some more work
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@ -31,7 +31,7 @@ the elements of $P$ either commute or anticommute.
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For $n$ qbits
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\begin{equation}
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P_n := \{\bigotimes\limits_{i=0}^{n-1} p_i | p_i \in P\}
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P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i | p_i \in P\right\}
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\end{equation}
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is called the multilocal Pauli group on $n$ qbits.
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@ -204,4 +204,94 @@ product of $n$ Pauli matrices. This has led to the simulation using stabilizer t
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\cite{gottesman_aaronson2008}.
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Interestingly also measurements are dynamics covered by the stabilizers.
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When an observable $g_i \in \{\pm X_i, \pm Y_i \pm Z_i\}$ is measured
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When an observable $g_a \in \{\pm X_a, \pm Y_a \pm Z_a\}$ acting on qbit $a$ is measured
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one has to consider the projector
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\begin{equation}
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P_{g_a,s} = \frac{I + (-1)^s g_a}{2}
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\end{equation}
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If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with probability $1$
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and the stabilizers are left unchanged:
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\begin{equation}
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\begin{aligned}
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\ket{\psi'} &= \frac{I + g_a}{2}\ket{\psi} \\
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&= \frac{I + g_a}{2}S^{(i)} \ket{\psi} \\
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&= S^{(i)} \frac{I + g_a}{2}\ket{\psi} \\
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&= S^{(i)}\ket{\psi'} \\
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\end{aligned}
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\end{equation}
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As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$.
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If $g_a$ does not commute with all stabilizers the following lemma gives
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the result of the measurement.
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\begin{lemma}
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\label{lemma:stab_measurement}
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Let $J := \{ S^{(i)} | [g_a, S^{(i)}] \neq 0\} \neq \{\}$. When measuring
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$\frac{I + (-1)^s g_a}{2} $
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$1$ and $0$ are obtained with probability $\frac{1}{2}$ and after choosing
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a $j \in J$ the new state $\ket{\psi'}$ is stabilized by
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\begin{equation}
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\langle \{(-1)^s g_a\} \cup \{S^{(i)} S^{(j)} | S^{(i)} \in J \setminus \{S^{(j)}\} \} \cup J^c\rangle
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\end{equation}
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\end{lemma}
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\begin{proof}
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As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators,
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$S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
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\begin{equation}
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\begin{aligned}
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P(s=+1) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\
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&= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\
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&= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
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&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\
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&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
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&= P(s=-1)
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\end{aligned}
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\notag
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\end{equation}
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With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$.
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Further for $S^{(i)},S^{(j)} \in J$
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\begin{equation}
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\begin{aligned}
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\frac{I + (-1)^sg_a}{2}\ket{\psi} &= \frac{I + (-1)^sg_a}{2}S^{(j)}S^{(i)} \ket{\psi} \\
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&= S^{(j)}\frac{I + (-1)^{s + 1}g_a}{2}S^{(i)} \ket{\psi} \\
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&= S^{(j)}S^{(i)}\frac{I + (-1)^{s + 2}g_a}{2}\ket{\psi} \\
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&= S^{(j)}S^{(i)}\frac{I + (-1)^sg_a}{2}\ket{\psi}
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\end{aligned}
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\notag
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\end{equation}
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the state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, and by
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$S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$.
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\end{proof}
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\subsection{The VOP-free Graph States}
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This section will discuss the vertex operator(VOP)-free graph states. Why they are called
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vertex operator-free will be clear in the following section about graph states.
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\begin{definition}
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The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$.
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In the following $V = \{0, ..., n-1\}$ will be used.
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$E$ is the set of edges $E = \left\{\{i, j\} | i,i \in V, i \neq j\right\}$.
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\end{definition}
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This definition of a graph is way less general than the definition of a mathematical graph.
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Using this definition will however allow to avoid an extensive list of constraints on the
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mathematical graph that are implied in this definition.
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\begin{definition}
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For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are
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\begin{equation}
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K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_i
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\end{equation}
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for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is the state stabilized by
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$\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$.
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\end{definition}
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