@ -59,6 +64,7 @@ Or by the composition of all gate applications up to this point: $(f_l \circ f_{
\section{A Bit of Randomness}
\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}.
Again, we start with simple one bit systems. Later, we'll see how to expand the following methods to full bit vectors/registers. In the deterministic single bit model above, the state transition of a bit $b$ in step $t$ is defined by $f_t(b)\in\{0,1\}$. Now, the transition function (or gate) is simply allowed to flip an unfair coin and either output 0 or 1 for heads or tails respectively. Of course, the state of $b$ prior to the transition should have an effect on the computation. That is, why we allow different (unfair) coins for either $b =0$ or $b =1$. To distinguish between deterministic and probabilistic transition functions, we will denote the latter by $\ptrans(b)\in\{0,1\}$. Or to reformulate this idea: Depending on the value of $b$, the output of $\ptrans(b)$ follows one of two Bernoulli trials. There are 4 possible transitions with probabilities $p_{00}$, $p_{01}$, $p_{10}$ and $p_{11}$, where $p_{ij}$ is the probability of $b$ transitioning form $i$ to $j$. Obviously, $\sum_j p_{ij}=1$ always needs to be satisfied.
@ -108,6 +114,7 @@ A simple calculation verifies that
\end{align*}
and thus $\ptrans$ preserves valid superpositions, which finally makes predictions of the full computation through all steps possible. In line with the fully deterministic model the state of $b$ at time $t$ can be described by:
\begin{equation}
\label{eq:deterministic_register_at_time_t}
\begin{aligned}
b_t &= \begin{cases}
\ptrans_t\parens*{b_{t-1}}&\text{if}\quad t > 0 \\
@ -118,8 +125,9 @@ and thus $\ptrans$ preserves valid superpositions, which finally makes predictio
\end{equation}
\subsection{Collapsing Superpositions}
\label{sec:superposition}
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}
\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:
@ -131,7 +139,7 @@ As mentioned above, a superposition can not immediately be evaluated. It rather
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}
\begin{definition}[Collapse of Superposition]
A state in superposition of basis states $\mathbf{E}=\parensc*{E_1, E_2, \dots, E_n}$ can be evaluated by collapsing it on one of its basis states. This is done by a measuring operator
@ -143,9 +151,42 @@ Let's consider the experiment of rolling a dice. Of course, for the observable \
\end{remark}
\subsection{Bit Registers in Superposition}
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 $\mathbf{B}=\parensc*{\mathbf{0}, \mathbf{1}}^n =\parensc*{\mathbf{1}, \mathbf{2}, \dots, \mathbf{N}}$.
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{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:
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
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 $\sum_{j=1}^N\sum_{i=1}^N p_i p_{ij}=\sum_{j=1}^N P\parens{\ptrans(\mathbf{b})=\mathbf{j}}=1$
\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}
\mathbf{b}_t &= \begin{cases}
\ptrans_t\parens*{\mathbf{b}_{t-1}}&\text{if}\:\: t > 0 \\
Tom Krüger
1 year ago