Amplitude Estimation (Intro)

Amplitude amplification can boost a success probability, but it does not tell us what that probability is. Quantum amplitude estimation answers the "how much?" question: it extracts the value $a = \sin^2\!\theta$ directly, with precision that improves quadratically faster than classical sampling.

From Phase to Amplitude

Recall the amplitude-amplification operator

$Q = -\mathcal{A}\,S_0\,\mathcal{A}^\dagger\,S_\chi,$

where $\mathcal{A}|0\rangle = \cos\theta\,|\text{bad}\rangle + \sin\theta\,|\text{good}\rangle$. This operator is a rotation by $2\theta$ in the two-dimensional good/bad plane, so its eigenvalues are $e^{\pm 2i\theta}$. Writing these as $e^{2\pi i\phi}$, the phases are

$\phi = \frac{\theta}{\pi} \quad \text{and} \quad \phi = 1 - \frac{\theta}{\pi}.$

Quantum Phase Estimation (QPE) can estimate $\phi$ to $m$ bits of precision using $m$ counting qubits and $O(2^m)$ controlled applications of $Q$. Once $\hat{\phi}$ is in hand, the amplitude follows:

$\hat{a} = \sin^2(\pi\hat{\phi}).$

This is amplitude estimation: the circuit reads the phase of $Q$, and arithmetic converts it to the desired probability.

A Minimal Example

To keep the circuit small enough to run in a browser, we replace $Q$ with the simpler phase gate $S$ on a single state qubit. $S$ has the same eigenvalue structure as $Q$:

$S|1\rangle = e^{i\pi/2}|1\rangle = e^{2\pi i \cdot (1/4)}|1\rangle, \quad \phi = \tfrac{1}{4}.$

With $\phi = 1/4$ the estimated amplitude is

$\hat{a} = \sin^2\!\bigl(\pi \cdot \tfrac{1}{4}\bigr) = \sin^2\!\tfrac{\pi}{4} = \tfrac{1}{2}.$

Circuit Layout

We use 3 qubits: qubits 0 and 1 form the 2-bit counting register (qubit 0 = most significant bit), and qubit 2 is the state qubit.

1. Prepare qubit 2 in $|1\rangle$ (the eigenstate of $S$): apply $X$ to qubit 2.
2. Superpose the counting qubits: $H$ on qubits 0 and 1.
3. Phase kickback — the core of QPE. Each counting qubit $j$ controls $U^{2^j}$ acting on the state qubit:
   • Qubit 1 (LSB, weight $2^0 = 1$) controls $S^1$: a controlled phase of $\pi/2$.
   • Qubit 0 (MSB, weight $2^1 = 2$) controls $S^2 = Z$: a controlled phase of $\pi$.
4. Inverse QFT on qubits 0 and 1 decodes the phase into the computational basis.
5. Measure all qubits.

Tracing the Arithmetic

After the phase kickbacks and before the inverse QFT, the counting-register state is

$\frac{1}{2}\bigl(|00\rangle + i|01\rangle - |10\rangle - i|11\rangle\bigr) = \mathrm{QFT}|01\rangle.$

The inverse QFT collapses this exactly to $|01\rangle$. Combined with the state qubit remaining in $|1\rangle$, the full output is

$|011\rangle \quad (q_0 = 0,\; q_1 = 1,\; q_2 = 1).$

Reading the counting register as a binary fraction: $0 \cdot 2^{-1} + 1 \cdot 2^{-2} = \tfrac{1}{4} = \hat{\phi}$. Applying the formula $\hat{a} = \sin^2(\pi\hat{\phi}) = \sin^2(\pi/4) = \tfrac{1}{2}$ recovers the amplitude exactly.

Try it

The starter code has the state preparation, Hadamards, and the inverse QFT already in place. Add the two controlled-phase gates that implement the phase kickback. The grader checks the distribution and expects probability 1 on the outcome $|011\rangle$.

Sign in on the full site to ask questions and join the discussion.