Mark/handouts
1
0
Fork 0
Code Pull Requests 2 Actions Packages Releases Activity

Added Tectonic.toml

Browse Source
This commit is contained in:
Mark
2024-02-17 08:57:20 -08:00
parent 7d4a27d57f
commit f590e0f252
9 changed files with 38 additions and 0 deletions
Download Patch File Download Diff File Expand all files Collapse all files

591
Advanced/Introduction to Quantum/src/parts/01 bits.tex Normal file
View File

@ -0,0 +1,591 @@
\section{Probabilistic Bits}
\definition{}
As we already know, a \textit{classical bit} may take the values \texttt{0} and \texttt{1}. \par
We can model this with a two-sided coin, one face of which is labeled \texttt{0}, and the other, \texttt{1}. \par
\vspace{2mm}
Of course, if we toss such a \say{bit-coin,} we'll get either \texttt{0} or \texttt{1}. \par
We'll denote the probability of getting \texttt{0} as $p_0$, and the probability of getting \texttt{1} as $p_1$. \par
As with all probabilities, $p_0 + p_1$ must be equal to 1.
\vfill
\definition{}
Say we toss a \say{bit-coin} and don't observe the result. We now have a \textit{probabilistic bit}, with a probability $p_0$
of being \texttt{0}, and a probability $p_1$ of being \texttt{1}.
\vspace{2mm}
We'll represent this probabilistic bit's \textit{state} as a vector:
$\left[\begin{smallmatrix}
p_0 \\ p_1
\end{smallmatrix}\right]$ \par
We do \textbf{not} assume this coin is fair, and thus $p_0$ might not equal $p_1$.
\note{
This may seem a bit redundant: since $p_0 + p_1$, we can always calculate one probability given the other. \\
We'll still include both probabilities in the state vector, since this provides a clearer analogy to quantum bits.
}
\vfill
\definition{}
The simplest probabilistic bit states are of course $[0]$ and $[1]$, defined as follows:
\begin{itemize}
\item $[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
\item $[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$
\end{itemize}
That is, $[0]$ represents a bit that we known to be \texttt{0}, \par
and $[1]$ represents a bit we know to be \texttt{1}.
\vfill
\definition{}
$[0]$ and $[1]$ form a \textit{basis} for all possible probabilistic bit states: \par
Every other probabilistic bit can be written as a \textit{linear combination} of $[0]$ and $[1]$:
\begin{equation*}
\begin{bmatrix} p_0 \\ p_1 \end{bmatrix}
=
p_0 \begin{bmatrix} 1 \\ 0 \end{bmatrix} +
p_1 \begin{bmatrix} 0 \\ 1 \end{bmatrix}
=
p_0 [0] + p_1 [1]
\end{equation*}
\vfill
\pagebreak
\problem{}
Every possible state of a probabilistic bit is a two-dimensional vector. \par
Draw all possible states on the axis below.
\begin{center}
\begin{tikzpicture}[scale = 2.0]
\fill[color = black] (0, 0) circle[radius=0.05];
\node[below left] at (0, 0) {$\left[\begin{smallmatrix} 0 \\ 0 \end{smallmatrix}\right]$};
\draw[->] (0, 0) -- (1.2, 0);
\node[right] at (1.2, 0) {$p_0$};
\fill[color = oblue] (1, 0) circle[radius=0.05];
\node[below] at (1, 0) {$[0]$};
\draw[->] (0, 0) -- (0, 1.2);
\node[above] at (0, 1.2) {$p_1$};
\fill[color = oblue] (0, 1) circle[radius=0.05];
\node[left] at (0, 1) {$[1]$};
\end{tikzpicture}
\end{center}
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale = 2.0]
\fill[color = black] (0, 0) circle[radius=0.05];
\node[below left] at (0, 0) {$\left[\begin{smallmatrix} 0 \\ 0 \end{smallmatrix}\right]$};
\draw[ored, -, line width = 2] (0, 1) -- (1, 0);
\draw[->] (0, 0) -- (1.2, 0);
\node[right] at (1.2, 0) {$p_0$};
\fill[color = oblue] (1, 0) circle[radius=0.05];
\node[below] at (1, 0) {$[0]$};
\draw[->] (0, 0) -- (0, 1.2);
\node[above] at (0, 1.2) {$p_1$};
\fill[color = oblue] (0, 1) circle[radius=0.05];
\node[left] at (0, 1) {$[1]$};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\pagebreak
\section{Measuring Probabilistic Bits}
\definition{}
As we noted before, a probabilistic bit represents a coin we've tossed but haven't looked at. \par
We do not know whether the bit is \texttt{0} or \texttt{1}, but we do know the probability of both of these outcomes. \par
\vspace{2mm}
If we \textit{measure} (or \textit{observe}) a probabilistic bit, we see either \texttt{0} or \texttt{1}---and thus our
knowledge of its state is updated to either $[0]$ or $[1]$, since we now certainly know what face the coin landed on.
\vspace{2mm}
Since measurement changes what we know about a probabilistic bit, it changes the probabilistic bit's state.
When we measure a bit, it's state \textit{collapses} to either $[0]$ or $[1]$, and the original state of the
bit vanishes. We \textit{cannot} recover the state $[x_0, x_1]$ from a measured probabilistic bit.
\definition{Multiple bits}
Say we have two probabilistic bits, $x$ and $y$, \par
with states
$[x]=[ x_0, x_1]$
and
$[y]=[y_0, y_1]$
\vspace{2mm}
The \textit{compound state} of $[x]$ and $[y]$ is exactly what it sounds like: \par
It is the probabilistic two-bit state $\ket{xy}$, where the probabilities of the first bit are
determined by $[x]$, and the probabilities of the second are determined by $[y]$.
\problem{}<firstcompoundstate>
Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nicefrac{1}{4}]$. \par
\begin{itemize}[itemsep = 1mm]
\item If we measure $x$ and $y$ simultaneously, \par
what is the probability of getting each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}?
\item If we measure $y$ first and observe \texttt{1}, \par
what is the probability of getting each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}?
\end{itemize}
\note[Note]{$[x]$ and $[y]$ are column vectors, but I've written them horizontally to save space.}
\vfill
\problem{}
With $x$ and $y$ defined as above, find the probability of measuring each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}.
\vfill
\problem{}
Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nicefrac{1}{4}]$. \par
What is the probability that $x$ and $y$ produce different outcomes?
\vfill
\pagebreak
\section{Tensor Products}
\definition{Tensor Products}
The \textit{tensor product} of two vectors is defined as follows:
\begin{equation*}
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
\otimes
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
=
\begin{bmatrix}
x_1
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\\[4mm]
x_2
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
x_1y_1 \\[1mm]
x_1y_2 \\[1mm]
x_2y_1 \\[1mm]
x_2y_2 \\[0.5mm]
\end{bmatrix}
\end{equation*}
That is, we take our first vector, multiply the second
vector by each of its components, and stack the result.
You could think of this as a generalization of scalar
mulitiplication, where scalar mulitiplication is a
tensor product with a vector in $\mathbb{R}^1$:
\begin{equation*}
a
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
=
\begin{bmatrix}
a_1
\end{bmatrix}
\otimes
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
=
\begin{bmatrix}
a_1
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
a_1y_1 \\[1mm]
a_1y_2
\end{bmatrix}
\end{equation*}
\problem{}
Say $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. \par
What is the dimension of $x \otimes y$?
\vfill
\problem{}<basistp>
What is the pairwise tensor product
$
\Bigl\{
\left[
\begin{smallmatrix}
1 \\ 0 \\ 0
\end{smallmatrix}
\right],
\left[
\begin{smallmatrix}
0 \\ 1 \\ 0
\end{smallmatrix}
\right],
\left[
\begin{smallmatrix}
0 \\ 0 \\ 1
\end{smallmatrix}
\right]
\Bigr\}
\otimes
\Bigl\{
\left[
\begin{smallmatrix}
1 \\ 0
\end{smallmatrix}
\right],
\left[
\begin{smallmatrix}
0 \\ 1
\end{smallmatrix}
\right]
\Bigr\}
$?
\note{in other words, distribute the tensor product between every pair of vectors.}
\vfill
\problem{}
What is the \textit{span} of the vectors we found in \ref{basistp}? \par
In other words, what is the set of vectors that can be written as linear combinations of the vectors above?
\vfill
Look through the above problems and convince yourself of the following fact: \par
If $a$ is a basis of $A$ and $b$ is a basis of $B$, $a \otimes b$ is a basis of $A \times B$. \par
\note{If you don't understand what this says, ask an instructor. \\ This is the reason we did the last few problems!}
\begin{instructornote}
\textbf{The idea here is as follows:}
If $a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$,
the values $ab$ can take are
$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$.
\vspace{2mm}
The same is true of any other state set: if $a$ takes values in $A$ and $b$ takes values in $B$, \par
the compound state $(a,b)$ takes values in $A \times B$.
\vspace{2mm}
We would like to do the same with probabilistic bits. \par
Given bits $\ket{a}$ and $\ket{b}$, how should we represent the state of $\ket{ab}$?
\end{instructornote}
\pagebreak
\problem{}
Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nicefrac{1}{4}]$. \par
What is $[x] \otimes [y]$? How does this relate to \ref{firstcompoundstate}?
\vfill
\problem{}
The compound state of two vector-form bits is their tensor product. \par
Compute the following. Is the result what we'd expect?
\begin{itemize}
\item $[0] \otimes [0]$
\item $[0] \otimes [1]$
\item $[1] \otimes [0]$
\item $[1] \otimes [1]$
\end{itemize}
\hint{
Remember that
$[0] = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
and
$[1] = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
}
\vfill
\problem{}<fivequant>
Of course, writing $[0] \otimes [1]$ is a bit excessive. We'll shorten this notation to $[01]$. \par
\vspace{2mm}
In fact, we could go further: if we wanted to write the set of bits $[1] \otimes [1] \otimes [0] \otimes [1]$, \par
we could write $[1101]$---but a shorter alternative is $[13]$, since $13$ is \texttt{1101} in binary.
\vspace{2mm}
Write $[5]$ as three-bit probabilistic state. \par
\begin{solution}
$[5] = [101] = [1] \otimes [0] \otimes [1] = [0,0,0,0,0,1,0,0]^T$ \par
Notice how we're counting from the top, with $[000] = [1,0,...,0]$ and $[111] = [0, ..., 0, 1]$.
\end{solution}
\vfill
\problem{}
Write the three-bit states $[0]$ through $[7]$ as column vectors. \par
\hint{You do not need to compute every tensor product. Do a few and find the pattern.}
\vfill
\pagebreak
\section{Operations on Probabilistic Bits}
Now that we can write probabilistic bits as vectors, we can represent operations on these bits
with linear transformations---in other words, as matrices.
\definition{}
Consider the NOT gate, which operates as follows: \par
\begin{itemize}
\item $\text{NOT}[0] = [1]$
\item $\text{NOT}[1] = [0]$
\end{itemize}
What should NOT do to a probabilistic bit $[x_0, x_1]$? \par
If we return to our coin analogy, we can think of the NOT operation as
flipping a coin we have already tossed, without looking at it's state.
Thus,
\begin{equation*}
\text{NOT} \begin{bmatrix}
x_0 \\ x_1
\end{bmatrix} = \begin{bmatrix}
x_1 \\ x_0
\end{bmatrix}
\end{equation*}
\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black!65!white}
Matrix multiplication works as follows:
\begin{equation*}
AB =
\begin{bmatrix}
1 & 2 \\
3 & 4 \\
\end{bmatrix}
\begin{bmatrix}
a_0 & b_0 \\
a_1 & b_1 \\
\end{bmatrix}
=
\begin{bmatrix}
1a_0 + 2a_1 & 1b_0 + 2b_1 \\
3a_0 + 4a_1 & 3b_0 + 4b_1 \\
\end{bmatrix}
\end{equation*}
Note that this is very similar to multiplying each column of $B$ by $A$. \par
The product $AB$ is simply $Ac$ for every column $c$ in $B$:
\begin{equation*}
Ac_0 =
\begin{bmatrix}
1 & 2 \\
3 & 4 \\
\end{bmatrix}
\begin{bmatrix}
a_0 \\ a_1
\end{bmatrix}
=
\begin{bmatrix}
1a_0 + 2a_1 \\
3a_0 + 4a_1
\end{bmatrix}
\end{equation*}
This is exactly the first column of the matrix product. \par
Also, note that each element of $Ac_0$ is the dot product of a row in $A$ and a column in $c_0$.
\end{ORMCbox}
\problem{}
Compute the following product:
\begin{equation*}
\begin{bmatrix}
1 & 0.5 \\ 0 & 1
\end{bmatrix}
\begin{bmatrix}
3 \\ 2
\end{bmatrix}
\end{equation*}
\vfill
\generic{Remark:}
Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangably.
\pagebreak
\problem{}
Find the matrix that represents the NOT operation on one probabilistic bit.
\begin{solution}
\begin{equation*}
\begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\problem{Extension by linearity}
Say we have an arbitrary operation $A$. \par
If we know how $A$ acts on $[1]$ and $[0]$, can we compute $A[x]$ for an arbitrary state $[x]$? \par
Say $[x] = [x_0, x_1]$.
\begin{itemize}
\item What is the probability we observe $0$ when we measure $x$?
\item What is the probability that we observe $M[0]$ when we measure $Mx$?
\end{itemize}
\vfill
\problem{}<linearextension>
Write $M[x_0, x_1]$ in terms of $M[0]$, $M[1]$, $x_0$, and $x_1$.
\begin{solution}
\begin{equation*}
M \begin{bmatrix}
x_0 \\ x_1
\end{bmatrix}
=
x_0 M \begin{bmatrix}
1 \\ 0
\end{bmatrix}
+
x_1 M \begin{bmatrix}
0 \\ 1
\end{bmatrix}
=
x_0 M [0] +
x_1 M [1]
\end{equation*}
\end{solution}
\vfill
\generic{Remark:}
Every matrix represents a \textit{linear} map, so the following is always true:
\begin{equation*}
A \times (px + qy) = pAx + qAy
\end{equation*}
\ref{linearextension} is just a special case of this fact.
\pagebreak

347
Advanced/Introduction to Quantum/src/parts/02 qubit.tex Normal file
View File

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

145
Advanced/Introduction to Quantum/src/parts/03 two qubits.tex Normal file
View File

@ -0,0 +1,145 @@
\section{Two Qubits}
\definition{}
Just as before, we'll represent multi-quibit states as linear combinations of multi-qubit basis states. \par
For example, a two-qubit state $\ket{ab}$ is the four-dimensional unit vector
\begin{equation}
\begin{bmatrix}
a \\ b \\ c \\ d
\end{bmatrix}
= a \ket{00} + b\ket{01} + c\ket{10} + d\ket{11}
\end{equation}
As always, multi-qubit states are unit vectors. \par
Thus, $a^2 + b^2 + c^2 + d^2 = 1$ in the two-bit case above.
\problem{}
Say we have two qubits $\ket{\psi}$ and $\ket{\varphi}$. \par
Show that $\ket{\psi} \otimes \ket{\varphi}$ is always a unit vector (and is thus a valid quantum state).
\vfill
\definition{Measurement II}<measureii>
Measurement of a two-qubit state works just like measurement of a one-qubit state: \par
If we measure $a\ket{00} + b\ket{01} + c\ket{10} + d\ket{11}$, \par
we get one of the four basis states with the following probabilities:
\begin{itemize}
\item $\mathcal{P}(\ket{00}) = a^2$
\item $\mathcal{P}(\ket{01}) = b^2$
\item $\mathcal{P}(\ket{10}) = c^2$
\item $\mathcal{P}(\ket{11}) = d^2$
\end{itemize}
Of course, the sum of all the above probabilities is $1$.
\problem{}
Consider the two-qubit state
$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$
\begin{itemize}[itemsep=2mm]
\item If we measure both bits of $\ket{\psi}$ simultaneously, \par
what is the probability of getting each of $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$?
\item If we measure the ONLY the first qubit, what is the probability we get $\ket{0}$? How about $\ket{1}$? \par
\hint{There are two basis states in which the first qubit is $\ket{0}$.}
\item Say we measured the second bit and read $\ket{1}$. \par
If we now measure the first bit, what is the probability of getting $\ket{0}$?
\end{itemize}
\vfill
\pagebreak
\problem{}
Again, consider the two-qubit state
$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$ \par
If we measure the first qubit of $\ket{\psi}$ and get $\ket{0}$, what is the resulting state of $\ket{\psi}$? \par
What would the state be if we'd measured $\ket{1}$ instead?
\vfill
\problem{}
Consider the three-qubit state $\ket{\psi} = c_0\ket{000} + c_1\ket{001} + ... + c_7 \ket{111}$. \par
Say we measure the first two qubits and get $\ket{00}$. What is the resulting state of $\ket{\psi}$?
\begin{solution}
We measure $\ket{00}$ with probability $c_0^2 + c_1^2$, and $\ket{\psi}$ collapses to
\begin{equation*}
\frac{c_0\ket{000} + c_1\ket{001}}{\sqrt{c_0^2 + c_1^2}}
\end{equation*}
\end{solution}
\vfill
\pagebreak
\definition{Entanglement}
Some product states can be factored into a tensor product of individual qubit states. For example,
\begin{equation*}
\frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr)
= \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes
\frac{1}{\sqrt{2}}\bigl( \ket{0} - \ket{1} \bigr)
\end{equation*}
Such states are called \textit{product states.} States that aren't product states are called \textit{entangled} states.
\problem{}
Factor the following product state:
\begin{equation*}
\frac{1}{2\sqrt{2}} \bigl(\sqrt{3}\ket{00} - \sqrt{3}\ket{01} + \ket{10} - \ket{11}\bigr)
\end{equation*}
\begin{solution}
\begin{equation*}
\frac{1}{2\sqrt{2}} \biggl(\sqrt{3}\ket{00} - \sqrt{3}\ket{01} + \ket{10} - \ket{11}\biggr)
= \biggl( \frac{\sqrt{3}}{2}\ket{0} + \frac{1}{2}\ket{1} \biggr) \otimes
\biggl(\frac{1}{\sqrt{2}}\ket{0} - \frac{1}{\sqrt{2}}\ket{1} \biggr)
\end{equation*}
\end{solution}
\vfill
\problem{}
Show that the following is an entangled state.
\begin{equation*}
\frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}
\end{equation*}
\begin{solution}
$
\left[
\begin{smallmatrix}
a_0 \\ a_1
\end{smallmatrix}
\right]
\otimes
\left[
\begin{smallmatrix}
b_0 \\ b_1
\end{smallmatrix}
\right]
=
a_0b_0\ket{00} + a_0b_1\ket{01} + a_1b_0\ket{10} + a_1b_1\ket{11}
$
\vspace{2mm}
So, we have that $a_1b_1 = a_0b_0 = \sqrt{2}^{-1}$ \par
But $a_0b_1 = a_1b_0 = 0$, so one of $a_0$ and $b_1$ must be zero. \par
We thus have a contradiction.
\end{solution}
\vfill
\pagebreak

311
Advanced/Introduction to Quantum/src/parts/04 logic gates.tex Normal file
View File

@ -0,0 +1,311 @@
\section{Logic Gates}
\definition{Matrices}
Throughout this handout, we've been using matrices. Again, recall that every linear map may be written as a matrix,
and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
map, we can write it as follows:
\begin{equation*}
f\left(
\ket{x}
\right)
=
\begin{bmatrix}
m_1 & m_2 \\
m_3 & m_4
\end{bmatrix}
\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
=
\left[
\begin{matrix}
m_1x_1 + m_2x_2 \\
m_3x_1 + m_4x_2
\end{matrix}
\right]
\end{equation*}
\definition{}
Before we discussing multi-qubit quantum gates, we need to review to classical logic. \par
Of course, a classical logic gate is a linear map from $\mathbb{B}^m$ to $\mathbb{B}^n$
\problem{}<notgatex>
The \texttt{not} gate is a map from $\mathbb{B}$ to $\mathbb{B}$ defined by the following table: \par
\begin{itemize}
\item $X\ket{0} = \ket{1}$
\item $X\ket{1} = \ket{0}$
\end{itemize}
Write the \texttt{not} gate as a matrix that operates on single-bit vector states. \par
That is, find a matrix $X$ so that
$
X\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]
= \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]
$
and
$
X\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]
= \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]
$ \par
\begin{solution}
\begin{equation*}
X = \begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\problem{}
The \texttt{and} gate is a map $\mathbb{B}^2 \to \mathbb{B}$ defined by the following table:
\begin{center}
\begin{tabular}{ c | c | c }
\hline
\texttt{a} & \texttt{b} & \texttt{a} and \texttt{b} \\
\hline
0 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 1
\end{tabular}
\end{center}
Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
\begin{solution}
\begin{equation*}
A = \begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{equation*}
\begin{instructornote}
Because of the way we represent bits here, we also have the following property: \par
The columns of $A$ correspond to the output for each input---i.e, $A$ is just a table of outputs. \par
\vspace{2mm}
For example, if we look at the first column of $A$ (which is $[1, 0]$), we see: \par
$A\ket{00} = A[1,0,0,0] = [1,0] = \ket{0}$
\vspace{2mm}
Also with the last column (which is $[0,1]$): \par
$A\ket{00} = A[0,0,0,1] = [0,1] = \ket{1}$
\end{instructornote}
\end{solution}
\vfill
\pagebreak
\generic{Remark:}
The way a quantum circuit handles information is a bit different than the way a classical circuit does.
We usually think of logic gates as \textit{functions}: they consume one set of bits, and return another:
\begin{center}
\begin{tikzpicture}[circuit logic US, scale=2]
\node[and gate] (and) at (0,-0.8) {\tiny\texttt{and}};
\draw[->] ([shift={(-0.5, 0)}] and.input 1) node[left] {\texttt{input A}} -- ([shift={(-0.25, 0)}]and.input 1);
\draw[->] ([shift={(-0.5, 0)}] and.input 2) node[left] {\texttt{input B}} -- ([shift={(-0.25, 0)}]and.input 2);
\draw ([shift={(-0.25, 0)}] and.input 1) -- (and.input 1);
\draw ([shift={(-0.25, 0)}] and.input 2) -- (and.input 2);
\draw[->] (and.output) -- ([shift={(0.5, 0)}] and.output) node[right] {\texttt{output}};
\end{tikzpicture}
\end{center}
This model, however, won't work for quantum logic. If we want to understand quantum gates, we need to see them
not as \textit{functions}, but as \textit{transformations}. This distinction is subtle, but significant:
\begin{itemize}
\item functions \textit{consume} a set of inputs and \textit{produce} a set of outputs
\item transformations \textit{change} a set of objects, without adding or removing any elements
\end{itemize}
\vspace{2mm}
Our usual logic circuit notation models logic gates as functions---we thus can't use it. \par
We'll need a different diagram to draw quantum circuits. \par
\vfill
First, we'll need a set of bits. For this example, we'll use two, drawn in a vertical array. \par
We'll also add a horizontal time axis, moving from left to right:
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {\texttt{0}};
\node[qubit] (b) at (0, -1) {\texttt{1}};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {\texttt{0}};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {\texttt{1}};
\draw[
color = oblue,
->>,
line width = 0.5mm
] (-1,-1.5) -- (5, -1.5);
\node[fill=white, text=oblue] at (2, -1.5) {\texttt{time axis}};
\node[left, gray] at (-1, -0.5) {State of each bit at start};
\node[right, gray] at (5, -0.5) {State of each bit at end};
\draw[
->,
color = gray,
line width = 0.2mm,
rounded corners = 2mm
]
(-1, -0.5) -- (-0.8, -0.5) -- (-0.8, 0) --(a)
;
\draw[
->,
color = gray,
line width = 0.2mm,
rounded corners = 2mm
]
(-1, -0.5) -- (-0.8, -0.5) -- (-0.8, -1) -- (b)
;
\draw[
<-,
color = gray,
line width = 0.2mm,
rounded corners = 2mm
]
(4.2, 0) -- (4.8, 0) -- (4.8, -0.5) -- (5, -0.5)
;
\draw[
<-,
color = gray,
line width = 0.2mm,
rounded corners = 2mm
]
(4.2, -1) -- (4.8, -1) -- (4.8, -0.5) -- (5, -0.5)
;
\end{tikzpicture}
\end{center}
In the diagram above, we didn't change our bits---so the labels at the start match those at the end.
\vfill
Thus, our circuit forms a grid, with bits ordered vertically and time horizontally. \par
If we want to change our state, we draw transformations as vertical boxes. \par
Every column represents a single transformation on the entire state:
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {\texttt{1}};
\node[qubit] (b) at (0, -1) {\texttt{0}};
\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {\texttt{0}};
\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {\texttt{1}};
\ghostqubox{a}{1}{b}{2}{$T_1$}
\ghostqubox{a}{2}{b}{3}{$T_2$}
\ghostqubox{a}{3}{b}{4}{$T_3$}
\end{tikzpicture}
\end{center}
Note that the transformations above span the whole state. This is important: \par
we cannot apply transformations to individual bits---we always transform the \textit{entire} state.
\vfill
\pagebreak
\generic{Setup:}
Say we want to invert the first bit of a two-bit state. That is, we want a transformation $T$ so that \par
\begin{center}
\begin{tikzpicture}[scale=0.8]
\node[qubit] (a) at (0, 0) {\texttt{a}};
\node[qubit] (b) at (0, -1) {\texttt{b}};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {\texttt{not a}};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {\texttt{b}};
\qubox{a}{1.5}{b}{2.5}{$T$}
\end{tikzpicture}
\end{center}
In other words, we want a matrix $T$ satisfying the following equalities:
\begin{itemize}
\item $T\ket{00} = \ket{10}$
\item $T\ket{01} = \ket{11}$
\item $T\ket{10} = \ket{00}$
\item $T\ket{11} = \ket{01}$
\end{itemize}
\problem{}
Find the matrix that corresponds to the above transformation. \par
\hint{
Remember that
$\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$ and
$\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$ \\
Also, we found earlier that $X = \left[\begin{smallmatrix} 0 && 1 \\ 1 && 0 \end{smallmatrix}\right]$,
and of course $I = \left[\begin{smallmatrix} 1 && 0 \\ 0 && 1 \end{smallmatrix}\right]$.
}
\begin{solution}
\begin{equation*}
T = \begin{bmatrix}
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\generic{Remark:}
We could draw the above transformation as a combination $X$ and $I$ (identity) gate:
\begin{center}
\begin{tikzpicture}[scale=0.8]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
\qubox{a}{1}{a}{2}{$X$}
\qubox{b}{1}{b}{2}{$I$}
\end{tikzpicture}
\end{center}
We can even omit the $I$ gate, since we now know that transformations affect the whole state: \par
\begin{center}
\begin{tikzpicture}[scale=0.8]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
\qubox{a}{1}{a}{2}{$X$}
\end{tikzpicture}
\end{center}
We're now done: this is how we draw quantum circuits.
Don't forget that transformations \textit{always} affect the whole state---even if our diagram doesn't explicitly state this.
\pagebreak
% TODO:
% distributive property of tensor product
% quantum gate algebra

234
Advanced/Introduction to Quantum/src/parts/05 quantum gates.tex Normal file
View File

@ -0,0 +1,234 @@
\section{Quantum Gates}
In the previous section, we stated that a quantum gate is a linear map. \par
Let's complete that definition.
\definition{}
A quantum gate is a \textit{orthonormal matrix}, which means any gate $G$
satisfies $GG^\text{T} = I$. \par
This implies the following: \par
\begin{itemize}
\item $G$ is square \par
\note{
If we think of $G$ as a map, this means that $G$ has as many inputs as it has outputs. \\
This is to be expected: we stated earlier that quantum gates do not destroy or create qubits.
}
\item $G$ preserves lengths; i.e $|x| = |Gx|$. \par
\note{This ensures that $G\ket{\psi}$ is always a valid state.}
\end{itemize}
(You will prove all these properties in any introductory linear algebra course. \\
This isn't a lesson on linear algebra, so you may take them as given today.)
\definition{}
Let $\mathbb{U} \subset \mathbb{R}^2$ be the set of points in the unit circle. \par
We can restate the above definition as follows: \par
A quantum gate is an invertible map from $\mathbb{U}^n$ to $\mathbb{U}^n$.
\definition{}<qgateislinear>
Let $G$ be a quantum gate. \par
Since quantum gates are, by definition, \textit{linear} maps,
the following holds: \par
\begin{equation*}
G\bigl(a_0 \ket{0} + a_1\ket{1}\bigr) = a_0G\ket{0} + a_1G\ket{1}
\end{equation*}
\problem{}<cnot>
Consider the \textit{controlled not} (or \textit{cnot}) gate, defined by the following table: \par
\begin{itemize}
\item $\text{X}_\text{c}\ket{00} = \ket{00}$
\item $\text{X}_\text{c}\ket{01} = \ket{11}$
\item $\text{X}_\text{c}\ket{10} = \ket{10}$
\item $\text{X}_\text{c}\ket{11} = \ket{01}$
\end{itemize}
In other words, the cnot gate inverts its first bit if its second bit is $\ket{1}$. \par
Find the matrix that applies the cnot gate.
\begin{solution}
\begin{equation*}
\text{X}_\text{c} = \left[\begin{smallmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{smallmatrix}\right]
\end{equation*}
\vspace{4mm}
If $\ket{a}$ is $\ket{0}$, $\ket{a} \otimes \ket{b}$ is
$
\left[
\begin{smallmatrix}
\left[
\begin{smallmatrix}
b_1 \\ b_2
\end{smallmatrix}
\right]
\\ 0 \\ 0
\end{smallmatrix}
\right]
$, and the \say{not} portion of the matrix is ignored.
\vspace{4mm}
If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is
$
\left[
\begin{smallmatrix}
0 \\ 0 \\
\left[
\begin{smallmatrix}
b_1 \\ b_2
\end{smallmatrix}
\right]
\end{smallmatrix}
\right]
$, and the \say{identity} portion of the matrix is ignored.
The state of $\ket{a}$ is always preserved, since it's determined by the position of
$\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ in the tensor product.
If $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on top, $\ket{a}$ is $\ket{0}$,
and if $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on the bottom, $\ket{a}$ is $\ket{1}$.
\end{solution}
\vfill
\pagebreak
\problem{}<applycnot>
Evaluate the following:
\begin{equation*}
\text{X}_\text{C}
\Bigl(
\frac{1}{2}\ket{00} +
\frac{1}{2}\ket{01} -
\frac{1}{2}\ket{10} -
\frac{1}{2}\ket{11}
\Bigr)
\end{equation*}
\vfill
\problem{}
If we measure the result of \ref{applycnot}, what are the probabilities of getting each state?
\vfill
\problem{}
Finally, modify the original cnot gate so that the roles of its bits are reversed: \par
$\text{X}_\text{c, flipped} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$.
\begin{solution}
\begin{equation*}
\text{X}_\text{c, flipped} = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\pagebreak
\definition{}
The \textit{Hadamard Gate} is given by the following matrix: \par
\begin{equation*}
H = \frac{1}{\sqrt{2}}\begin{bmatrix}
1 & 1 \\
1 & -1
\end{bmatrix}
\end{equation*}
\note{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal.}
\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black!65!white}
Matrix multiplication works as follows:
\begin{equation*}
AB =
\begin{bmatrix}
1 & 2 \\
3 & 4 \\
\end{bmatrix}
\begin{bmatrix}
a_0 & b_0 \\
a_1 & b_1 \\
\end{bmatrix}
=
\begin{bmatrix}
1a_0 + 2a_1 & 1b_0 + 2b_1 \\
3a_0 + 4a_1 & 3b_0 + 4b_1 \\
\end{bmatrix}
\end{equation*}
Note that this is very similar to multiplying each column of $B$ by $A$. \par
The product $AB$ is simply $Ac$ for every column $c$ in $B$:
\begin{equation*}
Ac_0 =
\begin{bmatrix}
1 & 2 \\
3 & 4 \\
\end{bmatrix}
\begin{bmatrix}
a_0 \\ a_1
\end{bmatrix}
=
\begin{bmatrix}
1a_0 + 2a_1 \\
3a_0 + 4a_1
\end{bmatrix}
\end{equation*}
This is exactly the first column of the matrix product. \par
Also, note that each element of $Ac_0$ is the dot product of a row in $A$ and a column in $c_0$.
\end{ORMCbox}
\problem{}
What is $HH$? \par
Using this result, find $H^{-1}$.
\begin{solution}
$HH = I$, so $H^{-1} = H$
\end{solution}
\vfill
\problem{}
What geometric transformation does $H$ apply to the unit circle? \par
\hint{Rotation or reflection? How much, or about which axis?}
\vfill
\problem{}
What are $H\ket{0}$ and $H\ket{1}$? \par
Are these states entangled?
\begin{solution}
$H\ket{0} = \frac{1}{\sqrt{2}}\bigl(\ket{0} + \ket{1}\bigr)$ and $H\ket{1} = \frac{1}{\sqrt{2}}\bigl(\ket{0} - \ket{1}\bigr)$ \par
Both of these are entangled states.
\end{solution}
\vfill
\pagebreak