path: root/source/know/concept/bells-theorem/index.md

---
title: "Bell's theorem"
date: 2021-03-28
categories:
- Physics
- Quantum mechanics
- Quantum information
layout: "concept"
---

**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](/know/concept/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 only multiply $A$ and $B$ for shared $\lambda$-values:
this is what makes it a **local** hidden variable:

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

From this, two inequalities can be derived,
which both prove Bell's theorem.


## Bell inequality

If $\vec{a} = \vec{b}$, then we know that $A$ and $B$ always have opposite spins:

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

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 which 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.


## CHSH inequality

The **Clauser-Horne-Shimony-Holt** or simply **CHSH inequality**
takes a slightly different approach, and is more useful in practice.

Consider four spin directions, two for $A$ called $\vec{a}_1$ and $\vec{a}_2$,
and two for $B$ called $\vec{b}_1$ and $\vec{b}_2$.
Let us introduce the following abbreviations:

$$\begin{aligned}
    A_1 &= A(\vec{a}_1, \lambda)
    \qquad \quad
    A_2 = A(\vec{a}_2, \lambda)
    \\
    B_1 &= B(\vec{b}_1, \lambda)
    \qquad \quad
    B_2 = B(\vec{b}_2, \lambda)
\end{aligned}$$

From the definition of the expectation value,
we know that the difference is given by:

$$\begin{aligned}
    \Expval{A_1 B_1} - \Expval{A_1 B_2}
    = \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \Big) \dd{\lambda}
\end{aligned}$$

We introduce some new terms and rearrange the resulting expression:

$$\begin{aligned}
    \Expval{A_1 B_1} - \Expval{A_1 B_2}
    &= \int \rho(\lambda) \Big( A_1 B_1 - A_1 B_2 \pm A_1 B_1 A_2 B_2 \mp A_1 B_1 A_2 B_2 \Big) \dd{\lambda}
    \\
    &= \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
    - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
\end{aligned}$$

Taking the absolute value of both sides
and invoking the triangle inequality then yields:

$$\begin{aligned}
    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
    &= \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
    - \!\int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
    \\
    &\le \bigg|\! \int \rho(\lambda) A_1 B_1 \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda} \!\bigg|
    + \bigg|\! \int \rho(\lambda) A_1 B_2 \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda} \!\bigg|
\end{aligned}$$

Using the fact that the product of $A$ and $B$ is always either $-1$ or $+1$,
we can reduce this to:

$$\begin{aligned}
    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
    &\le \int \rho(\lambda) \Big| A_1 B_1 \Big| \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
    + \!\int \rho(\lambda) \Big| A_1 B_2 \Big| \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
    \\
    &\le \int \rho(\lambda) \Big( 1 \pm A_2 B_2 \Big) \dd{\lambda}
    + \!\int \rho(\lambda) \Big( 1 \pm A_2 B_1 \Big) \dd{\lambda}
\end{aligned}$$

Evaluating these integrals gives us the following inequality,
which holds for both choices of $\pm$:

$$\begin{aligned}
    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
    &\le 2 \pm \Expval{A_2 B_2} \pm \Expval{A_2 B_1}
\end{aligned}$$

We should choose the signs such that the right-hand side is as small as possible, that is:

$$\begin{aligned}
    \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big|
    &\le 2 \pm \Big( \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big)
    \\
    &\le 2 - \Big| \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
\end{aligned}$$

Rearranging this and once again using the triangle inequality,
we get the CHSH inequality:

$$\begin{aligned}
    2
    &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} \Big| + \Big| \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
    \\
    &\ge \Big| \Expval{A_1 B_1} - \Expval{A_1 B_2} + \Expval{A_2 B_2} + \Expval{A_2 B_1} \Big|
\end{aligned}$$

The quantity on the right-hand side is sometimes called the **CHSH quantity** $S$,
and measures the correlation between the spins of $A$ and $B$:

$$\begin{aligned}
    \boxed{
        S \equiv \Expval{A_2 B_1} + \Expval{A_2 B_2} + \Expval{A_1 B_1} - \Expval{A_1 B_2}
    }
\end{aligned}$$

The CHSH inequality places an upper bound on the magnitude of $S$
for LHV-based theories:

$$\begin{aligned}
    \boxed{
        |S| \le 2
    }
\end{aligned}$$


## Tsirelson's bound

Quantum physics can violate the CHSH inequality, but by how much?
Consider the following two-particle operator,
whose expectation value is the CHSH quantity, i.e. $S = \expval{\hat{S}}$:

$$\begin{aligned}
    \hat{S}
    = \hat{A}_2 \otimes \hat{B}_1 + \hat{A}_2 \otimes \hat{B}_2 + \hat{A}_1 \otimes \hat{B}_1 - \hat{A}_1 \otimes \hat{B}_2
\end{aligned}$$

Where $\otimes$ is the tensor product,
and e.g. $\hat{A}_1$ is the Pauli matrix for the $\vec{a}_1$-direction.
The square of this operator is then given by:

$$\begin{aligned}
    \hat{S}^2
    = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_1 \hat{B}_2
    + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1^2 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_1 \hat{B}_2
    \\
    + &\hat{A}_2^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_2^2 \otimes \hat{B}_2^2
    + \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_2 \hat{A}_1 \otimes \hat{B}_2^2
    \\
    + &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_1^2 + \hat{A}_1 \hat{A}_2 \otimes \hat{B}_1 \hat{B}_2
    + \hat{A}_1^2 \otimes \hat{B}_1^2 - \hat{A}_1^2 \otimes \hat{B}_1 \hat{B}_2
    \\
    - &\hat{A}_1 \hat{A}_2 \otimes \hat{B}_2 \hat{B}_1 - \hat{A}_1 \hat{A}_2 \otimes \hat{B}_2^2
    - \hat{A}_1^2 \otimes \hat{B}_2 \hat{B}_1 + \hat{A}_1^2 \otimes \hat{B}_2^2
    \\
    = \quad &\hat{A}_2^2 \otimes \hat{B}_1^2 + \hat{A}_2^2 \otimes \hat{B}_2^2 + \hat{A}_1^2 \otimes \hat{B}_1^2 + \hat{A}_1^2 \otimes \hat{B}_2^2
    \\
    + &\hat{A}_2^2 \otimes \acomm{\hat{B}_1}{\hat{B}_2} - \hat{A}_1^2 \otimes \acomm{\hat{B}_1}{\hat{B}_2}
    + \acomm{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_1^2 - \acomm{\hat{A}_1}{\hat{A}_2} \otimes \hat{B}_2^2
    \\
    + &\hat{A}_1 \hat{A}_2 \otimes \comm{\hat{B}_1}{\hat{B}_2} - \hat{A}_2 \hat{A}_1 \otimes \comm{\hat{B}_1}{\hat{B}_2}
\end{aligned}$$

Spin operators are unitary, so their square is the identity,
e.g. $\hat{A}_1^2 = \hat{I}$. Therefore $\hat{S}^2$ reduces to:

$$\begin{aligned}
    \hat{S}^2
    &= 4 \: (\hat{I} \otimes \hat{I}) + \comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}
\end{aligned}$$

The *norm* $\norm{\hat{S}^2}$ of this operator
is the largest possible expectation value $\expval{\hat{S}^2}$,
which is the same as its largest eigenvalue.
It is given by:

$$\begin{aligned}
    \Norm{\hat{S}^2}
    &= 4 + \Norm{\comm{\hat{A}_1}{\hat{A}_2} \otimes \comm{\hat{B}_1}{\hat{B}_2}}
    \\
    &\le 4 + \Norm{\comm{\hat{A}_1}{\hat{A}_2}} \Norm{\comm{\hat{B}_1}{\hat{B}_2}}
\end{aligned}$$

We find a bound for the norm of the commutators by using the triangle inequality, such that:

$$\begin{aligned}
    \Norm{\comm{\hat{A}_1}{\hat{A}_2}}
    = \Norm{\hat{A}_1 \hat{A}_2 - \hat{A}_2 \hat{A}_1}
    \le \Norm{\hat{A}_1 \hat{A}_2} + \Norm{\hat{A}_2 \hat{A}_1}
    \le 2 \Norm{\hat{A}_1 \hat{A}_2}
    \le 2
\end{aligned}$$

And $\norm{\comm{\hat{B}_1}{\hat{B}_2}} \le 2$ for the same reason.
The norm is the largest eigenvalue, therefore:

$$\begin{aligned}
    \Norm{\hat{S}^2}
    \le 4 + 2 \cdot 2
    = 8
    \quad \implies \quad
    \Norm{\hat{S}}
    \le \sqrt{8}
    = 2 \sqrt{2}
\end{aligned}$$

We thus arrive at **Tsirelson's bound**,
which states that quantum mechanics can violate
the CHSH inequality by a factor of $\sqrt{2}$:

$$\begin{aligned}
    \boxed{
        |S|
        \le 2 \sqrt{2}
    }
\end{aligned}$$

Importantly, this is a *tight* bound,
meaning that there exist certain spin measurement directions
for which Tsirelson's bound becomes an equality, for example:

$$\begin{aligned}
    \hat{A}_1 = \hat{\sigma}_z
    \qquad
    \hat{A}_2 = \hat{\sigma}_x
    \qquad
    \hat{B}_1 = \frac{\hat{\sigma}_z + \hat{\sigma}_x}{\sqrt{2}}
    \qquad
    \hat{B}_2 = \frac{\hat{\sigma}_z - \hat{\sigma}_x}{\sqrt{2}}
\end{aligned}$$

Using the fact that $\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$,
it can then be shown that $S = 2 \sqrt{2}$ in this case.



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