Project: Estimate a Phase Precisely

Quantum phase estimation (QPE) extracts the phase $\phi$ of a unitary's eigenvalue by encoding it into a register of counting qubits and then reading it out with an inverse QFT. The central trade-off is simple: adding one more counting qubit halves the smallest phase difference that the circuit can distinguish.

The precision formula

With $n$ counting qubits the register can represent $2^n$ evenly spaced values in $[0, 1)$. The precision (the spacing between adjacent representable phases) is

Each extra qubit adds one bit to the binary fraction that encodes $\phi$:

| $n$ | Precision $\delta\phi$ | Representable values | |-----|------------------------|----------------------| | 1 | $1/2$ | 2 | | 2 | $1/4$ | 4 | | 3 | $1/8$ | 8 | | 4 | $1/16$ | 16 |

When $\phi$ is an exact $n$-bit binary fraction the algorithm returns the correct integer $k = \phi \cdot 2^n$ with probability 1. When $\phi$ falls between two representable values the probability spreads, but the closest integer still receives at least $4/\pi^2 \approx 40\%$ of the probability mass (Nielsen and Chuang, §5.2).

Circuit structure

The QPE circuit has three stages:

1. Hadamard layer. Apply $H$ to every counting qubit to create a uniform superposition.

2. Phase kicks. Each counting qubit $j$ is given a phase gate $P(\lambda_j)$ with $\lambda_j = 2\pi \cdot 2^j \cdot \phi$. This encodes the binary digits of $\phi$ directly into the phases of the counting register:

   $$|+\rangle_j \;\longrightarrow\; \frac{|0\rangle + e^{2\pi i \cdot 2^j \phi}\,|1\rangle}{\sqrt{2}}.$$

3. Inverse QFT. Decodes the phase-encoded register back to the computational basis state $|k\rangle$ where $k = \phi \cdot 2^n$.

A worked example: $\phi = 5/8$

Binary expansion: $5/8 = 0.101_2$. With $n = 3$ qubits the precision is $1/8$, and $5/8$ is an exact 3-bit fraction, so QPE should return the integer $k = 5$ (binary $|101\rangle$) with probability 1.

Phase kicks for $\phi = 5/8$:

• Qubit 0: $\lambda_0 = 2\pi \cdot 1 \cdot 5/8 = 5\pi/4$
• Qubit 1: $\lambda_1 = 2\pi \cdot 2 \cdot 5/8 = 5\pi/2$
• Qubit 2: $\lambda_2 = 2\pi \cdot 4 \cdot 5/8 = 5\pi \equiv \pi \pmod{2\pi}$

After the inverse QFT the three counting qubits collapse to $|101\rangle$. A 2-qubit register could not represent $5/8$ — the nearest values are $1/2$ and $3/4$ — so those extra qubits are genuinely necessary.

Try it

The starter code has the Hadamard layer, the phase kicks for qubits 0 and 2, and the full inverse QFT. Add the missing phase kick for qubit 1. When your circuit is correct, the distribution grader will show all probability on outcome $|101\rangle$ (decimal 5).

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