Building H from Rotations

The Hadamard gate $H$ looks like a special object in most introductions, defined by its matrix and left at that. But it is also a perfectly ordinary rotation — one that can be expressed as a short sequence of $R_Y$ and $R_Z$ gates. Understanding this connection deepens your intuition for the Bloch sphere and is the first step toward decomposing arbitrary single-qubit unitaries.

H as a Bloch-sphere rotation

On the Bloch sphere every single-qubit gate is a rotation. The Hadamard gate corresponds to a rotation by $\pi$ around the axis $\hat{n} = \tfrac{1}{\sqrt{2}}(\hat{x} + \hat{z})$ , the diagonal between the $x$ - and $z$ -axes. Applying it once maps north pole to equator and back, which is why $H^2 = I$ .

Decomposing H into RY and RZ

Any rotation around a tilted axis can be decomposed into a sequence of axis-aligned rotations. One compact decomposition of $H$ is:

H = e^{i\phi}\, R_Y\!\left(\tfrac{\pi}{2}\right)\, R_Z(\pi),

where $e^{i\phi}$ is an unobservable global phase.

We can verify this by multiplying the matrices. The two rotation matrices are

R_Y\!\left(\tfrac{\pi}{2}\right) = \begin{pmatrix} \cos\tfrac{\pi}{4} & -\sin\tfrac{\pi}{4} \\[4pt] \sin\tfrac{\pi}{4} & \cos\tfrac{\pi}{4} \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix}, \qquad R_Z(\pi) = \begin{pmatrix} e^{-i\pi/2} & 0 \\ 0 & e^{i\pi/2} \end{pmatrix} = \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix}.

Their product (applied right-to-left, $R_Z$ first then $R_Y$ ) is

R_Y\!\left(\tfrac{\pi}{2}\right) R_Z(\pi) = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -i & 0 \\ 0 & i \end{pmatrix} = \frac{1}{\sqrt{2}}\begin{pmatrix} -i & -i \\ -i & i \end{pmatrix} = \frac{-i}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} = -i\,H.

The scalar factor $-i$ is a global phase — it has no effect on any measurement outcome or on the action of the gate as a unitary operation. Up to that phase, the two-gate sequence is exactly $H$ .

Why this matters

No real quantum processor ships a native Hadamard gate; hardware exposes a small set of rotation primitives. Compilers translate higher-level gates — including $H$ — into sequences of native rotations. The decomposition above is exactly the kind of identity a quantum compiler uses when targeting a backend whose native single-qubit gate is, say, $R_Z$ and $R_Y$ (or $R_Z$ and $\sqrt{X}$ ).

Try it

Implement $H$ using only rotation gates. The grader checks the full unitary matrix up to global phase — doing nothing will fail because the identity differs from $H$ .

Run your code to see the quantum state.