trureturingGitHub MATHEMATICAL ATLAS

RESEARCH / Focused target

Spin-1 tensor spin-K/2 precession separable bound

Literature status: not recheckedRepository record: proved

RELEASED FOUNDATIONS / PROPOSED CONNECTION

Research connections

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Problem

The question

Khoi-Nguyen Huynh-Vu, Lin Htoo Zaw and Valerio Scarani, Certification of genuine multipartite entanglement in spin ensembles with measurements of total angular momentum, arXiv:2311.00806v2, §III, Conjecture 2, Eq. (32), PDF p. 8:

The separable bound for {ȷ̃, ȷ̃′} = {1, K/2} with K ≥ 7 is Psep_K({1, K/2}) = ½ [1 + 2^{−(K−1)} binom(K−1, (K−1)/2) (K−1)/(K+1)].

The protocol takes odd K, J_k = cos(2πk/K)J_x + sin(2πk/K)J_y, and Q_K = (1/K) Σ_k pos(J_k), where 2 pos acts on an eigenvalue m as 1+sgn(m), including half weight at zero. Eqs. (7) and (10) express the separable bound as the maximum over unit product pure states. The two factors here have dimensions 3 and K+1. Preregistration: https://github.com/the-omega-institute/trureturing/issues/11873.

Motivation

Our foothold

D5/S3/Quantum/Entanglement/PrecessionSpinOneSeparableBound.result proves the quantified product-pure form of Conjecture 2. The public spin matrices, precession observable and spectral projector follow the source definitions; the compression is a proved identity, not a definition of the observable. The literature anchor is Library/QuantumStates/huynhvu2024precession.md.

Gap

Missing bridges

The paper states that it could not prove Eq. (32), and reports agreement with numerics through K=101. The preregistered literature reading checks the paper and the related works arXiv:2311.00805, 2405.17966, 2411.03132, 2401.16147, 2401.14328 and 2509.03166. It records no proof of Eq. (32) in that searched scope; this is not a claim of literature priority. The published journal version is not verified here.

Route

Proposed approach

The spectral projectors use Mathlib’s Matrix.IsHermitian.cfc with positiveWeight and Real.sign; the finite spectrum requires no continuity hypothesis. The proof constructs the binomial/Krawtchouk eigenbasis of J_x and its inverse by polynomial homogenization. A root-of-unity filter computes the rotation average. The half-integer sign decomposition retains the nonzero endpoint contribution (−1)^L c_K, with K=2L+1.

For a unit spin-1 state a, the compression of 2Q_K−I is supported on indices {0,1,2,K−2,K−1,K}. Its off-diagonal block is (−1)^L c_K M/(K+1), where y=|a₁|², w=conj(a₀)a₁+conj(a₁)a₂, z=conj(a₀)a₂ and

$$ M=\begin{pmatrix} 1-Ky & \sqrt{2K},w & \sqrt{K(K-1)/2},z\ \sqrt{2K},w & -z & 0\ \sqrt{K(K-1)/2},z & 0 & 0 \end{pmatrix}. $$

The squared Frobenius bound uses |w|²≤2y(1−y) and |z|²≤(1−y)²/4. Its scalar remainder is

$$ f_K(y)=(1-Ky)^2+8Ky(1-y)+\frac{K^2-K+1}{4}(1-y)^2-(K-1)^2. $$

For K≥7, its quadratic coefficient (5K²−33K+1)/4 is nonnegative; f_K(0)≤0 and f_K(1)=0 imply f_K(y)≤0 on [0,1]. Cauchy–Schwarz then bounds the full product quadratic form. The spin-1 middle state and a phase-adjusted superposition of the two endpoint states attain that bound. Every auxiliary proof is local to the single public result.

Falsifier

What would falsify this route

The claim would fail if a unit product state at any odd K≥7 exceeded the formula, or if the stated value were not attained. IsGreatest proves both the universal upper bound and attainment. Numerical examples are outside the proof premises. Even K and K<7 are outside this statement.

Evidence

Evidence to collect

The canonical module is D5/S3/Quantum/Entanglement/PrecessionSpinOneSeparableBound.lean; its Scribe mirror displays every public definition and claim/result. The imports are pinned Mathlib; there are no frozen D5 prerequisites. The Lean kernel axiom closure of result is propext, Classical.choice and Quot.sound.

Boundary computations use NumPy 2.2.5 / SciPy 1.16.0 with the literal spin matrices and spectral projector. Run the following program with python3. It diagonalizes Q compressed by $a=(\sin\alpha/\sqrt2,\cos\alpha,\sin\alpha/\sqrt2)$ and maximizes only this one-parameter family. Its outputs are numerical lower bounds, not certified global maxima. Eigenvalues within $10^{-10}$ of zero receive the half weight; this numerical tolerance is not part of the Lean definition.

import numpy as np
from scipy.optimize import minimize_scalar

def spin(n):
    j = n / 2
    raising = np.zeros((n + 1, n + 1), dtype=complex)
    for r in range(1, n + 1):
        m = j - r
        raising[r - 1, r] = np.sqrt(j * (j + 1) - m * (m + 1))
    return (raising + raising.conj().T) / 2, (raising - raising.conj().T) / (2j)

def boundary(k):
    ax, ay = spin(2)
    bx, by = spin(k)
    x = np.kron(ax, np.eye(k + 1)) + np.kron(np.eye(3), bx)
    y = np.kron(ay, np.eye(k + 1)) + np.kron(np.eye(3), by)
    q = np.zeros_like(x)
    for r in range(k):
        angle = 2 * np.pi * r / k
        values, vectors = np.linalg.eigh(np.cos(angle) * x + np.sin(angle) * y)
        weights = np.where(values > 1e-10, 1, np.where(values < -1e-10, 0, 0.5))
        q += (vectors * weights) @ vectors.conj().T / k
    def score(alpha):
        a = np.array([np.sin(alpha) / np.sqrt(2), np.cos(alpha), np.sin(alpha) / np.sqrt(2)])
        compressed = np.einsum('p,piqj,q->ij', a.conj(), q.reshape(3, k + 1, 3, k + 1), a)
        return np.linalg.eigvalsh(compressed)[-1]
    optimum = minimize_scalar(lambda alpha: -score(alpha), bounds=(0, np.pi / 2),
                              method='bounded', options={'xatol': 1e-14})
    print(k, 'alpha', optimum.x, 'score', -optimum.fun, 'middle_score', score(0))

for k in (3, 5):
    boundary(k)

Triage

Scope assessment

Conjecture 2 is proved in its preregistered product-pure form. proof_shape: content; admission_basis: open-problem-resolution (#11873; Proved); utility: none. This is an unbounded analytic proof, not a bounded enumeration or certified finite instance. Registration is paused under CLAUDE.md §3.9. No atom or coverage edge is used.

What the settlement shows

  • Computed source discrepancy at the excluded parameters: The arXiv:2311.00806v2 HTML immediately after Eq. (32) states that the separable bound is 1/2 for K=3 and K=5. The Evidence program gives a K=3 product score of 0.6603513979123848 and a K=5 middle-state score of 0.625. These computed lower bounds disagree with that HTML statement. The PDF rendering of this sentence is unverified. This discrepancy does not affect the proved domain, odd K≥7, and supplies no kernel-certified small-parameter refutation.
  • Proved in this module: The decisive mechanism is the exact six-index compression, including the sign operator's endpoint term, followed by the Frobenius estimate and an explicit attaining state. The equality holds for every odd K≥7, rather than just the paper's numerical range. The middle-state construction establishes sharpness for this product score; it does not classify all optimizers.
  • Computed with the Evidence command: K=3 gives a product score 0.6603513979123848 at α=0.6945473082666525, exceeding the extrapolated formula 0.625. This explains why a statement covering all odd K≥3 would need a different bound. At K=3 the two sets of three edge indices also overlap; the disjoint six-index block argument does not apply. The value is a numerical witness, not a Lean refutation or a certified claim of global optimality.
  • Computed by exact rational coefficient expansion: At K=5 the coefficients of f_K, in ascending powers of y, are (−39/4,39/2,−39/4), so f_5(y)=−(39/4)(1−y)²≤0. Command: python3 -c 'from fractions import Fraction as F; k=5; print(F(-3*k*k+7*k+1,4), F(-k*k+13*k-1,2), F(5*k*k-33*k+1,4))'. The calculation uses exact fractions.Fraction coefficients ((−3K²+7K+1)/4,(−K²+13K−1)/2,(5K²−33K+1)/4). Thus the nonnegative-quadratic-coefficient shortcut requires K≥7, but the scalar Frobenius estimate itself also permits K=5. The numerical middle-state score at K=5 is 0.625. Open: extending the matrix compression and attainment proofs to K=5; this module's local identities require L≥3 and do not establish that extension.
  • Open: Other spin pairs need a new sign decomposition and compression estimate. The spin-1 factor is used through its three eigenlevels; the present proof supplies no uniform bound for other first spins, nor a classification of even-K protocols.
  • Proved in this module / source interpretation: For the pair {1,K/2} the exact threshold in Eq. (32) is available without a numerical cutoff, so the source's criterion that exceeding the separable bound certifies entanglement has this analytic threshold. Open: a separate formalization of mixed separable density matrices and their convex reduction to product pure states. Eq. (10) already supplies that source reading. The paper's results for other spin pairs and its other conjectures receive no additional formal conclusion here.

ASSUMED-UNVERIFIED

Unverified assumptions

The external literature reading is the preregistration's searched scope; completeness and the published journal version are not verified. The source-to-Lean interpretation and mixed-state convex reduction are semantic review obligations, distinct from the kernel-checked pure-product statement. Numerical boundary values are floating-point computations; they carry neither exact certification nor a global-optimality claim. The PDF rendering of the HTML sentence following Eq. (32) about K=3 and K=5 is unverified.

Dossier & anchor history

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