Minor edits

Advanced/Introduction to Quantum/src/parts/03 two qubits.tex

@@ -89,7 +89,7 @@ Some product states can be factored into a tensor product of individual qubit st
 \begin{equation*}
 	\frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr)
 	= \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes
-		\frac{1}{\sqrt{2}}\bigl( \ket{0} - \ket{1} \bigr)
+		\frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr)
 \end{equation*}
 Such states are called \textit{product states.} States that aren't product states are called \textit{entangled} states.
 

Advanced/Introduction to Quantum/src/parts/04 logic gates.tex

@@ -76,6 +76,7 @@ The \texttt{and} gate is a map $\mathbb{B}^2 \to \mathbb{B}$ defined by the foll
 \end{center}
 
 Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
+\hint{Remember, we write bits as vectors.}
 
 
 \begin{solution}

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

@@ -23,12 +23,6 @@ This implies the following: \par
 (You will prove all these properties in any introductory linear algebra course. \\
 This isn't a lesson on linear algebra, so you may take them as given today.)
 
-\definition{}
-Let $\mathbb{U} \subset \mathbb{R}^2$ be the set of points in the unit circle. \par
-We can restate the above definition as follows: \par
-A quantum gate is an invertible map from $\mathbb{U}^n$ to $\mathbb{U}^n$.
-
-
 \generic{Remark:}
 Let $G$ be a quantum gate. \par
 Since quantum gates are, by definition, \textit{linear} maps,
@@ -219,16 +213,5 @@ Using this result, find $H^{-1}$.
 What geometric transformation does $H$ apply to the unit circle? \par
 \hint{Rotation or reflection? How much, or about which axis?}
 
-\vfill
-
-\problem{}
-What are $H\ket{0}$ and $H\ket{1}$? \par
-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 entangled states.
-\end{solution}
-
 \vfill
 \pagebreak