Edits

Advanced/Introduction to Quantum/parts/00.00 bits.tex

@@ -8,7 +8,7 @@ To keep things simple, we'll use regular (usually called \textit{classical}) bit
 
 \definition{Binary Digits}
 $\mathbb{B}$ 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.}
+\note[Note]{We've seen $\mathbb{B}$ before---it's the set of integers mod 2.}
 
 \vspace{2mm}
 
@@ -28,7 +28,8 @@ 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 $\mathbb{B}^2$, the set
+	$\{(\texttt{0},\texttt{0}), (\texttt{0},\texttt{1}), (\texttt{1},\texttt{0}), (\texttt{1},\texttt{1})\}$
 	\item $\mathbb{B}^n$, the set of all possible states of $n$ bits.
 \end{itemize}
 
@@ -61,35 +62,35 @@ We'll therefore say that \texttt{0} and \texttt{1} \textit{orthogonal} (or equiv
 
 \vspace{2mm}
 
-We can draw $\vec{0}$ and $\vec{1}$ as perpendicular axis on a plane to represent this:
+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]
 		\fill[color = black] (0, 0) circle[radius=0.05];
 
 		\draw[->] (0, 0) -- (1.5, 0);
-		\node[right] at (1.5, 0) {$\vec{0}$ axis};
+		\node[right] at (1.5, 0) {$\vec{e}_0$ 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};
+		\node[above] at (0, 1.5) {$\vec{e}_1$ 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
+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 $[1, 0] = (1 \times \vec{0}) + (0 \times \vec{1})$, and is thus all $\vec{0}$ and no $\vec{1}$. \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
 
 \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
+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 $\texttt{x} = \vec{0} + \vec{1}$ explicitly. \\
+	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.
 }
 
@@ -100,10 +101,10 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts
 		\fill[color = black] (0, 0) circle[radius=0.05];
 
 		\draw[->] (0, 0) -- (1.5, 0);
-		\node[right] at (1.5, 0) {$\vec{0}$ axis};
+		\node[right] at (1.5, 0) {$\vec{e}_0$ axis};
 
 		\draw[->] (0, 0) -- (0, 1.5);
-		\node[above] at (0, 1.5) {$\vec{1}$ axis};
+		\node[above] at (0, 1.5) {$\vec{e}_1$ axis};
 
 		\fill[color = oblue] (1, 0) circle[radius=0.05];
 		\node[below] at (1, 0) {\texttt{0}};
@@ -141,7 +142,7 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts
 \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.
+Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, there's nothing in between.
 
 
 \vspace{8mm}
@@ -158,11 +159,11 @@ Our bit is fully $\vec{0}$ or fully $\vec{1}$. There's nothing in between.
 
 
 \definition{Orthonormal Basis}
-The unit vectors $\vec{0}$ and $\vec{1}$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. \par
+The unit vectors $\vec{e}_0$ and $\vec{e}_1$ 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}. \\
+	Note that $\vec{e}_0$ and $\vec{e}_1$ are orthonormal by \textit{definition}. \\
 	We don't have to prove anything, we simply defined them as such.
 } \par
 
@@ -215,11 +216,11 @@ Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them
 
 \null\hfill
 \begin{minipage}{0.48\textwidth}
-	\[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{0}) + (0 \times \vec{1}) \]
+	\[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{e}_0) + (0 \times \vec{e}_1) \]
 \end{minipage}
 \hfill
 \begin{minipage}{0.48\textwidth}
-	\[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{0}) + (1 \times \vec{1}) \]
+	\[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{e}_0) + (1 \times \vec{e}_1) \]
 \end{minipage}
 \hfill\null
 
@@ -250,10 +251,10 @@ Write \texttt{x} and \texttt{y} in the diagram below in terms of $\ket{0}$ and $
 	\fill[color = black] (0, 0) circle[radius=0.05];
 
 	\draw[->] (0, 0) -- (1.5, 0);
-	\node[right] at (1.5, 0) {$\vec{0}$ axis};
+	\node[right] at (1.5, 0) {$\vec{e}_0$ axis};
 
 	\draw[->] (0, 0) -- (0, 1.5);
-	\node[above] at (0, 1.5) {$\vec{1}$ axis};
+	\node[above] at (0, 1.5) {$\vec{e}_1$ axis};
 
 	\fill[color = oblue] (1, 0) circle[radius=0.05];
 	\node[below] at (1, 0) {$\ket{0}$};

Advanced/Introduction to Quantum/parts/00.01 two bits.tex

@@ -61,7 +61,7 @@ $\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \
 \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.
+the compound state $(a,b)$ takes values in $A \times B$.
 
 
 \vspace{4mm}
@@ -69,8 +69,6 @@ the compound state $(a,b)$ takes values in $A \times B$. This is trivial.
 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
 
 
@@ -303,7 +301,10 @@ Write $\ket{5}$ as three-bit state vector. \par
 
 \problem{}
 Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
-What do you see?
+\hint{
+	You do not need to compute every tensor product. \\
+	Do a few, you should quickly see the pattern.
+}
 
 
 

Advanced/Introduction to Quantum/parts/02.00 half a qubit.tex

@@ -165,13 +165,11 @@ Our measurement again returns either $\ket{0}$ or $\ket{1}$, with the following
 
 \vspace{2mm}
 
-In addition $\ket{\psi}$ \textit{collapses} to the state we measure: it instantly jumps to the state we measure, \par
+In addition, $\ket{\psi}$ \textit{collapses} when it is measured: it instantly changes its state to the result of the measurement,
-leaving no trace of its previous state. If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and
+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.
-\vspace{2mm}
-
 Quantum bits \textit{cannot} be measured without their state collapsing. \par
-We cannot certainly know the state of a qubit unless that state is $\ket{0}$ or $\ket{1}$.
 
 \pagebreak
 

Advanced/Introduction to Quantum/parts/03.00 logic gates.tex

@@ -107,7 +107,7 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
 
 \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 some set of bits, and return another:
+We usually think of logic gates as \textit{functions}: they consume one set of bits, and return another:
 
 
 \begin{center}
@@ -143,10 +143,10 @@ 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{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{0}};
+	\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {\texttt{1}};
 
 	\draw[
 		color = oblue,

Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex

@@ -12,12 +12,12 @@ This implies the following: \par
 \begin{itemize}
 	\item $G$ is square \par
 	\note{
-		If we think of $G$ as a map, this implies it has as many inputs as it has outputs. \\
+		If we think of $G$ as a map, this means that $G$ has as many inputs as it has outputs. \\
-		This makes sense---we stated earlier that quantum gates do not destroy or create qubits.
+		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!}
+	\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. \\
@@ -103,12 +103,6 @@ Find the matrix that applies the cnot gate.
 
 
 \vfill
-
-\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}.
-
-
 \pagebreak
 
 
@@ -133,7 +127,11 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti
 
 \vfill
 
+\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}.
 
+\pagebreak
 
 %\problem{}
 %Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
@@ -149,9 +147,6 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti
 %	\end{equation*}
 %\end{solution}
 
-\vfill
-\pagebreak
-
 %\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 $\ket{1}$.
@@ -238,11 +233,11 @@ Using this result, find $H^{-1}$.
 
 \problem{}
 What are $H\ket{0}$ and $H\ket{1}$? \par
-Are these states superpositions?
+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 superpositions.
+	Both of these are entangled states.
 \end{solution}
 
 \vfill