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implemented some of Simon's changes

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Daniel Knüttel 2020-02-24 12:05:45 +01:00
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thesis/chapters/stabilizer.tex
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@ -48,7 +48,7 @@ via the tensor product.
\label{def:stabilizer} \label{def:stabilizer}
An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff
\begin{enumerate} \begin{enumerate}
\item{$\forall i,j = 1, ..., N$ $[S^{(i)}, S^{(j)}] = 0$%: $S^{(i)}$ and $S^{(j)}$ commute \item{$\forall i,j = 1, ..., N$: $[S^{(i)}, S^{(j)}] = 0$%: $S^{(i)}$ and $S^{(j)}$ commute
} }
\item{$-I \notin S$} \item{$-I \notin S$}
\end{enumerate} \end{enumerate}
@ -87,7 +87,7 @@ properties in order to understand the properties of the group.
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 generators
of G of G
$$ \langle g_1, ..., g_m \rangle \equiv \langle g_i \rangle_{i=1,...,m}$$ \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 $g_i$
and $m$ is the smallest integer for which these statements hold. and $m$ is the smallest integer for which these statements hold.
@ -166,7 +166,7 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers.
\begin{definition} \begin{definition}
For $n$ qbits For $n$ qbits
\begin{equation} \begin{equation}
C_n := \left\{U \in SU(n) \middle| UpU^\dagger \in P_n \forall p \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. is called the Clifford group. $C_1 =: C_L$ is called the local Clifford group.
@ -190,7 +190,7 @@ $U$ but there exists a group for which $S'$ will be a set of stabilizers.
\item{See \cite[Theorem 10.6]{nielsen_chuang_2010}} \item{See \cite[Theorem 10.6]{nielsen_chuang_2010}}
\item{ \item{
One can easily verify that $\sqrt{iZ} \in C_L$ and $\sqrt{-iX} \in C_L$. One can easily verify that $\sqrt{iZ} \in C_L$ and $\sqrt{-iX} \in C_L$.
Further one can show easily that (up to a global phase) Further one can show that (up to a global phase)
$H = \sqrt{iZ} \sqrt{-iX}^3 \sqrt{iZ}$ and $S = \sqrt{iZ}^3$. $H = \sqrt{iZ} \sqrt{-iX}^3 \sqrt{iZ}$ and $S = \sqrt{iZ}^3$.
The length of the product can be seen when explicitly calculating The length of the product can be seen when explicitly calculating
@ -200,7 +200,7 @@ $U$ 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 $U \in C_n$ $S'$ is a set of 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 $n$ 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 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 efficient as only $n$ stabilizers have to be modified, each being just the tensor
@ -211,8 +211,8 @@ product of $n$ Pauli matrices. This has led to the simulation using stabilizer t
\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 an observable $g_a \in \{\pm X_a, \pm Y_a \pm Z_a\}$ acting on qbit $a$ is measured 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 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}
@ -239,7 +239,7 @@ the result of the measurement.
\label{lemma:stab_measurement} \label{lemma:stab_measurement}
Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$. When measuring Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$. When measuring
$\frac{I + (-1)^s g_a}{2} $ $\frac{I + (-1)^s g_a}{2} $
$1$ and $0$ are obtained with probability $\frac{1}{2}$ and after choosing $s=1$ 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 a $j \in J$ the new state $\ket{\psi'}$ is stabilized by
\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
@ -252,18 +252,18 @@ the result of the measurement.
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
P(s=+1) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\ P(s=0) &= \hbox{Tr}\left(\frac{I + g_a}{2}\ket{\psi}\bra{\psi}\right) \\
&= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\ &= \hbox{Tr}\left(\frac{I + g_a}{2}S^{(j)} \ket{\psi}\bra{\psi}\right)\\
&= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\ &= \hbox{Tr}\left(S^{(j)}\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\ &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}S^{(j)}\right)\\
&= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\ &= \hbox{Tr}\left(\frac{I - g_a}{2}\ket{\psi}\bra{\psi}\right)\\
&= P(s=-1) &= P(s=1)
\end{aligned} \end{aligned}
\notag \notag
\end{equation} \end{equation}
With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$. With $P(s=+1) + P(s=-1) = 1$ follows $P(s=+1) = \frac{1}{2} = P(s=-1)$.
Further for $S^{(i)},S^{(j)} \in J$ Further for $S^{(i)},S^{(j)} \in J$:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
@ -275,7 +275,7 @@ 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 $S^{(j)}S^{(i)}$ $i,j \in J$, and by
$S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$. $S^{(i)} \in J^c$. $(-1)^sg_a$ trivially stabilizes $\ket{\psi'}$.
\end{proof} \end{proof}
@ -289,7 +289,7 @@ vertex operator-free will be clear in the following section about graph states.
\label{def:graph} \label{def:graph}
The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$. The tuple $(V, E)$ is called a graph iff $V$ is a set of vertices with $|V| = n \in \mathbb{N}$.
In the following $V = \{0, ..., n-1\}$ will be used. In the following $V = \{0, ..., n-1\}$ will be used.
$E$ is the set of edges $E = \left\{\{i, j\} \middle| i,i \in V, i \neq j\right\}$. $E$ is the set of edges $E = \left\{\{i, j\} \middle| i,j \in V, i \neq j\right\}$.
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 called the neighbourhood
of $i$. of $i$.
@ -302,19 +302,19 @@ mathematical graph 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_i 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 the state stabilized by
$\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$. $\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$.
\end{definition} \end{definition}
It is clear that the $K_G^{(i)}$ multilocal Pauli operators. That they commute It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute
follows from the fact that $\{i,j\} \in E \Leftrightarrow \{j,i\} \in E$ so for two follows from the fact that $\{i,j\} \in E \Leftrightarrow \{j,i\} \in E$ therefore for two
operators $K_G^{(i)}$ and $K_G^{(j)}$ either $\{i, j\} \notin E$ so they commute trivially operators $K_G^{(i)}$ and $K_G^{(j)}$ either $\{i, j\} \notin E$ so they commute trivially.
if $\{i,j\} \in E$ $X_i$, $Z_j$ and $X_j$, $Z_i$ anticommute which yields that the If $\{i,j\} \in E$ $X_i$, $Z_j$ and $X_j$, $Z_i$ anticommute which yields that the
operators commute. operators commute.
This definition of a graph state might not seem to be quite straight forward 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 $\ket{\bar{G}}$ 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 is unique. The following lemma will provide a way to construct this state
from the graph. from the graph.
@ -331,16 +331,16 @@ from the graph.
\end{equation} \end{equation}
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}
Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$ $X_i \ket{+} = +1 \ket{+}$. Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$: $X_i \ket{+} = +1 \ket{+}$.
Set $\ket{\tilde{G}} := \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right)\ket{+}$. Set $\ket{\tilde{G}} := \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$.
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
K_G^{(i)} \ket{\tilde{G}} & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\left(\prod\limits_{\{l,j\} \in E} CZ_{l,j} \right) \ket{+} \\ K_G^{(i)} \ket{\tilde{G}} & = X_i \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right) \ket{+} \\
& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\ & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)X_i\prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
& = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) X_i \ket{+} \\ & = \left(\prod\limits_{\{i,j\} \in E} Z_j\right)\prod\limits_{\{l,k\} \in E}\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{+} \\
& = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + (-1)^{2\delta_{i,l}}\ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\ & = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + (-1)^{\delta_{i,l} + \sum\limits_{\{i,j\} \in E} \delta_{j,k}}\ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
& = \prod\limits_{\{l,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\right) \ket{+} \\ & = \prod\limits_{\{l,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
& = +1 \ket{\tilde{G}} & = +1 \ket{\tilde{G}}
\end{aligned} \end{aligned}
\end{equation} \end{equation}
@ -507,9 +507,15 @@ From now on $C_L = \langle H, S \rangle$ (disregarding a global phase) will be u
one can show (by construction) that $H, S$ generate a possible choice of $C_L$, as is 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 will be used in one specific operation on graph states. $C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which will be used in one specific operation on graph states.
$$ S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)$$ \begin{equation}
$$ \sqrt{-iX} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)$$ S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)
$$ \sqrt{-iZ} = \exp(-i\frac{\pi}{4})\left(\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right)$$ \end{equation}
\begin{equation}
\sqrt{-iX} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)
\end{equation}
\begin{equation}
\sqrt{-iZ} = \exp(-i\frac{\pi}{4})\left(\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right)
\end{equation}
\subsubsection{Dynamics of Graph States} \subsubsection{Dynamics of Graph States}