Quantum edits

Advanced/Introduction to Quantum/main.tex

@@ -44,6 +44,55 @@
 	\input{parts/02.00 half a qubit}
 	\input{parts/02.01 two halves}
 	\input{parts/03.00 logic gates}
-	%\input{parts/03.01 quantum gates}
+	\input{parts/03.01 quantum gates}
 
-\end{document}
+\end{document}
+
+\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
+
+
+
+
+
+
+
+
+
+% \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}

Advanced/Introduction to Quantum/parts/02.01 two halves.tex

@@ -108,5 +108,38 @@ Factor the following product state:
 \end{solution}
 
 
+\vfill
+
+\problem{}
+Show that the following is an entangled state.
+\begin{equation*}
+	\frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}
+\end{equation*}
+
+\begin{solution}
+	$
+	\left[
+	\begin{smallmatrix}
+		a_0 \\ a_1
+	\end{smallmatrix}
+	\right]
+	\otimes
+	\left[
+		\begin{smallmatrix}
+			b_0 \\ b_1
+		\end{smallmatrix}
+	\right]
+	=
+	a_0b_0\ket{00} + a_0b_1\ket{01} + a_1b_0\ket{10} + a_1b_1\ket{11}
+	$
+
+	\vspace{2mm}
+
+	So, we have that $a_1b_1 = a_0b_0 = \sqrt{2}^{-1}$ \par
+	But $a_0b_1 = a_1b_0 = 0$, so one of $a_0$ and $b_1$ must be zero. \par
+	We thus have a contradiction.
+\end{solution}
+
+
 \vfill
 \pagebreak

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

@@ -1,38 +1,78 @@
 \section{Quantum Gates}
 
+
+In the previous section, we stated that a quantum gate is a linear map. \par
+Let's complete that definition.
+
 \definition{}
-The quantum analog of a logic gate is a \textit{quantum gate,} \par
+A quantum gate is a \textit{orthonormal matrix}, which means any gate $G$
-which we'll define as a linear map from $\mathbb{B}^n$ to $\mathbb{B}^n$. \par
+satisfies $GG^\text{T} = I$. \par
-\note[Note]{This definition is temporary, we'll extend it soon.}
+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. \\
+		This makes sense---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!}
+\end{itemize}
+
+(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 $\mathbb{U}^n$.
 
 
+\definition{}<qgateislinear>
+Let $G$ be a quantum gate. \par
+Since quantum gates are, by definition, \textit{linear} maps,
+the following holds: \par
 
+\begin{equation*}
+	G\bigl(a_0 \ket{0} + a_1\ket{1}\bigr) = a_0G\ket{0} + a_1G\ket{1}
+\end{equation*}
 
 \problem{}
-Consider the CNOT (controlled not) gate. \par
+Consider the \textit{controlled not} (or \textit{cnot}) gate, defined by the following table: \par
-When applied to a two-bit state $\ket{ab}$, CNOT inverts $b$ iff $a$ is $\ket{1}$. \par
+\begin{itemize}
-Find the matrix that represents the CNOT gate. \par
+	\item $\text{X}_\text{c}\ket{00} = \ket{00}$
-\hint{what are the dimensions of this matrix?}
+	\item $\text{X}_\text{c}\ket{01} = \ket{11}$
+	\item $\text{X}_\text{c}\ket{10} = \ket{10}$
+	\item $\text{X}_\text{c}\ket{11} = \ket{01}$
+\end{itemize}
+In other words, the cnot gate inverts its first bit if its second bit is $\ket{1}$. \par
+Find the matrix that applies the cnot gate.
 
 \begin{solution}
 	\begin{equation*}
-		\text{CNOT} = \begin{bmatrix}
+		\text{CNOT} = \left[\begin{smallmatrix}
 			1 & 0 & 0 & 0 \\
 			0 & 1 & 0 & 0 \\
 			0 & 0 & 0 & 1 \\
 			0 & 0 & 1 & 0 \\
-		\end{bmatrix}
+		\end{smallmatrix}\right]
 	\end{equation*}
 
 	\vspace{4mm}
 
 	If $\ket{a}$ is $\ket{0}$, $\ket{a} \otimes \ket{b}$ is
 	$
-	\begin{bmatrix}
+	\left[
-		\begin{bmatrix}
+	\begin{smallmatrix}
+		\left[
+		\begin{smallmatrix}
 			b_1 \\ b_2
-		\end{bmatrix} \\ 0 \\ 0
+		\end{smallmatrix}
-	\end{bmatrix}
+		\right]
+		\\ 0 \\ 0
+	\end{smallmatrix}
+	\right]
 	$, and the \say{not} portion of the matrix is ignored.
 
 
@@ -40,12 +80,16 @@ Find the matrix that represents the CNOT gate. \par
 
 	If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is
 	$
-	\begin{bmatrix}
+	\left[
+	\begin{smallmatrix}
 		0 \\ 0 \\
-		\begin{bmatrix}
+		\left[
+		\begin{smallmatrix}
 			b_1 \\ b_2
-		\end{bmatrix}
+		\end{smallmatrix}
-	\end{bmatrix}
+		\right]
+	\end{smallmatrix}
+	\right]
 	$, and the \say{identity} portion of the matrix is ignored.
 
 
@@ -55,479 +99,141 @@ Find the matrix that represents the CNOT gate. \par
 	and if $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on the bottom, $\ket{a}$ is $\ket{1}$.
 \end{solution}
 
-\vfill
 
 
 
-
-
-
-
-
-
-\problem{}
-Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
-
-\begin{solution}
-	\begin{equation*}
-		\text{CNOT}_{\text{mod}} = \begin{bmatrix}
-			0 & 1 & 0 & 0 \\
-			1 & 0 & 0 & 0 \\
-			0 & 0 & 1 & 0 \\
-			0 & 0 & 0 & 1
-		\end{bmatrix}
-	\end{equation*}
-\end{solution}
-
-\vfill
-\pagebreak
-
-\iffalse
-\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 $\ket{1}$.
-
-
-\begin{solution}
-	\begin{equation*}
-		\text{CNOT}_{\text{flip}} = \begin{bmatrix}
-			1 & 0 & 0 & 0 \\
-			0 & 0 & 0 & 1 \\
-			0 & 0 & 1 & 0 \\
-			0 & 1 & 0 & 0 \\
-		\end{bmatrix}
-	\end{equation*}
-\end{solution}
-
-\vfill
-\fi
-
-
-
-
-
-
-
-
-
-\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
 
+\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}.
 
 
-
-
-
-
-
-
-\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}
-
-\vfill
 \pagebreak
 
 
 
-
+\problem{}<applycnot>
-
+Evaluate the following:
-
+\begin{equation*}
-
+	\text{X}_\text{C}
-
+	\Bigl(
-
+		\frac{1}{2}\ket{00} +
-
+		\frac{1}{2}\ket{01} -
-
+		\frac{1}{2}\ket{10} -
-
+		\frac{1}{2}\ket{11}
-
+	\Bigr)
-
+\end{equation*}
-The last thing we need is a way to draw complex sequences of gates. \par
-We already know how to do this with classical gates, \'a la logic circuit:
-
-\begin{center}
-\begin{tikzpicture}[circuit logic US, scale=1.5]
-	\node[and gate] (and) at (0,-0.8) {\tiny\texttt{and}};
-	\draw ([shift={(-0.5, 0)}] and.input 1) node[left] {\texttt{1}} -- (and.input 1);
-	\draw ([shift={(-0.5, 0)}] and.input 2) node[left] {\texttt{0}} -- (and.input 2);
-	\draw (and.output) -- ([shift={(0.5, 0)}] and.output) node[right] {\texttt{0}};
-\end{tikzpicture}
-\end{center}
-
-We draw quantum circuits in a very similar way. \par
-For example, here a simple three-bit circuit consisting of a CNOT gate on the first bit, \par
-controlled by the third. The first bit is inverted iff the third bit is $\ket{1}$:
-
-\begin{center}
-\begin{tikzpicture}[scale=1]
-	\node[qubit] (a) at (0, 0) {$\ket{0}$};
-	\node[qubit] (b) at (0, -1) {$\ket{0}$};
-	\node[qubit] (c) at (0, -2) {$\ket{1}$};
-
-	\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$};
-	\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$};
-	\draw[wire] (c) -- ([shift={(4, 0)}] c.center) node[qubit] {$\ket{1}$};
-
-	\draw[wire]
-		($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
-		($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
-	;
-
-	\draw[wirejoin]
-		($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
-		circle[radius=0.1] coordinate(dot)
-	;
-
-	\qubox{a}{1}{a}{3}{CNOT}
-\end{tikzpicture}
-\end{center}
-
-
-
-
-
-
-
-
-
-
-
-
-
-\problem{}
-Draw the CNOT gate as a classical logic circuit. \par
-\hint{This can be done with one gate.}
-
-\begin{solution}
-	\begin{center}
-	\begin{tikzpicture}[circuit logic US, scale=2]
-		\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
-		\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
-		\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
-		\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$a$};
-		\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$b$};
-		\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
-		\draw (dot) -- (dot -| 1,0) node[right] {$b_\text{out}$};
-		\draw (xor.output) -- (xor.output -| 1,0) node[right] {$a_\text{out}$};
-	\end{tikzpicture}
-	\end{center}
-\end{solution}
 
 
 \vfill
 
-Gate controls may be marked with a filled circle or an empty circle. \par
-Empty circles denote \textit{inverse controls,} which (of course) have an inverse effect. \par
-For example, the two circuits below are identical:
-
-\null\hfill
-\begin{minipage}{0.48\textwidth}
-	\begin{center}
-	\begin{tikzpicture}[scale=1]
-		\node[qubit] (a) at (0, 0) {$\ket{0}$};
-		\node[qubit] (b) at (0, -1) {$\ket{0}$};
-
-		\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[
-qubit] {$\ket{a}$};
-		\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
-qubit] {$\ket{b}$};
-
-		\draw[wire]
-			($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
-			($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
-		;
-
-		\draw[wireijoin]
-			($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
-			circle[radius=0.1] coordinate(dot)
-		;
-
-		\qubox{a}{1}{a}{3}{CNOT}
-	\end{tikzpicture}
-	\end{center}
-\end{minipage}
-\hfill
-\begin{minipage}{0.48\textwidth}
-	\begin{center}
-	\begin{tikzpicture}[scale=1]
-		\node[qubit] (a) at (0, 0) {$\ket{0}$};
-		\node[qubit] (b) at (0, -1) {$\ket{0}$};
-
-		\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[
-qubit] {$\ket{a}$};
-		\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
-qubit] {$\ket{b}$};
-
-		\draw[wire]
-			($([shift={(1.5,0)}] a)!0.5!([shift={(2.5,0)}] a)$) --
-			($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
-		;
-
-		\draw[wirejoin]
-			($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
-			circle[radius=0.1] coordinate(dot)
-		;
-		\qubox{a}{1}{a}{3}{CNOT}
-		\qubox{b}{0.5}{b}{1.5}{X}
-		\qubox{b}{2.5}{b}{3.5}{X}
-	\end{tikzpicture}
-	\end{center}
-\end{minipage}
-\hfill\null
-
-
-
-
-
-
-
-
-
 \problem{}
-What are $\ket{a}$ and $\ket{b}$ in the diagrams above?
+If we measure the result of \ref{applycnot}, what are the probabilities of getting each state?
 
-\begin{solution}
+\vfill
-	\begin{center}
-	\begin{tikzpicture}[scale=1]
-		\node[qubit] (a) at (0, 0) {$\ket{0}$};
-		\node[qubit] (b) at (0, -1) {$\ket{0}$};
 
-		\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[
-qubit] {$\ket{1}$};
-		\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[
-qubit] {$\ket{0}$};
 
-		\draw[wire]
-			($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
-			($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
-		;
 
-		\draw[wireijoin]
+%\problem{}
-			($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
+%Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
-			circle[radius=0.1] coordinate(dot)
+%
-		;
+%\begin{solution}
-
+%	\begin{equation*}
-		\qubox{a}{1}{a}{3}{CNOT}
+%		\text{CNOT}_{\text{mod}} = \begin{bmatrix}
-	\end{tikzpicture}
+%			0 & 1 & 0 & 0 \\
-	\end{center}
+%			1 & 0 & 0 & 0 \\
-\end{solution}
+%			0 & 0 & 1 & 0 \\
+%			0 & 0 & 0 & 1
+%		\end{bmatrix}
+%	\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 $\ket{1}$.
+%
+%
+%\begin{solution}
+%	\begin{equation*}
+%		\text{CNOT}_{\text{flip}} = \begin{bmatrix}
+%			1 & 0 & 0 & 0 \\
+%			0 & 0 & 0 & 1 \\
+%			0 & 0 & 1 & 0 \\
+%			0 & 1 & 0 & 0 \\
+%		\end{bmatrix}
+%	\end{equation*}
+%\end{solution}
+%
+%\vfill
 
 
-
+\definition{}
-
+The \textit{Hadamard Gate} $H$, is given by the following matrix: \par
-
+\begin{equation*}
-
+	H = \frac{1}{\sqrt{2}}\begin{bmatrix}
-
+		1 & 1 \\
-
+		1 & -1
-
+	\end{bmatrix}
-
+\end{equation*}
-As we noted before, quantum gates don't \textit{consume} bits, they \textit{transform} them. \par
+\note[Note]{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal}
-Thus, quantum circuits are drawn with a fixed set of bits, whose states change with time: \par
-\note[Note]{
-	In this diagram, CNOT and SWAP are drawn as $\oplus$ and \rotatebox[origin=c]{90}{$\leftrightarrows$} to save space.
-}
-
-\begin{center}
-\begin{tikzpicture}[scale=1]
-	\node[qubit] (a) at (0, 0) {$\ket{a}$};
-	\node[qubit] (b) at (0, -1) {$\ket{1}$};
-	\node[qubit] (c) at (0, -2) {$\ket{c}$};
-	\node[qubit] (d) at (0, -3) {$\ket{d}$};
-	\node[qubit] (e) at (0, -4) {$\ket{1}$};
-
-	\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$};
-	\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{b}$};
-	\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$};
-	\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$};
-	\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{e}$};
-
-
-	\draw[wire]
-		($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
-		($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
-	;
-	\draw[wirejoin]
-		($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
-		circle[radius=0.1] coordinate(dot)
-	;
-	\draw[wireijoin]
-		($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
-		circle[radius=0.1] coordinate(dot)
-	;
-	\qubox{a}{1}{a}{2}{T}
-
-	\qubox{a}{4}{a}{5}{X}
-
-	\draw[wire]
-		($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
-		($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
-	;
-	\draw[wireijoin]
-		($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
-		circle[radius=0.1] coordinate(dot)
-	;
-	\qubox{e}{2}{e}{3}{$\oplus$}
-
-	\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}}
-
-
-	\draw[wire]
-		($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
-		($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
-	;
-	\draw[wirejoin]
-		($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
-		circle[radius=0.1] coordinate(dot)
-	;
-	\draw[wirejoin]
-		($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
-		circle[radius=0.1] coordinate(dot)
-	;
-	\qubox{b}{5}{b}{6}{T}
-\end{tikzpicture}
-\end{center}
-
-In a quantum circuit, the ONLY way two bits can interact is through a gate. \par
-We cannot add \say{branches} in quantum circuits like we do in classical circuits:
-
-\begin{center}
-\begin{tikzpicture}[circuit logic US, scale=1.2]
-	\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
-	\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
-	\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
-	\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$x$};
-	\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$y$};
-	\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
-	\draw (dot) -- (dot -| 1,0) node[right] {$y_\text{out}$};
-	\draw (xor.output) -- (xor.output -| 1,0) node[right] {$x_\text{out}$};
-
-	\draw[->, line width = 1, ogrape]
-		([shift={(0.3,-0.5)}] dot) node[right] {This is a branch}
-		-| ([shift={(0,-0.2)}] dot)
-	;
-\end{tikzpicture}
-\end{center}
-
-
-
-
-
-
-
-
-
-
-
-
 
 
 \problem{}
-Find the values of $\ket{a}$ through $\ket{e}$ in the above circuit.
+What is $HH$? \par
+Using this result, find $H^{-1}$.
 
 \begin{solution}
-	\begin{center}
+	$HH = I$, so $H^{-1} = H$
-	\begin{tikzpicture}[scale=1]
-		\node[qubit] (a) at (0, 0) {$\ket{1}$};
-		\node[qubit] (b) at (0, -1) {$\ket{1}$};
-		\node[qubit] (c) at (0, -2) {$\ket{1}$};
-		\node[qubit] (d) at (0, -3) {$\ket{0}$};
-		\node[qubit] (e) at (0, -4) {$\ket{1}$};
-
-		\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[
-qubit] {$\ket{0}$};
-		\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[
-qubit] {$\ket{1}$};
-		\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[
-qubit] {$\ket{1}$};
-		\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[
-qubit] {$\ket{0}$};
-		\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[
-qubit] {$\ket{0}$};
-
-
-		\draw[wire]
-			($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
-			($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
-		;
-		\draw[wirejoin]
-			($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
-			circle[radius=0.1] coordinate(dot)
-		;
-		\draw[wireijoin]
-			($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
-			circle[radius=0.1] coordinate(dot)
-		;
-		\qubox{a}{1}{a}{2}{T}
-
-		\qubox{a}{4}{a}{5}{X}
-
-		\draw[wire]
-			($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
-			($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
-		;
-		\draw[wireijoin]
-			($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
-			circle[radius=0.1] coordinate(dot)
-		;
-		\qubox{e}{2}{e}{3}{$\oplus$}
-
-		\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}}
-
-
-		\draw[wire]
-			($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
-			($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
-		;
-		\draw[wirejoin]
-			($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
-			circle[radius=0.1] coordinate(dot)
-		;
-		\draw[wirejoin]
-			($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
-			circle[radius=0.1] coordinate(dot)
-		;
-		\qubox{b}{5}{b}{6}{T}
-	\end{tikzpicture}
-	\end{center}
 \end{solution}
 
-
 \vfill
-\pagebreak
+
+\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black}
+	Matrix multiplication works as follows:
+
+	\begin{equation*}
+		AB =
+		\begin{bmatrix}
+			1 & 2 \\
+			3 & 4 \\
+		\end{bmatrix}
+		\begin{bmatrix}
+			a_0 & b_0 \\
+			a_1 & b_1 \\
+		\end{bmatrix}
+		=
+		\begin{bmatrix}
+			1a_0 + 2a_1 & 1b_0 + 2b_1 \\
+			3a_0 + 4a_1 & 3b_0 + 4b_1 \\
+		\end{bmatrix}
+	\end{equation*}
+
+
+	Note that this is very similar to multiplying each column of $B$ by $A$. \par
+	The product $AB$ is simply $Ac$ for every column $c$ in $B$:
+
+	\begin{equation*}
+		Ac_0 =
+		\begin{bmatrix}
+			1 & 2 \\
+			3 & 4 \\
+		\end{bmatrix}
+		\begin{bmatrix}
+			a_0 \\ a_1
+		\end{bmatrix}
+		=
+		\begin{bmatrix}
+			1a_0 + 2a_1 \\
+			3a_0 + 4a_1
+		\end{bmatrix}
+	\end{equation*}
+
+	This is exactly the first column of the matrix product.
+\end{ORMCbox}
+
+\pagebreak