Categories: Physics, Quantum information, Quantum mechanics.

**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:

\[\begin{aligned} \ket{\Psi^{-}} = \frac{1}{\sqrt{2}} \Big( \ket{\uparrow \downarrow} - \ket{\downarrow \uparrow} \Big) \end{aligned}\]

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.

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.

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:

\[\begin{aligned} \boxed{ |S| \le 2 } \end{aligned}\]

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.

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*Quantum information: lecture notes*, 2021, unpublished.

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