did some work

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Daniel Knüttel 2019-10-17 19:32:25 +02:00
parent 37cb33b5b4
commit 519e58610f
4 changed files with 34 additions and 21 deletions

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@ -23,3 +23,4 @@ clean:
-rm main.out -rm main.out
-rm main.pdf -rm main.pdf
-rm main.toc -rm main.toc
-rm main.bbl

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@ -10,39 +10,25 @@ $\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ whic
is helpful in the following discussion. is helpful in the following discussion.
A gate operating on a qbit is a unitary operator $G \in U(2)$. One can show that A gate operating on a qbit is a unitary operator $G \in U(2)$. One can show that
$\forall G \in U(2)$ $G$ can be expressed as a product of unitary generator matrices, $\forall G \in U(2)$ $G$ can be arbitrarily good as a product of unitary generator matrices\cite[Chapter 4.3]{kaye_ea2007},
common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ with common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ with
\label{ref:singleqbitgates} \label{ref:singleqbitgates}
$$X := \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right) $$ $$X := \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right) $$
$$Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) $$ $$Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) $$
$$H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right) $$ $$H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right) $$
$$R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right) $$ $$R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)$$
\subsection{$N$ Qbit Systems} \subsection{$N$ Qbit Systems}
\label{ref:nqbitsystems} \label{ref:nqbitsystems}
A system of $N$ identical qbits is described by the kronecker product of the
single qbit states. A nice definition for this product is:
\begin{definition}
Let $M$, $N$ be complex matrices, $M = \left(\begin{array}{cc} M_1 & M_2 \\ M_3 & M_4 \end{array}\right)$
. Then the kronecker product is defined as
$$M \otimes N = \left(\begin{array}{cc} M_1 \otimes N & M_2 \otimes N \\ M_3 \otimes N & M_4 \otimes N \end{array}\right)$$
Where for any $c \in \mathbb{C}$ $ c \otimes N := cN$.
\end{definition}
\begin{postulate} \begin{postulate}
A $N$ qbit quantum mechanical state is the kronecker product of the $N$ single qbit A $N$ qbit quantum mechanical state is the tensor product\cite[Definition 14.3]{wuest1995} of the $N$ single qbit
state. states. The $N$ qbit operators are the tensor product of the $N$ single qbit operators.
\end{postulate} \end{postulate}
Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$ Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$
be the basis of the one-qbit systems. Then two-qbit states are defined as be the basis of the one-qbit systems. Then two-qbit basis states are
$$ \ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)$$ $$ \ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)$$
$$ \ket{1} := \ket{0b01} := \ket{0}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right)$$ $$ \ket{1} := \ket{0b01} := \ket{0}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right)$$
@ -57,14 +43,15 @@ $$ \ket{\psi} = \sum\limits_{i = 0}^{2^N - 1} c_i \ket{i} $$
$$ \sum\limits_{i = 0}^{2^N - 1} |c_i|^2 = 1$$ $$ \sum\limits_{i = 0}^{2^N - 1} |c_i|^2 = 1$$
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. to generate an arbitrary $N$ qbit gate\cite[Chapter 4.3]{kaye_ea2007}.
$$ CX(1, 0) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$ $$ CX(1, 0) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$
$$ CZ(1, 0) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)$$ $$ CZ(1, 0) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)$$
Where $1$ is the act-qbit and $0$ the control-qbit. Where $1$ is the act-qbit and $0$ the control-qbit. In words $CX$ ($CZ$) apply an $X$ ($Z$) gate on the act-qbit,
if the control-qbit is set.
\subsection{Measurement} \subsection{Measurement}
@ -79,3 +66,13 @@ Where $1$ is the act-qbit and $0$ the control-qbit.
\end{postulate} \end{postulate}
Measuring a qbit will also yield a classical result $0$ or $1$ with the respective probabilities. Measuring a qbit will also yield a classical result $0$ or $1$ with the respective probabilities.
\begin{corrolary}
In general the measurement of a qbit is not invertible, in particular it cannot be represented as a
unitary operator.
\end{corrolary}
\begin{proof}
The measuerment in not injective: Measuring both
$\ket{0}$ and $\frac{1}{\sqrt{2}}(\ket{0} + \ket{1}$ (can) map to $\ket{0}$.
\end{proof}

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@ -30,3 +30,17 @@
author={Simon Anders and Hans J. Briegel}, author={Simon Anders and Hans J. Briegel},
year=2005 year=2005
} }
@book{
wuest1995,
title={Höhere Mathematik Für Physiker Teil 1},
author={Rainer Wüst},
year=1995,
publisher={de Gruyter}
}
@book{
kaye_ea2007,
title={An Introduction to Quantum Computing},
author={Phillip Kaye, Raymond Laflamme and Michelle Mosca},
year=2007,
publisher={Oxford University Press}
}

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@ -14,6 +14,7 @@
\newtheorem{definition}{Definition} \newtheorem{definition}{Definition}
\newtheorem{postulate}{Postulate} \newtheorem{postulate}{Postulate}
\newtheorem{corrolary}{Corrolary}
\title{Development of an Extensible Quantum Computing \title{Development of an Extensible Quantum Computing
Simulator with a Focus on Simulation in the Graph Formalism } Simulator with a Focus on Simulation in the Graph Formalism }