Higgs Doublet and Spontaneous Symmetry Breaking

1. The Need for Spontaneous Symmetry Breaking

Before we dive in, let's clarify why we can't just add mass terms by hand. A mass term for a gauge boson, like $\frac{1}{2}{m}^2{W}_{\mu }{W}^{\mu }$, is not invariant under the $SU(2{)}_L$ gauge transformations we established. Adding such a term would explicitly break the fundamental symmetry of our theory, destroying its mathematical consistency and renormalizability.

The solution is a beautiful concept called Spontaneous Symmetry Breaking (SSB). The idea is that the laws of physics (the Lagrangian) can possess a symmetry that is not respected by the ground state (the vacuum) of the system.

To build your intuition, the following video explains the general idea of SSB using a simple real scalar field before we tackle the more complex electroweak case.

2. The Higgs Doublet and Its Potential

To break the $SU(2{)}_L\times U(1{)}_Y$ symmetry, we must introduce a new field that transforms non-trivially under it. This is the Higgs field, which is postulated to be a complex scalar field organized as a doublet under $SU(2{)}_L$.

Specifically, we assign it the following quantum numbers under $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$:

• Higgs Field $H$: $(1,2{)}_{+1/2}$

This means the Higgs is a color singlet, an $SU(2{)}_L$ doublet (with weak isospin $T=1/2$), and carries weak hypercharge $Y=1/2$. As a complex doublet, it consists of four real scalar fields.

The Lagrangian for the Higgs field is:

The dynamics are governed by the kinetic term and a potential $V(H)$. For SSB to occur, this potential must have a specific form:

Here, $\lambda$ must be positive for the potential to be bounded from below. The crucial part is the sign of ${\mu }^2$:

• If ${\mu }^2>0$, the potential has a single minimum at $H=0$. The vacuum is symmetric, and no symmetry breaking occurs.
• If ${\mu }^2<0$, the minimum is no longer at $H=0$. The potential takes on the shape of a "Mexican hat" or "sombrero".

The minimum of the potential now lies on a circle (or more accurately, a 3-sphere in the 4D field space) defined by:

The quantity $v$ is the vacuum expectation value (VEV) of the Higgs field, a fundamental parameter of the Standard Model with an experimentally determined value of approximately 246 GeV. The following reading formalizes the Higgs Lagrangian and its role in the Standard Model.

3. Breaking $SU(2{)}_L\times U(1{)}_Y\to U(1{)}_{EM}$

Since the vacuum can be any state satisfying $H^{†}H=v^2/2$, the universe must "choose" one specific ground state to settle into. This choice breaks the symmetry. By gauge freedom, we can rotate our coordinates in field space to align this chosen vacuum along a specific direction without loss of generality. The standard convention is:

This non-zero VEV is the source of electroweak symmetry breaking. The original $SU(2{)}_L\times U(1{)}_Y$ symmetry is now broken because the vacuum state $⟨H{⟩}_0$ is no longer invariant under all of its transformations.

Which part of the symmetry survives? A symmetry operation is unbroken if it leaves the vacuum state invariant. Let's find the generator(s) that annihilate the vacuum. An infinitesimal transformation on $H$ is given by:

Applying this to our chosen vacuum:

Let's evaluate the action of each generator:

• ${\sigma }^1\left(\begin{array}{c}0\\ v\end{array}\right)=\left(\begin{array}{c}v\\ 0\end{array}\right)\ne 0$. The symmetry associated with $T^1$ is broken.
• ${\sigma }^2\left(\begin{array}{c}0\\ v\end{array}\right)=\left(\begin{array}{c}-iv\\ 0\end{array}\right)\ne 0$. The symmetry associated with $T^2$ is broken.
• For the diagonal generators: $$\left({\alpha }_3\frac{{\sigma }^3}{2}+\beta \frac{1}{2}1\right)\left(\begin{array}{c}0\\ v\end{array}\right)=\left(\left(\begin{array}{cc}{\alpha }_3/2& 0\\ 0& -{\alpha }_3/2\end{array}\right)+\left(\begin{array}{cc}\beta /2& 0\\ 0& \beta /2\end{array}\right)\right)\left(\begin{array}{c}0\\ v\end{array}\right)=\left(\begin{array}{c}0\\ (-\frac{{\alpha }_3}{2}+\frac{\beta }{2})v\end{array}\right)$$

For this to be zero, we require ${\alpha }_3=\beta$. This means there is a linear combination of generators that leaves the vacuum invariant. The corresponding generator is:

This is precisely the electric charge generator! The vacuum state $⟨H{⟩}_0$ has zero electric charge, and the $U(1)$ symmetry generated by $Q$ remains unbroken. This is the $U(1{)}_{EM}$ of electromagnetism.

The following resources provide a rigorous derivation of this central result.

4. Goldstone's Theorem and the "Eaten" Bosons

We started with a Higgs doublet containing 4 real scalar fields. After SSB, we can parameterize the fluctuations around the vacuum. One fluctuation corresponds to moving up and down the "walls" of the Mexican hat potential—this field will be massive and is the physical Higgs boson. The other three fluctuations correspond to moving around the "valley" at the bottom of the potential.

Goldstone's Theorem states that for every spontaneously broken continuous global symmetry, a massless scalar particle—a Goldstone boson—appears. We broke 3 generators ($T^1,T^2$, and the combination orthogonal to $Q$), so we should expect 3 massless Goldstone bosons.

However, in a local gauge theory, something remarkable happens: these would-be Goldstone bosons are "eaten" by the gauge bosons corresponding to the broken symmetries. A massless vector boson (like the photon) has only 2 degrees of freedom (its transverse polarizations). A massive vector boson needs 3 degrees of freedom (two transverse and one longitudinal). The Goldstone bosons provide precisely these missing longitudinal degrees of freedom, allowing the $W^{\pm }$ and $Z$ bosons to become massive.

This video segment provides an excellent non-mathematical explanation of this crucial final step.

Conclusion

Key Takeaways:

• To generate mass without explicitly violating gauge symmetry, we introduce a complex $SU(2{)}_L$ doublet scalar field, the Higgs field, with weak hypercharge $Y=1/2$.
• The Higgs potential is chosen such that its minimum is at a non-zero value, forcing the Higgs field to acquire a vacuum expectation value (VEV), $v$.
• The choice of a specific vacuum state is not symmetric under the full $SU(2{)}_L\times U(1{)}_Y$ group, causing Spontaneous Symmetry Breaking.
• The vacuum remains invariant under transformations generated by $Q=T^3+Y$. This leaves the $U(1{)}_{EM}$ of electromagnetism as the single unbroken gauge symmetry.
• Of the four initial Higgs field components, one becomes the massive physical Higgs boson. The other three become the longitudinal modes of the $W^{+}$, $W^{-}$, and $Z$ bosons, allowing them to acquire mass.

We have now established the full mechanism. We started with a symmetric theory with four massless gauge bosons and ended with a broken symmetry that predicts three massive vector bosons ($W^{\pm },Z$) and one massless one (the photon, $\gamma$).

Preview of the Next Lesson:

In the next lesson, we will make this concrete. We will substitute the Higgs field, expanded around its VEV, back into the Higgs kinetic term $(D_{\mu }H{)}^{†}(D^{\mu }H)$ and read off the resulting mass terms for the gauge bosons. This will allow us to derive the masses of the W and Z bosons in terms of the VEV ($v$) and the gauge coupling constants ($g$ and $g^{\prime }$).