Lab

Any pure state of one qubit can be written, up to an unobservable global phase, with two angles:

$$\lvert\psi\rangle = \cos\tfrac{\theta}{2}\,\lvert 0\rangle + e^{i\varphi}\sin\tfrac{\theta}{2}\,\lvert 1\rangle$$

Read $(\theta, \varphi)$ as polar and azimuthal angles and every state becomes a point on a unit sphere: $\lvert 0\rangle$ at the north pole, $\lvert 1\rangle$ at the south, and equal superpositions around the equator.

The useful fact is that every single-qubit gate is a rotation of that sphere. Up to global phase,

$$U = e^{-i\alpha\,\hat{n}\cdot\vec{\sigma}/2}$$

rotates the Bloch vector by angle $\alpha$ about the axis $\hat n$. That is all the simulator above does — it never stores amplitudes, only the arrow, and turns it with Rodrigues' formula:

Gate	Axis	Angle
X, Y, Z	x, y, z	π
H	(x + z)/√2	π
S	z	π/2
T	z	π/4

The probability of measuring 0 is the height of the arrow: $P(0) = \cos^2\tfrac{\theta}{2} = \tfrac{1 + z}{2}$. Try H then T repeatedly: the state walks around the equator and the probabilities never change — the information is all in the phase.

Limits: one qubit, pure states only. Entanglement has no picture here — two entangled qubits don't each have their own arrow. For that, see qsim-rs.