Can you also explain this strange notation. Already the first equation is not clear to me. What does it mean? It's an average of what? What's in the numerator? In this way it's indeed a mystery compared to quantum theory, which is not a mystery but the solution to the mystery of the observed behavior of microscopic particles as well as the then ununderstandable stability of macrocsopic matter surrounding us.
That said, let me come to your AJP preprint. I'll got through it as I'd be a referee.
Section I is confusing and doesn't make sense to me to begin with.You should explain Mermin's apparatus to make your paper self-consistent. You don't explain it but rather open several other topics (4D spacetime views and Fermat's principle) which are completely unrelated to the "conundrum of entanglement". Since Bell it's the more clear that quantum theory is not the mystery but the solution to describe the behavior of subatomic particles, in this case spin-entangled states of two particles.
As a referee, I'd suggest to cancel Sect. I and use Sect. II as the introduction, explaining clearly Mermin's apparatus. You should explain what's entangled. It's the spins of the two particles emitted from the middle box. It doesn't make sense to say "two particles are entangled" in QM. You have to say which observables are entangled. Figs. 3 and 4 are unexplained. What are they good for? To make the paper understandable to at least a physics student who has heard the QM 1 lecture, you should just explain the experiment in terms of standard QT, i.e., say that the two spin-1/2 particles are prepared in the pure ##j=0##, ##j_3=0## state represented by the state vector
$$|\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle - |-1/2,1/2 \rangle),$$
where the notation for the two-particle spin states is the usual one, i.e.,
$$|\sigma_{z1},\sigma_{z2} \rangle \equiv |\sigma_{z 1} \rangle \otimes |\sigma_{z2} \rangle.$$
To make your paper as mysterious as you can you don't even tell this your reader anywhere.
What you describe then is completely ununderstandable to me. It doesn't reflect at all what QT predicts to be measured in A's and B's measurements. It's not clearly explained. You can calculate it easily of course. You simply quote the result in Eqs. (2) and (3) without clearly saying what's measured. Obviously what's meant is that A and B choose a plane (say the ##xy## plane for simplicity since due to the total isotropy of the entangled state it doesn't matter anyway which plane they choose). Then with two unit vectors ##\vec{n}(\alpha)=(\cos \alpha,\sin \alpha,0)## what's measured are the spin components of A's and B's particles in directions ##\vec{n}(\alpha)## and ##\vec{n}(\beta)## respectively. The probabilities quoted in Eqs. (2) and (3) are then, written in standard notation
$$P(\sigma_{1\alpha},\sigma_{2\beta})=|\langle \vec{n}(\alpha) \cdot \vec{\sigma}_1,\vec{n}(\beta) \cdot \vec{\sigma}_2|\Psi \rangle|^2.$$
On the left-hand side of the equation I denoted spin components in direction ##\alpha## in the above defined sense as ##\sigma_{\alpha}=\vec{n}(\alpha) \cdot \vec{\sigma})##. I'll use this abbreviation from now on.
Of course the possible outcomes for each single-particle spin component are ##\pm 1/2##, and you give the correct probs. for all four possible simultaneous outcomes in Eqs. (2) to (3). But why don't you give this simple explanation rather than the very complicated description so far?
Fig. 6 and its caption is absolutely enigmatic to me. I still don't get the meaning of the words "angular momentum is conserved on average" should mean for unaligned measurements, i.e., for ##\alpha-\beta \neq 0## or ##\pi##. In which sense should there be angular-momentum conservation be measured. I've brought this argument again and again already several times in this thread, and it's not answered. It doesn't even make sense in a classical context to check angular momentum conservation of a system by measuring components of angular momenta on different parts of the system in different directions! Also what's represented in this space-time diagram? Measurement outcomes of A's and B's measurements? Why do I need a space-time diagram to depict this?
I've no clue what ##\langle \alpha,\beta \rangle## should mean either. What's summed over? I can only guess it is
$$\langle 4 \sigma_{1 \alpha} \sigma_{2 \alpha} \rangle=\sum_{\sigma_{1 \alpha},\sigma_{2 \alpha} =-1/2}^{+1/2} 4 \sigma_{1 \alpha} \sigma_{2 \alpha} P(\sigma_{1 \alpha}, \sigma_{2 \alpha}).$$
Then at least I can reproduce Eq. (4).
That Alice's and Bob's "spin angular momenta cancel on average" is the next mysterious statement. Do you mean that for any single-particle spin component the average is 0? That's of course true due to the complete isotropy of the spin-singlet state. Of course, this follows also from the probabilities given by Eqs. (2) and (3). Of course, everything is completely determined by the probabilities (2) and (3). So to translate the very complicated text, what you claim is that in some way you can get these quantum probabilities by a not precisely defined "principle of angular-momentum conservation on average"? I cannot invisage how I can make sense of that, although so far I could make some conjectures about what you wanted to say. As I repeatedly said, I've no clue what the fact that in this setup the single-particle spin components have a 0 expectation value to do with angular-momentum conservation.
That's trivial for the physical situation I guessed you really want to described, given the completely isotropic preparation of the two-particle state (the ##j=0## state). Formally you get the statistics of the single-particle spins by "tracing out the other particle", and this leads to
$$\hat{\rho}=\frac{1}{2} \hat{1}=\frac{1}{2} \left (|\sigma_{\alpha}=1/2 \rangle \langle \sigma_{\alpha}=1/2| + |\sigma_{\alpha}=-1/2 \rangle \langle \sigma_{\alpha}=-1/2| \right )$$
for any ##\alpha \in [0,2 \pi)##.
I'd be very interested, how your referee reports come out from AJP...
