Bell's inequality for angles 120°

Question

In 1964, John Bell first derived the original Bell inequality, $|E(a,b)-E(a,c)|\leq1+E(b,c)$. Here $a,b,c$ are three different possible spin measurement directions, and $E$ is the measured correlation, meaning the expectation value of the product of the signs of the two outcomes, given the specified measurement settings. (For example : there are two detectors A and B, and spin is measured in units of $\hbar/2$,now if for a particular measurement A measures +1 and B measures -1 then their product is -1, if we consider many runs of the experiment and caculate the average of the product for directions $a$ and $b$ which is $P(a,b)$, then quantum mechanics predict $P(a,b)= -a.b$.)

From this form of Bell's inequality, consider starting with a singlet spin state, and defining these three measurement directions such that the angle between each of the vectors of spin measurements are 120°. Then, I find that the correlations are $E(a,b)=-a.b= 1/2=E(a,c)=E(b,c)$. Thus $|\frac{1}{2}-\frac{1}{2}|\leq 1+\frac{1}{2}$ means $0\leq3/2$ which satisfies Bell's inequality, instead of violating it.

And yet this example is often used to show a violation of Bell's inequality. In the linked article, the calculation agrees with mine, showing that for any two different measurements, the outcomes have the same sign 75% of the time and a different sign 25% of the time. So the correlation functions $E(a,b)$, etc., are all indeed 0.5. Please help explain in what way this system violates the Bell inequality.

Answer 1

Bell's inequality is not necessarily violated for all possible angles between measurement directions - it may be that measurements at some angles meet Bell's inequality, as in the example you have given.

However, we know that Bell's inequality is violated for some angles between measurements. A common example is when the measurements are at $0$, $45$ and $90$ degrees - see this Wikipedia article.

Your confusion may arise from the fact that some explanations of Bell's theorem use a set up of detectors in which $P(a,b) = \cos^2 (\frac {\theta} 2)$; this set up is different from that in Bell's original paper, where $P(a,b) = - \cos \theta$.

Answer 2

The correct answer, already given in a comment, isn't that there is a factor of two in the angles; rather, it's the fact that the quanta magazine used a different inequality than the one originally formulated by Bell. More specifically, they used an inequality due to Mermin, which you can also read more about here.

In Mermin's scenario, we have three random variables for Alice and Bob each: $A_j$, $B_k$, and $j,k = 1,2,3$. These variables correspond to the different settings for Alice and Bob, and take on the values $\pm 1$. We are interested in how often Alice and Bob give the same answer, which is given by the product of their random variables. The sum of these products is lower bounded:

$$ \left|\sum_{jk}A_jB_k\right| \geq 1 $$

since there are nine terms in the sum, and the smallest value is when four (five) are +1 (-1) or vice versa. Let's look at the case where

$$ \sum_{jk}A_jB_k \leq -1. $$

Let the joint expectation values be $E(A_j,B_k) = E(A_j B_k)$. Then

$$ E(\sum_{jk}A_jB_k) = \sum_{j}E(A_j,B_j) + \sum_{j\neq k} E(A_j,B_k) \leq -1. $$

If we require anti-correlated outcomes when Alice and Bob choose the same setting ($E(A_j,B_j) = -1$) this can be written

$$ \sum_{j\neq k} E(A_j,B_k) \leq 2 $$

If you do the same for the other sign in the absolute value you get $\sum_{j\neq k} E(A_j,B_k) \geq 4$. Taking the settings $A_i = B_i = \{0^{\circ},120^{\circ},240^{\circ}\}$ we see that the quantum prediction violates the inequality

$$ 4 \nleq \sum_{j\neq k} E(A_j,B_k) = 6\cdot\frac{1}{2} = 3 \nleq 2. $$

If you want to write the inequality on a form even more similar to the original Bell inequality you can use $E(a,b) = E(b,a)$ (this follows from the requirement of anti-correlation for same settings) to simplify it to

$$ E(a,b) + E(a,c) + E(b,c) \leq 1 \ \text{or}\ 2\leq E(a,b) + E(a,c) + E(b,c) \\ \iff |E(a,b) + E(a,c) + E(b,c) - \frac{3}{2}| \geq \frac{1}{2}. $$

Note that neither this form of the inequality, nor Bell's original one, are experimentally testable since they use perfect anti-correlation for equal settings as part of the derivation.

Edit: I actually don't think the bound $E(a,b) + E(a,c) + E(b,c) \geq 2$ is necessary. It's easy to show that there is a strategy that satisfies $E(a,b) + E(a,c) + E(b,c) = 1$, and if there was one that satisfied $E(a,b) + E(a,c) + E(b,c) = 2$ a convex combination of them could violate the inequality. Maybe someone can show this in a more direct way.

Answer 3

This experiment you describe can indeed be used to show that something non-classical is going on. However, as you noticed, you can't see that non-classicality by looking at Bell's original inequality from his 1964 paper. (Note that this is an inequality which is barely used anymore; these days the gold standard is the CHSH inequality.) True, unlike CHSH, that original inequality had three settings, as does your example. But the non-classicality of your example isn't apparent until you also analyze the additional three (perfect) anti-correlations $E(a,a)=-1$, $E(b,b)=-1$ and $E(c,c)=-1$. With these perfect anti-correlations in the mix, you can indeed show that there is no classical account of these experiments.

If you're asking about the relevant inequality which these correlations violate, the original point of this very example was to avoid that sort of technical argument. Instead, Mermin wanted to show that the assumptions behind Bell's original Inequality (and the CHSH, and others) must be violated without using inequalities. The argument simply shows that there is no classical explanation for the experiment. In other words, given the usual assumptions behind Bell's Theorem, there is simply no set of hidden states which will allow both the correlations you calculated as well as the perfect anti-correlations $E=-1$ for the same settings.

EDIT: The wrong paper was cited before; this set-up was from Mermin, and indeed the experiment has its own wikipedia page.