Single-Qubit Phase Estimation

The full Quantum Phase Estimation (QPE) circuit can use many ancilla qubits to achieve high precision, but the essential ideas are already present when the counting register contains just one qubit. In that simplest case the entire algorithm fits in two qubits and four gates, yet it already demonstrates phase kickback and the inverse QFT working together to extract an eigenphase.

The two-qubit QPE circuit

Let $U$ be a unitary with eigenvector $|u\rangle$ satisfying

U|u\rangle = e^{2\pi i \varphi}|u\rangle, \qquad \varphi \in [0,1).

We allocate two qubits:

Qubit 0 — the single counting (ancilla) qubit, initialised to $|0\rangle$ .
Qubit 1 — the eigenstate register, initialised to $|u\rangle$ .

The circuit has three stages.

Stage 1 — Superpose the counting qubit.

H|0\rangle = \tfrac{1}{\sqrt{2}}\bigl(|0\rangle + |1\rangle\bigr).

The full state is now $\tfrac{1}{\sqrt{2}}\bigl(|0\rangle + |1\rangle\bigr) \otimes |u\rangle$ .

Stage 2 — Controlled- $U$ (phase kickback).

Apply $U$ to the eigenstate register only when the counting qubit is $|1\rangle$ . Because $|u\rangle$ is an eigenstate, this does not entangle the two registers in amplitude — it merely multiplies the $|1\rangle$ amplitude of the counting qubit by the eigenvalue $e^{2\pi i\varphi}$ :

\tfrac{1}{\sqrt{2}}\bigl(|0\rangle \cdot |u\rangle + |1\rangle \cdot e^{2\pi i\varphi}|u\rangle\bigr) = \tfrac{1}{\sqrt{2}}\bigl(|0\rangle + e^{2\pi i\varphi}|1\rangle\bigr) \otimes |u\rangle.

The eigenstate register is unchanged; the phase has been kicked back into the counting qubit.

Stage 3 — Inverse QFT on 1 qubit.

The 1-qubit QFT is just $H$ , so the inverse QFT is also $H$ . Applying $H$ to the counting qubit:

H\Bigl(\tfrac{|0\rangle + e^{2\pi i\varphi}|1\rangle}{\sqrt{2}}\Bigr) = \frac{(1+e^{2\pi i\varphi})|0\rangle + (1-e^{2\pi i\varphi})|1\rangle}{2}.

Measuring the counting qubit now yields:

P(0) = \left|\frac{1+e^{2\pi i\varphi}}{2}\right|^2 = \cos^2(\pi\varphi), \qquad P(1) = \sin^2(\pi\varphi).

A worked example: the $Z$ gate

The $Z$ gate acts on its eigenstate $|1\rangle$ as

Z|1\rangle = -|1\rangle = e^{i\pi}|1\rangle = e^{2\pi i \cdot 1/2}|1\rangle,

so $\varphi = 1/2$ . Plugging into the formula above:

P(0) = \cos^2\!\left(\pi \cdot \tfrac{1}{2}\right) = \cos^2\!\left(\tfrac{\pi}{2}\right) = 0, \qquad P(1) = \sin^2\!\left(\tfrac{\pi}{2}\right) = 1.

The counting qubit is measured as 1 with certainty, encoding the 1-bit binary fraction $0.1_2 = 1/2 = \varphi$ exactly. Because $\varphi = 1/2$ is representable in one bit there is no approximation error.

In the circuit this means:

c.x(1) — prepare the eigenstate $|1\rangle$ on qubit 1.
c.h(0) — put the counting qubit into superposition.
c.cz(0, 1) — controlled- $Z$ performs the phase kickback (CZ multiplies $|11\rangle$ by $-1$ , which is exactly controlled- $Z$ acting on the $|1\rangle$ eigenstate of $Z$ ).
c.h(0) — inverse QFT undoes the superposition and converts the phase to a basis state.
c.measure(0) — read out the estimated phase bit.

Try it

Complete the circuit below. The eigenstate $|1\rangle$ is already prepared and the first Hadamard is in place. Your task is to add the controlled-Z (the controlled- $U$ step) and the second Hadamard (the inverse QFT), then measure the counting qubit. The grader checks the output distribution: all probability should land on bitstring 11 (counting qubit = 1, eigenstate qubit = 1, with qubit 0 most significant).

Run your code to see the quantum state.

The two-qubit QPE circuit

A worked example: the ZZZ gate

Try it

A worked example: the $Z$ gate