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diff --git a/source/know/concept/ghz-paradox/index.md b/source/know/concept/ghz-paradox/index.md index a59ccfe..9951883 100644 --- a/source/know/concept/ghz-paradox/index.md +++ b/source/know/concept/ghz-paradox/index.md @@ -12,43 +12,46 @@ layout: "concept" The **Greenberger-Horne-Zeilinger** or **GHZ paradox** is an alternative proof of [Bell's theorem](/know/concept/bells-theorem/) that does not use inequalities, -but the three-particle entangled **GHZ state** $$\Ket{\mathrm{GHZ}}$$ instead, +but the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$ instead, $$\begin{aligned} \boxed{ - \Ket{\mathrm{GHZ}} - = \frac{1}{\sqrt{2}} \Big( \Ket{000} + \Ket{111} \Big) + \ket{\mathrm{GHZ}} + = \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big) } \end{aligned}$$ -Where $$\Ket{0}$$ and $$\Ket{1}$$ are qubit states, -for example, the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$. +Where $$\ket{0}$$ and $$\ket{1}$$ are qubit states, +specifically the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$. If we now apply certain products of the Pauli matrices $$\hat{\sigma}_x$$ and $$\hat{\sigma}_y$$ -to the three particles, we find: +as [quantum gates](/know/concept/quantum-gate/) +to this three-particle state, we find: $$\begin{aligned} - \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \Ket{\mathrm{GHZ}} - &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} - + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \Big) + \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} + + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \Big) \\ - &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes \Ket{1} \otimes \Ket{1} + \Ket{0} \otimes \Ket{0} \otimes \Ket{0} \Big) - = \Ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes \ket{1} \otimes \ket{1} + \ket{0} \otimes \ket{0} \otimes \ket{0} \Big) \\ - \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \Ket{\mathrm{GHZ}} - &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_y \Ket{0} \otimes \hat{\sigma}_y \Ket{0} - + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \Big) + &= \ket{\mathrm{GHZ}} \\ - &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes i \Ket{1} \otimes i \Ket{1} + \Ket{0} \otimes i \Ket{0} \otimes i \Ket{0} \Big) - = - \Ket{\mathrm{GHZ}} + \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \ket{\mathrm{GHZ}} + &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_y \ket{0} \otimes \hat{\sigma}_y \ket{0} + + \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_y \ket{1} \otimes \hat{\sigma}_y \ket{1} \Big) + \\ + &= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes i \ket{1} \otimes i \ket{1} + \ket{0} \otimes i \ket{0} \otimes i \ket{0} \Big) + \\ + &= - \ket{\mathrm{GHZ}} \end{aligned}$$ In other words, the GHZ state is a simultaneous eigenstate of these composite operators, with eigenvalues $$+1$$ and $$-1$$, respectively. -Let us introduce two other product operators, -such that we have a set of four observables, -for which $$\Ket{\mathrm{GHZ}}$$ gives these eigenvalues: +Let us introduce two more operators in the same way, +so that we have a set of four observables, +for which $$\ket{\mathrm{GHZ}}$$ gives these eigenvalues: $$\begin{aligned} \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x |