The Exercise

The Hadamard gate $H$ on a single qubit can be built from rotations about the $\hat x$ and $\hat y$ axes. A rotation by $\pi/2$ about $\hat y$, followed by a rotation by $\pi$ about $\hat x$, gives $H$ up to the global phase $e^{i\pi/2} = i$, a factor common to every state and without observable effect.^✦

The library states the decomposition for the matrices of the gates, with the rotations built as exponentials of the Pauli matrices defined below.

Gates and Rotations of a Qubit

The Hadamard matrix $H$ is the complex $2 \times 2$ matrix $$H = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}.$$(1)^✦

The Pauli matrices $X$, $Y$ and $Z$ are the complex $2 \times 2$ matrices $$X = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \qquad Y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \qquad Z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},$$(2) with $i$ the imaginary unit.

Let $\vec\sigma = (X, Y, Z)$ be the Pauli vector. For a real vector $\vec n = (n_x, n_y, n_z)$ in three dimensions, the matrix $\vec n \cdot \vec\sigma$ is the complex $2 \times 2$ matrix $n_x X + n_y Y + n_z Z$.^✦ The rotation $R_{\vec n}(\theta)$ about the axis $\vec n$ by a real angle $\theta$ is the complex $2 \times 2$ matrix $$R_{\vec n}(\theta) = \exp\!\left(-\frac{i\theta}{2}\, \vec n \cdot \vec\sigma\right).$$(3)^✦ For a unit vector $\hat n$ this is the rotation of a qubit by the angle $\theta$ about $\hat n$.

The rotation $R_x(\theta)$ by an angle $\theta$ about the $\hat x$ axis is $R_{\hat n}(\theta)$ with $\hat n = (1, 0, 0)$, and the rotation $R_y(\theta)$ about the $\hat y$ axis is $R_{\hat n}(\theta)$ with $\hat n = (0, 1, 0)$: $$R_x(\theta) = e^{-i\theta X/2}, \qquad R_y(\theta) = e^{-i\theta Y/2}.$$(4)

The Claim

In a product of gates the one that acts first stands on the right, next to the state.

What Is Checked

The claim is stated over the Hadamard matrix (1) and the rotations $R_x$ and $R_y$ of (4).