Edits

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

@@ -290,8 +290,7 @@ We could draw the above transformation as a combination $X$ and $I$ (identity) g
 \end{tikzpicture}
 \end{center}
 
-We can even omit the $I$ gate, since we now know that transforms affect the whole state. \par
-Of course, empty spaces always imply an $I$ gate.
+We can even omit the $I$ gate, since we now know that transforms affect the whole state: \par
 \begin{center}
 \begin{tikzpicture}[scale=0.8]
 	\node[qubit] (a) at (0, 0) {$\ket{0}$};

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

@@ -105,9 +105,8 @@ Find the matrix that applies the cnot gate.
 \vfill
 
 \generic{Remark:}
-Note that a quantum gate is fully defined by the place it maps
-our basis states $\ket{0}$ and $\ket{1}$ (or,$\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates).
-This directly follows from \ref{qgateislinear}.
+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
@@ -172,28 +171,18 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti
 %\vfill
 
 
+
 \definition{}
-The \textit{Hadamard Gate} $H$, is given by the following matrix: \par
+The \textit{Hadamard Gate} is given by the following matrix: \par
 \begin{equation*}
 	H = \frac{1}{\sqrt{2}}\begin{bmatrix}
 		1 & 1 \\
 		1 & -1
 	\end{bmatrix}
 \end{equation*}
-\note[Note]{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal}
+\note{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal.}
 
-
-\problem{}
-What is $HH$? \par
-Using this result, find $H^{-1}$.
-
-\begin{solution}
-	$HH = I$, so $H^{-1} = H$
-\end{solution}
-
-\vfill
-
-\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black}
+\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black!65!white}
 	Matrix multiplication works as follows:
 
 	\begin{equation*}
@@ -236,4 +225,25 @@ Using this result, find $H^{-1}$.
 	This is exactly the first column of the matrix product.
 \end{ORMCbox}
 
+
+\problem{}
+What is $HH$? \par
+Using this result, find $H^{-1}$.
+
+\begin{solution}
+	$HH = I$, so $H^{-1} = H$
+\end{solution}
+
+\vfill
+
+\problem{}
+What are $H\ket{0}$ and $H\ket{1}$? \par
+Are these states superpositions?
+
+\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.
+\end{solution}
+
+\vfill
 \pagebreak