Rotation gates: Rx, Ry, Rz

Every gate you’ve met so far is a fixed rotation: $H$ rotates by a specific angle, the Paulis rotate by exactly $\pi$, $S$ and $T$ rotate by $\pi/2$ and $\pi/4$. There’s no dial. You can’t ask for “half a Hadamard” or “a 37-degree X.”

The rotation gates fix that. There are three of them — one for each axis — and each one takes an angle $\theta$ as a parameter:

$R_x(\theta) \qquad R_y(\theta) \qquad R_z(\theta)$

They rotate the Bloch-sphere state by angle $\theta$ around the corresponding axis. Change $\theta$, get a different gate.

The formulas

The matrix forms come from exponentiating the Pauli gates:

$R_x(\theta) = e^{-i\theta X / 2} = \cos\!\frac{\theta}{2}\,I - i\sin\!\frac{\theta}{2}\,X$

$R_y(\theta) = e^{-i\theta Y / 2} = \cos\!\frac{\theta}{2}\,I - i\sin\!\frac{\theta}{2}\,Y$

$R_z(\theta) = e^{-i\theta Z / 2} = \cos\!\frac{\theta}{2}\,I - i\sin\!\frac{\theta}{2}\,Z$

Don’t let the exponentials intimidate you. Each one is a 2x2 matrix you could compute by hand. For example:

$R_x(\theta) = \begin{pmatrix} \cos(\theta/2) & -i\sin(\theta/2) \\ -i\sin(\theta/2) & \cos(\theta/2) \end{pmatrix}$

The thing you should notice is the $\theta/2$ in the formula. That’s the half-angle convention from Lesson 2.2 of the Bloch sphere — rotations on the Bloch sphere are by angle $\theta$, but the corresponding rotation of the state vector (which lives in a different space) is by angle $\theta/2$.

So $R_x(2\pi)$ is a full rotation on the Bloch sphere (the state should come back to where it started) but the matrix formula gives:

$R_x(2\pi) = \cos(\pi)\, I - i\sin(\pi)\, X = -I$

A minus sign! The state vector has picked up a global phase of $-1$, which is invisible (it’s a global phase) but still mathematically real. You have to go around twice — $R_x(4\pi)$ — to get back to exactly $+I$. This is the famous spin-$\frac{1}{2}$ spinor property, and it’s responsible for a lot of weirdness in quantum mechanics.

Drag the slider

Things to try:

1. Pick $R_x$ and drag the angle from $0$ to $\pi$. Watch the state on the Bloch sphere rotate from $|0\rangle$ (north pole) smoothly down to $|1\rangle$ (south pole), passing through the $+y$ equator point at $\theta = \pi/2$.
2. Switch to $R_y$ and do the same. Same kind of rotation, but now it goes through the $+x$ equator point instead.
3. Switch to $R_z$, starting from $|0\rangle$. Drag the angle. The state stays at $|0\rangle$! Why?
4. Starting from $|0\rangle$, apply $R_x(\pi/2)$. Where does the state end up? What’s the corresponding Cartesian point on the Bloch sphere?
5. Snap $\theta$ to $\pi$. The gate is now $R_x(\pi)$, which should equal $X$ (up to a global phase). Check the state — it’s $|1\rangle$, confirming that $R_x(\pi) \approx X$.

The last point is the key realization: the discrete gates you met earlier are all special cases of the rotation gates at specific angles.

The connection to Pauli gates

$R_x(\pi) = -iX \qquad R_y(\pi) = -iY \qquad R_z(\pi) = -iZ$

Apart from a global phase of $-i$ (which doesn’t matter physically), each rotation at angle $\pi$ is the corresponding Pauli. Similarly:

$R_z(\pi/2) = e^{-i\pi/4}\, S \qquad R_z(\pi/4) = e^{-i\pi/8}\, T$

$S$ and $T$ are also just $R_z$ rotations at smaller angles. Once you have $R_x$, $R_y$, $R_z$ as continuous gates, you can recover all the discrete gates as special cases.

Rotation gates are what real hardware runs

On most quantum processors (superconducting, trapped-ion, neutral-atom), you don’t get to just “apply H” or “apply X” directly. The hardware implements gates by pulsing microwaves or lasers at the qubit, and those pulses naturally produce continuous rotations — often $R_z$ is “free” (you just relabel your frame) and $R_x$ or $R_y$ is done with a real pulse.

So when you write circuit.h(q) in Qiskit, the hardware compiler decomposes it into a sequence of rotations and executes those. Rotations are the native language.

What’s next

You now have continuous control over single-qubit rotations. In the next lesson we’ll see that this is enough to reach every single-qubit state and every single-qubit gate — rotation gates are “universal” for one qubit.