From 519e58610f958bd419802bdb8b47a8a8213b48b7 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?Daniel=20Kn=C3=BCttel?= Date: Thu, 17 Oct 2019 19:32:25 +0200 Subject: [PATCH] did some work --- thesis/Makefile | 1 + thesis/chapters/introduction_qc.tex | 39 +++++++++++++---------------- thesis/main.bib | 14 +++++++++++ thesis/main.tex | 1 + 4 files changed, 34 insertions(+), 21 deletions(-) diff --git a/thesis/Makefile b/thesis/Makefile index d9c9da1..4cd2f38 100644 --- a/thesis/Makefile +++ b/thesis/Makefile @@ -23,3 +23,4 @@ clean: -rm main.out -rm main.pdf -rm main.toc + -rm main.bbl diff --git a/thesis/chapters/introduction_qc.tex b/thesis/chapters/introduction_qc.tex index 66d2b5b..5c0d6aa 100644 --- a/thesis/chapters/introduction_qc.tex +++ b/thesis/chapters/introduction_qc.tex @@ -10,39 +10,25 @@ $\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ whic is helpful in the following discussion. 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 \label{ref:singleqbitgates} $$X := \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right) $$ $$Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right) $$ $$H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right) $$ -$$R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right) $$ +$$R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)$$ \subsection{$N$ Qbit Systems} \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} -A $N$ qbit quantum mechanical state is the kronecker product of the $N$ single qbit -state. + A $N$ qbit quantum mechanical state is the tensor product\cite[Definition 14.3]{wuest1995} of the $N$ single qbit +states. The $N$ qbit operators are the tensor product of the $N$ single qbit operators. \end{postulate} Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$ -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 $$ \ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)$$ $$ \ket{1} := \ket{0b01} := \ket{0}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right)$$ @@ -57,14 +43,15 @@ $$ \ket{\psi} = \sum\limits_{i = 0}^{2^N - 1} c_i \ket{i} $$ $$ \sum\limits_{i = 0}^{2^N - 1} |c_i|^2 = 1$$ 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}. $$ CX(1, 0) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0 \end{array}\right)$$ $$ CZ(1, 0) = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & -1 \end{array}\right)$$ -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} @@ -79,3 +66,13 @@ Where $1$ is the act-qbit and $0$ the control-qbit. \end{postulate} 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} diff --git a/thesis/main.bib b/thesis/main.bib index c73d0b0..dce6d16 100644 --- a/thesis/main.bib +++ b/thesis/main.bib @@ -30,3 +30,17 @@ author={Simon Anders and Hans J. Briegel}, 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} +} diff --git a/thesis/main.tex b/thesis/main.tex index aed4be6..8f1b5d5 100644 --- a/thesis/main.tex +++ b/thesis/main.tex @@ -14,6 +14,7 @@ \newtheorem{definition}{Definition} \newtheorem{postulate}{Postulate} +\newtheorem{corrolary}{Corrolary} \title{Development of an Extensible Quantum Computing Simulator with a Focus on Simulation in the Graph Formalism }