Watrous’s no-disentanglers conjecture; QMA versus QMA(2) relative to an oracle

For every pair of constants ε,δ≥0\varepsilon,\delta\ge 0 satisfying ε+δ<1\varepsilon+\delta<1, every (ε,δ)(\varepsilon,\delta)-disentangler with output space KK and input space HH must satisfy dim⁡H=2Ω(dim⁡K)\dim H=2^{\Omega(\dim K)}; equivalently, no such disentangler can have input dimension subexponential in the dimension of its output space.

References

Primary source

arXiv

Progress summary

Refreshed
Claimed progress

A new preprint claims to settle the conjecture for quantum oracles, while the corresponding unconditional complexity question remains open.

Watrous’s conjecture asks whether approximate procedures that remove entanglement must use exponentially larger input dimension, a result that would separate the power of one versus two unentangled quantum proofs. The latest report claims this has been settled relative to a quantum oracle.

Known results

  • Watrous (2008) conjectured that every (ε,δ)(\varepsilon,\delta)-disentangler with fixed ε,δ<1\varepsilon,\delta<1 requires dim⁡H=2Ω(dim⁡K)\dim H=2^{\Omega(\dim K)}.
  • Watrous (2008) proved the exact case ε=δ=0\varepsilon=\delta=0: no finite-dimensional perfect disentangler exists when dim⁡K≥2\dim K\ge 2.
  • A 2024 preprint proved exponential bounds for strengthened disentanglers under ε+δ<1\varepsilon+\sqrt{\delta}<1, but did not establish the stated oracle separation.

September 2, 2026 claimed oracle separation

A report dated September 2, 2026 says the authors of a new preprint prove a relative separation QMA≠QMA(2)\mathrm{QMA}\ne\mathrm{QMA}(2) and an exponential lower bound for (ε,δ)(\varepsilon,\delta)-disentanglers whenever ε+δ<1\varepsilon+\delta<1. This is a claimed resolution in the oracle setting, but it is not independently verified here and does not imply an unconditional separation.

Current status (as of September 2026): The claimed quantum-oracle separation and corresponding disentangler lower bound are reported but unverified; the unconditional question whether QMA=QMA(2)\mathrm{QMA}=\mathrm{QMA}(2) remains open.

Sources

Solutions 0

No solutions have been posted yet.