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The quartic bracket is (H2+ϕ0ϕ0+2ϕ+ϕ−)2(H^2+\phi^0\phi^0+2\phi^+\phi^-)^2(H2+ϕ0ϕ0+2ϕ+ϕ−)2

Proved
SMHiggsPotential.quartic_bracket_eq_sq

by Lucas · Sep 30, 2026 · Mathlib 0df444a (Lean v4.33.1)

higgsparticle-physicsstandard-model

For all real H,ϕ0,pH,\phi^0,pH,ϕ0,p (with p=ϕ+ϕ−p=\phi^+\phi^-p=ϕ+ϕ− in the application),

H4+(ϕ0)4+4p2+4(ϕ0)2p+4H2p+2(ϕ0)2H2=(H2+(ϕ0)2+2p)2.H^4+(\phi^0)^4+4p^2+4(\phi^0)^2p+4H^2p+2(\phi^0)^2H^2=(H^2+(\phi^0)^2+2p)^2.H4+(ϕ0)4+4p2+4(ϕ0)2p+4H2p+2(ϕ0)2H2=(H2+(ϕ0)2+2p)2.

This is the bracket of the quartic term of box 2.

Preamble
import Mathlib
import Definitions.Def_SMHiggsPotential_Defs
open Complex
Formal statement
namespace SMHiggsPotential

theorem quartic_bracket_eq_sq (H φ0 p : ℝ) :
    H ^ 4 + φ0 ^ 4 + 4 * p ^ 2 + 4 * φ0 ^ 2 * p + 4 * H ^ 2 * p + 2 * φ0 ^ 2 * H ^ 2 =
      (H ^ 2 + φ0 ^ 2 + 2 * p) ^ 2 := by sorry

end SMHiggsPotential
Source
Standard Model Lagrangian in Veltman's conventions (Feynman-'t Hooft gauge), as printed in the five-box chart supplied by the proposer (after M. Veltman, *Diagrammatica: The Path to Feynman Diagrams*, Cambridge University Press, 1994). Box 2, the non-derivative scalar terms: −12mh2H2-\frac12 m_h^2H^2−21​mh2​H2, the βh[… ]\beta_h[\dots]βh​[…] term, +2M4g2αh+\frac{2M^4}{g^2}\alpha_h+g22M4​αh​, −gα[H3+Hϕ0ϕ0+2Hϕ+ϕ−]-g\alpha[H^3+H\phi^0\phi^0+2H\phi^+\phi^-]−gα[H3+Hϕ0ϕ0+2Hϕ+ϕ−] and −18g2αh[H4+… ]-\frac18g^2\alpha_h[H^4+\dots]−81​g2αh​[H4+…].
Read-back

What the Lean code literally says, in plain math · Aristotle (Harmonic) - drafting agent, NON-BLIND (not an independent auditor)

NON-BLIND READ-BACK — NOT INDEPENDENT TESTIMONY. This read-back was written by the same agent (Aristotle, by Harmonic) that drafted the Lean statement, with full knowledge of the intended meaning and of the source image. It was not produced by a blind, independent auditor, and no reviewer should treat it as independent evidence that the formalization is faithful.

Binders. Three arbitrary real numbers H,ϕ0,pH,\phi^0,pH,ϕ0,p; no hypotheses (ppp may be negative).

Claim.

H4+(ϕ0)4+4p2+4(ϕ0)2p+4H2p+2(ϕ0)2H2=(H2+(ϕ0)2+2p)2.H^4+(\phi^0)^4+4p^2+4(\phi^0)^2p+4H^2p+2(\phi^0)^2H^2=(H^2+(\phi^0)^2+2p)^2.H4+(ϕ0)4+4p2+4(ϕ0)2p+4H2p+2(ϕ0)2H2=(H2+(ϕ0)2+2p)2.

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