From bd33aeb739a41840cc9049cc241be3df2ab0d501 Mon Sep 17 00:00:00 2001 From: Mark Date: Thu, 8 Feb 2024 09:21:22 -0800 Subject: [PATCH] Edits --- .../parts/00.00 bits.tex | 37 ++++++++++--------- .../parts/00.01 two bits.tex | 9 +++-- .../parts/02.00 half a qubit.tex | 8 ++-- .../parts/03.00 logic gates.tex | 6 +-- .../parts/03.01 quantum gates.tex | 23 +++++------- 5 files changed, 39 insertions(+), 44 deletions(-) diff --git a/Advanced/Introduction to Quantum/parts/00.00 bits.tex b/Advanced/Introduction to Quantum/parts/00.00 bits.tex index 37b996c..c0f3595 100644 --- a/Advanced/Introduction to Quantum/parts/00.00 bits.tex +++ b/Advanced/Introduction to Quantum/parts/00.00 bits.tex @@ -8,7 +8,7 @@ To keep things simple, we'll use regular (usually called \textit{classical}) bit \definition{Binary Digits} $\mathbb{B}$ is the set of binary digits. In other words, $\mathbb{B} = \{\texttt{0}, \texttt{1}\}$. \par -\note[Note]{We've seen $\mathbb{B}$ before: It's $(\mathbb{Z}_2, +)$, the addition group mod 2.} +\note[Note]{We've seen $\mathbb{B}$ before---it's the set of integers mod 2.} \vspace{2mm} @@ -28,7 +28,8 @@ In this handout, we'll often see the following sets: \begin{itemize} \item $\mathbb{R}^2$, a two-dimensional plane \item $\mathbb{R}^n$, an n-dimensional space - \item $\mathbb{B}^2$, the set $\{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$ + \item $\mathbb{B}^2$, the set + $\{(\texttt{0},\texttt{0}), (\texttt{0},\texttt{1}), (\texttt{1},\texttt{0}), (\texttt{1},\texttt{1})\}$ \item $\mathbb{B}^n$, the set of all possible states of $n$ bits. \end{itemize} @@ -61,35 +62,35 @@ We'll therefore say that \texttt{0} and \texttt{1} \textit{orthogonal} (or equiv \vspace{2mm} -We can draw $\vec{0}$ and $\vec{1}$ as perpendicular axis on a plane to represent this: +We can draw $\vec{e}_0$ and $\vec{e}_1$ as perpendicular axis on a plane to represent this: \begin{center} \begin{tikzpicture}[scale=1.5] \fill[color = black] (0, 0) circle[radius=0.05]; \draw[->] (0, 0) -- (1.5, 0); - \node[right] at (1.5, 0) {$\vec{0}$ axis}; + \node[right] at (1.5, 0) {$\vec{e}_0$ axis}; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {\texttt{0}}; \draw[->] (0, 0) -- (0, 1.5); - \node[above] at (0, 1.5) {$\vec{1}$ axis}; + \node[above] at (0, 1.5) {$\vec{e}_1$ axis}; \fill[color = oblue] (0, 1) circle[radius=0.05]; \node[left] at (0, 1) {\texttt{1}}; \end{tikzpicture} \end{center} -The point marked $1$ is at $[0, 1]$. It is no parts $\vec{0}$, and all parts $\vec{1}$. \par +The point marked $1$ is at $[0, 1]$. It is no parts $\vec{e}_0$, and all parts $\vec{e}_1$. \par Of course, we can say something similar about the point marked $0$: \par -It is at $[1, 0] = (1 \times \vec{0}) + (0 \times \vec{1})$, and is thus all $\vec{0}$ and no $\vec{1}$. \par +It is at $[1, 0] = (1 \times \vec{e}_0) + (0 \times \vec{e}_1)$, and is thus all $\vec{e}_0$ and no $\vec{e}_1$. \par \vspace{2mm} Naturally, the coordinates $[0, 1]$ and $[1, 0]$ denote how much of each axis a point \say{contains.} \par -We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts $\vec{0}$ and $\vec{1}$: \par +We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts $\vec{e}_0$ and $\vec{e}_1$: \par \note[Note]{ - We could also write $\texttt{x} = \vec{0} + \vec{1}$ explicitly. \\ + We could also write $\texttt{x} = \vec{e}_0 + \vec{e}_1$ explicitly. \\ I've drawn \texttt{x} as a point on the left, and as a sum on the right. } @@ -100,10 +101,10 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts \fill[color = black] (0, 0) circle[radius=0.05]; \draw[->] (0, 0) -- (1.5, 0); - \node[right] at (1.5, 0) {$\vec{0}$ axis}; + \node[right] at (1.5, 0) {$\vec{e}_0$ axis}; \draw[->] (0, 0) -- (0, 1.5); - \node[above] at (0, 1.5) {$\vec{1}$ axis}; + \node[above] at (0, 1.5) {$\vec{e}_1$ axis}; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {\texttt{0}}; @@ -141,7 +142,7 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts \vspace{4mm} But \texttt{x} isn't a member of $\mathbb{B}$, it's not a valid state. \par -Our bit is fully $\vec{0}$ or fully $\vec{1}$. There's nothing in between. +Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, there's nothing in between. \vspace{8mm} @@ -158,11 +159,11 @@ Our bit is fully $\vec{0}$ or fully $\vec{1}$. There's nothing in between. \definition{Orthonormal Basis} -The unit vectors $\vec{0}$ and $\vec{1}$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. \par +The unit vectors $\vec{e}_0$ and $\vec{e}_1$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. \par \note{ \say{ortho-} means \say{orthogonal}; normal means \say{normal,} which means length $= 1$. \\ }{ - Note that $\vec{0}$ and $\vec{1}$ are orthonormal by \textit{definition}. \\ + Note that $\vec{e}_0$ and $\vec{e}_1$ are orthonormal by \textit{definition}. \\ We don't have to prove anything, we simply defined them as such. } \par @@ -215,11 +216,11 @@ Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them \null\hfill \begin{minipage}{0.48\textwidth} - \[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{0}) + (0 \times \vec{1}) \] + \[ \ket{0} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} = (1 \times \vec{e}_0) + (0 \times \vec{e}_1) \] \end{minipage} \hfill \begin{minipage}{0.48\textwidth} - \[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{0}) + (1 \times \vec{1}) \] + \[ \ket{1} = \begin{bmatrix} 0 \\ 1 \end{bmatrix} = (0 \times \vec{e}_0) + (1 \times \vec{e}_1) \] \end{minipage} \hfill\null @@ -250,10 +251,10 @@ Write \texttt{x} and \texttt{y} in the diagram below in terms of $\ket{0}$ and $ \fill[color = black] (0, 0) circle[radius=0.05]; \draw[->] (0, 0) -- (1.5, 0); - \node[right] at (1.5, 0) {$\vec{0}$ axis}; + \node[right] at (1.5, 0) {$\vec{e}_0$ axis}; \draw[->] (0, 0) -- (0, 1.5); - \node[above] at (0, 1.5) {$\vec{1}$ axis}; + \node[above] at (0, 1.5) {$\vec{e}_1$ axis}; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {$\ket{0}$}; diff --git a/Advanced/Introduction to Quantum/parts/00.01 two bits.tex b/Advanced/Introduction to Quantum/parts/00.01 two bits.tex index ecd7b00..8f144d0 100644 --- a/Advanced/Introduction to Quantum/parts/00.01 two bits.tex +++ b/Advanced/Introduction to Quantum/parts/00.01 two bits.tex @@ -61,7 +61,7 @@ $\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \ \vspace{1mm} 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 $A \times B$. This is trivial. +the compound state $(a,b)$ takes values in $A \times B$. \vspace{4mm} @@ -69,8 +69,6 @@ the compound state $(a,b)$ takes values in $A \times B$. This is trivial. We would like to do the same in vector notation. Given $\ket{a}$ and $\ket{b}$, \par how should we represent the state of $\ket{ab}$? - -\vfill \pagebreak @@ -303,7 +301,10 @@ Write $\ket{5}$ as three-bit state vector. \par \problem{} Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par -What do you see? +\hint{ + You do not need to compute every tensor product. \\ + Do a few, you should quickly see the pattern. +} diff --git a/Advanced/Introduction to Quantum/parts/02.00 half a qubit.tex b/Advanced/Introduction to Quantum/parts/02.00 half a qubit.tex index aaf65df..7a1c6d2 100644 --- a/Advanced/Introduction to Quantum/parts/02.00 half a qubit.tex +++ b/Advanced/Introduction to Quantum/parts/02.00 half a qubit.tex @@ -165,13 +165,11 @@ Our measurement again returns either $\ket{0}$ or $\ket{1}$, with the following \vspace{2mm} -In addition $\ket{\psi}$ \textit{collapses} to the state we measure: it instantly jumps to the state we measure, \par -leaving no trace of its previous state. If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and +In addition, $\ket{\psi}$ \textit{collapses} when it is measured: it instantly changes its state to the result of the measurement, +leaving no trace of its previous state. \par +If we measure $\ket{\psi}$ and get $\ket{1}$, $\ket{\psi}$ becomes $\ket{1}$---and it will remain in that state until it is changed. -\vspace{2mm} - Quantum bits \textit{cannot} be measured without their state collapsing. \par -We cannot certainly know the state of a qubit unless that state is $\ket{0}$ or $\ket{1}$. \pagebreak diff --git a/Advanced/Introduction to Quantum/parts/03.00 logic gates.tex b/Advanced/Introduction to Quantum/parts/03.00 logic gates.tex index 7ec920a..6e1e7b4 100644 --- a/Advanced/Introduction to Quantum/parts/03.00 logic gates.tex +++ b/Advanced/Introduction to Quantum/parts/03.00 logic gates.tex @@ -107,7 +107,7 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par \generic{Remark:} The way a quantum circuit handles information is a bit different than the way a classical circuit does. -We usually think of logic gates as \textit{functions}: they consume some set of bits, and return another: +We usually think of logic gates as \textit{functions}: they consume one set of bits, and return another: \begin{center} @@ -143,10 +143,10 @@ We'll also add a horizontal time axis, moving from left to right: \begin{center} \begin{tikzpicture}[scale=1] \node[qubit] (a) at (0, 0) {\texttt{0}}; - \node[qubit] (b) at (0, -1) {\texttt{0}}; + \node[qubit] (b) at (0, -1) {\texttt{1}}; \draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {\texttt{0}}; - \draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {\texttt{0}}; + \draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {\texttt{1}}; \draw[ color = oblue, 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 056804a..69b886c 100644 --- a/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex +++ b/Advanced/Introduction to Quantum/parts/03.01 quantum gates.tex @@ -12,12 +12,12 @@ 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. + If we think of $G$ as a map, this means that $G$ has as many inputs as it has outputs. \\ + This is to be expected: 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!} + \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. \\ @@ -103,12 +103,6 @@ Find the matrix that applies the cnot gate. \vfill - -\generic{Remark:} -As we just saw, a quantum gate is fully defined by the place it maps our basis states $\ket{0}$ and $\ket{1}$ \par -(or, $\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates). This directly follows from \ref{qgateislinear}. - - \pagebreak @@ -133,7 +127,11 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti \vfill +\generic{Remark:} +As we just saw, a quantum gate is fully defined by the place it maps our basis states $\ket{0}$ and $\ket{1}$ \par +(or, $\ket{00...0}$ through $\ket{11...1}$ for multi-qubit gates). This directly follows from \ref{qgateislinear}. +\pagebreak %\problem{} %Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied. @@ -149,9 +147,6 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti % \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 $\ket{1}$. @@ -238,11 +233,11 @@ Using this result, find $H^{-1}$. \problem{} What are $H\ket{0}$ and $H\ket{1}$? \par -Are these states superpositions? +Are these states entangled? \begin{solution} $H\ket{0} = \frac{1}{\sqrt{2}}\bigl(\ket{0} + \ket{1}\bigr)$ and $H\ket{1} = \frac{1}{\sqrt{2}}\bigl(\ket{0} - \ket{1}\bigr)$ \par - Both of these are superpositions. + Both of these are entangled states. \end{solution} \vfill