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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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2 changed files with 40 additions and 34 deletions
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6
thesis/chapters/quantum_computing.tex
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@ -59,7 +59,7 @@ and will be used in some calculations later:
\label{ref:many_qbits}
\begin{postulate}
A $n$-qbit quantum mechanical state is the tensor product\cite[Definition 14.3]{wuest1995} of the $n$ one-qbit
A $n$-qbit quantum mechanical state is the tensor product \cite[Definition 14.3]{wuest1995} of the $n$ one-qbit
states.
\end{postulate}
@ -146,7 +146,7 @@ In Definition \ref{def:CU} $i$ is called the act-qbit and $j$ the control-qbit.
$CU$ applies the gate $U$ to the act-qbit if the control-qbit is in its $\ket{1}$ state.
One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough
to generate an arbitrary $n$-qbit gate\cite[Chapter 4.3]{kaye_ea2007}\cite{barenco_ea_1995}.
to generate an arbitrary $n$-qbit gate \cite[Chapter 4.3]{kaye_ea2007}\cite{barenco_ea_1995}.
The matrix representation of $CX$ and $CZ$ for two qbits is given by:
\begin{equation}
@ -251,5 +251,5 @@ The eigenvalue of $T$ can now be estimated by using the phase estimation circuit
Where $T\ket{\varphi} = \exp(2\pi i\varphi) \ket{\varphi}$ and the measurement result $\tilde{\varphi} = \frac{x}{2^n}$ is an
estimation for $\varphi$. If a success rate of $1-\epsilon$ and an accuracy of $| \varphi - \tilde{\varphi} | < 2^{-m}$
is wanted, then $N = m + \log_2(2 + \frac{1}{2\epsilon})$ qbits are required\cite{nielsen_chuang_2010}\cite{lehner2019}.
is wanted, then $N = m + \log_2(2 + \frac{1}{2\epsilon})$ qbits are required \cite{nielsen_chuang_2010}\cite{lehner2019}.

68
thesis/chapters/stabilizer.tex
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@ -48,7 +48,7 @@ via the tensor product.
\label{def:stabilizer}
An abelian subgroup $S = \{S^{(0)}, ..., S^{(N)}\}$ of $P_n$ is called a set of stabilizers iff
\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{$-I \notin S$}
\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
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 $g_i$
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}
For $n$ qbits
\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}
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{
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 $S = \sqrt{iZ}^3$.
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{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
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
@ -211,8 +211,8 @@ product of $n$ Pauli matrices. This has led to the simulation using stabilizer t
\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
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}
@ -239,7 +239,7 @@ the result of the measurement.
\label{lemma:stab_measurement}
Let $J := \left\{ S^{(i)} \middle| [g_a, S^{(i)}] \neq 0\right\} \neq \{\}$. When measuring
$\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
\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
@ -252,18 +252,18 @@ the result of the measurement.
\begin{equation}
\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(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}\right)\\
&= P(s=-1)
&= P(s=1)
\end{aligned}
\notag
\end{equation}
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{aligned}
@ -275,21 +275,21 @@ 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
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 $\ket{\psi'}$.
\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
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 $|V| = n \in \mathbb{N}$.
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
of $i$.
@ -302,19 +302,19 @@ mathematical graph 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_i
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
$\langle K_G^{(i)} \rangle_{i = 0, ..., n-1}$.
\end{definition}
It is clear that the $K_G^{(i)}$ multilocal Pauli operators. That they commute
follows from the fact that $\{i,j\} \in E \Leftrightarrow \{j,i\} \in E$ so for two
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
It is clear that the $K_G^{(i)}$ are multilocal Pauli operators. That they commute
follows from the fact that $\{i,j\} \in E \Leftrightarrow \{j,i\} \in E$ therefore for two
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
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 $\ket{\bar{G}}$
is unique. The following lemma will provide a way to construct this state
from the graph.
@ -331,16 +331,16 @@ from the graph.
\end{equation}
\end{lemma}
\begin{proof}
Let $\ket{+} := \left(\prod\limits_{l \in V} H_l\right) \ket{0}$ as before. Note that for any $X_i$ $X_i \ket{+} = +1 \ket{+}$.
Set $\ket{\tilde{G}} := \left(\prod\limits_{\{i,j\} \in E} CZ_{i,j} \right)\ket{+}$.
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 $\ket{\tilde{G}} := \left(\prod\limits_{\{l,k\} \in E} CZ_{l,k} \right)\ket{+}$.
\begin{equation}
\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{+} \\
& = \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)\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{+} \\
& = \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,j\} \in E}\left( \ket{0}\bra{0}_j \otimes I_l + \ket{1}\bra{1}_j \otimes Z_l\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,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,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,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,k\} \in E}\left( \ket{0}\bra{0}_k \otimes I_l + \ket{1}\bra{1}_k \otimes Z_l\right) \ket{+} \\
& = +1 \ket{\tilde{G}}
\end{aligned}
\end{equation}
@ -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
$C_L = \langle \sqrt{-iX}, \sqrt{-iZ}\rangle$ which will be used in one specific operation on graph states.
$$ S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)$$
$$ \sqrt{-iX} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)$$
$$ \sqrt{-iZ} = \exp(-i\frac{\pi}{4})\left(\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right)$$
\begin{equation}
S = \left(\begin{array}{cc} 1 & 0 \\ 0 & i \end{array}\right)
\end{equation}
\begin{equation}
\sqrt{-iX} = \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & i \\ i & 1 \end{array}\right)
\end{equation}
\begin{equation}
\sqrt{-iZ} = \exp(-i\frac{\pi}{4})\left(\begin{array}{cc} 1 & 0 \\ 0 & -i \end{array}\right)
\end{equation}
\subsubsection{Dynamics of Graph States}
@ -605,7 +611,7 @@ clearing $o_b$ one can retry to clear $o_a$.
In any case at least one vertex operator has been cleared. If both vertex operators have been
cleared Case 1 will be applied, if there is just one cleared vertex operator it is sure that it
is the vertex with non-operand neighbours. Using this one can still apply a $CZ$: Without loss of generality
assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form\cite{andersbriegel2005}
assume that $a$ has non-operand neighbours and $b$ does not. Now the state $\ket{G}$ has the form \cite{andersbriegel2005}
\begin{equation}
\begin{aligned}