RY for Real Amplitudes

Every state on the Bloch sphere can be written in the form $\cos(\theta/2)|0\rangle + e^{i\phi}\sin(\theta/2)|1\rangle$. When you only need real, non-negative amplitudes — a common starting point for algorithms like Grover search — the $e^{i\phi}$ factor plays no role, and you can reach the target state with a single gate: the RY rotation.

What RY does

The $R_Y(\theta)$ gate rotates the Bloch sphere by angle $\theta$ about the $y$-axis:

Acting on the ground state $|0\rangle = \binom{1}{0}$, the result is

Both amplitudes are real and the probabilities sum to one because $\cos^2(\theta/2) + \sin^2(\theta/2) = 1$.

A few special cases are worth memorising:

| $\theta$ | $P(0)$ | $P(1)$ | Resulting state | |---|---|---|---| | $0$ | $1$ | $0$ | $\|0\rangle$ | | $\pi/2$ | $1/2$ | $1/2$ | $\cos(\pi/4)\|0\rangle + \sin(\pi/4)\|1\rangle \approx 0.707\|0\rangle + 0.707\|1\rangle = \|{+}\rangle$ | | $\pi$ | $0$ | $1$ | $\|1\rangle$ | | $2\pi$ | $1$ | $0$ | $-\|0\rangle$ (global phase, same as $\|0\rangle$ physically) |

The $\theta = \pi/2$ case gives the same 50/50 split as the Hadamard, yet their matrices are different. Both prepare $|+\rangle$ (up to a global phase), confirming that the Born-rule probabilities capture everything measurable.

Choosing the angle

To hit a target amplitude $\alpha$ on $|0\rangle$ (with $0 \le \alpha \le 1$), solve

For example, to put 75% of the probability into $|0\rangle$ you need $\alpha = \sqrt{3/4} = \tfrac{\sqrt{3}}{2}$, which gives $\theta/2 = \pi/6$ and $\theta = \pi/3$.

Try it

Use $R_Y$ to prepare the state $\cos(\pi/6)|0\rangle + \sin(\pi/6)|1\rangle$. The grader checks the full statevector, so the angle must be exact.

After pressing Run, look at the amplitudes: $|0\rangle$ should carry $\approx 0.866 = \cos(\pi/6)$ and $|1\rangle$ should carry $\approx 0.500 = \sin(\pi/6)$, matching $P(0) = 3/4$ and $P(1) = 1/4$.

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