Discrete vs Continuous States

Quantum mechanics describes states in a Hilbert space — a complete, complex inner-product space. That Hilbert space can be finite-dimensional or infinite-dimensional, and the contrast between the two cases is at the heart of the difference between a qubit and a particle with a wavefunction.

The qubit: a two-dimensional Hilbert space

A qubit lives in $\mathbb{C}^2$ , the simplest non-trivial quantum system. Its most general pure state is the superposition

$|\psi\rangle = \alpha\,|0\rangle + \beta\,|1\rangle,$

where $\alpha, \beta \in \mathbb{C}$ are the amplitudes and the normalisation condition demands

$|\alpha|^2 + |\beta|^2 = 1.$

The Born rule is immediate: a measurement in the computational basis yields outcome $0$ with probability $|\alpha|^2$ and outcome $1$ with probability $|\beta|^2$ . Because the Hilbert space has dimension $N = 2$ , there are exactly two possible outcomes, and the state is fully specified by two complex numbers (with one real constraint from normalisation and one irrelevant global phase, leaving two real degrees of freedom). More generally, an $n$ -qubit register lives in $\mathbb{C}^{2^n}$ , with $2^n$ discrete basis states.

The particle on a line: an infinite-dimensional Hilbert space

A particle moving freely in one dimension has a position that can take any value in $\mathbb{R}$ . Its state is encoded not in a finite list of amplitudes but in a wavefunction $\psi(x)$ , a complex-valued function over a continuous variable. Think of $\psi(x)$ as an uncountably infinite column of amplitudes — one complex number for every point $x$ .

The Hilbert space is $L^2(\mathbb{R})$ , the set of all square-integrable functions on the real line. Its inner product is

$\langle \phi | \psi \rangle = \int_{-\infty}^{\infty} \phi^*(x)\,\psi(x)\, dx,$

the continuous analogue of the discrete sum above. Normalisation reads

$\langle \psi | \psi \rangle = \int_{-\infty}^{\infty} |\psi(x)|^2\, dx = 1.$

Because the dimension is infinite there is no finite list of numbers that represents the state exactly. You need the entire function. (A technical aside: although the position label $x$ ranges over an uncountable continuum, $L^2(\mathbb{R})$ is separable — it has a countable orthonormal basis, for example the harmonic-oscillator eigenfunctions $\{h_n(x)\}_{n=0}^{\infty}$ . So its Hilbert-space dimension is countably infinite, $\aleph_0$ , the same as the discrete space $\ell^2$ of square-summable sequences. The continuum lives in the labels, not in the dimension.)

The Born rule in each language

The translation between the two cases is direct:

| Feature | Qubit (discrete) | Particle on a line (continuous) | |---------|-----------------|----------------------------------| | State space | $\mathbb{C}^N$ | $L^2(\mathbb{R})$ | | State vector | $|\psi\rangle = \sum_{k} \alpha_k |k\rangle$ | $\psi(x)$ | | Amplitude for outcome | $\alpha_k$ | $\psi(x)$ | | Probability rule | $P(k) = |\alpha_k|^2$ | $P(a \le x \le b) = \int_a^b |\psi(x)|^2\, dx$ | | Normalisation | $\sum_k |\alpha_k|^2 = 1$ | $\int_{-\infty}^{\infty} |\psi(x)|^2\, dx = 1$ |

In the continuous case you cannot ask "what is the probability of finding the particle at exactly $x = 0$ ?" The answer is always zero, because a single point has measure zero. The physically meaningful quantity is the probability density $|\psi(x)|^2$ , which has units of $\text{length}^{-1}$ in one dimension. Probability lives in integrals, not in the value of $\psi$ at a point.

Why this matters for quantum computing

Most of the circuits you build in this course operate entirely in the finite-dimensional setting: $n$ qubits, $2^n$ basis states, complex amplitudes, discrete measurement outcomes. The infinite- dimensional wavefunction language is needed when you want to describe how a physical qubit — say, an electron spin — connects to the underlying quantum mechanics of a real particle.

Spin- $\tfrac{1}{2}$ is the cleanest bridge: the spin degree of freedom of an electron is a genuine two-level system with Hilbert space $\mathbb{C}^2$ , so the qubit formalism is not an approximation for spin — it is exact. By contrast, the position and momentum of the same electron live in $L^2(\mathbb{R})$ , requiring the continuous formalism. In practice, a full description of the electron couples both Hilbert spaces via a tensor product, but for quantum-computing purposes the spin (qubit) part is almost always handled in isolation.