Metadata
slug: champagne-johnston-macdonald-pipes-2022-symmetric-witness-spectrum
bibkey: champagnejohnstonmacdonaldpipes2022spectral
doi: 10.1016/j.laa.2022.05.004
url: https://arxiv.org/abs/2108.10405v1
triage: theorem
motivation_gids:
- D5/S3/Quantum/Entanglement/AbsoluteSeparability/SymmetricQubitWitnessSpectrum.result
The eigenvalues of symmetric two-qubit entanglement witnesses satisfy µ3 ≥ −√(µ1µ2)
Problem
G. Champagne, N. Johnston, M. MacDonald and L. Pipes, Spectral properties of symmetric quantum states and symmetric entanglement witnesses, Linear Algebra Appl. 649, 273–300 (2022), arXiv:2108.10405v1, study the eigenvalues of symmetric entanglement witnesses.
- Witnesses. A symmetric entanglement witness is a Hermitian W on C^d ∨ C^d with (v ⊗ v)*W(v ⊗ v) ≥ 0 for all v (§2.3); positive semidefinite W are included (footnote 1).
- Theorem 3. For d = 2 and real µ1 ≥ µ2 ≥ µ3, the inequalities (a) µ2 ≥ 0 and (b) µ3 ≥ −√(µ1µ2) are sufficient for the existence of a symmetric witness with these eigenvalues, and necessary for decomposable ones. Every symmetric witness satisfies (a) and a weaker inequality (c).
- The conjecture. After Theorem 3 and in §6(c) (page 20): “We conjecture that inequalities (a) and (b) from that theorem provide the correct characterization.”
The verbatim statements are in the literature note.
Issue #17536 fixes the reading:
- C²∨C² is identified with C³ through the orthonormal Dicke basis e₀₀, (e₀₁ + e₁₀)/√2, e₁₁, in which v ⊗ v has coordinates (v₀², √2 v₀v₁, v₁²).
- “Eigenvalues µ1 ≥ µ2 ≥ µ3” means that the characteristic polynomial is (X − µ1)(X − µ2)(X − µ3).
- The conjecture: for all real µ1 ≥ µ2 ≥ µ3, a symmetric witness with these eigenvalues exists if and only if µ2 ≥ 0 and µ3 ≥ −√(µ1µ2).
The conjecture is proved in this reading.
Motivation
The spectrum of a witness decides which entangled states it can detect from spectral data alone. For ordinary two-qubit witnesses the spectra are known (Johnston–Patterson, arXiv:1708.05901). For symmetric witnesses, which are positive only on the product states v ⊗ v, the source obtained the exact spectral region only for decomposable witnesses, and Figure 2 shows the gap between (b) and (c).
Gap
Issue #17536 records the literature check before any Lean:
- arXiv v1 is the only version;
- the nine citing works with arXiv records listed by Semantic Scholar were read at every in-text citation of the source, and none states or settles the conjecture;
- the repository had no entry for the paper.
not-found-in-searched-scope.
Route
- The null cone. With J = [[0, 0, −1], [0, 1, 0], [−1, 0, 0]], xᵀJx = x₁² − 2x₀x₂. A vector x ∈ C³ is a coherent vector (v₀², √2 v₀v₁, v₁²) if and only if xᵀJx = 0.
- Transport to the eigenbasis. The witness and the diagonal matrix of its eigenvalues have the same characteristic polynomial, so W = U D U* with U unitary. G = UᵀJU is symmetric and unitary, and the witness condition becomes: zᵀGz = 0 implies Σᵢ dᵢ|zᵢ|² ≥ 0.
- Two roots. Put the negative eigenvalue −c at index 2 and t = |G₂₂|. Then t > 0, because otherwise e₂ would be a null vector with negative expectation. For each u ∈ C², the cone equation in z₂ has two roots whose product is uᵀAu/G₂₂, where A is the upper-left 2 × 2 corner of G. Positivity at both roots gives c|uᵀAu| ≤ t(µ1|u₀|² + µ2|u₁|²).
- A 2 × 2 bound. If A is complex symmetric and |uᵀAu| ≤ p|u₀|² + q|u₁|² for all u, then p, q ≥ 0 and |det A| ≤ pq. The proof is polarization followed by two test pairs.
- The corner. For a unitary G, adj G = (det G)·G*, so |det A| = |G₂₂| = t.
- Conclusion. Steps 3–5 give c²t ≤ t²µ1µ2, hence c² ≤ tµ1µ2 ≤ µ1µ2, that is, µ3 ≥ −√(µ1µ2). Step 4 also gives µ2 ≥ 0. If µ3 ≥ 0, both inequalities are immediate.
- Sufficiency. W = diag(µ1, µ3, µ2) in the Dicke basis has coherent expectation (√µ1|v₀|² − √µ2|v₁|²)² + 2(µ3 + √(µ1µ2))|v₀|²|v₁|² ≥ 0.
Falsifier
The proof would not settle the conjecture if:
- the module’s witnesses were narrower than all Hermitian W on C²∨C² that are positive on v ⊗ v, for example by excluding positive semidefinite or degenerate W;
- the eigenvalue condition were weaker than the characteristic polynomial;
- one direction of the characterization were missing.
The module quantifies over every Hermitian 3 × 3 complex matrix in the Dicke basis with nonnegative coherent expectation, fixes its characteristic polynomial, and proves both directions.
Evidence
The canonical source is D5/S3/Quantum/Entanglement/AbsoluteSeparability/SymmetricQubitWitnessSpectrum.lean (396 lines).
- It defines
coherent,IsSymmetricWitnessandclaim, and provesresult : claim. - It uses Mathlib’s spectral theorem and the eigenvalue comparison
Matrix.IsHermitian.eigenvalues_eq_eigenvalues_iff. It also uses two theorems made public for this settlement:QuadraticRadii.quadratic_factors, the factorization of a complex quadratic;diagonal_quadratic_formofCanonicalRelativeModularModel, the real quadratic form of a real diagonal matrix, lifted from two identical local facts of that module.
- It uses only the standard axioms
propext,Classical.choiceandQuot.sound; nosorry,native_decide, or new axiom. - The escape registrations:
Reg/D5/S3/Quantum/Entanglement/AbsoluteSeparability/SymmetricQubitWitnessSpectrum.leanreads out the lower bound −√(µ1µ2) on µ3 inclaim. Its rejected realization, −√(µ1µ2) − 1, fails at µ1 = µ2 = 0 and µ3 = −1, where it would require a witness thatresultexcludes. The registration isdeclared_validated.QuadraticRadii.quadratic_factorsis registered inReg/D5/S3/Zeros/PochhammerDeformation/QuadraticRadii.lean; it reads out the second root w of the factorization. The registration isdeclared_validated.diagonal_quadratic_formis registered inReg/D5/S3/Quantum/Divergence/CanonicalRelativeModularModel.lean; it reads out the weighted sum Σᵢ dᵢ‖yᵢ‖². The registration isdeclared_validated.
Frozen state: Golden/Frozen/state/D5/S3/Quantum/Entanglement/AbsoluteSeparability/SymmetricQubitWitnessSpectrum.lean.json, statement sha256:f1e7ea21bf0528a2415971810622185e9420fc12e36d7fbf25e459e6c7da9455.
Declaration statements:
result:sha256:733da05943859dbf1271dfb62293583372e136b762dc5d45a1405fc582338fa0;claim:sha256:0e69a98c72918ad682272bc38fcdac99144cf9e09169957a1804dfbd3e095c98.
Triage
Tier 1 conjecture of a 2022 paper, preregistered in issue #17536 before any Lean.
theorem; resolution proved.
| declaration | proof_shape | escape_witness | admission_basis |
|---|---|---|---|
| result | bind-only | none: after inlining, the null cone, the spectral transport, the two-root bound, the 2 × 2 determinant bound, the corner identity and the diagonal witness are instantiations of pinned Mathlib and frozen D5 results followed by normalization | open-problem-resolution |
| the null-cone, root, bilinear, determinant, spectral, corner and diagonal-witness helpers (private) | bind-only (consumed on the proof path of result) | none | consumed helpers |
Every other public declaration is a definition used by a public statement. The module’s 9 private theorems are bind-only helpers consumed on the proof path of result, and its 4 private definitions and its compiler-generated declarations also lie on the proof path of result. The bind-only form is admitted under the open-problem-resolution basis of CLAUDE.md §3.2; no escape witness is claimed. The utility is none. There is no digestion atom.
What the settlement shows
Proved by result: for all real µ1 ≥ µ2 ≥ µ3, a symmetric entanglement witness on C²∨C² with eigenvalues µ1, µ2, µ3 exists if and only if µ2 ≥ 0 and µ3 ≥ −√(µ1µ2). This is the exact characterization asked in §6(c).
Argued, not formalized.
- The proof gives more than the bound. Let ν be a unit eigenvector of the negative eigenvalue −c, and let t = |νᵀJν| ∈ (0, 1] be its value on the cone form. Then c² ≤ tµ1µ2. Equality µ3 = −√(µ1µ2) therefore forces t = 1: in coefficient form [[x, y/√2], [y/√2, z]], t = |y² − 2xz| = 2|det|, so the negative eigenvector is a maximally entangled symmetric state. The extremal diagonal witness has ν = (e₀₁ + e₁₀)/√2, for which t = 1.
- With Theorem 3 of the source, the eigenvalue region of all symmetric witnesses on C²∨C² equals that of the decomposable ones.
Open.
- The spectra of symmetric witnesses on C^d ∨ C^d for d ≥ 3.
- The source’s other open problems: antisymmetric witnesses (§6(a)), Schmidt-number witnesses (§6(b)), and the minimal set of matrices for absolute symmetric PPT (§6(d)).
Effect on the paper. The conjecture after Theorem 3 holds. Inequality (c) of Theorem 3 can be replaced by (b) for every symmetric witness, and (a) with (b) characterize the spectra exactly.
ASSUMED-UNVERIFIED
The bounded literature check does not establish exhaustive worldwide novelty or priority. The journal version was not read; its agreement with arXiv v1 is unverified. The statements in “Argued, not formalized” are not formalized.