Rotations RX, RY, RZ

The Bloch sphere is more than a picture — it is a map of every possible single-qubit state, and the rotation gates $R_X$ , $R_Y$ , $R_Z$ are the tools you use to move freely across its surface.

The rotation matrices

Each rotation gate is parameterized by an angle $\theta$ and rotates the Bloch vector by that angle around the named axis. Their matrices are exponentials of the Pauli operators:

R_X(\theta) = e^{-i\theta X/2} = \cos\!\tfrac{\theta}{2}\,I - i\sin\!\tfrac{\theta}{2}\,X = \begin{pmatrix} \cos\tfrac{\theta}{2} & -i\sin\tfrac{\theta}{2} \\ -i\sin\tfrac{\theta}{2} & \cos\tfrac{\theta}{2} \end{pmatrix},

R_Y(\theta) = e^{-i\theta Y/2} = \cos\!\tfrac{\theta}{2}\,I - i\sin\!\tfrac{\theta}{2}\,Y = \begin{pmatrix} \cos\tfrac{\theta}{2} & -\sin\tfrac{\theta}{2} \\ \sin\tfrac{\theta}{2} & \cos\tfrac{\theta}{2} \end{pmatrix},

R_Z(\theta) = e^{-i\theta Z/2} = \cos\!\tfrac{\theta}{2}\,I - i\sin\!\tfrac{\theta}{2}\,Z = \begin{pmatrix} e^{-i\theta/2} & 0 \\ 0 & e^{i\theta/2} \end{pmatrix}.

All three have the same pattern: rotating by $\theta = 2\pi$ returns the original state up to a global phase of $-1$ (unobservable), and rotating by $\theta = \pi$ implements a Pauli gate up to global phase (e.g. $R_X(\pi) = -iX$ ).

A concrete example: RY(π/2) on |0⟩

Apply $R_Y(\pi/2)$ to $|0\rangle = \binom{1}{0}$ :

R_Y\!\left(\tfrac{\pi}{2}\right)|0\rangle = \begin{pmatrix} \cos\tfrac{\pi}{4} & -\sin\tfrac{\pi}{4} \\ \sin\tfrac{\pi}{4} & \cos\tfrac{\pi}{4} \end{pmatrix} \binom{1}{0} = \frac{1}{\sqrt{2}}\binom{1}{1} = \frac{|0\rangle + |1\rangle}{\sqrt{2}} = |+\rangle.

This is precisely the same state as $H|0\rangle$ — no surprise, because $H$ is itself a rotation by $\pi$ around the diagonal $X{+}Z$ axis.

How RX and RZ behave

$R_X(\theta)$ rotates around the $x$ -axis. Starting from $|0\rangle$ (north pole) it sweeps toward $|1\rangle$ through a path that picks up imaginary components: at $\theta = \pi$ you get $-i|1\rangle$ , and at $\theta = \pi/2$ the qubit lands on the equator at the $-y$ pole of the Bloch sphere — the state is $\tfrac{1}{\sqrt{2}}(|0\rangle - i|1\rangle)$ , which has a factor of $-i$ on the $|1\rangle$ amplitude.

$R_Z(\theta)$ rotates around the $z$ -axis. Because $|0\rangle$ sits on the north pole of the $z$ -axis, $R_Z(\theta)|0\rangle = e^{-i\theta/2}|0\rangle$ — a global phase only. The state $|+\rangle = \tfrac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$ , however, is on the equator, so $R_Z$ moves it visibly: $R_Z(\theta)|+\rangle = \tfrac{1}{\sqrt{2}}(e^{-i\theta/2}|0\rangle + e^{i\theta/2}|1\rangle)$ .

Together $R_X$ , $R_Y$ , and $R_Z$ (or any two of them plus a phase gate) are sufficient to express every single-qubit unitary — they form a universal single-qubit gate set.

Try it

Apply $R_Y(\pi/2)$ to qubit 0. After running, inspect the statevector: the two amplitudes should both be $\approx 0.707 = \tfrac{1}{\sqrt{2}}$ (no imaginary parts).

Run your code to see the quantum state.