Hadamard as a Rotation

Every single-qubit gate is a rotation of the Bloch sphere. The Pauli gates $X$ , $Y$ , and $Z$ rotate 180° about the $x$ -, $y$ -, and $z$ -axes respectively. The Hadamard gate $H$ fits the same pattern — but its rotation axis is the diagonal direction halfway between $+x$ and $+z$ .

The rotation axis

The normalised axis is $\hat{n} = \frac{1}{\sqrt{2}}(1, 0, 1)$ , pointing equally along $+x$ and $+z$ . A rotation by angle $\theta$ about $\hat{n}$ is described by

R_{\hat{n}}(\theta) = \cos\!\frac{\theta}{2}\, I - i\sin\!\frac{\theta}{2}\,(\hat{n} \cdot \vec{\sigma}),

where $\vec{\sigma} = (X, Y, Z)$ is the vector of Pauli matrices. For $H$ , the angle is $\theta = \pi$ (180°), so $\cos(\pi/2) = 0$ and $\sin(\pi/2) = 1$ , which gives

H = -i\,(\hat{n} \cdot \vec{\sigma}) = -i \cdot \frac{X + Z}{\sqrt{2}} = -i \cdot \frac{1}{\sqrt{2}}\begin{pmatrix}1 & 1 \\ 1 & -1\end{pmatrix}.

The prefactor $-i$ is a global phase and has no observable effect, so we conventionally write

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

What the rotation does geometrically

Because $H$ rotates 180° about the axis $\hat{n}$ , it swaps any two points that are mirror images of each other across $\hat{n}$ . In particular:

The north pole $|0\rangle$ (the $+z$ point) is sent to the $+x$ point, which is $|+\rangle$ .
The $+x$ point $|+\rangle$ is sent back to $|0\rangle$ .
The south pole $|1\rangle$ is sent to the $-x$ point, which is $|-\rangle$ .

This explains a key identity: applying $H$ twice returns every state to itself ( $H^2 = I$ ), because two 180° rotations about the same axis cancel.

The Z–X basis swap

One elegant consequence of the Hadamard rotation axis is that $H$ exchanges the $Z$ eigenstates with the $X$ eigenstates:

H|0\rangle = |{+}\rangle, \quad H|1\rangle = |{-}\rangle, \quad H|{+}\rangle = |0\rangle, \quad H|{-}\rangle = |1\rangle.

Conjugation by $H$ therefore converts a $Z$ measurement into an $X$ measurement and vice versa. This is not coincidence — it follows directly from the rotation axis being exactly midway between the two axes, so the 180° rotation swaps them.

Try it

Implement the Hadamard gate. The grader compares the full unitary matrix, so an empty circuit will not pass — you must place the correct rotation.

Run your code to see the quantum state.