did some work
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@ -23,3 +23,4 @@ clean:
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-rm main.out
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-rm main.out
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-rm main.pdf
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-rm main.pdf
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-rm main.toc
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-rm main.toc
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-rm main.bbl
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@ -10,7 +10,7 @@ $\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ whic
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is helpful in the following discussion.
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is helpful in the following discussion.
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A gate operating on a qbit is a unitary operator $G \in U(2)$. One can show that
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A gate operating on a qbit is a unitary operator $G \in U(2)$. One can show that
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$\forall G \in U(2)$ $G$ can be expressed as a product of unitary generator matrices,
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$\forall G \in U(2)$ $G$ can be arbitrarily good as a product of unitary generator matrices\cite[Chapter 4.3]{kaye_ea2007},
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common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ with
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common choices for the generators are $ X, H, R_{\phi}$ and $Z, H, R_{\phi}$ with
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\label{ref:singleqbitgates}
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\label{ref:singleqbitgates}
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@ -22,27 +22,13 @@ $$R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)
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\subsection{$N$ Qbit Systems}
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\subsection{$N$ Qbit Systems}
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\label{ref:nqbitsystems}
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\label{ref:nqbitsystems}
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A system of $N$ identical qbits is described by the kronecker product of the
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single qbit states. A nice definition for this product is:
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\begin{definition}
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Let $M$, $N$ be complex matrices, $M = \left(\begin{array}{cc} M_1 & M_2 \\ M_3 & M_4 \end{array}\right)$
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. Then the kronecker product is defined as
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$$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)$$
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Where for any $c \in \mathbb{C}$ $ c \otimes N := cN$.
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\end{definition}
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\begin{postulate}
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\begin{postulate}
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A $N$ qbit quantum mechanical state is the kronecker product of the $N$ single qbit
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A $N$ qbit quantum mechanical state is the tensor product\cite[Definition 14.3]{wuest1995} of the $N$ single qbit
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state.
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states. The $N$ qbit operators are the tensor product of the $N$ single qbit operators.
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\end{postulate}
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\end{postulate}
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Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$
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Let $\ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ and $\ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \end{array} \right)$
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be the basis of the one-qbit systems. Then two-qbit states are defined as
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be the basis of the one-qbit systems. Then two-qbit basis states are
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$$ \ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)$$
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$$ \ket{0} := \ket{0b00} := \ket{0}_s \otimes \ket{0}_s := \left(\begin{array}{c} 1 \\ 0 \\ 0 \\ 0 \end{array} \right)$$
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$$ \ket{1} := \ket{0b01} := \ket{0}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right)$$
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$$ \ket{1} := \ket{0b01} := \ket{0}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 1 \\ 0 \\ 0 \end{array} \right)$$
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@ -57,14 +43,15 @@ $$ \ket{\psi} = \sum\limits_{i = 0}^{2^N - 1} c_i \ket{i} $$
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$$ \sum\limits_{i = 0}^{2^N - 1} |c_i|^2 = 1$$
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$$ \sum\limits_{i = 0}^{2^N - 1} |c_i|^2 = 1$$
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One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough
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One can show that the gates in \ref{ref:singleqbitgates} together with an entanglement gate, such as $CX$ or $CZ$ are enough
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to generate an arbitrary $N$ qbit gate.
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to generate an arbitrary $N$ qbit gate\cite[Chapter 4.3]{kaye_ea2007}.
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$$ 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)$$
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$$ 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)$$
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$$ 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)$$
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$$ 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)$$
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Where $1$ is the act-qbit and $0$ the control-qbit.
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Where $1$ is the act-qbit and $0$ the control-qbit. In words $CX$ ($CZ$) apply an $X$ ($Z$) gate on the act-qbit,
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if the control-qbit is set.
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\subsection{Measurement}
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\subsection{Measurement}
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@ -79,3 +66,13 @@ Where $1$ is the act-qbit and $0$ the control-qbit.
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\end{postulate}
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\end{postulate}
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Measuring a qbit will also yield a classical result $0$ or $1$ with the respective probabilities.
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Measuring a qbit will also yield a classical result $0$ or $1$ with the respective probabilities.
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\begin{corrolary}
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In general the measurement of a qbit is not invertible, in particular it cannot be represented as a
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unitary operator.
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\end{corrolary}
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\begin{proof}
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The measuerment in not injective: Measuring both
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$\ket{0}$ and $\frac{1}{\sqrt{2}}(\ket{0} + \ket{1}$ (can) map to $\ket{0}$.
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\end{proof}
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@ -30,3 +30,17 @@
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author={Simon Anders and Hans J. Briegel},
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author={Simon Anders and Hans J. Briegel},
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year=2005
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year=2005
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}
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}
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@book{
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wuest1995,
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title={Höhere Mathematik Für Physiker Teil 1},
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author={Rainer Wüst},
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year=1995,
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publisher={de Gruyter}
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}
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@book{
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kaye_ea2007,
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title={An Introduction to Quantum Computing},
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author={Phillip Kaye, Raymond Laflamme and Michelle Mosca},
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year=2007,
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publisher={Oxford University Press}
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}
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@ -14,6 +14,7 @@
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\newtheorem{definition}{Definition}
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\newtheorem{definition}{Definition}
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\newtheorem{postulate}{Postulate}
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\newtheorem{postulate}{Postulate}
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\newtheorem{corrolary}{Corrolary}
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\title{Development of an Extensible Quantum Computing
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\title{Development of an Extensible Quantum Computing
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Simulator with a Focus on Simulation in the Graph Formalism }
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Simulator with a Focus on Simulation in the Graph Formalism }
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