PeterDonis said:
I agree that it seems like there must be some underlying mechanism that produces the correlations that violate the Bell inequalities, and "nonlocal connection" is as good a name for this unknown mechanism as any. But the fact remains that the mechanism is unknown (and even our belief that there must be some such mechanism might possibly be wrong).
I couldn't agree more with what you said in the part I deleted. Thanks for telling
@DrChinese what I'm telling him for ages in other words. Maybe it helps!
The only point is that you don't need any "unknown mechanism". It's all described by standard QT (and of course relativistic QFT, which is just a special case of QT with the special feature that it's in accordance with the SRT space-time structure and all causality constraints implied by this space-time structure) entanglement.
Entanglement occurs naturally when you have two separable parts of a system, e.g., two particles. Here separable means that you can prepare product states ##|\psi_1 \rangle \otimes |\psi_2 \rangle##, e.g., when you have to dinstinguishable (sic!) particles. Then the common Hilbert space is ##\mathcal{H}_1 \otimes \mathcal{H}_2##, which consists of course not only of product states but all linear combinations thereof. Of course, the product states are part of the Hilbert space, and thus it is possible to prepare the two-particle system in such product states. These are by definition the states, for which the particles are NOT entangled. Einstein called this very nicely "separability", i.e., there are states.
But as is easy to see, the separable states (i.e., the product states) are quite special, and indeed interactions between the two particles lead to linear combinations which cannot be written as product states, and then you call the parts of the system "entangled". That's the origin of the violation of the Bell inequalities and thus the stronger-than-classical correlations of far-distant parts on an entangled quantum system.
To start with you always use particles created in some local process and usually the entanglement is due to some conservation law:
E.g., in the original EPR argument you can think of two (asymptotic) free particles originating from a particle decay. The original particle has a pretty well defined momentum. Now suppose it decays at some time ##t=0## in some region (e.g., take an ##\alpha##-decaying nucleus in a cloud chamber, decaying to the ##\alpha## particle and the daughter nucleus). The resulting asymptotic free state is a free ##\alpha## particle and the doughter nucleus with pretty well defined total momentum and a pretty well defined relative position. Note that these to observables are compatible, and you can prepare the two-particle system in a product state of these two observables,
$$\Psi(r,P)=\psi_{\text{rel}}(r) \tilde{\psi}_{\text{CM}}(P).$$
This you can Fourier transform to the product state
$$\Psi(r,R)=\psi_{\text{rel}}(r) \psi_{\text{CM}}(R),$$
where ##\psi_{\text{CM}}(R)## is a pretty broad distribution.
Now it's easy to transform this to the single-particle positions. You simply need to set
$$r=x_1-x_2, \quad R=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}.$$
Wrt. to these observables you end up in the entangled state
$$\Phi(x_1,x_2)=\Psi(x_1-x_2,(m_1 x_1+m_2 x_2)/(m_1+m_2))=\psi_{\text{rel}}(x_1-x_2) \psi_{\text{cm}}[(m_1 x_1+m_2 x_2)/(m_1+m_2)].$$
Through the usual manipulations it's easy to see that the probability distributions for both ##x_1## and ##x_2## alone are broad, i.e., both positions are pretty much indetermined. The same holds true for the momenta (which you get by Fourier transforming ##\Phi## wrt. ##x_1## and ##x_2## of course).
If you determine by a measurement, which can be as accurate as you wish, the position of particle 1, then you get also a narrow distribution for the position of particle 2 (since ##\psi_{\text{rel}}(x_1-x_2)## is narrowly peaked). The same is true in momentum space: Determining ##p_1## well also determines ##p_2##. Of course you can never determine both ##x_1## and ##p_1## well at the same time, because the precise position measurement prevents the momentum to be determined well too and vice versa.
As you see, there's no need for any further "mechanism" to describe the correlations due to entanglement (i.e., "inseparability") than the known quantum-dynamical rules, and entanglement is the rule rather than the exception.
If it comes to indistinguishable particles, it's even difficult to define separable states. For bosons the two-boson state with the two particles in the same state is an example. All other two-particle states are entangled in the one or the other observable due to the necessity of Bose symmetrization (most conveniently taken account of by using creation and annhilation field operators).
2. I mostly agree with your statement, except you skip the situation in which one measurement unambiguously occurs first. So here's how I would summarize: