some more work

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Daniel Knüttel 2020-02-19 12:36:56 +01:00
parent 1a5b8f3ad1
commit 587af80d14
3 changed files with 61 additions and 23 deletions

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@ -142,8 +142,7 @@ $CU$ applies the gate $U$ to the act-qbit if the control-qbit is in its $\ket{1}
One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough 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 (this is quickly verified by applying the The matrix representation of $CX$ and $CZ$ for two qbits is given by
matrices to the basis states)
\begin{equation} \begin{equation}
CX_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right) CX_{1, 0} = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)
@ -216,6 +215,37 @@ the gate, for instance the circuit for $CZ_{2, 1}CX_{2,0}$ is
} }
\] \]
Several qbits can be abbreviated by writing a slash on the qbit line:
\[
\Qcircuit @C=1em @R=.7em {
& {/} \qw & \qw & \gate{Z^{\otimes n}} & \qw \\
}
\]
\subsection{Quantum Algorithms} \subsection{Quantum Algorithms}
The great hope behind quantum computing is that it will speed up some computations
exponentially using algorithms that utilize the laws of quantum mechanics. Current
algorithms are based upon quantum search and quantum fourier transform \cite{nielsen_chuang_2010}.
The latter is particular interesting for physical problems as it is used in the phase estimation
algorithm that can be used to analyze the spectrum of the transfer matrix
\begin{equation}
T = \exp(itH)
\end{equation}
The eigenvalue of $T$ can now be estimated by using the phase estimation circuit:
\[
\Qcircuit @C=1em @R=.7em {
& \lstick{\ket{0}} & {/^N} \qw & \gate{H^{\otimes n}} & \ctrl{1} & \gate{FT^\dagger} & \qw & \meter & \rstick{x}\\
& \lstick{\ket{\varphi}} & {/} \qw & \qw & \gate{T} & \qw & {/} \qw & \qw& \rstick{\ket{\varphi}} \\
}
\]
Where $T\ket{\varphi} = \exp(2\pi i\varphi) \ket{\varphi}$ and the measurement result $\tilde{\varphi} = \frac{x}{2^n}$ is an
estimate for $\varphi$. If a success rate of $1-\epsilon$ and an accuracy of $| \varphi - \tilde{\varphi} | < 2^{-m}$
is wanted $N = m + \log_2(2 + \frac{1}{2\epsilon})$ qbits are required\cite{nielsen_chuang_2010}\cite{lehner2019}.

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@ -31,7 +31,7 @@ the elements of $P$ either commute or anticommute.
For $n$ qbits For $n$ qbits
\begin{equation} \begin{equation}
P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i | p_i \in P\right\} P_n := \left\{\bigotimes\limits_{i=0}^{n-1} p_i \middle| p_i \in P\right\}
\end{equation} \end{equation}
is called the multilocal Pauli group on $n$ qbits. is called the multilocal Pauli group on $n$ qbits.
@ -68,12 +68,12 @@ via the tensor product as do the (anti-)commutator relationships.
\begin{enumerate} \begin{enumerate}
\item{$(iI)^2 = (-iI)^2 = -I$. Which contradicts the definition of $S$.} \item{$(iI)^2 = (-iI)^2 = -I$. Which contradicts the definition of $S$.}
\item{From the definition of $S$ ($G_n$ respectively) follows that any \item{From the definition of $S$ ($G_n$ respectively) follows that any
$S^{(i)} \in S$ has the form $\pm i^l (\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j)$ where $S^{(i)} \in S$ has the form $\pm i^l \left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$ where
$\tilde{p}_j \in \{X, Y, Z, I\}$ and $l \in \{0, 1\}$. As $(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j)$ $\tilde{p}_j \in \{X, Y, Z, I\}$ and $l \in \{0, 1\}$. As $\left(\bigotimes\limits_{j=0}^{n-1} \tilde{p}_j\right)$
is hermitian $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly. is hermitian $(S^{(i)})^2$ is either $+I$ or $-I$. As $-I \notin S$ $(S^{(i)})^2 = I$ follows directly.
} }
\item{Following the argumentation above $(S^{(i)})^2 = -I \Leftrightarrow l=1$ \item{Following the argumentation above $\left(S^{(i)}\right)^2 = -I \Leftrightarrow l=1$
therefore $(S^{(i)})^2 = -I \Leftrightarrow (S^{(i)})^\dagger \neq (S^{(i)})$.} therefore $\left(S^{(i)}\right)^2 = -I \Leftrightarrow \left(S^{(i)}\right)^\dagger \neq S^{(i)}$.}
\end{enumerate} \end{enumerate}
\end{proof} \end{proof}
@ -108,7 +108,7 @@ can be diagonalized simultaneously. This motivates and justifies the following d
For a set of stabilizers $S$ the vector space For a set of stabilizers $S$ the vector space
\begin{equation} \begin{equation}
V_S := \{\ket{\psi} | S^{(i)}\ket{\psi} = +1\ket{\psi} \forall S^{(i)} \in S\} V_S := \left\{\ket{\psi} \middle| S^{(i)}\ket{\psi} = +1\ket{\psi} \forall S^{(i)} \in S\right\}
\end{equation} \end{equation}
is called the space of stabilizer states associated with $S$ and one says is called the space of stabilizer states associated with $S$ and one says
@ -124,7 +124,7 @@ result:
\begin{theorem} \begin{theorem}
\label{thm:unique_s_state} \label{thm:unique_s_state}
For a $n$ qbit system and a set $S = \langle S^{(i)} \rangle_{i=1, ..., n}$ 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 a trivial phase unique 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 $S$. state $\ket{\psi}$ that is stabilized by $S$.
@ -157,14 +157,14 @@ however some statements that can still be made:
\end{aligned} \end{aligned}
\end{equation} \end{equation}
Note that in \ref{def:stabilizer} it has been demanded that stabilizers are a 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 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. $U$ 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) | UpU^\dagger \in P_n \forall p \in P_n\right\} C_n := \left\{U \in SU(n) \middle| UpU^\dagger \in P_n \forall p \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.
@ -235,12 +235,12 @@ the result of the measurement.
\begin{lemma} \begin{lemma}
\label{lemma:stab_measurement} \label{lemma:stab_measurement}
Let $J := \{ S^{(i)} | [g_a, S^{(i)}] \neq 0\} \neq \{\}$. 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 $1$ and $0$ 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 \{S^{(i)} S^{(j)} | S^{(i)} \in J \setminus \{S^{(j)}\} \} \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}
@ -287,9 +287,9 @@ 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 $E = \left\{\{i, j\} | i,i \in V, i \neq j\right\}$. $E$ is the set of edges $E = \left\{\{i, j\} \middle| i,i \in V, i \neq j\right\}$.
For a vertex $i$ $n_i := \left\{j \in V | \{i, j\} \in E\right\}$ 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 $i$.
\end{definition} \end{definition}
@ -436,11 +436,11 @@ that will be used later.
= 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 $\{K_G^{(i)} | i \notin n_a\} \cup \{S^{(i)} | i\in n_a\}$ and Because $\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{S^{(i)} \middle| i\in n_a\right\}$ and
$\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}$ are both $n$ commuting $\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}$ are both $n$ commuting
multi-local Pauli operators where the $S^{(i)}$ can be generated from the $K_{G'}^{(i)}$ 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 $K_{G'}^{(j)}$ and $\ket{\bar{G}'}$ is a $+1$ eigenstate of $K_{G'}^{(j)}$
$\langle\{K_G^{(i)} | i \notin n_a\} \cup \{K_{G'}^{(i)} | i\in n_a \}\rangle$ $\langle\left\{K_G^{(i)} \middle| i \notin n_a\right\} \cup \left\{K_{G'}^{(i)} \middle| i\in n_a \right\}\rangle$
are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given are the stabilizers of $\ket{\bar{G}'}$ and the associated graph is changed as given
in the third equation. in the third equation.
\end{proof} \end{proof}
@ -665,8 +665,8 @@ so it is easier to list the operators that anticommute:
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
A_{\pm X_a} &= \left\{j | \{j, a\} \in E\right\}\\ A_{\pm X_a} &= \left\{j \middle| \{j, a\} \in E\right\}\\
A_{\pm Y_a} &= \left\{j | \{j, a\} \in E\right\} \cup \{a\} \\ A_{\pm Y_a} &= \left\{j \middle| \{j, a\} \in E\right\} \cup \{a\} \\
A_{\pm Z_a} &= \{a\}\\ A_{\pm Z_a} &= \{a\}\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
@ -695,8 +695,8 @@ $K_G^{(a)}$ is chosen. The graph is changed according to
\begin{equation} \begin{equation}
\begin{aligned} \begin{aligned}
E'_{Z} &= E \setminus \left\{\{i,a\} | i \in V\right\}\\ E'_{Z} &= E \setminus \left\{\{i,a\} \middle| i \in V\right\}\\
E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} | i \in V\right\}\\ E'_{Y} &= E \Delta (n_a \otimes n_a) \setminus \left\{\{i,a\} \middle| i \in V\right\}\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}
@ -714,7 +714,7 @@ For $g_a = X_a$ one has to choose a $b \in n_a$ and the transformations are
E'_{X} = E &\Delta (n_b \otimes n_a) \\ E'_{X} = E &\Delta (n_b \otimes n_a) \\
& \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\ & \Delta ((n_b \cap n_a) \otimes (n_b \cap n_a)) \\
& \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\ & \Delta (\{b\} \otimes (n_a \setminus \{b\})) \\
& \setminus \left\{\{i,a\} | i \in V\right\}\\ & \setminus \left\{\{i,a\} \middle| i \in V\right\}\\
\end{aligned} \end{aligned}
\end{equation} \end{equation}

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@ -147,3 +147,11 @@
author={M. Hein, J. Eisert, H.J. Briegel}, author={M. Hein, J. Eisert, H.J. Briegel},
note={https://arxiv.org/abs/quant-ph/0307130v7} note={https://arxiv.org/abs/quant-ph/0307130v7}
} }
@article{
lehner2019,
title={Lecture Notes on Quantum Computing SS 2019},
year=2019,
author={Christoph Lehner},
note={Unpublished Work}
}