A dimension-free lower bound on γ₃

An 8.65% improvement over the published constant — prover round only, not yet adversarially checked.

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Background

A state with non-positive partial transpose may still be undistillable — a possibility raised independently by DiVincenzo, Shor, Smolin, Terhal and Thapliyal [3] and by Dür, Cirac, Lewenstein and Bruß [4], and unresolved for twenty-five years. It is Problem 5 in the list of Horodecki, Rudnicki and Życzkowski [2].

In July 2026 the two-copy case fell. Bharti, Gajjala and Haug [1] prove a sharp dimension-free partial-trace inequality and deduce that a Werner state ρα is two-copy distillable iff α < −1/2, settling Problem 5 at two copies. They also construct explicit constants γk with α ≥ −γk implying k-copy undistillability in every dimension, which moves the live frontier to three copies.

Worth noting for this page: their own abstract records that the initial proofs were machine-generated and then verified and rewritten by the authors — the same division of labour as here.

The problem

Determine the largest γ₃ such that α ≥ −γ₃ implies three-copy undistillability of ρα in every local dimension. Equivalently, at the endpoint α = −1/2, decide whether the endpoint partial-trace form q₃ is nonnegative on every operator of rank at most two.

The published value is γ₃ = 1/6 [1]; the two-copy answer is 1/2.

The bound

The route is a lemma bounding the three-term sum, A₁ + A₂ + A₃ ≤ 4N + T, from a balanced-frame estimate together with the swap majorization F₁ + F₂ + F₃ ≺ 2I + F₁F₂F₃. This gives q₃(−t) ≥ (1 − 6t + 3t² − 2t³)N, and the largest root of the cubic yields

γ₃ ≥ (1 + 31/3 − 32/3)/2 = 0.181083…

dimension-free, against the published 1/6 = 0.1666… — an increase of about 8.65%.

The empirics point much higher

Minimising q₃(−½, ·)/‖C‖² over operators of rank at most two returns exactly zero at d = 2, 3, 4, 5, 6, with no violation anywhere and the α-scan strictly positive above −½. Every equality witness is rank-two with equal singular values, and most are genuinely three-spread, with operator-Schmidt rank four across every cut. That suggests

γ₃ = 1/2, sharp

matching the two-copy threshold of [1] — that is, the third copy buys nothing at the endpoint. That, rather than the increment above, is the prize.

Status: not adversarially verified

The verification round did not run; the workflow terminated on a usage limit after the prover pass. What has been checked by hand: the swap-majorization minimum eigenvalue is exactly zero at d = 2, 3; the lemma holds on 12k random rank-two draws at d ≤ 4; the chain bound holds; the arithmetic for the constant is exact. That is spot-checking, not adversarial verification, and on this record the distinction is the whole point.

What remains open

  1. Verify or break the lemma. The balanced-frame step is the one to attack. If it survives, the increment is publishable; if it does not, that is worth recording too.
  2. Prove γ₃ = 1/2. The evidence is uniform across d = 2…6 and the equality witnesses are highly structured, which usually means an exact argument exists.
  3. All k. If the third copy buys nothing, the question is whether any finite k does — which is the original NPT bound-entanglement problem [2,3,4].

References

  1. K. Bharti, R. Gajjala and T. Haug, Two-copy nondistillability of Werner states: sharp partial-trace inequalities, arXiv:2607.24479.
  2. P. Horodecki, Ł. Rudnicki and K. Życzkowski, Five open problems in quantum information, arXiv:2002.03233.
  3. D. P. DiVincenzo, P. W. Shor, J. A. Smolin, B. M. Terhal and A. V. Thapliyal, Evidence for bound entangled states with negative partial transpose, quant-ph/9910026.
  4. W. Dür, J. I. Cirac, M. Lewenstein and D. Bruß, Distillability and partial transposition in bipartite systems, quant-ph/9910022.