Entanglement as determined correlation

Question

I imagine that at the moment of entanglement the wave function collapses and the quantum states of the 2 entangled particles will be set, so that there will be no superposition after the moment of entanglement.

So that what appears to us as magical link of the random states between two particles, is just a pre-determined correlated states with no random probability of states.

In this lecture the speaker claims that entanglement and measurements are same phenomenon: Google Tech Talk by Ron Garret.

So if we assume that entanglement and measurements are the same thing, then we can argue that wave function collapses at moment of entanglement.

Entanglement is a property of a quantum state (whether there exist uniquely defined pure states for the subsystems or not), measurement is an operation you can do on a quantum system. They cannot be "the same thing" because the refer to different categories (property vs. operation), so I do not understand the question. — ACuriousMind, Feb 17 '17 at 13:11
Previous discussions of Ron Garret's Google Tech Talk on this site: Is quantum entanglement functionally equivalent to a measurement? and Does the following experiment disprove the Copenhagen interpretation?. — Emilio Pisanty, Feb 17 '17 at 15:21

score 1 · Answer 1 · answered Feb 17 '17 at 15:27

This sort of 'pre-determined correlation' is precisely the sort of correlation that is ruled out by Bell's theorem. If the wavefunction were to 'collapse' (whatever that actually means, particularly since the whole point of CHSH-like games is that at the point of separation you have yet to decide which observable you are going to measure, out of a set of incompatible measurements) at the moment the boxes are separated, you would have exactly a local-hidden-variables theory.

This is a reasonable model of reality, at least to our classical eyes. However, it is restricted by Bell's theorem in the amount of correlation it can exhibit, and experimental measurements conclusively show more correlation than that. So: no, this doesn't work.

Entanglement as determined correlation

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