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thesis/chapters/stabilizer.tex
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@ -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}