Minor edits

This commit is contained in:
Mark 2024-02-21 16:54:42 -08:00
parent 70c3680506
commit 579e05e2c3
Signed by: Mark
GPG Key ID: C6D63995FE72FD80
4 changed files with 65 additions and 11 deletions

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@ -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}

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@ -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

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@ -20,19 +20,17 @@
singlenumbering
]{../../../resources/ormc_handout}
\usepackage{../../../resources/macros}
\def\ket#1{\left|#1\right\rangle}
\def\bra#1{\left\langle#1\right|}
\usepackage{units}
\input{tikzset}
\uptitlel{Advanced 2}
\uptitler{Winter 2022}
\title{Intro to Quantum Computing I}
\subtitle{Prepared by \githref{Mark} on \today{}}
\def\ket#1{\left|#1\right\rangle}
\def\bra#1{\left\langle#1\right|}
% TODO: spend more time on probabalistic bits.
% This could even be its own handout, especially

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@ -21,13 +21,11 @@
shortwarning
]{../../../resources/ormc_handout}
\usepackage{../../../resources/macros}
\def\ket#1{\left|#1\right\rangle}
\def\bra#1{\left\langle#1\right|}
\usepackage{units}
\input{tikzset}
\def\ket#1{\left|#1\right\rangle}
\def\bra#1{\left\langle#1\right|}
\uptitlel{Advanced 2}
\uptitler{Winter 2022}
@ -35,6 +33,13 @@
\subtitle{Prepared by \githref{Mark} on \today{}}
% Hack to keep numbering from previous handout
\makeatletter
\setcounter{\@problemcounter}{34}
\setcounter{section_counter}{6}
\makeatother
\begin{document}
\maketitle