Checkpoint: Estimate a Phase

This checkpoint brings together every building block from the module — phase kickback, controlled- $U$ powers, and the inverse QFT — into a single working Quantum Phase Estimation (QPE) circuit.

The target problem

The $S$ gate (phase gate) acts on its eigenstate $|1\rangle$ as

$S|1\rangle = \begin{pmatrix}1 & 0 \\ 0 & i\end{pmatrix}\begin{pmatrix}0 \\ 1\end{pmatrix} = i|1\rangle = e^{2\pi i \cdot 1/4}|1\rangle.$

The eigenphase is $\varphi = 1/4$ . Written in two-bit binary, $\varphi = 0.01_2$ , so a two-qubit QPE circuit should produce the bitstring $01$ in the counting register with certainty.

Circuit structure

A QPE circuit for $n = 2$ counting qubits and a single-qubit target has four stages.

Stage 1 — Prepare the eigenstate. Apply $X$ to the target qubit (qubit 2), turning $|0\rangle$ into the eigenstate $|1\rangle$ .

Stage 2 — Superpose the counting register. Apply $H$ to each counting qubit (qubits 0 and 1), creating the uniform superposition

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

Stage 3 — Phase kickback. Each counting qubit $k$ controls $S^{2^k}$ on the target:

Qubit 0 (MSB): controls $S^{2^1} = S^2 = P(\pi) = Z$ , implemented as cp(PI, 0, 2).
Qubit 1 (LSB): controls $S^{2^0} = S^1 = P(\pi/2)$ , implemented as cp(PI/2, 1, 2).

In standard QPE, the MSB counting qubit controls the highest power of $U$ so that the resulting superposition of the counting register equals the quantum Fourier transform of the phase integer. Because $|1\rangle$ is an eigenstate of $S$ , each controlled- $S^{2^k}$ does not entangle the registers — it only multiplies the $|1\rangle$ component of the respective counting qubit by the phase factor $e^{2\pi i \cdot 2^k \varphi}$ . After kickback, the counting register holds

$\frac{1}{2}\bigl(|00\rangle + e^{i\pi/2}|01\rangle + e^{i\pi}|10\rangle + e^{i3\pi/2}|11\rangle\bigr) = \frac{1}{2}\sum_{j=0}^{3} e^{2\pi i \varphi j}|j\rangle.$

This is exactly $\mathrm{QFT}_4|1\rangle$ , the quantum Fourier transform of the state $|1\rangle$ (integer 1, encoding $\varphi = 1/2^2$ ).

Stage 4 — Inverse QFT. Applying $\mathrm{QFT}_4^{-1}$ undoes the Fourier transform and collapses the counting register to $|01\rangle$ . The 2-qubit inverse QFT circuit is

$\mathrm{SWAP}(0,1) \;\to\; H_1 \;\to\; CP(-\pi/2)_{0\to 1} \;\to\; H_0.$

After the inverse QFT, measuring the counting register gives 01 with probability 1, so reading off $\varphi = 1/2^2 = 1/4$ is exact.

Try it

Implement the full QPE circuit. The grader checks the output distribution: the combined state of all three qubits should place all probability on the bitstring 011 (counting register 01, target qubit 1 — qubit 2 remains in $|1\rangle$ since phase kickback leaves an eigenstate unchanged up to global phase).

Run your code to see the quantum state.