# Is entanglement a classical phenomena (2)? [closed]

The answer to this question seems to be yes, because you can simulate it with a classical computer and thus by a local classical theory (rule 110 CA) (see this question). However most people disagree with this fact (that is classical), and I a would like to understand why. Why, for instance, is always the main argument against emergent quantum mechanics, such as the famous experiments with oil droplets on a vibrating fluid?

• No. It is not a phenomenon that is described in classical physics books. You know... classical physics... that's the false description of nature that we used to teach exclusively until the early 20th century. Is that your question? :-) May 13, 2015 at 21:08
• no, my question is the general meaning of classical as local, and nonrandom.
– user66432
May 13, 2015 at 21:09
• That's not the definition of classical physics, which for sure is neither local (immediate action at infinite distance by potentials) nor non-random (see the assumptions in statistical mechanics). I think you need to take a step back and ask yourself what you are really worried about. What are you really worried about? May 13, 2015 at 21:12
• @brucesmitherson That would be because it's quite unclear. Please try to elaborate and/or explain at even more clearly than you've done thusfar! :)
– Danu
May 13, 2015 at 21:14
• You can simulate any quantum system on a classical computer (albeit at the cost of exponential slowdown). That does not make all quantum systems classical. May 13, 2015 at 21:33

No, entanglement is not a classical phenomenon. That is because its very definition is quantum:

Let $H_1, H_2$ be quantum spaces of states of two system. A quantum state $\chi\in H_1\otimes H_2$ of the combined system is called separable if it is a simple tensor, i.e. if there are $\phi\in H_1,\psi\in H_2$ such that $\chi = \phi\otimes\psi$. A state is called entangled if it is not separable.

And that's it. Classically, you cannot mimick this because the combined space of classical configuration spaces $Q_1,Q_2$ is the product $Q_1\times Q_2$, where by definition there are $q_1,q_2$ such that $q = (q_1,q_2)$ for every $q\in Q_1\times Q_2$.

In words, this is saying the following: To every classical state of a system there are uniquely specified states of the subsystems. For a quantum system, the non-entangled states are precisely the states where there are uniquely specified states of the subsystem from which they arise.

It has to be noted that not every entangled state shows "unclassical" behaviour in the sense that you cannot "see" (in the sense of, "model with classical theories") the same kind of correlation or result in some classical system. Entanglement is, in and of itself, a statement about the theory of states of systems, not about correlations and observations.

• Curiously, one could imagine building Newtonian machines that simulate entanglement, since the propagation of information is instantaneous in classical mechanics... I wonder if that will throw the OP a bone? May 13, 2015 at 21:21
• I assume it is relativistic.
– user66432
May 13, 2015 at 21:23
• @brucesmitherson: Whether your theory is relativistic or not does not really play a role. The combined phase space of a system is the product of the individual phase spaces, and you can do a Hamiltonian formulation even of GR, so whether your "classical" theory is Newtonian, SR or GR does not (at least in a way that would be to me evident) influence what I wrote. May 13, 2015 at 21:25
• This is indeed the initial definition. But the question could be "is there a hidden variable deterministic system that can have entangled-like features?". Bohm QM is such an approach and people from this school of thoughts have proposed entanglement experiments. Jun 8, 2015 at 19:14
• I killed a bunch of comments that were delving into the personal (and edited one, too). Please focus on the physics and remember to "Be nice" at all time. Jun 8, 2015 at 21:07