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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.
Starting model: Higgs boson and Yukawa mass matricesLeft: the down-sector flavour axes d and s; the solid arrow is the light mass eigenvector of M_d = (9, k; k, b) MeV at θ_d = -12.7°, the dashed arrow the up-sector mass axis at θ_u = 0.0°. Right: bars of |V_ud| and |V_us| for θ_C = θ_d − θ_u = -12.7°, beside the measured 0.224; m_d = 4.72 MeV.solid: down-sector mass axis d′, θd = −12.7°dashed: up-sector mass axis u′, θᵤ = 0.0°Md = ( 9 19 ; 19 89 ) MeVsddashed line: measured 0.224θC = θd − θᵤ = −12.7°|Vud||Vᵤₛ|
Start with a picture. Change the model in the lab below.

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.

1

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.

2

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 GeV

Couplings 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/v

Yukawa 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 = 0

Mass 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| = 0.0038

Yukawa 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 input
3

Interactive lab

Predict first. Explore the model. Compare the result.

Explore the model

19 MeV
−40 to 40 MeV; steps of 1.
89 MeV
30 to 150 MeV; steps of 1.
0.0 °
−45 to 45 °; steps of 0.5.

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.

Interactive Physics V modelLeft: the down-sector flavour axes d and s; the solid arrow is the light mass eigenvector of M_d = (9, k; k, b) MeV at θ_d = -12.7°, the dashed arrow the up-sector mass axis at θ_u = 0.0°. Right: bars of |V_ud| and |V_us| for θ_C = θ_d − θ_u = -12.7°, beside the measured 0.224; m_d = 4.72 MeV.solid: down-sector mass axis d′, θd = −12.7°dashed: up-sector mass axis u′, θᵤ = 0.0°Md = ( 9 19 ; 19 89 ) MeVsddashed line: measured 0.224θC = θd − θᵤ = −12.7°|Vud||Vᵤₛ|
Left: the down-sector flavour axes d and s; the solid arrow is the light mass eigenvector of Md = (9, k; k, b) MeV at θd = −12.7°, the dashed arrow the up-sector mass axis at θᵤ = 0.0°. Right: bars of |Vud| and |Vᵤₛ| for θC = θd − θᵤ = −12.7°, beside the measured 0.224; md = 4.72 MeV.

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 °

Baseline 19; choose −40 to 40, in steps of 1.
4. Compared with the saved baseline, this result will…

This experiment is kept only while this page is open. It does not change saved course progress.

4

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.

5

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.

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.

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.

6

Quick check

Select an answer and explain it before checking.

A colleague changes basis for the right-handed down-type quarks, dR → W dR, with W an arbitrary 3 × 3 unitary matrix, and re-derives the mass spectrum and the CKM matrix. What changes?
7

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.

Extra challenging questions

Connect several ideas, explore supplied data, and compare your reasoning with a worked solution.