\section{Superdense Coding}

Consider the following entangled two-qubit states, called the \textit{bell states}:
\begin{itemize}
	\item $\ket{\Phi^+} = \frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}$
	\item $\ket{\Phi^-} = \frac{1}{\sqrt{2}}\ket{00} - \frac{1}{\sqrt{2}}\ket{11}$
	\item $\ket{\Psi^+} = \frac{1}{\sqrt{2}}\ket{01} + \frac{1}{\sqrt{2}}\ket{10}$
	\item $\ket{\Psi^-} = \frac{1}{\sqrt{2}}\ket{01} - \frac{1}{\sqrt{2}}\ket{10}$
\end{itemize}

\problem{}
The probabilistic bits we get when measuring any of the above may be called \textit{anticorrelated bits}. \par
If we measure the first bit of any of these states and observe $1$, what is the resulting compound state? \par
What if we observe $0$ instead? \par
Do you see why we can call these bits anticorrelated?

\vfill

\problem{}
Show that the bell states are orthogonal \par
\hint{Dot product}

\vfill

\problem{}<bellmeasure>
Say we have a pair of qubits in one of the four bell states. \par
How can we find out which of the four states we have, with certainty? \par
\hint{$H\ket{+} = \ket{0}$, and $H\ket{-} = \ket{1}$}

\begin{solution}
	$X_\text{c}\ket{\Phi^+} = \ket{+0}$ and $(H \otimes I)\ket{+0} = \ket{00}$ \par
	$X_\text{c}\ket{\Psi^+} = \ket{+1}$ and $(H \otimes I)\ket{+1} = \ket{01}$ \par
	$X_\text{c}\ket{\Phi^-} = \ket{-0}$ and $(H \otimes I)\ket{-0} = \ket{10}$ \par
	$X_\text{c}\ket{\Psi^-} = \ket{-1}$ and $(H \otimes I)\ket{-1} = \ket{11}$ \par
\end{solution}


\vfill
\pagebreak

\definition{}
The $Z$ gate is defined as follows: \par
\begin{equation*}
	Z\begin{bmatrix}
		\psi_0 \\ \psi_1
	\end{bmatrix}
	=
	\begin{bmatrix}
		\psi_0 \\ -\psi_1
	\end{bmatrix}
\end{equation*}

\problem{}
Suppose that Alice and Bob are each in possession of one qubit. \par
These two qubits are entangled, and have the compound state $\ket{\Phi^+}$. \par
\note[Note]{We could say that they each have \say{half} of $\ket{\Phi^+}$.}
How can Alice send a two-bit classical state
(i.e, one of the four values \texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}) \par
to Bob by only sending one qubit?

\begin{solution}
	Alice can turn any bell state into any other by applying operations to her qubit. \par
	Once she does so, Bob may use the procedure in \ref{bellmeasure} to read one of four states.

	\null\hfill
	\begin{minipage}{0.3\textwidth}
		\begin{center}
			\begin{tikzpicture}[scale = 1]
				\node[qubit] (a) at (0, 0) {};
				\node[qubit] (b) at (0, -1) {};
				\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
				\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};
				\node[right] at (0, -0.5) {$\ket{\Phi^+}$};
				\node[left] at (4, -0.5) {$\ket{\Phi^-}$};

				\qubox{a}{1.5}{a}{2.5}{$Z$}
			\end{tikzpicture}
		\end{center}
	\end{minipage}
	\hfill
	\begin{minipage}{0.3\textwidth}
		\begin{center}
			\begin{tikzpicture}[scale = 1]
				\node[qubit] (a) at (0, 0) {};
				\node[qubit] (b) at (0, -1) {};
				\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
				\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};
				\node[right] at (0, -0.5) {$\ket{\Phi^+}$};
				\node[left] at (4, -0.5) {$\ket{\Psi^+}$};

				\qubox{a}{1.5}{a}{2.5}{$X$}
			\end{tikzpicture}
		\end{center}
	\end{minipage}
	\hfill
	\begin{minipage}{0.3\textwidth}
		\begin{center}
			\begin{tikzpicture}[scale = 1]
				\node[qubit] (a) at (0, 0) {};
				\node[qubit] (b) at (0, -1) {};
				\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
				\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};
				\node[right] at (0, -0.5) {$\ket{\Phi^+}$};
				\node[left] at (4, -0.5) {$\ket{\Psi^-}$};

				\qubox{a}{1}{a}{2}{$X$}
				\qubox{a}{2}{a}{3}{$Z$}
			\end{tikzpicture}
		\end{center}
	\end{minipage}
	\hfill\null

	\vspace{4mm}
	\linehack{}

	The complete circuit is shown below. Double lines indicate classical bits.


	\begin{center}
		\begin{tikzpicture}[scale = 1]
			\node[qubit] (a) at (0, 0) {$a$};
			\node[qubit] (b) at (0, -1) {$b$};
			\node[qubit] (c) at (0, -2) {$\ket{\Phi^+_\text{A}}$};
			\node[qubit] (d) at (0, -3) {$\ket{\Phi^+_\text{B}}$};
			\draw[wire, double] (a) -- ([shift={(9, 0)}] a.center) node[qubit] {};
			\draw[wire, double] (b) -- ([shift={(9, 0)}] b.center) node[qubit] {};
			\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {};
			\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {};

			\draw[wire, double]
				([shift={(7, 0)}] c.center)
				-- ([shift={(9, 0)}] c.center)
				node[qubit] {$a$}
			;

			\draw[wire, double]
				([shift={(7, 0)}] d.center)
				-- ([shift={(9, 0)}] d.center)
				node[qubit] {$b$}
			;


			\draw[wire, double]
				($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$) --
				($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
			;
			\draw[wirejoin]
				($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
				circle[radius=0.1] coordinate(dot)
			;

			\qubox{c}{1}{c}{2}{$X$}

			\draw[wire, double]
				($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
				($([shift={(2,0)}] c)!0.5!([shift={(3,0)}] c)$)
			;
			\draw[wirejoin]
				($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$)
				circle[radius=0.1] coordinate(dot)
			;

			\qubox{c}{2}{c}{3}{$Z$}



			\draw[wire]
				($([shift={(4,0)}] c)!0.5!([shift={(5,0)}] c)$) --
				($([shift={(4,0)}] d)!0.5!([shift={(5,0)}] d)$)
			;
			\draw[wirejoin]
				($([shift={(4,0)}] c)!0.5!([shift={(5,0)}] c)$)
				circle[radius=0.1] coordinate(dot)
			;

			\qubox{d}{4}{d}{5}{$X$}
			\qubox{c}{5}{c}{6}{$H$}



			\qubox{c}{6.3}{c}{8}{measure}
			\qubox{d}{6.3}{d}{8}{measure}
		\end{tikzpicture}
	\end{center}
\end{solution}


\vfill

\generic{Remark:}
Superdense coding consumes a pre-shared entangled pair to transmit two bits of information.
This entanglement may \textit{not} be re-used---it is destroyed when Bob measures the final qubit states.

\pagebreak

