\section{The Stabilizer Formalism and VOP-Free Graph States} \subsection{Stabilizers and Stabilizer States} This chapter discusses the stabilizer formalism that was introduced by Gottesman\cite{gottesman1997} for quantum error correction but soon proved to be a useful tool to describe a subset of states: the stabilizer states which can be simulated in polynomial time \cite{gottesman2008}. \begin{definition} \begin{equation} p \in P_n \Rightarrow p = \bigotimes\limits_{i=0}^n p_i \\ \forall i: p_i \in P := \{\pm 1, \pm i\} \cdot \{I, X, Y, Z\} \end{equation} Where $X = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right)$, $Y = \left(\begin{array}{cc} 0 & i \\ -i & 0\end{array}\right)$ and $Z = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right)$ are the Pauli matrices and $I$ is the identity. $p \in P_n$ is called a multi-local Pauli operator. \end{definition} \begin{definition} For a group $G$, $g_1, ..., g_n$ are called the generators of $G$ iff $\forall g \in G: g = \prod\limits_{i \in I} g_i$ for a subsed $I$ of $\{1, ..., n\}$. We write $G = \langle g_i \rangle_i$ if G is generated by the $g_i$. The generators $g_i$ are chosen to be the smallest set of generators of $G$. \end{definition} \begin{definition} \label{def:stabilizer} For a $n$ qbit state $\ket{\psi}$ $\langle S_i \rangle_i$ is called the stabilizer of $\ket{\psi}$ if \begin{enumerate} \item{$\forall i = 1, ..., n$ $S_i \in P_n$} \item{$\forall i,j = 1, ..., n$ $[S_i, S_j] = 0$ $S_i$ and $S_j$ commute} \item{$\forall i = 1, ..., n$ $S_i\ket{\psi} = +1 \ket{\psi}$} \end{enumerate} \end{definition} \begin{lemma} For every $\langle S_i \rangle_i$ fulfilling the first two conditions in definition \ref{def:stabilizer} there exists a (up to a global phase) unique state $\ket{\psi}$ fulfilling the third condition. This state is called stabilizer state. \end{lemma} \begin{proof} All multi-local Pauli operators are hermitian (observables in terms of quantum mechanics), as the $S_i$ commute they have a common set of eigenstates. Because each $S_i$ has eigenvalues $+1, -1$, there exist $2^n$ eigenstates, one state $\ket{\psi}$ with eigenvalue $+1$ for all $S_i$. As the dimension of $n$ qbits is $2^n$ the state $\ket{psi}$ is unique up to a global phase. \end{proof} One can study the dynamics of stabilizer states using only the stabilizers\cite{nielsen_chuang_2010}. Two important cases are the unitary transformation of a state and the measurement of a qbit. When applying a unitary gate to a stabilizer state $\ket{\psi}$ the resulting state will in general be no stabilizer state anymore, however there exists a group of transformations that map stabilizers to other stabilizers: the Clifford group. \begin{definition} \begin{equation} C_n := \{U \in SU(2) | UpU^\dagger \in P_n \forall p \in P_n\} \end{equation} is called the Clifford group on $n$ qbits. $C_1$ is called the local Clifford group. \end{definition} The properties of this group will be discussed later, for the time being is existence is enough. \begin{lemma} Let $\ket{\psi}$ be stabilized by $\langle S_i \rangle_i$, then $U\ket{\psi}$ is stabilized by $\langle US_iU^\dagger \rangle_i$. \end{lemma} \begin{proof} $$ U\ket{\psi} = US_i\ket{\psi} = US_iU^\dagger U\ket{\psi}$$ So $U\ket{\psi}$ is a $+1$ eigenstate of $US_iU^\dagger$. \end{proof} This is an important insight that is used for simulations\cite{gottesman_aaronson2008}, as updating the $n$ stabilizers that are a tensor product of $n$ Pauli matrices scales with roughly $\mathcal{O}(n^2)$ instead of $\mathcal{O}(2^n)$ for the state vector approach. Every qbit can be measured in the $X, Y$ or $Z$ basis which is a projection using \begin{equation} \frac{I + (-1)^s g_a}{2} \end{equation} Where $g_a \in \{Z_a, Y_a, X_a\}$ and $s \in \{0, 1\}$. How the stabilizers change when measuring a qbit is given by the following lemma: \begin{lemma} \label{lemma:stab_measurement} Let $J := \{ S_i | [g_a, S_i] \neq 0\}$. Then \begin{enumerate} \item{If $J = \{\}$, one value is measured with probability $1$ and the stabilizers are unchanged.} \item{If $J \neq \{\}$, $1$ and $0$ are measured with probability $\frac{1}{2}$ and after choosing a $j \in J$ the new state $\ket{\psi'}$ is stabilized by \begin{equation} \langle \{(-1)^s g_a\} \cup \{S_i S_j | i \in J \setminus \{j\} \} \cup \{S_i | i \in J^c\}\rangle \end{equation}} \end{enumerate} \end{lemma} \begin{proof} \begin{enumerate} \item{As $g_a$ commutes with all stabilizers, $\ket{\psi}$ is an eigenstate of $g_a$, so the result of the measurement is deterministic and $\ket{\psi}$ is left unchanged.} \item{As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators, $S_i$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then \begin{equation} \begin{aligned} P(s=+1) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\ &= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\ &= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\ &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\ &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\ &= P(s=-1) \end{aligned} \notag \end{equation} With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$. Further for $S^{(i)},S^{(j)} \in J$ \begin{equation} \begin{aligned} \frac{I + (-1)^sg_a}{2}\ket{\psi} &= \frac{I + (-1)^sg_a}{2}S^{(j)}S^{(i)} \ket{\psi} \\ &= S^{(j)}\frac{I + (-1)^{s + 1}g_a}{2}S^{(i)} \ket{\psi} \\ &= S^{(j)}S^{(i)}\frac{I + (-1)^{s + 2}g_a}{2}\ket{\psi} \\ &= S^{(j)}S^{(i)}\frac{I + (-1)^sg_a}{2}\ket{\psi} \end{aligned} \notag \end{equation} the state after measurement is stabilized by $S^{(j)}S^{(i)} i,j \in J$, and by $S^{(i)} \in J^c\setminus\{a\}$. $g_a$ trivially stabilizes $\ket{\psi'}$. } \end{enumerate} \end{proof} \subsection{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 $\bar{G} = (V, E)$ 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. $\ket{\overline{G}}$ is the $+1$ 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 eigenstates (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} \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} \begin{definition} Let $\bar{G} = (V, E)$ be a graph as in definition \ref{def:vop_free_g_state}. For a vertex $i \in V$ $n_i := \{j \in V | \{i,j\} \in E\}$ is called the neighbourhood of $i$. \end{definition} \begin{lemma} Let $\bar{G}$, $\ket{\bar{G}}$, $K_G^{(i)}$ be as in definition \ref{def:vop_free_g_state} and for a vertex $a \in V$ set \begin{equation} M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j} \end{equation} Then the graph $\bar{G}'$ associated with $\ket{\bar{G}'} = M_a\ket{\bar{G}}$ is changed according to the following equations: \begin{equation} \begin{aligned} n_a' &= n_a \\ n_j' &= n_j, \hbox{ if } j \notin n_a\\ n_j' &= (n_j \cup n_a) \setminus (n_j \cap n_a), \hbox{ if } j \in n_a \end{aligned} \end{equation} I.e. the neighbourhood of $a$ is toggled. \end{lemma} \begin{proof} $\ket{\bar{G}'}$ is stabilized by $\langle M_a K_G^{(i)} M_a^\dagger \rangle_i$, so it is sufficient to study how the $ K_G^{(i)}$ change under $M_a$. At first note that $[K_G^{(a)}, M_a] = 0$ and $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$, so the first two equations follow trivially. For $j \in n_a$ set \begin{equation} \begin{aligned} S^{(j)} &= M_a K_G^{(j)} M_a^\dagger \\ &= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right) \sqrt{iZ_j} K_G^{(j)} \sqrt{iZ_j}^\dagger \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right) \sqrt{-iX_a}^\dagger \\ &= \sqrt{-iX_a} \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}\right)\sqrt{iZ_j} X_j \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) Z_a \sqrt{iZ_j}^\dagger \left(\prod\limits_{l \in n_a \setminus \{j\}} \sqrt{iZ_l}^\dagger\right) \sqrt{-iX_a}^\dagger \\ &= \sqrt{iZ_j} X_j\sqrt{iZ_j}^\dagger\left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) \sqrt{-iX_a} Z_a \sqrt{-iX_a}^\dagger \\ &= (-1)^2 Y_j Y_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right)\\ &= (-1)i^2 Z_j X_a X_j Z_a \left(\prod\limits_{m \in n_j \setminus \{a\}} Z_m\right) \end{aligned} \end{equation} One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenvalue of the new $K_{G'}^{(i)}$. It is clear that $\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$. To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a = \{j\} \cup I$ and $n_j = \{a\} \cup J$. Then \begin{equation} \begin{aligned} S^{(j)} &= Z_j X_a X_j Z_a \prod\limits_{l \in J} Z_l\\ &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in J} Z_l\right) \left(\prod\limits_{l \in I}Z_l\right) \left(\prod\limits_{l \in I}Z_l\right) \\ &= Z_j X_a X_j Z_a \left(\prod\limits_{l \in ((I\cup J) \setminus (I\cap J))} Z_L\right) \left(\prod\limits_{l \in I}Z_l\right) \\ &= K_{G'}^{(a)} K_{G'}^{(j)} \\ &= K_{G}^{(a)} K_{G'}^{(j)} \end{aligned} \end{equation} Using this ine can show that $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$: \begin{equation} \ket{\bar{G}'} = S^{(j)}\ket{\bar{G}'} = K_{G}^{(a)} K_{G'}^{(j)}\ket{\bar{G}'} = K_{G'}^{(j)}K_{G}^{(a)}\ket{\bar{G}'} = K_{G'}^{(j)}\ket{\bar{G}'} \end{equation} Because $\{K_G^{(i)} | i \notin n_a\} \cup \{S^{(i)} | i\in n_a\}$ and $\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}$ are both $n$ commuting multi-local Pauli operators where the $S^{(i)}$ can be generated from the $K_{G'}^{(i)}$ and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$ $\langle\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}\rangle$ are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given in the third equation. \end{proof} These insights can be used to understand how measurement works on the vop-free graph state \cite{nielsen_chuang_2010}: Consider a state $\ket{\psi}$ that is stabilized by $g_1, ... g_n$ and a Pauli operator $g \in \{Z_a, Y_a, X_a\}$ that is to be measured. Recalling lemma \ref{lemma:stab_measurement} the following relations follow immideately: \begin{enumerate} \item{For $g = X_a$ if $a$ is an isolated qbit $+1$ is measured with probability $1$ and the state $\ket{\bar{G}}$ is unchanged.} \item{For $g = X_a$ and $a$ is non-isolated, choose $b \in n_a$ and the new stabilizers are \begin{equation} \langle \{(-1)^sX_a\} \cup \{K_G^{(b)}K_G^{(i)} | i \in n_a \setminus \{b\}\} \cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle \end{equation}} \item{For $g = Z_a$ the new stabilizers are \begin{equation} \langle \{(-1)^sZ_a\} \cup \{K_G^{(i)} | i \in V \setminus n_a\} \rangle \end{equation}} \item{For $g = Y_a$ the new stabilizers are \begin{equation} \langle \{(-1)^sY_a\} \cup \{K_G^{(a)}K_G^{(i)} | i \in n_a\} \cup \{K_G^{(i)} | i \in V \setminus n_a \setminus \{a\}\} \rangle \end{equation}} \end{enumerate} The states after the measurement are in general no vop-free graph states anymore, the following discussion will allow to construct new vop-free graph states and Clifford transformations from the vop-free graph state to the resulting state. In the case $g = X_a$ and $n_a = \{\}$ the graph is obviously unchanged. \begin{lemma} \label{lemma:Z_measurement} \begin{enumerate} \item{For a result $+Z_a$ the new state is $\ket{+_Z}_a \otimes \ket{\bar{G}'}$ with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$. } \item{For a result $-Z_a$ the new state is $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$ with $\ket{\bar{G}'}$ being stabilized by $K_G^{(i)}$ for $i \notin n_a$ and $Z_aK_G^{(i)}$ for $i \in n_a$ and $U = \prod\limits_{i \in n_a}Z_i$.} \end{enumerate} \end{lemma} \begin{proof} \begin{enumerate} \item{ It is trivial that $Z_a$ and $K_G^{(i)} i \neq a$ stabilize $\ket{+_Z}_a \otimes \ket{\bar{G}'}$. $Z_aK_G^{(i)}$ for $i \in n_a$ do not act on $a$, so $\ket{\bar{G}'}$ is well-defined. } \item{ The state $\ket{-_Z}_a \otimes \ket{\bar{G}'}$ is the $-1$ eigenstate of $K_G^{(i)}$ for all $i \in n_a$. This can be corrected by transforming $K_G^{(i)}$ to $-K_G^{(i)} = Z_i K_G^{(i)} Z_i^\dagger$ which stabilizes the state $U\ket{-_Z}_a \otimes \ket{\bar{G}'}$. } \end{enumerate} \end{proof} \begin{lemma} When $Y_a$ is measured with a result $s \in \{0, 1\}$ the state after the measurement is \begin{equation} \ket{(-1)^s_Y}_a \otimes U_{Y,s} \ket{\bar{G}'} \end{equation} \begin{equation} U_{Y,s} = \prod\limits_{l \in n_a} \sqrt{(-1)^s iZ_l} \end{equation} Where $\ket{\bar{G}'}$ is associated with the graph $G' = (V \setminus \{a\}, E')$ with $E'$ being changed according to $\forall i \in n_a$ $n_i' = (n_i \cup n_a) \setminus (n_i \cap n_a) \setminus \{a\}$. \end{lemma} \begin{proof} It is known that $(-1)^s Y_a$ has to stabilize the state after measurement, further $\forall j \in n_a$ $S^{(j)} = K_G^{(a)} K_G^{(j)}$, are the stabilizers of the new state around the measured qbit and the stabilizers $K_G^{(j)}, j \notin n_a$ are unchanged. Let $j \in n_a$ and $n_a =: \{j\} \cup I$, $n_j =: \{a\} \cup J$. \begin{equation} \begin{aligned} S^{(i)} &= K_G^{(a)} K_G^{(j)} \\ &= X_a \left(\prod\limits_{l \in n_a} Z_l\right)X_j\left(\prod\limits_{l\in n_j} Z_l\right) \\ &= X_a Z_a Z_j X_j \left(\prod\limits_{l \in I} Z_l\right)\left(\prod\limits_{l \in J} Z_l\right)\\ &= X_a Z_a Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)} Z_l\right) \\ &= iY_a Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J) \setminus \{a\}} Z_l\right) \\ &= (-1)^si S^{(a)} Z_j X_j \left(\prod\limits_{l \in (I \cup J)\setminus(I\cap J)\setminus\{a\}} Z_l\right) \\ &= S^{(a)} \sqrt{(-1)^siZ_j} K_{G'}^{(j)} \sqrt{(-1)^siZ_j}^\dagger \end{aligned} \end{equation} with $G'$ as above, the rest of the argument is analogous to lemma \ref{lemma:Z_measurement}. \end{proof}