The QPE Circuit

Quantum Phase Estimation (QPE) turns an unobservable global phase into a measurable integer stored in a binary register. This lesson shows how the three building blocks — a Hadamard layer, controlled-$U$ powers, and an inverse Quantum Fourier Transform — fit together into a single circuit.

The goal

Given a unitary $U$ and one of its eigenvectors $|u\rangle$ with

QPE writes an $n$-bit binary approximation of $\varphi$ into a dedicated counting register of $n$ qubits. If $\varphi$ is exactly representable as an $n$-bit fraction, the result is exact.

Circuit structure

The circuit uses two registers:

• Counting register: $n$ qubits, all initialised to $|0\rangle$. These will hold the phase.
• Eigenstate register: one or more qubits initialised to the eigenstate $|u\rangle$.

Stage 1 — Hadamard layer

Apply $H$ to every qubit in the counting register to create the uniform superposition

Stage 2 — Controlled-$U$ powers

Index the counting qubits $j = 0, 1, \ldots, n-1$ from MSB to LSB. Qubit $j$ controls the operation $U^{2^{n-1-j}}$ on the eigenstate register.

Because $U|u\rangle = e^{2\pi i\varphi}|u\rangle$, repeated application gives $U^{2^k}|u\rangle = e^{2\pi i\varphi\cdot 2^k}|u\rangle$.

A controlled-$U^{2^k}$ imprints the phase $e^{2\pi i\varphi\cdot 2^k}$ on the control qubit only when that qubit is $|1\rangle$. After all $n$ controlled operations the counting register is

This tensor product is the Quantum Fourier Transform of $|\tilde\varphi\rangle$, where $\tilde\varphi$ is the integer $2^n\varphi$ (exact when $\varphi$ is an $n$-bit fraction).

Stage 3 — Inverse QFT

Applying $QFT^\dagger$ to the counting register maps the Fourier-encoded state back to the computational-basis state $|\tilde\varphi\rangle$. A measurement then reads out the $n$-bit integer whose binary value gives the phase.

Worked example: $U = S$, $n = 2$ counting qubits

The $S$ gate has matrix $\operatorname{diag}(1, e^{i\pi/2})$, so $S|1\rangle = e^{i\pi/2}|1\rangle = e^{2\pi i\cdot 1/4}|1\rangle$. The phase is $\varphi = 1/4$, which in 2-bit binary is $0.01_2$, encoding the integer $\tilde\varphi = 1$.

Registers: $q_0$ (MSB counting), $q_1$ (LSB counting), $q_2$ (eigenstate, prepared as $|1\rangle$).

After Stage 1, Hadamard on both counting qubits:

Stage 2 applies $CS^{2^0} = CS$ controlled by $q_1$, then $CS^{2^1} = CZ$ controlled by $q_0$. Since $S|1\rangle = i|1\rangle$ and $Z|1\rangle = -|1\rangle$:

The counting register $\tfrac{1}{2}(|0\rangle-|1\rangle)(|0\rangle+i|1\rangle)$ equals $QFT_4|01\rangle$, because $QFT|j\rangle = \tfrac{1}{\sqrt{4}}\sum_{k=0}^{3}e^{2\pi ijk/4}|k\rangle$, giving $QFT|1\rangle = \tfrac{1}{2}(1,\,i,\,-1,\,-i)^T$ in the basis $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, which matches the tensor product above.

Stage 3 ($QFT^\dagger$ on $q_0, q_1$) returns the counting register to $|01\rangle$. Measuring all three qubits yields the bitstring $011$ with certainty, and the top two bits $01 = 1/4 \cdot 4$ correctly encode $\varphi = 1/4$.

Try it

Build the complete QPE circuit for $U = S$ with 2 counting qubits. The starter code prepares the eigenstate $|1\rangle$ on $q_2$. Add Stage 1, Stage 2, Stage 3, and a final measurement. The grader checks the output distribution, which should place all probability on the bitstring $011$.

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