Categories: Physics, Quantum information, Quantum mechanics.

The Greenberger-Horne-Zeilinger or GHZ paradox is an alternative proof of Bell’s 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.

1. N. Brunner, Quantum information theory: lecture notes, 2019, unpublished.
2. J.B. Brask, Quantum information: lecture notes, 2021, unpublished.