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+---
+title: "Bell's theorem"
+firstLetter: "B"
+publishDate: 2021-03-28
+categories:
+- Physics
+- Quantum mechanics
+- Quantum information
+
+date: 2021-03-28T14:41:32+02:00
+draft: false
+markup: pandoc
+---
+
+# 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](/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.