some more changes by Simon
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@ -331,22 +331,38 @@ from the graph.
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\end{equation}
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\end{lemma}
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\begin{proof}
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Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$.
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Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before.
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Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$.
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In the following discussion the direction $\prod\limits_{\{l,k\} \in E} := \prod\limits_{\{l,k\} \in E, l < k}$
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is introduced as the graph is undirected and edges must not be handled twice.
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Set $\ket{\tilde{G}} := \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$.
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\begin{equation}
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\begin{aligned}
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K_G^{(i)} \ket{\tilde{G}} & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right) \ket{+} \\
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& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
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& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_k \otimes Z_l\right) X_i \ket{+} \\
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& = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l} + \sum\limits_{\{i,j\} \in E} \delta_{j,k}}\ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
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K_G^{(i)} \ket{\tilde{G}}
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& = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)
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\left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right) \ket{+} \\
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& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,k\} \in E}
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\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
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& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,k\} \in E}
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\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_k \otimes Z_l\right) X_i \ket{+} \\
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\end{aligned}
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\end{equation}
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As $X,Z$ anticommute. $X_i$ can now be absorbed into $\ket{+}$. The next step is a bit tricky:
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A $Z_j$ can be absorbed into a $\ket{0}\bra{0}_j$ giving no phase or into a $\ket{1}\bra{1}_j$ yielding
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a phase of $-1$. If there is no projector on $j$ the $Z_j$ can be commuted to the next projector.
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It is guaranteed that a projector on $j$ exists by the definition of $\ket{\tilde{G}}$.
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\begin{equation}
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\begin{aligned}
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K_G^{(i)} \ket{\tilde{G}}
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& = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l} + \delta_{j,k}}\ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
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& = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
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& = +1 \ket{\tilde{G}}
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\end{aligned}
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\end{equation}
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as $X, Z$ anticommute and $Z\ket{1} = -1\ket{1}$.
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The $\delta_{i,l} + \delta_{j,k}$ is either $0$ or $2$ by the definitions of $K_G^{(i)}$ and $\ket{\tilde{G}}$.
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\end{proof}
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\subsubsection{Dynamics of the VOP-free Graph States}
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@ -357,8 +373,8 @@ resulting in a multiplication of $Z_j$ to $K_G^{(i)}$ and $Z_i$ to $K_G^{(j)}$.
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is done by using the symmetric set difference:
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\begin{definition}
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For to finite sets $A,B$ the symmetric set difference $\Delta$ is
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defined as
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For two finite sets $A,B$ the symmetric set difference $\Delta$ is
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defined as:
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\begin{equation}
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A \Delta B = (A \cup B) \setminus (A \cap B)
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@ -366,7 +382,7 @@ is done by using the symmetric set difference:
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\end{definition}
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Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$.
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Another transformation on the VOP-free graph states is for a vertex $a \in V$
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Another transformation on the VOP-free graph states is for a vertex $a \in V$:
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\begin{equation}
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M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}
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@ -379,7 +395,7 @@ that will be used later.
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\label{lemma:M_a}
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When applying $M_a$ to a state $\ket{\bar{G}}$ the new state
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$\ket{\bar{G}'}$ is again a VOP-free graph state and the
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graph is updated according to
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graph is updated according to:
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\begin{equation}
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\begin{aligned}
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n_a' &= n_a \\
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@ -413,10 +429,10 @@ that will be used later.
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\end{aligned}
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\end{equation}
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One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenvalue
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One can now construct a new set of $K_{G'}^{(i)}$ s.t. $M_a \ket{\bar{G}}$ is the $+1$ eigenstate
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of the new $K_{G'}^{(i)}$. It is clear that $\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$.
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To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a = \{j\} \cup I$ and $n_j = \{a\} \cup J$.
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Then
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Then follows:
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\begin{equation}
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\begin{aligned}
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@ -431,7 +447,7 @@ that will be used later.
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\end{aligned}
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\end{equation}
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Using this ine can show that $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$:
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Using this one can show that $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$:
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\begin{equation}
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\ket{\bar{G}'} = S^{(j)}\ket{\bar{G}'} = K_{G}^{(a)} K_{G'}^{(j)}\ket{\bar{G}'}
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@ -443,7 +459,7 @@ that will be used later.
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multi-local Pauli operators where the $S^{(i)}$ can be generated from the $K_{G'}^{(i)}$
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and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$
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$\langle\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}\rangle$
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are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given
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are the stabilizers of $\ket{\bar{G}'}$. Therefore the associated graph is changed as given
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in the third equation.
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\end{proof}
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@ -451,12 +467,12 @@ that will be used later.
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The definition of a VOP-free graph state above raises an obvious question:
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Can any stabilizer state be described using just a graph?
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The answer is quite simple: No. The most simple cases are the single qbit
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stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is a simple extension
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The answer is straight forward: No. The most simple cases are the single qbit
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stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is an extension
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to the VOP-free graph states that allows the representation of an arbitrary
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stabilizer state. The proof that indeed any state can be represented is
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just constructive. As seen in theorem \ref{thm:clifford_group_approx} any $c \in C_n$
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can be constructed from $CZ$ and $C_L$ and in the following discussion it will become
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purely constructive. As seen in theorem \ref{thm:clifford_group_approx} any $c \in C_n$
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can be constructed from $CZ$ and $C_L$. In the following discussion it will become
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clear that both $C_L$ and $CZ$ can be applied to a general graph state.
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\subsubsection{Graph States and Vertex Operators}
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@ -467,7 +483,7 @@ clear that both $C_L$ and $CZ$ can be applied to a general graph state.
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A tuple $(V, E, O)$ is called the graphical representation of a stabilizer state
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if $(V, E)$ is a graph as in Definition \ref{def:graph} and $O = \{o_1, ..., o_n\}$ where $o_i \in C_L$.
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The state $\ket{G}$ is defined by the eigenvalue relation
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The state $\ket{G}$ is defined by the eigenvalue relation:
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\begin{equation}
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+1 \ket{G} = \left(\prod\limits_{j=1}^no_j\right) K_G^{(i)} \left(\prod\limits_{j=1}^no_j\right)^\dagger \ket{G}
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@ -485,8 +501,8 @@ Recalling the dynamics of stabilizer states the following relation follows immed
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The great advantage of this representation of a stabilizer state is its space requirement:
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Instead of storing $n^2$ $P_1$ matrices only some vertices (which often are implicit),
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the edges and some vertex operators ($n$ matrices) have to be stored. The following theorem
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will improve this even further: instead of $n$ matrices it is enough to store $n$ integers
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representing the vertex operators is enough:
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will improve this even further: instead of $n$ matrices it is sufficient to store $n$ integers
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representing the vertex operators:
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\begin{theorem}
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$C_L$ has $24$ degrees of freedom.
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@ -498,14 +514,14 @@ representing the vertex operators is enough:
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of $X,Z$ only.
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As the transformations are unitary they preserve eigenvalues, so $X$ can be mapped
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to $\pm X, \pm Y, \pm Z$ which gives $6$ degrees of freedom, the image of $Z$ has to
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to $\pm X, \pm Y, \pm Z$ which gives $6$ degrees of freedom. Furthermore the image of $Z$ has to
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anti-commute with the image of $X$ so $Z$ has four possible images under the transformation.
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This gives another $4$ degrees of freedom and a total of $24$.
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\end{proof}
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From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be used,
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one can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is
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$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which will be used in one specific operation on graph states.
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From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be used.
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One can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is
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$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific operation on graph states.
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\begin{equation}
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S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)
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@ -525,7 +541,7 @@ basically a stabilizer tableaux that might require less memory than the tableaux
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CHP. The true power of this formalism is seen when studying its dynamics. The simplest case
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is a local Clifford operator $c_j$ acting on a qbit $j$: The stabilizers of are changed to
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$\langle c_j S^{(i)} c_j^\dagger\rangle_i$. Using the definition of the graphical representation
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it is clear that just the vertex operators are changed and the new vertex operators are given by
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it is clear that just the vertex operators are changed and the new vertex operators are given by:
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\begin{equation}
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\begin{aligned}
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@ -550,11 +566,11 @@ are changed to $E' = E \Delta \left\{\{a,b\}\right\}$.
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The two qbits are isolated: From the definition of the graph state it is clear that
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any isolated clique of the graph can be treated independently. Therefore the two isolated qbits
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can be treated as an independent state and the set of two qbit stabilizer states is finite. An
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upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$: with or without
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upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$: With and without
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an edge between the qbits and $24$ Clifford operators on each vertex.
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All those states and the resulting state after a $CZ$ application can be computed and while doing so one
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gets another interesting result that will be useful later: If one vertex has the vertex operator $I$ the
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All those states and the resulting state after a $CZ$ application can be computed which leads to
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another interesting result that will be useful later: If one vertex has the vertex operator $I$ the
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resulting state can be chosen such that at least one of the vertex operators is $I$ again and in particular
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the identity on the vertex can be preserved under the application of a $CZ$.
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