Qubit: $A$ is positive definite (ellipsoid) $\iff c_-<c<c_+$

Proved

HardyFiveAxioms.qubitA_posDef_iff

by Lucas · Oct 2, 2026 · Mathlib 0df444a (Lean v4.33.1)

linear-algebramathematical-physicsquantum-foundationsquantum-information

Let $0\le a,b\le1$ and $c\in\mathbb R$ , and let $A=A(a,b,c)$ be the matrix (83). Then $A$ has three positive eigenvalues, i.e. it is positive definite and the surface $\vec v^{\,T}A\vec v=\tfrac12$ is an ellipsoid, if and only if

c_-<c<c_+ .

This is Eq. (86). Hardy argues that pure states must lie on an ellipsoid, and this yields exactly the quantum constraint (36) on $c$ .

Formalization Note "Three positive eigenvalues" is encoded as Matrix.PosDef (symmetric with $x^TAx>0$ for $x\neq0$ ), which is equivalent for real symmetric matrices.

Preamble

import Mathlib
import Definitions.Def_hardy2001_qubit

Formal statement

namespace HardyFiveAxioms

/-- Hardy 2001, Section 8.6, Eq. (86): for `0 ≤ a, b ≤ 1`, the matrix `A` has three positive
eigenvalues (the surface `v⃗ᵀ A v⃗ = 1/2` is an ellipsoid) if and only if `c₋ < c < c₊`. -/
theorem qubitA_posDef_iff (a b c : ℝ) (ha₀ : 0 ≤ a) (ha₁ : a ≤ 1) (hb₀ : 0 ≤ b) (hb₁ : b ≤ 1) :
    (qubitA a b c).PosDef ↔ cMinus a b < c ∧ c < cPlus a b := by sorry

end HardyFiveAxioms

Source

L. Hardy, *Quantum Theory From Five Reasonable Axioms*, arXiv:quant-ph/0101012v4 (2001), https://arxiv.org/abs/quant-ph/0101012, pp. 21–22, Section 8.6, Eqs. (83)–(86)

Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic)

NON-BLIND READ-BACK — NOT INDEPENDENT TESTIMONY. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement below, with full knowledge of the source paper and of the intended meaning. It was not produced by a blind, independent auditor, so it may share the drafter's blind spots. Reviewers must not treat it as independent confirmation of faithfulness; compare the Lean code against the source directly.

Let $a,b,c$ be real numbers with $0\le a\le1$ and $0\le b\le1$ . Let $A$ be the real symmetric $3\times3$ matrix with diagonal entries $\tfrac12$ and off-diagonal entries $a-\tfrac12$ (positions 1,2), $b-\tfrac12$ (positions 1,3) and $c-\tfrac12$ (positions 2,3). Then $A$ is positive definite ( $A$ is symmetric and $x^TAx>0$ for every nonzero $x\in\mathbb R^3$ ) if and only if $c_-<c$ and $c<c_+$ , where $c_\pm=1-a-b+2ab\pm2\sqrt{ab(1-a)(1-b)}$ .

Human review

Endorsed by Shuze Chen · Oct 8, 2026
Confirmed by the moderator at approval.
Endorsed by Lucas · Oct 8, 2026
Confirmed by the mission captain (proposal self-audit).

View graph

Qubit: AAA is positive definite (ellipsoid) ⟺ c−<c<c+\iff c_-<c<c_+⟺c−​<c<c+​

Qubit: $A$ is positive definite (ellipsoid) $\iff c_-<c<c_+$