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diff --git a/source/know/concept/chsh-inequality/index.md b/source/know/concept/chsh-inequality/index.md new file mode 100644 index 0000000..984bae6 --- /dev/null +++ b/source/know/concept/chsh-inequality/index.md @@ -0,0 +1,273 @@ +--- +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. |