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Diffstat (limited to 'source/know/concept/bells-theorem')
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1 files changed, 29 insertions, 237 deletions
diff --git a/source/know/concept/bells-theorem/index.md b/source/know/concept/bells-theorem/index.md index a01bf9e..1589a7a 100644 --- a/source/know/concept/bells-theorem/index.md +++ b/source/know/concept/bells-theorem/index.md @@ -17,13 +17,13 @@ Suppose that we have two spin-1/2 particles, called $$A$$ and $$B$$, in an entangled [Bell state](/know/concept/bell-state/): $$\begin{aligned} - \Ket{\Psi^{-}} - = \frac{1}{\sqrt{2}} \Big( \Ket{\uparrow \downarrow} - \Ket{\downarrow \uparrow} \Big) + \ket{\Psi^{-}} + = \frac{1}{\sqrt{2}} \Big( \ket{\uparrow \downarrow} - \ket{\downarrow \uparrow} \Big) \end{aligned}$$ 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 $$\Ket{\downarrow}$$. +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 $$B$$. @@ -69,21 +69,29 @@ $$\begin{aligned} \end{aligned}$$ 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: +Note that we multiply $$A$$ and $$B$$ at the same $$\lambda$$-value, +hence it is a *local* hidden variable: $$\begin{aligned} - \Expval{A_a B_b} - = \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda} + \expval{A_a B_b} + \equiv \int \rho(\lambda) \: A(\vec{a}, \lambda) \: B(\vec{b}, \lambda) \dd{\lambda} \end{aligned}$$ -From this, two inequalities can be derived, -which both prove Bell's theorem. +From this, we can make several predictions about LHV theories, +which turn out to disagree with various theoretical +and experimental results in quantum mechanics. +The two most famous LHV predictions are +the **Bell inequality** and +the [CHSH inequality](/know/concept/chsh-inequality/). + ## Bell inequality -If $$\vec{a} = \vec{b}$$, then we know that $$A$$ and $$B$$ always have opposite spins: +We present Bell's original proof of his theorem. +If $$\vec{a} = \vec{b}$$, then we know that +measuring $$A$$ and $$B$$ gives them opposite spins, +because they start in the entangled state $$\ket{\Psi^{-}}$$: $$\begin{aligned} A(\vec{a}, \lambda) @@ -94,7 +102,7 @@ $$\begin{aligned} The expectation value of the product can therefore be rewritten as follows: $$\begin{aligned} - \Expval{A_a B_b} + \expval{A_a B_b} = - \int \rho(\lambda) \: A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} \end{aligned}$$ @@ -102,7 +110,7 @@ Next, we introduce an arbitrary third direction $$\vec{c}$$, and use the fact that $$( A(\vec{b}, \lambda) )^2 = 1$$: $$\begin{aligned} - \Expval{A_a B_b} - \Expval{A_a B_c} + \expval{A_a B_b} - \expval{A_a B_c} &= - \int \rho(\lambda) \Big( A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) - A(\vec{a}, \lambda) \: A(\vec{c}, \lambda) \Big) \dd{\lambda} \\ &= - \int \rho(\lambda) \Big( 1 - A(\vec{b}, \lambda) \: A(\vec{c}, \lambda) \Big) A(\vec{a}, \lambda) \: A(\vec{b}, \lambda) \dd{\lambda} @@ -114,7 +122,7 @@ Taking the absolute value of the whole left, and of the integrand on the right, we thus get: $$\begin{aligned} - \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big| + \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} \\ @@ -122,24 +130,24 @@ $$\begin{aligned} \end{aligned}$$ Since $$\rho(\lambda)$$ is a normalized probability density function, -we arrive at the **Bell inequality**: +we arrive at the Bell inequality: $$\begin{aligned} \boxed{ - \Big| \Expval{A_a B_b} - \Expval{A_a B_c} \Big| - \le 1 + \Expval{A_b B_c} + \Big| \expval{A_a B_b} - \expval{A_a B_c} \Big| + \le 1 + \expval{A_b B_c} } \end{aligned}$$ Any theory involving an LHV $$\lambda$$ must obey this inequality. -The problem, however, is that quantum mechanics dictates the expectation values -for the state $$\Ket{\Psi^{-}}$$: +The problem, however, is that quantum mechanics dictates +the expectation values for the state $$\ket{\Psi^{-}}$$: $$\begin{aligned} - \Expval{A_a B_b} = - \vec{a} \cdot \vec{b} + \expval{A_a B_b} = - \vec{a} \cdot \vec{b} \end{aligned}$$ -Finding directions which violate the Bell inequality is easy: +Finding directions that 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 $$0.293$$, @@ -147,222 +155,6 @@ 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, |