
Nuclear, particle & quantum-information sequence · Interactive lesson
Higgs boson and Yukawa mass matrices
Substitute Φ = (0, (v + h)/√2)ᵀ everywhere and the model hands you three things at once: the boson's mass from the curvature of the potential, couplings that are masses divided by v, and fermion mass matrices that need two unitaries to diagonalise — together with an honest tally of which numbers were put in by hand.
UPV-18.06 · Lesson 6 of 9 in this unit. Position is not a completion score.Nuclear, particle & quantum-information sequence. These units continue into nuclear and particle physics and quantum information. Follow your university’s prerequisites, mathematical depth and assessment; this independent pathway is not a universal programme specification.
Build the model
Every coupling of h is a mass over v; the masses themselves are inputs.
The key idea
The Higgs boson is the radial excitation h of the doublet about its vacuum value v; since every mass in the model is proportional to v, every coupling of h is that mass divided by v, and the fermion masses are the singular values of arbitrary Yukawa matrices Yf v/√2.
Example. With mₕ = 125.2 GeV and v = 246.2 GeV, λ = mₕ²/(2v²) = 0.129; the top couples to h with strength mₜ/v = 172.6/246.2 = 0.70, the electron with 0.000511/246.2 = 2.1 × 10⁻⁶.
The complete physical model and assumptions
Once three of the doublet's four real components have been eaten, what remains is a single real scalar h riding on the vacuum value: Φ = (0, (v + h)/√2)ᵀ in unitary gauge. Substituting that one expression everywhere is the whole content of this topic. In the potential it turns the curvature at the minimum into a mass, mₕ² = 2λv², so the measured 125 GeV fixes λ ≈ 0.13 and predicts nothing. In the gauge kinetic term it turns each boson mass term M² into M²(1 + h/v)², so h couples to W and Z with a strength set by their masses. And in the only gauge-invariant, renormalisable term that can join a left-handed doublet to a right-handed singlet, the Yukawa term, it produces a mass matrix Mf = Yf v/√2 together with a coupling to h that is the same matrix divided by v. Because Yf is a general complex 3 × 3 matrix its diagonalisation needs two unitaries, one per chirality, and the up- and down-type left-handed rotations need not agree: their mismatch, VCKM = ULu† ULd, is the only place the flavour basis leaves a trace. The cost is honesty about inputs. Thirteen of the Standard Model's free parameters live in this one term, every Yukawa is fitted to a measured mass, and the first-generation ones have never been measured through the Higgs at all.
1Put v + h into the potential and read off the mass
Three of the doublet's four real fields are gauge artefacts once SU(2)L × U(1)Y breaks, so choose unitary gauge and write Φ = (0, (v + h)/√2)ᵀ with h real. Then Φ†Φ = (v + h)²/2 and V = μ²Φ†Φ + λ(Φ†Φ)² becomes μ²(v + h)²/2 + λ(v + h)⁴/4. Requiring the term linear in h to vanish gives v² = −μ²/λ; the quadratic term is then λv² h², so ½mₕ² = λv² and mₕ² = 2λv² = −2μ². The cubic and quartic pieces, λv h³ + (λ/4)h⁴, are the self-couplings of h. Measurement runs opposite to the algebra: mₕ = 125.20 GeV and v = 246.22 GeV give λ = mₕ²/(2v²) = 0.129 and |μ| = mₕ/√2 = 88.5 GeV, so both potential parameters are fitted. The trilinear coefficient λv = 31.8 GeV is the one genuine prediction of the potential, and double-Higgs production has so far bounded it only to within a few times that value.
2Every coupling of h is a mass over v
Every mass in the model is a coupling times v, so replace v by v + h in each mass term and the coupling of h falls out. For a boson the mass enters squared: |DμΦ|² contains MW² W⁺μW⁻μ (1 + h/v)² and ½MZ² ZμZμ (1 + h/v)², so the terms linear in h are (2MW²/v) h W⁺W⁻ and (MZ²/v) h ZZ, which are 52.5 GeV and 33.8 GeV for MW = 80.37 GeV and MZ = 91.19 GeV. The Z term lacks the factor 2 only because its two fields are identical; its vertex factor is 2MZ²/v = 67.5 GeV, the same form as the W's 2MW²/v. For a fermion the mass enters once: yf (v + h)/√2 f̄f gives mf f̄f plus (mf/v) h f̄f, a dimensionless coupling of 0.70 for the top and 2.1 × 10⁻⁶ for the electron. On log axes of vertex factor against mass the fermion points mf/v lie on a line of slope one and the boson points 2MV²/v on a line of slope two, one common v fixing both intercepts. The ATLAS and CMS κ-fits place t, b, τ, W and Z on those lines to about ten per cent, see the muon at the three-σ level, bound the charm to within a few times its predicted value, and say nothing direct about e, u or d.
3The Yukawa term is the only bridge from left to right
A Dirac mass m f̄L fR is forbidden before breaking: fL sits in an SU(2) doublet with one hypercharge and fR in a singlet with another, so the product is not gauge invariant. The doublet Φ is what repairs it. Take QL with Y = 1/3, Φ with Y = 1 and dR with Y = −2/3: Q̄L Φ dR is an SU(2) singlet with hypercharge −1/3 + 1 − 2/3 = 0. The up quark needs the conjugate doublet Φ̃ = iσ₂Φ*, which carries Y = −1: −1/3 − 1 + 4/3 = 0 again. Leptons use L̄L Φ eR with 1 + 1 − 2 = 0. These three structures, each multiplied by an arbitrary complex 3 × 3 matrix Yf in generation space, are the only renormalisable gauge-invariant terms that pair left with right, and there is no νR in the model to pair with, so no neutrino mass appears. Setting Φ to its vacuum value leaves mass matrices Mf = Yf v/√2 that are in general neither Hermitian nor symmetric: eighteen real numbers in each sector, with no principle fixing any of them.
4Diagonalise with two unitaries: masses are singular values
Because Mf is a general complex matrix, its eigenvalues are not its masses and one similarity transformation will not do. The right object is the singular value decomposition M = UL D UR†, with UL and UR unitary and D real, non-negative and diagonal: UL is the eigenvector matrix of the Hermitian, positive M M†, UR that of M†M, and both share the eigenvalues mᵢ². The two chiralities rotate independently, fL′ = UL† fL and fR′ = UR† fR, which is legitimate because the kinetic terms f̄L iγμ∂μ fL and f̄R iγμ∂μ fR are each invariant under any unitary rotation of their own field. In practice you call numpy.linalg.svd(M), which returns UL, the singular values and UR†, and the whole verification is checking that UL D UR† reproduces M to machine precision. The 2 × 2 toy in the worked examples shows the mechanism by hand: a non-symmetric matrix with entries near 5 and 90 MeV returns md = 4.70 MeV, mₛ = 93.0 MeV and a left rotation of 13.0°. Should a negative eigenvalue appear, it is absorbed as a sign in UR and the mass is still the positive singular value.
5The mismatch of the left-handed rotations is the CKM matrix
Rotating to the mass basis is harmless everywhere except one vertex. A neutral current f̄L γμ fL becomes f̄L′ UL† UL γμ fL′ = f̄L′ γμ fL′, diagonal again because UL is unitary; the same holds for right-handed fields and for the couplings to Z, photon and gluon. That is the GIM mechanism: no tree-level flavour change in neutral currents. The coupling to h stays diagonal for a different reason: it is the mass matrix itself divided by v, so the same two unitaries that diagonalise M diagonalise it too. The charged current ūL γμ dL mixes two sectors rotated by different matrices, so it becomes ūL′ (ULu† ULd) γμ dL′, and VCKM = ULu† ULd survives as the misalignment between the up and down left-handed rotations. URu and URd never appear, since the W couples only to left-handed fields, and in the lepton sector the massless neutrino can simply be redefined by ULe, so no leptonic mixing exists in the model as written, although neutrino oscillations show that real neutrinos have mass and do mix. A 3 × 3 unitary matrix has nine real parameters; rephasing the six quark fields removes five, one overall phase being invisible, leaving three angles and one phase — Wolfenstein's λ = 0.225, A = 0.83, ρ̄ = 0.16 and η̄ = 0.35 — and that single phase is the only source of CP violation in the quark sector.
6Count what is input and what is predicted
Tally what this one Lagrangian term costs. Yu and Yd together hold 36 real numbers; the freedom to rotate QL, uR and dR by U(3)³ removes 27 minus the single unbroken baryon-number phase, leaving 36 − 26 = 10 physical parameters: six quark masses, three CKM angles and one phase. The lepton sector adds three charged masses, so thirteen of the Standard Model's nineteen parameters are Yukawa data, every one fitted and none derived; the Higgs mass itself supplies λ rather than following from it. What the model does predict, and what is actually tested, is that every coupling of h is the corresponding mass over one common v, that VCKM is unitary, and that neutral currents are flavour-diagonal. Even that test is partial: the couplings to t, b, τ, W and Z are measured, the muon is emerging, the charm is loosely bounded, and for u, d and e the coupling to the Higgs has never been observed. Each first-generation Yukawa in the tables is a number inferred from a mass, not a measurement of the Higgs.
Key equations
Read the units, assumptions and limits before substituting.
Unitary gauge and the Higgs mass
Φ = (0, (v + h)/√2)ᵀmₕ² = 2λv² = −2μ²
Turns the measured mass into the quartic: λ is read off, never predicted, and λv = 31.8 GeV is the still-unmeasured h³ coupling.
v = 246.22 GeV and mₕ = 125.20 GeV give λ = 0.129 and |μ| = mₕ/√2 = 88.5 GeVCouplings of h scale with mass
L ⊃ (2MW²/v) h W⁺W⁻ + (MZ²/v) h ZZ − (mf/v) h f̄f
One common v fixes every coupling from a mass — the straight line on the coupling–mass plot that the LHC κ-fits test.
boson coefficients in GeV (52.5 and 33.8), fermion coefficient dimensionless; vertex factors 2MW²/v and 2MZ²/vYukawa term, the only left–right bridge
−LY = Q̄L Yd Φ dR + Q̄L Yu Φ̃ uR + L̄L Ye Φ eR + h.c.Φ̃ = iσ₂Φ*
The only renormalisable gauge-invariant way to pair L with R, so all fermion mass enters through Φ; with no νR there is no neutrino mass.
Yf are 3 × 3 complex and dimensionless; hypercharges sum to zero, −1/3 + 1 − 2/3 = 0Mass matrix and its singular value decomposition
Mf = Yf v/√2 = UL D UR†D = diag(m₁, m₂, m₃)
Masses are singular values, not eigenvalues: a non-Hermitian M needs two unitaries, and UR is invisible to the W.
UL diagonalises M M†, UR diagonalises M†M;numpy.linalg.svd(M) returns UL, D and UR†CKM as the left-handed misalignment
VCKM = ULu† ULdneutral currents keep U†U = 1
The charged current is the only vertex where the two rotations fail to cancel; neutral currents stay flavour-diagonal (GIM).
unitary, 3 angles + 1 phase after 5 rephasings; |Vᵤₛ| = 0.2243, |Vcb| = 0.041, |Vub| = 0.0038Yukawa values from masses
yf = √2 mf / v
Five orders of magnitude of hierarchy the model carries but does not explain.
yₜ = 0.99, yb = 0.024, yτ = 0.010, yₑ = 2.9 × 10⁻⁶ — every one an inputInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Slide k from 0 to 19: the solid axis swings to −12.7° and md drops from 9 to 4.7 MeV. Then set θᵤ to −12.5° and |Vᵤₛ| collapses: only the mismatch between the sectors is observable. Push k past 28 and md passes through zero and grows again: masses are singular values, not signed eigenvalues.
Live results
- LIGHT MASS md
- 4.72 MeV
- HEAVY MASS mₛ
- 93.3 MeV
- θC = θd − θᵤ
- -12.7 °
- |Vᵤₛ| = |sin θC|
- 0.220
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Off-diagonal entry k of Md: 19 MeV; Heavy diagonal entry b of Md: 89 MeV; Up-sector rotation angle θᵤ: 0.0 °
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
"The Higgs gives particles their mass" slides into "the Standard Model predicts the masses", and it does not. Gauge invariance fixes the form of the Yukawa term, Q̄L Φ dR, but says nothing about its coefficient: each entry of Yu, Yd and Ye is a free complex number, and mf = yf v/√2 is read backwards from the measured mass, so quoting yₜ = 0.99 is bookkeeping, not a result. What the model does predict is the ratio of coupling to mass, 1/v, identical for every fermion — which is why the LHC coupling plot is a test and the mass spectrum is not. The Higgs mass is fitted the same way, through λ. And most of the mass you can weigh is not Yukawa at all: the proton's 938 MeV is the field and kinetic energy of confined quarks and gluons, with only about 9 MeV of it quark mass.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
Take mₕ = 125.20 GeV and v = 246.22 GeV. Find the quartic coupling λ, the magnitude of the mass parameter μ and the trilinear coefficient λv. Then compare the Higgs couplings of the top quark (mₜ = 172.6 GeV) and the electron (mₑ = 0.511 MeV): the dimensionless Yukawas and the ratio between them.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- In unitary gauge Φ = (0, (v + h)/√2)ᵀ, and V = μ²Φ†Φ + λ(Φ†Φ)² expands to a constant plus λv² h² + λv h³ + (λ/4) h⁴, so mₕ² = 2λv² and λ = mₕ²/(2v²).
- λ = 125.20²/(2 × 246.22²) = 15675.0/121248.6 = 0.1293.
- At the minimum v² = −μ²/λ, so μ² = −λv² = −mₕ²/2 = −7837.5 GeV² and |μ| = 88.5 GeV; the h³ coefficient is λv = 0.1293 × 246.22 = 31.8 GeV.
- yf = √2 mf/v: yₜ = 1.4142 × 172.6/246.22 = 0.991 and yₑ = 1.4142 × 5.11 × 10⁻⁴/246.22 = 2.94 × 10⁻⁶.
- The ratio yₜ/yₑ = mₜ/mₑ = 172.6/(5.11 × 10⁻⁴) = 3.38 × 10⁵: five orders of magnitude, every decade of it put in by hand.
Answer. λ = 0.129, |μ| = 88.5 GeV, λv = 31.8 GeV; yₜ = 0.99 and yₑ = 2.9 × 10⁻⁶, a ratio of 3.4 × 10⁵ that nothing in the model explains.
MediumExample 2
In a two-generation toy, choose the up-sector basis so that Mᵤ is already diagonal, and let the down-sector mass matrix in the flavour basis be Md = ( 4.58 −20.9 ; 1.06 90.6 ) MeV. Find the two down-type masses, the left-handed rotation ULd, and hence the Cabibbo angle and |Vᵤₛ|.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- Md is not symmetric, so its eigenvalues are not the masses. The masses are the singular values, the square roots of the eigenvalues of the Hermitian matrix Md Mdᵀ, and ULd is the eigenvector matrix of Md Mdᵀ.
- Md Mdᵀ: (1,1) = 4.58² + 20.9² = 20.98 + 436.81 = 457.79; (2,2) = 1.06² + 90.6² = 1.12 + 8208.36 = 8209.48; (1,2) = 4.58 × 1.06 − 20.9 × 90.6 = 4.85 − 1893.54 = −1888.69, all in MeV².
- For a real symmetric 2 × 2 the eigenvalues are mean ± half-difference: mean = (457.79 + 8209.48)/2 = 4333.64 and half-difference = √(3875.85² + 1888.69²) = √(1.85893 × 10⁷) = 4311.54, so λ₊ = 8645.18 MeV² and λ₋ = 22.10 MeV².
- Masses: mₛ = √8645.18 = 93.0 MeV and md = √22.10 = 4.70 MeV. The right rotation URd, which would diagonalise MdᵀMd, is never needed.
- Rotation angle: tan 2θ = 2(−1888.69)/(457.79 − 8209.48) = 0.4873, so 2θ = 26.0° and θ = 13.0°. The light eigenvector satisfies (457.79 − 22.10)x − 1888.69y = 0, so y/x = 0.2307 and it points at +13.0°: ULd = R(13.0°).
- With ULu = 1, V = ULu† ULd = R(13.0°), so |Vᵤₛ| = sin 13.0° = 0.225 and |Vud| = cos 13.0° = 0.974, against the measured 0.2243 and 0.9737.
numpy.linalg.svd(Md) returns the same three numbers.
Answer. md = 4.70 MeV, mₛ = 93.0 MeV, θC = 13.0°, |Vᵤₛ| = 0.225; the CKM angle is the mismatch of the two left rotations and URd never entered.
HardExample 3
Using Γ(h → f f̄) = Nc mₕ mf²/(8π v²) at leading order, with mₕ = 125.20 GeV, v = 246.22 GeV, mμ = 105.66 MeV, mₑ = 0.5110 MeV and a total width Γₕ = 4.1 MeV, compute Γ(h → μ⁺μ⁻) and its branching ratio, predict BR(h → e⁺e⁻), and find the largest electron coupling modifier κₑ = yₑ/yₑSM allowed by the limit BR(h → e⁺e⁻) < 3.0 × 10⁻⁴.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- The vertex is mf/v, so the width goes as mf²; leptons have Nc = 1, and the phase-space factor (1 − 4mf²/mₕ²)³⁄² differs from 1 by less than 10⁻⁵, so drop it.
- Denominator: 8π v² = 25.133 × 60624 = 1.5237 × 10⁶ GeV². Numerator: mₕ mμ² = 125.20 × 0.10566² = 125.20 × 0.011164 = 1.3977 GeV³.
- Γ(h → μμ) = 1.3977/(1.5237 × 10⁶) = 9.17 × 10⁻⁷ GeV = 0.917 keV, so BR(h → μμ) = 9.17 × 10⁻⁷/(4.1 × 10⁻³) = 2.24 × 10⁻⁴ — about two decays in ten thousand, which is what the LHC evidence for h → μμ is built on.
- Only the mass changes for the electron: Γ(h → ee)/Γ(h → μμ) = (mₑ/mμ)² = (0.5110/105.66)² = (4.836 × 10⁻³)² = 2.34 × 10⁻⁵, so BR(h → ee) = 2.24 × 10⁻⁴ × 2.34 × 10⁻⁵ = 5.2 × 10⁻⁹, a width of 0.021 eV.
- A modified coupling scales the width by κₑ²: κₑ² × 5.2 × 10⁻⁹ < 3.0 × 10⁻⁴ gives κₑ² < 5.7 × 10⁴ and κₑ < 240. The electron Yukawa could be two hundred times its Standard Model value without contradicting any Higgs measurement.
Answer. Γ(h → μμ) = 0.92 keV and BR = 2.2 × 10⁻⁴; BR(h → ee) = 5.2 × 10⁻⁹; the limit only requires κₑ < 240, so the first-generation Yukawa is not measured at all.
Quick check
Select an answer and explain it before checking.
Review & continue
The Higgs boson is the radial excitation h of the doublet about its vacuum value v; since every mass in the model is proportional to v, every coupling of h is that mass divided by v, and the fermion masses are the singular values of arbitrary Yukawa matrices Yf v/√2.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.