Minor edits

Advanced/Introduction to Quantum/src/parts/07 superdense.tex

@@ -163,7 +163,7 @@ to Bob by only sending one qubit?
 
 
 
-			\draw[wire, double]
+			\draw[wire]
 				($([shift={(4,0)}] c)!0.5!([shift={(5,0)}] c)$) --
 				($([shift={(4,0)}] d)!0.5!([shift={(5,0)}] d)$)
 			;
@@ -179,7 +179,6 @@ to Bob by only sending one qubit?
 
 			\qubox{c}{6.3}{c}{8}{measure}
 			\qubox{d}{6.3}{d}{8}{measure}
-
 		\end{tikzpicture}
 	\end{center}
 \end{solution}

Advanced/Introduction to Quantum/src/parts/08 teleport.tex

@@ -119,8 +119,60 @@ What should Bob do so that $\ket{\Phi^+_B}$ takes the state $\ket{\psi}$ had ini
 
 		\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
+