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+---
+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.