Open Problem 1 of Alhejji–Knill is false

A counterexample to the word-trace bridge lemma for fractional Schatten norms.

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Background

Quantum capacity is given by a regularised expression, and the regularisation is genuinely necessary: no finite number of channel uses suffices in general [4], capacity is non-additive even in very simple channels [3], and the general problem is undecidable [5]. Progress comes from finding structured families where the regularisation collapses.

The spin-alignment conjecture of Alhejji and Knill [1] is one such route, proved there for integer Schatten norms, classical states, and two-state mixtures. Carrying it to all Schatten norms — and so, by continuation, to the von Neumann entropy — would give a single-letter quantum capacity for platypus-type families. Among their concluding remarks is a bridge lemma that would supply exactly that extension.

The problem

Let A₀, A₁, B₀, B₁ be positive semidefinite with λ(A₀) = λ(B₀) and λ(A₁) = λ(B₁), and write Πs for the product along a binary word s. Suppose word-trace dominance holds:

tr Πs(A) ≥ tr Πs(B) for every word s. (H)

Does it follow that ‖A₀ + A₁‖p ≥ ‖B₀ + B₁‖p for every p ∈ [1, ∞)? (C)

The hypothesis constrains only integer moments; the conclusion is about fractional powers of a spectrum. Whether the former controls the latter is the whole content of the question — and since it is posed as a question, an instance satisfying (H) and violating (C) settles it.

A commuting counterexample

The answer is no. Take

A₀ = diag(176, 0, 64) · A₁ = diag(80, 49, 0) · B₀ = diag(64, 176, 0) · B₁ = A₁

Spectra match and all four are positive semidefinite. Everything commutes, so every word collapses to a product of powers and (H) reduces to one analytic statement, (64/176)k + (49/80)m ≤ 859/880 < 1. The conclusion fails at p = 3/2, where the two spectral sums are 4951 and 5103 — a comparison between integers, with no numerical tolerance anywhere.

So the failure is classical: non-commutativity plays no part in it.

A non-commuting 3×3 instance, found first, fails on the whole interval p ∈ (1, p) with *p = 1.97584…, and a separate argument rules out *d = 2, making qutrits minimal.

Verification

Exact rational or 100-digit arithmetic throughout. The 3×3 instance was re-derived from scratch by an independent agent with its own frame and cone argument, reaching the same resonance constant; hostile search covered every exact word to length 13 with no violation. The diagonal family came from a container-isolated run with no network and no sight of the earlier work.

What remains open

  1. A referee-grade write-up of the 3×3 case. Every scalar inequality was discharged symbolically, so this is a write-up obligation, not a gap — but no one has typeset it as a proof a referee could read linearly.
  2. The kernel lemma for the diagonal family, corroborated on thousands of draws but not yet proved symbolically. Since the family commutes this should be routine, and it would make the counterexample self-contained on half a page.
  3. The conjecture itself. This blocks the overlap-lemma route from integer to fractional Schatten norms. It does not refute spin alignment — in the instance above the A-side is not aligned — and the weaker compatible-marginal statement of [2] is still open.

References

  1. M. A. Alhejji and E. Knill, Towards a resolution of the spin alignment problem, arXiv:2307.06894.
  2. Z. Song and L. Chen, A counterexample to the strong spin alignment conjecture, arXiv:2603.25410.
  3. F. Leditzky, D. Leung, V. Siddhu, G. Smith and J. A. Smolin, Generic nonadditivity of quantum capacity in simple channels, arXiv:2202.08377.
  4. T. Cubitt, D. Elkouss, W. Matthews, M. Ozols, D. Pérez-García and S. Strelchuk, Unbounded number of channel uses are required to see quantum capacity, arXiv:1408.5115.
  5. A. Bhattacharyya, A. Mehta and Y. Zhao, On the undecidability of quantum channel capacities, arXiv:2601.22471.