diff --git a/Advanced/Introduction to Quantum/main.tex b/Advanced/Introduction to Quantum/main.tex
index 7ed68d9..e186af3 100755
--- a/Advanced/Introduction to Quantum/main.tex	
+++ b/Advanced/Introduction to Quantum/main.tex	
@@ -17,8 +17,7 @@
 % use the [solutions] flag to show solutions.
 \documentclass[
 	solutions,
-	singlenumbering,
-	unfinished
+	singlenumbering
 ]{../../resources/ormc_handout}
 \usepackage{../../resources/macros}
 
@@ -46,53 +45,25 @@
 	\input{parts/03.00 logic gates}
 	\input{parts/03.01 quantum gates}
 
-\end{document}
+	%\section{Superdense Coding}
+	%TODO
+	%\vfill
+	%\pagebreak
 
-\problem{}
-The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par
-Find its matrix.
-
-\begin{solution}
-	\begin{equation*}
-		\text{SWAP} = \begin{bmatrix}
-			1 & 0 & 0 & 0 \\
-			0 & 0 & 1 & 0 \\
-			0 & 1 & 0 & 0 \\
-			0 & 0 & 0 & 1 \\
-		\end{bmatrix}
-	\end{equation*}
-\end{solution}
-
-\vfill
+	%\section{Quantum Error Correction}
+	%TODO: no cloning theorem, bit flip code
+	%\vfill
+	%\pagebreak
 
 
+	%\section{Quantum Teleportation}
+	%TODO
+	%\vfill
+	%\pagebreak
+
+	%\section{One Real Qubit}
+	% No problems, appendix.
+	% bloch sphere, etc.
 
 
-
-
-
-
-
-% \problem{}
-% The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par
-% In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par
-% Find the $T$ gate's matrix. \par
-% \note{
-% 	This gate is particularly interesting because it's a \textit{universal quantum gate}: \\
-% 	like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates.
-% }
-%
-% \begin{solution}
-% 	\begin{equation*}
-% 		\text{T} = \begin{bmatrix}
-% 			1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
-% 			0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
-% 			0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
-% 			0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
-% 			0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
-% 			0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
-% 			0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
-% 			0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
-% 		\end{bmatrix}
-% 	\end{equation*}
-% \end{solution}
\ No newline at end of file
+\end{document}
\ No newline at end of file
diff --git a/Advanced/Introduction to Quantum/parts/03.00 logic gates.tex b/Advanced/Introduction to Quantum/parts/03.00 logic gates.tex
index 7ab5fde..7ec920a 100644
--- a/Advanced/Introduction to Quantum/parts/03.00 logic gates.tex	
+++ b/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) {$\ket{0}$};
diff --git a/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex b/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex
index a9572cc..056804a 100644
--- a/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex	
+++ b/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