746 lines
35 KiB
TeX
746 lines
35 KiB
TeX
% vim: ft=tex
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\section{The Stabilizer Formalism}
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The stabilizer formalism was originally introduced by Gottesman \cite{gottesman1997}
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for quantum error correction and is a useful tool to encode quantum information
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such that it is protected against noise. The prominent Shor code \cite{shor1995}
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is an example of a stabilizer code (although it was discovered before the stabilizer
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formalism was discovered), as are the 3-qbit bit-flip and phase-flip codes.
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It was only later that Gottesman and Knill discovered that stabilizer states can
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be simulated in polynomial time on a classical machine \cite{gottesman2008}. This
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performance has since been improved to $n\log(n)$ time on average \cite{andersbriegel2005}.
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\subsection{Stabilizers and Stabilizer States}
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\subsubsection{Local Pauli Group and Multilocal Pauli Group}
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\begin{definition}
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\begin{equation}
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P := \{\pm 1, \pm i\} \cdot \{I, X, Y, Z\}
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\end{equation}
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Is called the Pauli group.
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\end{definition}
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The group property of $P$ can be verified easily. Note that
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the elements of $P$ either commute or anticommute.
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\begin{definition}
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For $n$ qbits
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\begin{equation}
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P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i \middle| 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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\end{definition}
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The group property of $P_n$ and the (anti-)commutator relationships follow directly from its definition
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via the tensor product.
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%Further are $p \in P_n$ hermitian and have the eigenvalues $\pm 1$ for
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%$p \neq \pm I$, $+1$ for $p = I$ and $-1$ for $p = -I$.
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\subsubsection{Stabilizers}
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\begin{definition}
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\label{def:stabilizer}
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An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff
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\begin{enumerate}
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\item{$\forall i,j = 1, ..., N$: $[S^{(i)}, S^{(j)}] = 0$%: $S^{(i)}$ and $S^{(j)}$ commute
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}
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\item{$-I \notin S$}
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\end{enumerate}
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\end{definition}
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\begin{lemma}
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If $S$ is a set of stabilizers, the following statements follow
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directly:
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\begin{enumerate}
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\item{$\pm iI \notin S$}
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\item{$(S^{(i)})^2 = I$ for all $i$}
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\item{$S^{(i)}$ are hermitian for all $i$}
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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\begin{enumerate}
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\item{$(iI)^2 = (-iI)^2 = -I$. Which contradicts the definition of $S$.}
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\item{From the definition of $S$ ($G_n$ respectively) follows that any
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$S^{(i)} \in S$ has the form $\pm i^l \left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$ where
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$\tilde{p}_j \in \{X, Y, Z, I\}$ and $l \in \{0, 1\}$. As $\left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$
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is hermitian $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly.
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}
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\item{Following the argumentation above $\left(S^{(i)}\right)^2 = -I \Leftrightarrow l=1$
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therefore $\left(S^{(i)}\right)^2 = -I \Leftrightarrow \left(S^{(i)}\right)^\dagger \neq S^{(i)}$.}
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\end{enumerate}
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\end{proof}
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Considering all the elements of a group can be impractical for some calculations,
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the generators of a group are introduced. Often it is enough to discuss the generator's
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properties in order to understand the properties of the group.
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\begin{definition}
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For a finite group $G$ and some $m \in \mathbb{N}$ one denotes the generators
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of G
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\begin{equation} \langle g_1, ..., g_m \rangle \equiv \langle g_i \rangle_{i=1,...,m}\end{equation}
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where $g_i \in G$, every element in $G$ can be written as a product of the $g_i$
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and $m$ is the smallest integer for which these statements hold.
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\end{definition}
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In the following discussions $\langle S^{(i)} \rangle_{i=0, ..., n-1}$ will be used as
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the required properties of a set of stabilizers that can be studied on its
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generators.
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\subsubsection{Stabilizer States}
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One important basic property of quantum mechanics is that hermitian operators have real eigenvalues
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and eigenspaces which are associated with these eigenvalues. Finding these eigenvalues and eigenvectors
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is what one calls solving a quantum mechanical system. One of the most fundamental insights of
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quantum mechanics is that commuting operators have a common set of eigenvectors, i.e. they
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can be diagonalized simultaneously. This motivates and justifies the following definition.
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\begin{definition}
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For a set of stabilizers $S$ the vector space
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\begin{equation}
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V_S := \left\{\ket{\psi} \middle| S^{(i)}\ket{\psi} = +1\ket{\psi} \forall S^{(i)} \in S\right\}
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\end{equation}
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is called the space of stabilizer states associated with $S$ and one says
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$\ket{\psi}$ is stabilized by $S$.
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\end{definition}
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It is clear that to show the stabilization property of
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$S$ the proof for the generators is sufficient,
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as all the generators forming an element in $S$ can be absorbed into $\ket{\psi}$.
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The dimension of $V_S$ is not immediately clear. One can however show that
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for a set of stabilizers $\langle S^{(i)} \rangle_{i=1, ..., n-m}$ the dimension
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$dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important
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result:
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\begin{theorem}
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\label{thm:unique_s_state}
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For a $n$ qbit system and stabilizers $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ the stabilizer
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space $V_S$ has $dim V_S = 1$, in particular there exists an up to a trivial phase unique
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state $\ket{\psi}$ that is stabilized by $S$.
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Without proof.
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\end{theorem}
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In the following discussions for $n$ qbits a set $S = \langle S^{(i)} \rangle_{i=1,...,n}$
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of $n$ independent stabilizers will be assumed.
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\subsubsection{Dynamics of Stabilizer States}
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Consider a $n$ qbit state $\ket{\psi}$ that is the $+1$ eigenstate of $S = \langle S^{(i)} \rangle_{i=1,...,n}$
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and a unitary transformation $U$ that describes the dynamics of the system, i.e.
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\begin{equation}
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\ket{\psi'} = U \ket{\psi}
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\end{equation}
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It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$ anymore. There are
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however some statements that can still be made:
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\begin{equation}
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\begin{aligned}
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\ket{\psi'} &= U \ket{\psi} \\
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&= U S^{(i)} \ket{\psi} \\
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&= U S^{(i)} U^\dagger U\ket{\psi} \\
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&= U S^{(i)} U^\dagger \ket{\psi'} \\
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&= S^{\prime(i)} \ket{\psi'} \\
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\end{aligned}
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\end{equation}
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Note that in Definition \ref{def:stabilizer} it has been demanded that stabilizers are a
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subgroup of the multilocal Pauli operators. This does not hold true for an arbitrary
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$U$ but there exists a group for which $S'$ will be a set of stabilizers.
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\begin{definition}
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For $n$ qbits
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\begin{equation}
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C_n := \left\{U \in SU(2^n) \middle| \forall p \in P_n: UpU^\dagger \in P_n \right\}
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\end{equation}
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is called the Clifford group. $C_1 =: C_L$ is called the local Clifford group.
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\end{definition}
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\begin{theorem}
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\label{thm:clifford_group_approx}
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\begin{enumerate}
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\item{$C_L$ can be generated using only $H$ and $S$.}
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\item{$C_L$ can be generated from $\sqrt{iZ} = \exp(\frac{i\pi}{4}) S^\dagger$
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and $\sqrt{-iX} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc} 1 & -i \\ -i & 1 \end{array}\right)$.
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Also $C_L$ is generated by a product of at most $5$ matrices $\sqrt{iZ}$, $\sqrt{-iX}$.
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}
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\item{$C_n$ can be generated using $C_L$ and $CZ$ or $CX$.}
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\end{enumerate}
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\end{theorem}
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\begin{proof}
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\begin{enumerate}
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\item{See \cite[Theorem 10.6]{nielsen_chuang_2010}}
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\item{
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One can easily verify that $\sqrt{iZ} \in C_L$ and $\sqrt{-iX} \in C_L$.
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Further one can show that (up to a global phase)
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$H = \sqrt{iZ} \sqrt{-iX}^3 \sqrt{iZ}$ and $S = \sqrt{iZ}^3$.
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The length of the product can be seen when explicitly calculating
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$C_L$.
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}
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\item{See \cite[Theorem 10.6]{nielsen_chuang_2010}}
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\end{enumerate}
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\end{proof}
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This is quite an important result: As under a transformation $U \in C_n$ $S' = U^\dagger S U$ is a set of
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$n$ independent stabilizers and $\ket{\psi'}$ is stabilized by $S'$ one can consider
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the dynamics of the stabilizers instead of the actual state. This is considerably more
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efficient as only $n$ stabilizers have to be modified, each being just the tensor
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product of $n$ Pauli matrices. This has led to the simulation using stabilizer tableaux
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\cite{gottesman_aaronson2008}.
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\subsubsection{Measurements on Stabilizer States}
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\label{ref:meas_stab}
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Interestingly also measurements are dynamics covered by the stabilizers.
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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 $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$. When measuring
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$\frac{I + (-1)^s g_a}{2} $
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$s=1$ and $s=0$ 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 \left\{S^{(i)} S^{(j)} \middle| S^{(i)} \in J \setminus \{S^{(j)}\} \right\} \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=0) &= \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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\subsubsection{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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\label{def:graph}
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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 $E = \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$.
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For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is called the neighbourhood
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of $i$.
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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_j
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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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It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute
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follows from the fact that $\{i,j\} \in E \Leftrightarrow \{j,i\} \in E$ therefore for two
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operators $K_G^{(i)}$ and $K_G^{(j)}$ either $\{i, j\} \notin E$ so they commute trivially.
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If $\{i,j\} \in E$ $X_i$, $Z_j$ and $X_j$, $Z_i$ anticommute which yields that the
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operators commute.
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This definition of a graph state might not seem to be straight forward
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but recalling theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$
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is unique. The following lemma will provide a way to construct this state
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from the graph.
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\begin{lemma}
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For a graph $G = (V, E)$ the associated state $\ket{\bar{G}}$ is
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constructed using
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\begin{equation}
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\begin{aligned}
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\ket{\bar{G}} &= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right)\left(\prod\limits_{i \in V} H_i\right) \ket{0} \\
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&= \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+} \\
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\end{aligned}
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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.
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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}}
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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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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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This representation gives an immediate result to how the stabilizers $\langle K_G^{(i)} \rangle_i$ change
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under the $CZ$ gate: When applying $CZ_{i,j}$ on $G = (V, E)$ the edge $\{i,j\}$ is toggled,
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resulting in a multiplication of $Z_j$ to $K_G^{(i)}$ and $Z_i$ to $K_G^{(j)}$. Toggling edges
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is done by using the symmetric set difference:
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\begin{definition}
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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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\end{equation}
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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 for a vertex $a \in V$ is:
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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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\end{equation}
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This transformation toggles the neighbourhood of $a$ which is an operation
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that will be used later.
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\begin{lemma}
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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
|
|
graph is updated according to:
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
n_a' &= n_a \\
|
|
n_j' &= n_j, \hbox{ if } j \notin n_a\\
|
|
n_j' &= n_j \Delta n_a, \hbox{ if } j \in n_a
|
|
\end{aligned}
|
|
\end{equation}
|
|
\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 $+1$ eigenstate
|
|
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 follows:
|
|
|
|
\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 one 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 $\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{S^{(i)} \middle| i\in n_a\right\}$ and
|
|
$\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}$ 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 $K_{G'}^{(j)}$
|
|
$\langle\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}\rangle$
|
|
are the stabilizers of $\ket{\bar{G}'}$. Therefore the associated graph is changed as given
|
|
in the third equation.
|
|
\end{proof}
|
|
|
|
\subsection{Graph States}
|
|
|
|
The definition of a VOP-free graph state above raises an obvious question:
|
|
Can any stabilizer state be described using just a graph?
|
|
The answer is straight forward: No. The most simple cases are the single qbit
|
|
stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is an extension
|
|
to the VOP-free graph states that allows the representation of an arbitrary
|
|
stabilizer state. The proof that indeed any state can be represented is
|
|
purely constructive. As seen in theorem \ref{thm:clifford_group_approx} any $c \in C_n$
|
|
can be constructed from $CZ$ and $C_L$. In the following discussion it will become
|
|
clear that both $C_L$ and $CZ$ can be applied to a general graph state.
|
|
|
|
\subsubsection{Graph States and Vertex Operators}
|
|
\label{ref:g_states_vops}
|
|
|
|
\begin{definition}
|
|
\label{def:g_state}
|
|
A tuple $(V, E, O)$ is called the graphical representation of a stabilizer state
|
|
if $(V, E)$ is a graph as in Definition \ref{def:graph} and $O = \{o_1, ..., o_n\}$ where $o_i \in C_L$.
|
|
|
|
The state $\ket{G}$ is defined by the eigenvalue relation:
|
|
|
|
\begin{equation}
|
|
+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}
|
|
\end{equation}
|
|
|
|
$o_i$ are called the vertex operators of $\ket{G}$.
|
|
\end{definition}
|
|
|
|
Recalling the dynamics of stabilizer states the following relation follows immediately:
|
|
|
|
\begin{equation}
|
|
\ket{G} = \left(\prod\limits_{j=1}^no_j\right) \ket{\bar{G}}
|
|
\end{equation}
|
|
|
|
The great advantage of this representation of a stabilizer state is its space requirement:
|
|
Instead of storing $n^2$ $P_1$ matrices only some vertices (which often are implicit),
|
|
the edges and some vertex operators ($n$ matrices) have to be stored. The following theorem
|
|
will improve this even further: instead of $n$ matrices it is sufficient to store $n$ integers
|
|
representing the vertex operators:
|
|
|
|
\begin{theorem}
|
|
$C_L$ has $24$ degrees of freedom.
|
|
\end{theorem}
|
|
\begin{proof}
|
|
It is clear that $\forall a \in C_L$ a is a group isomorphism $P_1 \circlearrowleft$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$.
|
|
Therefore $a$ will preserve the (anti-)commutator relations of $P$.
|
|
Further note that $Y = iXZ$, so one has to consider the anti-commutator relations
|
|
of $X,Z$ only.
|
|
|
|
As the transformations are unitary they preserve eigenvalues, so $X$ can be mapped
|
|
to $\pm X, \pm Y, \pm Z$ which gives $6$ degrees of freedom. Furthermore the image of $Z$ has to
|
|
anti-commute with the image of $X$ so $Z$ has four possible images under the transformation.
|
|
This gives another $4$ degrees of freedom and a total of $24$.
|
|
\end{proof}
|
|
|
|
From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be used.
|
|
One can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is
|
|
$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific operation on graph states.
|
|
|
|
\begin{equation}
|
|
S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)
|
|
\end{equation}
|
|
\begin{equation}
|
|
\sqrt{-iX} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)
|
|
\end{equation}
|
|
\begin{equation}
|
|
\sqrt{-iZ} = \exp(-i\frac{\pi}{4})\left(\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right)
|
|
\end{equation}
|
|
|
|
|
|
\subsubsection{Dynamics of Graph States}
|
|
|
|
So far the graphical representation of stabilizer states is just another way to store
|
|
basically a stabilizer tableaux that might require less memory than the tableaux used in
|
|
CHP. The true power of this formalism is seen when studying its dynamics. The simplest case
|
|
is a local Clifford operator $c_j$ acting on a qbit $j$: The stabilizers of are changed to
|
|
$\langle c_j S^{(i)} c_j^\dagger\rangle_i$. Using the definition of the graphical representation
|
|
it is clear that just the vertex operators are changed and the new vertex operators are given by:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
o_i' &= o_i &\mbox{if } i \neq j\\
|
|
o_i' &= c o_i c^\dagger &\mbox{if } i = j\\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
The action of a $CZ$ gate on the state $(V, E, O)$ is in most cases less trivial.
|
|
Let $i \neq j$ be two qbits, now consider the action of $CZ_{a,b}$ on $(V, E, O)$.
|
|
The cases given here follow the implementation of a $CZ$ application in \cite{pyqcs},
|
|
the respective paragraphs from \cite{andersbriegel2005} are given in italic.
|
|
|
|
|
|
\textbf{Case 1}(\textit{Case 1})\textbf{:}\\
|
|
Both $o_a$ and $o_b$ commute with $CZ_{a,b}$. This is the case for exactly
|
|
four vertex operators: $\mathcal{Z} = \left\{I, Z, S, S^\dagger\right\}$.
|
|
In this case the CZ can be pulled past the vertex operators and just the edges
|
|
are changed to $E' = E \Delta \left\{\{a,b\}\right\}$.
|
|
|
|
\textbf{Case 2}(\textit{Sub-Sub-Case 2.2.1})\textbf{:}\\
|
|
The two qbits are isolated: From the definition of the graph state it is clear that
|
|
any isolated clique of the graph can be treated independently. Therefore the two isolated qbits
|
|
can be treated as an independent state and the set of two qbit stabilizer states is finite. An
|
|
upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$: With and without
|
|
an edge between the qbits and $24$ Clifford operators on each vertex.
|
|
|
|
All those states and the resulting state after a $CZ$ application can be computed which leads to
|
|
another interesting result that will be useful later: If one vertex has the vertex operator $I$ the
|
|
resulting state can be chosen such that at least one of the vertex operators is $I$ again and in particular
|
|
the identity on the vertex can be preserved under the application of a $CZ$.
|
|
|
|
\textbf{Case 3}(\textit{Case 2, but one sub-case has been handled})\textbf{:}\\
|
|
At least one vertex operator does not commute with $CZ$ and at least one vertex
|
|
has non-operand neighbours. In this case one can try to clear the vertex operators
|
|
which will succeed for at least one qbit.
|
|
|
|
The transformation given in
|
|
Lemma \ref{lemma:M_a} is used to "clear" the vertex operators. Recalling that
|
|
the transformation $M_j$ toggles the neighbourhood of vertex $j$ gives substance
|
|
to the following theorem:
|
|
|
|
\begin{theorem}
|
|
A graph state $\ket{G}$ associated with $(V, E, O)$ is invariant when
|
|
simultaneously toggling the neighbourhood of a non-isolated qbit $j$
|
|
and right-multiplying $M_j^\dagger$ to the vertex operators in the sense
|
|
that $\sqrt{-iX}^\dagger = \sqrt{iX}$ is right-multiplied to $o_j$ and
|
|
$\sqrt{iZ}^\dagger = \sqrt{-iZ}$ is right-multiplied to $o_l$ for all
|
|
neighbours $l$ of $j$.
|
|
|
|
Without proof.
|
|
\end{theorem}
|
|
|
|
As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and $\sqrt{iZ}$.
|
|
This yields an algorithm to reduce the vertex operator of a non-isolated qbit $j$ to the identity.
|
|
The combined operation of toggling the neighbourhood of $j$ and right-multiplying
|
|
$M_j^\dagger$ is called $L_j$ transformation and it transforms $(V, E, O)$ into a so-called
|
|
local Clifford equivalent graphical representation. The algorithm is given by the following steps:
|
|
|
|
\begin{enumerate}
|
|
\item{Check whether $n_a \setminus \{b\} \neq \{\}$. If this condition does not hold
|
|
the vertex operator on $a$ cannot be cleared. }
|
|
\item{Express the vertex operator $o_a$ as a product of $\sqrt{-iX}$ and $\sqrt{iZ}$.}
|
|
\item{Move from right to left through the product:
|
|
\begin{enumerate}
|
|
\item{If the current matrix is $\sqrt{-iX}$ apply $L_a$.}
|
|
\item{If the current matrix is $\sqrt{iZ}$ select a vertex
|
|
$j \in n_a \setminus \{b\}$ (which is possible as it has been checked before)
|
|
and apply $L_j$.}
|
|
\end{enumerate}
|
|
}
|
|
\end{enumerate}
|
|
|
|
This algorithm has the important properties that if the algorithm succeeds
|
|
$o_a = I$ and $o_b$ has picked up powers of $\sqrt{iZ}^\dagger$.
|
|
If $b$ has non-operand neighbours after clearing the vertex operators on $a$ the vertex operators on $b$
|
|
can be cleared using the same algorithm which gives $o_b = I$ and $o_a = (\sqrt{iZ}^\dagger)^l = S^l \in \mathcal{Z}$
|
|
for some $l \in \{1, ..., 5\}$. Therefore Case 1 can now be applied.
|
|
|
|
If $o_a$ could not be cleared $o_b$ can be cleared using the same procedure and after
|
|
clearing $o_b$ one can retry to clear $o_a$.
|
|
|
|
In any case at least one vertex operator has been cleared. If both vertex operators have been
|
|
cleared Case 1 will be applied. If there is just one cleared vertex operator it
|
|
is the vertex with non-operand neighbours. Using this one can still apply a $CZ$: Without loss of generality
|
|
assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \cite{andersbriegel2005}:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
\ket{G} &= \left(\prod\limits_{o_i \in O} o_i\right) \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j}\right) \ket{+}_n \\
|
|
&= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) o_b (CZ_{a,b})^s \ket{+}_n \\
|
|
&= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) \ket{+}_{n-2} \otimes \left(o_b (CZ_{a,b})^s \ket{+}_2\right) \\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
As $o_b$ commutes with all operators but $CZ_{a,b}$ and $s \in \{0, 1\}$ indicates whether there is an edge between $a$ and $b$.
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
CZ_{a,b}\ket{G} &= CZ_{a,b}\left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) o_b (CZ_{a,b})^s \ket{+} \\
|
|
&= \left(\prod\limits_{o_i \in O \setminus \{o_b\}} o_i \right) \left(\prod\limits_{\{i,j\} \in E \setminus \{a,b\}} CZ_{i,j}\right) \ket{+}_{n-2} \otimes \left(CZ_{a,b} o_b (CZ_{a,b})^s \ket{+}_2\right) \\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
This allows to re-use the method in Case 2 to apply the $CZ$ while keeping the $o_a = I$.
|
|
|
|
As it has been shown how both $CZ$ and $C_L$ act on a graph state $\ket{G}$ and the resulting state
|
|
is a graph state as well this proofs constructively that
|
|
the graphical representation of a stabilizer state is indeed able to represent any stabilizer state.
|
|
If one wants to do computations using this formalism it is however also necessary to perform measurements.
|
|
|
|
\subsubsection{Measurements on Graph States}
|
|
|
|
Recalling \ref{ref:meas_stab} it is clear that one has to compute the commutator of
|
|
the observable $g_a = Z_a$ with the stabilizers to get the probability amplitudes
|
|
which is a quite expensive computation in theory. It is possible to simplify
|
|
the problem by pulling the observable behind the vertex operators. For this consider
|
|
the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
P_{a,s} \ket{\psi} &= P_{a,s} \left(\prod\limits_{o_i \in O} o_i \right) \ket{\bar{G}} \\
|
|
&= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right)P_a o_a \ket{\bar{G}} \\
|
|
&= \left(\prod\limits_{o_i \in O \setminus \{o_a\}}o_i \right) o_a o_a^\dagger P_a o_a \ket{\bar{G}} \\
|
|
&= \left(\prod\limits_{o_i \in O} o_i \right) o_a^\dagger P_a o_a \ket{\bar{G}} \\
|
|
&= \left(\prod\limits_{o_i \in O} o_i \right) \tilde{P}_{a,s} \ket{\bar{G}} \\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
This transformed projector has the important property that it still is a Pauli projector
|
|
as $o_a$ is a Clifford operator:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
\tilde{P}_{a,s} &= o_a^\dagger P_a o_a \\
|
|
&= o_a^\dagger \frac{I + (-1)^sZ_a}{2} o_a \\
|
|
&= \frac{I + (-1)^s o_a^\dagger Z_a o_a}{2} \\
|
|
&= \frac{I + (-1)^s \tilde{g}_a}{2} \\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is enough to study the measurements
|
|
of any Pauli operator on the vertex operator free graph states. The commutators of the observable
|
|
with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute
|
|
and it is easier to list the operators that anticommute:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
A_{\pm X_a} &= \left\{j \middle| \{j, a\} \in E\right\}\\
|
|
A_{\pm Y_a} &= \left\{j \middle| \{j, a\} \in E\right\} \cup \{a\} \\
|
|
A_{\pm Z_a} &= \{a\}\\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a (-X_a)$
|
|
is measured the result $s=0(1)$ is obtained with probability $1$ and $(V, E, O)$ is unchanged.
|
|
In any other case the results $s=1$ and $s=0$ have probability $\frac{1}{2}$ and both
|
|
graph and vertex operators have to be updated. Further it is clear that measurements of $-\tilde{g}_a$
|
|
and $\tilde{g}_a$ are related by just inverting the result $s$.
|
|
|
|
The calculations to obtain the transformation on graph and vertex operators are lengthy and follow
|
|
the scheme of Lemma \ref{lemma:M_a}. \cite[Section IV]{hein_eisert_briegel2008} also contains
|
|
the steps required to obtain the following results:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
U_{Z,s} &= \left(\prod\limits_{b \in n_a} Z_b^s\right) X_a^s H_a \\
|
|
U_{Y,s} &= \prod\limits_{b \in n_a} \sqrt{(-1)^{1-s} iZ_b} \sqrt{(-1)^{1-s} iZ_a}\\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}.
|
|
The second part makes sure that the qbit $a$ is diagonal in measured observable and has the correct eigenvalue.
|
|
When comparing with Lemma \ref{lemma:stab_measurement} in both cases $Y,Z$ the anticommuting stabilizer
|
|
$K_G^{(a)}$ is chosen. The graph is changed according to:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
E'_{Z} &= E \setminus \left\{\{i,a\} \middle| i \in V\right\}\\
|
|
E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} \middle| i \in V\right\}\\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
|
|
For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
U_{X,0} &= \sqrt{iY_b} \prod\limits_{c \in n_a \setminus n_b \setminus \{b\}} Z_c \\
|
|
U_{X,1} &= \sqrt{-iY_b} \prod\limits_{c \in n_b \setminus n_a \setminus \{a\}} Z_c \\
|
|
\end{aligned}
|
|
\end{equation}
|
|
\begin{equation}
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\begin{aligned}
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E'_{X} = E &\Delta (n_b \otimes n_a) \\
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& \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\
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& \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\
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& \setminus \left\{\{i,a\} \middle| i \in V\right\}\\
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\end{aligned}
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\end{equation}
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