814 lines
37 KiB
TeX
814 lines
37 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
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\cite{gottesman1997} for quantum error correction and is a useful tool to
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encode quantum information such that it is protected against noise. The
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prominent Shor code \cite{shor1995} is an example of a stabilizer code
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(although it was discovered before the stabilizer formalism was discovered), as
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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
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can be simulated in polynomial time on a classical machine
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\cite{gottesman2008}. This performance has since been improved to $n\log(n)$
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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 I, \pm X, \pm Y, \pm Z, \pm iI, \pm iX, \pm iY, \pm iZ\}
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\end{equation}
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with the matrix product is called the Pauli group \cite{andersbriegel2005}.
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\end{definition}
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The group property of $P$ can be verified easily. Note that the elements of $P$
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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 \cite{andersbriegel2005}.
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\end{definition}
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The group property of $P_n$ and the (anti-)commutator relationships follow
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directly from its definition 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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The discussion below follows the argumentation given in \cite{nielsen_chuang_2010}.
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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 and unitary $(S^{(i)})^2$ 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
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calculations, the generators of a group are introduced. Often it is enough to
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discuss the generator's properties in order to understand the properties of the
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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
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generators of G
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\begin{equation} \langle g_1, ..., g_m \rangle \equiv \langle g_i
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\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
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$g_i$ 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
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used as the required properties of a set of stabilizers that can be studied on
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its generators.
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\subsubsection{Stabilizer States}
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One important basic property of quantum mechanics is that hermitian operators
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have real eigenvalues and eigenspaces which are associated with these
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eigenvalues. Finding these eigenvalues and eigenvectors is what one calls
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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
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eigenvectors, i.e. they can be diagonalized simultaneously. This motivates and
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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 $S$ \cite{nielsen_chuang_2010}.
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\end{definition}
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It is clear that to show the stabilization property of $S$ the proof for the
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generators is sufficient, as all the generators forming an element in $S$ can
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be absorbed into $\ket{\psi}$. The dimension of $V_S$ is not immediately
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clear. One can however show that for a set of stabilizers $\langle S^{(i)}
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\rangle_{i=1, ..., n-m}$ the dimension $\dim V_S = 2^m$ \cite[Chapter
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10.5]{nielsen_chuang_2010}. This yields the following important result:
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\begin{theorem} \label{thm:unique_s_state} For a $n$ qbit system and
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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
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a trivial phase unique state $\ket{\psi}$ that is stabilized by $S$.
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Without proof. \end{theorem}
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In the following discussions for $n$ qbits a set $S = \langle S^{(i)}
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\rangle_{i=1,...,n}$ 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 $S
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= \langle S^{(i)} \rangle_{i=1,...,n}$ and a unitary transformation $U$ that
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describes the dynamics of the system, i.e.
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\begin{equation} \ket{\psi'} = U \ket{\psi} \end{equation}
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It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$
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anymore. There are however some statements that can still be made
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\cite{nielsen_chuang_2010}:
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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
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stabilizers are a subgroup of the multilocal Pauli operators. This does not
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hold true for an arbitrary $U$ but there exists a group for which $S'$ will be
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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
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group \cite{andersbriegel2005}.
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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 $\sqrt{-iX} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc} 1 & -i
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\\ -i & 1 \end{array}\right)$.
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Also $C_L$ is generated by a product of at most $5$ matrices
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$\sqrt{iZ}$, $\sqrt{-iX}$. }
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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 $U \in C_n$ $S'
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= U^\dagger S U$ is a set of $n$ independent stabilizers and $\ket{\psi'}$ is
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stabilized by $S'$ one can consider the dynamics of the stabilizers instead of
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the actual state. This is considerably more efficient as only $n$ stabilizers
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have to be modified, each being just the tensor product of $n$ Pauli matrices.
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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} \label{ref:meas_stab}
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Interestingly also measurements are dynamics covered by the stabilizers. When
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an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is
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measured 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
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probability $1$ 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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\cite{nielsen_chuang_2010}.
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If $g_a$ does not commute with all stabilizers the following lemma gives the
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result of the measurement.
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\begin{lemma}
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\label{lemma:stab_measurement} Let $J := \left\{ S^{(i)} \middle| [g_a,
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S^{(i)}] \neq 0\right\} \neq \{\}$ and $J^c := \left\{S^{(i)} \middle|
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S^{(i)} \notin J \right\}$. When measuring $\frac{I + (-1)^s g_a}{2} $ $s=1$
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and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing a $j
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\in J$ the new state $\ket{\psi'}$ is stabilized by \cite{nielsen_chuang_2010}
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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
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operators, $S^{(i)}$ 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 $S^{(j)}S^{(i)}$ $i,j \in J$,
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and by $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$
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\cite{nielsen_chuang_2010}.
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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
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are called vertex operator-free will be clear in the following section about
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graph states.
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\begin{definition} \label{def:graph} The tuple $(V, E)$ is called a graph iff
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$V$ is a set of vertices with $|V| = n \in \mathbb{N}$ elements. In the
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following $V = \{0, ..., n-1\}$ will be used. $E$ is the set of edges $E
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\subset \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$.
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For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is
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called the neighbourhood of $i$ \cite{hein_eisert_briegel2008}.
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\end{definition}
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This definition of a graph is way less general than the definition of a graph
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in graph theory. Using this definition will however allow to avoid an
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extensive list of constraints on the graph from graph theory that are implied
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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
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the state stabilized by $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$
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\cite{hein_eisert_briegel2008}.
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\end{definition}
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It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they
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commute is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$
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\begin{equation}
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\begin{aligned}
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K_G^{(a)} K_G^{(b)} &= X_a \left(\prod\limits_{i \in n_a} Z_i\right)
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X_b \left(\prod\limits_{j\in n_b} Z_j\right)\\
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&= X_a \left(\prod\limits_{i \in \setminus \{b\}} Z_i\right) Z_b
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X_b \left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right) Z_a\\
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&= X_a Z_b X_b Z_a
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\left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right)
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\left(\prod\limits_{i \in \setminus \{b\}} Z_i\right)\\
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&= -X_b Z_b X_a Z_a
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\left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right)
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\left(\prod\limits_{i \in \setminus \{b\}} Z_i\right)\\
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&= X_b Z_a X_a Z_b
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\left(\prod\limits_{j\in n_b\setminus \{b\}} Z_j\right)
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\left(\prod\limits_{i \in \setminus \{b\}} Z_i\right)\\
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&= K_G^{(b)} K_G^{(a)}.\\
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\end{aligned}
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\end{equation}
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This definition of a graph state might not seem to be straight forward but
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recalling Theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$ is
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unique. The following lemma will provide a way to construct this state from the
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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 \cite{hein_eisert_briegel2008}
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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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FIXME: This
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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 $X_i$: $X_i \ket{+} = +1 \ket{+}$. In the following
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discussion the direction $\prod\limits_{\{l,k\} \in E} :=
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\prod\limits_{\{l,k\} \in E, l < k}$ is introduced as the graph is
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undirected and edges must not be handled twice. Set $\ket{\tilde{G}} :=
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\left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$.
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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
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step is a bit tricky: A $Z_j$ can be absorbed into a $\ket{0}\bra{0}_j$
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giving no phase or into a $\ket{1}\bra{1}_j$ yielding a phase of $-1$. If
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there is no projector on $j$ the $Z_j$ can be commuted to the next
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projector. It is guaranteed that a projector on $j$ exists by the
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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{+} \\
|
|
& = +1 \ket{\tilde{G}}
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
The $\delta_{i,l} + \delta_{j,k}$ is either $0$ or $2$ by the definitions
|
|
of $K_G^{(i)}$ and $\ket{\tilde{G}}$.
|
|
\end{proof}
|
|
|
|
\subsubsection{Dynamics of the VOP-free Graph States}
|
|
|
|
This representation gives an immediate result to how the stabilizers $\langle
|
|
K_G^{(i)} \rangle_i$ change under the $CZ$ gate: When applying $CZ_{i,j}$ on $G
|
|
= (V, E)$ the edge $\{i,j\}$ is toggled, resulting in a multiplication of $Z_j$
|
|
to $K_G^{(i)}$ and $Z_i$ to $K_G^{(j)}$. Toggling edges is done by using the
|
|
symmetric set difference:
|
|
|
|
\begin{definition}
|
|
For two finite sets $A,B$ the symmetric set difference $\Delta$ is
|
|
defined as \cite{hein_eisert_briegel2008}
|
|
|
|
\begin{equation}
|
|
A \Delta B = (A \cup B) \setminus (A \cap B).
|
|
\end{equation}
|
|
\end{definition}
|
|
|
|
Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$.
|
|
Another transformation on the VOP-free graph states for a vertex $a \in V$ is
|
|
\cite{andersbriegel2005}
|
|
|
|
\begin{equation}
|
|
M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}.
|
|
\end{equation}
|
|
|
|
This transformation toggles the neighbourhood of $a$ which is an operation
|
|
that will be used later\cite{andersbriegel2005}.
|
|
|
|
\begin{lemma}
|
|
\label{lemma:M_a}
|
|
When applying $M_a$ to a state $\ket{\bar{G}}$ the new state
|
|
$\ket{\bar{G}'}$ is again a VOP-free graph state and the
|
|
graph is updated according to\cite{andersbriegel2005}:
|
|
\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 $M_a^2 \alpha K_G^{(a)} \Rightarrow [K_G^{(a)},
|
|
M_a] = 0$. Further $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$,
|
|
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 $K_{G'}^{(i)}$ such that $M_a
|
|
\ket{\bar{G}}$ is the $+1$ eigenstate of the new $K_{G'}^{(i)}$. Because
|
|
$\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$ it is clear that
|
|
$\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$. To construct the
|
|
$K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a =: \{j\} \cup F$ and $n_j =:
|
|
\{a\} \cup D$. Then follows:
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
S^{(j)} &= Z_j X_a X_j Z_a \prod\limits_{l \in D} Z_l\\
|
|
&= Z_j X_a X_j Z_a \left(\prod\limits_{l \in D} Z_l\right)
|
|
\left(\prod\limits_{l \in F}Z_l\right)
|
|
\left(\prod\limits_{l \in F}Z_l\right) \\
|
|
&= Z_j X_a X_j Z_a \left(\prod\limits_{l \in ((F\cup D) \setminus (F\cap D))} Z_L\right)
|
|
\left(\prod\limits_{l \in F}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 $\ket{G}$ \cite{andersbriegel2005}.
|
|
\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 $n^2$ $P$ 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 \cite{andersbriegel2005}.
|
|
\end{theorem}
|
|
\begin{proof} It is clear that $\forall a \in C_L$ a is a group isomorphism $P
|
|
\rightarrow P$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. 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 \cite{andersbriegel2005}.
|
|
|
|
\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\cite{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 are changed to $\langle c_j S^{(i)}
|
|
c_j^\dagger\rangle_i$. Using the definition of the graphical representation one
|
|
sees 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. Most of the discussion follows
|
|
the one given in \cite{andersbriegel2005} closely.
|
|
|
|
|
|
\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 one can
|
|
derive 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 $2\cdot24^2$: A factor of two for the
|
|
options with and withoit 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 (i.e. neighbours that are neither $a$ nor $b$) 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 $j$ \cite{andersbriegel2005}.
|
|
|
|
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, which 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 $a$, then 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}
|
|
|
|
This is adapted from \cite{andersbriegel2005}; measurement results and updating
|
|
the graph after a measurement is described in \cite{hein_eisert_briegel2008}.
|
|
|
|
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}
|
|
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Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is
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enough to study the measurements of any Pauli operator on the vertex operator
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free graph states. The commutators of the observable with the $K_G^{(i)}$ are
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|
quite easy to compute. Note that Pauli matrices either commute or anticommute
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|
and it is easier to list the operators that anticommute
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|
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|
\begin{equation}
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\begin{aligned}
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A_{\pm X_a} &= \left\{j \middle| \{j, a\} \in E\right\}\\
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A_{\pm Y_a} &= \left\{j \middle| \{j, a\} \in E\right\} \cup \{a\} \\
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A_{\pm Z_a} &= \{a\}.\\
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|
\end{aligned}
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|
\end{equation}
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|
|
|
This gives one immediate result: if a qbit $a$ is isolated and the operator
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|
$\tilde{g}_a = X_a (-X_a)$ is measured the result $s=0(1)$ is obtained with
|
|
probability $1$ and $(V, E, O)$ is unchanged. In any other case the results
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|
$s=1$ 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}
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|
U_{Z,s} &= \left(\prod\limits_{b \in n_a} Z_b^s\right) X_a^s H_a \\
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|
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}
|
|
\begin{aligned}
|
|
E'_{X} = E &\Delta (n_b \otimes n_a) \\
|
|
& \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\
|
|
& \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\
|
|
& \setminus \left\{\{i,a\} \middle| i \in V\right\}.\\
|
|
\end{aligned}
|
|
\end{equation}
|
|
|