From 689d085e6f93dc232855fb251d96b27c6404a5bb Mon Sep 17 00:00:00 2001 From: mark Date: Tue, 6 Feb 2024 16:49:40 -0800 Subject: [PATCH] Quantum edits --- Advanced/Introduction to Quantum/main.tex | 53 +- .../parts/02.01 two halves.tex | 33 + .../parts/03.01 quantum gates.tex | 624 +++++------------- 3 files changed, 249 insertions(+), 461 deletions(-) diff --git a/Advanced/Introduction to Quantum/main.tex b/Advanced/Introduction to Quantum/main.tex index 7ef168d..7ed68d9 100755 --- a/Advanced/Introduction to Quantum/main.tex +++ b/Advanced/Introduction to Quantum/main.tex @@ -44,6 +44,55 @@ \input{parts/02.00 half a qubit} \input{parts/02.01 two halves} \input{parts/03.00 logic gates} - %\input{parts/03.01 quantum gates} + \input{parts/03.01 quantum gates} -\end{document} \ No newline at end of file +\end{document} + +\problem{} +The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par +Find its matrix. + +\begin{solution} + \begin{equation*} + \text{SWAP} = \begin{bmatrix} + 1 & 0 & 0 & 0 \\ + 0 & 0 & 1 & 0 \\ + 0 & 1 & 0 & 0 \\ + 0 & 0 & 0 & 1 \\ + \end{bmatrix} + \end{equation*} +\end{solution} + +\vfill + + + + + + + + + +% \problem{} +% The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par +% In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par +% Find the $T$ gate's matrix. \par +% \note{ +% This gate is particularly interesting because it's a \textit{universal quantum gate}: \\ +% like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates. +% } +% +% \begin{solution} +% \begin{equation*} +% \text{T} = \begin{bmatrix} +% 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ +% 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ +% 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ +% 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ +% 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ +% 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ +% 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ +% 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ +% \end{bmatrix} +% \end{equation*} +% \end{solution} \ No newline at end of file diff --git a/Advanced/Introduction to Quantum/parts/02.01 two halves.tex b/Advanced/Introduction to Quantum/parts/02.01 two halves.tex index f977f99..b7bc1e4 100644 --- a/Advanced/Introduction to Quantum/parts/02.01 two halves.tex +++ b/Advanced/Introduction to Quantum/parts/02.01 two halves.tex @@ -108,5 +108,38 @@ Factor the following product state: \end{solution} +\vfill + +\problem{} +Show that the following is an entangled state. +\begin{equation*} + \frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11} +\end{equation*} + +\begin{solution} + $ + \left[ + \begin{smallmatrix} + a_0 \\ a_1 + \end{smallmatrix} + \right] + \otimes + \left[ + \begin{smallmatrix} + b_0 \\ b_1 + \end{smallmatrix} + \right] + = + a_0b_0\ket{00} + a_0b_1\ket{01} + a_1b_0\ket{10} + a_1b_1\ket{11} + $ + + \vspace{2mm} + + So, we have that $a_1b_1 = a_0b_0 = \sqrt{2}^{-1}$ \par + But $a_0b_1 = a_1b_0 = 0$, so one of $a_0$ and $b_1$ must be zero. \par + We thus have a contradiction. +\end{solution} + + \vfill \pagebreak \ No newline at end of file diff --git a/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex b/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex index de71f22..a9572cc 100644 --- a/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex +++ b/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex @@ -1,38 +1,78 @@ \section{Quantum Gates} + +In the previous section, we stated that a quantum gate is a linear map. \par +Let's complete that definition. + \definition{} -The quantum analog of a logic gate is a \textit{quantum gate,} \par -which we'll define as a linear map from $\mathbb{B}^n$ to $\mathbb{B}^n$. \par -\note[Note]{This definition is temporary, we'll extend it soon.} +A quantum gate is a \textit{orthonormal matrix}, which means any gate $G$ +satisfies $GG^\text{T} = I$. \par +This implies the following: \par + +\begin{itemize} + \item $G$ is square \par + \note{ + If we think of $G$ as a map, this implies it has as many inputs as it has outputs. \\ + This makes sense---we stated earlier that quantum gates do not destroy or create qubits. + } + + \item $G$ preserves lengths; i.e $|x| = |Gx|$. \par + \note{This ensures that $G\ket{\psi}$ is always a valid state!} +\end{itemize} + +(You will prove all these properties in any introductory linear algebra course. \\ +This isn't a lesson on linear algebra, so you may take them as given today.) + +\definition{} +Let $\mathbb{U} \subset \mathbb{R}^2$ be the set of points in the unit circle. \par +We can restate the above definition as follows: \par +A quantum gate is an invertible map from $\mathbb{U}^n$ to $\mathbb{U}^n$. +\definition{} +Let $G$ be a quantum gate. \par +Since quantum gates are, by definition, \textit{linear} maps, +the following holds: \par +\begin{equation*} + G\bigl(a_0 \ket{0} + a_1\ket{1}\bigr) = a_0G\ket{0} + a_1G\ket{1} +\end{equation*} \problem{} -Consider the CNOT (controlled not) gate. \par -When applied to a two-bit state $\ket{ab}$, CNOT inverts $b$ iff $a$ is $\ket{1}$. \par -Find the matrix that represents the CNOT gate. \par -\hint{what are the dimensions of this matrix?} +Consider the \textit{controlled not} (or \textit{cnot}) gate, defined by the following table: \par +\begin{itemize} + \item $\text{X}_\text{c}\ket{00} = \ket{00}$ + \item $\text{X}_\text{c}\ket{01} = \ket{11}$ + \item $\text{X}_\text{c}\ket{10} = \ket{10}$ + \item $\text{X}_\text{c}\ket{11} = \ket{01}$ +\end{itemize} +In other words, the cnot gate inverts its first bit if its second bit is $\ket{1}$. \par +Find the matrix that applies the cnot gate. \begin{solution} \begin{equation*} - \text{CNOT} = \begin{bmatrix} + \text{CNOT} = \left[\begin{smallmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ - \end{bmatrix} + \end{smallmatrix}\right] \end{equation*} \vspace{4mm} If $\ket{a}$ is $\ket{0}$, $\ket{a} \otimes \ket{b}$ is $ - \begin{bmatrix} - \begin{bmatrix} + \left[ + \begin{smallmatrix} + \left[ + \begin{smallmatrix} b_1 \\ b_2 - \end{bmatrix} \\ 0 \\ 0 - \end{bmatrix} + \end{smallmatrix} + \right] + \\ 0 \\ 0 + \end{smallmatrix} + \right] $, and the \say{not} portion of the matrix is ignored. @@ -40,12 +80,16 @@ Find the matrix that represents the CNOT gate. \par If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is $ - \begin{bmatrix} + \left[ + \begin{smallmatrix} 0 \\ 0 \\ - \begin{bmatrix} + \left[ + \begin{smallmatrix} b_1 \\ b_2 - \end{bmatrix} - \end{bmatrix} + \end{smallmatrix} + \right] + \end{smallmatrix} + \right] $, and the \say{identity} portion of the matrix is ignored. @@ -55,479 +99,141 @@ Find the matrix that represents the CNOT gate. \par and if $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on the bottom, $\ket{a}$ is $\ket{1}$. \end{solution} -\vfill - - - - - - -\problem{} -Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied. - -\begin{solution} - \begin{equation*} - \text{CNOT}_{\text{mod}} = \begin{bmatrix} - 0 & 1 & 0 & 0 \\ - 1 & 0 & 0 & 0 \\ - 0 & 0 & 1 & 0 \\ - 0 & 0 & 0 & 1 - \end{bmatrix} - \end{equation*} -\end{solution} - -\vfill -\pagebreak - -\iffalse -\problem{} -Finally, modify the original CNOT gate so that the roles of its bits are reversed: \par -$\text{CNOT}_{\text{flip}} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$. - - -\begin{solution} - \begin{equation*} - \text{CNOT}_{\text{flip}} = \begin{bmatrix} - 1 & 0 & 0 & 0 \\ - 0 & 0 & 0 & 1 \\ - 0 & 0 & 1 & 0 \\ - 0 & 1 & 0 & 0 \\ - \end{bmatrix} - \end{equation*} -\end{solution} - -\vfill -\fi - - - - - - - - - -\problem{} -The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par -Find its matrix. - -\begin{solution} - \begin{equation*} - \text{SWAP} = \begin{bmatrix} - 1 & 0 & 0 & 0 \\ - 0 & 0 & 1 & 0 \\ - 0 & 1 & 0 & 0 \\ - 0 & 0 & 0 & 1 \\ - \end{bmatrix} - \end{equation*} -\end{solution} - \vfill +\generic{Remark:} +Note that a quantum gate is fully defined by the place it maps +our basis states $\ket{0}$ and $\ket{1}$ (or,$\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates). +This directly follows from \ref{qgateislinear}. - - - - - - -\problem{} -The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par -In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par -Find the $T$ gate's matrix. \par -\note{ - This gate is particularly interesting because it's a \textit{universal quantum gate}: \\ - like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates. -} - -\begin{solution} - \begin{equation*} - \text{T} = \begin{bmatrix} - 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ - 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ - 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ - 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ - 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ - 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ - 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ - 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ - \end{bmatrix} - \end{equation*} -\end{solution} - -\vfill \pagebreak - - - - - - - - - - - -The last thing we need is a way to draw complex sequences of gates. \par -We already know how to do this with classical gates, \'a la logic circuit: - -\begin{center} -\begin{tikzpicture}[circuit logic US, scale=1.5] - \node[and gate] (and) at (0,-0.8) {\tiny\texttt{and}}; - \draw ([shift={(-0.5, 0)}] and.input 1) node[left] {\texttt{1}} -- (and.input 1); - \draw ([shift={(-0.5, 0)}] and.input 2) node[left] {\texttt{0}} -- (and.input 2); - \draw (and.output) -- ([shift={(0.5, 0)}] and.output) node[right] {\texttt{0}}; -\end{tikzpicture} -\end{center} - -We draw quantum circuits in a very similar way. \par -For example, here a simple three-bit circuit consisting of a CNOT gate on the first bit, \par -controlled by the third. The first bit is inverted iff the third bit is $\ket{1}$: - -\begin{center} -\begin{tikzpicture}[scale=1] - \node[qubit] (a) at (0, 0) {$\ket{0}$}; - \node[qubit] (b) at (0, -1) {$\ket{0}$}; - \node[qubit] (c) at (0, -2) {$\ket{1}$}; - - \draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$}; - \draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$}; - \draw[wire] (c) -- ([shift={(4, 0)}] c.center) node[qubit] {$\ket{1}$}; - - \draw[wire] - ($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) -- - ($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$) - ; - - \draw[wirejoin] - ($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$) - circle[radius=0.1] coordinate(dot) - ; - - \qubox{a}{1}{a}{3}{CNOT} -\end{tikzpicture} -\end{center} - - - - - - - - - - - - - -\problem{} -Draw the CNOT gate as a classical logic circuit. \par -\hint{This can be done with one gate.} - -\begin{solution} - \begin{center} - \begin{tikzpicture}[circuit logic US, scale=2] - \node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}}; - \draw (xor.input 1) ++(-0.5, 0) coordinate (start); - \draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin); - \draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$a$}; - \draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$b$}; - \filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot); - \draw (dot) -- (dot -| 1,0) node[right] {$b_\text{out}$}; - \draw (xor.output) -- (xor.output -| 1,0) node[right] {$a_\text{out}$}; - \end{tikzpicture} - \end{center} -\end{solution} +\problem{} +Evaluate the following: +\begin{equation*} + \text{X}_\text{C} + \Bigl( + \frac{1}{2}\ket{00} + + \frac{1}{2}\ket{01} - + \frac{1}{2}\ket{10} - + \frac{1}{2}\ket{11} + \Bigr) +\end{equation*} \vfill -Gate controls may be marked with a filled circle or an empty circle. \par -Empty circles denote \textit{inverse controls,} which (of course) have an inverse effect. \par -For example, the two circuits below are identical: - -\null\hfill -\begin{minipage}{0.48\textwidth} - \begin{center} - \begin{tikzpicture}[scale=1] - \node[qubit] (a) at (0, 0) {$\ket{0}$}; - \node[qubit] (b) at (0, -1) {$\ket{0}$}; - - \draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[ -qubit] {$\ket{a}$}; - \draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[ -qubit] {$\ket{b}$}; - - \draw[wire] - ($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) -- - ($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$) - ; - - \draw[wireijoin] - ($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$) - circle[radius=0.1] coordinate(dot) - ; - - \qubox{a}{1}{a}{3}{CNOT} - \end{tikzpicture} - \end{center} -\end{minipage} -\hfill -\begin{minipage}{0.48\textwidth} - \begin{center} - \begin{tikzpicture}[scale=1] - \node[qubit] (a) at (0, 0) {$\ket{0}$}; - \node[qubit] (b) at (0, -1) {$\ket{0}$}; - - \draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[ -qubit] {$\ket{a}$}; - \draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[ -qubit] {$\ket{b}$}; - - \draw[wire] - ($([shift={(1.5,0)}] a)!0.5!([shift={(2.5,0)}] a)$) -- - ($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$) - ; - - \draw[wirejoin] - ($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$) - circle[radius=0.1] coordinate(dot) - ; - \qubox{a}{1}{a}{3}{CNOT} - \qubox{b}{0.5}{b}{1.5}{X} - \qubox{b}{2.5}{b}{3.5}{X} - \end{tikzpicture} - \end{center} -\end{minipage} -\hfill\null - - - - - - - - - \problem{} -What are $\ket{a}$ and $\ket{b}$ in the diagrams above? +If we measure the result of \ref{applycnot}, what are the probabilities of getting each state? -\begin{solution} - \begin{center} - \begin{tikzpicture}[scale=1] - \node[qubit] (a) at (0, 0) {$\ket{0}$}; - \node[qubit] (b) at (0, -1) {$\ket{0}$}; +\vfill - \draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[ -qubit] {$\ket{1}$}; - \draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[ -qubit] {$\ket{0}$}; - \draw[wire] - ($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) -- - ($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$) - ; - \draw[wireijoin] - ($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$) - circle[radius=0.1] coordinate(dot) - ; - - \qubox{a}{1}{a}{3}{CNOT} - \end{tikzpicture} - \end{center} -\end{solution} +%\problem{} +%Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied. +% +%\begin{solution} +% \begin{equation*} +% \text{CNOT}_{\text{mod}} = \begin{bmatrix} +% 0 & 1 & 0 & 0 \\ +% 1 & 0 & 0 & 0 \\ +% 0 & 0 & 1 & 0 \\ +% 0 & 0 & 0 & 1 +% \end{bmatrix} +% \end{equation*} +%\end{solution} \vfill \pagebreak +%\problem{} +%Finally, modify the original CNOT gate so that the roles of its bits are reversed: \par +%$\text{CNOT}_{\text{flip}} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$. +% +% +%\begin{solution} +% \begin{equation*} +% \text{CNOT}_{\text{flip}} = \begin{bmatrix} +% 1 & 0 & 0 & 0 \\ +% 0 & 0 & 0 & 1 \\ +% 0 & 0 & 1 & 0 \\ +% 0 & 1 & 0 & 0 \\ +% \end{bmatrix} +% \end{equation*} +%\end{solution} +% +%\vfill - - - - - - - - -As we noted before, quantum gates don't \textit{consume} bits, they \textit{transform} them. \par -Thus, quantum circuits are drawn with a fixed set of bits, whose states change with time: \par -\note[Note]{ - In this diagram, CNOT and SWAP are drawn as $\oplus$ and \rotatebox[origin=c]{90}{$\leftrightarrows$} to save space. -} - -\begin{center} -\begin{tikzpicture}[scale=1] - \node[qubit] (a) at (0, 0) {$\ket{a}$}; - \node[qubit] (b) at (0, -1) {$\ket{1}$}; - \node[qubit] (c) at (0, -2) {$\ket{c}$}; - \node[qubit] (d) at (0, -3) {$\ket{d}$}; - \node[qubit] (e) at (0, -4) {$\ket{1}$}; - - \draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$}; - \draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{b}$}; - \draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$}; - \draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$}; - \draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{e}$}; - - - \draw[wire] - ($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) -- - ($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$) - ; - \draw[wirejoin] - ($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$) - circle[radius=0.1] coordinate(dot) - ; - \draw[wireijoin] - ($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$) - circle[radius=0.1] coordinate(dot) - ; - \qubox{a}{1}{a}{2}{T} - - \qubox{a}{4}{a}{5}{X} - - \draw[wire] - ($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) -- - ($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$) - ; - \draw[wireijoin] - ($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$) - circle[radius=0.1] coordinate(dot) - ; - \qubox{e}{2}{e}{3}{$\oplus$} - - \qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}} - - - \draw[wire] - ($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) -- - ($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$) - ; - \draw[wirejoin] - ($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) - circle[radius=0.1] coordinate(dot) - ; - \draw[wirejoin] - ($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$) - circle[radius=0.1] coordinate(dot) - ; - \qubox{b}{5}{b}{6}{T} -\end{tikzpicture} -\end{center} - -In a quantum circuit, the ONLY way two bits can interact is through a gate. \par -We cannot add \say{branches} in quantum circuits like we do in classical circuits: - -\begin{center} -\begin{tikzpicture}[circuit logic US, scale=1.2] - \node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}}; - \draw (xor.input 1) ++(-0.5, 0) coordinate (start); - \draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin); - \draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$x$}; - \draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$y$}; - \filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot); - \draw (dot) -- (dot -| 1,0) node[right] {$y_\text{out}$}; - \draw (xor.output) -- (xor.output -| 1,0) node[right] {$x_\text{out}$}; - - \draw[->, line width = 1, ogrape] - ([shift={(0.3,-0.5)}] dot) node[right] {This is a branch} - -| ([shift={(0,-0.2)}] dot) - ; -\end{tikzpicture} -\end{center} - - - - - - - - - - - - +\definition{} +The \textit{Hadamard Gate} $H$, is given by the following matrix: \par +\begin{equation*} + H = \frac{1}{\sqrt{2}}\begin{bmatrix} + 1 & 1 \\ + 1 & -1 + \end{bmatrix} +\end{equation*} +\note[Note]{Note that we divide by $\sqrt{2}$, since $H$ must be orthonormal} \problem{} -Find the values of $\ket{a}$ through $\ket{e}$ in the above circuit. +What is $HH$? \par +Using this result, find $H^{-1}$. \begin{solution} - \begin{center} - \begin{tikzpicture}[scale=1] - \node[qubit] (a) at (0, 0) {$\ket{1}$}; - \node[qubit] (b) at (0, -1) {$\ket{1}$}; - \node[qubit] (c) at (0, -2) {$\ket{1}$}; - \node[qubit] (d) at (0, -3) {$\ket{0}$}; - \node[qubit] (e) at (0, -4) {$\ket{1}$}; - - \draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[ -qubit] {$\ket{0}$}; - \draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[ -qubit] {$\ket{1}$}; - \draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[ -qubit] {$\ket{1}$}; - \draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[ -qubit] {$\ket{0}$}; - \draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[ -qubit] {$\ket{0}$}; - - - \draw[wire] - ($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) -- - ($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$) - ; - \draw[wirejoin] - ($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$) - circle[radius=0.1] coordinate(dot) - ; - \draw[wireijoin] - ($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$) - circle[radius=0.1] coordinate(dot) - ; - \qubox{a}{1}{a}{2}{T} - - \qubox{a}{4}{a}{5}{X} - - \draw[wire] - ($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) -- - ($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$) - ; - \draw[wireijoin] - ($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$) - circle[radius=0.1] coordinate(dot) - ; - \qubox{e}{2}{e}{3}{$\oplus$} - - \qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}} - - - \draw[wire] - ($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) -- - ($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$) - ; - \draw[wirejoin] - ($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) - circle[radius=0.1] coordinate(dot) - ; - \draw[wirejoin] - ($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$) - circle[radius=0.1] coordinate(dot) - ; - \qubox{b}{5}{b}{6}{T} - \end{tikzpicture} - \end{center} + $HH = I$, so $H^{-1} = H$ \end{solution} - \vfill -\pagebreak \ No newline at end of file + +\begin{ORMCbox}{Review: Matrix Multiplication}{black!10!white}{black} + Matrix multiplication works as follows: + + \begin{equation*} + AB = + \begin{bmatrix} + 1 & 2 \\ + 3 & 4 \\ + \end{bmatrix} + \begin{bmatrix} + a_0 & b_0 \\ + a_1 & b_1 \\ + \end{bmatrix} + = + \begin{bmatrix} + 1a_0 + 2a_1 & 1b_0 + 2b_1 \\ + 3a_0 + 4a_1 & 3b_0 + 4b_1 \\ + \end{bmatrix} + \end{equation*} + + + Note that this is very similar to multiplying each column of $B$ by $A$. \par + The product $AB$ is simply $Ac$ for every column $c$ in $B$: + + \begin{equation*} + Ac_0 = + \begin{bmatrix} + 1 & 2 \\ + 3 & 4 \\ + \end{bmatrix} + \begin{bmatrix} + a_0 \\ a_1 + \end{bmatrix} + = + \begin{bmatrix} + 1a_0 + 2a_1 \\ + 3a_0 + 4a_1 + \end{bmatrix} + \end{equation*} + + This is exactly the first column of the matrix product. +\end{ORMCbox} + +\pagebreak