Computational Basis States |0⟩ and |1⟩

Overview

Quantum computing uses a small set of standard reference states to describe qubits. The most important are the computational basis states $|0\rangle$ and $|1\rangle$. This page defines them formally, shows how to represent them as column vectors, and explains why “basis choice” matters.

Intuition

When you work with ordinary coordinates, you choose axes: $x$ and $y$. A point in the plane can be described by numbers only after you decide what counts as the $x$-direction and the $y$-direction.

A basis plays the same role for qubits. The symbols $|0\rangle$ and $|1\rangle$ name two reference directions. Once those directions are chosen, any state can be described by “coordinates” (amplitudes) relative to them.

In this page we focus on the computational basis because it corresponds to the most common measurement: you ask “is the qubit 0 or 1?” and the measurement returns one of those two outcomes.

Formal Description

The computational basis is a pair of orthonormal vectors. “Orthonormal” means two things:

• Each basis vector has length 1.
• The vectors are perpendicular (their inner product is 0).

In Dirac notation we write these vectors as $|0\rangle$ and $|1\rangle$.

A concrete way to represent them is as column vectors:

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

Once you fix a basis, any qubit state can be written as a linear combination of the basis vectors:

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

The phrase “basis choice matters” means that the same physical state can have different coordinates depending on what reference vectors you choose. In this tutorial series, the computational basis is the default reference unless explicitly stated otherwise.

Measurement in the computational basis produces outcomes labeled 0 and 1. If

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

then measuring in this basis yields 0 with probability $|\alpha|^2$ and 1 with probability $|\beta|^2$.

Worked Example

Let

$$|\psi\rangle = \tfrac{3}{5}|0\rangle + \tfrac{4}{5}|1\rangle.$$

This is normalized because $(\tfrac{3}{5})^2 + (\tfrac{4}{5})^2 = 1$. In column-vector form,

$$|\psi\rangle = \begin{pmatrix}3/5\\4/5\end{pmatrix}.$$

If you measure in the computational basis, the probability of outcome 0 is $(3/5)^2 = 9/25$, and the probability of outcome 1 is $(4/5)^2 = 16/25$.

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What’s Next

Now that we have a precise definition of the computational basis, we can define what a quantum state is in general: a state vector with complex amplitudes and a normalization rule. That broader view is the next page.