Understanding Bell’s theorem part 4: the counterfactual version

I was recently leafing through the great book Quantum Computation and Quantum Information, and noticed that the version of Bell’s theorem it presents is not any of those I wrote about in my three posts about Bell’s theorem, but rather the one I like the least, the counterfactual definiteness version. Annoyed, I checked another great book, Quantum Theory: Concepts and Methods, and saw that it also uses this god-forsaken version1. Nevertheless, I decided that the world has bigger problems to deal with and set it aside. Until I dropped by the Quantum Information lecture given by my boss, David Gross, and saw that he also prefers this undignified version. That’s it! The rise of fascism can wait. I need to set the record straight on counterfactual definiteness.

Before I start ranting about what I find so objectionable about it, I’ll present the proof of this version of Bell’s theorem the best I can. So, what is counterfactual definiteness? It is the assumption that not only the measurement you did in fact do has a definite answer, but also the measurement you did not do has a definite answer. If feels a lot like determinism, but it is not really the same thing, as the assumption is silent about how the result of the counterfactual measurement is determined, it just says that it is. To be more clear, let’s take a look at the data that comes from a real Bell test, the Delft experiment:2

N $x$ $y$ $a$ $b$
1 0 0 1 1
2 0 0 0 0
3 1 1 1 0
4 1 1 0 1
5 0 0 1 1
6 1 1 1 0
7 0 0 1 0
8 1 0 1 1
9 0 0 1 1
10 0 1 0 0

The first column indicates the rounds of the experiment, the $x$ and $y$ columns indicate the settings of Alice and Bob, and the $a$ and $b$ columns the results of their measurements. If one assumes counterfactual definiteness, then definite results must also exist for the measurements that were not made, for example in the first round there must exist results corresponding to the setting $x=1$ for Alice and $y=1$ for Bob. This data would then be just part of some more complete data table, for example this:

N $a_0$ $a_1$ $b_0$ $b_1$
1 1 0 1 1
2 0 1 0 1
3 1 1 0 0
4 1 0 1 1
5 1 1 1 1
6 1 1 0 0
7 1 1 0 0
8 1 1 1 0
9 1 0 1 0
10 0 0 1 0

In this table the column $a_0$ has the results of Alice’s measurements when her setting is $x=0$, and so on. The real data points, corresponding to the Delft experiment, are in black, and I filled in red the hypothetical results for the measurements that were not made.

What is the problem with assuming counterfactual definiteness, then? A complete table certainly exists. But it makes it possible to do something that wasn’t before: we can evaluate the entire CHSH game in every single round, instead of having to choose a single pair of settings. As a quick reminder, to win the CHSH game Alice and Bob must give the same answers when their settings are $(0,0)$, $(0,1)$, or $(1,0)$, and give different answers when their setting is $(1,1)$. In other words, they must have $a_0=b_0$, $a_0=b_1$, $a_1=b_0$, and $a_1 \neq b_1$. But if you try to satisfy all these equations simultaneously, you get that $a_0=b_0=a_1 \neq b_1 = a_0$, a contradiction. At most, you can satisfy 3 out of the 4 equations3. Then since in every row the score in the CHSH game is at most $3/4$, if we sample randomly from each row a pair of $a_x,b_y$ we have that
\[ \frac14(p(a_0=b_0) + p(a_0=b_1) + p(a_1=b_0) + p(a_1\neq b_1)) \le \frac34,\]
which is the CHSH inequality.

But if you select the actual Delft data from each row, the score will be $0.9$. Contradiction? Well, no, because you didn’t sample randomly, but just chose $1$ out of $4^{10}$ possibilities, which would happen with probability $1/4^{10} \approx 10^{-6}$ if you actually did it randomly. One can indeed violate the CHSH inequality by luck, it is just astronomically unlikely.

Proof presented, so now ranting: what is wrong with this version of the theorem? It is just so lame! It doesn’t even explicitly deal with the issue of locality, which is fundamental in all other versions of the theorem4! The conclusion that one takes from it, according to Asher Peres himself, is that “Unperformed experiments have no results”. To which the man in the street could reply “Well, duh, of course unperformed experiments have no results, why are you wasting my time with this triviality?”. It leaves the reader with the impression that they only need to give up the notion that unperformed experiments have results, and they are from then on safe from Bell’s theorem. But this is not true at all! The other proofs of Bell’s theorem still hold, so you still need to give up either determinism or no action at a distance, if you consider the simple version, or unconditionally give up local causality, if you consider the nonlocal version, or choose between generalised local causality and living in a single world, if you consider the Many-Worlds version.

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