Electroweak mixing and charged currents

Everything starts from one operator acting on one vector. The Higgs doublet has hypercharge Y = 1, so D_μ φ = (∂_μ − ig W^a_μ τ^a/2 − i(g′/2) B_μ)φ, and in unitary gauge the vacuum is φ₀ = (0, v/√2)ᵀ with v = 246.22 GeV. Let the matrix g W^a τ^a/2 + (g′/2)B act on (0, 1)ᵀ: the Pauli matrices return the column ((g/2)(W¹ − iW²), (−gW³ + g′B)/2)ᵀ. Its squared norm times v²/2 is the mass term: |D_μ φ₀|² = (v²/8)[g²(W¹)² + g²(W²)² + (gW³ − g′B)²]. Read it as ½ (field)ᵀ M² (field). The first two pieces are already diagonal, so W¹ and W² each get M² = g²v²/4. The third is a quadratic form in two fields and it is not diagonal — that is the whole neutral problem. Numerically v²/8 = 7578 GeV², so with g = 0.6528 the charged coefficient is 7578 × 0.4262 = 3230 GeV² = ½M_W². Nothing here was quantised: it is classical field theory, a Hermitian matrix sandwiched between a vector and itself.

W¹ and W² are real fields with the same mass, so any rotation among them is free; the useful one is complex. Define T± = T¹ ± iT² and W± = (W¹ ∓ iW²)/√2, and the identity W¹T¹ + W²T² = (W⁺T⁺ + W⁻T⁻)/√2 follows in one line. Because T⁺ raises T₃ by one unit, W⁺ absorbed by e_L (T₃ = −½) produces ν_L (T₃ = +½): the field carries charge +1 and the interaction changes flavour. Only left-chiral fields sit in doublets, so the current is ν̄_L γ^μ e_L = ½ ν̄ γ^μ(1 − γ⁵) e — vector minus axial, with P_L = (1 − γ⁵)/2 supplying the parity violation. The vertex factor is g/√2 and M_W = gv/2 = 0.6528 × 123.11 = 80.37 GeV. At q² ≪ M_W² the W propagator is −1/M_W², so two vertices give a contact strength g²/(2M_W²); matching Fermi's 4G_F/√2 gives G_F/√2 = g²/(8M_W²) = 1/(2v²), and 1/(√2 × 246.22²) = 1.166 × 10⁻⁵ GeV⁻² is the muon-decay value. The coupling g cancels: muon decay measures v, not g.

Write (v²/8)(gW³ − g′B)² as ½ (W³, B) M² (W³, B)ᵀ. Then M² = (v²/4) u uᵀ with u = (g, −g′)ᵀ: the outer product of one vector with itself. That is rank one, so det M² = 0 identically and one eigenvalue is exactly zero whatever g and g′ are — no tuning, no small parameter. Explicitly, with v²/4 = 15156 GeV², M² = [[6458.7, −3462.9], [−3462.9, 1856.6]] GeV², and 6458.7 × 1856.6 − 3462.9² = 0 to the rounding. The null vector is orthogonal to u, so it points along (g′, g); the other eigenvector is u itself and takes the entire trace, (v²/4)(g² + g′²) = 8315 GeV², so M_Z = 91.19 GeV. Why rank one? The vacuum (0, v/√2)ᵀ is annihilated by exactly one combination of the four generators, Q = T₃ + Y/2, and a single doublet can break at most three of them. Cauchy interlacing adds a bonus: the W³ diagonal entry g²v²/4 = M_W² must lie between the two eigenvalues, 0 ≤ M_W² ≤ M_Z², so M_W ≤ M_Z, with equality only at g′ = 0, before any angle is named.

Normalise the null vector: A = (g′W³ + gB)/√(g² + g′²). Setting sin θ_W = g′/√(g² + g′²) and cos θ_W = g/√(g² + g′²) makes A = sin θ_W W³ + cos θ_W B, and its orthogonal partner is Z = cos θ_W W³ − sin θ_W B: one rotation R(θ_W) of the (W³, B) plane, with tan θ_W = g′/g. Orthogonality is not optional — the kinetic terms −¼(∂W³)² − ¼(∂B)² must stay diagonal and unit-normalised after the change of basis. With g = 0.6528 and g′ = 0.3500: tan θ_W = 0.536, θ_W = 28.2°, sin²θ_W = 0.2233. Now push the rotation through the fermion coupling g T₃ W³ + g′(Y/2) B. Along A it reads (g sin θ_W T₃ + g′ cos θ_W Y/2) A = e (T₃ + Y/2) A with e = g sin θ_W = g′ cos θ_W = 0.308: the null vector couples to Q with one universal charge, which is the check that the rotation is right. Along Z it reads (g/cos θ_W)(T₃ − Q sin²θ_W) Z: the Z talks to neutrinos, which the photon cannot, with couplings fixed by the same angle and no new parameter.

The numerical route is four lines: M2 = (v*v/4) * np.outer(u, u) with u = [g, −gp]; w, V = np.linalg.eigh(M2); masses from np.√(w); and the angle from np.degrees(np.arctan2(V[0,0], V[1,0])), the ratio of the photon's W³ to B components. Three things to check rather than trust. First, eigh returns eigenvalues ascending, so the photon is index 0 and the Z is index 1; the zero comes back as roundoff of order 10⁻¹² GeV², possibly negative, so np.√hands you nan unless you clip it — confirm the rank with np.linalg.matrixᵣₐₙₖ(M2), never by testing a float against 0.0. Second, eigenvectors are defined only up to sign, so V[:,0] may arrive as −(sin θ_W, cos θ_W); fix the convention by making the B component positive before reading θ_W. Third, check the invariants: the trace must equal (v²/4)(g² + g′²) and the product of the eigenvalues must vanish. With the standard inputs the run prints M_Z = 91.19 GeV, θ_W = 28.2° and M_W/M_Z = 0.8813 = cos θ_W to four figures — the tree-level relation emerging from a diagonalisation that never assumed it.

Count inputs. The tree-level sector has three, g, g′ and v, and three low-energy handles, α, G_F and sin²θ_W, so both masses are predictions: M_W = √(πα/(√2G_F))/sin θ_W and M_Z = M_W/cos θ_W. The angle is measured independently at the Z pole from the ratio of vector to axial couplings, sin²θ_eff = 0.2315, so M_W = 91.19 × √(1 − 0.2315) = 79.94 GeV against the measured 80.37 GeV: 0.5% low, which is ρ = M_W²/(M_Z² cos²θ_W) = 1.011, one percent above the tree value. That percent is signal. The largest loop correction comes from the top–bottom doublet, whose splitting breaks the custodial symmetry that protected ρ = 1: Δρ = 3G_F mₜ²/(8√2π²) = 0.0093 for mₜ = 172.6 GeV. Read backwards, the miss measures mₜ, which is how LEP placed the top near 170 GeV before the Tevatron saw it. Two caveats: the on-shell scheme defines sin²θ_W ≡ 1 − M_W²/M_Z² = 0.2232, making the relation true by construction and moving the physics into 0.2232 versus 0.2315; and ρ = 1 at tree level is a doublet property, so its near-truth is evidence for the doublet.