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The Higgs Mechanism

The Higgs mechanism gives mass to the W±W^\pm and Z0Z^0 bosons while preserving gauge invariance. The Key idea: a scalar field acquires a non-zero vacuum expectation value (VEV), spontaneously Breaking the electroweak symmetry.

The Higgs potential:

V(ϕ)=μ2ϕ†ϕ+λ(ϕ†ϕ)2V(\phi) = \mu^2 \phi^\dagger\phi + \lambda(\phi^\dagger\phi)^2

With μ2<0\mu^2 \lt 0 and λ>0\lambda \gt 0. This is the Mexican hat potential: the minimum is at ∣ϕ∣=v/2|\phi| = v/\sqrt{2} where v=−μ2/λv = \sqrt{-\mu^2/\lambda}.

4.2 Worked Example: Mexican Hat Potential Analysis

Section titled “4.2 Worked Example: Mexican Hat Potential Analysis”
Example 4.1: Detailed analysis of spontaneous symmetry breaking

Consider the complex scalar field ϕ=(ϕ1+iϕ2)/2\phi = (\phi_1 + i\phi_2)/\sqrt{2} with the potential:

V(ϕ)=μ2∣ϕ∣2+λ∣ϕ∣4V(\phi) = \mu^2\lvert\phi\rvert^2 + \lambda\lvert\phi\rvert^4

With μ2<0\mu^2 \lt 0, λ>0\lambda \gt 0.

Step 1: Find the minimum. Setting ∂V/∂ϕi=0\partial V/\partial\phi_i = 0:

∂V∂ϕ1=ϕ1(μ2+λ(ϕ12+ϕ22))=0\frac{\partial V}{\partial\phi_1} = \phi_1(\mu^2 + \lambda(\phi_1^2 + \phi_2^2)) = 0 ∂V∂ϕ2=ϕ2(μ2+λ(ϕ12+ϕ22))=0\frac{\partial V}{\partial\phi_2} = \phi_2(\mu^2 + \lambda(\phi_1^2 + \phi_2^2)) = 0

The solutions are:

  • ϕ1=ϕ2=0\phi_1 = \phi_2 = 0: This is a local maximum since V"′=μ2<0V"' = \mu^2 \lt 0.
  • ϕ12+ϕ22=−μ2/λ≡v2\phi_1^2 + \phi_2^2 = -\mu^2/\lambda \equiv v^2: This is the circle of minima.

The VEV is v=−μ2/λv = \sqrt{-\mu^2/\lambda}. The symmetry is U(1)\mathrm{U}(1) (phase rotations ϕ→eiαϕ\phi \to e^{i\alpha}\phi), which has a continuous set of degenerate ground states.

Step 2: Expand around the vacuum. Choose the vacuum ⟨ϕ⟩=v/2\langle\phi\rangle = v/\sqrt{2} by Fixing the gauge. Parameterise:

ϕ(x)=12(v+h(x))\phi(x) = \frac{1}{\sqrt{2}}(v + h(x))

Where h(x)h(x) is a real scalar field (the “Higgs field” fluctuation).

Step 3: Compute the physical mass. Substituting into the potential:

V=μ2⋅12(v+h)2+λ⋅14(v+h)4V = \mu^2 \cdot \frac{1}{2}(v + h)^2 + \lambda \cdot \frac{1}{4}(v + h)^4

Expanding and keeping terms through quadratic order in hh:

V=μ2v22+λv44⏟constant+(μ2v+λv3)⏟=0 h+(μ22+3λv22)⏟mh2/2 h2+⋯V = \underbrace{\frac{\mu^2 v^2}{2} + \frac{\lambda v^4}{4}}_{\mathrm{constant} + \underbrace{(\mu^2 v + \lambda v^3)}_{= 0}\,h + \underbrace{\left(\frac{\mu^2}{2} + \frac{3\lambda v^2}{2}\right)}_{m_h^2/2}\,h^2 + \cdots}

Using v2=−μ2/λv^2 = -\mu^2/\lambdaThe coefficient of hh vanishes (as required at the minimum) And:

mh2=μ2+3λv2=−λv2+3λv2=2λv2m_h^2 = \mu^2 + 3\lambda v^2 = -\lambda v^2 + 3\lambda v^2 = 2\lambda v^2

The physical Higgs mass is mh=v2λm_h = v\sqrt{2\lambda}.

Step 4: Goldstone boson. The original complex field had two real degrees of freedom. After symmetry breaking, one (hh) becomes a massive particle. The other (the phase Fluctuation) is the Goldstone boson --- a massless mode corresponding to motion Along the circle of degenerate minima. In gauge theory, this Goldstone boson is “eaten” By the gauge field to become its longitudinal polarisation. ■\blacksquare

The Higgs field is an SU(2) doublet:

ϕ=(ϕ+ϕ0),⟨ϕ⟩0=12(0v)\phi = \begin{pmatrix} \phi^+ \\ \phi^0 \end{pmatrix}, \quad \langle\phi\rangle_0 = \frac{1}{\sqrt{2}}\begin{pmatrix} 0 \\ v \end{pmatrix}

Where v≈246v \approx 246 GeV.

Expanding around the VEV, the kinetic term ∣Dμϕ∣2|D_\mu\phi|^2 (with DμD_\mu the covariant derivative) Produces mass terms:

mW=12gv,mZ=12g2+g′2 v,mγ=0m_W = \frac{1}{2}gv, \quad m_Z = \frac{1}{2}\sqrt{g^2 + g'^2}\,v, \quad m_\gamma = 0

Where gg and g′g' are the SU(2) and U(1) coupling constants.

Proof of the mass relation. The covariant derivative is:

Dμ=∂μ−igτa2Wμa−ig′Y2BμD_\mu = \partial_\mu - ig\frac{\tau^a}{2}W^a_\mu - ig'\frac{Y}{2}B_\mu

The kinetic term ∣Dμϕ∣2|D_\mu\phi|^2 evaluated at the VEV gives:

∣Dμ⟨ϕ⟩0∣2=v28[g2(Wμ1)2+g2(Wμ2)2+(gWμ3−g′Bμ)2]\lvert D_\mu\langle\phi\rangle_0\rvert^2 = \frac{v^2}{8}\left[g^2(W^1_\mu)^2 + g^2(W^2_\mu)^2 + (gW^3_\mu - g'B_\mu)^2\right]

Identifying Wμ±=(Wμ1∓iWμ2)/2W^\pm_\mu = (W^1_\mu \mp iW^2_\mu)/\sqrt{2}The first two terms give:

12(gv2)2Wμ+W−μ  ⟹  mW=gv2\frac{1}{2}\left(\frac{gv}{2}\right)^2 W^+_\mu W^{-\mu} \implies m_W = \frac{gv}{2}

The last term, after the rotation to Z0Z^0 and AAGives:

mZ=v2g2+g′2,mγ=0■m_Z = \frac{v}{2}\sqrt{g^2 + g'^2}, \quad m_\gamma = 0 \quad \blacksquare

The ratio:

mWmZ=cos⁡θW\frac{m_W}{m_Z} = \cos\theta_W

Defines the Weinberg angle θW≈28.7∘\theta_W \approx 28.7^\circ.

Fermions acquire mass through Yukawa couplings to the Higgs field:

LYukawa=−yfψˉLϕψR+h.c.\mathcal{L}_{\mathrm{Yukawa} = -y_f \bar{\psi}_L \phi \psi_R + \mathrm{h}.c.}

After symmetry breaking, this gives mf=yfv/2m_f = y_f v/\sqrt{2}. The Yukawa couplings yfy_f are free Parameters of the Standard Model, not predicted by the theory.

Example 4.2: Yukawa coupling of the top quark

The top quark has mass mt≈173m_t \approx 173 GeV/c2c^2. The Yukawa coupling is:

y_t = \frac{\sqrt{2}\,m_t}{v} = \frac{\sqrt{2} \times 173\;\mathrm{GeV}{246\;\mathrm{GeV} \approx 0.994}}

This is remarkably close to 1 --- the top quark has the largest Yukawa coupling of all Fermions and is the only fermion with a coupling of order unity. For comparison:

  • Electron: ye=me2/v≈2.9×10−6y_e = m_e\sqrt{2}/v \approx 2.9 \times 10^{-6}
  • Muon: yμ≈6.1×10−4y_\mu \approx 6.1 \times 10^{-4}
  • Bottom quark: yb≈0.024y_b \approx 0.024

The enormous range of Yukawa couplings (over five orders of magnitude) is the flavour Problem and remains unexplained within the Standard Model.

After gauge fixing (unitary gauge), the four degrees of freedom of the Higgs doublet become:

  • Three Goldstone bosons absorbed by W±W^\pm and Z0Z^0 (giving them their longitudinal polarisation states).
  • One physical scalar particle HH with mass mH=2λ v≈125m_H = \sqrt{2\lambda}\,v \approx 125 GeV/c2c^2.

The Higgs boson was discovered at the LHC by ATLAS and CMS on July 4, 2012, with mH≈125m_H \approx 125 GeV/c2c^2. The discovery confirmed the mechanism of electroweak symmetry Breaking and earned the 2013 Nobel Prize for Englert and Higgs.