Categories: Physics, Quantum information, Quantum mechanics.

# Bell’s theorem

Bell’s theorem states that the laws of quantum mechanics cannot be explained by theories built on so-called local hidden variables (LHVs).

Suppose that we have two spin-1/2 particles, called $A$ and $B$, in an entangled Bell state:

\begin{aligned} \ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big( \ket{\uparrow \downarrow} - \ket{\downarrow \uparrow} \Big) \end{aligned}

Since they are entangled, if we measure the $z$-spin of particle $A$, and find e.g. $\ket{\uparrow}$, then particle $B$ immediately takes the opposite state $\ket{\downarrow}$. The point is that this collapse is instant, regardless of the distance between $A$ and $B$.

Einstein called this effect “action-at-a-distance”, and used it as evidence that quantum mechanics is an incomplete theory. He said that there must be some hidden variable $\lambda$ that determines the outcome of measurements of $A$ and $B$ from the moment the entangled pair is created. However, according to Bell’s theorem, he was wrong.

To prove this, let us assume that Einstein was right, and some $\lambda$, which we cannot understand, let alone calculate or measure, controls the results. We want to know the spins of the entangled pair along arbitrary directions $\vec{a}$ and $\vec{b}$, so the outcomes for particles $A$ and $B$ are:

\begin{aligned} A(\vec{a}, \lambda) = \pm 1 \qquad \quad B(\vec{b}, \lambda) = \pm 1 \end{aligned}

Where $\pm 1$ are the eigenvalues of the Pauli matrices in the chosen directions $\vec{a}$ and $\vec{b}$:

\begin{aligned} \hat{\sigma}_a &= \vec{a} \cdot \vec{\sigma} = a_x \hat{\sigma}_x + a_y \hat{\sigma}_y + a_z \hat{\sigma}_z \\ \hat{\sigma}_b &= \vec{b} \cdot \vec{\sigma} = b_x \hat{\sigma}_x + b_y \hat{\sigma}_y + b_z \hat{\sigma}_z \end{aligned}

Whether $\lambda$ is a scalar or a vector does not matter; we simply demand that it follows an unknown probability distribution $\rho(\lambda)$:

\begin{aligned} \int \rho(\lambda) \dd{\lambda} = 1 \qquad \quad \rho(\lambda) \ge 0 \end{aligned}

The product of the outcomes of $A$ and $B$ then has the following expectation value. Note that we multiply $A$ and $B$ at the same $\lambda$-value, hence it is a local hidden variable:

\begin{aligned} \expval{A_a B_b} \equiv \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda} \end{aligned}

From this, we can make several predictions about LHV theories, which turn out to disagree with various theoretical and experimental results in quantum mechanics. The two most famous LHV predictions are the Bell inequality and the CHSH inequality.

## Bell inequality

We present Bell’s original proof of his theorem. If $\vec{a} = \vec{b}$, then we know that measuring $A$ and $B$ gives them opposite spins, because they start in the entangled state $\ket{\Psi^{-}}$:

\begin{aligned} A(\vec{a}, \lambda) = A(\vec{b}, \lambda) = - B(\vec{b}, \lambda) \end{aligned}

The expectation value of the product can therefore be rewritten as follows:

\begin{aligned} \expval{A_a B_b} = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} \end{aligned}

Next, we introduce an arbitrary third direction $\vec{c}$, and use the fact that $( A(\vec{b}, \lambda) )^2 = 1$:

\begin{aligned} \expval{A_a B_b} - \expval{A_a B_c} &= - \int \rho(\lambda) \Big( A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) - A(\vec{a}, \lambda) \: A(\vec{c}, \lambda) \Big) \dd{\lambda} \\ &= - \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} \end{aligned}

Inside the integral, the only factors that can be negative are the last two, and their product is $\pm 1$. Taking the absolute value of the whole left, and of the integrand on the right, we thus get:

\begin{aligned} \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big| &\le \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) \: \Big| A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \Big| \dd{\lambda} \\ &\le \int \rho(\lambda) \dd{\lambda} - \int \rho(\lambda) A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \dd{\lambda} \end{aligned}

Since $\rho(\lambda)$ is a normalized probability density function, we arrive at the Bell inequality:

\begin{aligned} \boxed{ \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big| \le 1 + \expval{A_b B_c} } \end{aligned}

Any theory involving an LHV $\lambda$ must obey this inequality. The problem, however, is that quantum mechanics dictates the expectation values for the state $\ket{\Psi^{-}}$:

\begin{aligned} \expval{A_a B_b} = - \vec{a} \cdot \vec{b} \end{aligned}

Finding directions that violate the Bell inequality is easy: for example, if $\vec{a}$ and $\vec{b}$ are orthogonal, and $\vec{c}$ is at a $\pi/4$ angle to both of them, then the left becomes $0.707$ and the right $0.293$, which clearly disagrees with the inequality, meaning that LHVs are impossible.

1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.
2. J.B. Brask, Quantum information: lecture notes, 2021, unpublished.