From 0e1044478b23381c1d34eff0b0e35fb9cf15d071 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Wed, 27 Nov 2019 10:48:40 +0100 Subject: [PATCH] did some work on the vertex operator free state --- thesis/chapters/graph_simulator.tex | 24 +++++++++++++++++++++++- 1 file changed, 23 insertions(+), 1 deletion(-) diff --git a/thesis/chapters/graph_simulator.tex b/thesis/chapters/graph_simulator.tex index faadef5..c044661 100644 --- a/thesis/chapters/graph_simulator.tex +++ b/thesis/chapters/graph_simulator.tex @@ -240,11 +240,33 @@ were derived from the vertex operator-free graph states. \end{equation} as $X$, $Z$ anticommute. - \end{proof} +\begin{lemma} + \begin{equation} + \ket{\overline{G}} = \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right) \left(\prod\limits_{l \in V} H_l\right) \ket{0} + \end{equation} + In particular definitions \ref{def:vop_free_g_state} and \ref{def:graph_state} are consistent, when there are no + vertex operators on the graph state $\ket{G}$. +\end{lemma} +\begin{proof} + Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$ $X_i \ket{+} = +1 \ket{+}$. + Set $\ket{\tilde{G}} := \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right)\ket{+}$. + \begin{equation} + \begin{aligned} + K_G^{(i)} \ket{\tilde{G}} & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\left(\prod\limits_{\{l,j\} \in E} CZ_{l,j} \right) \ket{+} \\ + & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\ + & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) X_i \ket{+} \\ + & = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{2\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\ + & = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\ + & = +1 \ket{\tilde{G}} + \end{aligned} + \end{equation} + + as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$. +\end{proof}