347 lines
10 KiB
TeX
347 lines
10 KiB
TeX
\section{One Qubit}
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Quantum bits (or \textit{qubits}) are very similar to probabilistic bits, but have one major difference: \par
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probabilities are replaced with \textit{amplitudes}.
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\vspace{2mm}
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Of course, a qubit can take the values \texttt{0} and \texttt{1}, which are denoted $\ket{0}$ and $\ket{1}$. \par
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Like probabilistic bits, a quantum bit is written as a linear combination of $\ket{0}$ and $\ket{1}$:
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\begin{equation*}
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\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}
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\end{equation*}
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Such linear combinations are called \textit{superpositions}.
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\vspace{2mm}
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The $\ket{~}$ you see in the expressions above is called a \say{ket,} and denotes a column vector. \par
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$\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one.} This is called bra-ket notation. \par
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\note[Note]{$\bra{0}$ is called a \say{bra,} but we won't worry about that for now.}
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\vspace{2mm}
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This is very similiar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
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As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
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and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
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\vspace{8mm}
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Recall that probabilistic bits are subject to the restriction that $p_0 + p_1 = 1$. \par
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Quantum bits have a similar condition: $\psi_0^2 + \psi_1^2 = 1$. \par
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Note that this implies that $\psi_0$ and $\psi_1$ are both in $[-1, 1]$: \par
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Quantum amplitudes may be negative, but probabilistic bit probabilities cannot.
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\vspace{2mm}
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If we plot the set of valid quantum states on our plane, we get a unit circle centered at the origin:
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\begin{center}
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\begin{tikzpicture}[scale=1.5]
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\draw[dashed] (0,0) circle(1);
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[->] (0, 0) -- (1.2, 0);
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below right] at (1, 0) {$\ket{0}$};
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\draw[->] (0, 0) -- (0, 1.2);
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[above left] at (0, 1) {$\ket{1}$};
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\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
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\node[above right] at (0.87, 0.5) {$\ket{\psi}$};
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\end{tikzpicture}
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\end{center}
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Recall that the set of probabilistic bits forms a line instead:
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\begin{center}
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\begin{tikzpicture}[scale = 1.5]
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\fill[color = black] (0, 0) circle[radius=0.05];
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\node[below left] at (0, 0) {$\left[\begin{smallmatrix} 0 \\ 0 \end{smallmatrix}\right]$};
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\draw[ored, -, line width = 2] (0, 1) -- (1, 0);
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\draw[->] (0, 0) -- (1.2, 0);
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\node[right] at (1.2, 0) {$p_0$};
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below] at (1, 0) {$[0]$};
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\draw[->] (0, 0) -- (0, 1.2);
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\node[above] at (0, 1.2) {$p_1$};
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[left] at (0, 1) {$[1]$};
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\end{tikzpicture}
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\end{center}
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\problem{}
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In the above unit circle, the counterclockwise angle from $\ket{0}$ to $\ket{\psi}$ is $30^\circ$\hspace{-1ex}. \par
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Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$.
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\vfill
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\pagebreak
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\definition{Measurement I}
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Just like a probabilistic bit, we must observed $\ket{0}$ or $\ket{1}$ when we measure a qubit. \par
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If we were to measure $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$, we'd observe either $\ket{0}$ or $\ket{1}$, \par
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with the following probabilities:
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\begin{itemize}[itemsep = 2mm, topsep = 2mm]
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\item $\mathcal{P}(\ket{1}) = \psi_1^2$
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\item $\mathcal{P}(\ket{0}) = \psi_0^2$
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\end{itemize}
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\note{Note that $\mathcal{P}(\ket{0}) + \mathcal{P}(\ket{1}) = 1$.}
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\vspace{2mm}
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As before, $\ket{\psi}$ \textit{collapses} when it is measured: its state becomes that which we observed in our measurement,
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leaving no trace of the previous superposition. \par
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\problem{}
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\begin{itemize}
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\item What is the probability we observe $\ket{0}$ when we measure $\ket{\psi}$? \par
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\item What can we observe if we measure $\ket{\psi}$ a second time? \par
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\item What are these probabilities for $\ket{\varphi}$?
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\end{itemize}
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\begin{center}
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\begin{tikzpicture}[scale=1.5]
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\draw[dashed] (0,0) circle(1);
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[->] (0, 0) -- (1.2, 0);
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below right] at (1, 0) {$\ket{0}$};
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\draw[->] (0, 0) -- (0, 1.2);
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[above left] at (0, 1) {$\ket{1}$};
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\draw[dotted] (0, 0) -- (0.87, 0.5);
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\draw[color=gray,->] (0.5, 0.0) arc (0:30:0.5);
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\node[right, color=gray] at (0.47, 0.12) {$30^\circ$};
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\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
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\node[above right] at (0.87, 0.5) {$\ket{\psi}$};
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\draw[dotted] (0, 0) -- (-0.707, -0.707);
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\draw[color=gray,->] (0.25, 0.0) arc (0:-135:0.25);
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\node[below, color=gray] at (0.2, -0.2) {$135^\circ$};
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\fill[color = ored] (-0.707, -0.707) circle[radius=0.05];
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\node[below left] at (-0.707, -0.707) {$\ket{\varphi}$};
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\end{tikzpicture}
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\end{center}
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\vfill
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As you may have noticed, we don't need two coordinates to fully define a quibit's state. \par
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We can get by with one coordinate just as well.
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Instead of referring to each state using its cartesian coordinates $\psi_0$ and $\psi_1$, \par
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we can address it using its \textit{polar angle} $\theta$, measured from $\ket{0}$ counterclockwise:
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\begin{center}
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\begin{tikzpicture}[scale=1.5]
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\draw[dashed] (0,0) circle(1);
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[dotted] (0, 0) -- (0.707, 0.707);
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\draw[color=gray,->] (0.5, 0.0) arc (0:45:0.5);
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\node[above right, color=gray] at (0.5, 0) {$\theta$};
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\draw[->] (0, 0) -- (1.2, 0);
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below right] at (1, 0) {$\ket{0}$};
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\draw[->] (0, 0) -- (0, 1.2);
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[above left] at (0, 1) {$\ket{1}$};
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\fill[color = ored] (0.707, 0.707) circle[radius=0.05];
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\node[above right] at (0.707, 0.707) {$\ket{\psi}$};
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\end{tikzpicture}
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\end{center}
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\problem{}
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Find $\psi_0$ and $\psi_1$ in terms of $\theta$ for an arbitrary qubit $\psi$.
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\vfill
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\pagebreak
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\problem{}
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Consider the following qubit states:
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\null\hfill\begin{minipage}{0.48\textwidth}
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\begin{equation*}
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\ket{+} = \frac{\ket{0} + \ket{1}}{\sqrt{2}}
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\end{equation*}
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\end{minipage}\hfill\begin{minipage}{0.48\textwidth}
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\begin{equation*}
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\ket{-} = \frac{\ket{0} - \ket{1}}{\sqrt{2}}
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\end{equation*}
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\end{minipage}\hfill\null
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\begin{itemize}
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\item Where are these on the unit circle?
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\item What are their polar angles?
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\item What are the probabilities of observing $\ket{0}$ and $\ket{1}$ when measuring $\ket{+}$ and $\ket{-}$?
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\end{itemize}
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\vfill
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\begin{center}
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\begin{tikzpicture}[scale = 2.5]
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\draw[dashed] (0,0) circle(1);
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\fill[color = black] (0, 0) circle[radius=0.05];
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\draw[->] (0, 0) -- (1.2, 0);
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\fill[color = oblue] (1, 0) circle[radius=0.05];
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\node[below right] at (1, 0) {$\ket{0}$};
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\draw[->] (0, 0) -- (0, 1.2);
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\fill[color = oblue] (0, 1) circle[radius=0.05];
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\node[above left] at (0, 1) {$\ket{1}$};
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\end{tikzpicture}
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\end{center}
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\vfill
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\vfill
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\pagebreak
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\section{Operations on One Qubit}
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We may apply transformations to qubits just as we apply transformations to probabilistic bits.
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Again, we'll represent transformations as $2 \times 2$ matrices, since we want to map
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one qubit state to another. \par
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\note{In other words, we want to map elements of $\mathbb{R}^2$ to elements of $\mathbb{R}^2$.} \par
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We will call such maps \textit{quantum gates,} since they are the quantum equivalent of classical logic gates.
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\vspace{2mm}
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There are two conditions a valid quantum gate $G$ must satisfy:
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\begin{itemize}[itemsep = 1mm]
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\item For any valid state $\ket{\psi}$, $G\ket{\psi}$ is a valid state. \par
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Namely, $G$ must preserve the length of any vector it is applied to. \par
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Recall that the set of valid quantum states is the set of unit vectors in $\mathbb{R}^2$
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\item Any quantum gate must be \textit{invertible}. \par
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We'll skip this condition for now, and return to it later.
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\end{itemize}
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In short, a quantum gate is a linear map that maps the unit circle to itself. \par
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There are only two kinds of linear maps that do this: reflections and rotations.
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\problem{}
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The $X$ gate is the quantum analog of the \texttt{not} gate, defined by the following table:
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\begin{itemize}
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\item $X\ket{0} = \ket{1}$
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\item $X\ket{1} = \ket{0}$
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\end{itemize}
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Find the matrix $X$.
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\begin{solution}
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\begin{equation*}
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\begin{bmatrix}
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0 & 1 \\ 1 & 0
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\end{bmatrix}
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\end{equation*}
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\end{solution}
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\vfill
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\problem{}
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What is $X\ket{+}$ and $X\ket{-}$? \par
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\hint{Remember that all matrices are linear maps. What does this mean?}
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\begin{solution}
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$X\ket{+} = \ket{+}$ and $X\ket{-} = -\ket{-}$ (that is, negative ket-minus). \par
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Most notably, remember that $G(a\ket{0} + b\ket{1}) = aG\ket{0} + bG\ket{1}$
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\end{solution}
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\vfill
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\problem{}
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In terms of geometric transformations, what does $X$ do to the unit circle?
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\begin{solution}
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It is a reflection about the $45^\circ$ axis.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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Let $Z$ be a quantum gate defined by the following table: \par
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\begin{itemize}
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\item $Z\ket{0} = \ket{0}$,
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\item $Z\ket{1} = -\ket{1}$.
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\end{itemize}
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What is the matrix $Z$? What are $Z\ket{+}$ and $Z\ket{-}$? \par
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What is $Z$ as a geometric transformation?
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\vfill
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\problem{}
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Is the map $B$ defined by the table below a valid quantum gate?
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\begin{itemize}
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\item $B\ket{0} = \ket{0}$
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\item $B\ket{1} = \ket{+}$
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\end{itemize}
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\hint{Find a $\ket{\psi}$ so that $B\ket{\psi}$ is not a valid qubit state}
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\begin{solution}
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$B\ket{+} = \frac{1 + \sqrt{2}}{2}\ket{0} + \frac{1}{2}\ket{1}$, which has a non-unit length of $\frac{\sqrt{2} + 1}{\sqrt{2}}$.
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\end{solution}
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\vfill
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\problem{Rotation}
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As we noted earlier, any rotation about the center is a valid quantum gate. \par
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Let's derive all transformations of this form.
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\begin{itemize}[itemsep = 1mm]
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\item Let $U_\phi$ be the matrix that represents a counterclockwise rotation of $\phi$ degrees. \par
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What is $U\ket{0}$ and $U\ket{1}$?
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\item Find the matrix $U_\phi$ for an arbitrary $\phi$.
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\end{itemize}
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\vfill
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\problem{}
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Say we have a qubit that is either $\ket{+}$ or $\ket{-}$. We do not know which of the two states it is in. \par
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Using one operation and one measurement, how can we find out, for certain, which qubit we received? \par
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\vfill
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\pagebreak |