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@ -163,7 +163,7 @@ to Bob by only sending one qubit?
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\draw[wire, double]
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\draw[wire]
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($([shift={(4,0)}] c)!0.5!([shift={(5,0)}] c)$) --
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($([shift={(4,0)}] d)!0.5!([shift={(5,0)}] d)$)
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;
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@ -179,7 +179,6 @@ to Bob by only sending one qubit?
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\qubox{c}{6.3}{c}{8}{measure}
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\qubox{d}{6.3}{d}{8}{measure}
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\end{tikzpicture}
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\end{center}
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\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
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\end{tikzpicture}
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\end{center}
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Note how similar this is to the superdense coding circuit.
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\end{solution}
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\vfill
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\pagebreak
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\problem{}
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With an informal proof, show that it is not possible to use superdense coding to send
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more than two classical bits through an entangled two-qubit quantum state.
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\begin{solution}
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If superdense coding was any more efficient, we could repeatedly apply superdense coding and quantum teleporation,
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to compress an arbitrary number of bits into two \say{seed} bits.
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\linehack{}
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\textbf{Even worse, this would allow faster-than-light communication:} \par
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Because the seed message is only 4 bits, Alice has decent odds of just
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guessing it. She'll guess wrong and trash the message the majority of the
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time but, by using an error correcting code, she can tell whether or not
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the guess was correct or she trashed the message. And by repeating the protocol
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enough times, we can increase the odds of the message being received arbitrarily
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close to certainty.
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\note[Note]{
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I'm implicitly assuming that if Alice uses the wrong seed, she gets a totally random message---or
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at least a message that isn't guaranteed to follow the error correction scheme better than chance would.
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The alternative, where Alice receives noise that's uncorrelated with the message and yet somehow satisfies
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arbitrary error correction schemes, is waaay too magical for me to even consider.
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}
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\vspace{2mm}
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Suppose Alice and Bob perform the iterated-ultradense-encode-and-guess process 100 times.
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That gives a failure rate of $(\nicefrac{15}{16})^{100} \approx 0.5\%$.
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Sure it's a hundred times more work than just sending the 4 bits, and less likely to succeed to boot,
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but the new protocol \textit{doesn't require any bits to be physically transmitted}.
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There's no signalling delay!
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\vspace{2mm}
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In fact, Alice could even perform the decoding process \textit{before} Bob did the encoding.
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But we're already so far into \say{everything is clearly broken} territory that creating time travel paradoxes is overkill.
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\vspace{5mm}
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From \url{https://algassert.com/2016/05/29/ultra-dense-coding-allows-ftl.html}
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\end{solution}
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\vfill
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\pagebreak
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@ -20,19 +20,17 @@
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singlenumbering
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]{../../../resources/ormc_handout}
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\usepackage{../../../resources/macros}
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\def\ket#1{\left|#1\right\rangle}
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\def\bra#1{\left\langle#1\right|}
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\usepackage{units}
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\input{tikzset}
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\uptitlel{Advanced 2}
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\uptitler{Winter 2022}
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\title{Intro to Quantum Computing I}
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\subtitle{Prepared by \githref{Mark} on \today{}}
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\def\ket#1{\left|#1\right\rangle}
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\def\bra#1{\left\langle#1\right|}
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% TODO: spend more time on probabalistic bits.
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% This could even be its own handout, especially
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@ -21,13 +21,11 @@
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shortwarning
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]{../../../resources/ormc_handout}
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\usepackage{../../../resources/macros}
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\def\ket#1{\left|#1\right\rangle}
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\def\bra#1{\left\langle#1\right|}
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\usepackage{units}
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\input{tikzset}
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\def\ket#1{\left|#1\right\rangle}
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\def\bra#1{\left\langle#1\right|}
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\uptitlel{Advanced 2}
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\uptitler{Winter 2022}
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@ -35,6 +33,13 @@
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\subtitle{Prepared by \githref{Mark} on \today{}}
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% Hack to keep numbering from previous handout
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\makeatletter
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\setcounter{\@problemcounter}{34}
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\setcounter{section_counter}{6}
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\makeatother
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\begin{document}
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\maketitle
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