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diff --git a/content/know/concept/ghz-paradox/index.pdc b/content/know/concept/ghz-paradox/index.pdc new file mode 100644 index 0000000..4b872d8 --- /dev/null +++ b/content/know/concept/ghz-paradox/index.pdc @@ -0,0 +1,121 @@ +--- +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. |