From acfb95831ecd6290dff5008470e45eec3e6007fe Mon Sep 17 00:00:00 2001 From: Mark Date: Sun, 18 Feb 2024 19:48:17 -0800 Subject: [PATCH] Edits --- .../src/parts/01 bits.tex | 56 +++++++++---------- .../src/parts/05 quantum gates.tex | 10 ++-- .../Introduction to Quantum/src/week 1.tex | 7 +++ 3 files changed, 39 insertions(+), 34 deletions(-) diff --git a/Advanced/Introduction to Quantum/src/parts/01 bits.tex b/Advanced/Introduction to Quantum/src/parts/01 bits.tex index 242dd5f..4c21c54 100644 --- a/Advanced/Introduction to Quantum/src/parts/01 bits.tex +++ b/Advanced/Introduction to Quantum/src/parts/01 bits.tex @@ -174,11 +174,6 @@ Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nic -\problem{} -With $x$ and $y$ defined as above, find the probability of measuring each of \texttt{00}, \texttt{01}, \texttt{10}, and \texttt{11}. - -\vfill - \problem{} Say $[x] = [\nicefrac{2}{3}, \nicefrac{1}{3}]$ and $[y] = [\nicefrac{3}{4}, \nicefrac{1}{4}]$. \par What is the probability that $x$ and $y$ produce different outcomes? @@ -324,27 +319,30 @@ In other words, what is the set of vectors that can be written as linear combina \vfill -Look through the above problems and convince yourself of the following fact: \par -If $a$ is a basis of $A$ and $b$ is a basis of $B$, $a \otimes b$ is a basis of $A \times B$. \par -\note{If you don't understand what this says, ask an instructor. \\ This is the reason we did the last few problems!} - -\begin{instructornote} - \textbf{The idea here is as follows:} - - If $a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$, - the values $ab$ can take are - $\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$. - - \vspace{2mm} - - The same is true of any other state set: if $a$ takes values in $A$ and $b$ takes values in $B$, \par - the compound state $(a,b)$ takes values in $A \times B$. - - \vspace{2mm} - - We would like to do the same with probabilistic bits. \par - Given bits $\ket{a}$ and $\ket{b}$, how should we represent the state of $\ket{ab}$? -\end{instructornote} +% This is wrong, but there's something here. +% maybe fix later? +% +%Look through the above problems and convince yourself of the following fact: \par +%If $a$ is a basis of $A$ and $b$ is a basis of $B$, $a \otimes b$ is a basis of $A \times B$. \par +%\note{If you don't understand what this says, ask an instructor. \\ This is the reason we did the last few problems!} +% +%\begin{instructornote} +% \textbf{The idea here is as follows:} +% +% If $a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$, +% the values $ab$ can take are +% $\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$. +% +% \vspace{2mm} +% +% The same is true of any other state set: if $a$ takes values in $A$ and $b$ takes values in $B$, \par +% the compound state $(a,b)$ takes values in $A \times B$. +% +% \vspace{2mm} +% +% We would like to do the same with probabilistic bits. \par +% Given bits $\ket{a}$ and $\ket{b}$, how should we represent the state of $\ket{ab}$? +%\end{instructornote} \pagebreak @@ -453,7 +451,7 @@ Consider the NOT gate, which operates as follows: \par \end{itemize} What should NOT do to a probabilistic bit $[x_0, x_1]$? \par If we return to our coin analogy, we can think of the NOT operation as -flipping a coin we have already tossed, without looking at it's state. +flipping a coin we have already tossed, without looking at its state. Thus, \begin{equation*} \text{NOT} \begin{bmatrix} @@ -544,8 +542,8 @@ Find the matrix that represents the NOT operation on one probabilistic bit. \problem{Extension by linearity} -Say we have an arbitrary operation $A$. \par -If we know how $A$ acts on $[1]$ and $[0]$, can we compute $A[x]$ for an arbitrary state $[x]$? \par +Say we have an arbitrary operation $M$. \par +If we know how $M$ acts on $[1]$ and $[0]$, can we compute $M[x]$ for an arbitrary state $[x]$? \par Say $[x] = [x_0, x_1]$. \begin{itemize} \item What is the probability we observe $0$ when we measure $x$? diff --git a/Advanced/Introduction to Quantum/src/parts/05 quantum gates.tex b/Advanced/Introduction to Quantum/src/parts/05 quantum gates.tex index f9a193f..0cb05bc 100644 --- a/Advanced/Introduction to Quantum/src/parts/05 quantum gates.tex +++ b/Advanced/Introduction to Quantum/src/parts/05 quantum gates.tex @@ -42,11 +42,11 @@ the following holds: \par 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 $\text{X}_\text{c}\ket{11} = \ket{01}$ + \item $\text{X}_\text{c}\ket{01} = \ket{01}$ + \item $\text{X}_\text{c}\ket{10} = \ket{11}$ + \item $\text{X}_\text{c}\ket{11} = \ket{10}$ \end{itemize} -In other words, the cnot gate inverts its first bit if its second bit is $\ket{1}$. \par +In other words, the cnot gate inverts its second bit if its first bit is $\ket{1}$. \par Find the matrix that applies the cnot gate. \begin{solution} @@ -127,7 +127,7 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti \vfill -\problem{} +\problem{} Finally, modify the original cnot gate so that the roles of its bits are reversed: \par $\text{X}_\text{c, flipped} \ket{ab}$ should invert $\ket{a}$ iff $\ket{b}$ is $\ket{1}$. diff --git a/Advanced/Introduction to Quantum/src/week 1.tex b/Advanced/Introduction to Quantum/src/week 1.tex index aa26178..9fa8953 100755 --- a/Advanced/Introduction to Quantum/src/week 1.tex +++ b/Advanced/Introduction to Quantum/src/week 1.tex @@ -34,6 +34,13 @@ \subtitle{Prepared by \githref{Mark} on \today{}} +% TODO: spend more time on probabalistic bits. +% This could even be its own handout, especially +% for younger classes! + +% Why are qubits amplitudes instead of probabilities? +% (Asher question) + \begin{document} \maketitle