diff --git a/Advanced/Introduction to Quantum/src/parts/06 hxh.tex b/Advanced/Introduction to Quantum/src/parts/06 hxh.tex new file mode 100644 index 0000000..cd99e92 --- /dev/null +++ b/Advanced/Introduction to Quantum/src/parts/06 hxh.tex @@ -0,0 +1,421 @@ +\section{HXH} + +Let's return to the quantum circuit diagrams we discussed a few pages ago. \par +Keep in mind that we're working with quantum gates and proper half-qubits---not classical bits, as we were before. + +\definition{Controlled Inputs} +A \textit{control input} or \textit{inverted control input} may be attached to any gate. \par +These are drawn as filled and empty circles in our circuit diagrams: + +\null\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=0.8] + \node[qubit] (a) at (0, 0) {$\ket{0}$}; + \node[qubit] (b) at (0, -1) {$\ket{0}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{0}$}; + \draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$}; + + \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{a}{1}{a}{2}{$X$} + + \node[gray] at ($([shift={(0,-0.75)}] dot)$) {Non-inverted control input}; + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=0.8] + \node[qubit] (a) at (0, 0) {$\ket{0}$}; + \node[qubit] (b) at (0, -1) {$\ket{0}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{1}$}; + \draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{0}$}; + + \draw[wire] + ($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) -- + ($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$) + ; + \draw[wireijoin] + ($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$) + circle[radius=0.1] coordinate(dot) + ; + \qubox{a}{1}{a}{2}{$X$} + + \node[gray] at ($([shift={(0,-0.75)}] dot)$) {Inverted control input}; + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill\null +\vspace{2mm} + +An $X$ gate with a (non-inverted) control input behaves like an $X$ gate if \textit{all} its control inputs are $\ket{1}$, +and like $I$ otherwise. An $X$ gate with an inverted control inputs does the opposite, behaving like $I$ if its input is $\ket{1}$ +and like $X$ otherwise. The two circuits above illustrate this fact---take a look at their inputs and outputs. + +\vspace{2mm} +Of course, we can give a gate multiple controls. \par +An $X$ gate with multiplie controls behaves like an $X$ gate if... +\begin{itemize} + \item all non-inverted controls are $\ket{1}$, and + \item all inverted controls are $\ket{0}$ +\end{itemize} +...and like $I$ otherwise. + +\problem{} +What are the final states of the qubits in the diagram below? + +\begin{center} + \begin{tikzpicture}[scale = 1.0] + \node[qubit] (a) at (0, 0) {$\ket{1}$}; + \node[qubit] (b) at (0, -1) {$\ket{0}$}; + \node[qubit] (c) at (0, -2) {$\ket{1}$}; + \node[qubit] (d) at (0, -3) {$\ket{0}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {?}; + \draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {?}; + \draw[wire] (c) -- ([shift={(3, 0)}] c.center) node[qubit] {?}; + \draw[wire] (d) -- ([shift={(3, 0)}] d.center) node[qubit] {?}; + + \draw[wire] + ($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) -- + ($([shift={(1,0)}] d)!0.5!([shift={(2,0)}] d)$) + ; + \draw[wirejoin] + ($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) + circle[radius=0.1] coordinate(dot) + ; + \draw[wireijoin] + ($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$) + circle[radius=0.1] coordinate(dot) + ; + \draw[wirejoin] + ($([shift={(1,0)}] d)!0.5!([shift={(2,0)}] d)$) + circle[radius=0.1] coordinate(dot) + ; + \qubox{b}{1}{b}{2}{$X$} + \end{tikzpicture} +\end{center} + +\vfill +\pagebreak + +\problem{} +Consider the diagram below, with one controlled $X$ gate: \par +\note[Note]{The CNOT gate from \ref{cnot} is a controlled $X$ gate.} + +\begin{center} + \begin{tikzpicture}[scale=0.8] + \node[qubit] (a) at (0, 0) {$\ket{a}$}; + \node[qubit] (b) at (0, -1) {$\ket{b}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {}; + \draw[wire] (b) -- ([shift={(3, 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{a}{1}{a}{2}{$X$} + \end{tikzpicture} +\end{center} + +Find a matrix $\text{X}_\text{c}$ that represents this gate, so that $\text{X}_\text{c}\ket{ab}$ works as expected. + +\begin{solution} + \begin{equation*} + \text{X}_\text{c} = \begin{bmatrix} + 1 & 0 & 0 & 0 \\ + 0 & 0 & 0 & 1 \\ + 0 & 0 & 1 & 0 \\ + 0 & 1 & 0 & 0 + \end{bmatrix} + \end{equation*} + Note that this is also the solution to \ref{cnotflipped}. +\end{solution} + +\vfill + +\problem{} +Now, evaluate the following. Remember that +$\ket{+} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} + \ket{1}\Bigr)$ and +$\ket{-} = \frac{1}{\sqrt{2}}\Bigl(\ket{0} - \ket{1}\Bigr)$ + +\null\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=0.8] + \node[qubit] (a) at (0, 0) {$\ket{0}$}; + \node[qubit] (b) at (0, -1) {$\ket{1}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$}; + \draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$}; + + \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{a}{1}{a}{2}{$X$} + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=0.8] + \node[qubit] (a) at (0, 0) {$\ket{0}$}; + \node[qubit] (b) at (0, -1) {$\ket{+}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$}; + \draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$}; + + \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{a}{1}{a}{2}{$X$} + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill\null + +\vspace{5mm} + +\null\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=0.8] + \node[qubit] (a) at (0, 0) {$\ket{-}$}; + \node[qubit] (b) at (0, -1) {$\ket{1}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$}; + \draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$}; + + \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{a}{1}{a}{2}{$X$} + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=0.8] + \node[qubit] (a) at (0, 0) {$\ket{+}$}; + \node[qubit] (b) at (0, -1) {$\ket{-}$}; + + \draw[wire] (a) -- ([shift={(3, 0)}] a.center) node[qubit] {$\ket{a}$}; + \draw[wire] (b) -- ([shift={(3, 0)}] b.center) node[qubit] {$\ket{b}$}; + + \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{a}{1}{a}{2}{$X$} + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill\null + +\hint{ + Note that some of these states are entangled. The circuit diagrams are a bit misleading: + we can't write an entangled state as two distinct qubits! + + \vspace{2mm} + + So, don't try to find $\ket{a}$ and $\ket{b}$. \par + 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. +} + +\begin{solution} + In all the below equations, let $\tau = \frac{1}{\sqrt{2}}$. + \begin{itemize}[itemsep = 1mm] + \item $ + \text{X}_\text{c}\ket{01} + = \ket{11} + $ + + \item $ + \text{X}_\text{c}\ket{0+} + = \tau\ket{00} + \tau\ket{11} + $ \note[Note]{This state is entangled!} + + \item $ + \text{X}_\text{c}\ket{-1} + = -\tau\ket{10} + \tau\ket{11} + = (-\ket{-}) \otimes \ket{1} + $ + + \item $ + \text{X}_\text{c}\ket{+-} + = \frac{1}{2}(\ket{00} - \ket{01} + \ket{10} - \ket{11}) + = \ket{+-} + $ + \end{itemize} +\end{solution} + +\vfill +\pagebreak + +\generic{Remark:} +Now, consider the following circuit: + +\begin{center} + \begin{tikzpicture}[scale=0.8] + \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] {$\ket{a}$}; + \draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{b}$}; + + \qubox{b}{1}{b}{2}{$H$} + \qubox{b}{3}{b}{4}{$H$} + + \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} + +We already know that $H$ is its own inverse: $HH = I$. \par +Applying $H$ to a qubit twice does not change its state. + +\note{ + Recall that $H = \frac{1}{\sqrt{2}}\left[\begin{smallmatrix} 1 & 1 \\ 1 & -1 \end{smallmatrix}\right]$ +} + + +\vspace{2mm} + +So, we might expect that the two circuits below are equivalent: \par +After all, we $H$ the second bit, use it to control an $X$ gate, and then $H$ it back to its previous state. + +\null\hfill +\begin{minipage}{0.48\textwidth}\begin{center} + \begin{tikzpicture}[scale=0.8] + \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] {$\ket{a}$}; + \draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{b}$}; + + \qubox{b}{1}{b}{2}{$H$} + \qubox{b}{3}{b}{4}{$H$} + + \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 +\begin{minipage}{0.48\textwidth}\begin{center} + \begin{tikzpicture}[scale=0.8] + \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] {$\ket{c}$}; + \draw[wire] (b) -- ([shift={(5, 0)}] b.center) node[qubit] {$\ket{d}$}; + + \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 + +\vspace{2mm} + +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 diff --git a/Advanced/Introduction to Quantum/src/parts/07 superdense.tex b/Advanced/Introduction to Quantum/src/parts/07 superdense.tex new file mode 100644 index 0000000..558b466 --- /dev/null +++ b/Advanced/Introduction to Quantum/src/parts/07 superdense.tex @@ -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{} +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 + diff --git a/Advanced/Introduction to Quantum/src/parts/08 teleport.tex b/Advanced/Introduction to Quantum/src/parts/08 teleport.tex new file mode 100644 index 0000000..0533738 --- /dev/null +++ b/Advanced/Introduction to Quantum/src/parts/08 teleport.tex @@ -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 +