179 lines
5.5 KiB
TeX
179 lines
5.5 KiB
TeX
\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
|
|
|