From 17823d4fda488c2cdd097a0102c60b71967bfe4e Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Mon, 25 Nov 2019 21:37:48 +0100 Subject: [PATCH] some more work on the graph formalism --- thesis/chapters/graph_simulator.tex | 72 +++++++++++++++++++++++++++-- 1 file changed, 67 insertions(+), 5 deletions(-) diff --git a/thesis/chapters/graph_simulator.tex b/thesis/chapters/graph_simulator.tex index 4bf551b..faadef5 100644 --- a/thesis/chapters/graph_simulator.tex +++ b/thesis/chapters/graph_simulator.tex @@ -187,24 +187,86 @@ Where every $o_i$ acts on the $i$-th qbit. One can show that any stabilizer state can be realized as a graph state (for instance in \cite{schlingenmann2001}). +\subsubsection{The Vertex Operator-Free Graph States} + +In order to understand some essential transformations of graph states it is necessary +to study the vertex operator-free graph states first, partially because the graph states as used in this paper +were derived from the vertex operator-free graph states. + +\begin{definition} + \label{def:vop_free_g_state} + A $n$ qbit vertex operator-free graph state $\ket{\overline{G}}$ is associated with a graph $(V, E)$ + by the $n$ operators + + \begin{equation} + K^{(i)}_G := X_i \left(\prod\limits_{\{i, j\} \in E} Z_j\right) + \end{equation} + + for all $i \in V$ where for some operator $O$ $O_i$ indicates that it acts on the $i$-th qbit. + + A state $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$. +\end{definition} + +\begin{corrolary} + All $K^{(i)}_G$ commute and are hermitian. Therefore they have a common set of eigen states + (in particular definition \ref{def:vop_free_g_state} is well defined). + In terms of quantum mechanics $K^{(i)}_G$ are observables. + + Further as $\ket{\overline{G}}$ is a $+1$ eigenstate of all $n$ $K^{(i)}_G$ which are + multi-local Pauli operators, $\{K^{(i)}_G | i \in \{0, ..., n-1\}\}$ is the stabilizer + of $\ket{\overline{G}}$ and $\ket{\overline{G}}$ is a stabilizer state. +\end{corrolary} + +\begin{proof} + As $X_i$ and $Z_i$ are hermitian their product is hermitian. + + Consider the case $\{i,j\} \notin E$ first: + \begin{equation} + \begin{aligned} + \left[K^{(i)}_G, K^{(j)}_G\right] = \left[X_i \prod\limits_{\{i, n\} \in E} Z_n, X_j \prod\limits_{\{j, m\} \in E} Z_m\right] = 0 + \end{aligned} + \end{equation} + + As operators acting on different qbits commute. The case $\{i,j\} \in E$ is slightly less trivial: + \begin{equation} + \begin{aligned} + \left[K^{(i)}_G, K^{(j)}_G\right] &= \left[X_i \left(\prod\limits_{\{i, n\} \in E, n \neq j} Z_n\right) Z_j, X_j \left(\prod\limits_{\{j, m\} \in E, m \neq i} Z_m\right) Z_i\right] \\ + &= \left[X_i Z_j \prod\limits_n Z_n, X_j Z_i \prod\limits_m Z_m\right]\\ + &= \left(X_i Z_j X_j Z_i - X_j Z_i X_i Z_j\right) \prod\limits_n Z_n \prod\limits_m Z_m \\ + &= \left(Z_j X_j X_i Z_i - X_j Z_j Z_i X_i\right) \prod\limits_n Z_n \prod\limits_m Z_m \\ + &= \left((-1)^2X_j Z_j Z_i X_i - X_j Z_j Z_i X_i\right) \prod\limits_n Z_n \prod\limits_m Z_m \\ + &= 0 + \end{aligned} + \end{equation} + + as $X$, $Z$ anticommute. + +\end{proof} + + + + + + + \subsection{Operations on the Graph State} \subsubsection{Single Qbit Gates} Recalling \eqref{eq:g_state} -Makes it clear that for any single qbit gate $o \in C_L$ with $o^{(k)}$ being the gate +Makes it clear that for any single qbit gate $g \in C_L$ with $g_k$ being the gate acting on qbit $k$ the state changes according to \begin{equation} \begin{aligned} - o^{(k)} \ket{G} &= o^{(k)} \left(\bigotimes\limits_{i=0}^{n-1} o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right) \ket{+} \\ - &= \left(\bigotimes\limits_{i=0}^{n-1} o^{\delta_{i,k}}o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right)\ket{+} + g_k \ket{G} &= g_k \left(\bigotimes\limits_{i=0}^{n-1} o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right) \ket{+} \\ + &= \left(\bigotimes\limits_{i=0}^{n-1} g_k^{\delta_{i,k}}o_{i} \right)\left(\bigotimes\limits_{\{i, j\} \in E} CZ_{i,j} \right)\ket{+} \end{aligned} \end{equation} -meaning that the graph state $(V, E, O)$ changes to $(V, E, \{o_0, ..., o_{k-1}, oo_k, o_{k+1}, ..., o_{n-1}\})$ -as $C_L$ is almost a group the element $oo_k \in C_L$ up to a global phase that is disregarded. All the results +meaning that the graph state $(V, E, O)$ changes to $(V, E, \{o_0, ..., o_{k-1}, go_k, o_{k+1}, ..., o_{n-1}\})$ +as $C_L$ is almost a group the element $go_k \in C_L$ up to a global phase that is disregarded. All the results of $C_L \times C_L \rightarrow C_L, a,b \mapsto ab$ have been precomputed in a lookup table and the vertex operators are updated according to that lookup table.