\section{Half a Qubit}

First, a pair of definitions. We've used both these terms implicitly in the previous section,
but we'll need to introduce proper definitions before we continue.

\definition{}
A \textit{linear combination} of two vectors $u$ and $v$ is the sum $au + bv$ for scalars $a$ and $b$. \par
\note[Note]{In other words, a linear combination is exactly what it sounds like.}

\definition{}
A \textit{normalized vector} (also called a \textit{unit vector}) is a vector with length 1.

\vspace{4mm}

\begin{tcolorbox}[
	enhanced,
	breakable,
	colback=white,
	colframe=ored,
	boxrule=0.6mm,
	arc=0mm,
	outer arc=0mm,
]
	\color{ored}
	\begingroup
		\large\centering
		\textbf{Disclaimer:} \par
	\endgroup
	\vspace{1ex}
	The \say{qubits} we're about to define aren't \textit{really} qubits. The proper definition is a bit more
	complicated, but don't worry about that yet. For now, take what I say as truth---we'll get to
	the complex definition soon enough.

	\vspace{2mm}

	The information provided in this handout does not, and is not intended to, constitute legal advice.
	All information, content, and material in this document is for general informational purposes only.
\end{tcolorbox}


\generic{Remark:}
Just like a classical bit, a \textit{quantum bit} (or \textit{qubit}) can take the values $\ket{0}$ and $\ket{1}$. \par
However, \texttt{0} and \texttt{1} aren't the only states a qubit may have.

\vspace{2mm}

We'll make sense of quantum bits by extending the \say{vectored} bit representation we developed in the previous section.
First, let's look at a diagram we drew a few pages ago:

\begin{ORMCbox}{Time Travel (Page 2)}{black!10!white}{black!65!white}
	A classical bit takes states in $\{\texttt{0}, \texttt{1}\}$, picking one at a time. \par
	We'll represent \texttt{0} and \texttt{1} as perpendicular unit vectors $\ket{0}$ and $\ket{1}$,
	show below.

	\begin{center}
		\begin{tikzpicture}[scale=1.5]
			\fill[color = black] (0, 0) circle[radius=0.05];

			\draw[->] (0, 0) -- (1.2, 0);
			\node[right] at (1.2, 0) {$\ket{0}$};
			\fill[color = oblue] (1, 0) circle[radius=0.05];
			\node[below] at (1, 0) {\texttt{0}};

			\draw[->] (0, 0) -- (0, 1.2);
			\node[above] at (0, 1.2) {$\ket{1}$};
			\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 $\ket{0}$, and all parts $\ket{1}$. \par
	Of course, we can say something similar about the point marked $0$: \par
	It is at $[1, 0] = (1 \times \ket{0}) + (0 \times \ket{1})$, and is thus all $\ket{0}$ and no $\ket{1}$. \par
\end{ORMCbox}

The diagram in the box above can also be used to describe the state of a qubit. \par
Like classical bits, qubits have the \textit{basis states} $\ket{0}$ and $\ket{1}$. \par
Unlike classical bits, qubits may take values that are some combination of both.


\vspace{2mm}

Namely, every possible state of a qubit is a \textit{normalized linear combination} of $\ket{0}$ and $\ket{1}$. \par
Such states are called \textit{superpositions} of $\ket{0}$ and $\ket{1}$, since they partially contain both states.


\vfill
\pagebreak














\definition{}
The state of a quantum bit is the column unit vector
$
	\ket{\psi}
	= \left[\begin{smallmatrix} a \\ b \end{smallmatrix}\right]
	= a\ket{0} + b\ket{1}$ for $a, b \in \mathbb{R}
$. \par

Note that the length of $\ket{\psi}$ must always be $1$, which is the same as saying that $a^2 + b^2 = 1$.

\vspace{2mm}
If we plot the set of valid quantum states on our plane, we get a unit circle centered at the origin: \par

\begin{center}
	\begin{tikzpicture}[scale=1.5]
		\draw[dashed] (0,0) circle(1);

		\fill[color = black] (0, 0) circle[radius=0.05];

		\draw[->] (0, 0) -- (1.2, 0);
		\fill[color = oblue] (1, 0) circle[radius=0.05];
		\node[below right] at (1, 0) {$\ket{0}$};

		\draw[->] (0, 0) -- (0, 1.2);
		\fill[color = oblue] (0, 1) circle[radius=0.05];
		\node[above left] at (0, 1) {$\ket{1}$};

		\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
		\node[above right] at (0.87, 0.5) {$\ket{\psi}$};
	\end{tikzpicture}
\end{center}

\problem{}
In the above diagram, the counterclockwise angle from $\ket{0}$ to $\ket{\psi}$ is $30^\circ$\hspace{-1ex}. \par
Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$.


\vfill


\definition{Measurement I}
Although a qubit may have many states, it must be $\ket{0}$ or $\ket{1}$ when we measure it. \par

\vspace{2mm}

As a trivial example, say $\ket{\psi}$ = $\ket{0}$. \par
If we were to measure $\ket{\psi}$, we'd get $\ket{0}$, and the state of the qubit wouldn't change.

\vspace{2mm}

However, something interesting happens when $\ket{\psi} = a\ket{0} + b\ket{1}$. \par
Our measurement again returns either $\ket{0}$ or $\ket{1}$, with the following probabilities: \par
\begin{itemize}[itemsep = 2mm, topsep = 2mm]
	\item $\mathcal{P}(\ket{1}) = a^2$
	\item $\mathcal{P}(\ket{0}) = b^2$
\end{itemize}
\note{
	Note that $\mathcal{P}(\ket{0}) + \mathcal{P}(\ket{1}) = 1$. \\
	As you already know, this is true of any probability function.
}


\vspace{2mm}

In addition, $\ket{\psi}$ \textit{collapses} when it is measured: it instantly changes its state to the result of the measurement,
leaving no trace of its previous state. \par
If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and
it will remain in that state until it is changed.
Quantum bits \textit{cannot} be measured without their state collapsing. \par

\pagebreak













\problem{}
\begin{itemize}
	\item What is the probability we get $\ket{0}$ if we measure $\ket{\psi_0}$? \par
	\item What outcomes can we get if we measure it a second time? \par
	\item What are these probabilities for $\ket{\psi_1}$?
\end{itemize}



\begin{center}
	\begin{tikzpicture}[scale=1.5]
		\draw[dashed] (0,0) circle(1);

		\fill[color = black] (0, 0) circle[radius=0.05];

		\draw[->] (0, 0) -- (1.2, 0);
		\fill[color = oblue] (1, 0) circle[radius=0.05];
		\node[below right] at (1, 0) {$\ket{0}$};

		\draw[->] (0, 0) -- (0, 1.2);
		\fill[color = oblue] (0, 1) circle[radius=0.05];
		\node[above left] at (0, 1) {$\ket{1}$};

		\draw[dotted] (0, 0) -- (0.87, 0.5);
		\draw[color=gray,->] (0.5, 0.0) arc (0:30:0.5);
		\node[right, color=gray] at (0.47, 0.12) {$30^\circ$};
		\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
		\node[above right] at (0.87, 0.5) {$\ket{\psi_0}$};


		\draw[dotted] (0, 0) -- (-0.707, -0.707);
		\draw[color=gray,->] (0.25, 0.0) arc (0:-135:0.25);
		\node[below, color=gray] at (0.2, -0.2) {$135^\circ$};
		\fill[color = ored] (-0.707, -0.707) circle[radius=0.05];
		\node[below left] at (-0.707, -0.707) {$\ket{\psi_1}$};
	\end{tikzpicture}
\end{center}



\vfill

As you may have noticed, we don't need two coordinates to fully define a quibit's state. \par
We can get by with one coordinate just as well.

\vspace{2mm}

Instead of referring to each state using its cartesian coordinates $a$ and $b$, \par
we can address it using its \textit{polar angle} $\theta$, measured from $\ket{0}$ counterclockwise:

\begin{center}
	\begin{tikzpicture}[scale=1.5]
		\draw[dashed] (0,0) circle(1);
		\fill[color = black] (0, 0) circle[radius=0.05];

		\draw[dotted] (0, 0) -- (0.707, 0.707);
		\draw[color=gray,->] (0.5, 0.0) arc (0:45:0.5);
		\node[above right, color=gray] at (0.5, 0) {$\theta$};

		\draw[->] (0, 0) -- (1.2, 0);
		\fill[color = oblue] (1, 0) circle[radius=0.05];
		\node[below right] at (1, 0) {$\ket{0}$};

		\draw[->] (0, 0) -- (0, 1.2);
		\fill[color = oblue] (0, 1) circle[radius=0.05];
		\node[above left] at (0, 1) {$\ket{1}$};

		\fill[color = ored] (0.707, 0.707) circle[radius=0.05];
		\node[above right] at (0.707, 0.707) {$\ket{\psi}$};
	\end{tikzpicture}
\end{center}


\problem{}
Find $a$ and $b$ in terms of $\theta$ for an arbitrary qubit.


\vfill
\pagebreak