Reading the Phase Register

After the two earlier stages of Quantum Phase Estimation — superposing the counting register and kicking back the phase through controlled-$U$ powers — the counting register holds a superposition of the form

This is exactly the Quantum Fourier Transform of the computational-basis state $|2^n \varphi\rangle$ (when $\varphi$ is exactly representable in $n$ bits). To decode the phase we must undo the QFT — i.e. apply the inverse QFT — and then measure.

Decoding with the inverse QFT

Applying the inverse QFT (IQFT) to the counting register maps the QFT-encoded superposition back to a single computational-basis state:

Measuring then yields the $n$-bit binary string for $2^n \varphi$ with probability 1. The decimal value of that string divided by $2^n$ is $\varphi$.

For $n = 2$ counting qubits, the IQFT is a small four-gate circuit on qubits $q_0$ (MSB) and $q_1$ (LSB):

1. SWAP $q_0$ and $q_1$ — undoes the bit-reversal implicit in the QFT.
2. $H$ on $q_1$ — Hadamard on the (now-leading) qubit.
3. $CP(-\pi/2)$ controlled on $q_0$, targeting $q_1$ — conjugate of the half-turn phase.
4. $H$ on $q_0$ — final Hadamard.

A worked example: $U = Z$, $\varphi = 1/2$

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

so the eigenphase is $\varphi = 1/2$. In binary this is $0.10_2$, meaning $2^2 \cdot \varphi = 2$ and the expected measurement result on two counting qubits is $|10\rangle$ (q0 = 1, q1 = 0).

With two counting qubits the controlled-unitary stage applies:

• $U^{2^0} = Z$ controlled on $q_1$ (LSB), since $q_1$ accumulates the least-significant phase bit.
• $U^{2^1} = Z^2 = I$ controlled on $q_0$ (MSB) — the identity, so it can be omitted.

After these controlled operations the counting register is in the state $|{+}\rangle_{q_0} \otimes |{-}\rangle_{q_1}$, which is exactly $\text{QFT}|10\rangle$. The IQFT maps it back to $|10\rangle$, and measurement reads out 10 — confirming $\varphi = 2/4 = 1/2$.

Try it

The starter code builds the QPE circuit up through the controlled-$Z$ kickback step. Add the 2-qubit IQFT on $q_0$ and $q_1$ so the distribution peaks at $|101\rangle$ (counting register $|10\rangle$ plus eigenstate $|1\rangle$), confirming $\varphi = 1/2$.

After running, the Probabilities panel should show 100% on the single state 101. The first two bits 10 give $\varphi = 1 \cdot 2^{-1} + 0 \cdot 2^{-2} = 1/2$, matching $Z$'s eigenphase.

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