---
title: "CHSH inequality"
sort_title: "CHSH inequality"
date: 2023-02-05
categories:
- Physics
- Quantum mechanics
- Quantum information
layout: "concept"
---

The **Clauser-Horne-Shimony-Holt (CHSH) inequality**
is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/),
which takes a slightly different approach
and is more useful in practice.

Suppose there is a local hidden variable (LHV) $$\lambda$$
with an unknown probability density $$\rho$$:

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

Given two spin-1/2 particles $$A$$ and $$B$$,
measuring their spins along arbitrary directions $$\vec{a}$$ and $$\vec{b}$$
would give each an eigenvalue $$\pm 1$$. We write this as:

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

If $$A$$ and $$B$$ start in an entangled [Bell state](/know/concept/bell-state/),
e.g. $$\ket{\Psi^{-}}$$, then we expect a correlation between their measurements results.
The product of the outcomes of $$A$$ and $$B$$ is:

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

So far, we have taken the same path as for proving Bell's inequality,
but for the CHSH inequality we must now diverge.



## Deriving the inequality

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:

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

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 the spin eigenvalues of $$A$$ and $$B$$
is always either $$-1$$ or $$+1$$ for all directions,
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:

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

Fundamental quantum mechanics says that
$$\Expval{A_a B_b} = - \vec{a} \cdot \vec{b}$$,
so $$S = 2 \sqrt{2}$$ 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.