The Exercise

The states $|+\rangle$ and $|-\rangle$ of a qubit are the eigenstates of the Pauli operator $X$. Whether two states differ by a relative phase depends on the basis.

A Qubit

A quantum system $S$ is a physical system described by its state space, a finite dimensional complex Hilbert space.^✦ The qubit is the two level quantum system whose state space is $\mathbb{C}^2$ itself, with its standard inner product.^✦ A pure state $|\psi\rangle$ of a quantum system $S$ is a vector of unit norm, $\lVert \psi \rVert = 1$, in the state space of $S$.^✦

The pure states $|+\rangle$ and $|-\rangle$ of the qubit are the column vectors of $\mathbb{C}^2$ whose entries are the normalization constant $1/\sqrt{2}$ or its negative: $$|+\rangle = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \end{pmatrix}, \qquad |-\rangle = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \end{pmatrix}.$$(1)

Relative Phase

Two pure states $|\psi\rangle$ and $|\varphi\rangle$ of a quantum system $S$ differ by a relative phase in a family of pure states $|e_k\rangle$ of $S$ if for each $k$ there is a real number $\theta_k$ with $$\langle e_k | \psi \rangle = e^{i\theta_k} \, \langle e_k | \varphi \rangle .$$(2)^✦

The basis $\{|+\rangle, |-\rangle\}$ of the qubit is the family of two pure states $$|e_0\rangle = |+\rangle, \qquad |e_1\rangle = |-\rangle .$$(3)^✦

The Claim

What Is Checked

The claim is stated over the qubit, its pure states $|+\rangle$ and $|-\rangle$ given by the vectors (1), the basis (3) that they form, and the relation (2) of differing by a relative phase in a family of pure states.

About these constructions the library proves that both vectors (1) have unit norm, $(1/\sqrt{2})^2 + (\pm 1/\sqrt{2})^2 = 1$, so that $|+\rangle$ and $|-\rangle$ are pure states of the qubit.