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BPP versus BQP

Determine whether BPP=BQP\mathsf{BPP}=\mathsf{BQP}, or prove that efficient bounded-error quantum computation is strictly more powerful than efficient randomized classical computation.

Source-reported statusAs of 2026-09-04Source & evidence ↓

Problem statement

Determine whether BPP=BQP\mathsf{BPP}=\mathsf{BQP}, or prove that efficient bounded-error quantum computation is strictly more powerful than efficient randomized classical computation.

Lean 4

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Background

This is a sharp equality question, even though an unconditional separation is a long-horizon challenge. Since P⊆BPP⊆BQP⊆PSPACE\mathsf{P}\subseteq\mathsf{BPP}\subseteq\mathsf{BQP}\subseteq\mathsf{PSPACE}, proving BPP≠BQP\mathsf{BPP}\ne\mathsf{BQP} would also prove P≠PSPACE\mathsf{P}\ne\mathsf{PSPACE}.

What is known: Bernstein–Vazirani's recursive Fourier sampling and Simon's problem give oracle/query-model evidence. Raz–Tal proved a classical-oracle separation of BQP from the entire polynomial hierarchy. These are relativized results, not unrelativized class separations. Shor's polynomial-time factoring algorithm is further evidence of quantum advantage, but an exponential advantage for factoring is not proved: no superpolynomial classical lower bound is known.

The question compares the same finite-input, bounded-error computation model; alternative data-access or output conventions belong in the methods discussion A16. Related: A2, A11.

Progress & source assessment

Source catalogue review: 4 September 2026. Improved indicates partial progress, including qualified preprint claims.

References
  1. [1]Bernstein & Vazirani, 'Quantum complexity theory', SIAM J. Comput. 26 (1997) link
  2. [2]Raz & Tal, 'Oracle separation of BQP and PH', STOC 2019 link

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