Demystifier said:
The point is, if you don't take into account the effect of measurement (intermediate collapse), then the correlator you compute does not correspond to the measured correlation. See e.g.
https://arxiv.org/abs/1610.03161
Oh, now I see your point. If ##A(t_i)## and ##A(t_j)## don't commute, then ##\left<A(t_j)A(t_i)\right>## is not the correct expression for the correlator. I agree with that. In principle, one could construct an example as in the Bell-CHSH setting, where the set ##\{A_1, A_2, A_3, A_4\}## (where ##A_i = A(t_i)##) could be decomposed into two sets ##\{A_1,A_3\}## and ##\{A_2,A_4\}## of non-commuting operators that commute with each other. In such a situation the correlators would be the measured ones and my previous CHSH argument would go through, which is already enough to demonstrate the inviability of Kolmogorov probability. However, this is of course a very special situation and in general, one can't expect such a decomposition into commuting sets. Let's discuss the general case:
In the general case, one needs the joint probability distribution ##P_{ij}(u,v)## on ##\mathrm{spec}(A_j)\times\mathrm{spec}(A_i)## in order to compute the correlator ##C_{ij} = \sum_{u,v\in\mathrm{spec}(A)} u v P_{ij}(u,v)##. This probability distribution is given by ##P_{ij}(u,v) = \left<\pi_j(u)\pi_i(v)\Psi,\pi_j(u)\pi_i(v)\Psi\right>##, where the ##\pi_i(u)## are the corresponding projectors, but it is only a legitimate probability distribution that obeys Kolmogorov's axioms if the histories ##\pi_j(u)\odot \pi_i(v)## are consistent, e.g. if some decoherence has happened. In that case, we have ##C_{ij} = \sum_{u,v\in\mathrm{spec}(A)} u v\left<\pi_j(u)\pi_i(v)\Psi,\pi_j(u)\pi_i(v)\Psi\right>##, but since different sets of consistent histories are still incompatible in general, these correlators will still violate the CHSH inequality in general.
If ##A_i## and ##A_j## commute, this correlator reduces to the usual expression:
$$C_{ij} = \sum_{u,v\in\mathrm{spec}(A)} u v\left<\pi_j(u)\pi_i(v)\Psi,\pi_j(u)\pi_i(v)\Psi\right>$$
$$= \sum_{u,v\in\mathrm{spec}(A)} u v\left<\Psi,\pi_i(u)\pi_j(v)\pi_j(u)\pi_i(v)\Psi\right> = \sum_{u,v\in\mathrm{spec}(A)} u v\left<\Psi,\pi_j(u)^2\pi_i(v)^2\Psi\right>$$
$$= \sum_{u,v\in\mathrm{spec}(A)} u v\left<\Psi,\pi_j(u)\pi_i(v)\Psi\right>= \left<\Psi,\left(\sum_u u\pi_j(u)\right)\left(\sum_v v\pi_i(v)\right)\Psi\right>$$
$$= \left<\Psi,A_jA_i\Psi\right> = \left<A_j A_i\right>$$