I see superdeterminism come up as one way that local realism and all that other good stuff can be preserved, but it also seems like it's a minority view. I've been trying to understand why, and really, what I need is just a clarification on what superdeterminism implies. Does it

a) only state that the universe is deterministic, with everything set in stone from the Big Bang onward, or

b) state that it's deterministic in what appears to be a very unlikely way, considering the results of Bell's Theorem experiments. Like, it'd have to be the equivalent of every time you reach into a scrabble bag, you happen to pull out "QUANTUM".

I don't understand the Bell's Theorem experiments well enough to ascertain whether this is what is claimed about superdeterminism. If it implies that things are pre-correlated in what seems to be a wildly unlikely manner, that's very different than if it just implies that things are pre-correlated, and I would understand the prevailing skepticism. Otherwise, not so much.

  • $\begingroup$ Superdeterminism is like saying the moon continuously and deterministically blinks out of existence, but the initial conditions of the universe have been set up so that this only happens exactly at the moments you're not looking at it. $\endgroup$
    – knzhou
    Jul 13 at 15:57
  • $\begingroup$ That is, superdeterminism is a much much much stronger assumption than mere determinism (though some disingenuous popularizers try to elide the difference), and stronger in a way that feels weird to almost all physicists. $\endgroup$
    – knzhou
    Jul 13 at 15:57
  • $\begingroup$ physics.stackexchange.com/questions/106725/… $\endgroup$
    – alanf
    Jul 13 at 17:30

The gist of nonlocality à la Bell is that there are conditional probability distributions $p(ab|xy)$, where $a,b$ and $x,y$ denote possible measurement outcomes and possible choice of measurements, respectively, which do not admit a "local realistic explanation", that is, a decomposition of the form $$p(ab|xy) = \sum_\lambda p(\lambda) p_\lambda(a|x)p_\lambda(b|y).\tag 1$$ Some quantum systems produce probability distributions which behave like this (i.e. cannot be decomposed like (1)), hence quantum mechanics not being local realistic etc.

One major assumption in (1) is that you still assume that the choices of measurement bases, $x,y$, are not correlated. This means that you assume that Alice and Bob "choose freely" how to interact and observe the systems they are given. "Superdeterminism" is what you get lifting this assumption. If you assume that Alice and Bob's choices of measurements were also determined beforehand (e.g. maybe they organised before moving apart and decided on what they'll measure), then the problem becomes trivial: any possible observation can be explained deterministically.

Let me stress that this is not even really about quantum mechanics. I'm saying that any possible observation (in the two-party scenario we consider here), quantum mechanical or not, can be explained deterministically with a "superdeterministic hidden variable".

The trivial "proof" of this is that, if you consider the measurement choices $x,y$ as also determined beforehand, you effectively remove the locality aspect from the equation. If the measurement choices are also correlated/chosen beforehand, then the fact that Alice and Bob are spatially separated is effectively vacuus. You just have a joint probability distribution $p(abxy)$, which you can always describe using some deterministic "classical" theory.

Of course, the downside of such a "superdeterministic explanation" is that it involves some sort of "cosmic conspiracy theory". Sure, it might just so happen that the weird correlations observed by Alice and Bob are actually due to the way they chose to measure their apparatuses being determined a priori together with some corresponding deterministic outcome, but how would that actually happen? If you were to devise a theory explaining that, it would be yes fully classical, but also likely quite weird and convoluted.


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