diff --git a/Advanced/Introduction to Quantum/main.tex b/Advanced/Introduction to Quantum/main.tex index f665a73..3077329 100755 --- a/Advanced/Introduction to Quantum/main.tex +++ b/Advanced/Introduction to Quantum/main.tex @@ -45,8 +45,6 @@ \input{parts/04 two halves} \input{parts/05 logic gates} \input{parts/06 quantum gates} - %\input{parts/03.00 logic gates} - %\input{parts/03.01 quantum gates} %\section{Superdense Coding} %TODO diff --git a/Advanced/Introduction to Quantum/parts/00 vectors.tex b/Advanced/Introduction to Quantum/parts/00 vectors.tex index 9218941..71f6708 100644 --- a/Advanced/Introduction to Quantum/parts/00 vectors.tex +++ b/Advanced/Introduction to Quantum/parts/00 vectors.tex @@ -1,4 +1,4 @@ -\section*{Prerequisite: Vector Basics} +\section*{Part 0: Vector Basics} \definition{Vectors} An $n$-dimensional \textit{vector} is an element of $\mathbb{R}^n$. In this handout, we'll write vectors as columns. \par @@ -90,6 +90,103 @@ What is the dot product of two orthogonal vectors? \vfill \pagebreak +\definition{Linear combinations} +A \textit{linear combination} of two or more vectors $v_1, v_2, ..., v_k$ is the weighted sum +\begin{equation*} + a_1v_1 + a_2v_2 + ... + a_kv_k +\end{equation*} +where $a_i$ are arbitrary real numbers. + + +\definition{Linear dependence} +We say a set of vectors $\{v_1, v_2, ..., v_k\}$ is \textit{linearly independent} if we can write $0$ as a nontrivial +linear combination of these vectors. For example, the following set is linearly dependent +\begin{equation*} + \Bigl\{ + \left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right], + \left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right], + \left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right] + \Bigr\} +\end{equation*} +Since $ +\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right] + +\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right] - +2 \left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right] += 0 +$. A graphical representation of this is below. + +\null\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=1] + \fill[color = black] (0, 0) circle[radius=0.05]; + + \node[right] at (1, 0) {$\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$}; + \node[above] at (0, 1) {$\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$}; + + \draw[->] (0, 0) -- (1, 0); + \draw[->] (0, 0) -- (0, 1); + \draw[->] (0, 0) -- (0.5, 0.5); + \node[above right] at (0.5, 0.5) {$\left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right]$}; + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill +\begin{minipage}{0.48\textwidth} + \begin{center} + \begin{tikzpicture}[scale=1] + \fill[color = black] (0, 0) circle[radius=0.05]; + + \node[below] at (0.5, 0) {$\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$}; + \node[right] at (1, 0.5) {$\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$}; + + \draw[->] (0, 0) -- (0.95, 0); + \draw[->] (1, 0) -- (1, 0.95); + \draw[->] (1, 1) -- (0.55, 0.55); + \draw[->] (0.5, 0.5) -- (0.05, 0.05); + \node[above left] at (0.5, 0.5) {$-2\left[\begin{smallmatrix} 0.5 \\ 0.5 \end{smallmatrix}\right]$}; + \end{tikzpicture} + \end{center} +\end{minipage} +\hfill\null + +\problem{} +Find a linearly independent set of vectors in $\mathbb{R}^3$ + +\vfill + + + +\definition{Coordinates} +Say we have a set of linearly independent vectors $B = \{b_1, ..., b_k\}$. \par +We can write linear combinations of $B$ as \textit{coordinates} with respect to this set: + +\vspace{2mm} + +If we have a vector $v = x_1b_1 + x_2b_2 + ... + x_kb_k$, we can write $v = (x_1, x_2, ..., x_k)$ with respect to $B$. + + +\vspace{4mm} + +For example, take +$B = \biggl\{ + \left[\begin{smallmatrix} 1 \\ 0 \\ 0 \end{smallmatrix}\right], + \left[\begin{smallmatrix} 0 \\ 1 \\ 0\end{smallmatrix}\right], + \left[\begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix}\right] +\biggr\}$ and $v = \left[\begin{smallmatrix} 8 \\ 3 \\ 9 \end{smallmatrix}\right]$ + +The coordinates of $v$ with respect to $B$ are, of course, $(8, 3, 9)$. + +\problem{} +What are the coordinates of $v$ with respect to the basis +$B = \biggl\{ + \left[\begin{smallmatrix} 1 \\ 0 \\ 1 \end{smallmatrix}\right], + \left[\begin{smallmatrix} 0 \\ 1 \\ 0\end{smallmatrix}\right], + \left[\begin{smallmatrix} 0 \\ 0 \\ -1 \end{smallmatrix}\right] +\biggr\}$? + + + %For example, the set $\{[1,0,0], [0,1,0], [0,0,1]\}$ (which we usually call $\{x, y, z\})$ %forms an orthonormal basis of $\mathbb{R}^3$. Every element of $\mathbb{R}^3$ can be written as a linear combination of these vectors: diff --git a/Advanced/Introduction to Quantum/parts/01 bits.tex b/Advanced/Introduction to Quantum/parts/01 bits.tex index 0c5732b..2873349 100644 --- a/Advanced/Introduction to Quantum/parts/01 bits.tex +++ b/Advanced/Introduction to Quantum/parts/01 bits.tex @@ -55,14 +55,9 @@ What is the size of $\mathbb{B}^n$? % NOTE: this is time-travelled later in the handout. % if you edit this, edit that too. -\cgeneric{Remark} +\generic{Remark:} Consider a single classical bit. It takes states in $\{\texttt{0}, \texttt{1}\}$, picking one at a time. \par -The states \texttt{0} and \texttt{1} are fully independent. They are completely disjoint; they share no parts. \par -We'll therefore say that \texttt{0} and \texttt{1} \textit{orthogonal} (or equivalently, \textit{perpendicular}). \par - -\vspace{2mm} - -We can draw $\vec{e}_0$ and $\vec{e}_1$ as perpendicular axis on a plane to represent this: +We'll write the states \texttt{0} and \texttt{1} as orthogonal unit vectors, labeled $\vec{e}_0$ and $\vec{e}_1$: \begin{center} \begin{tikzpicture}[scale=1.5] @@ -84,27 +79,21 @@ We can draw $\vec{e}_0$ and $\vec{e}_1$ as perpendicular axis on a plane to repr 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 $[1, 0] = (1 \times \vec{e}_0) + (0 \times \vec{e}_1)$, and is thus all $\vec{e}_0$ and no $\vec{e}_1$. \par +\note[Note]{$[0, 1]$ and $[1, 0]$ are coordinates in the basis $\{\vec{e}_0, \vec{e}_1\}$} \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 $\vec{e}_0$ and $\vec{e}_1$: \par -\note[Note]{ - 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. -} +We could, of course, mark the point \texttt{x} at $[1, 1]$ which is equal parts $\vec{e}_0$ and $\vec{e}_1$: \par -\null\hfill -\begin{minipage}{0.48\textwidth} \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) {$\vec{e}_0$ axis}; + \node[right] at (1.5, 0) {$\vec{e}_0$}; \draw[->] (0, 0) -- (0, 1.5); - \node[above] at (0, 1.5) {$\vec{e}_1$ axis}; + \node[above] at (0, 1.5) {$\vec{e}_1$}; \fill[color = oblue] (1, 0) circle[radius=0.05]; \node[below] at (1, 0) {\texttt{0}}; @@ -117,38 +106,14 @@ We could, of course, mark the point \texttt{x} at $[1, 1]$, which is equal parts \node[above right] at (1, 1) {\texttt{x}}; \end{tikzpicture} \end{center} -\end{minipage} -\hfill -\begin{minipage}{0.48\textwidth} - \begin{center} - \begin{tikzpicture}[scale=1.5] - \fill[color = black] (0, 0) circle[radius=0.05]; - \fill[color = oblue] (1, 0) circle[radius=0.05]; - \node[below] at (1, 0) {\texttt{0}}; - - \fill[color = oblue] (0, 1) circle[radius=0.05]; - \node[left] at (0, 1) {\texttt{1}}; - - \draw[dashed, color = gray, ->] (0, 0) -- (0.9, 0.0); - \draw[dashed, color = gray, ->] (1, 0.1) -- (1, 0.9); - \fill[color = oblue] (1, 1) circle[radius=0.05]; - \node[above right] at (1, 1) {\texttt{x}}; - \end{tikzpicture} - \end{center} -\end{minipage} -\hfill\null \vspace{4mm} -But \texttt{x} isn't a member of $\mathbb{B}$, it's not a valid state. \par -Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, there's nothing in between. \par -\note{ - Note that the unit vectors $\vec{e}_0$ and $\vec{e}_1$ form an \textit{orthonormal basis} of the plane $\mathbb{R}^2$. -} +But \texttt{x} isn't a member of $\mathbb{B}$---it's not a state that a classical bit can take. \par +By our current definitions, the \textit{only} valid states of a bit are $\texttt{0} = [1, 0]$ and $\texttt{1} = [0, 1]$. \vfill - \pagebreak @@ -169,7 +134,7 @@ Our bit is fully $\vec{e}_0$ or fully $\vec{e}_1$. By our current definitions, t \definition{Vectored Bits} This brings us to what we'll call the \textit{vectored representation} of a bit. \par -Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them into their components: \par +Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them into their $\vec{e}_0$ and $\vec{e}_1$ components: \par \null\hfill \begin{minipage}{0.48\textwidth} @@ -186,10 +151,11 @@ Instead of writing our bits as just \texttt{0} and \texttt{1}, we'll break them This may seem needlessly complex---and it is, for classical bits. \par We'll see why this is useful soon enough. -\generic{One more thing:} +\vspace{4mm} + The $\ket{~}$ you see in the two expressions above is called a \say{ket,} and denotes a column vector. \par -$\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one.} \par -This is called bra-ket notation. $\bra{0}$ is called a \say{bra,} but we won't worry about that for now. +$\ket{0}$ is pronounced \say{ket zero,} and $\ket{1}$ is pronounced \say{ket one.} This is called bra-ket notation. \par +\note[Note]{$\bra{0}$ is called a \say{bra,} but we won't worry about that for now.} diff --git a/Advanced/Introduction to Quantum/parts/02 two bits.tex b/Advanced/Introduction to Quantum/parts/02 two bits.tex index b0be2d0..2454d3c 100644 --- a/Advanced/Introduction to Quantum/parts/02 two bits.tex +++ b/Advanced/Introduction to Quantum/parts/02 two bits.tex @@ -1,74 +1,44 @@ \section{Two Bits} -How do we represent multi-bit states using vectors? \par -Unfortunately, this is hard to visualize---but the idea is simple. + + +\problem{} +As we already know, the set of states a single bit can take is $\mathbb{B} = \{\texttt{0}, \texttt{1}\}$. \par +What is the set of compound states \textit{two} bits can take? How about $n$ bits? \par +\hint{Cartesian product.} + + +\vspace{5cm} +Of course, \ref{compoundclassicalbits} is fairly easy: \par +If $a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$, +the values $ab$ can take are +$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$. -\problem{} -What is the set of possible states of two bits (i.e, $\mathbb{B}^2$)? - - -\vspace{2cm} - - - - - - -\cgeneric{Remark} -When we have two bits, we have four orthogonal states: -$\overrightarrow{00}$, $\overrightarrow{01}$, $\overrightarrow{10}$, and $\overrightarrow{11}$. \par -We need four dimensions to draw all of these vectors, so I can't provide a picture... \par -but the idea here is the same as before. - - - - - - - -\problem{} -Write $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$ as column vectors \par -with respect to the orthonormal basis $\{\overrightarrow{00}, \overrightarrow{01}, \overrightarrow{10}, \overrightarrow{11}\}$. - -\vfill - - - - - - - - - -\cgeneric{Remark} -So, we represent each possible state as an axis in an $n$-dimensional space. \par -A set of $n$ bits gives us $2^n$ possible states, which forms a basis in $2^n$ dimensions. - -\vspace{1mm} - -Say we now have two seperate bits: $\ket{a}$ and $\ket{b}$. \par -How do we represent their compound state? \par - -\vspace{4mm} - -If we return to our usual notation, this is very easy: -$a$ is in $\{\texttt{0}, \texttt{1}\}$ and $b$ is in $\{\texttt{0}, \texttt{1}\}$, \par -so the possible compound states of $ab$ are -$\{\texttt{0}, \texttt{1}\} \times \{\texttt{0}, \texttt{1}\} = \{\texttt{00}, \texttt{01}, \texttt{10}, \texttt{11}\}$ - -\vspace{1mm} +\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} -\vspace{4mm} +We would like to do the same in vector notation. Given bits $\ket{a}$ and $\ket{b}$, +how should we represent the state of $\ket{ab}$? We'll spend the rest of this section solving this problem. -We would like to do the same in vector notation. Given $\ket{a}$ and $\ket{b}$, \par -how should we represent the state of $\ket{ab}$? + + +\problem{} +When we have two bits, we have four orthogonal states: +$\overrightarrow{00}$, $\overrightarrow{01}$, $\overrightarrow{10}$, and $\overrightarrow{11}$. \par + +\vspace{2mm} + +Write $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$ as column vectors \par +with respect to the orthonormal basis $\{\overrightarrow{00}, \overrightarrow{01}, \overrightarrow{10}, \overrightarrow{11}\}$. + +\vfill \pagebreak @@ -83,8 +53,7 @@ how should we represent the state of $\ket{ab}$? \definition{Tensor Products} -The \textit{tensor product} between two vectors -is defined as follows: +The \textit{tensor product} of two vectors is defined as follows: \begin{equation*} \begin{bmatrix} x_1 \\ x_2 @@ -233,9 +202,12 @@ $? \problem{} What is the \textit{span} of the vectors we found in \ref{basistp}? \par -In other words, what is the set of vectors that can be written as weighted sums of the vectors above? +In other words, what is the set of vectors that can be written as linear combinations of the vectors above? \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 $A \times B$. \pagebreak @@ -275,9 +247,7 @@ Compute the following. Is the result what we'd expect? \problem{} -Of course, writing $\ket{0} \otimes \ket{1}$ is a bit excessive. \par -We'll shorten this notation to $\ket{01}$. \par -Thus, the two-bit kets we saw on the previous page are, by definition, tensor products. +Of course, writing $\ket{0} \otimes \ket{1}$ is a bit excessive. We'll shorten this notation to $\ket{01}$. \par \vspace{2mm} @@ -302,10 +272,7 @@ Write $\ket{5}$ as three-bit state vector. \par \problem{} Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par -\hint{ - You do not need to compute every tensor product. \\ - Do a few, you should quickly see the pattern. -} +\hint{You do not need to compute every tensor product. Do a few and find the pattern.} diff --git a/Advanced/Introduction to Quantum/parts/03 half a qubit.tex b/Advanced/Introduction to Quantum/parts/03 half a qubit.tex index 4bdc728..4c68264 100644 --- a/Advanced/Introduction to Quantum/parts/03 half a qubit.tex +++ b/Advanced/Introduction to Quantum/parts/03 half a qubit.tex @@ -1,17 +1,5 @@ \section{Half a Qubit} -First, a pair of definitions. We've used both these terms implicitly in the previous section, -but we'll need to introduce proper definitions before we continue. - -\definition{} -A \textit{linear combination} of two vectors $u$ and $v$ is the sum $au + bv$ for scalars $a$ and $b$. \par -\note[Note]{In other words, a linear combination is exactly what it sounds like.} - -\definition{} -A \textit{normalized vector} (also called a \textit{unit vector}) is a vector with length 1. - -\vspace{4mm} - \begin{tcolorbox}[ enhanced, breakable, @@ -38,7 +26,7 @@ A \textit{normalized vector} (also called a \textit{unit vector}) is a vector wi \end{tcolorbox} -\cgeneric{Remark} +\generic{Remark:} Just like a classical bit, a \textit{quantum bit} (or \textit{qubit}) can take the values $\ket{0}$ and $\ket{1}$. \par However, \texttt{0} and \texttt{1} aren't the only states a qubit may have. @@ -47,7 +35,7 @@ However, \texttt{0} and \texttt{1} aren't the only states a qubit may have. We'll make sense of quantum bits by extending the \say{vectored} bit representation we developed in the previous section. First, let's look at a diagram we drew a few pages ago: -\begin{ORMCbox}{Time Travel (Page 2)}{black!10!white}{black!65!white} +\begin{ORMCbox}{Time Travel (Page 5)}{black!10!white}{black!65!white} A classical bit takes states in $\{\texttt{0}, \texttt{1}\}$, picking one at a time. \par We'll represent \texttt{0} and \texttt{1} as perpendicular unit vectors $\ket{0}$ and $\ket{1}$, show below. @@ -169,7 +157,7 @@ In addition, $\ket{\psi}$ \textit{collapses} when it is measured: it instantly c 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. -Quantum bits \textit{cannot} be measured without their state collapsing. \par +Quantum bits cannot be measured without their state collapsing. \par \pagebreak @@ -187,7 +175,7 @@ Quantum bits \textit{cannot} be measured without their state collapsing. \par \problem{} \begin{itemize} - \item What is the probability we get $\ket{0}$ if we measure $\ket{\psi_0}$? \par + \item What is the probability we get $\ket{0}$ when we measure $\ket{\psi_0}$? \par \item What outcomes can we get if we measure it a second time? \par \item What are these probabilities for $\ket{\psi_1}$? \end{itemize} diff --git a/Advanced/Introduction to Quantum/parts/05 logic gates.tex b/Advanced/Introduction to Quantum/parts/05 logic gates.tex index abbdbd0..6e1e7b4 100644 --- a/Advanced/Introduction to Quantum/parts/05 logic gates.tex +++ b/Advanced/Introduction to Quantum/parts/05 logic gates.tex @@ -105,7 +105,7 @@ Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par \vfill \pagebreak -\cgeneric{Remark} +\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 one set of bits, and return another: @@ -275,7 +275,7 @@ Find the matrix that corresponds to the above transformation. \par \vfill -\cgeneric{Remark} +\generic{Remark:} We could draw the above transformation as a combination $X$ and $I$ (identity) gate: \begin{center} \begin{tikzpicture}[scale=0.8] diff --git a/Advanced/Introduction to Quantum/parts/06 quantum gates.tex b/Advanced/Introduction to Quantum/parts/06 quantum gates.tex index 40369c2..9dee549 100644 --- a/Advanced/Introduction to Quantum/parts/06 quantum gates.tex +++ b/Advanced/Introduction to Quantum/parts/06 quantum gates.tex @@ -127,7 +127,7 @@ If we measure the result of \ref{applycnot}, what are the probabilities of getti \vfill -\cgeneric{Remark} +\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}. @@ -217,7 +217,8 @@ The \textit{Hadamard Gate} is given by the following matrix: \par \end{bmatrix} \end{equation*} - This is exactly the first column of the matrix product. + This is exactly the first column of the matrix product. \par + Also, note that each element of $Ac_0$ is the dot product of a row in $A$ and a column in $c_0$. \end{ORMCbox}