Preparing an Arbitrary State

Every pure single-qubit state corresponds to exactly one point on the Bloch sphere. The north pole is $|0\rangle$ , the south pole is $|1\rangle$ , and every other point encodes a superposition. Two angles fully specify a point: the polar angle $\theta$ (latitude from the north pole) and the azimuthal angle $\phi$ (longitude). Parametrically,

|\psi\rangle = \cos\!\tfrac{\theta}{2}\,|0\rangle + e^{i\phi}\sin\!\tfrac{\theta}{2}\,|1\rangle.

The factor of $1/2$ on the angles is a convention: it ensures the antipodal points $\theta = 0$ and $\theta = \pi$ give the orthogonal pair $|0\rangle$ and $|1\rangle$ .

Reaching any state with two rotations

To prepare $|\psi(\theta, \phi)\rangle$ starting from $|0\rangle$ , apply two gates in sequence:

$R_Y(\theta)$ rotates around the $y$ -axis and sets the amplitude ratio:

R_Y(\theta)|0\rangle = \cos\!\tfrac{\theta}{2}\,|0\rangle + \sin\!\tfrac{\theta}{2}\,|1\rangle.

Both amplitudes are real after this step — the state lives on the meridian $\phi = 0$ .

$R_Z(\phi)$ rotates around the $z$ -axis and dials in the relative phase:

R_Z(\phi)\bigl(\alpha|0\rangle + \beta|1\rangle\bigr) = e^{-i\phi/2}\alpha\,|0\rangle + e^{i\phi/2}\beta\,|1\rangle.

Up to a global phase $e^{-i\phi/2}$ (which is unobservable), this multiplies the $|1\rangle$ component by $e^{i\phi}$ , placing the state at longitude $\phi$ on the Bloch sphere.

The combined action is

R_Z(\phi)\,R_Y(\theta)|0\rangle = e^{-i\phi/2}\bigl(\cos\!\tfrac{\theta}{2}\,|0\rangle + e^{i\phi}\sin\!\tfrac{\theta}{2}\,|1\rangle\bigr),

which equals the target state up to the global phase $e^{-i\phi/2}$ . Global phases are physically equivalent because they cancel in every Born-rule probability and every expectation value.

A worked example

Suppose the target is $\theta = \pi/3$ (polar angle) and $\phi = \pi/3$ (azimuthal angle):

|\psi\rangle = \cos\!\tfrac{\pi}{6}\,|0\rangle + e^{i\pi/3}\sin\!\tfrac{\pi}{6}\,|1\rangle = \tfrac{\sqrt{3}}{2}\,|0\rangle + \tfrac{1}{2}e^{i\pi/3}|1\rangle.

The probabilities are $P(0) = \cos^2(\pi/6) = 3/4$ and $P(1) = \sin^2(\pi/6) = 1/4$ , which come from the amplitude magnitudes $\tfrac{\sqrt{3}}{2} \approx 0.866$ and $\tfrac{1}{2} = 0.5$ .

To realise this state, first apply $R_Y(\pi/3)$ to set the amplitude ratio, then $R_Z(\pi/3)$ to attach the phase $e^{i\pi/3}$ to the $|1\rangle$ component.

Try it

Prepare the state with polar angle $\theta = \pi/3$ and azimuthal angle $\phi = \pi/3$ :

|\psi\rangle = \cos\!\tfrac{\pi}{6}\,|0\rangle + e^{i\pi/3}\sin\!\tfrac{\pi}{6}\,|1\rangle.

Use $R_Y$ to fix the amplitudes, then $R_Z$ to set the phase. The grader checks the full statevector.

Run your code to see the quantum state.

After pressing Run, the statevector panel should show the $|0\rangle$ amplitude as a complex number near $0.75 - 0.433i$ (magnitude $\approx 0.866$ ) and the $|1\rangle$ amplitude near $0.433 + 0.25i$ (magnitude $0.5$ , argument $\pi/6 \approx 0.524$ rad). The relative phase between the two amplitudes is $\pi/3$ , which is the azimuthal angle $\phi$ of the target state. The probabilities tab should show $P(0) = 0.75$ and $P(1) = 0.25$ .