superposition

main.tex

@@ -4,7 +4,7 @@
 \usepackage[utf8]{inputenc}
 \usepackage{lmodern}
 
-\usepackage{amsmath, amssymb, amsfonts}
+\usepackage{amsmath, amsthm, amssymb, amsfonts}
 \usepackage{mathtools}
 \usepackage{physics}
 
@@ -13,6 +13,14 @@
 
 \newcommand{\ptrans}{\delta}
 \DeclarePairedDelimiter{\parens}{\lparen}{\rparen}
+\DeclarePairedDelimiter{\parensc}{\{}{\}}
+
+\DeclareMathOperator{\affinehull}{\text{aff}}
+\DeclareMathOperator{\R}{\mathbb{R}}
+
+\newtheorem{theorem}{Theorem}
+\newtheorem{lemma}{Lemma}
+\newtheorem{definition}{Definition}
 
 \begin{document}
 
@@ -80,7 +88,7 @@ p^t_{b=x} = \begin{cases}
 \end{gather}
 It is important to note, that $\mathbf{0}$ and $\mathbf{1}$ in \cref{eq:exp_state_single_bit} are not the scalar values of $b$. They define abstract objects denoting the fact that $b$ is in state $0$ or $1$, so they are just arbitrary labels. For instance, same states could also be labeled $\{\mathbf{T}, \mathbf{F}\}$ or $\{\top, \bot\}$. But if $\mathbf{0}$ and $\mathbf{1}$ are some kind of abstract object and not scalar value, how can \cref{eq:exp_state_single_bit} be evaluated? As of now it can't. Later we will define representations of these abstract stats, which are closed under addition and scalar multiplication, making \cref{eq:exp_state_single_bit} also (a representation of) an abstract state.
 
-From \cref{eq:exp_state_single_bit}, we will now derive a standard form of our random bit $b$. We don't view $b$ as being either in state $\mathbf{0}$ OR $\mathbf{1}$ anymore. From now on, we think of $b$ as being in $\mathbf{0}$ AND $\mathbf{1}$ simultaneously with certain probabilities $p_{b=0}$ and $p_{b=1}$, so $b$ is in a superposition:
+From \cref{eq:exp_state_single_bit}, we will now derive a standard form of our random bit $b$. We don't view $b$ as being either in state $\mathbf{0}$ OR $\mathbf{1}$ anymore. From now on, we think of $b$ as being in $\mathbf{0}$ AND $\mathbf{1}$ simultaneously with certain probabilities $p_{b=0}$ and $p_{b=1}$, The one bit system $b$ is in a \emph{superposition} of two \emph{basis states} $\mathbf{0}$ and $\mathbf{1}$:
 $$
 b = p_0 \mathbf{0} + p_1 \mathbf{1} \quad , p_0 + p_1 = 1
 $$
@@ -108,8 +116,27 @@ and thus $\ptrans$ preserves valid superpositions, which finally makes predictio
 \end{aligned}
 \end{equation}
 
+\subsection{Collapsing Superpositions}
+Extending this formalism to bit registers is actually fairly straight forward. Systems can be in superposition of arbitrary many basis states. But first, it is time to talk a bit more about the concept of superposition. 
+\begin{definition}{Superposition of Probabilities}
+  If $\mathbf{E} \coloneqq \parensc*{E_1, E_2, \dots, E_n}$ is the set of all possible outcomes of an experiment, then a superposition of probable outcomes is defined by:
+  \begin{equation}
+    E \coloneqq \sum_{i=1}^n p_i E_i \quad \text{with}\:\: p_i = P\parens*{E_i} \:\text{and}\:\: \sum_{i=1}^n p_i = 1
+  \end{equation}  
+\end{definition}
+
+As mentioned above, a superposition can not immediately be evaluated. It rather should be seen as a mathematical object holding incomplete knowledge about a certain property of some (stochastic) process, described by a random distribution $(p_i)_{i=1}^n$. Too actually evaluate a superposition, the missing information needs to be filled in by some kind of extra process e.g. performing an experiment, measuring an observable. After this extra information is filled in the property under consideration is fully known and the superposition \emph{collapses} to one of the actually realizable outcomes in $\mathbf{E}$. In this model a system can be in an uncertain state which only can be made concrete by some external influence like measuring an observable. This sounds quite abstract and especially the fact that a measurement could alter the state of a real physical system seems quite counterintuitive, but we will later see that this principle is actually grounded in reality.
+
+Let's consider the experiment of rolling a dice. Of course, for the observable \emph{number of eyes} the expected outcomes are $\mathbf{E} = \parensc{1, 2, \dots, 6}$. While the dice is still in the cup and in the state of being shaken number of eyes can not be reasonably determined, even if a transparent cup is being used. The dice is in a superposition $E = \sum_{i=1}^6 \frac{1}{6} \mathbf{i}$ of showing all numbers of eyes 1 to 6 with uniform probability $\frac{1}{6}$. In order to determine the number of eyes thrown, the dice needs to rest on a solid base, such that one side is evidently showing up. So by \emph{throwing the dice} we interfere with the system by stopping to shake the cup and placing the dice on a solid base (table). With the dice now laying on the table it is clearly showing only one number of eyes. The superposition collapsed!
+
+\begin{definition}{Collapse of Superposition}
+  
+\end{definition}
+
 \subsection{Bit Registers in Superposition}
 
+
+
 \section{Introducing: Linear Algebra}
 
 \section{Making it Quantum}