DarMM said:
The only problem is that quantum mechanics involves non-classical correlations. That is correlations outside the polytope given by assuming that your variables all belong in a single sample space. You can show (Kochen-Specker, Colbeck-Renner, etc) that theories with correlations outside of this polytope by necessity lack a dynamical account for their outcomes or correlations.
I just thought I'd put an example of the proof of this here if people enjoy it. Consider ##X## and ##Z## polarization measurements on two particles. All measurements have outcomes ##\{0,1\}##. I'll call the observers ##A## and ##B##. Imagine we find they are correlated as follows:
| ##X_A## | ##Z_A## |
##X_B## | ##=## | ##=## |
##Z_B## | ##=## | ##\neq## |
i.e. if they both perform an ##X## measurement the results will be equal.
Now consider the chance that ##A## obtains ##0## when they measure ##X_A##:
$$p\left(0|X_A\right)$$
From no-signalling this doesn't depend on the ##B## measurement, so we'll just take it to be ##X_B##, then
$$p\left(0|X_A\right) = p\left(00|X_A X_B\right) + p\left(01|X_A X_B\right)$$
Of course the second term is zero so:
$$p\left(0|X_A\right) = p\left(00|X_A X_B\right)$$
Since this is purely based on the correlation array it doesn't matter if we include any other arbitrary collection of events ##e## that occurred prior to the measurements:
$$p\left(0|X_A , e\right) = p\left(00|X_A X_B , e\right)$$
If we then focus on the chance for an ##X_B## measurement to produce zero we get a similar result:
$$p\left(0|X_B , e\right) = p\left(00|X_A X_B , e\right)$$
And thus we have:
$$p\left(0|X_A , e\right) - p\left(0|X_B , e\right) = 0$$
Iterating through a few different combinations of measurements we get three more equations like this for other sets of outcomes, thus in total we have:
<br />
p\left(0|X_A , e\right) - p\left(0|X_B , e\right) = 0\\<br />
p\left(0|X_B , e\right) - p\left(0|Z_A , e\right) = 0\\<br />
p\left(0|Z_A , e\right) - p\left(0|Z_B , e\right) = 0\\<br />
p\left(0|Z_B , e\right) - p\left(1|X_A , e\right) = 0<br />
These cancel off against each other to give us:
$$p\left(0|X_A , e\right) - p\left(1|X_A , e\right) = 0$$
Since we have ##p\left(1|X_A , e\right) = 1 - p\left(0|X_A , e\right) ## this gives us:
$$p\left(0|X_A , e\right) = \frac{1}{2}$$
So the outcome of an ##X_A## measurement cannot be deterministic. With this you can show none of the other outcomes can be deterministic either.
The correlations I used here are supra-quantum, i.e. stronger than those in quantum mechanics. Ekert and Renner proved that the same holds true in QM (
https://www.nature.com/articles/nature13132?draft=journal, note they use information theoretic language so phrase it in terms of privacy).
The correlations are too strong for individual outcomes to be deterministic.
If you try the same with classical correlations the equations come out underdetermined thus the solutions have a free parameter ##\lambda## which can be adjusted to give deterministic solutions for the correlations.