--- title: "GHZ paradox" firstLetter: "G" publishDate: 2021-03-29 categories: - Physics - Quantum mechanics - Quantum information date: 2021-03-29T15:15:41+02:00 draft: false markup: pandoc --- # GHZ Paradox 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.