\section{One Bit} Before we discuss quantum computation, we first need to construct a few tools. \par To keep things simple, we'll use regular (usually called \textit{classical}) bits for now. \definition{Binary Digits} $\mathbb{B}$ is the set of binary digits. In other words, $\mathbb{B} = \{\texttt{0}, \texttt{1}\}$. \par \note[Note]{We've seen $\mathbb{B}$ before---it's the set of integers mod 2.} \vspace{2mm} \definition{Cartesian Products} Let $A$ and $B$ be sets. \par The \textit{cartesian product} $A \times B$ is the set of all pairs $(a, b)$ where $a \in A$ and $b \in B$. \par As usual, we can write $A \times A \times A$ as $A^3$. \par \vspace{2mm} In this handout, we'll often see the following sets: \begin{itemize} \item $\mathbb{R}^2$, a two-dimensional plane \item $\mathbb{R}^n$, an n-dimensional space \item $\mathbb{B}^2$, the set $\{(\texttt{0},\texttt{0}), (\texttt{0},\texttt{1}), (\texttt{1},\texttt{0}), (\texttt{1},\texttt{1})\}$ \item $\mathbb{B}^n$, the set of all possible states of $n$ bits. \end{itemize} \problem{} What is the size of $\mathbb{B}^n$? \vfill \pagebreak % NOTE: this is time-travelled later in the handout. % if you edit this, edit that too. \cgeneric{Remark} Consider a single classical bit. It takes states in $\{\texttt{0}, \texttt{1}\}$, picking one at a time. \par The states \texttt{0} and \texttt{1} are fully independent. They are completely disjoint; they share no parts. \par We'll therefore say that \texttt{0} and \texttt{1} \textit{orthogonal} (or equivalently, \textit{perpendicular}). \par \vspace{2mm} We can draw $\vec{e}_0$ and $\vec{e}_1$ as perpendicular axis on a plane to represent this: \begin{center} \begin{tikzpicture}[scale=1.5] \fill[color = black] (0, 0) circle[radius=0.05]; \draw[->] (0, 0) -- (1.5, 0); \node[right] at (1.5, 0) {$\vec{e}_0$ axis}; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {\texttt{0}}; \draw[->] (0, 0) -- (0, 1.5); \node[above] at (0, 1.5) {$\vec{e}_1$ axis}; \fill[color = oblue] (0, 1) circle[radius=0.05]; \node[left] at (0, 1) {\texttt{1}}; \end{tikzpicture} \end{center} The point marked $1$ is at $[0, 1]$. It is no parts $\vec{e}_0$, and all parts $\vec{e}_1$. \par Of course, we can say something similar about the point marked $0$: \par It is at $[1, 0] = (1 \times \vec{e}_0) + (0 \times \vec{e}_1)$, and is thus all $\vec{e}_0$ and no $\vec{e}_1$. \par \vspace{2mm} Naturally, the coordinates $[0, 1]$ and $[1, 0]$ denote how much of each axis a point \say{contains.} \par We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts $\vec{e}_0$ and $\vec{e}_1$: \par \note[Note]{ We could also write $\texttt{x} = \vec{e}_0 + \vec{e}_1$ explicitly. \\ I've drawn \texttt{x} as a point on the left, and as a sum on the right. } \null\hfill \begin{minipage}{0.48\textwidth} \begin{center} \begin{tikzpicture}[scale=1.5] \fill[color = black] (0, 0) circle[radius=0.05]; \draw[->] (0, 0) -- (1.5, 0); \node[right] at (1.5, 0) {$\vec{e}_0$ axis}; \draw[->] (0, 0) -- (0, 1.5); \node[above] at (0, 1.5) {$\vec{e}_1$ axis}; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {\texttt{0}}; \fill[color = oblue] (0, 1) circle[radius=0.05]; \node[left] at (0, 1) {\texttt{1}}; \draw[dashed, color = gray, ->] (0, 0) -- (0.9, 0.9); \fill[color = oblue] (1, 1) circle[radius=0.05]; \node[above right] at (1, 1) {\texttt{x}}; \end{tikzpicture} \end{center} \end{minipage} \hfill \begin{minipage}{0.48\textwidth} \begin{center} \begin{tikzpicture}[scale=1.5] \fill[color = black] (0, 0) circle[radius=0.05]; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {\texttt{0}}; \fill[color = oblue] (0, 1) circle[radius=0.05]; \node[left] at (0, 1) {\texttt{1}}; \draw[dashed, color = gray, ->] (0, 0) -- (0.9, 0.0); \draw[dashed, color = gray, ->] (1, 0.1) -- (1, 0.9); \fill[color = oblue] (1, 1) circle[radius=0.05]; \node[above right] at (1, 1) {\texttt{x}}; \end{tikzpicture} \end{center} \end{minipage} \hfill\null \vspace{4mm} But \texttt{x} isn't a member of $\mathbb{B}$, it's not a valid state. \par Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, there's nothing in between. \par \note{ Note that the unit vectors $\vec{e}_0$ and $\vec{e}_1$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. } \vfill \pagebreak \definition{Vectored Bits} This brings us to what we'll call the \textit{vectored representation} of a bit. \par Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them into their components: \par \null\hfill \begin{minipage}{0.48\textwidth} \[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{e}_0) + (0 \times \vec{e}_1) \] \end{minipage} \hfill \begin{minipage}{0.48\textwidth} \[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{e}_0) + (1 \times \vec{e}_1) \] \end{minipage} \hfill\null \vspace{2mm} This may seem needlessly complex---and it is, for classical bits. \par We'll see why this is useful soon enough. \generic{One more thing:} The $\ket{~}$ you see in the two expressions above is called a \say{ket,} and denotes a column vector. \par $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one.} \par This is called bra-ket notation. $\bra{0}$ is called a \say{bra,} but we won't worry about that for now. \problem{} Write \texttt{x} and \texttt{y} in the diagram below in terms of $\ket{0}$ and $\ket{1}$. \par \begin{center} \begin{tikzpicture}[scale=1.5] \fill[color = black] (0, 0) circle[radius=0.05]; \draw[->] (0, 0) -- (1.5, 0); \node[right] at (1.5, 0) {$\vec{e}_0$ axis}; \draw[->] (0, 0) -- (0, 1.5); \node[above] at (0, 1.5) {$\vec{e}_1$ axis}; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {$\ket{0}$}; \fill[color = oblue] (0, 1) circle[radius=0.05]; \node[left] at (0, 1) {$\ket{1}$}; \draw[dashed, color = gray, ->] (0, 0) -- (0.9, 0.9); \fill[color = ored] (1, 1) circle[radius=0.05]; \node[above right] at (1, 1) {\texttt{x}}; \draw[dashed, color = gray, ->] (0, 0) -- (-0.9, 0.9); \fill[color = ored] (-1, 1) circle[radius=0.05]; \node[above right] at (-1, 1) {\texttt{y}}; \end{tikzpicture} \end{center} \vfill \pagebreak