# The H2 Wiki

## Introduction

Bell’s theorem is a mathematical result which proves that quantum physics is incompatible with what are called “local hidden-variable theories”. John Stewart Bell came up with a mathematical inequality that “local hidden-variable theories” must satisfy, yet it was subsequently demonstrated by experiment that quantum systems can violate the inequality. One philosophical interpretation is that the physical world is non-local: actions in one part of the physical universe can instantaneously affect other parts of the universe. Einstein called this “spooky action at a distance”.

A first undergraduate course in quantum mechanics is enough to understand Bell’s theorem but nonetheless I found the standard treatments completely incomprehensible, not mathematically incomprehensible but incomprehensible in their meaning. I struggled to internalise the meaning of the inequality or why it should be surprising that it could be violated. This article presents the treatment that I find most accessible. I hope you find it as fascinating as I do!

## The CHSH inequality

We will look at the CHSH inequality because it is slightly more general than Bell’s original inequality. It states that if there are physical observables $$a$$, $$b$$, $$a'$$ and $$b'$$, each of which can take on only values $$-1$$ or $$+1$$, then $|E(a, b) - E(a, b’) + E(a’, b) + E(a’, b’)| \leq 2$ where each $$E(x, y)$$ is called the “quantum correlation” between an observation $$x$$ and an observation $$y$$. “Quantum correlation” sounds mysterious but it’s just the expectation of the product of $$x$$ and $$y$$: $E(x, y) = \mathbb{E}(xy)$

It is of little benefit that the definition of “quantum correlation” is technically simple. I still have no intuition about it! The CHSH inequality is equally mysterious. I have no intuition about whether it is something that could be violated or not. I have much more intuition about probabilities than “quantum correlations”. Luckily the “quantum correlation” can be rewritten simply in terms of probabilities. After doing so the CHSH inequality will become very natural! We start by observing that, because $$x$$ and $$y$$ can take on only values $$+1$$ and $$-1$$, $E(x, y) = \mathbb{E}(xy) = \mathbb{P}(x = y) - \mathbb{P}(x \ne y) = 1 - 2\mathbb{P}(x \ne y)$ so we can rewrite the CHSH inequality as $|1 - 2\mathbb{P}(a \ne b) - 1 + 2\mathbb{P}(a \ne b’) + 1 - 2\mathbb{P}(a’ \ne b) + 1 - 2\mathbb{P}(a’ \ne b’)| \leq 2$ and after some algebra we conclude that $\mathbb{P}(a \ne b’) \leq \mathbb{P}(a \ne b) + \mathbb{P}(a’ \ne b) + \mathbb{P}(a’ \ne b’)$

Now this I can understand! If $$a \ne b'$$ then at least one of $$a \ne b$$, $$a' \ne b$$ and $$a' \ne b'$$ must hold. This is elementary probability theory of the most basic order. Of course it’s true!

Is this form of the inequality sufficient to show that the predictions of quantum mechanics differ from the predictions of basic probability theory? Remarkably yes. There are simple experiments on quantum systems that do not obey this inequality.

Specifically, we can perform the following. Write $\newcommand{\bra}{\left\langle{#1}\right|} \newcommand{\ket}{\left|{#1}\right\rangle} \newcommand{\halfpi}{\frac{\pi}{2}} \ket{\theta} = \left(\begin{array}{c} \cos\theta\\\sin\theta\ \end{array}\right)$ for the state of a photon polarised at angle $$\theta$$. A fundamental physical property of these states is that they satisfy the relationship $\left\langle \theta_1 | \theta_2 \right\rangle = \cos(\theta_1 - \theta_2)$ The self-adjoint operator $S_{\phi} = \ket{\phi}\bra{\phi} - \ket{\phi + \halfpi}\bra{\phi + \halfpi}$ represents a measurement of the polarisation at angle $$\phi$$. Specifically $S_\phi \ket{\phi} = \ket{\phi} \;\;\;\;\;\;\;\; S_\phi \ket{\phi + \halfpi} = -\ket{\phi + \halfpi}$ That is, the value of an observation is $$+1$$ if the photon is indeed polarised at angle $$\phi$$, and $$-1$$ if the photon is polarised at angle $$\phi + \pi/2$$. Now consider a pair of photons with entangled polarisation $\newcommand{\V}{0} \newcommand{\onesqrttwo}{\frac{1}{\sqrt{2}}} \newcommand{\H}{\halfpi} \psi = \onesqrttwo \left(\ket{\V}\ket{\V} +\ket{\H}\ket{\H}\right)$

Direct calculation shows that this state is, with respect to a general orthonormal basis, \begin{multline} \psi = \onesqrttwo\cos(\theta - \theta’) \left(\ket{\theta} \ket{\theta’} + \ket{\theta + \halfpi} \ket{\theta’ + \halfpi}\right) \\ + \onesqrttwo\sin(\theta - \theta’) \left(\ket{\theta} \ket{\theta’ + \halfpi} - \ket{\theta + \halfpi} \ket{\theta’}\right) \end{multline} Therefore if we were to measure the first photon with $$S_\theta$$ and the second photon with $$S_{\theta'}$$ then we would obtain the same outcome (i.e. both $$+1$$ or both $$-1$$) in each case with probability $$\cos^2(\theta-\theta')$$ and differing outcomes (i.e. one $$+1$$ and the other $$-1$$) with probability $$\sin^2(\theta-\theta')$$.

Thus if we choose

• $$a$$ to be the observation of $$S_0$$
• $$b$$ to be the observation of $$S_{\pi/8}$$
• $$a'$$ to be the observation of $$S_{\pi/4}$$
• $$b'$$ to be the observation of $$S_{3\pi/8}$$

then $\mathbb{P}(a \ne b’) = \sin^2(3\pi/8) = \frac{2 + \sqrt{2}}{4} > 3 / 4$ but $\mathbb{P}(a \ne b) + \mathbb{P}(a’ \ne b) + \mathbb{P}(a’ \ne b’) = 3\sin^2(\pi/8) = 3 \left(\frac{2 - \sqrt{2}}{4}\right) < 3 / 4$ so the CHSH inequality would be violated. Indeed this experiment has been performed in practice! There is very strong evidence that physical reality is non-local.