From d6df56a51b815c2d935948b0fbac2f280a4f05a3 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Mon, 17 Feb 2020 18:49:25 +0100 Subject: [PATCH] a few more lines --- thesis/chapters/stabilizer.tex | 16 +++++++++++++++- 1 file changed, 15 insertions(+), 1 deletion(-) diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex index c0d20cd..8763c5d 100644 --- a/thesis/chapters/stabilizer.tex +++ b/thesis/chapters/stabilizer.tex @@ -659,4 +659,18 @@ as $o_a$ is a Clifford operator: \end{equation} Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. It is therefore enough to study the measurements -of any Pauli operator on the vertex operator free graph states. +of any Pauli operator on the vertex operator free graph states. The commutators of the observable +with the $K_G^{(i)}$ are quite easy to compute, note that Pauli matrices wither commute or anticommute +so it is easier to list the operators that anticommute: + +\begin{equation} + \begin{aligned} + A_{\pm X_a} &= \left\{j | \{j, a\} \in E\right\}\\ + A_{\pm Y_a} &= \left\{j | \{j, a\} \in E\right\} \cup \{a\} \\ + A_{\pm Z_a} &= \{a\}\\ + \end{aligned} +\end{equation} + +This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a$ +is measured the result $+1$ is obtained with probability $1$ and $(V, E, O)$ is unchanged. +