How is superdeterminism a loophole to Bell's theorem?

Question

I read wiki about superdeterminism and Bell's theorem and the lecture on free will, and I struggle to understand what free will has to do with the Bell's theorem.

From my understanding, for Bell's inequality to be true, one requires local realism and that the hidden variables associated with Alice, Bob and detector ($a$, $b$ and $\lambda$) to be uncorrelated, from which one can derive the CHSH inequality, which is almost the same thing as Bell's inequality. Therefore, if the Bell's inequality is violated we have to either give up local realism or the absence of correlation between $a$, $b$ and $\lambda$. Gerard 't Hooft in his Cellular Automaton interpretation of quantum mechanics on p. 43 derives the conditional probability function for $\lambda$, given $a$ and $b$ (in certain interpretation of the variables): $$ P( \lambda|a, b)=\frac{1}{2}|\sin(4\lambda-2a-2b)|.\tag{1} $$
From the said above, it would seem, that whenever one wants to build a classical theory that matches quantum measurements, they would have to make sure the conditional probability (1) holds no matter what. To me, this is the gist of the problem, since the chaotic behavior of the external noise usually kills any correlations between spatially separated systems over time, given the noise actually influences the hidden variables. And if the noise does not influence $a$, $b$ and $\lambda$, then they are constants of motion and the outcomes of the experiment are going to be always the same! I cannot even think of what kind of system would have such bizarre property that the individual variables are influenced by noise, but the mutual correlation would not be destroyed.

Of course, one could hypothetically solve the equations of motion backwards and actually obtain the configuration of noise that would produce the desired correlation. But then, the predictive power of such theory would be close to zero, since even though the noise preserved the correlation till now (for which we performed the fitting), it is not guarantied to preserve it in the future.

Where does "free will" comes into play here? There seem to be no point where it would be invoked into the argument.

Answer 1

You never get to observe the probability distribution directly. You only make individual measurements and build up statistics by combining these measurements. If we assume that the measurements you make are independent trials, then sampling many times from the experiment will give you a sense of the underlying distribution.

But, if for some reason, the measurements were correlated, it could be that you are not really sampling from the "true" distribution. For example, if you are a puppet being controlled by some higher being engaging in a conspiracy, you could simply be carrying out a sequence of measurements that were carefully selected to give results that look as though they were random samples from the quantum mechanical probability distribution, but in fact were deterministic.

My personal feeling is that it is not worth thinking for very long about superdeterminism.

Answer 2

Mathematically, the only type of correlations that can be made in a local Einstein-Bell realistic world are of the form

$$C(\mathbf{a}, \mathbf{b}) = \int_{\lambda \in \Lambda} A(\mathbf{a}, \lambda) B(\mathbf{b}, \lambda)\ d\mu_\Lambda$$

where we integrate with respect to a probability measure $\mu_\Lambda$ on the state of objective system states $\Lambda$, where we are extracting the spin components in the directions of $\mathbf{a}$ and $\mathbf{b}$ respectively for the two particles. However, to match the predictions of quantum theory, we must have

$$C(\mathbf{a}, \mathbf{b}) = -\mathbf{a} \cdot \mathbf{b}$$

This cannot be written as an integral of the form given, because a product of two scalar functions of single vectors cannot equal a single scalar function of two vectors generally. Namely, $\mathbf{a} \cdot \mathbf{b}$ does not factor as

$$\mathbf{a} \cdot \mathbf{b} = f(\mathbf{a})\ g(\mathbf{b})$$

for individual functions $f: \mathrm{Vec}(\mathbb{R}, 3) \rightarrow \mathbb{R}$ and $g: \mathrm{Vec}(\mathbb{R}, 3) \rightarrow \mathbb{R}$. You need to mix vector components together in a way that this type of multiplication cannot do.

But this only is a problem so long as there's no mixing factor in the integral: One way to have one is to have $A$ and $B$, the probabilities that measurements at the two different particles will yield a result given their objective states, depend on both $\mathbf{a}$ and $\mathbf{b}$. This is non-local realism, because now the local measurement represented by $A$ must somehow yank distant information from the one represented by $B$, and conversely.

The other way, though, is to say that $\lambda$ is correlated to $\mathbf{a}$ and $\mathbf{b}$ themselves. Either this means merely setting the instruments alone changes the particles, or that the measurement settings themselves are somehow influenced, whether by the particles or by some other event in the past. This last approach is the superdeterministic approach.

Answer 3

I've read this article by Sabine Hossenfleder (review by Gerald Hooft!), and it is a nice recent review of the Bell's theorem, superdeterminism and arguments for it and against it, as well as examples of some superdeterministic models. The "free will" was indeed claimed to be "red herring", and it was effectively stated that the whole business of superdeterminism is about how to make the hidden variable of the source $\lambda$ dependent on the on the hidden variables of Alice and Bob $a$ and $b$ (or the other way around).

Unfortunately, the problem of the hidden variables being influenced by noise but the correlations not being influenced by noise was not raised at all. I am not sure if it is because there is no good response, or because the author is not aware about the problem

Answer 4

When you talk about “noise” in the superdeterminists view (at least in my view) you are talking about something with a very large but finite number of external factors. A random number generator is never truly random, just meant to capture as much entropy from its local system as possible. It is impossible to design something that does anything random, no matter how much noise you throw at it.

The word “random” is the true conundrum to me, and the implication is that “spontaneous” processes are where the truly interesting problems are found. Why does an electron spontaneously emit a photon when it chooses to? What is alpha decay? I’ve heard people explain it as caused by quantum tunneling and the inherent randomness of the cosmos, but that doesn’t make any sense to me. There must be some tiny machine in there, not dice. Those are the hidden variables we should be looking for. Spontaneous emission seems the most approachable thing to study.

And I personally don’t understand why one dismisses this as undermining science or the independence of the observer. Perhaps it cannot be known, but I am convinced that it is a mechanism fundamental to the universe. If there were 2 identical universes that electron would emit that photon at the exact same instant.

Since nothing can be truly random, everything in the universe is necessarily correlated.