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2025-01-19 20:24:51 -08:00
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30 changed files with 105 additions and 49 deletions

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@ -32,7 +32,7 @@
\def\bra#1{\left\langle#1\right|}
% TODO: spend more time on probabalistic bits.
% TODO: spend more time on probabalastic bits.
% This could even be its own handout, especially
% for younger classes!

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@ -498,7 +498,7 @@ Compute the following product:
\generic{Remark:}
Also, recall that every matrix is linear map, and that every linear map may be written as a matrix. \par
We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangably.
We often use the terms \textit{matrix}, \textit{transformation}, and \textit{linear map} interchangeably.
\pagebreak

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@ -18,7 +18,7 @@ $\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one
\note[Note]{$\bra{0}$ is called a \say{bra,} but we won't worry about that for now.}
\vspace{2mm}
This is very similiar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
This is very similar to the \say{box} $[~]$ notation we used for probabilistic bits. \par
As before, we will write $\ket{0} = \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$
and $\ket{1} = \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.

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@ -64,7 +64,7 @@ and like $X$ otherwise. The two circuits above illustrate this fact---take a loo
\vspace{2mm}
Of course, we can give a gate multiple controls. \par
An $X$ gate with multiplie controls behaves like an $X$ gate if...
An $X$ gate with multiple controls behaves like an $X$ gate if...
\begin{itemize}
\item all non-inverted controls are $\ket{1}$, and
\item all inverted controls are $\ket{0}$

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@ -1,7 +1,8 @@
\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.
Quantum teleportation 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
@ -131,7 +132,7 @@ With an informal proof, show that it is not possible to use superdense coding to
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,
If superdense coding was any more efficient, we could repeatedly apply superdense coding and quantum teleportation,
to compress an arbitrary number of bits into two \say{seed} bits.
\linehack{}