Over-Rotation

Grover's algorithm works by rotating the quantum state toward the marked item, a little further each iteration. If you stop at the right moment — after the optimal number of steps — the probability of finding the answer is nearly 1. But if you keep rotating, you overshoot: the amplitude of the marked state shrinks again, and eventually the probability distribution returns to something close to uniform. This is over-rotation.

The amplitude geometry

The state at any point lives in a 2D plane spanned by the marked basis state $|w\rangle$ and the uniform superposition of all unmarked states $|s'\rangle$ . Starting from the uniform superposition $|s\rangle$ , Grover's algorithm rotates toward $|w\rangle$ by a fixed angle $2\theta$ per iteration, where

$\sin\theta = \frac{1}{\sqrt{N}}.$

After $k$ iterations the amplitude on the marked state is $\sin((2k+1)\theta)$ , so the success probability is

$P_k = \sin^2\!\bigl((2k+1)\theta\bigr).$

Over-rotation for N = 4

With only $N = 4$ items (two qubits), $\sin\theta = 1/2$ , so $\theta = \pi/6$ . The optimal number of iterations is

$k^* = \left\lfloor \frac{\pi}{4\theta} \right\rfloor = \left\lfloor \frac{\pi}{4 \cdot \pi/6} \right\rfloor = \left\lfloor \frac{3}{2} \right\rfloor = 1.$

One iteration is already perfect:

$P_1 = \sin^2\!\left(3 \cdot \frac{\pi}{6}\right) = \sin^2\!\left(\frac{\pi}{2}\right) = 1.$

Now apply a second iteration. The angle advances to $5\theta = 5\pi/6$ :

$P_2 = \sin^2\!\left(\frac{5\pi}{6}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}.$

The success probability has collapsed from 100% back to 25% — exactly what you would get from a random guess! The state has rotated all the way past the target and landed near the opposite side of the plane.

The three unmarked states each end up with amplitude $-1/2$ , probability $1/4$ as well, so the output distribution is perfectly uniform across all four states. Running the circuit and measuring gives no useful information at all.

Why this matters in practice

For large $N$ , the optimal number of iterations is $k^* \approx \tfrac{\pi}{4}\sqrt{N}$ . Every additional iteration beyond $k^*$ rotates the state away from the target. Because $k^*$ must be known in advance (it depends on $N$ and the number of solutions $M$ ), running the algorithm blindly with too many iterations is as bad as running it with too few. Algorithms for the case where $M$ is unknown, such as quantum counting, address this by first estimating $M$ with quantum phase estimation.

Try it

Build the two-iteration Grover circuit for $N = 4$ , marking $|11\rangle$ with a CZ oracle. After two full iterations the probability should be spread uniformly across all four basis states, confirming the over-rotation effect. The grader checks the output distribution.

Run your code to see the quantum state.

Inspect the histogram: each of the four outcomes should have probability close to $0.25$ . Compare this to the one-iteration result, where all probability sits on $|11\rangle = |3\rangle$ .