Edits

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

@@ -174,11 +174,6 @@ Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nic
 
 
 
-\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?
@@ -324,27 +319,30 @@ In other words, what is the set of vectors that can be written as linear combina
 
 \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}
+% This is wrong, but there's something here.
+% maybe fix later?
+%
+%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
 
@@ -453,7 +451,7 @@ Consider the NOT gate, which operates as follows: \par
 \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.
+flipping a coin we have already tossed, without looking at its state.
 Thus,
 \begin{equation*}
 	\text{NOT} \begin{bmatrix}
@@ -544,8 +542,8 @@ Find the matrix that represents the NOT operation on one probabilistic bit.
 
 
 \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 we have an arbitrary operation $M$. \par
+If we know how $M$ acts on $[1]$ and $[0]$, can we compute $M[x]$ 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 $x$?

Advanced/Introduction to Quantum/src/parts/05 quantum gates.tex

@@ -42,11 +42,11 @@ the following holds: \par
 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 $\text{X}_\text{c}\ket{11} = \ket{01}$
+	\item $\text{X}_\text{c}\ket{01} = \ket{01}$
+	\item $\text{X}_\text{c}\ket{10} = \ket{11}$
+	\item $\text{X}_\text{c}\ket{11} = \ket{10}$
 \end{itemize}
-In other words, the cnot gate inverts its first bit if its second bit is $\ket{1}$. \par
+In other words, the cnot gate inverts its second bit if its first bit is $\ket{1}$. \par
 Find the matrix that applies the cnot gate.
 
 \begin{solution}
@@ -127,7 +127,7 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti
 
 \vfill
 
-\problem{}
+\problem{}<cnotflipped>
 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 $\ket{1}$.
 

Advanced/Introduction to Quantum/src/week 1.tex

@@ -34,6 +34,13 @@
 \subtitle{Prepared by \githref{Mark} on \today{}}
 
 
+% TODO: spend more time on probabalistic bits.
+% This could even be its own handout, especially
+% for younger classes!
+
+% Why are qubits amplitudes instead of probabilities?
+% (Asher question)
+
 \begin{document}
 
 	\maketitle