# Can quantum entanglement be simulated on a digital computer to any degree of precision?

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## Solution 1

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. 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! 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.

## Solution 2

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.

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Updated on August 01, 2022

• Admin over 1 year

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?

• DanielSank over 8 years
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."
• Admin over 8 years
@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.
• CuriousOne over 8 years
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.
• Admin over 8 years
@curious one: did you really understand the question? the point is than then that the interpretation of Bell's theroem must be wrong
• CuriousOne over 8 years
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.
• Admin over 8 years
@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
• CuriousOne over 8 years
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.
• gatsu over 8 years
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.
• CuriousOne over 8 years
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.
• Admin over 8 years
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.
• CuriousOne over 8 years
@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.
• ACuriousMind over 8 years
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)
• image357 over 8 years
@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.
• CuriousOne over 8 years
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.
• image357 over 8 years
@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 over 8 years
@CuriousOne: CAPITAL LETTERS won't make your arguments more valid xD
• ACuriousMind over 8 years
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?)
• CuriousOne over 8 years
@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.
• image357 over 8 years
@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 over 8 years
@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.
• Mark Mitchison over 8 years
@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.
• PyRulez over 8 years
@CuriousOne There wouldn't be random number generation, since you just list all the possibilities, not sample them.
• PyRulez over 8 years
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?
• CuriousOne over 8 years
@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?
• Rococo over 8 years
Yes, that's basically the same statement.
• Slaviks over 8 years
@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
• Mark Mitchison over 8 years
@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.)
• Slaviks over 8 years
@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?
• Mark Mitchison over 8 years
@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 over 8 years
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!
• Slaviks over 8 years
@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.
• Mark Mitchison over 8 years
@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.