Self-Assessment Quiz: Quantum Phase Estimation

Twenty questions on the QPE circuit, precision, eigenstate preparation, and the algorithms QPE sits beneath. Aim for 16+.

Question 1

QPE estimates $\varphi$ where:

A) $U|\psi\rangle = e^{2\pi i\varphi}|\psi\rangle$ B) $U|\psi\rangle = \varphi|\psi\rangle$ C) $|\psi\rangle = \varphi|0\rangle$ D) $\varphi$ is a measurement outcome

Question 2

With $t$ counting qubits, QPE resolves $\varphi$ to:

A) $t$ bits of precision B) $2^t$ bits C) 1 bit D) $\log t$ bits

Question 3

The counting register is prepared in:

A) $|0\rangle^{\otimes t}$ B) Uniform superposition via $H^{\otimes t}$ C) A Bell state D) $|1\rangle^{\otimes t}$

Question 4

The controlled operations applied are:

A) Controlled-$U^{2^j}$ for $j = 0,\ldots,t-1$ B) Controlled-$U$ only C) Controlled-$H$ D) Uncontrolled $U$

Question 5

The final step before measurement is:

A) The QFT B) The inverse QFT C) A Hadamard layer only D) A SWAP

Question 6

The phase kicked onto counting qubit $j$ is:

A) $e^{2\pi i\varphi}$ B) $e^{2\pi i 2^j\varphi}$ C) $e^{2\pi i\varphi/2^j}$ D) $\varphi$

Question 7

Achieving precision $\varepsilon$ requires roughly:

A) $t = \log_2(1/\varepsilon)$ counting qubits B) $t = 1/\varepsilon$ C) $t = \varepsilon$ D) $t = 2^{1/\varepsilon}$

Question 8

The number of applications of $U$ scales as:

A) $O(t)$ B) $O(2^t) = O(1/\varepsilon)$ C) $O(1)$ D) $O(\log t)$

Question 9

This makes QPE's cost in terms of precision:

A) Exponentially better than sampling B) Linear in $1/\varepsilon$ — the Heisenberg limit C) Constant D) Quadratic in $\varepsilon$

Question 10

If the input is not an eigenstate but a superposition of eigenstates:

A) QPE fails B) QPE returns one eigenphase, sampled with probability given by the overlap C) QPE returns the average phase D) QPE returns zero

Question 11

That property is exploited in:

A) Quantum chemistry, where the input approximates a ground state B) Grover's algorithm C) Teleportation D) QKD

Question 12

If the ground-state overlap is $|\langle\psi_0|\psi\rangle|^2 = 0.01$, you expect success:

A) Always B) About 1% of runs, so ~100 repetitions C) Never D) 99% of runs

Question 13

Iterative phase estimation:

A) Uses one counting qubit repeatedly with classical feedback B) Requires more qubits C) Is less accurate in principle D) Does not use $U$

Question 14

Its main advantage is:

A) Fewer qubits, at the cost of mid-circuit measurement and feed-forward B) Fewer applications of $U$ C) No measurement needed D) Higher precision limit

Question 15

QPE is the core subroutine of:

A) Shor's algorithm and quantum simulation eigenvalue estimation B) Teleportation C) Superdense coding D) BB84

Question 16

True or false: QPE requires an exact eigenstate to produce a meaningful answer.

Question 17

True or false: QPE's circuit depth grows exponentially with the number of counting qubits.

Question 18

True or false: QPE is practical on today's NISQ hardware for chemistry problems.

Question 19

Short answer. Explain why QPE achieves the Heisenberg limit while repeated sampling achieves only the standard quantum limit.

Question 20

Short answer. A team wants QPE for a molecular ground-state energy. List the two preparation problems they must solve first.

Answer Key

Q Ans Note
1 A $U$ must be unitary, so its eigenvalues are pure phases.
2 A $t$ bits, with success probability depending on how close $\varphi$ is to a $t$-bit binary fraction.
3 B Uniform superposition over all $2^t$ counting values.
4 A Doubling powers write successive binary digits of $\varphi$ into the phases.
5 B The inverse QFT decodes the Fourier-encoded phase into a binary number.
6 B Which is precisely why the powers double.
7 A Logarithmic in $1/\varepsilon$ — the qubit cost is cheap.
8 B The gate cost is $\sum_j 2^j = 2^t - 1$, linear in $1/\varepsilon$. Qubits are cheap; applications of $U$ are not.
9 B The Heisenberg limit, $O(1/\varepsilon)$, versus $O(1/\varepsilon^2)$ for naive repeated sampling.
10 B The counting register and eigenstate register become entangled; measuring collapses to one eigenphase.
11 A You need only an approximate ground state with decent overlap, not the exact one.
12 B Expected repetitions scale as the inverse overlap — and overlap typically decays exponentially with system size, which is the central practical obstacle.
13 A Reuses a single ancilla, feeding each measured bit forward to adjust the next rotation.
14 A It trades width for mid-circuit measurement and classical feed-forward — same total $U$ applications.
15 A Shor is QPE on modular multiplication; chemistry eigenvalue estimation is QPE on time evolution.
16 False An approximate eigenstate works; you get the corresponding eigenphase with probability equal to the squared overlap.
17 True The controlled-$U^{2^j}$ sequence means depth grows as $2^t$, which is why precision is expensive in depth even though it is cheap in qubits.
18 False Chemistry-scale QPE needs circuit depths far beyond NISQ coherence — which is exactly why VQE (Ch. 19) exists as a shallow-circuit substitute.
19 — Repeated sampling of a phase from $M$ independent measurements averages down shot noise as $1/\sqrt M$, giving $\varepsilon \sim 1/\sqrt M$ — the standard quantum limit. QPE instead applies $U$ coherently $2^t$ times, so the phase accumulates coherently as $2^j\varphi$ rather than being re-sampled. The accumulated signal grows linearly in the number of applications, giving $\varepsilon \sim 1/M$ — the Heisenberg limit, a quadratic improvement in the number of uses of $U$.
20 — (1) State preparation: produce a trial state with non-negligible overlap with the true ground state. Overlap generally decays exponentially with system size (the "orthogonality catastrophe"), so this is not a detail — it can dominate the total cost. (2) Hamiltonian simulation: implement controlled-$e^{-iHt}$ accurately enough, typically via Trotterization or qubitization, whose gate cost grows with the number of Hamiltonian terms and the required accuracy. Both must be solved before QPE's asymptotic advantage is realizable.