--- 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.