Quiz: Quantum Chemistry with VQE

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Quiz: Quantum Chemistry with VQE

20 questions. Answers with explanations follow.

1. A molecule has 7 spatial orbitals in a minimal basis. How many qubits does a Jordan–Wigner encoding require?

(a) 7
(b) 14
(c) 49
(d) 128

2. Chemical accuracy is:

(a) $1.6\times10^{-3}$ Hartree, ≈ 1 kcal/mol
(b) $1.6\times10^{-6}$ Hartree
(c) whatever precision the simulator reports
(d) 1 Hartree

3. §36.2 measured BeH₂ at 666 Hamiltonian terms and H₂O at 1,086. Both need 14 qubits. What explains the difference?

(a) H₂O has a larger basis set
(b) H₂O has more electrons, so more four-index integrals survive
(c) a bug in the Hamiltonian builder
(d) BeH₂ is linear and H₂O is bent

4. How does the Hamiltonian term count scale with qubit count?

(a) $\mathcal{O}(n)$
(b) $\mathcal{O}(n^2)$
(c) $\mathcal{O}(n^4)$
(d) $\mathcal{O}(2^n)$

5. Comparing Bravyi–Kitaev to Jordan–Wigner on LiH, §36.3 measured:

(a) BK has far fewer terms
(b) the same 631 terms; BK has lower Pauli weight
(c) BK needs fewer qubits
(d) the two are identical in every respect

6. Why does Jordan–Wigner produce Pauli strings of length $\mathcal{O}(n)$?

(a) it stores parity locally and pays in occupation
(b) it stores occupation locally and pays in parity, which needs a $Z$-string over all preceding modes
(c) it is an inefficient implementation
(d) fermions require long strings under any mapping

7. §36.3 found BK's improvement over JW at 12 qubits to be:

(a) dramatic — an order of magnitude
(b) modest — mean weight 6.16 → 5.62, max 12 → 10
(c) negative — BK was worse
(d) exactly zero

8. An ActiveSpace(n_electrons=10, n_orbitals=2) is rejected because:

(a) 10 is not an even number
(b) 2 orbitals is too few for any molecule
(c) 10 electrons cannot fit in 4 spin orbitals — the Pauli principle forbids it
(d) the module only supports closed-shell systems

9. ActiveSpace refuses an error_ha without a validated_against because:

(a) an error without a reference does not say what it was measured against
(b) the field is required by the dataclass
(c) errors must always be positive
(d) it makes serialization easier

10. ★ In §36.5, VQE's error against the (2e,2o) active space's exact answer was:

(a) $2.01\times10^{-2}$ Ha
(b) $2.04\times10^{-9}$ Ha
(c) $1.6\times10^{-3}$ Ha
(d) it did not converge

11. ★★ And the (2e,2o) active space's own error, against the full LiH Hamiltonian, was:

(a) zero — active spaces are exact by construction
(b) $2.04\times10^{-9}$ Ha
(c) $2.01\times10^{-2}$ Ha = 12.62 kcal/mol
(d) unmeasurable

12. ★★★ The ratio between those two errors is:

(a) about 1
(b) about 1,000
(c) about 9,870,104
(d) about $10^{20}$

13. The correct summary of §36.5 is:

(a) VQE failed
(b) VQE succeeded, and the approximation made to fit the device is ten million times larger than VQE's error
(c) the active space was well chosen
(d) chemical accuracy was reached

14. total_error() raises on an unvalidated active space rather than assuming zero because:

(a) zero is hard to represent
(b) a default of zero would be silently wrong by seven orders of magnitude, in the flattering direction
(c) raising is faster
(d) the active space is usually exactly zero

15. chemical_accuracy_claim returns REACHED_WITHIN_ACTIVE_SPACE when:

(a) the VQE error is inside chemical accuracy but the total error is not
(b) the total error is inside chemical accuracy
(c) the active space was never validated
(d) VQE failed to converge

16. Exact diagonalization of the full 12-qubit LiH Hamiltonian took:

(a) 22.996 seconds
(b) 6 hours
(c) 22.996 QPU-years
(d) it could not be done

17. ★★ The classical baseline a quantum chemistry method must actually beat is:

(a) full CI
(b) CCSD(T)
(c) Hartree–Fock
(d) whichever is most convenient

18. ★★ At the claimed 50-orbital crossover, §36.7 priced VQE at:

(a) $1.91\times10^{20}$ shots = 6.06 × 10⁸ QPU-years
(b) $1.91\times10^{6}$ shots
(c) about a week of device time
(d) it is not computable

19. ★★★ "The crossover is not where VQE becomes useful — it is where ______."

(a) hardware becomes available
(b) classical methods stop being exact
(c) the shot budget becomes affordable
(d) barren plateaus disappear

20. ★★★ The route that survives §36.7's arithmetic is:

(a) more optimizer steps
(b) a larger active space
(c) phase estimation on error-corrected hardware, where precision costs $\mathcal{O}(1/\epsilon)$ rather than $\mathcal{O}(1/\epsilon^2)$
(d) there is none

Answers

1. (b) 14. One qubit per spin orbital, so twice the spatial orbitals. §36.2 measured this on all four molecules — BeH₂ and H₂O both have 7 spatial orbitals and both take 14 qubits.

2. (a). $1.6\times10^{-3}$ Hartree ≈ 1 kcal/mol. It is a chemistry threshold rather than a numerical one: below it, computed energies predict measured reaction rates and equilibria. §36.1.

3. (b) H₂O has more electrons. Ten to BeH₂'s six. The electron count decides how many of the four-index two-electron integrals vanish, so the qubit count does not determine the term count. This matters because device capability is quoted in qubits and the shot bill is denominated in terms — a device sized in qubits is not a device sized for a molecule. §36.2.

4. (c) $\mathcal{O}(n^4)$, from the two-electron integrals $h_{pqrs}$ and their four orbital indices. The fit is rough — it over-predicts BeH₂ by 261% — and the exponent is what matters.

5. (b) the same 631 terms. The common claim that BK reduces the term count is wrong as stated. What BK reduces is Pauli weight, which sets the measurement circuit depth per term. §36.3.

6. (b). JW stores occupation locally — qubit $j$ is the occupation of spin-orbital $j$ — and pays for it in parity, which requires a $Z$-string over every preceding mode. So a single $a^\dagger_p a_q$ becomes a Pauli string of length $\mathcal{O}(n)$. BK stores a mixture so both are recoverable from $\mathcal{O}(\log n)$ qubits.

7. (b) modest. Mean 6.16 → 5.62, max 12 → 10. The $\mathcal{O}(n)$ versus $\mathcal{O}(\log n)$ separation needs more qubits to show, which is worth stating because the usual presentation implies it is dramatic at any size.

8. (c) the Pauli principle. Two orbitals is four spin orbitals, which holds at most four electrons.

9. (a). An error is only meaningful relative to what it was measured against. 0.0201 Ha vs (4e,6o) is a fact; 0.0201 Ha alone is a number with no referent, and Case Study 36.1 is what happens when the referent gets lost between two people who are each doing correct work.

10. (b) $2.04\times10^{-9}$ Ha, from three parameters. Genuinely excellent — the variational method is not the weak link.

11. (c) $2.01\times10^{-2}$ Ha = 12.62 kcal/mol, about 12.6× chemical accuracy.

12. (c) 9,870,104.

13. (b). Both halves are true and both matter. The optimizer converged beautifully on a problem that had already been made wrong in the third decimal place, by a classical decision taken before any quantum code ran.

14. (b). And note which direction the silent error runs: toward the flattering answer. That is not a coincidence — the flattering number is the one that stops the search.

15. (a). REACHED and REACHED_WITHIN_ACTIVE_SPACE are different strings because they are different claims. The second is usually what is true; the first is usually what gets said.

16. (a) 22.996 seconds. Milliseconds for the smaller spaces. Every molecule currently in reach of a quantum device is exactly solvable classically — Chapter 24 §24.7's claim, measured.

17. (b) CCSD(T). Nobody runs full CI past ~20 orbitals; it is exponential. CCSD(T) is $\mathcal{O}(n^7)$ and reaches chemical accuracy on most main-group chemistry at 100+ orbitals. Benchmarking against full CI is Chapter 21 §21.7's error with a molecule attached: comparing against the method nobody uses.

18. (a). And stacking every mitigation in the book — grouping, shadows, better ansätze, low-rank factorization — gives roughly 6 × 10⁵×, which is four orders of magnitude against a deficit of eight.

19. (b) classical methods stop being exact. Those are different points and they are conflated constantly. Past the crossover the classical answer is approximate, but CCSD(T) is approximate the way a well-characterized instrument is approximate — and it runs on hardware you already own.

20. (c) phase estimation on error-corrected hardware. The precision comes from circuit depth rather than sampling, which is why the $1/\epsilon^2$ becomes $1/\epsilon$. Chapter 22 §22.4 built that route and Chapter 25 §25.10 explained why it is gated on error correction. The honest reading of this chapter is that the shot budget is not an engineering problem with VQE — it is an argument for a different algorithm on different hardware.