Where the Masses Come From · Elementary Particle Physics

The mechanism predicts that a scalar exists and that its couplings are proportional to mass. It predicts neither the masses nor the scalar’s own, and that split is what the rest of the chapter tests.

🎯 Why this matters

A theory fixing ratios but not values is easier to test than one fixing everything. Ratios are what experiments measure best, so the mechanism can be checked without anybody having predicted a single mass.

Everything so far has assumed masses that the theory forbids. §9.1 showed why: a mass term joins $\psi_L$ to $\psi_R$ , and those live in different representations of SU(2), so the term is not gauge invariant and cannot be written. The gauge bosons are worse — an explicit $M^2W_\mu W^\mu$ destroys the symmetry outright, and with it the renormalizability ‘t Hooft proved in 1971.

This section is the BEH mechanism — how the masses get in anyway, without the Lagrangian’s symmetry being touched.

The idea, in the simplest possible case

Forget the group theory. Take one real scalar field with a potential:

V(\Phi) = \tfrac12\,\htmlClass{t-m}{\mu^2}\Phi^2 + \tfrac14\,\htmlClass{t-l}{\lambda}\Phi^4

(9.101)

Bettini p. 395. Two terms, because higher powers would spoil renormalizability. Everything depends on the SIGN of the first coefficient.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

a symmetric law with an asymmetric ground state

μ² = -1.00 — symmetry spontaneously brokenexplicit tilt = 0.00 — the analogue of a quark mass

ground state at φ =

-0.707

across the valley — massive

3.99

around it — the Goldstone mode

0.00

Broken — and notice the law did not change. The potential is still perfectly symmetric; the two minima prove it. But the ball must sit at one of them, and once it does, the symmetry is invisible in the ground state. Motion around the valley floor costs nothing: that flat direction is a massless Goldstone boson. Motion across it costs: that is the massive partner. In QCD the flat directions are the pseudoscalar octet and the massive one is the f₀.

Where the four degrees of freedom go

φ₁, φ₂	the charged pair	→ longitudinal states of W⁺ and W⁻	eaten
φ₄	the flat direction — the Goldstone mode	→ longitudinal state of the Z⁰	eaten
φ₃	the radial direction — costs energy to excite	→ the physical H	left over

Nothing is created and nothing is lost. A massless vector boson has two polarisation states; a massive one has three. So every boson that acquires mass must take exactly one scalar with it — and 3 taken + 1 left over = the 4 you started with. The photon stays massless because one generator of SU(2)⊗U(1) survives unbroken, and that surviving generator is electric charge.

The same widget as §6.9, now with the ledger that makes it the Higgs case. Slide μ² through zero: the law stays symmetric — the ring of minima proves it — but the ball must sit somewhere, and once it does the symmetry is invisible in the ground state. Motion around the valley costs nothing; motion across it costs.

For $\mu^2 < 0$ the minima sit at

\Phi_{\min} = \pm\sqrt{-\mu^2/\lambda} \equiv \pm\upsilon

and it is convenient to add a constant so the minimum energy is zero, giving the Mexican-hat form the widget draws:

V(\Phi) = \frac{\lambda}{4}\left(\Phi^2 - \upsilon^2\right)^2

💡 What this really says — nothing has an imaginary mass, and the vacuum is not empty

Two things trip everyone here, and both are worth stating plainly.

” $\mu$ imaginary” is not a particle with imaginary mass. A particle is a small oscillation of a field about a stable configuration. When $\mu^2 < 0$ the point $\Phi = 0$ is a maximum, so there are no small oscillations about it — there is nothing there to be a particle. Expand instead about the true minimum $\Phi = \upsilon$ , and the curvature there is $+2\lambda\upsilon^2$ , positive, giving a perfectly ordinary mass $M_H = \upsilon\sqrt{2\lambda}$ . The book says “there will be no physical object with imaginary mass” and that is why.

The ground state is not the state with zero field. This is the genuinely new idea. Ordinarily “vacuum” means “nothing here”. Here the vacuum is a configuration in which the scalar field has a non-zero value everywhere, and that value — the vacuum expectation value $\upsilon$ — is what every mass in the theory is proportional to.

An engineer will recognise the shape as bias. A transistor amplifier is linear only about an operating point that is not zero; the small-signal parameters — gain, impedance, cutoff — are all properties of the bias point, not of the device in isolation. Change the bias and every one of them changes. Here the bias is $\upsilon = 246$ GeV, and the “small-signal parameters” are the masses of every particle in the Standard Model.

The buckling column is the other standard image: load a vertical strut hard enough and it bows, in a direction that the perfectly axisymmetric problem never picked out. The equations are symmetric; the outcome is not; and asking which way it bowed is not a question the equations answer.

The real thing: four fields, three eaten

For the electroweak theory $\Phi$ is not one real field but a complex SU(2) doublet — four real components. Requiring the vacuum to be electrically neutral kills two of them, leaving a Mexican hat in the plane of the other two, and choosing a point on its circular valley breaks the symmetry.

🪜 From four scalars to three longitudinal states and one Higgs

Step 1 of 5 — start with the doublet

\Phi = \begin{pmatrix}\phi^+\\ \phi^0\end{pmatrix},\qquad \phi^+ = \frac{\phi_1 + i\phi_2}{\sqrt2},\quad \phi^0 = \frac{\phi_3 + i\phi_4}{\sqrt2}

Why you may do this: Two complex fields, so four real degrees of freedom. The superscripts are electric charges — the doublet must contain one charged and one neutral component for the hypercharge assignment of §9.1 to work.

Bettini pp. 396–398. The counting is the whole content, and it is exact.

⚙️ Engineer’s bridge — the Goldstone bosons are not destroyed — they are a change of basis

“The gauge bosons eat the Goldstone bosons” is the standard phrase and it sounds like a physical process. It is not; it is a statement about coordinates, and the book gives the right analogy in one paragraph about classical electromagnetism.

A photon travelling along $z$ has a four-potential with four components, but $A_z$ and $A_t$ are gauge-dependent and unobservable — only the two transverse components are physical. So “four components, two physical” was already true in QED, and nobody finds it disturbing.

The BEH mechanism is the same bookkeeping with a different split. Before breaking: three massless gauge fields with two physical states each (6), plus four scalars — but three combinations of those scalars are pure gauge and unobservable. After breaking: three massive gauge fields with three physical states each (9), plus one scalar. 6 + 4 = 10 = 9 + 1. The would-be Goldstone modes were never independently observable; the symmetry breaking just relabels which field they belong to.

This is the same move as choosing a gauge, and an engineer does it constantly: the state of a system does not change when you change coordinates, but which variables look independent does. A rotating machine analysed in the rotor frame has quantities that appear and disappear relative to the stator frame, and no energy went anywhere. What the BEH mechanism supplies is not new degrees of freedom, but a basis in which the physical content is manifest — and in that basis three gauge bosons are massive.

The corollary worth carrying: you cannot see a Goldstone boson here. Asking “where did they go” is asking about a coordinate artefact. What you can see is the one degree of freedom that was left over, and that is the Higgs boson .

Where it breaks: “they were never there” is language about a gauge choice, not about physics, and it misleads if pushed. In a different gauge the would-be Goldstone fields appear explicitly in the Lagrangian and cancel only in observables. More importantly, the degrees of freedom did not vanish: they are the longitudinal polarizations, and those are observable — the amplitude for longitudinally polarized WW scattering grows with energy and would violate unitarity near 1 TeV without the Higgs, which is exactly the argument that guaranteed the LHC would find something. A coordinate artefact cannot break your theory at a calculable energy. These can.

The VEV is not a free parameter — it is fixed by the Fermi constant, which has been known since the 1930s:

\htmlClass{t-v}{\upsilon} = \frac{1}{\sqrt{\sqrt2\,\htmlClass{t-g}{G_F}}} = 246\;\text{GeV}

(9.114)

Bettini p. 398. The scale of electroweak symmetry breaking, extracted from muon decay. Every mass in the Standard Model is this number times a dimensionless coupling.

Every symbol, one at a time

Hover or tap a symbol above — it lights up in the equation and its meaning, units and type appear here.

Nothing is created or destroyed — twelve degrees of freedom, rearranged

before breakingafter breakingW⁺, W⁻, Z — massless62 transverse polarizations eachγ — massless2Φ — a complex doubletφ⁺, φ⁰ → 4 real scalars4total12W⁺, W⁻, Z — MASSIVE93 polarizations each — a longitudinal one appearsγ — still massless2H — the survivor1total123 of the 4 are eatenthey become the longitudinal modesthe 4th stays, and is the particle you can find

A massless spin-1 field has 2 polarizations; a massive one has 3. The extra state has to come from somewhere, and it does.

Supplied — the step-through above does the counting correctly and in words, and the ledger form is what makes it impossible to misread. “Eaten” is not a metaphor for a scalar disappearing. A massless spin-1 field has two transverse polarizations and a massive one has three, so giving W⁺, W⁻ and Z mass requires three new states, and the doublet has exactly three to spare.

The totals match on both sides because degrees of freedom cannot be created by a change of variables — which is all spontaneous breaking is. The photon keeps two because the vacuum remains invariant under the generator of electric charge, and the one scalar left over is the Higgs boson: not the cause of the masses, but the piece of the field that was not used up making them.

Fermion masses: a second mechanism, in the same field

The gauge-boson masses came from the kinetic term. The fermion masses come from somewhere else entirely, and the book is careful to say so: “different terms in the Lagrangian are responsible”, and both must be tested separately.

The problem, restated: $m\bar e_Le_L$ and $m\bar e_Re_R$ vanish identically by the properties of the chiral projectors, and $m\bar e_Le_R$ is an isospin doublet, not a scalar, so it cannot appear in a gauge-invariant Lagrangian. But $\bar e_L\phi^0 e_R$ is an isoscalar, and its coefficient is the electron’s Yukawa coupling — the doublet index of $\Phi$ contracts with the doublet index of $e_L$ . So:

\mathcal{L}_e = \frac{1}{\sqrt2}f_e\left(\bar e_L\phi^0 e_R + \bar e_R\phi^0 e_L\right) \qquad\Longrightarrow\qquad m_e = \frac{1}{\sqrt2}f_e\,\upsilon

what the Yukawa couplings actually are

import numpy as np
GF, MH = 1.1663788e-5, 125.25
v = 1/np.sqrt(np.sqrt(2)*GF)
lam = MH**2/(2*v**2)
print("the scale of electroweak symmetry breaking")
print(f"  v = 1/sqrt(sqrt2 G_F) = {v:.2f} GeV")
print(f"  lambda = M_H^2/(2 v^2) = {lam:.4f}   (M_H = {MH} GeV)")

print("\nevery fermion mass is v times its own coupling, f = sqrt2 m / v:")
F = {}
for nm, m in (('electron', 0.5109989e-3), ('muon', 0.1056584), ('tau', 1.77686),
              ('b quark', 4.18), ('top', 172.69)):
    F[nm] = np.sqrt(2)*m/v
    print(f"  {nm:11s} {F[nm]:.3e}")
print(f"\n  the top Yukawa is {F['top']:.3f} -- ONE, to within a per cent.")
print(f"  the electron's is 3 x 10^-6.  the ratio is {F['top']/F['electron']:.1e}, and the theory")
print( "  predicts NEITHER, nor their ratio.  that is 9 free parameters for")
print( "  the charged fermions alone (12 with the CKM angles and phase).")

mu, md, mp = 2.16, 4.67, 938.27
print("\nQuestion 9.5 -- how much of your mass is the Higgs responsible for?")
print(f"  the proton is uud, and 2 m_u + m_d = {2*mu+md:.2f} MeV")
print(f"  m_p = {mp} MeV, so with ZERO u and d Yukawa couplings")
print(f"  the proton would still weigh about {mp-(2*mu+md):.0f} MeV        book: ~928")
print(f"\n  -> the BEH mechanism supplies {(2*mu+md)/mp*100:.1f}% of the mass of ordinary matter.")
print( "     the other 99% is the gluon field energy of Sec. 6.7.")

V = lam*v**4/4
print("\nthe vacuum is not empty, and its energy density is absurd:")
print(f"  V = lambda v^4 / 4 = {V:.2e} GeV^4 = {V*130.2:.1e} GeV per cubic fermi")
print( "  (this is the cosmological constant problem, and the book does not")
print( "   mention it: the observed vacuum energy is ~10^-47 GeV^4, some 55")
print( "   orders of magnitude smaller.)")

prints

the scale of electroweak symmetry breaking
v = 1/sqrt(sqrt2 G_F) = 246.22 GeV
lambda = M_H^2/(2 v^2) = 0.1294   (M_H = 125.25 GeV)

every fermion mass is v times its own coupling, f = sqrt2 m / v:
electron    2.935e-06
muon        6.069e-04
tau         1.021e-02
b quark     2.401e-02
top         9.919e-01

the top Yukawa is 0.992 -- ONE, to within a per cent.
the electron's is 3 x 10^-6.  the ratio is 3.4e+05, and the theory
predicts NEITHER, nor their ratio.  that is 9 free parameters for
the charged fermions alone (12 with the CKM angles and phase).

Question 9.5 -- how much of your mass is the Higgs responsible for?
the proton is uud, and 2 m_u + m_d = 8.99 MeV
m_p = 938.27 MeV, so with ZERO u and d Yukawa couplings
the proton would still weigh about 929 MeV        book: ~928

-> the BEH mechanism supplies 1.0% of the mass of ordinary matter.
   the other 99% is the gluon field energy of Sec. 6.7.

the vacuum is not empty, and its energy density is absurd:
V = lambda v^4 / 4 = 1.19e+08 GeV^4 = 1.5e+10 GeV per cubic fermi
(this is the cosmological constant problem, and the book does not
 mention it: the observed vacuum energy is ~10^-47 GeV^4, some 55
 orders of magnitude smaller.)

⚠️ “The Higgs gives mass to everything” is false, and the book says so once

Buried on p. 399 is a sentence worth pulling out: “contrary to what one often finds written, the Higgs boson is not the origin of all the mass, but simply of that of the gauge bosons and of the fundamental fermions.”

Run the numbers, as Question 9.5 asks. The proton is $uud$ , and the three current quark masses sum to 9 MeV out of 938. Set every light-quark Yukawa coupling to zero and the proton still weighs about 929 MeV.

So the BEH mechanism accounts for roughly 1 % of the mass of ordinary matter. The other 99 % is the energy of the gluon field confined inside the nucleon — §6.7’s story, and a completely different mechanism.

What the Higgs is responsible for is more interesting than “all mass” anyway: the $W$ and $Z$ masses, without which the weak interaction would have infinite range and the universe would be unrecognisable; and the electron mass, which sets the size of atoms. Question 9.6 pushes this further — with massless $u$ and $d$ the pion would be massless too, the nuclear force would have infinite range, and the deuteron would be about 1.5 Bohr radii across.

What the mechanism predicts, and what it does not

Bettini Eq. (9.117). Once M_H is known every coupling is fixed — which is what made the post-2012 programme a set of null tests rather than measurements.

coupling↕	value↕	what it means↕
$g_{Hff}$	$m_f/\upsilon$	proportional to the fermion mass — so the H prefers the heaviest thing it can reach
$g_{HVV}$	$2M_V^2/\upsilon$	proportional to the boson mass squared, because the boson masses come from a different term
$g_{HHVV}$	$2M_V^2/\upsilon^2$	the four-point vertex, from the same term
$g_{HHH}$	$3M_H^2/\upsilon$	the self-coupling — a direct probe of the shape of the potential, and still untested (§9.20)
$g_{HHHH}$	$3M_H^2/\upsilon^2$	and the four-Higgs vertex, far out of reach

The linear-versus-quadratic split in the first two rows is the sharpest experimental signature of the whole mechanism, and §9.19 measures it directly: plot the coupling against the mass and bosons and fermions should fall on two different power laws, both passing through the same $\upsilon$ .

Aside — how a massless photon couples to a Higgs at all

Fig. 9.41 shows $H\to\gamma\gamma$ with a branching ratio of $2\times10^{-3}$ — and it was one of the two discovery channels. But the Higgs couples to mass, and the photon has none.

The coupling is entirely indirect, through loops of the heaviest charged things available (Fig. 9.40): a $W$ loop and a top loop. Two features matter:

the $W$ loop is about five times the top loop, because $g_{HWW}$ goes as $M_W^2$ while $g_{Htt}$ goes as $m_t$ — the quadratic coupling wins;
the two amplitudes have opposite signs, as bosonic and fermionic loops always do. They partially cancel.

That destructive interference is why $H\to\gamma\gamma$ is rare — and why it is such a good place to look for new physics: any new heavy charged particle would add its own loop and shift the rate. A channel that exists only through virtual particles is a channel sensitive to particles nobody has seen.

And what the theory does not predict is the list that closes the section:

$\lambda$ is arbitrary, so $M_H$ is not predicted — the book says so flatly, and it is why §9.13’s search had to cover 114 GeV to 1 TeV;
every Yukawa coupling is arbitrary, all nine of them, spanning $3\times10^{-6}$ to 1. “The inability of the theory to give any prediction on purpose points to a serious limitation of the Standard Model”;
and neutrinos get no mass at all, because the recipe needs a right-chirality partner and none has ever been seen. That is not a small omission — it is the hole Chapter 10 falls into, since neutrinos oscillate and therefore have mass.

🔑 If you remember only three things

Fermion masses arrive by a second mechanism using the same field. Calling it one mechanism is a simplification that hides where the arbitrary numbers enter.
Four fields go in and three are eaten. The counting is the argument, and what survives is the one particle left over.
The photon stays massless because one combination is untouched. Masslessness here is a leftover of the arithmetic rather than a separate assumption.

Where this goes next

§9.13 searches for the scalar at LEP and the Tevatron and does not find it, but bounds it: $M_H > 114.4$ GeV from below, and a window excluded from above. §9.14 builds the machine that could cover the rest, and §9.15 is 4 July 2012.

Then §9.16–9.19 check that what was found is this — a scalar, with couplings proportional to mass for fermions and mass squared for bosons, on the two power laws this section predicts.

✅ Check yourself — where the masses come from

0/6 answered · 0 correct

1.The potential has μ² < 0. Does that mean a particle of imaginary mass?
2.Four real scalar degrees of freedom go in and one Higgs comes out. Where did the other three go?
3.Why does the photon stay massless while the W and Z do not?
4.Setting the u and d Yukawa couplings to zero, roughly what would the proton weigh?
5.g_Hff ∝ m_f but g_HVV ∝ M_V². Why the different powers, and why does it matter?
6.H → γγ has a branching ratio of 2×10⁻³ and was a discovery channel. How does a massless photon couple to a particle that couples to mass?