RX and the X Axis

The Bloch sphere pictures a qubit's state as a point on the surface of a unit sphere. The north pole is $|0\rangle$ , the south pole is $|1\rangle$ , and the equator holds all equal-weight superpositions. Any single-qubit gate is a rotation of this sphere. The $RX(\theta)$ gate is the specialised rotation that spins the Bloch vector around the $x$ axis by angle $\theta$ .

The matrix

The $RX$ rotation gate is defined as

RX(\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}.

The factor of $\tfrac{1}{2}$ appears because a rotation by angle $\theta$ on the Bloch sphere corresponds to a rotation by $\tfrac{\theta}{2}$ in the two-dimensional complex vector space of qubits (the spin- $\tfrac{1}{2}$ representation).

Substituting special values makes the connection concrete:

$\theta = 0$ : $RX(0) = I$ — no rotation, state unchanged.
$\theta = \pi$ : $\cos\tfrac{\pi}{2} = 0$ , $\sin\tfrac{\pi}{2} = 1$ , so $RX(\pi) = -iX$ . Acting on $|0\rangle$ gives $-i|1\rangle$ , which differs from $|1\rangle$ only by the global phase $-i$ — physically identical to $|1\rangle$ .
$\theta = \pi/2$ : $RX(\pi/2)$ rotates the north pole half-way to the equator, landing on the $+y$ axis of the Bloch sphere.

How it differs from X

The Pauli $X$ gate is exactly $iRX(\pi)$ : it flips $|0\rangle \leftrightarrow |1\rangle$ with no extra freedom. $RX(\theta)$ generalises this: by choosing $\theta$ you can reach any point along the arc from $|0\rangle$ to $|1\rangle$ on the Bloch sphere. This continuous control is what makes rotation gates central to building arbitrary single-qubit operations — any single-qubit unitary can be decomposed into three rotations (the Euler angle decomposition).

Try it

Apply $RX(\pi)$ to qubit 0. The grader checks the full state vector, so the amplitudes must match those predicted by the matrix above: $|0\rangle \to -i|1\rangle$ .

Run your code to see the quantum state.

After running, inspect the State Vector panel: you should see amplitude $-i \approx -0.000 - 1.000i$ on the $|1\rangle$ component, confirming the half-turn around the $x$ axis.