transition matrix

main.tex

@@ -5,6 +5,7 @@
 \usepackage{lmodern}
 
 \usepackage{amsmath, amsthm, amssymb, amsfonts}
+\usepackage[mathscr]{euscript}
 \usepackage{mathtools}
 \usepackage{physics}
 
@@ -17,6 +18,8 @@
 
 \DeclareMathOperator{\affinehull}{\text{aff}}
 \DeclareMathOperator{\R}{\mathbb{R}}
+\DeclareMathOperator{\spanspace}{\text{span}}
+\DeclareMathOperator{\Hom}{\text{hom}}
 
 \theoremstyle{plain}
 \newtheorem{theorem}{Theorem}
@@ -63,6 +66,8 @@ $$
 Or by the composition of all gate applications up to this point: $(f_l \circ f_{l-1} \circ \cdots \circ f_1)(\mathbf{b}_0)$. Actually, a composition of gates is also just another logical gate $F \coloneqq (f_l \circ f_{l-1} \circ \cdots \circ f_1) : \{0,1\}^n \to \{0,1\}^n$. If we are not interested in intermediate states, we can thus define our computation in the form of $\mathbf{b}_{\text{out}} \coloneqq F(\mathbf{b}_{\text{in}})`$, with $`F: \{0,1\}^n \to \{0,1\}^n$.
 
 \section{A Bit of Randomness}
+\label{sec:probabilistic_model}
+
 \subsection{Single Bits in Superposition}
 \label{sec:oneBitInSuperposition}
 Many real world problems are believed  to not be efficiently solvable on fully deterministic computers like the model described above (if $\mathbf{P} \neq \mathbf{NP}$). Fortunately, it turns out that if we allow for some randomness in our algorithms, we're often able to efficiently find solutions for such hard problems with sufficiently large success probabilities. Often times, the error probabilities can even be made exponentially small. For this reason, we also want to introduce randomness into our model. Algorithms or computational models harnessing the power of randomness are usually called \emph{probabilistic}.
@@ -154,17 +159,21 @@ Let's consider the experiment of rolling a dice. Of course, for the observable \
 Extending the probabilistic one-bit model from \cref{sec:oneBitInSuperposition} to bit registers is almost trivial given the definitions from \cref{sec:superposition}. A $n$-bit register can be in $N = 2^n$ possible states, giving rise to a superposition of $N$ basis states for probabilistic register states. 
 \begin{definition}
   \label{def:nbitRegister}
-  The state of a $n$-bit register in a probabilistic computation is defined by a superposition of all possible basis states $\mathbf{B} = \parensc*{\mathbf{0}, \mathbf{1}}^n = \parensc*{\mathbf{1}, \mathbf{2}, \dots, \mathbf{N}}$.
+  The state of a $n$-bit register in a probabilistic computation is defined by a superposition of all possible basis states $\mathbf{B} = \parensc*{\mathbf{0}, \mathbf{1}}^n = \parensc*{\mathbf{0}, \mathbf{1}, \dots, \mathbf{N-1}}$.
   \begin{equation}
-    \mathbf{b} \coloneqq \sum_{i=1}^N p_i \cdot \mathbf{i} \quad \:\text{with}\:\: P\parens*{\mathbf{b} = \mathbf{i}} = p_i
+    \mathbf{b} \coloneqq \sum_{i=0}^{N-1} p_i \cdot \mathbf{i} \quad \:\text{with}\:\: P\parens*{\mathbf{b} = \mathbf{i}} = p_i
   \end{equation}
 \end{definition}
-Similar to \cref{sec:oneBitInSuperposition} the transition function $\ptrans$ can be defined on its effect on basis states. For each transition the probabilites of transitioning from basis state $\mathbf{i}$ to basis state $\mathbf{j}$ must be defined. The mapping between states in superposition will then be defined linearly.
+\begin{remark}
+  It should be noted that the number representation $\parensc*{\mathbf{i}}_{i=0}^{N-1}$ is defined as the bit string $\{\mathbf{0}, \mathbf{1}\}^n$ in a base of 10. So it is just a shorter label for the state of a $n$-bit register and NOT a scalar value.
+\end{remark}
+Similar to \cref{sec:oneBitInSuperposition} the transition function $\ptrans$ can be defined on its effect on basis states. For each transition the probabilities of transitioning from basis state $\mathbf{i}$ to basis state $\mathbf{j}$ must be defined. The mapping between states in superposition will then be defined linearly.
 \begin{definition}
   \label{def:probabilisticTransitionFunction}
-  Let $\mathbf{b} = \sum_{i=1}^{N} p_i \mathbf{i}$ as defined in \cref{def:nbitRegister} and let $p_{\mathbf{ij}}$ the probability of transitioning form basis state $\mathbf{i}$ to basis state $\mathbf{j}$, then the transition function is defined by:
+  Let $\mathbf{b} = \sum_{i=0}^{N-1} p_i \mathbf{i}$ be a $n$-bit register as defined in \cref{def:nbitRegister} and let $p_{\mathbf{ij}}$ be the probability of transitioning form basis state $\mathbf{i}$ to basis state $\mathbf{j}$, then the transition function is defined by:
   \begin{equation}
-    \ptrans\parens*{\mathbf{b}} \coloneqq \sum_{i=1}^N p_i \ptrans(\mathbf{i}) = \sum_{i=1}^N \sum_{j=1}^N p_i p_{ij} \mathbf{j}
+    \label{eq:ptrans_on_register}
+    \ptrans\parens*{\mathbf{b}} \coloneqq \sum_{i=0}^{N-1} p_i \ptrans(\mathbf{i}) = \sum_{i=0}^{N-1} \sum_{j=0}^{N-1} p_i p_{ij} \mathbf{j}
   \end{equation}
 \end{definition}
 
@@ -173,11 +182,11 @@ Similar to \cref{sec:oneBitInSuperposition} the transition function $\ptrans$ ca
   A transition function as defined by \cref{def:probabilisticTransitionFunction} maps superposition to valid superpositions.
 \end{theorem}
 \begin{proof}
-  Let $\ptrans$ be a probabilistic transition function and let $\mathbf{b}$ a register state in superposition. By \cref{def:probabilisticTransitionFunction} we get $\ptrans(\mathbf{b}) = \sum_{i=1}^N \sum_{j=1}^N p_i p_{ij} \mathbf{j}$ a simple reordering leads to
+  Let $\ptrans$ be a probabilistic transition function and let $\mathbf{b}$ a register state in superposition. By \cref{def:probabilisticTransitionFunction} we get $\ptrans(\mathbf{b}) = \sum_{i=0}^{N-1} \sum_{j=0}^{N-1} p_i p_{ij} \mathbf{j}$ a simple reordering leads to
   $$
-  \ptrans(\mathbf{b}) = \sum_{j=1}^N \parens*{\sum_{i=1}^N p_i p_{ij}} \mathbf{j}
+  \ptrans(\mathbf{b}) = \sum_{j=0}^{N-1} \parens*{\sum_{i=0}^{N-1} p_i p_{ij}} \mathbf{j}
   $$
-  Obviously, $p_i p_{ij} = P\parens{\mathbf{b} = \mathbf{i}} P\parens{\ptrans(\mathbf{b}) = \mathbf{j} \:|\: \mathbf{b} = \mathbf{i}}$. It follows directly from the law of total probability that $\sum_{j=1}^N\sum_{i=1}^N p_i p_{ij} = \sum_{j=1}^N P\parens{\ptrans(\mathbf{b}) = \mathbf{j}} = 1$
+  Obviously, $p_i p_{ij} = P\parens{\mathbf{b} = \mathbf{i}} P\parens{\ptrans(\mathbf{b}) = \mathbf{j} \:|\: \mathbf{b} = \mathbf{i}}$. It follows directly from the law of total probability that $\sum_{j=0}^{N-1}\sum_{i=0}^{N-1} p_i p_{ij} = \sum_{j=0}^{N-1} P\parens{\ptrans(\mathbf{b}) = \mathbf{j}} = 1$
 \end{proof}
 A direct consequence of \cref{thm:superpositionsClosedUnderProbabilisticTransition} is that the space of probabilistic transition functions is also closed under composition. In accordance to \cref{eq:deterministic_register_at_time_t} the state of a register $\mathbf{b}$ in a probabilistic computation at time $t$ can be described by:
 \begin{equation}\begin{aligned}
@@ -187,7 +196,64 @@ A direct consequence of \cref{thm:superpositionsClosedUnderProbabilisticTransiti
   \end{cases} \\
                &= \parens*{\ptrans_t \circ \ptrans_{t-1} \circ \cdots \circ \ptrans_1}(\mathbf{b}_0)
 \end{aligned}\end{equation}
+
 \section{Introducing: Linear Algebra}
+The definitions of \cref{sec:probabilistic_model} fully describe a probabilistic computational model. Unfortunately, working with them can be quite cumbersome. This section will introduce an algebraic apparatus based on the definitions from above, with many helpful tools to describe computations and state evolutions. As some terminology and especially the linear properties of \cref{def:probabilisticTransitionFunction} already suggest the mathematical framework of choice will be linear algebra. Let's start by assessing the components of the model described above. We have:
+\begin{itemize}
+  \item States (in superposition)
+  \item State transitions
+  \item Measurements (collapse of superposition)
+\end{itemize}
+As it turns out, all three components and their interactions can be expressed in the language of linear algebra. Readers familiar with that field of mathematics probably already noticed that $\ptrans$ is a linear function and the space of states in superposition looks a lot like a vector space. 
+
+\subsection{The State Space}
+The defining property of a superposition is the probability distribution of its basis states. Given an enumeration all basis states the superposition is fully defined by the list of probabilities $\parens{p_0, p_1, \dots, p_{N-1}}$.
+\begin{definition}[State Spaces of Probabilistic Computations]
+  Given a state basis $\mathbf{B}$ of a $n$-bit register, the state space of probabilistic computations on this register is defined as:
+  \begin{equation*}
+    \mathbf{B}^n \coloneqq \parensc*{\mathbf{b} \sum_{i=0}^{N-1} p_i \mathbf{i} \:\middle|\: p_i \in \R_+ \:,\: \sum_{i=0}^{N-1} p_i = 1}
+  \end{equation*}
+\end{definition}
+
+\begin{definition}
+  The coordinate map is a linear function $\Psi_{\mathbf{B}} : \mathbf{B}^{n} \to \R^N$ mapping the state space to $\R^N$:
+  \begin{equation*}
+    \forall \mathbf{b} \in \mathbf{B}^n :\quad \Psi_{\mathbf{B}}\parens{\mathbf{b}} = \parens{p_0, p_1, \dots, p_{N-1}}^T = \sum_{i=1}^N p_i \mathbf{e}_i
+  \end{equation*}
+  Often $\Psi_{\mathbf{B}}(\mathbf{b})$ is called the coordinate vector of $\mathbf{b}$ with respect to the basis $\mathbf{B}$.
+\end{definition}
+
+\begin{lemma}
+  \label{thm:state_space_unit_sphere_surface_isomorphism}
+  The state space of probabilistic computations is isomorphic to the surface of the unit sphere in the first quadrant of $\R^N$. 
+  \begin{equation*}
+    \mathbf{B}^n \cong \parensc*{\mathbf{v} \in \R_+^N \:\middle|\: \norm{\mathbf{v}} = 1}
+  \end{equation*}
+\end{lemma}
+
+\begin{proof}
+  For an arbitrary state $\mathbf{b}$ the coordinate vector $\Psi_{\mathbf{B}}(b) = \mathbf{v}$ is the direction of a ray in the first quadrant of $\R^N$ starting from the origin. Rescaling $\mathbf{v}$ results in the point where this ray intersects the unit sphere $\varphi(\mathbf{v}) = \frac{\mathbf{v}}{\norm{\mathbf{v}}} = \mathbf{v}'$ which can be inverted by $\varphi^{-1}(\mathbf{v}') = \frac{\mathbf{v}'}{\norm{\mathbf{v}'}_1} = \mathbf{v}$. Thus, $\varphi \circ \varphi^{-1} = \varphi^{-1} \circ \varphi = \text{id}$ and
+  $$
+  \Psi_{\mathbf{B}}\parens*{\mathbf{B}^n}  = \parensc{\mathbf{v} \in \R_+^N \:|\: \norm{\mathbf{v}}_1 = 1}\cong \parensc{\mathbf{v} \in \R_+^N \:|\: \norm{\mathbf{v}} = 1}
+  $$
+\end{proof}
+
+\subsection{Transition Matrices}
+It follows directly from \cref{eq:exp_state_single_bit} that $\ptrans : \spanspace\parens{\mathbf{B}} \to \spanspace\parens{\mathbf{B}}$ is a linear transition on the space spanned by state basis $\mathbf{B}$ and \cref{thm:superpositionsClosedUnderProbabilisticTransition} even states that $\ptrans : \mathbf{B}^n \to \mathbf{B}^n$ and $\mathbf{B}^n$ is closed under $\ptrans$. It is well known, that the space of all linear maps $\Hom_{\R}\parens{V,W}$ between two finite-dimensional real vector spaces $V$ and $W$ is isomorphic to $\R^{\parens{\dim(W), \dim(V)}}$. So, there must exist an isomorphism between transition functions $\ptrans$ and $\R^{\parens{N,N}}$.
+\begin{theorem}%[see \cite{Knabner}]
+  \label{thm:probabilistic_matrix_representation}
+  Let $\mathbf{B} = \parensc{\mathbf{b}_i}_{i=1}^N$ be a $n$-bit state basis and $\mathscr{B} = \parensc{\mathbf{v}_j}_{j=1}^N$ a basis of $\R^N$, then there exists a matrix $A = (a_{ij}) \in \R^{\parens{N,N}}$ such that
+  \begin{itemize}
+    \item $\forall \mathbf{b}_i \in \mathbf{B} :\quad \ptrans(\mathbf{b}_i) = \sum_{j=1}^N a_{ji} \mathbf{v}_j$
+    \item $\ptrans\parens*{\sum_{i=1}^N x_i \mathbf{b}_i} = \sum_{j=1}^N y_j \mathbf{v}_j \iff A\parens{x_1, x_2, \dots, x_N}^T = \parens{y_1, y_2, \dots, y_N}^T$
+  \end{itemize}  
+\end{theorem}
+
+\begin{remark}
+  Usually it is custom to choose the standard basis $\parensc{\mathbf{e}_i}_{i=1}^N$ for $\R^N$, then \cref{thm:probabilistic_matrix_representation} describes how $A$ can be used to describe how $\ptrans$ affects basis states in coordinate space. The $j$-th column vector $A\mathbf{e}_j = \mathbf{a}^j = \parens{a_{1j}, a_{2j}, \dots, a_{Nj}}^T$ represents the probability distribution of $\ptrans(\mathbf{b_j})$. It follows that $a_{ij} = p_{ji}$, with $p_{ji}$ being the probability of transitioning from $\mathbf{b}_j$ to $\mathbf{b}_i$. Consequently, $A = P^T$ with $P = (p_{ij})$. 
+\end{remark}
+
+\subsection{Measurements}
 
 \section{Making it Quantum}
 \end{document}