Mark/handouts
Mark/handouts
1
0
Fork 0
Code Pull Requests 2 Actions Packages Releases Activity

Quantum edits

Browse Source
This commit is contained in:
Mark 2024-01-31 09:23:08 -08:00
parent 82064f890c
commit 4dd9645738
Signed by: Mark
GPG Key ID: C6D63995FE72FD80
7 changed files with 713 additions and 989 deletions
Show Stats Download Patch File Download Diff File Expand all files Collapse all files

9
Advanced/Introduction to Quantum/main.tex
View File

@ -38,6 +38,11 @@
\maketitle
\input{parts/0 bits}
\input{parts/1 gates}
\input{parts/00.00 bits}
\input{parts/00.01 two bits}
\input{parts/02.00 half a qubit}
\input{parts/02.01 two halves}
%\input{parts/03.00 gates}
%\input{parts/03.01 and a gate}
\end{document}

359
Advanced/Introduction to Quantum/parts/0 bits.tex → Advanced/Introduction to Quantum/parts/00.00 bits.tex
View File

@ -6,33 +6,18 @@ To keep things simple, we'll use regular (usually called \textit{classical}) bit
\definition{}
\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.}
\vspace{2mm}
Multiplication in $\mathbb{B}$ works just as you'd expect: \par
$
\texttt{0} \times \texttt{0} =
\texttt{0} \times \texttt{1} =
\texttt{1} \times \texttt{0} = \texttt{0}
$; and $\texttt{1} \times \texttt{1} = \texttt{1}$.
\vspace{2mm}
We'll treat addition a bit differently: \par
$0 + 0 = 0$ and $0 + 1 = 1$, but $1 + 1$, for our purposes, is undefined.
\definition{}
\definition{Cartesian Products}
Let $A$ and $B$ be sets. \par
The \textit{cartesian product} $A \times B$ is the set of all pairs $(a, b)$ where $a \in A$ and $b \in B$. \par
As usual, we can write $A \times A \times A$ as $A^3$. \par
@ -67,9 +52,8 @@ What is the size of $\mathbb{B}^n$?
% NOTE: this is time-travelled later in the handout.
% if you edit this, edit that too.
\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
@ -98,7 +82,7 @@ We can draw $\vec{0}$ and $\vec{1}$ as perpendicular axis on a plane to represen
The point marked $1$ is at $[0, 1]$. It is no parts $\vec{0}$, and all parts $\vec{1}$. \par
Of course, we can say something similar about the point marked $0$: \par
It is at $[1, 0] = (1 \times \vec{0}) + (2 \times \vec{1})$. In other words, all $\vec{0}$ and no $\vec{1}$. \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
\vspace{2mm}
@ -173,7 +157,7 @@ Our bit is fully $\vec{0}$ or fully $\vec{1}$. There's nothing in between.
\definition{}
\definition{Orthonormal Basis}
The unit vectors $\vec{0}$ and $\vec{1}$ 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 $= 1$. \\
@ -225,7 +209,7 @@ The tuple $[a,b,c]$ is called the \textit{coordinate} of a point with respect to
\definition{}
\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
@ -287,334 +271,5 @@ Write \texttt{x} and \texttt{y} in the diagram below in terms of $\ket{0}$ and $
\end{tikzpicture}
\end{center}
\vfill
\pagebreak
\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{}
What is the set of possible states of two bits (i.e, $\mathbb{B}^2$)?
\vspace{2cm}
\generic{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
\generic{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}
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.
\vspace{4mm}
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
\definition{}
The \textit{tensor product} between two vectors
is defined as follows:
\begin{equation*}
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
\otimes
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
=
\begin{bmatrix}
x_1
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\\[4mm]
x_2
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
x_1y_1 \\[1mm]
x_1y_2 \\[1mm]
x_2y_1 \\[1mm]
x_2y_2 \\[0.5mm]
\end{bmatrix}
\end{equation*}
That is, we take our first vector, multiply the second
vector by each of its components, and stack the result.
You could think of this as a generalization of scalar
mulitiplication, where scalar mulitiplication is a
tensor product with a vector in $\mathbb{R}^1$:
\begin{equation*}
a
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
=
\begin{bmatrix}
a_1
\end{bmatrix}
\otimes
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
=
\begin{bmatrix}
a_1
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
a_1y_1 \\[1mm]
a_1y_2
\end{bmatrix}
\end{equation*}
\vspace{2mm}
Also, note that the tensor product is very similar to the
Cartesian product: if we take $x$ and $y$ as sets, with
$x = \{x_1, x_2\}$ and $y = \{y_1, y_2\}$, the Cartesian product
contains the same elements as the tensor product---every possible
pairing of an element in $x$ with an element in $y$:
\begin{equation*}
x \times y = \{~(x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2y_2)~\}
\end{equation*}
In fact, these two operations are (in a sense) essentially identical. \par
Let's quickly demonstrate this.
\problem{}
Say $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. \par
What is the dimension of $x \otimes y$?
\vfill
\problem{}<basistp>
What is the pairwise tensor product
$
\Bigl\{
\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]
\Bigr\}
\otimes
\Bigl\{
\left[
\begin{smallmatrix}
1 \\ 0
\end{smallmatrix}
\right],
\left[
\begin{smallmatrix}
0 \\ 1
\end{smallmatrix}
\right]
\Bigr\}
$?
\note{in other words, distribute the tensor product between every pair of vectors.}
\vfill
\problem{}
The vectors we found in \ref{basistp} are a basis of what space? \par
\vfill
\pagebreak
\problem{}
The compound state of two vector-form bits is their tensor product. \par
Compute the following. Is the result what we'd expect?
\begin{itemize}
\item $\ket{0} \otimes \ket{0}$
\item $\ket{0} \otimes \ket{1}$
\item $\ket{1} \otimes \ket{0}$
\item $\ket{1} \otimes \ket{1}$
\end{itemize}
\hint{
Remember that the coordinates of
$\ket{0}$ are $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$,
and the coordinates of
$\ket{1}$ are $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
}
\vfill
\problem{}<fivequant>
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.
\vspace{2mm}
In fact, we could go further: if we wanted to write the set of bits $\ket{1} \otimes \ket{1} \otimes \ket{0} \otimes \ket{1}$, \par
we could write $\ket{1101}$---but a shorter alternative is $\ket{13}$, since $13$ is \texttt{1101} in binary.
\vspace{2mm}
Write $\ket{5}$ as three-bit state vector. \par
\begin{solution}
$\ket{5} = \ket{101} = \ket{1} \otimes \ket{0} \otimes \ket{1} = [0,0,0,0,0,1,0,0]^T$ \par
Notice how we're counting from the top, with $\ket{000} = [1,0,...,0]$ and $\ket{111} = [0, ..., 0, 1]$.
\end{solution}
\vfill
\problem{}
Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
What do you see?
\vfill
\pagebreak

314
Advanced/Introduction to Quantum/parts/00.01 two bits.tex Normal file
View File

@ -0,0 +1,314 @@
\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{}
What is the set of possible states of two bits (i.e, $\mathbb{B}^2$)?
\vspace{2cm}
\generic{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
\generic{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}
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.
\vspace{4mm}
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
\definition{Tensor Products}
The \textit{tensor product} between two vectors
is defined as follows:
\begin{equation*}
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
\otimes
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
=
\begin{bmatrix}
x_1
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\\[4mm]
x_2
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
x_1y_1 \\[1mm]
x_1y_2 \\[1mm]
x_2y_1 \\[1mm]
x_2y_2 \\[0.5mm]
\end{bmatrix}
\end{equation*}
That is, we take our first vector, multiply the second
vector by each of its components, and stack the result.
You could think of this as a generalization of scalar
mulitiplication, where scalar mulitiplication is a
tensor product with a vector in $\mathbb{R}^1$:
\begin{equation*}
a
\begin{bmatrix}
x_1 \\ x_2
\end{bmatrix}
=
\begin{bmatrix}
a_1
\end{bmatrix}
\otimes
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
=
\begin{bmatrix}
a_1
\begin{bmatrix}
y_1 \\ y_2
\end{bmatrix}
\end{bmatrix}
=
\begin{bmatrix}
a_1y_1 \\[1mm]
a_1y_2
\end{bmatrix}
\end{equation*}
\vspace{2mm}
Also, note that the tensor product is very similar to the
Cartesian product: if we take $x$ and $y$ as sets, with
$x = \{x_1, x_2\}$ and $y = \{y_1, y_2\}$, the Cartesian product
contains the same elements as the tensor product---every possible
pairing of an element in $x$ with an element in $y$:
\begin{equation*}
x \times y = \{~(x_1,y_1), (x_1,y_2), (x_2,y_1), (x_2y_2)~\}
\end{equation*}
In fact, these two operations are (in a sense) essentially identical. \par
Let's quickly demonstrate this.
\problem{}
Say $x \in \mathbb{R}^n$ and $y \in \mathbb{R}^m$. \par
What is the dimension of $x \otimes y$?
\vfill
\problem{}<basistp>
What is the pairwise tensor product
$
\Bigl\{
\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]
\Bigr\}
\otimes
\Bigl\{
\left[
\begin{smallmatrix}
1 \\ 0
\end{smallmatrix}
\right],
\left[
\begin{smallmatrix}
0 \\ 1
\end{smallmatrix}
\right]
\Bigr\}
$?
\note{in other words, distribute the tensor product between every pair of vectors.}
\vfill
\problem{}
The vectors we found in \ref{basistp} are a basis of what space? \par
\vfill
\pagebreak
\problem{}
The compound state of two vector-form bits is their tensor product. \par
Compute the following. Is the result what we'd expect?
\begin{itemize}
\item $\ket{0} \otimes \ket{0}$
\item $\ket{0} \otimes \ket{1}$
\item $\ket{1} \otimes \ket{0}$
\item $\ket{1} \otimes \ket{1}$
\end{itemize}
\hint{
Remember that the coordinates of
$\ket{0}$ are $\left[\begin{smallmatrix} 1 \\ 0 \end{smallmatrix}\right]$,
and the coordinates of
$\ket{1}$ are $\left[\begin{smallmatrix} 0 \\ 1 \end{smallmatrix}\right]$.
}
\vfill
\problem{}<fivequant>
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.
\vspace{2mm}
In fact, we could go further: if we wanted to write the set of bits $\ket{1} \otimes \ket{1} \otimes \ket{0} \otimes \ket{1}$, \par
we could write $\ket{1101}$---but a shorter alternative is $\ket{13}$, since $13$ is \texttt{1101} in binary.
\vspace{2mm}
Write $\ket{5}$ as three-bit state vector. \par
\begin{solution}
$\ket{5} = \ket{101} = \ket{1} \otimes \ket{0} \otimes \ket{1} = [0,0,0,0,0,1,0,0]^T$ \par
Notice how we're counting from the top, with $\ket{000} = [1,0,...,0]$ and $\ket{111} = [0, ..., 0, 1]$.
\end{solution}
\vfill
\problem{}
Write the three-bit states $\ket{0}$ through $\ket{7}$ as column vectors. \par
What do you see?
\vfill
\pagebreak

268
Advanced/Introduction to Quantum/parts/02.00 half a qubit.tex Normal file
View File

@ -0,0 +1,268 @@
\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,
colback=white,
colframe=ored,
boxrule=0.6mm,
arc=0mm,
outer arc=0mm,
]
\color{ored}
\begingroup
\large\centering
\textbf{Disclaimer:} \par
\endgroup
\vspace{1ex}
The \say{qubits} we're about to define aren't \textit{really} qubits. The proper definition is a bit more
complicated, but don't worry about that yet. For now, take what I say as truth---we'll get to
the complex definition soon enough.
\vspace{2mm}
The information provided in this handout does not, and is not intended to, constitute legal advice.
All information, content, and material in this document is for general informational purposes only.
\end{tcolorbox}
\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.
\vspace{2mm}
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{timetravel}
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.
\begin{center}
\begin{tikzpicture}[scale=1.5]
\fill[color = black] (0, 0) circle[radius=0.05];
\draw[->] (0, 0) -- (1.2, 0);
\node[right] at (1.2, 0) {$\ket{0}$};
\fill[color = oblue] (1, 0) circle[radius=0.05];
\node[below] at (1, 0) {\texttt{0}};
\draw[->] (0, 0) -- (0, 1.2);
\node[above] at (0, 1.2) {$\ket{1}$};
\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 $\ket{0}$, and all parts $\ket{1}$. \par
Of course, we can say something similar about the point marked $0$: \par
It is at $[1, 0] = (1 \times \ket{0}) + (0 \times \ket{1})$, and is thus all $\ket{0}$ and no $\ket{1}$. \par
\end{timetravel}
The diagram in the box above can also be used to describe the state of a qubit. \par
Like classical bits, qubits have the \textit{basis states} $\ket{0}$ and $\ket{1}$. \par
Unlike classical bits, qubits may take values that are some combination of both.
\vspace{2mm}
Namely, every possible state of a qubit is a \textit{normalized linear combination} of $\ket{0}$ and $\ket{1}$. \par
Such states are called \textit{superpositions} of $\ket{0}$ and $\ket{1}$, since they partially contain both states.
\vfill
\pagebreak
\definition{}
The state of a quantum bit is the column unit vector
$
\ket{\psi}
= \left[\begin{smallmatrix} a \\ b \end{smallmatrix}\right]
= a\ket{0} + b\ket{1}$ for $a, b \in \mathbb{R}
$. \par
Note that the length of $\ket{\psi}$ must always be $1$, which is the same as saying that $a^2 + b^2 = 1$.
\vspace{2mm}
If we plot the set of valid quantum states on our plane, we get a unit circle centered at the origin: \par
\begin{center}
\begin{tikzpicture}[scale=1.5]
\draw[dashed] (0,0) circle(1);
\fill[color = black] (0, 0) circle[radius=0.05];
\draw[->] (0, 0) -- (1.2, 0);
\fill[color = oblue] (1, 0) circle[radius=0.05];
\node[below right] at (1, 0) {$\ket{0}$};
\draw[->] (0, 0) -- (0, 1.2);
\fill[color = oblue] (0, 1) circle[radius=0.05];
\node[above left] at (0, 1) {$\ket{1}$};
\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
\node[above right] at (0.87, 0.5) {$\ket{\psi}$};
\end{tikzpicture}
\end{center}
\problem{}
In the above diagram, the counterclockwise angle from $\ket{0}$ to $\ket{\psi}$ is $30^\circ$\hspace{-1ex}. \par
Write $\ket{\psi}$ as a linear combination of $\ket{0}$ and $\ket{1}$.
\vfill
\definition{Measurement I}
Although a qubit may have many states, it must be $\ket{0}$ or $\ket{1}$ when we measure it. \par
\vspace{2mm}
As a trivial example, say $\ket{\psi}$ = $\ket{0}$. \par
If we were to measure $\ket{\psi}$, we'd get $\ket{0}$, and the state of the qubit wouldn't change.
\vspace{2mm}
However, something interesting happens when $\ket{\psi} = a\ket{0} + b\ket{1}$. \par
Our measurement again returns either $\ket{0}$ or $\ket{1}$, with the following probabilities: \par
\begin{itemize}[itemsep = 2mm, topsep = 2mm]
\item $\mathcal{P}(\ket{1}) = a^2$
\item $\mathcal{P}(\ket{0}) = b^2$
\end{itemize}
\note{
Note that $\mathcal{P}(\ket{0}) + \mathcal{P}(\ket{1}) = 1$. \\
As you already know, this is true of any probability function.
}
\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
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
\problem{}
\begin{itemize}
\item What is the probability we get $\ket{0}$ if 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}
\begin{center}
\begin{tikzpicture}[scale=1.5]
\draw[dashed] (0,0) circle(1);
\fill[color = black] (0, 0) circle[radius=0.05];
\draw[->] (0, 0) -- (1.2, 0);
\fill[color = oblue] (1, 0) circle[radius=0.05];
\node[below right] at (1, 0) {$\ket{0}$};
\draw[->] (0, 0) -- (0, 1.2);
\fill[color = oblue] (0, 1) circle[radius=0.05];
\node[above left] at (0, 1) {$\ket{1}$};
\draw[dotted] (0, 0) -- (0.87, 0.5);
\draw[color=gray,->] (0.5, 0.0) arc (0:30:0.5);
\node[right, color=gray] at (0.47, 0.12) {$30^\circ$};
\fill[color = ored] (0.87, 0.5) circle[radius=0.05];
\node[above right] at (0.87, 0.5) {$\ket{\psi_0}$};
\draw[dotted] (0, 0) -- (-0.707, -0.707);
\draw[color=gray,->] (0.25, 0.0) arc (0:-135:0.25);
\node[below, color=gray] at (0.2, -0.2) {$135^\circ$};
\fill[color = ored] (-0.707, -0.707) circle[radius=0.05];
\node[below left] at (-0.707, -0.707) {$\ket{\psi_1}$};
\end{tikzpicture}
\end{center}
\vfill
As you may have noticed, we don't need two coordinates to fully define a quibit's state. \par
We can get by with one coordinate just as well.
\vspace{2mm}
Instead of referring to each state using its cartesian coordinates $a$ and $b$, \par
we can address it using its \textit{polar angle} $\theta$, measured from $\ket{0}$ counterclockwise:
\begin{center}
\begin{tikzpicture}[scale=1.5]
\draw[dashed] (0,0) circle(1);
\fill[color = black] (0, 0) circle[radius=0.05];
\draw[dotted] (0, 0) -- (0.707, 0.707);
\draw[color=gray,->] (0.5, 0.0) arc (0:45:0.5);
\node[above right, color=gray] at (0.5, 0) {$\theta$};
\draw[->] (0, 0) -- (1.2, 0);
\fill[color = oblue] (1, 0) circle[radius=0.05];
\node[below right] at (1, 0) {$\ket{0}$};
\draw[->] (0, 0) -- (0, 1.2);
\fill[color = oblue] (0, 1) circle[radius=0.05];
\node[above left] at (0, 1) {$\ket{1}$};
\fill[color = ored] (0.707, 0.707) circle[radius=0.05];
\node[above right] at (0.707, 0.707) {$\ket{\psi}$};
\end{tikzpicture}
\end{center}
\problem{}
Find $a$ and $b$ in terms of $\theta$ for an arbitrary qubit.
\vfill
\pagebreak

112
Advanced/Introduction to Quantum/parts/02.01 two halves.tex Normal file
View File

@ -0,0 +1,112 @@
\section{Two Halves of a Qubit}
\definition{}
Just as before, we'll represent multi-quibit states as linear combinations of multi-qubit basis states. \par
For example, a two-qubit state $\ket{ab}$ is the four-dimensional unit vector
\begin{equation}
\begin{bmatrix}
a \\ b \\ c \\ d
\end{bmatrix}
= a \ket{00} + b\ket{01} + c\ket{10} + d\ket{11}
\end{equation}
As always, multi-qubit states are unit vectors. \par
Thus, $a^2 + b^2 + c^2 + d^2 = 1$ in the two-bit case above.
\problem{}
Say we have two qubits $\ket{\psi}$ and $\ket{\varphi}$. \par
Show that $\ket{\psi} \otimes \ket{\varphi}$ is always a unit vector (and is thus a valid quantum state).
\vfill
\definition{Measurement II}<measureii>
Measurement of a two-qubit state works just like measurement of a one-qubit state: \par
If we measure $a\ket{00} + b\ket{01} + c\ket{10} + d\ket{11}$, \par
we get one of the four basis states with the following probabilities:
\begin{itemize}
\item $\mathcal{P}(\ket{00}) = a^2$
\item $\mathcal{P}(\ket{01}) = b^2$
\item $\mathcal{P}(\ket{10}) = c^2$
\item $\mathcal{P}(\ket{11}) = d^2$
\end{itemize}
Of course, the sum of all the above probabilities is $1$.
\problem{}
Consider the two-qubit state
$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$
\begin{itemize}[itemsep=2mm]
\item If we measure both bits of $\ket{\psi}$ simultaneously, \par
what is the probability of getting each of $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$?
\item If we measure the ONLY the first qubit, what is the probability we get $\ket{0}$? How about $\ket{1}$? \par
\hint{There are two basis states in which the first qubit is $\ket{0}$.}
\item Say we measured the second bit and read $\ket{1}$. \par
If we now measure the first bit, what is the probability of getting $\ket{0}$?
\end{itemize}
\vfill
\pagebreak
\problem{}
Again, consider the two-qubit state
$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$ \par
If we measure the first qubit of $\ket{\psi}$ and get $\ket{0}$, what is the resulting state of $\ket{\phi}$? \par
What would the state be if we'd measured $\ket{1}$ instead?
\vfill
\problem{}
Consider the three-qubit state $\ket{\psi} = c_0\ket{000} + c_1\ket{001} + ... + c_7 \ket{111}$. \par
Say we measure the first two qubits and get $\ket{00}$. What is the resulting state of $\ket{\psi}$?
\begin{solution}
We measure $\ket{00}$ with probability $c_0^2 + c_1^2$, and $\ket{\psi}$ collapses to
\begin{equation*}
\frac{c_0\ket{000} + c_1\ket{001}}{\sqrt{c_0^2 + c_1^2}}
\end{equation*}
\end{solution}
\vfill
\pagebreak
\definition{Entanglement}
Some product states can be factored into a tensor product of individual qubit states. For example,
\begin{equation*}
\frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr)
= \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes
\frac{1}{\sqrt{2}}\bigl( \ket{0} - \ket{1} \bigr)
\end{equation*}
Such states are called \textit{product states.} States that aren't product states are called \textit{entangled} states.
\problem{}
Factor the following product state:
\begin{equation*}
\frac{1}{2\sqrt{2}} \bigl(\sqrt{3}\ket{00} - \sqrt{3}\ket{01} + \ket{10} - \ket{11}\bigr)
\end{equation*}
\begin{solution}
\begin{equation*}
\frac{1}{2\sqrt{2}} \biggl(\sqrt{3}\ket{00} - \sqrt{3}\ket{01} + \ket{10} - \ket{11}\biggr)
= \biggl( \frac{\sqrt{3}}{2}\ket{0} + \frac{1}{2}\ket{1} \biggr) \otimes
\biggl(\frac{1}{\sqrt{2}}\ket{0} - \frac{1}{\sqrt{2}}\ket{1} \biggr)
\end{equation*}
\end{solution}
\vfill
\pagebreak

635
Advanced/Introduction to Quantum/parts/1 gates.tex
View File

@ -1,635 +0,0 @@
\section{Logic Gates}
Now that we know how to write vectored bits, let's look at the ways we can change them.
\generic{Remark:}
A few weeks ago, we talked about matrices. Recall that every linear map may be written as a matrix,
and that every matrix represents a linear map. For example, if $f: \mathbb{R}^2 \to \mathbb{R}^2$ is a linear
map, we can write it as follows:
\begin{equation*}
f\left(
\ket{x}
\right)
=
\begin{bmatrix}
m_1 & m_2 \\
m_3 & m_4
\end{bmatrix}
\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}
=
\left[
\begin{matrix}
m_1x_1 + m_2x_2 \\
m_3x_1 + m_4x_2
\end{matrix}
\right]
\end{equation*}
\problem{}
The \say{not} gate is a map from $\mathbb{B}$ to $\mathbb{B}$ defined by the following table:
\begin{itemize}
\item $\text{not}(\texttt{1}) = \texttt{0}$
\item $\text{not}(\texttt{0}) = \texttt{1}$
\end{itemize}
Write the not gate as a matrix that operates on single-bit vector states. \par
That is, find a matrix $N$ so that $N\ket{0} = \ket{1}$ and $N\ket{1} = \ket{0}$.
\begin{solution}
\begin{equation*}
N = \begin{bmatrix}
0 & 1 \\ 1 & 0
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\problem{}
The \say{and} gate is a map $\mathbb{B}^2 \to \mathbb{B}$ defined by the following table:
\begin{center}
\begin{tabular}{ c | c | c }
\hline
\texttt{a} & \texttt{b} & \texttt{a} and \texttt{b} \\
\hline
0 & 0 & 0 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
1 & 1 & 1
\end{tabular}
\end{center}
Find a matrix $A$ so that $A\ket{\texttt{ab}}$ works as expected. \par
\hint{
What is the dimension of the input? What is the dimension of the desired output? \\
What do these two values tell us about the dimension of the matrix $A$?
}
\begin{solution}
\begin{equation*}
A = \begin{bmatrix}
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\pagebreak
\generic{Remark:}
The way a quantum computer handles information is a bit different than the way a classical computer does.
For spooky physics reasons, all quantum gates must be invertible. Naturally, this implies that the usual
logic gates we use (and, or, xor) aren't valid quantum gates: each of these takes two inputs (four states)
and produces one output (two states), losing information by mapping many inputs to the same output. \par
\note[Note]{The \say{not} gate is fine. It is its own inverse, after all!}
\vspace{2mm}
Although we are still using classical bits, we'll design our gates to be reversible. \par
One consequence of the \say{reversibility} rule is that all quantum gate matrices must be square. \par
(i.e, they must take the same number of inputs and outputs.) \par
\note[Note]{All invertible matrices are square, but not all square matrices are invertible.}
\vspace{2mm}
This is fairly intuitive: if we have more inputs than we have outputs, we inevitably lose information.
\vspace{2mm}
There's also a better way to think about this: rather than seeing quantum gates as \textit{functions}
that consume one set of bits and produce another, it's better to think of them as \textit{transformations}
we apply to an existing set of bits.
\problem{}
For example, take the CNOT (controlled not) gate. \par
When applied to a two-bit state $\ket{ab}$, CNOT inverts $b$ iff $a$ is $\ket{1}$. \par
Find the matrix that represents the CNOT gate. \par
\hint{what are the dimensions of this matrix?}
\begin{solution}
\begin{equation*}
\text{CNOT} = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{bmatrix}
\end{equation*}
\vspace{4mm}
If $\ket{a}$ is $\ket{0}$, $\ket{a} \otimes \ket{b}$ is
$
\begin{bmatrix}
\begin{bmatrix}
b_1 \\ b_2
\end{bmatrix} \\ 0 \\ 0
\end{bmatrix}
$, and the \say{not} portion of the matrix is ignored.
\vspace{4mm}
If $\ket{a}$ is $\ket{1}$, $\ket{a} \otimes \ket{b}$ is
$
\begin{bmatrix}
0 \\ 0 \\
\begin{bmatrix}
b_1 \\ b_2
\end{bmatrix}
\end{bmatrix}
$, and the \say{identity} portion of the matrix is ignored.
The state of $\ket{a}$ is always preserved, since it's determined by the position of
$\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ in the tensor product.
If $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on top, $\ket{a}$ is $\ket{0}$,
and if $\left[\begin{smallmatrix}b_1 \\ b_2\end{smallmatrix}\right]$ is on the bottom, $\ket{a}$ is $\ket{1}$.
\end{solution}
\vfill
\problem{}
Now, modify the CNOT gate so that it inverts $\ket{a}$ whenever it is applied.
\begin{solution}
\begin{equation*}
\text{CNOT}_{\text{mod}} = \begin{bmatrix}
0 & 1 & 0 & 0 \\
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\pagebreak
\iffalse
\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 $\ket{1}$.
\begin{solution}
\begin{equation*}
\text{CNOT}_{\text{flip}} = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\fi
\problem{}
The SWAP gate swaps two bits: $\text{SWAP}\ket{ab} = \ket{ba}$. \par
Find its matrix.
\begin{solution}
\begin{equation*}
\text{SWAP} = \begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\problem{}
The $T$ gate is a three-bit gate that inverts its right bit iff its left and middle inputs are both $\ket{1}$. \par
In other words, $T\ket{11x} = \ket{11}\ket{\text{not } x}$, and $T\ket{abx} = \ket{abx}$ for all other inputs. \par
Find the $T$ gate's matrix. \par
\note{
This gate is particularly interesting because it's a \textit{universal quantum gate}: \\
like NOR and NAND in classical logic, any quantum gate may emulated by only applying $T$ gates.
}
\begin{solution}
\begin{equation*}
\text{T} = \begin{bmatrix}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\
\end{bmatrix}
\end{equation*}
\end{solution}
\vfill
\pagebreak
The last thing we need is a way to draw complex sequences of gates. \par
We already know how to do this with classical gates:
\begin{center}
\begin{tikzpicture}[circuit logic US, scale=1.5]
\node[and gate] (and) at (0,-0.8) {\tiny\texttt{and}};
\draw ([shift={(-0.5, 0)}] and.input 1) node[left] {\texttt{1}} -- (and.input 1);
\draw ([shift={(-0.5, 0)}] and.input 2) node[left] {\texttt{0}} -- (and.input 2);
\draw (and.output) -- ([shift={(0.5, 0)}] and.output) node[right] {\texttt{0}};
\end{tikzpicture}
\end{center}
We draw quantum circuits in a very similar way. \par
For example, here a simple three-bit circuit consisting of a CNOT gate on the first bit, \par
controlled by the third. The first bit is X'd iff the third bit is $\ket{1}$:
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\node[qubit] (c) at (0, -2) {$\ket{1}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$};
\draw[wire] (c) -- ([shift={(4, 0)}] c.center) node[qubit] {$\ket{1}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
;
\draw[wirejoin]
($([shift={(1,0)}] c)!0.5!([shift={(3,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\end{tikzpicture}
\end{center}
\problem{}
Draw the CNOT gate as a classical logic circuit. \par
\hint{This can be done with one gate.}
\begin{solution}
\begin{center}
\begin{tikzpicture}[circuit logic US, scale=2]
\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$a$};
\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$b$};
\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
\draw (dot) -- (dot -| 1,0) node[right] {$b_\text{out}$};
\draw (xor.output) -- (xor.output -| 1,0) node[right] {$a_\text{out}$};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
Gate controls may be marked with a filled circle or an empty circle. \par
Empty circles denote \textit{inverse controls,} which (of course) have an inverse effect. \par
For example, the two circuits below are identical:
\null\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{a}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{b}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
;
\draw[wireijoin]
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill
\begin{minipage}{0.48\textwidth}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{a}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{b}$};
\draw[wire]
($([shift={(1.5,0)}] a)!0.5!([shift={(2.5,0)}] a)$) --
($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
;
\draw[wirejoin]
($([shift={(1.5,0)}] b)!0.5!([shift={(2.5,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\qubox{b}{0.5}{b}{1.5}{X};
\qubox{b}{2.5}{b}{3.5}{X};
\end{tikzpicture}
\end{center}
\end{minipage}
\hfill\null
\problem{}
What are $\ket{a}$ and $\ket{b}$ in the diagrams above?
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{0}$};
\node[qubit] (b) at (0, -1) {$\ket{0}$};
\draw[wire] (a) -- ([shift={(4, 0)}] a.center) node[qubit] {$\ket{1}$};
\draw[wire] (b) -- ([shift={(4, 0)}] b.center) node[qubit] {$\ket{0}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(3,0)}] a)$) --
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
;
\draw[wireijoin]
($([shift={(1,0)}] b)!0.5!([shift={(3,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{3}{CNOT};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\pagebreak
As we noted before, quantum gates are \textit{transformations,} modifying a set of bits. \par
Thus, quantum circuits are drawn with a fixed set of bits, whose states transform with time. \par
\note[Note]{
In this diagram, CNOT and SWAP are drawn as $\oplus$ and \rotatebox[origin=c]{90}{$\leftrightarrows$} to save space.
}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{a}$};
\node[qubit] (b) at (0, -1) {$\ket{1}$};
\node[qubit] (c) at (0, -2) {$\ket{c}$};
\node[qubit] (d) at (0, -3) {$\ket{d}$};
\node[qubit] (e) at (0, -4) {$\ket{1}$};
\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$};
\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{b}$};
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$};
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$};
\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{e}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
;
\draw[wirejoin]
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wireijoin]
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{2}{T};
\qubox{a}{4}{a}{5}{X};
\draw[wire]
($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
;
\draw[wireijoin]
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{e}{2}{e}{3}{$\oplus$};
\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}};
\draw[wire]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
;
\draw[wirejoin]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wirejoin]
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{b}{5}{b}{6}{T};
\end{tikzpicture}
\end{center}
In a quantum circuit, the ONLY way two bits can interact is through a gate. \par
We cannot add \say{branches} in quantum circuits like we do in classical circuits:
\begin{center}
\begin{tikzpicture}[circuit logic US, scale=1.2]
\node[xor gate] (xor) at (0, 0) {\tiny\texttt{xor}};
\draw (xor.input 1) ++(-0.5, 0) coordinate (start);
\draw (xor.input 1) ++(-0.25, 0) coordinate (startjoin);
\draw (xor.input 1) -- (xor.input 1 -| start) node[left] {$x$};
\draw (xor.input 2) -| (0,-0.25 -| startjoin) |- (0,-0.25, -| start) node[left] {$y$};
\filldraw (0,-0.25, -| startjoin) circle[radius=0.3mm] coordinate(dot);
\draw (dot) -- (dot -| 1,0) node[right] {$y_\text{out}$};
\draw (xor.output) -- (xor.output -| 1,0) node[right] {$x_\text{out}$};
\draw[->, line width = 1, ogrape]
([shift={(0.3,-0.5)}] dot) node[right] {This is a branch}
-| ([shift={(0,-0.2)}] dot)
;
\end{tikzpicture}
\end{center}
\problem{}
Find the values of $\ket{a}$ through $\ket{e}$ in the above circuit.
\begin{solution}
\begin{center}
\begin{tikzpicture}[scale=1]
\node[qubit] (a) at (0, 0) {$\ket{1}$};
\node[qubit] (b) at (0, -1) {$\ket{1}$};
\node[qubit] (c) at (0, -2) {$\ket{1}$};
\node[qubit] (d) at (0, -3) {$\ket{0}$};
\node[qubit] (e) at (0, -4) {$\ket{1}$};
\draw[wire] (a) -- ([shift={(7, 0)}] a.center) node[qubit] {$\ket{0}$};
\draw[wire] (b) -- ([shift={(7, 0)}] b.center) node[qubit] {$\ket{1}$};
\draw[wire] (c) -- ([shift={(7, 0)}] c.center) node[qubit] {$\ket{1}$};
\draw[wire] (d) -- ([shift={(7, 0)}] d.center) node[qubit] {$\ket{0}$};
\draw[wire] (e) -- ([shift={(7, 0)}] e.center) node[qubit] {$\ket{0}$};
\draw[wire]
($([shift={(1,0)}] a)!0.5!([shift={(2,0)}] a)$) --
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
;
\draw[wirejoin]
($([shift={(1,0)}] b)!0.5!([shift={(2,0)}] b)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wireijoin]
($([shift={(1,0)}] c)!0.5!([shift={(2,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{a}{1}{a}{2}{T};
\qubox{a}{4}{a}{5}{X};
\draw[wire]
($([shift={(2,0)}] e)!0.5!([shift={(3,0)}] e)$) --
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
;
\draw[wireijoin]
($([shift={(2,0)}] d)!0.5!([shift={(3,0)}] d)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{e}{2}{e}{3}{$\oplus$};
\qubox{b}{3}{c}{4}{\rotatebox[origin=c]{90}{$\leftrightarrows$}};
\draw[wire]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$) --
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
;
\draw[wirejoin]
($([shift={(5,0)}] a)!0.5!([shift={(6,0)}] a)$)
circle[radius=0.1] coordinate(dot)
;
\draw[wirejoin]
($([shift={(5,0)}] c)!0.5!([shift={(6,0)}] c)$)
circle[radius=0.1] coordinate(dot)
;
\qubox{b}{5}{b}{6}{T};
\end{tikzpicture}
\end{center}
\end{solution}
\vfill
\pagebreak

5
Advanced/Introduction to Quantum/tikzset.tex
View File

@ -32,6 +32,11 @@
}
% Macros
% Do NOT put a semicolon after qubox,
% it gives a "character ; not found" error.
% LaTeX is odd.
\def\qubox#1#2#3#4#5{
% 1: point ne
% 2: point ne x offset