Categories: Physics, Quantum information, Quantum mechanics.

# CHSH inequality

The Clauser-Horne-Shimony-Holt (CHSH) inequality is an alternative proof of Bell’s 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, 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.