Born Rule in Practice

Every time you measure a qubit, the quantum state you built — with all its amplitudes — collapses to a single classical outcome. The Born rule is the bridge from those amplitudes to the probabilities you can observe.

From amplitudes to probabilities

A one-qubit state is a superposition

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

where the complex numbers $\alpha$ and $\beta$ satisfy $|\alpha|^2 + |\beta|^2 = 1$. Measuring in the computational basis gives outcome $0$ with probability $|\alpha|^2$ and outcome $1$ with probability $|\beta|^2$. That is the Born rule.

For example, if $\alpha = \tfrac{1}{\sqrt{2}}$ and $\beta = \tfrac{1}{\sqrt{2}}$ (the $|+\rangle$ state), then

$P(0) = \left|\tfrac{1}{\sqrt{2}}\right|^2 = \tfrac{1}{2}, \qquad P(1) = \left|\tfrac{1}{\sqrt{2}}\right|^2 = \tfrac{1}{2}.$

As a less symmetric example, consider $\alpha = \cos(\pi/6) = \tfrac{\sqrt{3}}{2}$ and $\beta = \sin(\pi/6) = \tfrac{1}{2}$. Then

$P(0) = \left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{3}{4}, \qquad P(1) = \left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}.$

The probabilities sum to one, as required.

Preparing a state with Ry

The rotation gate $R_y(\theta)$ acts on $|0\rangle$ as

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

To reach $\alpha = \cos(\pi/6)$ and $\beta = \sin(\pi/6)$, set $\theta/2 = \pi/6$, giving $\theta = \pi/3$. Applying c.ry(Math.PI / 3, 0) places the qubit in exactly this state.

Try it

Prepare the state $\cos(\pi/6)|0\rangle + \sin(\pi/6)|1\rangle$ and measure it. The grader checks that $P(0) \approx 3/4$ and $P(1) \approx 1/4$.

Open the Probabilities tab after running to confirm the 3:1 ratio of bar heights.

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