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2
Advanced/Introduction to Quantum/main.tex
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@ -45,8 +45,6 @@
\input{parts/04 two halves}
\input{parts/05 logic gates}
\input{parts/06 quantum gates}
%\input{parts/03.00 logic gates}
%\input{parts/03.01 quantum gates}
%\section{Superdense Coding}
%TODO

99
Advanced/Introduction to Quantum/parts/00 vectors.tex
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@ -1,4 +1,4 @@
\section*{Prerequisite: Vector Basics}
\section*{Part 0: Vector Basics}
\definition{Vectors}
An $n$-dimensional \textit{vector} is an element of $\mathbb{R}^n$. In this handout, we'll write vectors as columns. \par
@ -90,6 +90,103 @@ What is the dot product of two orthogonal vectors?
\vfill
\pagebreak
\definition{Linear combinations}
A \textit{linear combination} of two or more vectors $v_1, v_2, ..., v_k$ is the weighted sum
\begin{equation*}
a_1v_1 + a_2v_2 + ... + a_kv_k
\end{equation*}
where $a_i$ are arbitrary real numbers.
\definition{Linear dependence}
We say a set of vectors $\{v_1, v_2, ..., v_k\}$ is \textit{linearly independent} if we can write $0$ as a nontrivial
linear combination of these vectors. For example, the following set is linearly dependent
\begin{equation*}
\Bigl\{
\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right],
\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right],
\left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right]
\Bigr\}
\end{equation*}
Since $
\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right] +
\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right] -
2 \left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right]
= 0
$. A graphical representation of this is below.
\null\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\begin{tikzpicture}[scale=1]
\fill[color = black] (0, 0) circle[radius=0.05];
\node[right] at (1, 0) {$\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$};
\node[above] at (0, 1) {$\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$};
\draw[->] (0, 0) -- (1, 0);
\draw[->] (0, 0) -- (0, 1);
\draw[->] (0, 0) -- (0.5, 0.5);
\node[above right] at (0.5, 0.5) {$\left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right]$};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\begin{tikzpicture}[scale=1]
\fill[color = black] (0, 0) circle[radius=0.05];
\node[below] at (0.5, 0) {$\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$};
\node[right] at (1, 0.5) {$\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$};
\draw[->] (0, 0) -- (0.95, 0);
\draw[->] (1, 0) -- (1, 0.95);
\draw[->] (1, 1) -- (0.55, 0.55);
\draw[->] (0.5, 0.5) -- (0.05, 0.05);
\node[above left] at (0.5, 0.5) {$-2\left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right]$};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill\null
\problem{}
Find a linearly independent set of vectors in $\mathbb{R}^3$
\vfill
\definition{Coordinates}
Say we have a set of linearly independent vectors $B = \{b_1, ..., b_k\}$. \par
We can write linear combinations of $B$ as \textit{coordinates} with respect to this set:
\vspace{2mm}
If we have a vector $v = x_1b_1 + x_2b_2 + ... + x_kb_k$, we can write $v = (x_1, x_2, ..., x_k)$ with respect to $B$.
\vspace{4mm}
For example, take
$B = \biggl\{
\left[\begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix}\right],
\left[\begin{smallmatrix} 0 \\ 1 \\ 0\end{smallmatrix}\right],
\left[\begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix}\right]
\biggr\}$ and $v = \left[\begin{smallmatrix} 8 \\ 3 \\ 9 \end{smallmatrix}\right]$
The coordinates of $v$ with respect to $B$ are, of course, $(8, 3, 9)$.
\problem{}
What are the coordinates of $v$ with respect to the basis
$B = \biggl\{
\left[\begin{smallmatrix} 1 \\ 0 \\ 1 \end{smallmatrix}\right],
\left[\begin{smallmatrix} 0 \\ 1 \\ 0\end{smallmatrix}\right],
\left[\begin{smallmatrix} 0 \\ 0 \\ -1 \end{smallmatrix}\right]
\biggr\}$?
%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:

60
Advanced/Introduction to Quantum/parts/01 bits.tex
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@ -55,14 +55,9 @@ What is the size of $\mathbb{B}^n$?
% NOTE: this is time-travelled later in the handout.
% if you edit this, edit that too.
\cgeneric{Remark}
\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 $\vec{e}_0$ and $\vec{e}_1$ as perpendicular axis on a plane to represent this:
We'll write the states \texttt{0} and \texttt{1} as orthogonal unit vectors, labeled $\vec{e}_0$ and $\vec{e}_1$:
\begin{center}
\begin{tikzpicture}[scale=1.5]
@ -84,27 +79,21 @@ We can draw $\vec{e}_0$ and $\vec{e}_1$ as perpendicular axis on a plane to repr
The point marked $1$ is at $[0, 1]$. It is no parts $\vec{e}_0$, and all parts $\vec{e}_1$. \par
Of course, we can say something similar about the point marked $0$: \par
It is at $[1, 0] = (1 \times \vec{e}_0) + (0 \times \vec{e}_1)$, and is thus all $\vec{e}_0$ and no $\vec{e}_1$. \par
\note[Note]{$[0, 1]$ and $[1, 0]$ are coordinates in the basis $\{\vec{e}_0, \vec{e}_1\}$}
\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 $\vec{e}_0$ and $\vec{e}_1$: \par
\note[Note]{
We could also write $\texttt{x} = \vec{e}_0 + \vec{e}_1$ explicitly. \\
I've drawn \texttt{x} as a point on the left, and as a sum on the right.
}
We could, of course, mark the point \texttt{x} at $[1, 1]$ which is equal parts $\vec{e}_0$ and $\vec{e}_1$: \par
\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) {$\vec{e}_0$ axis};
\node[right] at (1.5, 0) {$\vec{e}_0$};
\draw[->] (0, 0) -- (0, 1.5);
\node[above] at (0, 1.5) {$\vec{e}_1$ axis};
\node[above] at (0, 1.5) {$\vec{e}_1$};
\fill[color = oblue] (1, 0) circle[radius=0.05];
\node[below] at (1, 0) {\texttt{0}};
@ -117,38 +106,14 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts
\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 $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, there's nothing in between. \par
\note{
Note that the unit vectors $\vec{e}_0$ and $\vec{e}_1$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$.
}
But \texttt{x} isn't a member of $\mathbb{B}$---it's not a state that a classical bit can take. \par
By our current definitions, the \textit{only} valid states of a bit are $\texttt{0} = [1, 0]$ and $\texttt{1} = [0, 1]$.
\vfill
\pagebreak
@ -169,7 +134,7 @@ Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, t
\definition{Vectored Bits}
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
Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them into their $\vec{e}_0$ and $\vec{e}_1$ components: \par
\null\hfill
\begin{minipage}{0.48\textwidth}
@ -186,10 +151,11 @@ Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them
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:}
\vspace{4mm}
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.
$\ket{0}$ 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.}

105
Advanced/Introduction to Quantum/parts/02 two bits.tex
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@ -1,74 +1,44 @@
\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{}<compoundclassicalbits>
As we already know, the set of states a single bit can take is $\mathbb{B} = \{\texttt{0}, \texttt{1}\}$. \par
What is the set of compound states \textit{two} bits can take? How about $n$ bits? \par
\hint{Cartesian product.}
\vspace{5cm}
Of course, \ref{compoundclassicalbits} is fairly easy: \par
If $a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$,
the values $ab$ can take are
$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$.
\problem{}
What is the set of possible states of two bits (i.e, $\mathbb{B}^2$)?
\vspace{2cm}
\cgeneric{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
\cgeneric{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}
\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}
\vspace{4mm}
We would like to do the same in vector notation. Given bits $\ket{a}$ and $\ket{b}$,
how should we represent the state of $\ket{ab}$? We'll spend the rest of this section solving this problem.
We would like to do the same in vector notation. Given $\ket{a}$ and $\ket{b}$, \par
how should we represent the state of $\ket{ab}$?
\problem{}
When we have two bits, we have four orthogonal states:
$\overrightarrow{00}$, $\overrightarrow{01}$, $\overrightarrow{10}$, and $\overrightarrow{11}$. \par
\vspace{2mm}
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
\pagebreak
@ -83,8 +53,7 @@ how should we represent the state of $\ket{ab}$?
\definition{Tensor Products}
The \textit{tensor product} between two vectors
is defined as follows:
The \textit{tensor product} of two vectors is defined as follows:
\begin{equation*}
\begin{bmatrix}
x_1 \\ x_2
@ -233,9 +202,12 @@ $?
\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 weighted sums of the vectors above?
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 $A \times B$.
\pagebreak
@ -275,9 +247,7 @@ Compute the following. Is the result what we'd expect?
\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.
Of course, writing $\ket{0} \otimes \ket{1}$ is a bit excessive. We'll shorten this notation to $\ket{01}$. \par
\vspace{2mm}
@ -302,10 +272,7 @@ Write $\ket{5}$ as three-bit state vector. \par
\problem{}
Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
\hint{
You do not need to compute every tensor product. \\
Do a few, you should quickly see the pattern.
}
\hint{You do not need to compute every tensor product. Do a few and find the pattern.}

20
Advanced/Introduction to Quantum/parts/03 half a qubit.tex
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@ -1,17 +1,5 @@
\section{Half a Qubit}
First, a pair of definitions. We've used both these terms implicitly in the previous section,
but we'll need to introduce proper definitions before we continue.
\definition{}
A \textit{linear combination} of two vectors $u$ and $v$ is the sum $au + bv$ for scalars $a$ and $b$. \par
\note[Note]{In other words, a linear combination is exactly what it sounds like.}
\definition{}
A \textit{normalized vector} (also called a \textit{unit vector}) is a vector with length 1.
\vspace{4mm}
\begin{tcolorbox}[
enhanced,
breakable,
@ -38,7 +26,7 @@ A \textit{normalized vector} (also called a \textit{unit vector}) is a vector wi
\end{tcolorbox}
\cgeneric{Remark}
\generic{Remark:}
Just like a classical bit, a \textit{quantum bit} (or \textit{qubit}) can take the values $\ket{0}$ and $\ket{1}$. \par
However, \texttt{0} and \texttt{1} aren't the only states a qubit may have.
@ -47,7 +35,7 @@ However, \texttt{0} and \texttt{1} aren't the only states a qubit may have.
We'll make sense of quantum bits by extending the \say{vectored} bit representation we developed in the previous section.
First, let's look at a diagram we drew a few pages ago:
\begin{ORMCbox}{Time Travel (Page 2)}{black!10!white}{black!65!white}
\begin{ORMCbox}{Time Travel (Page 5)}{black!10!white}{black!65!white}
A classical bit takes states in $\{\texttt{0}, \texttt{1}\}$, picking one at a time. \par
We'll represent \texttt{0} and \texttt{1} as perpendicular unit vectors $\ket{0}$ and $\ket{1}$,
show below.
@ -169,7 +157,7 @@ In addition, $\ket{\psi}$ \textit{collapses} when it is measured: it instantly c
leaving no trace of its previous state. \par
If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and
it will remain in that state until it is changed.
Quantum bits \textit{cannot} be measured without their state collapsing. \par
Quantum bits cannot be measured without their state collapsing. \par
\pagebreak
@ -187,7 +175,7 @@ Quantum bits \textit{cannot} be measured without their state collapsing. \par
\problem{}
\begin{itemize}
\item What is the probability we get $\ket{0}$ if we measure $\ket{\psi_0}$? \par
\item What is the probability we get $\ket{0}$ when we measure $\ket{\psi_0}$? \par
\item What outcomes can we get if we measure it a second time? \par
\item What are these probabilities for $\ket{\psi_1}$?
\end{itemize}

4
Advanced/Introduction to Quantum/parts/05 logic gates.tex
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@ -105,7 +105,7 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
\vfill
\pagebreak
\cgeneric{Remark}
\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:
@ -275,7 +275,7 @@ Find the matrix that corresponds to the above transformation. \par
\vfill
\cgeneric{Remark}
\generic{Remark:}
We could draw the above transformation as a combination $X$ and $I$ (identity) gate:
\begin{center}
\begin{tikzpicture}[scale=0.8]

5
Advanced/Introduction to Quantum/parts/06 quantum gates.tex
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@ -127,7 +127,7 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti
\vfill
\cgeneric{Remark}
\generic{Remark:}
As we just saw, a quantum gate is fully defined by the place it maps our basis states $\ket{0}$ and $\ket{1}$ \par
(or, $\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates). This directly follows from \ref{qgateislinear}.
@ -217,7 +217,8 @@ The \textit{Hadamard Gate} is given by the following matrix: \par
\end{bmatrix}
\end{equation*}
This is exactly the first column of the matrix product.
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}