180 lines
5.5 KiB
TeX
180 lines
5.5 KiB
TeX
\section{Quantum Teleportation}
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Superdense coding lets us convert quantum bandwidth into classical bandwidth. \par
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Quantum teleportation does the opposite, using two classical bits and an entangled pair
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to transmit a quantum state.
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\generic{Setup:}
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Again, suppose Alice and Bob each have half of a $\ket{\Phi^+}$ state. \par
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We'll call the state Alice wants to teleport $\ket{\psi} = \psi_0\ket{0} + \psi_1\ket{1}$. \par
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\problem{}
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What is the three-qubit state $\ket{\psi}\ket{\Phi^+}$ in terms of $\psi_0$ and $\psi_1$?
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\vfill
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\problem{}
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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
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She then measures both qubits and sends the result to Bob.
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\begin{center}
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\begin{tikzpicture}[scale = 1]
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\node[qubit] (a) at (0, 0) {$\ket{\Phi^+_\text{A}}$};
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\node[qubit] (b) at (0, -1) {$\ket{\psi}$};
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\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
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\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{b}{2}{b}{3}{$H$}
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\qubox{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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What should Bob do so that $\ket{\Phi^+_B}$ takes the state $\ket{\psi}$ had initially?
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\begin{solution}
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\begin{itemize}
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\item
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If Bob receives \texttt{00}, he does nothing.
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\item
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If Bob receives \texttt{01}, he applies an $X$ gate to his qubit.
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\item
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If Bob receives \texttt{01}, he applies a $Z$ gate to his qubit.
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\item
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If Bob receives \texttt{11}, he applies $ZX$ to his qubit.
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\end{itemize}
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\linehack{}
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The complete circuit is shown below. Double lines indicate classical bits.
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\begin{center}
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\begin{tikzpicture}[scale = 1]
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\node[qubit] (a) at (0, -1) {$\ket{\Phi^+_\text{A}}$};
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\node[qubit] (b) at (0, -2) {$\ket{\Phi^+_\text{B}}$};
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\node[qubit] (c) at (0, 0) {$\ket{\psi}$};
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\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {};
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\draw[wire] (b) -- ([shift={(9, 0)}] b.center) node[qubit] {$\ket{\psi}$};
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\draw[wire] (c) -- ([shift={(5, 0)}] c.center) node[qubit] {};
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\draw[wire, double]
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([shift={(5, 0)}] a.center)
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-- ([shift={(9, 0)}] a.center)
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node[qubit] {}
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;
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\draw[wire, double]
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([shift={(5, 0)}] c.center)
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-- ([shift={(9, 0)}] c.center)
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node[qubit] {}
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;
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\draw[wire]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
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($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{c}{2}{c}{3}{$H$}
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\qubox{a}{1}{a}{2}{$X$}
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\qubox{a}{3.8}{a}{5.5}{measure}
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\qubox{c}{3.8}{c}{5.5}{measure}
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\draw[wire, double]
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($([shift={(6,0)}] a)!0.5!([shift={(7,0)}] a)$) --
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($([shift={(6,0)}] b)!0.5!([shift={(7,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(6,0)}] a)!0.5!([shift={(7,0)}] a)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{b}{6}{b}{7}{$X$}
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\draw[wire, double]
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($([shift={(7,0)}] b)!0.5!([shift={(8,0)}] b)$) --
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($([shift={(7,0)}] c)!0.5!([shift={(8,0)}] c)$)
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;
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\draw[wirejoin]
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($([shift={(7,0)}] c)!0.5!([shift={(8,0)}] c)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{b}{7}{b}{8}{$Z$}
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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 teleportation,
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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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