119 lines
4.9 KiB
TeX
119 lines
4.9 KiB
TeX
% vim: ft=tex
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\section{Introduction to Binary Quantum Computing}
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\subsection{Single Qbits}
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\label{ref:singleqbitsystems}
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A qbit is a two-level quantum mechanical system $ \{\ket{\uparrow} \equiv \ket{1}, \ket{\downarrow} \equiv \ket{0}\} $
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with $\braket{\uparrow}{\downarrow} = 0$. One can associate
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$\ket{\uparrow} \equiv \left(\begin{array}{c} 0 \\ 1\end{array} \right)$ and
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$\ket{\downarrow} \equiv \left(\begin{array}{c} 1 \\ 0 \end{array} \right)$ which
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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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$\forall G \in U(2)$ $G$ 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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\label{ref:singleqbitgates}
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\begin{equation}
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X := \left(\begin{array}{cc} 0 & 1 \\ 1 & 0\end{array}\right)
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\end{equation}
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\begin{equation}
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Z := \left(\begin{array}{cc} 1 & 0 \\ 0 & -1\end{array}\right)
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\end{equation}
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\begin{equation}
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H := \frac{1}{\sqrt{2}}\left(\begin{array}{cc} 1 & 1 \\ 1 & -1\end{array}\right)
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\end{equation}
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\begin{equation}
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R_{\phi} := \left(\begin{array}{cc} 1 & 0 \\ 0 & \exp(i\phi)\end{array}\right)
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\end{equation}
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\subsection{$N$ Qbit Systems}
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\label{ref:nqbitsystems}
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\begin{postulate}
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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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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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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 basis states are
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\begin{equation}
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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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\end{equation}
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\begin{equation}
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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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\end{equation}
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\begin{equation}
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\ket{2} := \ket{0b10} := \ket{1}_s \otimes \ket{0}_s := \left(\begin{array}{c} 0 \\ 0 \\ 1 \\ 0 \end{array} \right)
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\end{equation}
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\begin{equation}
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\ket{3} := \ket{0b11} := \ket{1}_s \otimes \ket{1}_s := \left(\begin{array}{c} 0 \\ 0 \\ 0 \\ 1 \end{array} \right)
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\end{equation}
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The $N$ qbit basis states can then be constructed in a similar manner.
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A general $N$ qbit state can then be written as a superposition of the
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basis states:
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\begin{equation}
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\ket{\psi} = \sum\limits_{i = 0}^{2^N - 1} c_i \ket{i}
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\end{equation}
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\begin{equation}
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\sum\limits_{i = 0}^{2^N - 1} |c_i|^2 = 1
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\end{equation}
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FIXME: rewrite this.
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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\cite[Chapter 4.3]{kaye_ea2007}.
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The matrix representation of $CX$ and $CZ$ for two qbits is given by
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\begin{equation}
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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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\end{equation}
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\begin{equation}
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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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\end{equation}
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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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The following notation\cite{dahlberg_ea2019} can be more handy when discussing more qbits:
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\begin{equation}\label{eq:CX_pr}
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CX(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes X_i
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\end{equation}
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\begin{equation}\label{eq:CZ_pr}
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CZ(i, j) = \ket{0}\bra{0}_j\otimes I_i + \ket{1}\bra{1}_j \otimes Z_i
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\end{equation}
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Where $i$ is the act-qbit, $j$ the control-qbit and $I_i$, $Z_i$ are the identity and Pauli $Z$ gate
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operating on qbit i. Note that $CX(i, j) = H(i) CZ(i, j) H(i)$ which will be important later.
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\subsection{Measurement}
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\begin{postulate}
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Let $\ket{\psi} = \alpha\ket{\phi_1} \otimes \ket{1}_n + \beta\ket{\phi_0} \otimes \ket{0}_n$ be a state
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where $\ket{1}_n, \ket{0}_n$ denote the $n$-th qbit state and $|\alpha|^2 + |\beta|^2 = 1$.
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Then the measurement of the $n$-th qbit will yield $\ket{\phi_1} \otimes \ket{1}_n$ with probability
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$|\alpha|^2$ and $\ket{\phi_0} \otimes \ket{0}_n$ with probability $|\beta|^2$.
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This is called collapse of the wave function.
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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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\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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Any unitary matrix $U$ has the inverse $U^\dagger \equiv U^{-1}$.
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\end{proof}
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