction relation)

Here is what I think is true:

  • If the objects of the theory are usual physical observables like position, momentum, angular momentum etc..., then it is not possible to construct a classical theory based only on propositions about these observables that eventually does not contradict outcomes of some experiments on quantum systems.

  • With the same set of objects, it is however possible to construct a quantum theory (in the previously defined sense) that gives rise to the right quantitative description of experiments of any kind on quantum systems (in virtue of Gleason's theorem).

Bell's theorem essentially acts at that level of description (for that type of ontology) and contributes to the first point essentially.

The twist comes from the fact that if the objects of the theory are the quantum state and probabilities of outcome of experiments of a particular system then it is possible to construct a classical theory of propositions about these objects that is in perfect agreement with experiments on quantum systems; and this is the Copenhagen Quantum Mechanics that we use everyday (equivalently the Everettian formulation is fine as well, although the space of objects is a bit smaller).

Simulations of quantum systems by a classical computer essentially compute and interpret the last theory which is again a (logically classical) theory of propositions about quantum state and probabilities of the system of interest. So this does not contradict, I think, Bell's theorem that does not rule out the same type of theories.

gatsu
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In short, yes it can be done but it isn't going to be exactly precise as quantum mechanics goes. People doing research in lattice QCD are doing this very sort of thing; however, the approaches that I've seen have been to program in known laws of physics and use vast amounts of computational power to simulate single particles. Now, this seems a bit counter-intuitive. After all, we see complex simulations all the time using several graphics engines for gaming and the detail of these simulations are becoming quite remarkable.

So back to quantum simulations and in particular quantum entanglement. I'm not going to sit here and argue (As many people like to do) classical physics and the finer points of quantum mechanics equations because I don't believe it's necessary here. There's no point because we're talking about processors that can simulate anything you want with any behavior you desire. Parallelization can simulate two things happening simultaneously communicating with each-other on a classical computer as two sets of instructions can be computed in parallel. As for measurement, since you're the programmer you get to determine what you can and cannot measure.

To my mind this isn't so much of a physics question as it is a programming question depending on what you expect to get out of it. Of course if you have a quantum computer then the methodology changes as there isn't any (Or perhaps only one layer of) abstraction between the physical process and the simulation. As for the actual measurements and/or two computers being separated at a distance, you would probably need a quantum computer (Or two of them) for the kind of precision you'd want but you can still simulate entanglement. I'll give some examples.

This doesn't have to be such a hard problem as in programming simulations. For instance, reference: List of QC_simulators where libraries can be found for simulating quantum entanglement (Just search the page for "Entanglement") in several languages.

I don't see it being non-computable on a classical computer in the sense that the simulation will simulate the quantum mechanical effect but to be clear, you won't actually be entangling two particles on a classical computer and be able to get the same measured results (Bell's inequality does apply here). Sure, you might program your simulation on a classical computer so that the particles are in their proper states but the act of measurement technically won't yield the same results as an actual act of measurement in a real life experiment.

To actually simulate entangled particles you would need a real quantum computer. Otherwise, Bell's inequality, well, wouldn't compute. There are various resources on quantum computing algorithms and programming quantum computers. I haven't researched them that much as I don't have access to a quantum computer but Google has setup a webGL based simulator using the gate method of QC which can be found here: Quantum Computing Playground which I've experimented with and is an interesting resource as are the libraries I previously linked.

In essence, the answer is both yes and no depending on what you're trying to get out of the simulation. Emulating quantum computations isn't like emulating a Nintendo. In that case they're both classical computers.

As a side-note, I would mention that the popular D-Wave computer doesn't use the gate model of quantum computing (Probably why there's been a lot of controversy surrounding their computer) but rather uses adiabatic quantum computing.

xendi
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If the emission of two entangled photons from nonlinear optical crystals does satisfy your definition of any QM model, then yes, it is possible to simulate it on a digital computer to any degree of precision. And I believe that it is more a question about how good undetermined are two randomization processes on the computer, first for the polarization of one of the two photons (the second photon has then simply the opposite polarisation state) and second for the destabilization of this states on the way from the source to the detectors.

We know that if we place the detectors close enough to the source we reach near to 100% the possible detection of 50% of the right state. In other words, placing a horizontal polarizer in front of the first detector and a vertical polarizer in front of the second detector in nearby 50% of all cases we get the photons on the detectors. This happens, if the way ist short. Then longer the way, then more outer influences destroy the directions of polarization of the two photons and the 50% get down. If we reach 40% for each detector, then we have a success of 80% compared with the ideal situation.

In both cases, the real process and the computer simulation, the states of the two photon is unknown until we measure it or until the computer displays the results. The mathematical probability function of our knowledge about this states collapses in the moment of seeing the result. But in the case of computer model one has to agree with the fact that the two photon states were generated before and once generated this has nothing to do with any probability. Coming to this point I ask, could it be, that the pair production works the same way. Once produced the states are fixed and only outer disturbances wash out the clear entanglement of these fixed states.

P.S.: I see clear the statements of well educated scientists that this could not be because it is never teached by this way. But this could not be a counter-argument to my answer. The answer only could be wrong if any of the steps in my answer could be falsificated. But be carefully, for the real process this is not possible since the process in the time between the photons emission and the photons detection is not viewable and my explanation is as good as the old one.

HolgerFiedler
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My problem with a positive answer would be that some local rule cellular automaton are Turing universal, which would imply that entanglement could be simulated by a model that uses a classical local rule. This seems wrong, doesn't it?

There is a classical, local, deterministic and realistic equivalent of QM, called bohmian mechanics. And no, it is NOT non-local, e.g. supports instantanious actions. Bell's locality condition is just non-sense. See this Link.

There is a nice playlist on youtube considering this topic: https://youtu.be/_6TNF854Xmo?list=PL7LbfRoKBR5OpRjt8toBOmzqGjH7zaM1m

Also, currently there are experiments done that support this interpretation of QM: https://www.quantamagazine.org/20140624-fluid-tests-hint-at-concrete-quantum-reality/

Furthermore there is a deep correspondance between classical statistical mechanics and the operator formulation of QM via Koopman-von-Neumann classical (statistical) mechanics: http://en.wikipedia.org/wiki/Koopman%E2%80%93von_Neumann_classical_mechanics

Strictly speaking, the only difference between usual classical (statistical) mechanics and QM is the fact that the particle's position and its momentum do not commute, that is a measurment on the former can change the later and vice versa. In ordinary classical mechanics this fact is blurred because one always interacts with heavy objects, for which this change is negligible.

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image357
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    You are fighting a fight that was lost 80 years ago. Measurements are not the reason why one can't measure position and momentum at the same time. QM is no more the theory of point particles than Newtonian mechanics ever was (we just teach it that way in high school because the kiddies can't understand continuums mechanics, yet). Single particle QM is not even self-consistent, you have to go to quantum field theory to get a self-consistent theory and then there are no more particles, but only quanta. – CuriousOne May 13 '15 at 20:44
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    The interpretation used has nothing to do with whether or not something can be simulated, has it? (If you think it has, please indicate in the answer, why) – ACuriousMind May 13 '15 at 21:18
  • @CuriousOne: So your only argument is: "But they say it's not this way?!". Seriously, I know that this battle was lost before I was born and placed ad acta when the only progress on that topic was due to philosophical considerations. Btw, I wasn't talking about QFT yet. Standard QM is about particles, still. – image357 May 13 '15 at 21:23
  • It's called QUANTUM mechanics for a reason, even when it's single particle quantum mechanics. What are you measuring in quantum mechanics? Particle or quanta? Be careful now, it's not a trick question. – CuriousOne May 13 '15 at 21:29
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    @ACuriousMind: As the OP already stated, if there is a single turing machine that relies on classical operations and which can calculate whatever QM should predict, this is essentially equivalent to reformulating QM to a classical theory under that turing machine's context. That is, "the world is essentially non-classical" is a false statement, since there exists a theory (realised by that turing machine) which can be described classically and predicts the same results as QM. My point to OP's question is that there is already a classical description. – image357 May 13 '15 at 21:32
  • @CuriousOne: CAPITAL LETTERS won't make your arguments more valid xD – image357 May 13 '15 at 21:34
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    I don't think "If a theory can be computed by a classical machine, then it is a classical theory" is obvious to be true. Even non-Bohmian QM is, operationally, just a bunch of linear algebra and PDE solving operations. Computers can certainly do that, too, why would you need a Bohmian description to get a Turing machine to spit out QM predictions? (E.g. there are algorithms for computing Feynman diagrams, and hence QFT predictions for which no "classical" or "realist" alternative is known (to me). Does this mean QFT is "classical"? What does "classical" even mean then?) – ACuriousMind May 13 '15 at 21:39
  • @Marcel: I was trying to be helpful since you seem to have missed the major clue in the naming of the theory. Newtonian mechanics, by the way, is also not called "particle mechanics" for a reason and it's almost the same reason. The concept of particle is simply not a central concept in either theory. It may be the central concept of poorly taught high school physics, and we may not be perfectly successful to eliminate that demon from poorly taught student's minds in university, but that doesn't change the facts about the way these theories work. – CuriousOne May 13 '15 at 21:39
  • @ACuriousMind: I'd say a classical theory is one that can be described by some reformulation/generalization of $F=ma$ (e.g. the relativistic equations of motion) for which the forces $F$ are given by external fields and the carges carried by the particle under description. I call that classical because this description shares a deep analogy to what we think of when we describe a solid object that we can actually see and touch. That saying, the point is that QM is in no way different to what we can see and touch, just a good generalization for small scales. – image357 May 13 '15 at 22:06
  • @ACuriousMind: Note that a turing machine whose realization relies on such classical descriptions, can in turn be used to give a (rather complicated) but still classical description of what it can calculate. The argument is: "There is a turing machine that can compute any QM system and end" $\rightarrow$ QM is essentially classical. However we know that the computability of some classical systems is not given: Take for instance a pendulum for which the exact frequency is non-computable in finite steps. Still there is a classical description. – image357 May 13 '15 at 22:16