Checkpoint: State Preparation

State preparation is the first step in almost every quantum algorithm: you must load the data or the initial conditions before any computation can take place. In module 8 you have built up two families of single-qubit tools — the $R_Y$ gate for tuning amplitude magnitudes, and the phase gate $P$ for adding relative phases. This checkpoint asks you to combine both.

The general single-qubit state

Any normalized single-qubit state can be written as

|\psi\rangle = \alpha\,|0\rangle + \beta\,|1\rangle, \qquad |\alpha|^2 + |\beta|^2 = 1.

If $\alpha$ and $\beta$ are both real and non-negative, a single $R_Y$ rotation suffices:

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

When $\beta$ carries a non-trivial complex phase, you need a second gate.

Setting magnitudes and phases independently

A clean two-step recipe exploits the fact that $R_Y$ leaves all amplitudes real, so any remaining phase can be added afterward with the phase gate $P(\lambda)$ :

P(\lambda) = \begin{pmatrix} 1 & 0 \\ 0 & e^{i\lambda} \end{pmatrix}.

$P(\lambda)$ leaves the $|0\rangle$ amplitude unchanged and multiplies the $|1\rangle$ amplitude by $e^{i\lambda}$ . The recipe is therefore:

Choose $\theta = 2\arccos|\alpha|$ and apply $R_Y(\theta)$ to set the magnitude ratio.
Choose $\lambda = \arg(\beta) - \arg(\alpha)$ and apply $P(\lambda)$ to install the relative phase.

For most real applications $\alpha$ is real and positive (so $\arg\alpha = 0$ ), which simplifies the phase choice to $\lambda = \arg(\beta)$ .

A worked example

Suppose the target is

|\psi\rangle = \frac{\sqrt{3}}{2}\,|0\rangle + \frac{i}{2}\,|1\rangle.

The magnitudes are $|\alpha| = \sqrt{3}/2$ and $|\beta| = 1/2$ .

Step 1 — set the magnitudes. Solve $\cos(\theta/2) = \sqrt{3}/2$ :

\frac{\theta}{2} = \frac{\pi}{6} \quad\Longrightarrow\quad \theta = \frac{\pi}{3}.

Applying $R_Y(\pi/3)$ gives $\frac{\sqrt{3}}{2}|0\rangle + \frac{1}{2}|1\rangle$ .

Step 2 — add the phase. The target phase on $|1\rangle$ is $i = e^{i\pi/2}$ , so $\lambda = \pi/2$ . Applying $P(\pi/2)$ multiplies the $|1\rangle$ amplitude by $e^{i\pi/2} = i$ :

P(\pi/2)\!\left(\frac{\sqrt{3}}{2}|0\rangle + \frac{1}{2}|1\rangle\right) = \frac{\sqrt{3}}{2}|0\rangle + \frac{i}{2}|1\rangle. \checkmark

You can verify normalization: $(\sqrt{3}/2)^2 + (1/2)^2 = 3/4 + 1/4 = 1$ .

Try it

Prepare the state $\frac{\sqrt{3}}{2}|0\rangle + \frac{i}{2}|1\rangle$ using two gates. The grader checks the full statevector (up to global phase), so both the amplitude magnitudes and the relative phase must be correct.

Run your code to see the quantum state.

After pressing Run, inspect the statevector readout: the $|0\rangle$ amplitude should be $\approx 0.866$ (real) and the $|1\rangle$ amplitude should be $\approx 0.500i$ (purely imaginary).