Quantum edits

Advanced/Introduction to Quantum/main.tex

@@ -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

Advanced/Introduction to Quantum/parts/00 vectors.tex

@@ -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:

Advanced/Introduction to Quantum/parts/01 bits.tex

@@ -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 write the states \texttt{0} and \texttt{1} as orthogonal unit vectors, labeled $\vec{e}_0$ and $\vec{e}_1$:
-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:
 
 \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
-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.
-}
 
-\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
+But \texttt{x} isn't a member of $\mathbb{B}$---it's not a state that a classical bit can take. \par
-Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, there's nothing in between. \par
+By our current definitions, the \textit{only} valid states of a bit are $\texttt{0} = [1, 0]$ and $\texttt{1} = [0, 1]$.
-\note{
-	Note that the unit vectors $\vec{e}_0$ and $\vec{e}_1$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$.
-}
 
 \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
+$\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one.} This is called bra-ket notation. \par
-This is called bra-ket notation. $\bra{0}$ is called a \say{bra,} but we won't worry about that for now.
+\note[Note]{$\bra{0}$ is called a \say{bra,} but we won't worry about that for now.}
 
 
 

Advanced/Introduction to Quantum/parts/02 two bits.tex

@@ -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{}
+\vspace{2mm}
-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}
 
 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
+The \textit{tensor product} of two vectors is defined as follows:
-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
+Of course, writing $\ket{0} \otimes \ket{1}$ is a bit excessive. We'll shorten this notation to $\ket{01}$. \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}
 
@@ -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{
+\hint{You do not need to compute every tensor product. Do a few and find the pattern.}
-	You do not need to compute every tensor product. \\
-	Do a few, you should quickly see the pattern.
-}
 
 
 

Advanced/Introduction to Quantum/parts/03 half a qubit.tex

@@ -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}

Advanced/Introduction to Quantum/parts/05 logic gates.tex

@@ -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]

Advanced/Introduction to Quantum/parts/06 quantum gates.tex

@@ -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}