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thesis/chapters/stabilizer.tex
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@ -2,15 +2,17 @@
\section{The Stabilizer Formalism} \section{The Stabilizer Formalism}
The stabilizer formalism was originally introduced by Gottesman \cite{gottesman1997} The stabilizer formalism was originally introduced by Gottesman
for quantum error correction and is a useful tool to encode quantum information \cite{gottesman1997} for quantum error correction and is a useful tool to
such that it is protected against noise. The prominent Shor code \cite{shor1995} encode quantum information such that it is protected against noise. The
is an example of a stabilizer code (although it was discovered before the stabilizer prominent Shor code \cite{shor1995} is an example of a stabilizer code
formalism was discovered), as are the 3-qbit bit-flip and phase-flip codes. (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 It was only later that Gottesman and Knill discovered that stabilizer states
be simulated in polynomial time on a classical machine \cite{gottesman2008}. This can be simulated in polynomial time on a classical machine
performance has since been improved to $n\log(n)$ time on average \cite{andersbriegel2005}. \cite{gottesman2008}. This performance has since been improved to $n\log(n)$
time on average \cite{andersbriegel2005}.
\subsection{Stabilizers and Stabilizer States} \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}. with the matrix product is called the Pauli group \cite{andersbriegel2005}.
\end{definition} \end{definition}
The group property of $P$ can be verified easily. Note that The group property of $P$ can be verified easily. Note that the elements of $P$
the elements of $P$ either commute or anticommute. either commute or anticommute.
\begin{definition} \begin{definition}
For $n$ qbits 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}. is called the multilocal Pauli group on $n$ qbits \cite{andersbriegel2005}.
\end{definition} \end{definition}
The group property of $P_n$ and the (anti-)commutator relationships follow directly from its definition The group property of $P_n$ and the (anti-)commutator relationships follow
via the tensor product. directly from its definition via the tensor product.
%Further are $p \in P_n$ hermitian and have the eigenvalues $\pm 1$ for %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 $p = -I$. %$p \neq \pm I$, $+1$ 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} \end{proof}
Considering all the elements of a group can be impractical for some calculations, Considering all the elements of a group can be impractical for some
the generators of a group are introduced. Often it is enough to discuss the generator's calculations, the generators of a group are introduced. Often it is enough to
properties in order to understand the properties of the group. discuss the generator's properties in order to understand the properties of the
group.
\begin{definition} \begin{definition}
For a finite group $G$ and some $m \in \mathbb{N}$ one denotes the generators For a finite group $G$ and some $m \in \mathbb{N}$ one denotes the
of G 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 $g_i$ where $g_i \in G$, every element in $G$ can be written as a product of the
and $m$ is the smallest integer for which these statements hold. $g_i$ and $m$ is the smallest integer for which these statements hold.
\end{definition} \end{definition}
In the following discussions $\langle S^{(i)} \rangle_{i=0, ..., n-1}$ will be used as In the following discussions $\langle S^{(i)} \rangle_{i=0, ..., n-1}$ will be
the required properties of a set of stabilizers that can be studied on its used as the required properties of a set of stabilizers that can be studied on
generators. its generators.
\subsubsection{Stabilizer States} \subsubsection{Stabilizer States}
One important basic property of quantum mechanics is that hermitian operators have real eigenvalues One important basic property of quantum mechanics is that hermitian operators
and eigenspaces which are associated with these eigenvalues. Finding these eigenvalues and eigenvectors have real eigenvalues and eigenspaces which are associated with these
is what one calls solving a quantum mechanical system. One of the most fundamental insights of eigenvalues. Finding these eigenvalues and eigenvectors is what one calls
quantum mechanics is that commuting operators have a common set of eigenvectors, i.e. they solving a quantum mechanical system. One of the most fundamental insights of
can be diagonalized simultaneously. This motivates and justifies the following definition. 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} \begin{definition}
For a set of stabilizers $S$ the vector space 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}. $\ket{\psi}$ is stabilized by $S$ \cite{nielsen_chuang_2010}.
\end{definition} \end{definition}
It is clear that to show the stabilization property of It is clear that to show the stabilization property of $S$ the proof for the
$S$ the proof for the generators is sufficient, generators is sufficient, as all the generators forming an element in $S$ can
as all the generators forming an element in $S$ can be absorbed into $\ket{\psi}$. be absorbed into $\ket{\psi}$. The dimension of $V_S$ is not immediately
The dimension of $V_S$ is not immediately clear. One can however show that clear. One can however show that for a set of stabilizers $\langle S^{(i)}
for a set of stabilizers $\langle S^{(i)} \rangle_{i=1, ..., n-m}$ the dimension \rangle_{i=1, ..., n-m}$ the dimension $\dim V_S = 2^m$ \cite[Chapter
$\dim V_S = 2^m$ \cite[Chapter 10.5]{nielsen_chuang_2010}. This yields the following important 10.5]{nielsen_chuang_2010}. This yields the following important result:
result:
\begin{theorem} \begin{theorem} \label{thm:unique_s_state} For a $n$ qbit system and
\label{thm:unique_s_state} stabilizers $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ the stabilizer
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
space $V_S$ has $\dim V_S = 1$, in particular there exists an up to a trivial phase unique a trivial phase unique state $\ket{\psi}$ that is stabilized by $S$.
state $\ket{\psi}$ that is stabilized by $S$.
Without proof. Without proof. \end{theorem}
\end{theorem}
In the following discussions for $n$ qbits a set $S = \langle S^{(i)} \rangle_{i=1,...,n}$ In the following discussions for $n$ qbits a set $S = \langle S^{(i)}
of $n$ independent stabilizers will be assumed. \rangle_{i=1,...,n}$ of $n$ independent stabilizers will be assumed.
\subsubsection{Dynamics of Stabilizer States} \subsubsection{Dynamics of Stabilizer States}
Consider a $n$ qbit state $\ket{\psi}$ that is the $+1$ eigenstate of $S = \langle S^{(i)} \rangle_{i=1,...,n}$ Consider a $n$ qbit state $\ket{\psi}$ that is the $+1$ eigenstate of $S
and a unitary transformation $U$ that describes the dynamics of the system, i.e. = \langle S^{(i)} \rangle_{i=1,...,n}$ and a unitary transformation $U$ that
describes the dynamics of the system, i.e.
\begin{equation} \begin{equation} \ket{\psi'} = U \ket{\psi} \end{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 It is clear that in general $\ket{\psi'}$ will not be stabilized by $S$
however some statements that can still be made \cite{nielsen_chuang_2010}: anymore. There are however some statements that can still be made
\cite{nielsen_chuang_2010}:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -161,9 +164,10 @@ however some statements that can still be made \cite{nielsen_chuang_2010}:
\end{aligned} \end{aligned}
\end{equation} \end{equation}
Note that in Definition \ref{def:stabilizer} it has been demanded that stabilizers are a Note that in Definition \ref{def:stabilizer} it has been demanded that
subgroup of the multilocal Pauli operators. This does not hold true for an arbitrary stabilizers are a subgroup of the multilocal Pauli operators. This does not
$U$ but there exists a group for which $S'$ will be a set of stabilizers. hold true for an arbitrary $U$ but there exists a group for which $S'$ will be
a set of stabilizers.
\begin{definition} \begin{definition}
For $n$ qbits 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\} C_n := \left\{U \in SU(2^n) \middle| \forall p \in P_n: UpU^\dagger \in P_n \right\}
\end{equation} \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} \end{definition}
\begin{theorem} \begin{theorem}
@ -179,10 +184,11 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers.
\begin{enumerate} \begin{enumerate}
\item{$C_L$ can be generated using only $H$ and $S$.} \item{$C_L$ can be generated using only $H$ and $S$.}
\item{$C_L$ can be generated from $\sqrt{iZ} = \exp(\frac{i\pi}{4}) S^\dagger$ \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 $CX$.} \item{$C_n$ can be generated using $C_L$ and $CZ$ or $CX$.}
\end{enumerate} \end{enumerate}
\end{theorem} \end{theorem}
@ -202,26 +208,26 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers.
\end{enumerate} \end{enumerate}
\end{proof} \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 This is quite an important result: As under a transformation $U \in C_n$ $S'
$n$ independent stabilizers and $\ket{\psi'}$ is stabilized by $S'$ one can consider = U^\dagger S U$ is a set of $n$ independent stabilizers and $\ket{\psi'}$ is
the dynamics of the stabilizers instead of the actual state. This is considerably more stabilized by $S'$ one can consider the dynamics of the stabilizers instead of
efficient as only $n$ stabilizers have to be modified, each being just the tensor the actual state. This is considerably more efficient as only $n$ stabilizers
product of $n$ Pauli matrices. This has led to the simulation using stabilizer tableaux 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}. \cite{gottesman_aaronson2008}.
\subsubsection{Measurements on Stabilizer States} \subsubsection{Measurements on Stabilizer States} \label{ref:meas_stab}
\label{ref:meas_stab}
Interestingly also measurements are dynamics covered by the stabilizers. Interestingly also measurements are dynamics covered by the stabilizers. When
When an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is measured an observable $g_a \in \{\pm X_a, \pm Y_a, \pm Z_a\}$ acting on qbit $a$ is
one has to consider the projector measured one has to consider the projector
\begin{equation} \begin{equation}
P_{g_a,s} = \frac{I + (-1)^s g_a}{2}. P_{g_a,s} = \frac{I + (-1)^s g_a}{2}.
\end{equation} \end{equation}
If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with probability $1$ If now $g_a$ commutes with all $S^{(i)}$ a result of $s=0$ is measured with
and the stabilizers are left unchanged: probability $1$ and the stabilizers are left unchanged:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -232,26 +238,26 @@ and the stabilizers are left unchanged:
\end{aligned} \end{aligned}
\end{equation} \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 If $g_a$ does not commute with all stabilizers the following lemma gives the
the result of the measurement. result of the measurement.
\begin{lemma} \begin{lemma}
\label{lemma:stab_measurement} \label{lemma:stab_measurement} Let $J := \left\{ S^{(i)} \middle| [g_a,
Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$ and S^{(i)}] \neq 0\right\} \neq \{\}$ and $J^c := \left\{S^{(i)} \middle|
$J^c := \left\{S^{(i)} \middle| S^{(i)} \notin J \right\}$. When measuring S^{(i)} \notin J \right\}$. When measuring $\frac{I + (-1)^s g_a}{2} $ $s=1$
$\frac{I + (-1)^s g_a}{2} $ and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing a $j
$s=1$ and $s=0$ are obtained with probability $\frac{1}{2}$ and after choosing \in J$ the new state $\ket{\psi'}$ is stabilized by \cite{nielsen_chuang_2010}
a $j \in J$ the new state $\ket{\psi'}$ is stabilized by \cite{nielsen_chuang_2010}
\begin{equation} \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. \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{equation}
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli operators, As $g_a$ is a Pauli operator and $S^{(i)} \in J$ are multi-local Pauli
$S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then operators, $S^{(i)}$ and $g_a$ anticommute. Choose a $S^{(j)} \in J$. Then
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -278,41 +284,44 @@ the result of the measurement.
\notag \notag
\end{equation} \end{equation}
The state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$, and by The state after measurement is stabilized by $S^{(j)}S^{(i)}$ $i,j \in J$,
$S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$ \cite{nielsen_chuang_2010}. and by $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$
\cite{nielsen_chuang_2010}.
\end{proof} \end{proof}
\subsection{The VOP-free Graph States} \subsection{The VOP-free Graph States}
\subsubsection{VOP-free Graph States} \subsubsection{VOP-free Graph States}
This section will discuss the vertex operator (VOP)-free graph states. Why they are called This section will discuss the vertex operator (VOP)-free graph states. Why they
vertex operator-free will be clear in the following section about graph states. are called vertex operator-free will be clear in the following section about
graph states.
\begin{definition} \begin{definition} \label{def:graph} The tuple $(V, E)$ is called a graph iff
\label{def:graph} $V$ is a set of vertices with $|V| = n \in \mathbb{N}$ elements. In the
The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$ elements. following $V = \{0, ..., n-1\}$ will be used. $E$ is the set of edges $E
In the following $V = \{0, ..., n-1\}$ will be used. \subset \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$.
$E$ 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 For a vertex $i$ $n_i := \left\{j \in V \middle| \{i, j\} \in E\right\}$ is
of $i$ \cite{hein_eisert_briegel2008}. called the neighbourhood of $i$ \cite{hein_eisert_briegel2008}.
\end{definition} \end{definition}
This definition of a graph is way less general than the definition of a graph in graph theory. This definition of a graph is way less general than the definition of a graph
Using this definition will however allow to avoid an extensive list of constraints on the in graph theory. Using this definition will however allow to avoid an
graph from graph theory that are implied in this definition. extensive list of constraints on the graph from graph theory that are implied
in this definition.
\begin{definition} \begin{definition}
For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are For a graph $G = (V = \{0, ..., n-1\}, E)$ the associated stabilizers are
\begin{equation} \begin{equation}
K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_j K_G^{(i)} := X_i \prod\limits_{\{i,j\} \in E} Z_j
\end{equation} \end{equation}
for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is the state stabilized by for all $i \in V$. The vertex operator free graph state $\ket{\bar{G}}$ is
$\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$ \cite{hein_eisert_briegel2008}. the state stabilized by $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$
\cite{hein_eisert_briegel2008}.
\end{definition} \end{definition}
It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they
is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$ commute is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -334,10 +343,10 @@ is clear if $\{a,b\} \notin E$. If $\{a, b\} \in E$
\end{equation} \end{equation}
This definition of a graph state might not seem to be straight forward This definition of a graph state might not seem to be straight forward but
but recalling Theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$ recalling Theorem \ref{thm:unique_s_state} it is clear that $\ket{\bar{G}}$ is
is unique. The following lemma will provide a way to construct this state unique. The following lemma will provide a way to construct this state from the
from the graph. graph.
\begin{lemma} \begin{lemma}
For a graph $G = (V, E)$ the associated state $\ket{\bar{G}}$ is 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. Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before.
Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$. Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$. In the following
In the following discussion the direction $\prod\limits_{\{l,k\} \in E} := \prod\limits_{\{l,k\} \in E, l < k}$ discussion the direction $\prod\limits_{\{l,k\} \in E} :=
is introduced as the graph is undirected and edges must not be handled twice. \prod\limits_{\{l,k\} \in E, l < k}$ is introduced as the graph is
Set $\ket{\tilde{G}} := \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$. 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{equation}
\begin{aligned} \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{+} \\ \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{aligned}
\end{equation} \end{equation}
As $X,Z$ anticommute. $X_i$ can now be absorbed into $\ket{+}$. The next step is a bit tricky: As $X,Z$ anticommute. $X_i$ can now be absorbed into $\ket{+}$. The next
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 step is a bit tricky: A $Z_j$ can be absorbed into a $\ket{0}\bra{0}_j$
a phase of $-1$. If there is no projector on $j$ the $Z_j$ can be commuted to the next projector. giving no phase or into a $\ket{1}\bra{1}_j$ yielding a phase of $-1$. If
It is guaranteed that a projector on $j$ exists by the definition of $\ket{\tilde{G}}$. 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{equation}
\begin{aligned} \begin{aligned}
@ -385,15 +397,17 @@ from the graph.
\end{aligned} \end{aligned}
\end{equation} \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} \end{proof}
\subsubsection{Dynamics of the VOP-free Graph States} \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 This representation gives an immediate result to how the stabilizers $\langle
under the $CZ$ gate: When applying $CZ_{i,j}$ on $G = (V, E)$ the edge $\{i,j\}$ is toggled, K_G^{(i)} \rangle_i$ change under the $CZ$ gate: When applying $CZ_{i,j}$ on $G
resulting in a multiplication of $Z_j$ to $K_G^{(i)}$ and $Z_i$ to $K_G^{(j)}$. Toggling edges = (V, E)$ the edge $\{i,j\}$ is toggled, resulting in a multiplication of $Z_j$
is done by using the symmetric set difference: to $K_G^{(i)}$ and $Z_i$ to $K_G^{(j)}$. Toggling edges is done by using the
symmetric set difference:
\begin{definition} \begin{definition}
For two finite sets $A,B$ the symmetric set difference $\Delta$ is 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} \end{definition}
Toggling an edge $\{i, j\}$ updates $E' = E \Delta \left\{\{i,j\}\right\}$. 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} \begin{equation}
M_a := \sqrt{-iX_a} \prod\limits_{j\in n_a} \sqrt{iZ_j}. 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{equation}
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
$\ket{\bar{G}'}$ is stabilized by $\langle M_a K_G^{(i)} M_a^\dagger \rangle_i$, so it is sufficient $\ket{\bar{G}'}$ is stabilized by $\langle M_a K_G^{(i)} M_a^\dagger
to study how the $ K_G^{(i)}$ change under $M_a$. \rangle_i$, so it is sufficient to study how the $ K_G^{(i)}$ change under
At first note that $M_a^2 \alpha K_G^{(a)} \Rightarrow [K_G^{(a)}, M_a] = 0$. $M_a$. At first note that $M_a^2 \alpha K_G^{(a)} \Rightarrow [K_G^{(a)},
Further $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$, 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 so the first two equations follow trivially. For $j \in n_a$ set
\begin{equation} \begin{equation}
@ -453,11 +468,12 @@ that will be used later\cite{andersbriegel2005}.
\end{aligned} \end{aligned}
\end{equation} \end{equation}
One can now construct a new set of $K_{G'}^{(i)}$ such that $M_a \ket{\bar{G}}$ is the $+1$ eigenstate One can now construct a new set of $K_{G'}^{(i)}$ such that $M_a
of the new $K_{G'}^{(i)}$. Because $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$ \ket{\bar{G}}$ is the $+1$ eigenstate of the new $K_{G'}^{(i)}$. Because
it is clear that $\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$. $\forall i\in V\setminus n_a$ $[K_G^{(i)}, M_a] = 0$ it is clear that
To construct the $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a =: \{j\} \cup F$ and $n_j =: \{a\} \cup D$. $\forall j \notin n_a$ $K_{G'}^{(j)} = K_G^{(j)}$. To construct the
Then follows: $K_{G'}^{(i)}$ let for some $j \in n_a$ $n_a =: \{j\} \cup F$ and $n_j =:
\{a\} \cup D$. Then follows:
\begin{equation} \begin{equation}
\begin{aligned} \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}'} = K_{G'}^{(j)}K_{G}^{(a)}\ket{\bar{G}'} = K_{G'}^{(j)}\ket{\bar{G}'}
\end{equation} \end{equation}
Because $\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{S^{(i)} \middle| i\in n_a\right\}$ and Because $\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{S^{(i)}
$\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}$ are both $n$ commuting \middle| i\in n_a\right\}$ and $\left\{K_G^{(i)} \middle| i \notin
multi-local Pauli operators where the $S^{(i)}$ can be generated from the $K_{G'}^{(i)}$ n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}$ are both $n$
and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$ commuting multi-local Pauli operators where the $S^{(i)}$ can be generated from
$\langle\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}\rangle$ the $K_{G'}^{(i)}$ and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$
are the stabilizers of $\ket{\bar{G}'}$. Therefore the associated graph is changed as given $\langle\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)}
in the third equation. \middle| i\in n_a \right\}\rangle$ are the stabilizers of $\ket{\bar{G}'}$.
Therefore the associated graph is changed as given in the third equation.
\end{proof} \end{proof}
\subsection{Graph States} \subsection{Graph States}
The definition of a VOP-free graph state above raises an obvious question: The definition of a VOP-free graph state above raises an obvious question: Can
Can any stabilizer state be described using just a graph? any stabilizer state be described using just a graph? The answer is straight
The answer is straight forward: No. The most simple cases are the single qbit forward: No. The most simple cases are the single qbit stated $\ket{0},\ket{1}$
stated $\ket{0},\ket{1}$ and $\ket{+_Y}, \ket{-_Y}$. But there is an extension and $\ket{+_Y}, \ket{-_Y}$. But there is an extension to the VOP-free graph
to the VOP-free graph states that allows the representation of an arbitrary states that allows the representation of an arbitrary stabilizer state. The
stabilizer state. The proof that indeed any state can be represented is proof that indeed any state can be represented is purely constructive. As seen
purely constructive. As seen in Theorem \ref{thm:clifford_group_approx} any $c \in C_n$ in Theorem \ref{thm:clifford_group_approx} any $c \in C_n$ can be constructed
can be constructed from $CZ$ and $C_L$. In the following discussion it will become from $CZ$ and $C_L$. In the following discussion it will become clear that both
clear that both $C_L$ and $CZ$ can be applied to a general graph state. $C_L$ and $CZ$ can be applied to a general graph state.
\subsubsection{Graph States and Vertex Operators} \subsubsection{Graph States and Vertex Operators}
\label{ref:g_states_vops} \label{ref:g_states_vops}
\begin{definition} \begin{definition}
\label{def:g_state} \label{def:g_state}
A tuple $(V, E, O)$ is called the graphical representation of a stabilizer state A tuple $(V, E, O)$ is called the graphical representation of a stabilizer
if $(V, E)$ is a graph as in Definition \ref{def:graph} and $O = \{o_1, ..., o_n\}$ where $o_i \in C_L$. 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: 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}. $o_i$ are called the vertex operators of $\ket{G}$ \cite{andersbriegel2005}.
\end{definition} \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} \begin{equation}
\ket{G} = \left(\prod\limits_{j=1}^no_j\right) \ket{\bar{G}} \ket{G} = \left(\prod\limits_{j=1}^no_j\right) \ket{\bar{G}}
\end{equation} \end{equation}
The great advantage of this representation of a stabilizer state is its space requirement: The great advantage of this representation of a stabilizer state is its space
Instead of storing $n^2$ $P$ matrices only some vertices (which often are implicit), requirement: Instead of storing $n^2$ $P$ matrices only some vertices (which
the edges and some vertex operators ($n$ matrices) have to be stored. The following theorem often are implicit), the edges and some vertex operators ($n$ matrices) have to
will improve this even further: instead of $n$ matrices it is sufficient to store $n$ integers be stored. The following theorem will improve this even further: instead of $n$
representing the vertex operators: matrices it is sufficient to store $n$ integers representing the vertex
operators:
\begin{theorem} \begin{theorem}
$C_L$ has $24$ degrees of freedom \cite{andersbriegel2005}. $C_L$ has $24$ degrees of freedom \cite{andersbriegel2005}.
\end{theorem} \end{theorem}
\begin{proof} \begin{proof} It is clear that $\forall a \in C_L$ a is a group isomorphism $P
It is clear that $\forall a \in C_L$ a is a group isomorphism $P \rightarrow P$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. \rightarrow P$: $apa^\dagger a p' a^\dagger = a pp'a^\dagger$. Therefore
Therefore $a$ will preserve the (anti-)commutator relations of $P$. $a$ will preserve the (anti-)commutator relations of $P$. Further note
Further note that $Y = iXZ$, so one has to consider the anti-commutator relations that $Y = iXZ$, so one has to consider the anti-commutator relations of
of $X,Z$ only. $X,Z$ only.
As the transformations are unitary they preserve eigenvalues, so $X$ can be mapped As the transformations are unitary they preserve eigenvalues, so $X$ can be
to $\pm X, \pm Y, \pm Z$ which gives $6$ degrees of freedom. Furthermore the image of $Z$ has to mapped to $\pm X, \pm Y, \pm Z$ which gives $6$ degrees of freedom.
anti-commute with the image of $X$ so $Z$ has four possible images under the transformation. Furthermore the image of $Z$ has to anti-commute with the image of $X$ so
This gives another $4$ degrees of freedom and a total of $24$. $Z$ has four possible images under the transformation. This gives another
\end{proof} $4$ 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. From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be
One can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is used. One can show (by construction) that $H, S$ generate a possible choice of
$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in one specific operation on graph states \cite{andersbriegel2005}. $C_L$, as is $C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which is required in
one specific operation on graph states \cite{andersbriegel2005}.
\begin{equation} \begin{equation}
S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right) 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} \subsubsection{Dynamics of Graph States}
So far the graphical representation of stabilizer states is just another way to store So far the graphical representation of stabilizer states is just another way to
basically a stabilizer tableaux that might require less memory than the tableaux used in store basically a stabilizer tableaux that might require less memory than the
CHP\cite{CHP}. The true power of this formalism is seen when studying its dynamics. The simplest case tableaux used in CHP\cite{CHP}. The true power of this formalism is seen when
is a local Clifford operator $c_j$ acting on a qbit $j$: The stabilizers are changed to studying its dynamics. The simplest case is a local Clifford operator $c_j$
$\langle c_j S^{(i)} c_j^\dagger\rangle_i$. Using the definition of the graphical representation acting on a qbit $j$: The stabilizers are changed to $\langle c_j S^{(i)}
one sees that just the vertex operators are changed and the new vertex operators are given by c_j^\dagger\rangle_i$. 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{equation}
\begin{aligned} \begin{aligned}
@ -575,30 +598,33 @@ one sees that just the vertex operators are changed and the new vertex operators
\end{aligned} \end{aligned}
\end{equation} \end{equation}
The action of a $CZ$ gate on the state $(V, E, O)$ is in most cases less trivial. The action of a $CZ$ gate on the state $(V, E, O)$ is in most cases less
Let $i \neq j$ be two qbits, now consider the action of $CZ_{a,b}$ on $(V, E, O)$. trivial. Let $i \neq j$ be two qbits, now consider the action of $CZ_{a,b}$ on
The cases given here follow the implementation of a $CZ$ application in \cite{pyqcs}, $(V, E, O)$. The cases given here follow the implementation of a $CZ$
the respective paragraphs from \cite{andersbriegel2005} are given in italic. application in \cite{pyqcs}, the respective paragraphs from
Most of the discussion follows the one given in \cite{andersbriegel2005} closely. \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{:}\\ \textbf{Case 1}(\textit{Case 1})\textbf{:}\\
Both $o_a$ and $o_b$ commute with $CZ_{a,b}$. This is the case for exactly Both $o_a$ and $o_b$ commute with $CZ_{a,b}$. This is the case for exactly four
four vertex operators: $\mathcal{Z} = \left\{I, Z, S, S^\dagger\right\}$. vertex operators: $\mathcal{Z} = \left\{I, Z, S, S^\dagger\right\}$. In this
In this case the CZ can be pulled past the vertex operators and just the edges case the CZ can be pulled past the vertex operators and just the edges are
are changed to $E' = E \Delta \left\{\{a,b\}\right\}$. changed to $E' = E \Delta \left\{\{a,b\}\right\}$.
\textbf{Case 2}(\textit{Sub-Sub-Case 2.2.1})\textbf{:}\\ \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 The two qbits are isolated: From the definition of the graph state one can
any isolated clique of the graph can be treated independently. Therefore the two isolated qbits derive that any isolated clique of the graph can be treated independently.
can be treated as an independent state and the set of two qbit stabilizer states is finite. An Therefore the two isolated qbits can be treated as an independent state and the
upper bound to the number of two qbit stabilizer states is given by $2\cdot24^2$: set of two qbit stabilizer states is finite. An upper bound to the number of
A factor of two for the options with and withoit an edge between the two qbit stabilizer states is given by $2\cdot24^2$: A factor of two for the
qbits and $24$ Clifford operators on each vertex. 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 All those states and the resulting state after a $CZ$ application can be
another interesting result that will be useful later: If one vertex has the vertex operator $I$ the computed which leads to another interesting result that will be useful later:
resulting state can be chosen such that at least one of the vertex operators is $I$ again and in particular 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 $CZ$. 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{:}\\ \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 this case one can try to clear the vertex operators which will succeed for at
least one qbit. least one qbit.
The transformation given in The transformation given in Lemma \ref{lemma:M_a} is used to "clear" the vertex
Lemma \ref{lemma:M_a} is used to "clear" the vertex operators. Recalling that operators. Recalling that the transformation $M_j$ toggles the neighbourhood of
the transformation $M_j$ toggles the neighbourhood of vertex $j$ gives substance vertex $j$ gives substance to the following theorem:
to the following theorem:
\begin{theorem} \begin{theorem}
A graph state $\ket{G}$ associated with $(V, E, O)$ is invariant when A graph state $\ket{G}$ associated with $(V, E, O)$ is invariant when
@ -623,11 +648,13 @@ to the following theorem:
Without proof. Without proof.
\end{theorem} \end{theorem}
As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$ and $\sqrt{iZ}$. As stated in \ref{ref:g_states_vops} $C_L$ is also generated by $\sqrt{-iX}$
This yields an algorithm to reduce the vertex operator of a non-isolated qbit $j$ to the identity. and $\sqrt{iZ}$. This yields an algorithm to reduce the vertex operator of
The combined operation of toggling the neighbourhood of $j$ and right-multiplying a non-isolated qbit $j$ to the identity. The combined operation of toggling
$M_j^\dagger$ is called $L_j$ transformation, which transforms $(V, E, O)$ into a so-called the neighbourhood of $j$ and right-multiplying $M_j^\dagger$ is called $L_j$
local Clifford equivalent graphical representation. The algorithm is given by the following steps: 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} \begin{enumerate}
\item{Check whether $n_a \setminus \{b\} \neq \{\}$. If this condition does not hold \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{aligned}
\end{equation} \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{equation}
\begin{aligned} \begin{aligned}
@ -707,8 +735,8 @@ operators. For this consider the projector $P_{a,s} = \frac{I + (-1)^sZ_a}{2}$
\end{aligned} \end{aligned}
\end{equation} \end{equation}
This transformed projector has the important property that it still is a Pauli projector This transformed projector has the important property that it still is a Pauli
as $o_a$ is a Clifford operator projector as $o_a$ is a Clifford operator
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -719,9 +747,10 @@ as $o_a$ is a Clifford operator
\end{aligned} \end{aligned}
\end{equation} \end{equation}
Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is enough to study the measurements Where $\tilde{g}_a \in \{+1, -1\}\cdot\{X_a, Y_a, Z_a\}$. Therefore, it is
of any Pauli operator on the vertex operator free graph states. The commutators of the observable enough to study the measurements of any Pauli operator on the vertex operator
with the $K_G^{(i)}$ are quite easy to compute. Note that Pauli matrices either commute or anticommute 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 and it is easier to list the operators that anticommute
\begin{equation} \begin{equation}
@ -732,15 +761,17 @@ and it is easier to list the operators that anticommute
\end{aligned} \end{aligned}
\end{equation} \end{equation}
This gives one immediate result: if a qbit $a$ is isolated and the operator $\tilde{g}_a = X_a (-X_a)$ This gives one immediate result: if a qbit $a$ is isolated and the operator
is measured the result $s=0(1)$ is obtained with probability $1$ and $(V, E, O)$ is unchanged. $\tilde{g}_a = X_a (-X_a)$ is measured the result $s=0(1)$ is obtained with
In any other case the results $s=1$ and $s=0$ have probability $\frac{1}{2}$ and both probability $1$ and $(V, E, O)$ is unchanged. In any other case the results
graph and vertex operators have to be updated. Further it is clear that measurements of $-\tilde{g}_a$ $s=1$ and $s=0$ have probability $\frac{1}{2}$ and both graph and vertex
and $\tilde{g}_a$ are related by just inverting the result $s$. 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 calculations to obtain the transformation on graph and vertex operators are
the scheme of Lemma \ref{lemma:M_a}. \cite[Section IV]{hein_eisert_briegel2008} also contains lengthy and follow the scheme of Lemma \ref{lemma:M_a}. \cite[Section
the steps required to obtain the following results IV]{hein_eisert_briegel2008} also contains the steps required to obtain the
following results
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -749,10 +780,11 @@ the steps required to obtain the following results
\end{aligned} \end{aligned}
\end{equation} \end{equation}
These transformations split it two parts: the first is a result of Lemma \ref{lemma:stab_measurement}. These transformations split it two parts: the first is a result of Lemma
The second part makes sure that the qbit $a$ is diagonal in measured observable and has the correct eigenvalue. \ref{lemma:stab_measurement}. The second part makes sure that the qbit $a$ is
When comparing with Lemma \ref{lemma:stab_measurement} in both cases $Y,Z$ the anticommuting stabilizer diagonal in measured observable and has the correct eigenvalue. When comparing
$K_G^{(a)}$ is chosen. The graph is changed according to 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{equation}
\begin{aligned} \begin{aligned}