From 75aeb837666b13f8ad73c6e3bcb223bb977f0be9 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Mon, 9 Mar 2020 17:32:07 +0100 Subject: [PATCH] some reformatting --- thesis/chapters/stabilizer.tex | 464 ++++++++++++++++++--------------- 1 file changed, 248 insertions(+), 216 deletions(-) diff --git a/thesis/chapters/stabilizer.tex b/thesis/chapters/stabilizer.tex index e6f7dbc..e2807ac 100644 --- a/thesis/chapters/stabilizer.tex +++ b/thesis/chapters/stabilizer.tex @@ -2,15 +2,17 @@ \section{The Stabilizer Formalism} -The stabilizer formalism was originally introduced by Gottesman \cite{gottesman1997} -for quantum error correction and is a useful tool to encode quantum information -such that it is protected against noise. The prominent Shor code \cite{shor1995} -is an example of a stabilizer code (although it was discovered before the stabilizer -formalism was discovered), as are the 3-qbit bit-flip and phase-flip codes. +The stabilizer formalism was originally introduced by Gottesman +\cite{gottesman1997} for quantum error correction and is a useful tool to +encode quantum information such that it is protected against noise. The +prominent Shor code \cite{shor1995} is an example of a stabilizer code +(although it was discovered before the stabilizer formalism was discovered), as +are the 3-qbit bit-flip and phase-flip codes. -It was only later that Gottesman and Knill discovered that stabilizer states can -be simulated in polynomial time on a classical machine \cite{gottesman2008}. This -performance has since been improved to $n\log(n)$ time on average \cite{andersbriegel2005}. +It was only later that Gottesman and Knill discovered that stabilizer states +can be simulated in polynomial time on a classical machine +\cite{gottesman2008}. This performance has since been improved to $n\log(n)$ +time on average \cite{andersbriegel2005}. \subsection{Stabilizers and Stabilizer States} @@ -24,8 +26,8 @@ performance has since been improved to $n\log(n)$ time on average \cite{andersbr with the matrix product is called the Pauli group \cite{andersbriegel2005}. \end{definition} -The group property of $P$ can be verified easily. Note that -the elements of $P$ either commute or anticommute. +The group property of $P$ can be verified easily. Note that the elements of $P$ +either commute or anticommute. \begin{definition} For $n$ qbits @@ -37,8 +39,8 @@ the elements of $P$ either commute or anticommute. is called the multilocal Pauli group on $n$ qbits \cite{andersbriegel2005}. \end{definition} -The group property of $P_n$ and the (anti-)commutator relationships follow directly from its definition -via the tensor product. +The group property of $P_n$ and the (anti-)commutator relationships follow +directly from its definition via the tensor product. %Further are $p \in P_n$ hermitian and have the eigenvalues $\pm 1$ for %$p \neq \pm I$, $+1$ for $p = I$ and $-1$ for $p = -I$. @@ -81,31 +83,35 @@ The discussion below follows the argumentation given in \cite{nielsen_chuang_201 \end{proof} -Considering all the elements of a group can be impractical for some calculations, -the generators of a group are introduced. Often it is enough to discuss the generator's -properties in order to understand the properties of the group. +Considering all the elements of a group can be impractical for some +calculations, the generators of a group are introduced. Often it is enough to +discuss the generator's properties in order to understand the properties of the +group. \begin{definition} - For a finite group $G$ and some $m \in \mathbb{N}$ one denotes the generators - of G + For a finite group $G$ and some $m \in \mathbb{N}$ one denotes the + generators of G - \begin{equation} \langle g_1, ..., g_m \rangle \equiv \langle g_i \rangle_{i=1,...,m}\end{equation} + \begin{equation} \langle g_1, ..., g_m \rangle \equiv \langle g_i + \rangle_{i=1,...,m}\end{equation} - where $g_i \in G$, every element in $G$ can be written as a product of the $g_i$ - and $m$ is the smallest integer for which these statements hold. + where $g_i \in G$, every element in $G$ can be written as a product of the + $g_i$ and $m$ is the smallest integer for which these statements hold. \end{definition} -In the following discussions $\langle S^{(i)} \rangle_{i=0, ..., n-1}$ will be used as -the required properties of a set of stabilizers that can be studied on its -generators. +In the following discussions $\langle S^{(i)} \rangle_{i=0, ..., n-1}$ will be +used as the required properties of a set of stabilizers that can be studied on +its generators. \subsubsection{Stabilizer States} -One important basic property of quantum mechanics is that hermitian operators have real eigenvalues -and eigenspaces which are associated with these eigenvalues. Finding these eigenvalues and eigenvectors -is what one calls solving a quantum mechanical system. One of the most fundamental insights of -quantum mechanics is that commuting operators have a common set of eigenvectors, i.e. they -can be diagonalized simultaneously. This motivates and justifies the following definition. +One important basic property of quantum mechanics is that hermitian operators +have real eigenvalues and eigenspaces which are associated with these +eigenvalues. Finding these eigenvalues and eigenvectors is what one calls +solving a quantum mechanical system. One of the most fundamental insights of +quantum mechanics is that commuting operators have a common set of +eigenvectors, i.e. they can be diagonalized simultaneously. This motivates and +justifies the following definition. \begin{definition} For a set of stabilizers $S$ the vector space @@ -118,38 +124,35 @@ can be diagonalized simultaneously. This motivates and justifies the following d $\ket{\psi}$ is stabilized by $S$ \cite{nielsen_chuang_2010}. \end{definition} -It is clear that to show the stabilization property of -$S$ the proof for the generators is sufficient, -as all the generators forming an element in $S$ can be absorbed into $\ket{\psi}$. -The dimension of $V_S$ is not immediately clear. One can however show that -for a set of stabilizers $\langle S^{(i)} \rangle_{i=1, ..., n-m}$ the dimension -$\dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important -result: +It is clear that to show the stabilization property of $S$ the proof for the +generators is sufficient, as all the generators forming an element in $S$ can +be absorbed into $\ket{\psi}$. The dimension of $V_S$ is not immediately +clear. One can however show that for a set of stabilizers $\langle S^{(i)} +\rangle_{i=1, ..., n-m}$ the dimension $\dim V_S = 2^m$ \cite[Chapter +10.5]{nielsen_chuang_2010}. This yields the following important result: -\begin{theorem} - \label{thm:unique_s_state} - For a $n$ qbit system and stabilizers $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ the stabilizer - space $V_S$ has $\dim V_S = 1$, in particular there exists an up to a trivial phase unique - state $\ket{\psi}$ that is stabilized by $S$. +\begin{theorem} \label{thm:unique_s_state} For a $n$ qbit system and + stabilizers $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ the stabilizer + space $V_S$ has $\dim V_S = 1$, in particular there exists an up to + a trivial phase unique state $\ket{\psi}$ that is stabilized by $S$. - Without proof. -\end{theorem} + Without proof. \end{theorem} -In the following discussions for $n$ qbits a set $S = \langle S^{(i)} \rangle_{i=1,...,n}$ -of $n$ independent stabilizers will be assumed. +In the following discussions for $n$ qbits a set $S = \langle S^{(i)} +\rangle_{i=1,...,n}$ of $n$ independent stabilizers will be assumed. \subsubsection{Dynamics of Stabilizer States} -Consider a $n$ qbit state $\ket{\psi}$ that is the $+1$ eigenstate of $S = \langle S^{(i)} \rangle_{i=1,...,n}$ -and a unitary transformation $U$ that describes the dynamics of the system, i.e. +Consider a $n$ qbit state $\ket{\psi}$ that is the $+1$ eigenstate of $S += \langle S^{(i)} \rangle_{i=1,...,n}$ and a unitary transformation $U$ that +describes the dynamics of the system, i.e. -\begin{equation} - \ket{\psi'} = U \ket{\psi} -\end{equation} +\begin{equation} \ket{\psi'} = U \ket{\psi} \end{equation} -It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$ anymore. There are -however some statements that can still be made \cite{nielsen_chuang_2010}: +It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$ +anymore. There are however some statements that can still be made +\cite{nielsen_chuang_2010}: \begin{equation} \begin{aligned} @@ -161,9 +164,10 @@ however some statements that can still be made \cite{nielsen_chuang_2010}: \end{aligned} \end{equation} -Note that in Definition \ref{def:stabilizer} it has been demanded that stabilizers are a -subgroup of the multilocal Pauli operators. This does not hold true for an arbitrary -$U$ but there exists a group for which $S'$ will be a set of stabilizers. +Note that in Definition \ref{def:stabilizer} it has been demanded that +stabilizers are a subgroup of the multilocal Pauli operators. This does not +hold true for an arbitrary $U$ but there exists a group for which $S'$ will be +a set of stabilizers. \begin{definition} For $n$ qbits @@ -171,7 +175,8 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers. C_n := \left\{U \in SU(2^n) \middle| \forall p \in P_n: UpU^\dagger \in P_n \right\} \end{equation} - is called the Clifford group. $C_1 =: C_L$ is called the local Clifford group \cite{andersbriegel2005}. + is called the Clifford group. $C_1 =: C_L$ is called the local Clifford + group \cite{andersbriegel2005}. \end{definition} \begin{theorem} @@ -179,10 +184,11 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers. \begin{enumerate} \item{$C_L$ can be generated using only $H$ and $S$.} \item{$C_L$ can be generated from $\sqrt{iZ} = \exp(\frac{i\pi}{4}) S^\dagger$ - and $\sqrt{-iX} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc} 1 & -i \\ -i & 1 \end{array}\right)$. + and $\sqrt{-iX} = \frac{1}{\sqrt{2}} \left(\begin{array}{cc} 1 & -i + \\ -i & 1 \end{array}\right)$. - Also $C_L$ is generated by a product of at most $5$ matrices $\sqrt{iZ}$, $\sqrt{-iX}$. - } + Also $C_L$ is generated by a product of at most $5$ matrices + $\sqrt{iZ}$, $\sqrt{-iX}$. } \item{$C_n$ can be generated using $C_L$ and $CZ$ or $CX$.} \end{enumerate} \end{theorem} @@ -202,26 +208,26 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers. \end{enumerate} \end{proof} -This is quite an important result: As under a transformation $U \in C_n$ $S' = U^\dagger S U$ is a set of -$n$ independent stabilizers and $\ket{\psi'}$ is stabilized by $S'$ one can consider -the dynamics of the stabilizers instead of the actual state. This is considerably more -efficient as only $n$ stabilizers have to be modified, each being just the tensor -product of $n$ Pauli matrices. This has led to the simulation using stabilizer tableaux +This is quite an important result: As under a transformation $U \in C_n$ $S' += U^\dagger S U$ is a set of $n$ independent stabilizers and $\ket{\psi'}$ is +stabilized by $S'$ one can consider the dynamics of the stabilizers instead of +the actual state. This is considerably more efficient as only $n$ stabilizers +have to be modified, each being just the tensor product of $n$ Pauli matrices. +This has led to the simulation using stabilizer tableaux \cite{gottesman_aaronson2008}. -\subsubsection{Measurements on Stabilizer States} -\label{ref:meas_stab} +\subsubsection{Measurements on Stabilizer States} \label{ref:meas_stab} -Interestingly also measurements are dynamics covered by the stabilizers. -When an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is measured -one has to consider the projector +Interestingly also measurements are dynamics covered by the stabilizers. When +an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is +measured one has to consider the projector -\begin{equation} - P_{g_a,s} = \frac{I + (-1)^s g_a}{2}. +\begin{equation} +P_{g_a,s} = \frac{I + (-1)^s g_a}{2}. \end{equation} -If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with probability $1$ -and the stabilizers are left unchanged: +If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with +probability $1$ and the stabilizers are left unchanged: \begin{equation} \begin{aligned} @@ -232,26 +238,26 @@ and the stabilizers are left unchanged: \end{aligned} \end{equation} -As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$ \cite{nielsen_chuang_2010}. +As the state that is stabilized by $S$ is unique $\ket{\psi'} = \ket{\psi}$ +\cite{nielsen_chuang_2010}. -If $g_a$ does not commute with all stabilizers the following lemma gives -the result of the measurement. +If $g_a$ does not commute with all stabilizers the following lemma gives the +result of the measurement. \begin{lemma} - \label{lemma:stab_measurement} - Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$ and - $J^c := \left\{S^{(i)} \middle| S^{(i)} \notin J \right\}$. When measuring - $\frac{I + (-1)^s g_a}{2} $ - $s=1$ and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing - a $j \in J$ the new state $\ket{\psi'}$ is stabilized by \cite{nielsen_chuang_2010} + \label{lemma:stab_measurement} Let $J := \left\{ S^{(i)} \middle| [g_a, + S^{(i)}] \neq 0\right\} \neq \{\}$ and $J^c := \left\{S^{(i)} \middle| +S^{(i)} \notin J \right\}$. When measuring $\frac{I + (-1)^s g_a}{2} $ $s=1$ +and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing a $j +\in J$ the new state $\ket{\psi'}$ is stabilized by \cite{nielsen_chuang_2010} \begin{equation} \langle \{(-1)^s g_a\} \cup \left\{S^{(i)} S^{(j)} \middle| S^{(i)} \in J \setminus \{S^{(j)}\} \right\} \cup J^c \rangle. \end{equation} \end{lemma} \begin{proof} - As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators, - $S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then + As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli + operators, $S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then \begin{equation} \begin{aligned} @@ -278,41 +284,44 @@ the result of the measurement. \notag \end{equation} - The state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, and by - $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$ \cite{nielsen_chuang_2010}. + The state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, + and by $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$ + \cite{nielsen_chuang_2010}. \end{proof} \subsection{The VOP-free Graph States} \subsubsection{VOP-free Graph States} -This section will discuss the vertex operator (VOP)-free graph states. Why they are called -vertex operator-free will be clear in the following section about graph states. +This section will discuss the vertex operator (VOP)-free graph states. Why they +are called vertex operator-free will be clear in the following section about +graph states. -\begin{definition} - \label{def:graph} - The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$ elements. - In the following $V = \{0, ..., n-1\}$ will be used. - $E$ is the set of edges $E \subset \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$. +\begin{definition} \label{def:graph} The tuple $(V, E)$ is called a graph iff + $V$ is a set of vertices with $|V| = n \in \mathbb{N}$ elements. In the + following $V = \{0, ..., n-1\}$ will be used. $E$ is the set of edges $E + \subset \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$. - For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is called the neighbourhood - of $i$ \cite{hein_eisert_briegel2008}. + For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is +called the neighbourhood of $i$ \cite{hein_eisert_briegel2008}. \end{definition} -This definition of a graph is way less general than the definition of a graph in graph theory. -Using this definition will however allow to avoid an extensive list of constraints on the -graph from graph theory that are implied in this definition. +This definition of a graph is way less general than the definition of a graph +in graph theory. Using this definition will however allow to avoid an +extensive list of constraints on the graph from graph theory that are implied +in this definition. \begin{definition} For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are \begin{equation} K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_j \end{equation} - for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is the state stabilized by - $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$ \cite{hein_eisert_briegel2008}. + for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is + the state stabilized by $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$ + \cite{hein_eisert_briegel2008}. \end{definition} -It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute -is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$ +It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they +commute is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$ \begin{equation} \begin{aligned} @@ -334,10 +343,10 @@ is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$ \end{equation} -This definition of a graph state might not seem to be straight forward -but recalling Theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$ -is unique. The following lemma will provide a way to construct this state -from the graph. +This definition of a graph state might not seem to be straight forward but +recalling Theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$ is +unique. The following lemma will provide a way to construct this state from the +graph. \begin{lemma} For a graph $G = (V, E)$ the associated state $\ket{\bar{G}}$ is @@ -355,10 +364,11 @@ from the graph. Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. - Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$. - In the following discussion the direction $\prod\limits_{\{l,k\} \in E} := \prod\limits_{\{l,k\} \in E, l < k}$ - is introduced as the graph is undirected and edges must not be handled twice. - Set $\ket{\tilde{G}} := \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$. + Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$. In the following + discussion the direction $\prod\limits_{\{l,k\} \in E} := + \prod\limits_{\{l,k\} \in E, l < k}$ is introduced as the graph is + undirected and edges must not be handled twice. Set $\ket{\tilde{G}} := + \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$. \begin{equation} \begin{aligned} @@ -371,10 +381,12 @@ from the graph. \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{+} \\ \end{aligned} \end{equation} - As $X,Z$ anticommute. $X_i$ can now be absorbed into $\ket{+}$. The next step is a bit tricky: - 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 - a phase of $-1$. If there is no projector on $j$ the $Z_j$ can be commuted to the next projector. - It is guaranteed that a projector on $j$ exists by the definition of $\ket{\tilde{G}}$. + As $X,Z$ anticommute. $X_i$ can now be absorbed into $\ket{+}$. The next + step is a bit tricky: 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 a phase of $-1$. If + there is no projector on $j$ the $Z_j$ can be commuted to the next + projector. It is guaranteed that a projector on $j$ exists by the + definition of $\ket{\tilde{G}}$. \begin{equation} \begin{aligned} @@ -385,15 +397,17 @@ from the graph. \end{aligned} \end{equation} - The $\delta_{i,l} + \delta_{j,k}$ is either $0$ or $2$ by the definitions of $K_G^{(i)}$ and $\ket{\tilde{G}}$. + 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: +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 @@ -405,7 +419,8 @@ is done by using the symmetric set difference: \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} +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}. @@ -428,10 +443,10 @@ that will be used later\cite{andersbriegel2005}. \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 $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$, + $\ket{\bar{G}'}$ 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} @@ -453,11 +468,12 @@ that will be used later\cite{andersbriegel2005}. \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: + 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} @@ -479,34 +495,36 @@ that will be used later\cite{andersbriegel2005}. = 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. + 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. +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 $o_i \in C_L$. + 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: @@ -517,36 +535,39 @@ clear that both $C_L$ and $CZ$ can be applied to a general graph state. $o_i$ are called the vertex operators of $\ket{G}$ \cite{andersbriegel2005}. \end{definition} -Recalling the dynamics of stabilizer states the following relation follows immediately: +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: +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. +\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} + 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}. +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) @@ -561,12 +582,14 @@ $C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific \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 +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} @@ -575,30 +598,33 @@ one sees that just the vertex operators are changed and the new vertex operators \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. +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 $E' = E \Delta \left\{\{a,b\}\right\}$. +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. +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 +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{:}\\ @@ -607,10 +633,9 @@ 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: +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 @@ -623,11 +648,13 @@ to the following theorem: 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: +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 @@ -668,7 +695,8 @@ non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \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 $b$. +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} @@ -707,8 +735,8 @@ operators. For this consider the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$ \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 +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} @@ -719,9 +747,10 @@ as $o_a$ is a Clifford operator \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 +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} @@ -732,15 +761,17 @@ and it is easier to list the operators that anticommute \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 $s$. +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 +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} @@ -749,10 +780,11 @@ the steps required to obtain the following results \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 +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}