\section{Quantum Teleportation}

Superdense coding lets us convert quantum bandwidth into classical bandwidth. \par
Quantum teleporation does the opposite, using two classical bits and an entangled pair to transmit a quantum state.

\generic{Setup:}
Again, suppose Alice and Bob each have half of a $\ket{\Phi^+}$ state. \par
We'll call the state Alice wants to teleport $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$. \par

\problem{}
What is the three-qubit state $\ket{\psi}\ket{\Phi^+}$ in terms of $\psi_0$ and $\psi_1$?

\vfill

\problem{}
To teleport $\ket{\psi}$, Alice applies the following circuit to her two qubits, where $\ket{\Phi^+_\text{A}}$ is her half of $\ket{\Phi^+}$. \par
She then measures both qubits and sends the result to Bob.

\begin{center}
	\begin{tikzpicture}[scale = 1]
		\node[qubit] (a) at (0, 0) {$\ket{\Phi^+_\text{A}}$};
		\node[qubit] (b) at (0, -1) {$\ket{\psi}$};
		\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
		\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};

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

		\qubox{b}{2}{b}{3}{$H$}
		\qubox{a}{1}{a}{2}{$X$}
	\end{tikzpicture}
\end{center}
What should Bob do so that $\ket{\Phi^+_B}$ takes the state $\ket{\psi}$ had initially?

\begin{solution}
	\begin{itemize}
		\item
		If Bob receives \texttt{00}, he does nothing.

		\item
		If Bob receives \texttt{01}, he applies an $X$ gate to his qubit.

		\item
		If Bob receives \texttt{01}, he applies a $Z$ gate to his qubit.

		\item
		If Bob receives \texttt{11}, he applies $ZX$ to his qubit.
	\end{itemize}

	\linehack{}

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


	\begin{center}
		\begin{tikzpicture}[scale = 1]
			\node[qubit] (a) at (0, -1) {$\ket{\Phi^+_\text{A}}$};
			\node[qubit] (b) at (0, -2) {$\ket{\Phi^+_\text{B}}$};
			\node[qubit] (c) at (0, 0) {$\ket{\psi}$};
			\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {};
			\draw[wire] (b) -- ([shift={(9, 0)}] b.center) node[qubit] {$\ket{\psi}$};
			\draw[wire] (c) -- ([shift={(5, 0)}] c.center) node[qubit] {};

			\draw[wire, double]
				([shift={(5, 0)}] a.center)
				-- ([shift={(9, 0)}] a.center)
				node[qubit] {}
			;

			\draw[wire, double]
				([shift={(5, 0)}] c.center)
				-- ([shift={(9, 0)}] c.center)
				node[qubit] {}
			;

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

			\qubox{c}{2}{c}{3}{$H$}
			\qubox{a}{1}{a}{2}{$X$}

			\qubox{a}{3.8}{a}{5.5}{measure}
			\qubox{c}{3.8}{c}{5.5}{measure}


			\draw[wire, double]
				($([shift={(6,0)}] a)!0.5!([shift={(7,0)}] a)$) --
				($([shift={(6,0)}] b)!0.5!([shift={(7,0)}] b)$)
			;
			\draw[wirejoin]
				($([shift={(6,0)}] a)!0.5!([shift={(7,0)}] a)$)
				circle[radius=0.1] coordinate(dot)
			;

			\qubox{b}{6}{b}{7}{$X$}


			\draw[wire, double]
			($([shift={(7,0)}] b)!0.5!([shift={(8,0)}] b)$) --
			($([shift={(7,0)}] c)!0.5!([shift={(8,0)}] c)$)
			;
			\draw[wirejoin]
				($([shift={(7,0)}] c)!0.5!([shift={(8,0)}] c)$)
				circle[radius=0.1] coordinate(dot)
			;
			\qubox{b}{7}{b}{8}{$Z$}

		\end{tikzpicture}
	\end{center}

	Note how similar this is to the superdense coding circuit.
\end{solution}

\vfill
\pagebreak

\problem{}
With an informal proof, show that it is not possible to use superdense coding to send
more than two classical bits through an entangled two-qubit quantum state.

\begin{solution}
	If superdense coding was any more efficient, we could repeatedly apply superdense coding and quantum teleporation,
	to compress an arbitrary number of bits into two \say{seed} bits.

	\linehack{}

	\textbf{Even worse, this would allow faster-than-light communication:} \par

	Because the seed message is only 4 bits, Alice has decent odds of just
	guessing it. She'll guess wrong and trash the message the majority of the
	time but, by using an error correcting code, she can tell whether or not
	the guess was correct or she trashed the message. And by repeating the protocol
	enough times, we can increase the odds of the message being received arbitrarily
	close to certainty.

	\note[Note]{
		I'm implicitly assuming that if Alice uses the wrong seed, she gets a totally random message---or
		at least a message that isn't guaranteed to follow the error correction scheme better than chance would.
		The alternative, where Alice receives noise that's uncorrelated with the message and yet somehow satisfies
		arbitrary error correction schemes, is waaay too magical for me to even consider.
	}

	\vspace{2mm}

	Suppose Alice and Bob perform the iterated-ultradense-encode-and-guess process 100 times.
	That gives a failure rate of $(\nicefrac{15}{16})^{100} \approx 0.5\%$.
	Sure it's a hundred times more work than just sending the 4 bits, and less likely to succeed to boot,
	but the new protocol \textit{doesn't require any bits to be physically transmitted}.
	There's no signalling delay!

	\vspace{2mm}

	In fact, Alice could even perform the decoding process \textit{before} Bob did the encoding.
	But we're already so far into \say{everything is clearly broken} territory that creating time travel paradoxes is overkill.


	\vspace{5mm}


	From \url{https://algassert.com/2016/05/29/ultra-dense-coding-allows-ftl.html}
\end{solution}


\vfill
\pagebreak

