Qubit

Qubit (or quantum bit) is the basic unit of quantum information. The Hilbert space of a single qubit is $\mathbb{C}^2$, meaning two complex amplitudes to describe it, not one.

A classical bit can only be 0 or 1. A qubit can be in a superposition: a combination of $\lvert 0\rangle$ and $\lvert 1\rangle$ with complex coefficients called probability amplitudes. The state evolves according to the Schrödinger equation and collapses to a definite 0 or 1 only when measured.

A general qubit state is a superposition over the two computational basis states, with the normalization condition saying the total probability must be 1:

$$\lvert\psi\rangle = \begin{pmatrix}a\\b\end{pmatrix} = a\underbrace{\begin{pmatrix}1\\ 0\end{pmatrix}}_{\lvert 0\rangle} + b\underbrace{\begin{pmatrix}0\\1\end{pmatrix}}_{\lvert 1\rangle} = a\lvert 0\rangle + b\lvert 1\rangle, \qquad |a|^2 + |b|^2 = 1$$

There are infinitely many single-qubit states but three orthonormal bases are standard, corresponding to the three axes of the Bloch sphere. In each basis the two states are perfectly sharp for that observable and spread out for the others.

The Z-basis states $\lvert 0\rangle$ and $\lvert 1\rangle$ are the eigenstates of the Pauli Z gate. They behave exactly like a classical bit: each has a definite value with no superposition. Every qubit state can be written as a linear combination of them.

$$\lvert 0\rangle = \begin{pmatrix}1\\0\end{pmatrix} \qquad \lvert 1\rangle = \begin{pmatrix}0\\1\end{pmatrix}$$

The X-basis states $\lvert +\rangle$ and $\lvert -\rangle$ are the eigenstates of the Pauli X gate. They look “spread out” in the computational basis, because each is an equal superposition of $\lvert 0\rangle$ and $\lvert 1\rangle$. In the X basis they are perfectly sharp. You get from the computational basis to the X basis by applying a Hadamard.

$$\lvert +\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\1\end{pmatrix} \qquad \lvert -\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\{-1}\end{pmatrix}$$

The Y-basis states $\lvert +i\rangle$ and $\lvert -i\rangle$ are the eigenstates of the Pauli Y gate. Their coefficients are complex, so they are the first states on this page with imaginary entries. You get from the computational basis to the Y basis by applying $SH$ ($H$ first, then $S$).

$$\lvert +i\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix} \qquad \lvert -i\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\{-i}\end{pmatrix}$$

Measuring a qubit in the Z basis gives outcome 0 or 1 and collapses the state. For $\lvert\psi\rangle = a\lvert 0\rangle + b\lvert 1\rangle$, the Born rule gives the probabilities:

$$P(0) = |a|^2 \qquad P(1) = |b|^2$$

Result 0 collapses the state to $\lvert 0\rangle$; result 1 collapses it to $\lvert 1\rangle$. The probability amplitudes are destroyed by measurement and cannot be recovered. Measuring in the X or Y basis works the same way with different outcome states: to measure in the X basis, apply $H$ before measuring in Z; to measure in the Y basis, apply $S^\dagger H$ before measuring in Z.

Every pure qubit state corresponds to a unique point on the surface of the Bloch sphere, a unit sphere in $\mathbb{R}^3$. The parametrisation uses two angles $\theta \in [0, \pi]$ and $\phi \in [0, 2\pi)$:

$$\lvert\psi\rangle = \cos\tfrac{\theta}{2}\lvert 0\rangle + e^{i\phi}\sin\tfrac{\theta}{2}\lvert 1\rangle$$

The north pole ($\theta = 0$) is $\lvert 0\rangle$; the south pole ($\theta = \pi$) is $\lvert 1\rangle$. The equator ($\theta = \pi/2$) holds all equal-amplitude superpositions: $\lvert +\rangle$ at $\phi = 0$, $\lvert -\rangle$ at $\phi = \pi$, $\lvert +i\rangle$ at $\phi = \pi/2$, $\lvert -i\rangle$ at $\phi = 3\pi/2$. Single-qubit gates act as rotations of the Bloch sphere.

Single-qubit gates are $2\times 2$ unitary matrices acting on $\mathbb{C}^2$. The three Pauli gates and the Hadamard are the most common.

$$X = \begin{pmatrix}0&1\\1&0\end{pmatrix} \qquad Y = \begin{pmatrix}0&{-i}\\i&0\end{pmatrix} \qquad Z = \begin{pmatrix}1&0\\0&{-1}\end{pmatrix} \qquad H = \frac{1}{\sqrt{2}}\begin{pmatrix}1&1\\1&{-1}\end{pmatrix} \qquad S = \begin{pmatrix}1&0\\0&i\end{pmatrix} \qquad T = \begin{pmatrix}1&0\\0&e^{i\pi/4}\end{pmatrix}$$

X flips $\lvert 0\rangle \leftrightarrow \lvert 1\rangle$ (the quantum NOT gate). Z applies a $-1$ phase to $\lvert 1\rangle$ and leaves $\lvert 0\rangle$ unchanged. Y is equivalent to $iXZ$. H maps $\lvert 0\rangle \to \lvert +\rangle$ and $\lvert 1\rangle \to \lvert -\rangle$, converting between the Z and X bases. S is the square root of Z ($S^2 = Z$); T is the square root of S ($T^2 = S$). For a full list see Qubit gates.

# Requires: pip install qiskit qiskit-aer
# Run: python qubit.py
# Prepares |+⟩ and samples 1000 shots; expect roughly equal counts of '0' and '1'.
from qiskit import QuantumCircuit
from qiskit_aer import AerSimulator
 
qc = QuantumCircuit(1, 1)
qc.h(0)        # Hadamard: |0⟩ → |+⟩ = equal superposition
qc.measure(0, 0)
 
counts = AerSimulator().run(qc, shots=1000).result().get_counts()
print(counts)  # {'0': ~500, '1': ~500}