The Higgs Mechanism

Spontaneously breaking a global symmetry produces massless Goldstone bosons. Spontaneously breaking a gauge symmetry does something else: the Goldstone modes are gauge artifacts, and rather than propagating as massless scalars they are absorbed by the gauge fields, which become massive. This is the Higgs mechanism, the only known way to give a gauge boson a mass without spoiling the gauge invariance that makes the theory consistent. This lesson works the mechanism first for a single abelian gauge field, then for the full $SU(2{)}_L\times U(1{)}_Y$ of the electroweak theory, deriving the $W$ and $Z$ masses, the massless photon, and the fermion masses from Yukawa couplings.

Throughout, $\hslash =c=1$. The Higgs doublet is written $\Phi$, its vacuum expectation value $v$, and the physical scalar excitation $h$.

The abelian Higgs model

Couple the complex scalar of the previous lesson to a $U(1)$ gauge field $A_{\mu }$ by promoting the global phase symmetry to a local one. The Lagrangian is

with $V(\varphi )=-{\mu }^2{\varphi }^{\ast }\varphi +\lambda ({\varphi }^{\ast }\varphi {)}^2$ and ${\mu }^2>0$, so the minimum is a circle $|\varphi |=v/\sqrt{2}$ with $v=\sqrt{{\mu }^2/\lambda }$. As before, choosing the vacuum breaks the symmetry. Parametrize the field about the vacuum in radial and phase pieces,

The phase $\theta (x)$ plays the role of a local gauge parameter, so it can be removed by a gauge transformation — the unitary gauge. Setting $\theta =0$ eliminates the Goldstone field from the Lagrangian entirely. The kinetic term of the scalar then contains

a mass term for the gauge field, $\frac{1}{2}{m}_A^2{A}_{\mu }{A}^{\mu }$ with

The gauge boson has become massive by absorbing the Goldstone mode. The degree-of- freedom count balances exactly: before breaking, a massless gauge field carries two polarizations and the complex scalar carries two real components, four in all; after breaking, a massive gauge field carries three polarizations and one real scalar $h$ survives, again four. The Goldstone boson has not disappeared — it has become the longitudinal polarization the gauge field lacked while massless.

The electroweak Higgs doublet

To break $SU(2{)}_L\times U(1{)}_Y$ down to the electromagnetic $U(1{)}_{\text{EM}}$, the scalar must be an $SU(2)$ doublet carrying hypercharge. The minimal choice is a single complex doublet with hypercharge $Y=1$,

four real scalar fields in all. For ${\mu }^2>0$ the potential is minimized on the sphere ${\Phi }^{†}\Phi =v^2/2$. The vacuum is chosen to lie entirely in the lower, electrically neutral component,

so that the vacuum carries no electric charge. This choice is what leaves the photon massless. The upper component ${\varphi }^{+}$ has $Q=T_3+Y/2=+\frac{1}{2}+\frac{1}{2}=1$ and the lower ${\varphi }^0$ has $Q=-\frac{1}{2}+\frac{1}{2}=0$; a charged vacuum would break electromagnetism, which is not observed, so the neutral direction is the physical one.

The unbroken generator is the specific combination that annihilates the vacuum. Acting on $⟨\Phi ⟩$ with the electric-charge operator $Q=T_3+Y/2=\frac{1}{2}{\sigma }^3+\frac{1}{2}$,

so $Q$ leaves the vacuum invariant: $U(1{)}_{\text{EM}}$ survives and its gauge boson, the photon, stays massless. The other three combinations of generators move the vacuum and are broken, producing three Goldstone bosons — exactly the number needed to give mass to three gauge bosons.

Gauge-boson masses

The gauge-boson masses come from the covariant-derivative term $(D_{\mu }\Phi {)}^{†}(D^{\mu }\Phi )$ evaluated at the vacuum, with

Substituting $\Phi \to ⟨\Phi ⟩$ and reading off the terms quadratic in the gauge fields gives, after collecting $W^1,W^2$ into $W^{\pm }$ and diagonalizing the neutral $W^3$–$B$ block,

The three broken directions have given mass to $W^{+},W^{-},Z$; the unbroken $U(1{)}_{\text{EM}}$ direction leaves the photon massless. Dividing the two massive results reproduces the tree-level relation of the previous lesson,

so the mass ratio and the mixing angle are the same information. The single physical scalar remaining after the three Goldstones are eaten is the Higgs boson $h$, with mass

Its mass depends on the quartic coupling $\lambda$, which is not fixed by any other measurement, so $M_h$ is a genuinely free parameter of the theory — the one number the Higgs mechanism cannot predict, and which had to be measured directly.

The electroweak scale

The value of $v$ is not predicted, but it is fixed by low-energy weak data. Matching the $W$-exchange amplitude to Fermi's four-fermion theory relates the Fermi constant $G_F$ to the $W$ mass and coupling,

using $M_W=gv/2$. Solving for $v$ with the measured $G_F=1.166\times 10^{-5}{\text{GeV}}^{-2}$,

This is the electroweak scale, the only dimensionful parameter in the classical Standard Model Lagrangian. Every gauge-boson and fermion mass is $v$ times a dimensionless coupling: $M_W=gv/2$, $M_Z=v\sqrt{g^2+g^{\prime 2}}/2$, and the fermion masses below. If $v$ were zero, every elementary particle would be massless.^tong-vev

Fermion masses and Yukawa couplings

A Dirac mass term $m\bar{\psi }\psi =m({\bar{\psi }}_L{\psi }_R+{\bar{\psi }}_R{\psi }_L)$ pairs a left-handed field with a right-handed one. In the electroweak theory the two have different $SU(2)$ and hypercharge assignments — ${\psi }_L$ is part of a doublet, ${\psi }_R$ a singlet — so a bare mass term is not gauge invariant. Mass must instead come from a gauge-invariant coupling to the Higgs doublet, the Yukawa coupling. For the electron,

which is an $SU(2)$ singlet because ${\bar{L}}_L$ (a doublet) contracts with $\Phi$ (a doublet), and hypercharge balances. When $\Phi$ takes its vacuum value $⟨\Phi ⟩=(0,v/\sqrt{2})$, this becomes a mass term with

Every fermion mass is its Yukawa coupling times $v/\sqrt{2}$. The couplings themselves span six orders of magnitude, from $y_t\approx 1$ for the top quark down to $y_e\approx 3\times 10^{-6}$ for the electron, and the Standard Model does not explain this hierarchy — it takes the Yukawa couplings as inputs. What it does predict is that the coupling of each fermion to the physical Higgs boson $h$ is proportional to that fermion's mass, since both descend from the same term: the heavier the particle, the more strongly it couples to the Higgs, a relation tested directly in Higgs decays.

Summary

The Higgs mechanism gives gauge bosons mass without breaking gauge invariance: a spontaneously broken gauge symmetry has no physical Goldstone bosons, and the would-be Goldstone modes become the longitudinal polarizations of the gauge fields. In the abelian model a single complex scalar makes the photon-like field massive, $m_A=qv$. In the electroweak theory a single Higgs doublet with vacuum $⟨\Phi ⟩=(0,v/\sqrt{2})$ breaks $SU(2{)}_L\times U(1{)}_Y\to U(1{)}_{\text{EM}}$; three of its four scalars are eaten by the $W^{\pm }$ and $Z$, giving $M_W=gv/2$ and $M_Z=M_W/\cos {\theta }_W$, while the photon stays massless because the vacuum is electrically neutral. The fourth scalar is the Higgs boson, $M_h=\sqrt{2\lambda }v$, with $\lambda$ free. Fermions get mass from Yukawa couplings to the same doublet, $m_f=y_{f}v/\sqrt{2}$, so every mass is $v\approx 246\text{GeV}$ times a dimensionless coupling and the coupling of each fermion to the Higgs is proportional to its mass. What remains is to find the particle: the discovery and properties of the Higgs boson.¹²