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Consider the following standard case of an entanglement experiment:

  1. We prepared two electrons A and B with states

$$\lvert A\rangle=a_1\lvert \uparrow\rangle+a_2\lvert \downarrow\rangle$$ $$\lvert B\rangle=b_1\lvert \uparrow\rangle+b_2\lvert \downarrow\rangle$$ that are not entangled. If we measure any one of them, the spin obtained will be $\uparrow$ or $\downarrow$ with probability determined by the coefficients $a_i$ an $b_j$ for $i,j=1,2$

We can easily construct the product state by

  1. We then brought them close together, and then applied a suitable magnetic field with hamitonian $\hat{H}$. This rotates the state so that it is no longer a product state, thus entangling the two electrons.

$$e^{i\hat{H}t}\lvert AB\rangle=\lvert A'B'\rangle$$

  1. We then isolate them, and now perform a measurement on $\lvert A'B'\rangle$ on each of the electrons, return to compare results, in order to determine the observable $\sigma_{nAB}=(\vec{\sigma}\cdot \hat{n}_A)\otimes (\vec{\sigma}\cdot \hat{n}_B)$, where $\hat{n}_A\cdot \hat{n}_B=\cos\theta$ and $\theta$ is the angle between the two directions of the unit vectors in space. This will give one of the many outcomes on the pairs of spins as predicted by $\langle \sigma_{nAB}\rangle$.

  2. By repeating steps 1-3 many times, we noticed the correlations obey the usual correlation dependence on $\cos\theta$ found in Bell's Theorem.

Now compare the above with the following qualitative "watered down" analogy

  1. We prepared two marbles A and B, both colorless. Each carry an instruction of probability which reads as:

    When touched, turn into (R)ed $r_A$ amount of the time or (Y)ellow $y_A$ amount of the time

similarly for B. Should we decide to touch either of them, they will change color, and we found there is no correlation between the colors R and Y.

  1. We brought the marbles close together, leave that for some known amount of time t. The two marbles are still colorless.

  2. We then isolate them. We then touch one of the marbles (A or B). Suppose for this trial we got A turned R. We then go back to where B is kept, and we found it turned Y.

  3. Repeat Steps 1-3 many times and we found that R and Y are correlated in a way that agrees with the resulting probability of what happens when objects with some given instruction of probability (inferred during the preparation of the marbles) is being place close together for time t.

Comparing between these two descriptions, it seems to suggest the correlated outcomes that is predicated by the $\cos \theta$ relationship of the violation of Bell's Theorem was established at the moment when the two electron became entangled hence their state cannot be factorised into a product state anymore, and hence their probabilities also become correlated.

The measurement then pick out one of the outcomes of this correlated probability distribution (which is determined by the wavefunction) thus there is no need of the notion that measurement of one electron instantly changes the other.

We also knew that the time evolution of the wavefunction hence the probability of some observable of the state is deterministic and obeys a Schrödinger equation.

Therefore this suggests the wavefunction of the electron is an intrinsic property similar to charges that can be transformed and shared between other electrons, and entanglement introduce correlations in this intrinsic property.

Now since the wavefunction is not an observable, hence unmeasurable, and even its probability can only be inferred from an ensemble of states and not just by measurement of one state. Yet it pretty much controls all the possible outcomes we can obtain from the experiment, with deterministic time evolution, and it effectively span across spacelike distances between the subsystems, it seems to qualify as a non-local hidden variable.

Yet nobody refer to the wavefunction as a hidden variable.

Why is the wavefunction not considered a (non-local) hidden variable?

Qmechanic
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Secret
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    what would be the use of doing so? Hidden variables have been proposed in order to have the wavefunction emerge from an underlying deterministic local system, to model the "nonlocality" of the wavefunction ( as with the Bohm pilot wave for example). – anna v Oct 23 '16 at 10:33
  • See this SE answer where the wavefunction is considered a hidden variable: http://physics.stackexchange.com/questions/3158/why-is-quantum-entanglement-considered-to-be-an-active-link-between-particles/3433#3433 . I personally still don't understand: you still have a non-deterministic theory (as anna v comment) – isometry Oct 23 '16 at 15:35
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    I am not sure, it is possible my understanding might be $\psi$-ontic, because what I have is 1) wavefunction evolves determinstically, 2) measurement is probabilistic and such probabilities are given by the wavefunction (probabilistic outcomes), 3) Before measurement there are no predetermined values (nonrealist), 4) interaction that establish the entanglement correlates some parts of the wavefunction of the subsystems to form a wavefunction of a composite system that spans spacelike distance (nonlocalism), with relativity protected by no communication theorem, 5) Wavefunction is a physical, – Secret Oct 23 '16 at 17:07
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    unmeasurable intrinsic property of the system, whose existence can be inferred by the probablistic outcomes of the observables and how entanglement establishes the correlation within it during some interaction in the past, and not because of knowledge learnt form measurement (thus a nonlocal hidden variable?) – Secret Oct 23 '16 at 17:08
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    typo: point 2) missing part: The measurement can change the wavefunction just like any interaction – Secret Oct 23 '16 at 17:36
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    The wavefunction is not a hidden variable because it is not hidden: we can experimentally distinguish $|\psi\rangle$ from $|\phi\rangle$ if $|\langle \phi | \psi \rangle | < 1$ – Ruben Verresen Oct 24 '16 at 11:35
  • Your marble experiment does not violate Bell's Theorem (or any other theorem of classical probability) and is therefore quite entirely irrelevant to the issue here. The wave function does not count as a hidden variable because it cannot *determinstically predict the outcomes of all measurements. In your marble experiment, we can easily explain the result by invoking some determinstic (though mysterious) force. – WillO Apr 29 '17 at 15:47
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    It's largely irrelevant to the main question, but your claim in the first point 2 is incorrect. Applying a magnetic field will not entangle the particles because the hamiltonian is a sum of single-particle operators. – Emilio Pisanty Apr 30 '17 at 13:42

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