The Higgs field and spontaneous symmetry breaking

Write the doublet's four real components as one radius, ρ² = 2φ†φ, so the potential becomes a function of a single variable: V = (μ²/2)ρ² + (λ/4)ρ⁴. λ must be positive or V has no floor. With μ² > 0 the only stationary point is ρ = 0, a minimum, and the four components are four scalars of mass μ. With μ² < 0 the origin turns into a maximum and dV/dρ = μ²ρ + λρ³ = 0 puts the minimum at ρ = v = √(−μ²/λ), at depth Vₘᵢₙ = −μ⁴/(4λ). Nothing in the Lagrangian changed; the sign of one coefficient moved the ground state. Numbers: v = 246.22 GeV and λ = 0.1291 give μ² = −λv² = −7.83 × 10³ GeV², and Vₘᵢₙ = −1.19 × 10⁸ GeV⁴ = −(104 GeV)⁴. Every point with ρ = v is a minimum, and since ρ is a radius in four real dimensions the set of minima is a 3-sphere. That degeneracy is the symmetry showing itself in the vacuum: SU(2)_L × U(1)_Y carries one vacuum into another at no cost in energy.

Pick ⟨φ⟩ = (0, v/√2)ᵀ; any other point on the sphere is a gauge transform of it. A generator is unbroken if it annihilates the vacuum and broken if it does not. With Tᵃ = σᵃ/2 and Y = 1: T₁⟨φ⟩ = (v/2√2)(1, 0)ᵀ and T₂⟨φ⟩ = (v/2√2)(−i, 0)ᵀ are non-zero, so T₁ and T₂ are broken. T₃⟨φ⟩ = (0, −v/2√2)ᵀ and (Y/2)⟨φ⟩ = (0, v/2√2)ᵀ are each non-zero, but their sum vanishes: Q = T₃ + Y/2 = diag(1, 0) annihilates the vacuum, while T₃ − Y/2 = diag(0, −1) does not. Three broken generators, one surviving U(1). Because the survivor is Q, the vev is electrically neutral and the massless boson is the photon; Y = 1 is precisely the hypercharge that makes the surviving combination the same T₃ + Y/2 that already fixes the fermion charges. Goldstone's theorem then promises one massless mode per broken generator, three in all, and the doublet has exactly one component to spare.

Put the vacuum into the kinetic term with D_μ = ∂_μ − igW^a_μTᵃ − ig′(Y/2)B_μ. Since σ⋅W acting on (0, 1)ᵀ gives (W₁ − iW₂, −W₃)ᵀ, the constant part of D_μ⟨φ⟩ is −i(v/2√2)(g(W₁ − iW₂), −gW₃ + g′B)ᵀ, with modulus squared (v²/8)[g²(W₁² + W₂²) + (gW₃ − g′B)²]. The charged combinations W^± = (W₁ ∓ iW₂)/√2 turn the first bracket into M_W²W⁺W⁻ with M_W = gv/2. The second is a 2 × 2 mass matrix in the (W₃, B) basis, (v²/4)[[g², −gg′], [−gg′, g′²]], whose determinant is zero: one eigenvalue is 0, belonging to A = (g′W₃ + gB)/√(g² + g′²), and the other is M_Z² = (g² + g′²)v²/4 for Z = (gW₃ − g′B)/√(g² + g′²). Hence cos θ_W = g/√(g² + g′²) = M_W/M_Z. With g = 0.652, g′ = 0.357 and v = 246.2 GeV: M_W = 80.3 GeV, M_Z = 91.5 GeV and sin²θ_W = 0.231, against 80.4 and 91.2 GeV measured. The tree-level ρ = M_W²/(M_Z²cos²θ_W) = 1 is a property of the doublet; a triplet vev would give a different number.

Count before and after. Before breaking: four massless gauge fields with two polarisations each, plus four real scalars, 8 + 4 = 12. After: W⁺, W⁻ and Z massive with three polarisations each, one massless photon with two, and one real scalar, 9 + 2 + 1 = 12. The three Goldstone modes have not vanished; they have become the longitudinal polarisations that a massive vector needs and a massless one lacks. Write φ = exp(iπᵃTᵃ/v)(0, (v + h)/√2)ᵀ: the πᵃ are the Goldstones, and a gauge transformation with parameters −πᵃ/v removes them entirely. That choice is unitary gauge, φ = (0, (v + h)/√2)ᵀ, in which only physical fields appear but the W propagator carries a k_μk_ν/M_W² term that spoils power counting. In the R_ξ family the Goldstones stay in the Lagrangian with gauge-dependent masses, √ξ M_W for the charged pair and √ξ M_Z for the neutral one, and their unphysical poles cancel, together with the ghosts', against those of the gauge fields in every S-matrix element — the bookkeeping that lets the theory be proved renormalisable.

Set φ = (1/√2)(G₁ + iG₂, v + h + iG₃)ᵀ, so 2φ†φ = (v + h)² + G², and V = (μ²/2)(2φ†φ) + (λ/4)(2φ†φ)². Differentiate twice at h = G = 0. Along any Gᵢ the curvature is μ² + λv² = 0: three exactly flat directions, the Goldstones, massless because V has the same value all around the sphere. Along h the curvature is μ² + 3λv² = 2λv², so mₕ² = 2λv² = −2μ². Expanding further gives the self-couplings, −λv h³ − (λ/4)h⁴, so the trilinear coupling is λv = mₕ²/(2v). Numbers: mₕ = 125.1 GeV and v = 246.22 GeV give λ = 15 650/(2 × 60 624) = 0.129, |μ| = mₕ/√2 = 88.5 GeV, and a trilinear strength λv = 31.8 GeV. Read the direction of inference carefully: G_F fixed v, the Higgs discovery fixed λ, and only then were μ² and the self-couplings known. The model predicts the shape of V, not its numbers, and the h³ coupling is still untested to better than order unity, the h⁴ coupling not tested at all.

The unit's habit is to let NumPy repeat the algebra and then compare. Build Tᵃ = σᵃ/2 and Y/2 = ½, the vacuum vec = [0, v/√2], and confirm the generator test: np.linalg.norm(T @ vec) is non-zero for T₁, T₂ and T₃ − Y/2 and zero to machine precision for T₃ + Y/2. Then form M² = (v²/4) × [[g², −gg′], [−gg′, g′²]] and call np.linalg.eigh. The eigenvalues come back as 0 and (g² + g′²)v²/4; the eigenvector of the zero eigenvalue is (g′, g)/√(g² + g′²) = (sin θ_W, cos θ_W), whose angle from the B axis is θ_W = arctan(g′/g) = 28.7° for g′/g = 0.5475. Sanity checks: the zero eigenvalue must vanish to rounding error rather than merely be small, the angle must satisfy sin²θ_W = g′²/(g² + g′²) = 0.231, and M_Z from eigh must reproduce M_W/cos θ_W. Two non-zero eigenvalues mean the matrix is no longer the rank-one (v²/4)uuᵀ with u = (g, −g′)ᵀ, usually a factor of 2 mishandled on the off-diagonal; a zero mode along (g′, −g) instead of (g′, g) is the sign error: either the relative minus between gW₃ and g′B has been lost, or the vev sits in the upper component, where T₃ − Y/2 rather than Q survives.