Zafa Pi said:
So deterministic theories and CFD theories are the same for you? They are for me. Can't QM predict the perfect correlation?
I wish that people hadn't brought up CFD with respect to QM, because it seems like a complicated concept that doesn't clarify anything. But it seems to me that CFD is only with respect to certain types of changes in the actual history of the world: If I had taken a left turn at Albuquerque I would have ended up at El Paso. We imagine an alternate world that is exactly like ours, except that I made a different choice. So CFD only makes sense relative to certain localized turning points in the history of the world. To me, that implies that there are two different ways CFD might fail:
- There might be no such turning points. In a superdeterministic theory, it doesn't make sense to ask what would have happened if Alice had done something different than she did, because Alice's choice was set in stone long ago, and there are no alternative histories.
- There might be no unique answer to the question: What would have happened? In a nondeterministic theory, the outcomes of experiments are not determined, and so there is no answer to a question such as: What result would Alice have gotten if she had measured X?
CFD is not exactly the same as determinism, though. Superdeterministic theories don't obey CFD, because there are no alternatives to ask about. And a nondeterministic theory might obey CFD for certain turning points and certain consequences, even if they don't for all pairs of turning-point/consequence. For example, if I flip a coin to decide whether to go left or right, and I go right, there might be a counterfactual answer to the question: What would have happened to me if the coin had landed tails?
The 1964 Bell proof he assumed determinism as you said (hidden variables and CFD). Can you provide a Bell type theorem with proof that doesn't assume determinism?
Bell's inequality is provable without assuming determinism. The usual proof (assuming determinism) goes like this:
Assume that there are functions A(\alpha, \lambda) and B(\beta, \lambda) giving Alice's and Bob's results, respectively as a function of Alice's setting, \alpha, Bob's setting, \beta and the hidden variable \lambda. This framing assumes that the results are deterministic functions of \alpha, \beta, \lambda. That's not necessary to get the inequality.
Instead, let's assume that Alice's result is only probabilistically related to her setting and the hidden variable, and similarly for Bob's result. There are two possible results, which we'll say are +1 and -1. The quantity we're interested in is the correlation, E(\alpha, \beta), which is the average value, over many trials, of the product of their results, for fixed \alpha and \beta. We can compute this in the following way:
E(\alpha, \beta) = P_{++}(\alpha, \beta) - P_{+-}(\alpha, \beta) - P_{-+}(\alpha, \beta) + P_{--}(\alpha, \beta)
where P_{++}(\alpha, \beta) is the probability Alice gets +1 and Bob gets +1, P_{+-}(\alpha, \beta) is the probability Alice gets +1 and Bob gets -1, etc.
Now, a local explanation of the correlation in terms of hidden variables would involve the following three functions:
- P(\lambda): the probability density for \lambda
- P_A(\alpha, \lambda): the probability that Alice gets +1 (the probability that she gets -1 is just 1 - P_A(\alpha, \lambda))
- P_B(\beta, \lambda): the probability that Bob gets +1 (the probability that he gets -1 is just 1 - P_B(\beta, \lambda))
Determinism would imply that P_A and P_B are either 0 or 1 for each choice of \alpha, \beta, \lambda. But we're not assuming determinism. Then in terms of these functions:
- P_{++}(\alpha, \beta) = \int d\lambda P(\lambda) P_A(\alpha, \lambda) P_B(\beta, \lambda)
- P_{+-}(\alpha, \beta) = \int d\lambda P(\lambda) P_A(\alpha, \lambda) (1 - P_B(\beta, \lambda))
- P_{-+}(\alpha, \beta) = \int d\lambda P(\lambda) (1 - P_A(\alpha, \lambda)) P_B(\beta, \lambda)
- P_{--}(\alpha, \beta) = \int d\lambda P(\lambda) (1 - P_A(\alpha, \lambda)) (1 - P_B(\beta, \lambda))
So:
E(\alpha, \beta) = \int d\lambda P(\lambda) (4 P_A(\alpha, \lambda) P_B(\beta, \lambda) - 2 P_A(\alpha, \lambda) - 2 P_B(\beta, \lambda) + 1)
Now the trick is to rewrite P_A(\alpha, \lambda) = \frac{1}{2}(X_A(\alpha, \lambda) + 1), P_B(\beta, \lambda) = \frac{1}{2}(X_B(\beta, \lambda) + 1), where X_A(\alpha, \lambda) and X_B(\beta, \lambda) are both numbers between +1 and -1. In terms of these,
E(\alpha, \beta) = \int d\lambda P(\lambda) X_A(\alpha, \lambda) X_B(\beta, \lambda)
Now, following Bell (or maybe CHSH, who developed a related inequality) we compute the combination:
C(\alpha, \alpha', \beta, \beta') \equiv E(\alpha, \beta) + E(\alpha', \beta) + E(\alpha', \beta) - E(\alpha', \beta')
In terms of our hidden variable, \lambda, this becomes:
C(\alpha, \alpha', \beta, \beta') = \int d\lambda P(\lambda) C_\lambda(\alpha, \alpha', \beta, \beta')
where
C_\lambda(\alpha, \alpha', \beta, \beta') \equiv X_A(\alpha, \lambda) X_B(\beta, \lambda) + X_A(\alpha', \lambda) X_B(\beta, \lambda) + X_A(\alpha, \lambda) X_B(\beta', \lambda) - X_A(\alpha', \lambda) X_B(\beta', \lambda)
It might not be obvious, but since each of the Xs are between +1 and -1, it follows that C_\lambda(\alpha, \alpha', \beta, \beta') must lie in the range [-2, +2]. So we have:
\int d\lambda P(\lambda) (-2) \leq C(\alpha, \alpha', \beta, \beta') \leq \int d\lambda P(\lambda) (+2)
And since the integral over all \lambda of P(\lambda) is just 1, we conclude:
-2 \leq C(\alpha, \alpha', \beta, \beta') \leq +2
So that's the same inequality as CHSH derived (a slight variant of Bell's original inequality), and it did not assume that the results were deterministic.
I have several in mind that have coin flipping involved, but I would still call them deterministic. Maybe they're not??
As I said, we can't know whether coin flips are
really deterministic, or not. We can only say that they are deterministic or nondeterministic according to this or that model.