Added hxh, superdense, and teleport
This commit is contained in:
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Advanced/Introduction to Quantum/src/parts/06 hxh.tex
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421
Advanced/Introduction to Quantum/src/parts/06 hxh.tex
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\section{HXH}
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Let's return to the quantum circuit diagrams we discussed a few pages ago. \par
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Keep in mind that we're working with quantum gates and proper half-qubits---not classical bits, as we were before.
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\definition{Controlled Inputs}
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A \textit{control input} or \textit{inverted control input} may be attached to any gate. \par
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These are drawn as filled and empty circles in our circuit diagrams:
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\null\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{0}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
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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{a}{1}{a}{2}{$X$}
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\node[gray] at ($([shift={(0,-0.75)}] dot)$) {Non-inverted control input};
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$};
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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[wireijoin]
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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{a}{1}{a}{2}{$X$}
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\node[gray] at ($([shift={(0,-0.75)}] dot)$) {Inverted control input};
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill\null
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\vspace{2mm}
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An $X$ gate with a (non-inverted) control input behaves like an $X$ gate if \textit{all} its control inputs are $\ket{1}$,
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and like $I$ otherwise. An $X$ gate with an inverted control inputs does the opposite, behaving like $I$ if its input is $\ket{1}$
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and like $X$ otherwise. The two circuits above illustrate this fact---take a look at their inputs and outputs.
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\vspace{2mm}
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Of course, we can give a gate multiple controls. \par
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An $X$ gate with multiplie controls behaves like an $X$ gate if...
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\begin{itemize}
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\item all non-inverted controls are $\ket{1}$, and
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\item all inverted controls are $\ket{0}$
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\end{itemize}
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...and like $I$ otherwise.
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\problem{}
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What are the final states of the qubits in the diagram below?
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\begin{center}
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\begin{tikzpicture}[scale = 1.0]
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\node[qubit] (a) at (0, 0) {$\ket{1}$};
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\node[qubit] (b) at (0, -1) {$\ket{0}$};
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\node[qubit] (c) at (0, -2) {$\ket{1}$};
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\node[qubit] (d) at (0, -3) {$\ket{0}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {?};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {?};
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\draw[wire] (c) -- ([shift={(3, 0)}] c.center) node[qubit] {?};
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\draw[wire] (d) -- ([shift={(3, 0)}] d.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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($([shift={(1,0)}] d)!0.5!([shift={(2,0)}] d)$)
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;
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\draw[wirejoin]
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($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$)
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circle[radius=0.1] coordinate(dot)
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;
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\draw[wireijoin]
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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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\draw[wirejoin]
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($([shift={(1,0)}] d)!0.5!([shift={(2,0)}] d)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{b}{1}{b}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\vfill
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\pagebreak
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\problem{}
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Consider the diagram below, with one controlled $X$ gate: \par
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\note[Note]{The CNOT gate from \ref{cnot} is a controlled $X$ gate.}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{a}$};
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\node[qubit] (b) at (0, -1) {$\ket{b}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {};
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\draw[wire] (b) -- ([shift={(3, 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{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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Find a matrix $\text{X}_\text{c}$ that represents this gate, so that $\text{X}_\text{c}\ket{ab}$ works as expected.
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\begin{solution}
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\begin{equation*}
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\text{X}_\text{c} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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0 & 0 & 1 & 0 \\
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0 & 1 & 0 & 0
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\end{bmatrix}
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\end{equation*}
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Note that this is also the solution to \ref{cnotflipped}.
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\end{solution}
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\vfill
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\problem{}
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Now, evaluate the following. Remember that
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$\ket{+} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} + \ket{1}\Bigr)$ and
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$\ket{-} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} - \ket{1}\Bigr)$
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\null\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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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{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{+}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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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{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill\null
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\vspace{5mm}
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\null\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{-}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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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{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill
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\begin{minipage}{0.48\textwidth}
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{+}$};
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\node[qubit] (b) at (0, -1) {$\ket{-}$};
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\draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$};
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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{a}{1}{a}{2}{$X$}
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\end{tikzpicture}
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\end{center}
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\end{minipage}
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\hfill\null
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\hint{
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Note that some of these states are entangled. The circuit diagrams are a bit misleading:
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we can't write an entangled state as two distinct qubits!
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\vspace{2mm}
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So, don't try to find $\ket{a}$ and $\ket{b}$. \par
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Instead find $\ket{ab} = \psi_0\ket{00} + \psi_1\ket{01} + \psi_2\ket{10} + \psi_3\ket{11}$, and factor it into $\ket{a} \otimes \ket{b}$ if you can.
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}
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\begin{solution}
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In all the below equations, let $\tau = \frac{1}{\sqrt{2}}$.
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\begin{itemize}[itemsep = 1mm]
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\item $
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\text{X}_\text{c}\ket{01}
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= \ket{11}
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$
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\item $
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\text{X}_\text{c}\ket{0+}
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= \tau\ket{00} + \tau\ket{11}
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$ \note[Note]{This state is entangled!}
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\item $
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\text{X}_\text{c}\ket{-1}
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= -\tau\ket{10} + \tau\ket{11}
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= (-\ket{-}) \otimes \ket{1}
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$
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\item $
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\text{X}_\text{c}\ket{+-}
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= \frac{1}{2}(\ket{00} - \ket{01} + \ket{10} - \ket{11})
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= \ket{+-}
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$
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\end{itemize}
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\end{solution}
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\vfill
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\pagebreak
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\generic{Remark:}
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Now, consider the following circuit:
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\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
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\node[qubit] (b) at (0, -1) {$\ket{1}$};
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\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {$\ket{a}$};
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\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{b}$};
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\qubox{b}{1}{b}{2}{$H$}
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\qubox{b}{3}{b}{4}{$H$}
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\draw[wire]
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($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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;
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\draw[wirejoin]
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($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
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circle[radius=0.1] coordinate(dot)
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;
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\qubox{a}{2}{a}{3}{$X$}
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\end{tikzpicture}
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\end{center}
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We already know that $H$ is its own inverse: $HH = I$. \par
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Applying $H$ to a qubit twice does not change its state.
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\note{
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Recall that $H = \frac{1}{\sqrt{2}}\left[\begin{smallmatrix} 1 & 1 \\ 1 & -1 \end{smallmatrix}\right]$
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}
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\vspace{2mm}
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So, we might expect that the two circuits below are equivalent: \par
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After all, we $H$ the second bit, use it to control an $X$ gate, and then $H$ it back to its previous state.
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\null\hfill
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\begin{minipage}{0.48\textwidth}\begin{center}
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\begin{tikzpicture}[scale=0.8]
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\node[qubit] (a) at (0, 0) {$\ket{0}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{1}$};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {$\ket{a}$};
|
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\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{b}$};
|
||||
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||||
\qubox{b}{1}{b}{2}{$H$}
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||||
\qubox{b}{3}{b}{4}{$H$}
|
||||
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||||
\draw[wire]
|
||||
($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
|
||||
($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
|
||||
circle[radius=0.1] coordinate(dot)
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||||
;
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||||
\qubox{a}{2}{a}{3}{$X$}
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||||
|
||||
\end{tikzpicture}
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||||
\end{center}\end{minipage}
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||||
\hfill
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||||
\begin{minipage}{0.48\textwidth}\begin{center}
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||||
\begin{tikzpicture}[scale=0.8]
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||||
\node[qubit] (a) at (0, 0) {$\ket{0}$};
|
||||
\node[qubit] (b) at (0, -1) {$\ket{1}$};
|
||||
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||||
\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {$\ket{c}$};
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||||
\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{d}$};
|
||||
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||||
\draw[wire]
|
||||
($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
|
||||
($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(2,0)}] b)!0.5!([shift={(3,0)}] b)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
\qubox{a}{2}{a}{3}{$X$}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}\end{minipage}
|
||||
\hfill\null
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||||
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||||
\vspace{2mm}
|
||||
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||||
This, however, isn't the case: \par
|
||||
If we compute the state $\ket{ab}$ in the left circuit, we get $[0.5, ~0.5, -0.5, ~0.5]$ (which is entangled), \par
|
||||
but the state $\ket{cd}$ on the right is $\ket{11} = [0,0,0,1]$. \par
|
||||
\note{This is easy to verify with a few matrix multiplications.}
|
||||
|
||||
|
||||
\vspace{4mm}
|
||||
|
||||
How does this make sense? \par
|
||||
Remember that a two-bit quantum state is \textit{not} equivalent to a pair of one-qubit quantum states.
|
||||
We must treat a multi-qubit state as a single unit.
|
||||
|
||||
Recall that a two-bit state $\ket{ab}$ comes with four probabilities:
|
||||
$\mathcal{P}(\texttt{00})$, $\mathcal{P}(\texttt{01})$, $\mathcal{P}(\texttt{10})$, and $\mathcal{P}(\texttt{11})$.
|
||||
If we change the probabilities of only $\ket{a}$, \textit{all four of these change!}
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
Because of this fact, \say{controlled gates} may not work as you expect. They may seem
|
||||
to \say{read} their controlling qubit without affecting its state, but remember---a
|
||||
controlled gate still affects the \textit{entire} state. As we noted before, it is
|
||||
not possible to apply a transformation to one bit of a quantum state.
|
||||
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale=1]
|
||||
\node[qubit] (a) at (0, 0) {\texttt{1}};
|
||||
\node[qubit] (b) at (0, -1) {\texttt{0}};
|
||||
|
||||
\draw[wire] (a) -- ([shift={(5, 0)}] a.center) node[qubit] {\texttt{0}};
|
||||
\draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {\texttt{1}};
|
||||
|
||||
\ghostqubox{a}{1}{b}{2}{$T_1$}
|
||||
\ghostqubox{a}{2}{b}{3}{$T_2$}
|
||||
\ghostqubox{a}{3}{b}{4}{$T_3$}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
195
Advanced/Introduction to Quantum/src/parts/07 superdense.tex
Normal file
195
Advanced/Introduction to Quantum/src/parts/07 superdense.tex
Normal file
@ -0,0 +1,195 @@
|
||||
\section{Superdense Coding}
|
||||
|
||||
Consider the following entangled two-qubit states, called the \textit{bell states}:
|
||||
\begin{itemize}
|
||||
\item $\ket{\Phi^+} = \frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}$
|
||||
\item $\ket{\Phi^-} = \frac{1}{\sqrt{2}}\ket{00} - \frac{1}{\sqrt{2}}\ket{11}$
|
||||
\item $\ket{\Psi^+} = \frac{1}{\sqrt{2}}\ket{01} + \frac{1}{\sqrt{2}}\ket{10}$
|
||||
\item $\ket{\Psi^-} = \frac{1}{\sqrt{2}}\ket{01} - \frac{1}{\sqrt{2}}\ket{10}$
|
||||
\end{itemize}
|
||||
|
||||
\problem{}
|
||||
The probabilistic bits we get when measuring any of the above may be called \textit{anticorrelated bits}. \par
|
||||
If we measure the first bit of any of these states and observe $1$, what is the resulting compound state? \par
|
||||
What if we observe $0$ instead? \par
|
||||
Do you see why we can call these bits anticorrelated?
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}
|
||||
Show that the bell states are orthogonal \par
|
||||
\hint{Dot product}
|
||||
|
||||
\vfill
|
||||
|
||||
\problem{}<bellmeasure>
|
||||
Say we have a pair of qubits in one of the four bell states. \par
|
||||
How can we find out which of the four states we have, with certainty? \par
|
||||
\hint{$H\ket{+} = \ket{0}$, and $H\ket{-} = \ket{1}$}
|
||||
|
||||
\begin{solution}
|
||||
$X_\text{c}\ket{\Phi^+} = \ket{+0}$ and $(H \otimes I)\ket{+0} = \ket{00}$ \par
|
||||
$X_\text{c}\ket{\Psi^+} = \ket{+1}$ and $(H \otimes I)\ket{+1} = \ket{01}$ \par
|
||||
$X_\text{c}\ket{\Phi^-} = \ket{-0}$ and $(H \otimes I)\ket{-0} = \ket{10}$ \par
|
||||
$X_\text{c}\ket{\Psi^-} = \ket{-1}$ and $(H \otimes I)\ket{-1} = \ket{11}$ \par
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
\definition{}
|
||||
The $Z$ gate is defined as follows: \par
|
||||
\begin{equation*}
|
||||
Z\begin{bmatrix}
|
||||
\psi_0 \\ \psi_1
|
||||
\end{bmatrix}
|
||||
=
|
||||
\begin{bmatrix}
|
||||
\psi_0 \\ -\psi_1
|
||||
\end{bmatrix}
|
||||
\end{equation*}
|
||||
|
||||
\problem{}
|
||||
Suppose that Alice and Bob are each in possession of one qubit. \par
|
||||
These two qubits are entangled, and have the compound state $\ket{\Phi^+}$. \par
|
||||
How can Alice send a two-bit classical state
|
||||
(i.e, one of the four values \texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}) \par
|
||||
to Bob by only sending one qubit?
|
||||
|
||||
\begin{solution}
|
||||
Alice can turn any bell state into any other by applying operations to her qubit. \par
|
||||
Once she does so, Bob may use the procedure in \ref{bellmeasure} to read one of four states.
|
||||
|
||||
\null\hfill
|
||||
\begin{minipage}{0.3\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale = 1]
|
||||
\node[qubit] (a) at (0, 0) {};
|
||||
\node[qubit] (b) at (0, -1) {};
|
||||
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
|
||||
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};
|
||||
\node[right] at (0, -0.5) {$\ket{\Phi^+}$};
|
||||
\node[left] at (4, -0.5) {$\ket{\Phi^-}$};
|
||||
|
||||
\qubox{a}{1.5}{a}{2.5}{$Z$}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill
|
||||
\begin{minipage}{0.3\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale = 1]
|
||||
\node[qubit] (a) at (0, 0) {};
|
||||
\node[qubit] (b) at (0, -1) {};
|
||||
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
|
||||
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};
|
||||
\node[right] at (0, -0.5) {$\ket{\Phi^+}$};
|
||||
\node[left] at (4, -0.5) {$\ket{\Psi^+}$};
|
||||
|
||||
\qubox{a}{1.5}{a}{2.5}{$X$}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill
|
||||
\begin{minipage}{0.3\textwidth}
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale = 1]
|
||||
\node[qubit] (a) at (0, 0) {};
|
||||
\node[qubit] (b) at (0, -1) {};
|
||||
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {};
|
||||
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {};
|
||||
\node[right] at (0, -0.5) {$\ket{\Phi^+}$};
|
||||
\node[left] at (4, -0.5) {$\ket{\Psi^-}$};
|
||||
|
||||
\qubox{a}{1}{a}{2}{$X$}
|
||||
\qubox{a}{2}{a}{3}{$Z$}
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{minipage}
|
||||
\hfill\null
|
||||
|
||||
\vspace{4mm}
|
||||
\linehack{}
|
||||
|
||||
The complete circuit is shown below. Double lines indicate classical bits.
|
||||
|
||||
|
||||
\begin{center}
|
||||
\begin{tikzpicture}[scale = 1]
|
||||
\node[qubit] (a) at (0, 0) {$a$};
|
||||
\node[qubit] (b) at (0, -1) {$b$};
|
||||
\node[qubit] (c) at (0, -2) {$\ket{\Phi^+_\text{A}}$};
|
||||
\node[qubit] (d) at (0, -3) {$\ket{\Phi^+_\text{B}}$};
|
||||
\draw[wire, double] (a) -- ([shift={(9, 0)}] a.center) node[qubit] {};
|
||||
\draw[wire, double] (b) -- ([shift={(9, 0)}] b.center) node[qubit] {};
|
||||
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {};
|
||||
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {};
|
||||
|
||||
\draw[wire, double]
|
||||
([shift={(7, 0)}] c.center)
|
||||
-- ([shift={(9, 0)}] c.center)
|
||||
node[qubit] {$a$}
|
||||
;
|
||||
|
||||
\draw[wire, double]
|
||||
([shift={(7, 0)}] d.center)
|
||||
-- ([shift={(9, 0)}] d.center)
|
||||
node[qubit] {$b$}
|
||||
;
|
||||
|
||||
|
||||
\draw[wire, double]
|
||||
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$) --
|
||||
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
|
||||
\qubox{c}{1}{c}{2}{$X$}
|
||||
|
||||
\draw[wire, double]
|
||||
($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$) --
|
||||
($([shift={(2,0)}] c)!0.5!([shift={(3,0)}] c)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(2,0)}] a)!0.5!([shift={(3,0)}] a)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
|
||||
\qubox{c}{2}{c}{3}{$Z$}
|
||||
|
||||
|
||||
|
||||
\draw[wire, double]
|
||||
($([shift={(4,0)}] c)!0.5!([shift={(5,0)}] c)$) --
|
||||
($([shift={(4,0)}] d)!0.5!([shift={(5,0)}] d)$)
|
||||
;
|
||||
\draw[wirejoin]
|
||||
($([shift={(4,0)}] c)!0.5!([shift={(5,0)}] c)$)
|
||||
circle[radius=0.1] coordinate(dot)
|
||||
;
|
||||
|
||||
\qubox{d}{4}{d}{5}{$X$}
|
||||
\qubox{c}{5}{c}{6}{$H$}
|
||||
|
||||
|
||||
|
||||
\qubox{c}{6.3}{c}{8}{measure}
|
||||
\qubox{d}{6.3}{d}{8}{measure}
|
||||
|
||||
\end{tikzpicture}
|
||||
\end{center}
|
||||
\end{solution}
|
||||
|
||||
|
||||
\vfill
|
||||
|
||||
\generic{Remark:}
|
||||
Superdense coding consumes a pre-shared entangled pair to transmit two bits of information.
|
||||
This entanglement may \textit{not} be re-used---it is destroyed when Bob measures the final qubit states.
|
||||
|
||||
\pagebreak
|
||||
|
||||
126
Advanced/Introduction to Quantum/src/parts/08 teleport.tex
Normal file
126
Advanced/Introduction to Quantum/src/parts/08 teleport.tex
Normal file
@ -0,0 +1,126 @@
|
||||
\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}
|
||||
\end{solution}
|
||||
|
||||
\vfill
|
||||
\pagebreak
|
||||
|
||||
Reference in New Issue
Block a user