\title{A Brief Introduction to Quantum Computation}
\maketitle
\section{From Randomness to Qubits}
\section{A Simple Computational Model}
What are Qubits? That's usually the first question getting addressed in any introduction to quantum computing, for a good reason. If we want to construct a new computational model, we first need to define the most basic building block: a single \emph{bit} of information. In classical computer science, the decision on how to define this smallest building block of information seems quite straight forward. We just take the most basic logical fact: either something is \emph{true} or \emph{false}, either 1 or 0. We have a name for an object holding this information: a \textbf{Bit}. Let's envision a computational model based on logical gates. Such a gate has one or more inputs and an output, with each either being \emph{true} or \emph{false}. Now consider a bit $b$ and a gate $f : \{0, 1\}\to\{0, 1\}$. We have a \emph{bit} of information $b$ and can get another \emph{bit} of information $b' \coloneqq f(b)$. In a final third step, we introduce a timescale, which means that now our \emph{bit} of information is time dependent. It can have different values at different times. To make it easier, we choose a discrete timescale. Our Bit $b$ has a distinct value on each point on the timescale. A value of a bit can only be changed in between time steps, by applying a logical gate to it:
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 $`F: \{0,1\}^n \to\{0,1\}^n$.
\subsection{A bit of randomness}
\section{A Bit of Randomness}
\subsection{Single Bits in Superposition}
Many real world problems are believed (if $\mathbf{P}\neq\mathbf{NP}$) to don't be efficiently solvable on fully deterministic computers like the model described above. 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 $p(b)\in\{0,1\}$. Or to reformulate this idea: Depending on the value of $b$, the output of $p(b)$ follows one of two Bernoulli trials with success probabilities $0\leq p_0, p_1\leq1$.
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.
$$
\begin{aligned}
P(p(b) = 1 \:|\: b = 0) &= p_0 \\
P(p(b) = 1 \:|\: b = 1) &= p_1
p_{00}\coloneqq P(\ptrans(b) = 0 \:|\: b = 0) \\
p_{01}\coloneqq P(\ptrans(b) = 1 \:|\: b = 0) \\
p_{10}\coloneqq P(\ptrans(b) = 0 \:|\: b = 1) \\
p_{11}\coloneqq P(\ptrans(b) = 1 \:|\: b = 1) \\
\end{aligned}
$$
Note that we regain our deterministic transition function $f$ from above, if we restrict the probabilities: $p_0, p_1\in\{0,1\}$. At this point, we can randomize our computation from above as follows:
Note that we regain our deterministic transition function $f$ from $\ptrans$, if we restrict the probabilities: $p_{00}, p_{10}\in\{0,1\}$. At this point, we can randomize our computation from above as follows:
$$
\begin{matrix}
\text{Bit}& b &\stackrel{p_1}{\to}&b &\stackrel{p_2}{\to}&\cdots&\to&b &\stackrel{p_k}{\to}&b \\
\text{Bit}& b &\stackrel{\ptrans_1}{\to}&b &\stackrel{\ptrans_2}{\to}&\cdots&\to&b &\stackrel{\ptrans_k}{\to}&b \\
Let's have a look at the state of $b$ after the first transition. In the deterministic model, we know with certainty that at this point in time, $b$ will have the value $f_1(b)$. In a probabilistic model, we can not predict the value of $b$ at time $t_1$ with 100\% certainty. In the terminology of probability theory, a probabilistic state transition or even the whole computation would be an \emph{experiment} and the value of bit $b$ at time $t$ would be described by a \emph{random variable}$X_t$. Random variables are defined to take a value out of a set of predefined value options $\Omega=\{\omega_1, \dots, \omega_n\}$ with certain probabilities $p_1,\dots,p_n$ for each value. Only after we perform the experiment and \emph{observe} its outcome, we get a specific value $x_t$ of the random variable $X_t$. We say that $x_t$ is a \emph{random sample} or realization of $X_t$. If we don't want to or can't sample (perform) the experiment, we still could compute the \emph{expected value}$E(X_t)=\sum_i p_i\omega_i$ (if $\Omega$ mathematically allows for such operations).
Let's return to our example: The expected state of $b$ at time $t_1$ is $E(p_1(b)\:|\: b =0)=(1- p_0)\cdot0+ p_0\cdot1$, if we assume that $b$ was in state 0 at time $t_0$. We say that $b$ is in a (linear) \emph{superposition} of both states 0 and 1.
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.
Let's return to our example: Just as in the deterministic case we would like to predict the state of $b$ after the transition $\ptrans_t$. For this we want to calculate the expected state of b at time $t$. Let $p^t_{ij}$ be the transition probabilities of $\ptrans_t$, furthermore $p^t_{b=x}$ denotes the probability of $b$ being in state $x$ at time $t$. Now we have:
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:
Until now, we have not given an explicit definition of the transition function $\ptrans$, apart from describing its effect. This is partly the case because we were lacking a formalism to describe uncertain states, so there was no direct way to describe the output of $\ptrans\parens{b}$. The other big problem would have been the question of how to handle an uncertain input state. Building on the superposition formalism $\ptrans\parens*{b}$ can be defined as a linear function:
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}
\begin{aligned}
b_t &= \begin{cases}
\ptrans_t\parens*{b_{t-1}}&\text{if}\quad t > 0 \\
Tom Krüger
2 years ago