98 lines
2.4 KiB
TeX
Executable File
98 lines
2.4 KiB
TeX
Executable File
% Copyright (C) 2023 <Mark (mark@betalupi.com)>
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%
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% This program is free software: you can redistribute it and/or modify
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% it under the terms of the GNU General Public License as published by
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% the Free Software Foundation, either version 3 of the License, or
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% (at your option) any later version.
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%
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% You may have received a copy of the GNU General Public License
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% along with this program. If not, see <https://www.gnu.org/licenses/>.
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%
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%
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%
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% If you edit this, please give credit!
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% Quality handouts take time to make.
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% use the [nosolutions] flag to hide solutions,
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% use the [solutions] flag to show solutions.
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\documentclass[
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solutions,
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singlenumbering,
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unfinished
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]{../../resources/ormc_handout}
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\usepackage{../../resources/macros}
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\def\ket#1{\left|#1\right\rangle}
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\def\bra#1{\left\langle#1\right|}
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\usepackage{units}
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\input{tikzset}
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\uptitlel{Advanced 2}
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\uptitler{Winter 2022}
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\title{Intro to Quantum Computing}
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\subtitle{Prepared by \githref{Mark} on \today{}}
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\begin{document}
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\maketitle
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\input{parts/00.00 bits}
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\input{parts/00.01 two bits}
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\input{parts/02.00 half a qubit}
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\input{parts/02.01 two halves}
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\input{parts/03.00 logic gates}
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\input{parts/03.01 quantum gates}
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\end{document}
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\problem{}
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The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par
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Find its matrix.
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\begin{solution}
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\begin{equation*}
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\text{SWAP} = \begin{bmatrix}
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1 & 0 & 0 & 0 \\
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0 & 0 & 1 & 0 \\
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0 & 1 & 0 & 0 \\
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0 & 0 & 0 & 1 \\
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\end{bmatrix}
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\end{equation*}
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\end{solution}
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\vfill
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% \problem{}
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% The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par
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% In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par
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% Find the $T$ gate's matrix. \par
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% \note{
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% This gate is particularly interesting because it's a \textit{universal quantum gate}: \\
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% like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates.
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% }
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%
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% \begin{solution}
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% \begin{equation*}
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% \text{T} = \begin{bmatrix}
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% 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
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% 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
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% 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
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% 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
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% 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
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% 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
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% 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
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% 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
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% \end{bmatrix}
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% \end{equation*}
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% \end{solution} |