Superposition and the qubit

A classical bit is one of two values: 0 or 1. The bit's state is fully described by a single discrete choice. Every operation on classical bits — AND, OR, NOT, XOR — is a deterministic function from $\{0, 1\}^n$ to $\{0, 1\}^m$.

A qubit generalizes this. Its state lives in a 2-dimensional complex vector space. The two basis vectors are written

$|0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$

A general qubit state is a complex linear combination

$|\psi\rangle = \alpha |0\rangle + \beta |1\rangle, \quad \alpha, \beta \in \mathbb{C}, \quad |\alpha|^2 + |\beta|^2 = 1.$

The coefficients $\alpha$ and $\beta$ are called amplitudes. The normalization condition $|\alpha|^2 + |\beta|^2 = 1$ says the state is a unit vector — a point on the unit sphere in $\mathbb{C}^2$. A qubit's state space is continuous, not discrete.

If a qubit is in state $|\psi\rangle = \alpha |0\rangle + \beta |1\rangle$ and you measure it in the standard basis, you do not observe $\alpha$ or $\beta$. You observe one of two classical outcomes:

• Outcome 0 with probability $|\alpha|^2$.
• Outcome 1 with probability $|\beta|^2$.

The probabilities sum to 1 because of the normalization condition.

After the measurement, the qubit is no longer in superposition. Its new state is whichever basis vector was observed — $|0\rangle$ or $|1\rangle$. This is the measurement postulate, and it is structurally important: a qubit holds a continuous-valued state, but each measurement extracts only one classical bit and destroys the rest of the information.

A single qubit measured once produces a sample from a distribution; it does not let you read off $\alpha$ or $\beta$. Recovering the distribution requires preparing the same state many times and measuring each copy.

A useful geometric picture of the qubit's state space comes from rewriting an arbitrary state as

$|\psi\rangle = \cos(\theta/2) |0\rangle + e^{i\phi} \sin(\theta/2) |1\rangle,$

where $\theta \in [0, \pi]$ and $\phi \in [0, 2\pi)$. (One global phase has been absorbed into the convention because it is not physically observable.) The pair $(\theta, \phi)$ are spherical coordinates of a point on a unit sphere — the Bloch sphere.

• $|0\rangle$ is the north pole; $|1\rangle$ is the south pole.
• $\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ — equal superposition with phase 0 — sits on the equator.
• $\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ sits on the opposite point of the equator.

The Bloch sphere is the natural picture for single-qubit dynamics: gates are rotations of the sphere. It also makes one fact visible — there are continuously many qubit states (every point on the sphere), but measurement returns only a binary outcome.

A reasonable question: could a probabilistic classical bit (e.g., a coin with probability $p$ of heads) achieve the same effect? It cannot, and the reason is that quantum amplitudes are complex numbers, not non-negative real probabilities.

The difference shows up in interference. Suppose two paths each contribute amplitude $\alpha$ to a final state $|0\rangle$. The probability of measuring $|0\rangle$ depends on the sum:

$P(0) = |\alpha + \alpha|^2 = |2\alpha|^2 = 4|\alpha|^2.$

Now suppose the two paths have amplitudes $\alpha$ and $-\alpha$. Then

$P(0) = |\alpha + (-\alpha)|^2 = 0.$

The two paths cancel. A classical probabilistic system has no analog of cancellation — probabilities only add, they never subtract. Quantum algorithms are designed to use this constructive and destructive interference to make wrong answers cancel while right answers add. This is the central engine of every quantum-algorithmic speedup.

A system of $n$ qubits lives in a Hilbert space of dimension $2^n$, formed by the tensor product of $n$ copies of the single-qubit space:

$\mathcal{H} = \underbrace{\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \cdots \otimes \mathbb{C}^2}_{n \text{ times}}, \quad \dim \mathcal{H} = 2^n.$

The basis vectors are written $|x_1 x_2 \ldots x_n\rangle$ for $x_i \in \{0, 1\}$, and a general state is a complex superposition over all $2^n$ basis vectors:

$|\psi\rangle = \sum_{x \in \{0,1\}^n} \alpha_x |x\rangle, \quad \sum |\alpha_x|^2 = 1.$

The exponential. A 50-qubit state is described by $2^{50} \approx 10^{15}$ complex amplitudes. A 300-qubit state has more amplitudes than there are atoms in the observable universe. Classically simulating a general $n$-qubit state requires storing all those amplitudes — exponential resources.

A quantum computer represents and manipulates this state directly in physical hardware. The cost of doing so on the hardware is polynomial in $n$ (one physical qubit per logical qubit, before error correction). This exponential resource difference is what makes the possibility of quantum speedup interesting. Whether any specific problem yields a useful speedup is a separate algorithmic question, addressed in a later lesson.

Quantum operations between measurements are unitary linear maps — they preserve the norm of the state vector. A unitary $U$ is a complex matrix with $U^\dagger U = I$, where $U^\dagger$ is the conjugate transpose.

The Pauli matrices are the standard examples for one qubit:

$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$

$X$ is the quantum analog of classical NOT: $X|0\rangle = |1\rangle$, $X|1\rangle = |0\rangle$. $Z$ flips the relative phase of $|1\rangle$ without flipping the bit. The Hadamard gate

$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}$

maps $|0\rangle \mapsto \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|1\rangle \mapsto \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$ — it creates equal superposition. Single-qubit gates are continuous (parameterized by rotation angles); a quantum circuit is built by composing them along with multi-qubit gates.

A short proof: suppose a universal cloner existed — a unitary $U$ such that for every state $|\psi\rangle$,

$U(|\psi\rangle \otimes |0\rangle) = |\psi\rangle \otimes |\psi\rangle.$

Apply $U$ to two distinct non-orthogonal states $|\psi\rangle$ and $|\phi\rangle$. Take inner products:

$\langle \psi | \phi \rangle = \langle \psi | \phi \rangle \cdot \langle 0 | 0 \rangle = \langle \psi | \phi \rangle$

on the left, and

$\langle \psi | \phi \rangle \cdot \langle \psi | \phi \rangle = \langle \psi | \phi \rangle^2$

on the right. Equating gives $\langle \psi | \phi \rangle = \langle \psi | \phi \rangle^2$, so $\langle \psi | \phi \rangle$ is either 0 or 1. The states must be orthogonal or identical — contradicting non-orthogonality.

Consequence. An unknown quantum state cannot be copied. This rules out a class of attacks (eavesdropping that requires reading a state and forwarding a copy), supports quantum key distribution, and constrains how error correction can work: you cannot back up a qubit by duplicating it. The error-correction lesson returns to this constraint.

From the math established so far, several facts follow about single-qubit systems before any algorithm is run.

• You can prepare any state $|\psi\rangle$ from $|0\rangle$ using single-qubit gates.
• You can apply any continuous rotation on the Bloch sphere via combinations of $X$, $Y$, $Z$ rotations.
• You can measure in any basis (rotate before measuring in the standard basis).
• You cannot observe $\alpha$ and $\beta$ directly; one measurement returns one classical bit.
• You cannot copy an unknown state.
• You cannot distinguish two non-orthogonal states with certainty in a single measurement.
• You cannot perform an operation that is non-unitary between measurements; the state stays on the unit sphere of its Hilbert space.

This constrained operational profile is what every quantum algorithm has to work within. Useful computation has to come from the structure of multi-qubit superpositions and the interference between paths — both subjects of the next lesson, which moves from one qubit to many and introduces entanglement.