268 lines
7.7 KiB
TeX
268 lines
7.7 KiB
TeX
\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{timetravel}
|
|
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{timetravel}
|
|
|
|
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} to the state we measure: it instantly jumps to the state we measure, \par
|
|
leaving no trace of its previous state. 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.
|
|
\vspace{2mm}
|
|
|
|
Quantum bits \textit{cannot} be measured without their state collapsing. \par
|
|
We cannot certainly know the state of a qubit unless that state is $\ket{0}$ or $\ket{1}$.
|
|
|
|
\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 |