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---
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 instead
the three-particle entangled **GHZ state** $$\ket{\mathrm{GHZ}}$$:
$$\begin{aligned}
\boxed{
\ket{\mathrm{GHZ}}
\equiv \frac{1}{\sqrt{2}} \Big( \ket{000} + \ket{111} \Big)
}
\end{aligned}$$
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 $$\hat{\sigma}_y$$
as [quantum gates](/know/concept/quantum-gate/)
to this three-particle state, we find:
$$\begin{aligned}
\hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x \ket{\mathrm{GHZ}}
&= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_x \ket{0}
+ \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_x \ket{1} \Big)
\\
&= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes \ket{1} \otimes \ket{1} + \ket{0} \otimes \ket{0} \otimes \ket{0} \Big)
\\
&= \ket{\mathrm{GHZ}}
\\
\hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y \ket{\mathrm{GHZ}}
&= \frac{1}{\sqrt{2}} \Big( \hat{\sigma}_x \ket{0} \otimes \hat{\sigma}_y \ket{0} \otimes \hat{\sigma}_y \ket{0}
+ \hat{\sigma}_x \ket{1} \otimes \hat{\sigma}_y \ket{1} \otimes \hat{\sigma}_y \ket{1} \Big)
\\
&= \frac{1}{\sqrt{2}} \Big( \ket{1} \otimes i \ket{1} \otimes i \ket{1} + \ket{0} \otimes i \ket{0} \otimes i \ket{0} \Big)
\\
&= - \ket{\mathrm{GHZ}}
\end{aligned}$$
In other words, the GHZ state is a simultaneous eigenstate of these composite operators,
with eigenvalues $$+1$$ and $$-1$$, respectively.
Let us do the same for two more operators,
so that we have a set of four observables
for which $$\ket{\mathrm{GHZ}}$$ gives these eigenvalues:
$$\begin{aligned}
\hat{\sigma}_x \otimes \hat{\sigma}_x \otimes \hat{\sigma}_x
\quad &\implies \quad +1
\\
\hat{\sigma}_x \otimes \hat{\sigma}_y \otimes \hat{\sigma}_y
\quad &\implies \quad -1
\\
\hat{\sigma}_y \otimes \hat{\sigma}_x \otimes \hat{\sigma}_y
\quad &\implies \quad -1
\\
\hat{\sigma}_y \otimes \hat{\sigma}_y \otimes \hat{\sigma}_x
\quad &\implies \quad -1
\end{aligned}$$
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.
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