--- title: "GHZ paradox" sort_title: "GHZ paradox" date: 2021-03-29 categories: - Physics - Quantum mechanics - Quantum information 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, $$\begin{aligned} \boxed{ \Ket{\mathrm{GHZ}} = \frac{1}{\sqrt{2}} \Big( \Ket{000} + \Ket{111} \Big) } \end{aligned}$$ Where $$\Ket{0}$$ and $$\Ket{1}$$ are qubit states, for example, the eigenvalues of the Pauli matrix $$\hat{\sigma}_z$$. If we now apply certain products of the Pauli matrices $$\hat{\sigma}_x$$ and $$\hat{\sigma}_y$$ to the three particles, we find: $$\begin{aligned} \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \Ket{\mathrm{GHZ}} &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_x \Ket{0} + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_x \Ket{1} \Big) \\ &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes \Ket{1} \otimes \Ket{1} + \Ket{0} \otimes \Ket{0} \otimes \Ket{0} \Big) = \Ket{\mathrm{GHZ}} \\ \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \Ket{\mathrm{GHZ}} &= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \Ket{0} \otimes \hat{\sigma}_y \Ket{0} \otimes \hat{\sigma}_y \Ket{0} + \hat{\sigma}_x \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \otimes \hat{\sigma}_y \Ket{1} \Big) \\ &= \frac{1}{\sqrt{2}} \Big( \Ket{1} \otimes i \Ket{1} \otimes i \Ket{1} + \Ket{0} \otimes i \Ket{0} \otimes i \Ket{0} \Big) = - \Ket{\mathrm{GHZ}} \end{aligned}$$ 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: $$\begin{aligned} \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \quad &\implies \quad +1 \\ \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \quad &\implies \quad -1 \\ \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y \quad &\implies \quad -1 \\ \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x \quad &\implies \quad -1 \end{aligned}$$ According to any local hidden variable (LHV) theory, the measurement outcomes of the operators are predetermined, and the three particles $$A$$, $$B$$ and $$C$$ can be measured separately, or in other words, the eigenvalues can be factorized: $$\begin{aligned} \hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \quad &\implies \quad +1 = m_x^A m_x^B m_x^C \\ \hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \quad &\implies \quad -1 = m_x^A m_y^B m_y^C \\ \hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y \quad &\implies \quad -1 = m_y^A m_x^B m_y^C \\ \hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x \quad &\implies \quad -1 = m_y^A m_y^B m_x^C \end{aligned}$$ Where $$m_x^A = \pm 1$$ etc. Let us now multiply both sides of these four equations together: $$\begin{aligned} (+1) (-1) (-1) (-1) &= (m_x^A m_x^B m_x^C) (m_x^A m_y^B m_y^C) (m_y^A m_x^B m_y^C) (m_y^A m_y^B m_x^C) \\ -1 &= (m_x^A)^2 (m_x^B)^2 (m_x^C)^2 (m_y^A)^2 (m_y^B)^2 (m_y^C)^2 \end{aligned}$$ This is a contradiction: the left-hand side is $$-1$$, but all six factors on the right are $$+1$$. This means that we must have made an incorrect assumption along the way. Our only assumption was that we could factorize the eigenvalues, so that e.g. particle $$A$$ could be measured on its own without an "action-at-a-distance" effect on $$B$$ or $$C$$. However, because that leads us to a contradiction, we must conclude that action-at-a-distance exists, and that therefore all LHV-based theories are invalid. ## References 1. N. Brunner, *Quantum information theory: lecture notes*, 2019, unpublished. 2. J.B. Brask, *Quantum information: lecture notes*, 2021, unpublished.