Superposition with Arbitrary Weights

Not every useful qubit state has equal amplitudes. A superposition with arbitrary weights lets you control exactly how much probability mass sits on $|0\rangle$ versus $|1\rangle$ — and that control is at the heart of preparing initial states for quantum algorithms.

The general single-qubit state

Any normalized single-qubit state can be written as

|\psi\rangle = \alpha|0\rangle + \beta|1\rangle,

where $\alpha, \beta \in \mathbb{C}$ and the normalization condition

|\alpha|^2 + |\beta|^2 = 1

must hold so that the Born-rule probabilities sum to one. When the amplitudes are real and non-negative, this is equivalent to writing $\alpha = \cos(\theta/2)$ and $\beta = \sin(\theta/2)$ for some angle $\theta \in [0, \pi]$ , since $\cos^2(\theta/2) + \sin^2(\theta/2) = 1$ by the Pythagorean identity.

The Ry rotation

The $R_y(\theta)$ gate rotates the Bloch-sphere state vector about the $y$ -axis by angle $\theta$ . Its matrix is

R_y(\theta) = \begin{pmatrix} \cos(\theta/2) & -\sin(\theta/2) \\ \sin(\theta/2) & \cos(\theta/2) \end{pmatrix}.

Applying it to $|0\rangle = \binom{1}{0}$ gives

R_y(\theta)|0\rangle = \begin{pmatrix} \cos(\theta/2) \\ \sin(\theta/2) \end{pmatrix} = \cos\!\left(\tfrac{\theta}{2}\right)|0\rangle + \sin\!\left(\tfrac{\theta}{2}\right)|1\rangle.

So choosing $\theta$ determines the split. When $\theta = \pi/2$ , we get $\cos(\pi/4) = \sin(\pi/4) = 1/\sqrt{2}$ — the familiar 50/50 superposition from $|0\rangle$ . Note that $R_y(\pi/2)$ and the Hadamard gate $H$ agree on $|0\rangle$ but are different unitaries ( $H$ maps $|1\rangle \to (|0\rangle - |1\rangle)/\sqrt{2}$ , whereas $R_y(\pi/2)$ maps $|1\rangle \to (-|0\rangle + |1\rangle)/\sqrt{2}$ ).

Confirming normalization

Suppose you want the state $|\psi\rangle = \tfrac{\sqrt{3}}{2}|0\rangle + \tfrac{1}{2}|1\rangle$ . Check that it is normalized:

\left|\tfrac{\sqrt{3}}{2}\right|^2 + \left|\tfrac{1}{2}\right|^2 = \tfrac{3}{4} + \tfrac{1}{4} = 1. \checkmark

To prepare it with $R_y$ , set $\cos(\theta/2) = \sqrt{3}/2$ . Because $\cos(\pi/6) = \sqrt{3}/2$ , we have $\theta/2 = \pi/6$ , so $\theta = \pi/3$ . The measurement probabilities are $P(0) = 3/4$ and $P(1) = 1/4$ .

Try it

Prepare the state $|\psi\rangle = \tfrac{\sqrt{3}}{2}|0\rangle + \tfrac{1}{2}|1\rangle$ by choosing the right angle for an $R_y$ rotation. The grader checks the full statevector, so the amplitudes must match the target exactly (up to global phase).

Run your code to see the quantum state.

After running, inspect the statevector panel: you should see amplitudes close to $0.866 \approx \sqrt{3}/2$ on $|0\rangle$ and $0.5$ on $|1\rangle$ , with probabilities $0.75$ and $0.25$ respectively.