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% Copyright (C) 2023 <Mark (mark@betalupi.com)>
%
% This program is free software: you can redistribute it and/or modify
% it under the terms of the GNU General Public License as published by
% the Free Software Foundation, either version 3 of the License, or
% (at your option) any later version.
%
% You may have received a copy of the GNU General Public License
% along with this program. If not, see <https://www.gnu.org/licenses/>.
%
%
%
% If you edit this, please give credit!
% Quality handouts take time to make.
% use the [nosolutions] flag to hide solutions,
% use the [solutions] flag to show solutions.
\documentclass[
solutions,
singlenumbering
]{../../resources/ormc_handout}
\usepackage{../../resources/macros}
\def\ket#1{\left|#1\right\rangle}
\def\bra#1{\left\langle#1\right|}
\usepackage{units}
\input{tikzset}
\uptitlel{Advanced 2}
\uptitler{Winter 2022}
\title{Intro to Quantum Computing I}
\subtitle{Prepared by \githref{Mark} on \today{}}
\begin{document}
\maketitle
\input{parts/0 bits}
\input{parts/1 gates}
\end{document}

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\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{}
$\mathbb{B}$ 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 $(\mathbb{Z}_2, +)$, the addition group mod 2.}
\vspace{2mm}
Multiplication in $\mathbb{B}$ works just as you'd expect: \par
$
\texttt{0} \times \texttt{0} =
\texttt{0} \times \texttt{1} =
\texttt{1} \times \texttt{0} = \texttt{0}
$; and $\texttt{1} \times \texttt{1} = \texttt{1}$.
\vspace{2mm}
We'll treat addition a bit differently: \par
$0 + 0 = 0$ and $0 + 1 = 1$, but $1 + 1$, for our purposes, is undefined.
\definition{}
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 $\{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$
\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
\generic{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 $\vec{0}$ and $\vec{1}$ 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) {$\vec{0}$ 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) {$\vec{1}$ 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 $\vec{0}$, and all parts $\vec{1}$. \par
Of course, we can say something similar about the point marked $0$: \par
It is at $[1, 0] = (1 \times \vec{0}) + (2 \times \vec{1})$. In other words, all $\vec{0}$ and no $\vec{1}$. \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 $\vec{0}$ and $\vec{1}$: \par
\note[Note]{
We could also write $\texttt{x} = \vec{0} + \vec{1}$ 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) {$\vec{0}$ axis};
\draw[->] (0, 0) -- (0, 1.5);
\node[above] at (0, 1.5) {$\vec{1}$ 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 $\vec{0}$ or fully $\vec{1}$. There's nothing in between.
\vspace{8mm}
\definition{}
The unit vectors $\vec{0}$ and $\vec{1}$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. \par
\note{
\say{ortho-} means \say{orthogonal}; normal means \say{normal,} which means length $= 1$. \\
}{
Note that $\vec{0}$ and $\vec{1}$ are orthonormal by \textit{definition}. \\
We don't have to prove anything, we simply defined them as such.
} \par
\vspace{2mm}
There's much more to say about basis vectors, but we don't need all the tools of linear algebra here. \par
We just need to understand that a set of $n$ orthogonal unit vectors defines an $n$-dimensional space. \par
This is fairly easy to think about: each vector corresponds to an axis of the space, and every point
in that space can be written as a \textit{linear combination} (i.e, a weighted sum) of these basis vectors.
\vspace{2mm}
For example, the set $\{[1,0,0], [0,1,0], [0,0,1]\}$ (which we usually call $\{x, y, z\})$
forms an orthonormal basis of $\mathbb{R}^3$. Every element of $\mathbb{R}^3$ can be written as a linear combination of these vectors:
\begin{equation*}
\left[\begin{smallmatrix} a \\ b \\ c \end{smallmatrix}\right]
=
a \left[\begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix}\right] +
b \left[\begin{smallmatrix} 0 \\ 1 \\ 0 \end{smallmatrix}\right] +
c \left[\begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix}\right]
\end{equation*}
The tuple $[a,b,c]$ is called the \textit{coordinate} of a point with respect to this basis.
\vfill
\pagebreak
\definition{}
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}
\[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{0}) + (0 \times \vec{1}) \]
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{0}) + (1 \times \vec{1}) \]
\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) {$\vec{0}$ axis};
\draw[->] (0, 0) -- (0, 1.5);
\node[above] at (0, 1.5) {$\vec{1}$ 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
\section{Two Bits}
How do we represent multi-bit states using vectors? \par
Unfortunately, this is hard to visualize---but the idea is simple.
\problem{}
What is the set of possible states of two bits (i.e, $\mathbb{B}^2$)?
\vspace{2cm}
\generic{Remark:}
When we have two bits, we have four orthogonal states:
$\overrightarrow{00}$, $\overrightarrow{01}$, $\overrightarrow{10}$, and $\overrightarrow{11}$. \par
We need four dimensions to draw all of these vectors, so I can't provide a picture... \par
but the idea here is the same as before.
\problem{}
Write $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$ as column vectors \par
with respect to the orthonormal basis $\{\overrightarrow{00}, \overrightarrow{01}, \overrightarrow{10}, \overrightarrow{11}\}$.
\vfill
\generic{Remark:}
So, we represent each possible state as an axis in an $n$-dimensional space. \par
A set of $n$ bits gives us $2^n$ possible states, which forms a basis in $2^n$ dimensions.
\vspace{1mm}
Say we now have two seperate bits: $\ket{a}$ and $\ket{b}$. \par
How do we represent their compound state? \par
\vspace{4mm}
If we return to our usual notation, this is very easy:
$a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$, \par
so the possible compound states of $ab$ are
$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$
\vspace{1mm}
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 $A \times B$. This is trivial.
\vspace{4mm}
We would like to do the same in vector notation. Given $\ket{a}$ and $\ket{b}$, \par
how should we represent the state of $\ket{ab}$?
\vfill
\pagebreak
\definition{}
The \textit{tensor product} between 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*}
\vspace{2mm}
Also, note that the tensor product is very similar to the
Cartesian product: if we take $x$ and $y$ as sets, with
$x = \{x_1, x_2\}$ and $y = \{y_1, y_2\}$, the Cartesian product
contains the same elements as the tensor product---every possible
pairing of an element in $x$ with an element in $y$:
\begin{equation*}
x \times y = \{~(x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2y_2)~\}
\end{equation*}
In fact, these two operations are (in a sense) essentially identical. \par
Let's quickly demonstrate this.
\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{}
The vectors we found in \ref{basistp} are a basis of what space? \par
\vfill
\pagebreak
\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 $\ket{0} \otimes \ket{0}$
\item $\ket{0} \otimes \ket{1}$
\item $\ket{1} \otimes \ket{0}$
\item $\ket{1} \otimes \ket{1}$
\end{itemize}
\hint{
Remember that the coordinates of
$\ket{0}$ are $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$,
and the coordinates of
$\ket{1}$ are $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
}
\vfill
\problem{}<fivequant>
Of course, writing $\ket{0} \otimes \ket{1}$ is a bit excessive. \par
We'll shorten this notation to $\ket{01}$. \par
Thus, the two-bit kets we saw on the previous page are, by definition, tensor products.
\vspace{2mm}
In fact, we could go further: if we wanted to write the set of bits $\ket{1} \otimes \ket{1} \otimes \ket{0} \otimes \ket{1}$, \par
we could write $\ket{1101}$---but a shorter alternative is $\ket{13}$, since $13$ is \texttt{1101} in binary.
\vspace{2mm}
Write $\ket{5}$ as three-bit state vector. \par
\begin{solution}
$\ket{5} = \ket{101} = \ket{1} \otimes \ket{0} \otimes \ket{1} = [0,0,0,0,0,1,0,0]^T$ \par
Notice how we're counting from the top, with $\ket{000} = [1,0,...,0]$ and $\ket{111} = [0, ..., 0, 1]$.
\end{solution}
\vfill
\problem{}
Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
What do you see?
\vfill
\pagebreak

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\section{Logic Gates}
Now that we know how to write vectored bits, let's look at the ways we can change them.
\generic{Remark:}
A few weeks ago, we talked about matrices. 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*}
\problem{}
The \say{not} gate is a map from $\mathbb{B}$ to $\mathbb{B}$ defined by the following table:
\begin{itemize}
\item $\text{not}(\texttt{1}) = \texttt{0}$
\item $\text{not}(\texttt{0}) = \texttt{1}$
\end{itemize}
Write the not gate as a matrix that operates on single-bit vector states. \par
That is, find a matrix $N$ so that $N\ket{0} = \ket{1}$ and $N\ket{1} = \ket{0}$.
\begin{solution}
\begin{equation*}
N = \begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\problem{}
The \say{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
\hint{
What is the dimension of the input? What is the dimension of the desired output? \\
What do these two values tell us about the dimension of the matrix $A$?
}
\begin{solution}
\begin{equation*}
A = \begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\pagebreak
\generic{Remark:}
The way a quantum computer handles information is a bit different than the way a classical computer does.
For spooky physics reasons, all quantum gates must be invertible. Naturally, this implies that the usual
logic gates we use (and, or, xor) aren't valid quantum gates: each of these takes two inputs (four states)
and produces one output (two states), losing information by mapping many inputs to the same output. \par
\note[Note]{The \say{not} gate is fine. It is its own inverse, after all!}
\vspace{2mm}
Although we are still using classical bits, we'll design our gates to be reversible. \par
One consequence of the \say{reversibility} rule is that all quantum gate matrices must be square. \par
(i.e, they must take the same number of inputs and outputs.) \par
\note[Note]{All invertible matrices are square, but not all square matrices are invertible.}
\vspace{2mm}
This is fairly intuitive: if we have more inputs than we have outputs, we inevitably lose information.
\vspace{2mm}
There's also a better way to think about this: rather than seeing quantum gates as \textit{functions}
that consume one set of bits and produce another, it's better to think of them as \textit{transformations}
we apply to an existing set of bits.
\problem{}
For example, take the CNOT (controlled not) gate. \par
When applied to a two-bit state $\ket{ab}$, CNOT inverts $b$ iff $a$ is $\ket{1}$. \par
Find the matrix that represents the CNOT gate. \par
\hint{what are the dimensions of this matrix?}
\begin{solution}
\begin{equation*}
\text{CNOT} = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{bmatrix}
\end{equation*}
\vspace{4mm}
If $\ket{a}$ is $\ket{0}$, $\ket{a} \otimes \ket{b}$ is
$
\begin{bmatrix}
\begin{bmatrix}
b_1 \\ b_2
\end{bmatrix} \\ 0 \\ 0
\end{bmatrix}
$, and the \say{not} portion of the matrix is ignored.
\vspace{4mm}
If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is
$
\begin{bmatrix}
0 \\ 0 \\
\begin{bmatrix}
b_1 \\ b_2
\end{bmatrix}
\end{bmatrix}
$, 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
\problem{}
Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
\begin{solution}
\begin{equation*}
\text{CNOT}_{\text{mod}} = \begin{bmatrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\pagebreak
\iffalse
\problem{}
Finally, modify the original CNOT gate so that the roles of its bits are reversed: \par
$\text{CNOT}_{\text{flip}} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$.
\begin{solution}
\begin{equation*}
\text{CNOT}_{\text{flip}} = \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
\fi
\problem{}
The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par
Find its matrix.
\begin{solution}
\begin{equation*}
\text{SWAP} = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\problem{}
The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par
In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par
Find the $T$ gate's matrix. \par
\note{
This gate is particularly interesting because it's a \textit{universal quantum gate}: \\
like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates.
}
\begin{solution}
\begin{equation*}
\text{T} = \begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\pagebreak
The last thing we need is a way to draw complex sequences of gates. \par
We already know how to do this with classical gates:
\begin{center}
\begin{tikzpicture}[circuit logic US, scale=1.5]
\node[and gate] (and) at (0,-0.8) {\tiny\texttt{and}};
\draw ([shift={(-0.5, 0)}] and.input 1) node[left] {\texttt{1}} -- (and.input 1);
\draw ([shift={(-0.5, 0)}] and.input 2) node[left] {\texttt{0}} -- (and.input 2);
\draw (and.output) -- ([shift={(0.5, 0)}] and.output) node[right] {\texttt{0}};
\end{tikzpicture}
\end{center}
We draw quantum circuits in a very similar way. \par
For example, here a simple three-bit circuit consisting of a CNOT gate on the first bit, \par
controlled by the third. The first bit is X'd iff the third bit is $\ket{1}$:
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\node[qubit] (c) at (0, -2) {$\ket{1}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$};
\draw[wire] (c) -- ([shift={(4, 0)}] c.center) node[qubit] {$\ket{1}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
;
\draw[wirejoin]
($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\end{tikzpicture}
\end{center}
\problem{}
Draw the CNOT gate as a classical logic circuit. \par
\hint{This can be done with one gate.}
\begin{solution}
\begin{center}
\begin{tikzpicture}[circuit logic US, scale=2]
\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$a$};
\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$b$};
\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
\draw (dot) -- (dot -| 1,0) node[right] {$b_\text{out}$};
\draw (xor.output) -- (xor.output -| 1,0) node[right] {$a_\text{out}$};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
Gate controls may be marked with a filled circle or an empty circle. \par
Empty circles denote \textit{inverse controls,} which (of course) have an inverse effect. \par
For example, the two circuits below are identical:
\null\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{a}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{b}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
;
\draw[wireijoin]
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{a}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{b}$};
\draw[wire]
($([shift={(1.5,0)}] a)!0.5!([shift={(2.5,0)}] a)$) --
($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
;
\draw[wirejoin]
($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\qubox{b}{0.5}{b}{1.5}{X};
\qubox{b}{2.5}{b}{3.5}{X};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill\null
\problem{}
What are $\ket{a}$ and $\ket{b}$ in the diagrams above?
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
;
\draw[wireijoin]
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\pagebreak
As we noted before, quantum gates are \textit{transformations,} modifying a set of bits. \par
Thus, quantum circuits are drawn with a fixed set of bits, whose states transform with time. \par
\note[Note]{
In this diagram, CNOT and SWAP are drawn as $\oplus$ and \rotatebox[origin=c]{90}{$\leftrightarrows$} to save space.
}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{a}$};
\node[qubit] (b) at (0, -1) {$\ket{1}$};
\node[qubit] (c) at (0, -2) {$\ket{c}$};
\node[qubit] (d) at (0, -3) {$\ket{d}$};
\node[qubit] (e) at (0, -4) {$\ket{1}$};
\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$};
\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{b}$};
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$};
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$};
\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{e}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
;
\draw[wirejoin]
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wireijoin]
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{2}{T};
\qubox{a}{4}{a}{5}{X};
\draw[wire]
($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
;
\draw[wireijoin]
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{e}{2}{e}{3}{$\oplus$};
\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}};
\draw[wire]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
;
\draw[wirejoin]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wirejoin]
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{b}{5}{b}{6}{T};
\end{tikzpicture}
\end{center}
In a quantum circuit, the ONLY way two bits can interact is through a gate. \par
We cannot add \say{branches} in quantum circuits like we do in classical circuits:
\begin{center}
\begin{tikzpicture}[circuit logic US, scale=1.2]
\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$x$};
\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$y$};
\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
\draw (dot) -- (dot -| 1,0) node[right] {$y_\text{out}$};
\draw (xor.output) -- (xor.output -| 1,0) node[right] {$x_\text{out}$};
\draw[->, line width = 1, ogrape]
([shift={(0.3,-0.5)}] dot) node[right] {This is a branch}
-| ([shift={(0,-0.2)}] dot)
;
\end{tikzpicture}
\end{center}
\problem{}
Find the values of $\ket{a}$ through $\ket{e}$ in the above circuit.
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{1}$};
\node[qubit] (b) at (0, -1) {$\ket{1}$};
\node[qubit] (c) at (0, -2) {$\ket{1}$};
\node[qubit] (d) at (0, -3) {$\ket{0}$};
\node[qubit] (e) at (0, -4) {$\ket{1}$};
\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$};
\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{1}$};
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$};
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$};
\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{0}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
;
\draw[wirejoin]
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wireijoin]
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{2}{T};
\qubox{a}{4}{a}{5}{X};
\draw[wire]
($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
;
\draw[wireijoin]
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{e}{2}{e}{3}{$\oplus$};
\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}};
\draw[wire]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
;
\draw[wirejoin]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wirejoin]
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{b}{5}{b}{6}{T};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\pagebreak

53
Advanced/Introduction to Quantum/tikzset.tex Normal file
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@ -0,0 +1,53 @@
\usetikzlibrary{arrows.meta}
\usetikzlibrary{shapes.geometric}
\usetikzlibrary{calc}
\usetikzlibrary{circuits.logic.US}
\usetikzlibrary{plotmarks}
\tikzset{
gate/.style = {
draw,
rectangle,
fill = white,
line width = 0.35mm
},
qubit/.style = {
fill = \ORMCbgcolor,
line width = 0.35mm
},
wire/.style = {
line width = 1
},
wirejoin/.style = {
fill = oblue,
draw = oblue,
line width = 1.5
},
wireijoin/.style = {
fill = white,
draw = oblue,
line width = 1.5
},
}
% Macros
\def\qubox#1#2#3#4#5{
% 1: point ne
% 2: point ne x offset
% 3: point sw
% 4: point sw x offset
% 5: label text
\draw[
line width = 1,
fill = white,
draw = black
]
([shift={(#2 + 0.1, 0.4)}] #1.center)
-- ([shift={(#2 + 0.1, -0.4)}] #1.center |- #3.center)
-- ([shift={(#4 - 0.1, -0.4)}] #3.center)
-- ([shift={(#4 - 0.1, 0.4)}] #1.center -| #3.center)
-- cycle
;
\node at ($([shift={(#2,0)}] #1)!0.5!([shift={(#4,0)}] #3)$) {#5};
}