Global Symmetry Breaking and Goldstone's Theorem

What is Spontaneous Symmetry Breaking?

A system exhibits spontaneous symmetry breaking if its fundamental laws are invariant under a symmetry transformation, but its lowest-energy state (the vacuum) is not. The system must "choose" a particular ground state from a set of energetically equivalent options, and this choice breaks the symmetry.

A simple classical analogy is a vertical ruler under compression. The laws of physics governing the ruler are rotationally symmetric around its vertical axis. If you push down hard enough, it will buckle. When it buckles, it must fall in a specific direction, spontaneously breaking the rotational symmetry. All directions are equally probable, but one must be chosen.

Let's see how this idea is realized in a simple field theory.

Breaking a Discrete ${\mathbb{Z}}_2$ Symmetry

Consider a theory of a single real scalar field $\varphi$ with the Lagrangian:

where ${\mu }^2$ and $\lambda$ are positive constants. This Lagrangian is symmetric under the discrete transformation $\varphi \to -\varphi$.

The ground state of the classical theory corresponds to the field configuration that minimizes the potential energy $V(\varphi )$. To find the minima, we take the derivative of the potential:

This gives three possible values for the field in the ground state:

1. $\varphi =0$
2. $\varphi =\pm \sqrt{{\mu }^2/\lambda }\equiv \pm v$

To determine which are stable minima, we check the second derivative, $\frac{{\partial }^{2}V}{\partial {\varphi }^2}=-{\mu }^2+3\lambda {\varphi }^2$:

• At $\varphi =0$, $\frac{{\partial }^{2}V}{\partial {\varphi }^2}=-{\mu }^2<0$. This is a local maximum, an unstable equilibrium.
• At $\varphi =\pm v$, $\frac{{\partial }^{2}V}{\partial {\varphi }^2}=-{\mu }^2+3\lambda ({\mu }^2/\lambda )=+2{\mu }^2>0$. These are two degenerate minima, the true ground states.

The potential has a "double-well" shape. While the potential itself is symmetric, the vacuum is not. The system must settle into one of the two ground states, either $⟨\varphi ⟩=+v$ or $⟨\varphi ⟩=-v$. This non-zero vacuum expectation value (VEV) signals that the $\varphi \to -\varphi$ symmetry has been spontaneously broken.

The following video gives an excellent visual and conceptual introduction to this idea.

Breaking a Continuous U(1) Symmetry: The Mexican Hat Potential

The story becomes even more interesting when we consider a continuous symmetry. Let's take a complex scalar field $\varphi =\frac{1}{\sqrt{2}}({\varphi }_1+i{\varphi }_2)$ with the Lagrangian:

This Lagrangian is invariant under the global U(1) phase rotation $\varphi \to e^{i\alpha }\varphi$.

The potential, in terms of the real components, is $V({\varphi }_1,{\varphi }_2)=-\frac{{\mu }^2}{2}({\varphi }_1^2+{\varphi }_2^2)+\frac{\lambda }{4}({\varphi }_1^2+{\varphi }_2^2{)}^2$. The minima of this potential satisfy:

Instead of two discrete points, the ground states form a continuous circle of radius $v$ in the ${\varphi }_1-{\varphi }_2$ plane. This is the famous "Mexican hat" potential.

Once again, the vacuum must pick a point on this circle, for example $⟨{\varphi }_1⟩=v$ and $⟨{\varphi }_2⟩=0$. This choice breaks the U(1) symmetry, since rotating this vacuum state by a phase $\alpha$ moves it to a different point on the circle.

What are the physical consequences? Let's analyze the spectrum of particles by considering small fluctuations around our chosen vacuum. We define new fields, $\sigma (x)$ and $\pi (x)$, representing fluctuations around the vacuum state $(v,0)$:

$\sigma$ represents fluctuations in the "radial" direction (up the walls of the potential), and $\pi$ represents fluctuations in the "angular" direction (along the flat valley).

To find the masses of these particles, we can expand the potential $V$ around the vacuum $(v,0)$ up to quadratic terms in the fields $\sigma$ and $\pi$. This corresponds to finding the mass matrix from the second derivatives of the potential.
The mass-squared for a field is given by the curvature of the potential in that field's direction.

• Mass of the $\sigma$ field (radial direction): The potential is curved in the radial direction. Calculating the second derivative, we find $m_{\sigma }^2={\frac{{\partial }^{2}V}{\partial {\varphi }_1^2}|}_{{\varphi }_1=v,{\varphi }_2=0}=2{\mu }^2>0$. This corresponds to a massive particle.
• Mass of the $\pi$ field (angular direction): The potential is flat along the circle of minima. The curvature is zero. Calculating the second derivative, we find $m_{\pi }^2={\frac{{\partial }^{2}V}{\partial {\varphi }_2^2}|}_{{\varphi }_1=v,{\varphi }_2=0}=0$. This corresponds to a massless particle.

This is a general feature: fluctuations along the flat directions of the potential valley, which exist because of the continuous symmetry, are massless. These massless particles are called Goldstone bosons.

The following resource provides a very clear derivation and explanation of this phenomenon for both the U(1) and a more general O(N) case.

Goldstone's Theorem

The appearance of a massless boson whenever a continuous global symmetry is spontaneously broken is not an accident of our specific model. It is a completely general result, formalized by Goldstone's Theorem.

Goldstone's Theorem: For every continuous global symmetry generator that is spontaneously broken, there must exist a corresponding massless, spin-0 particle in the theory. These particles are the Goldstone bosons.

The number of Goldstone bosons is equal to the number of broken generators. If a symmetry group $G$ is broken down to a subgroup $H$, the number of Goldstone bosons is $\text{dim}(G)-\text{dim}(H)$.

Proof of Goldstone's Theorem

The proof is an elegant formalization of the intuition we gained from the Mexican hat potential. We will prove it for a classical field theory.

1. Consider a theory with a set of real scalar fields ${\varphi }_a$ and a potential $V(\varphi )$ that is invariant under a continuous symmetry group $G$. Let the infinitesimal action of a generator $T^A$ of the group's Lie algebra be:

   $${\varphi }_a\to {\varphi }_a^{\prime }={\varphi }_a+i{\alpha }^A(T^A{)}_{ab}{\varphi }_b$$

   where ${\alpha }^A$ are infinitesimal parameters.

2. The invariance of the potential means $V({\varphi }^{\prime })=V(\varphi )$. Expanding to first order in ${\alpha }^A$, this implies:

   $$\frac{\partial V}{\partial {\varphi }_a}(T^A{)}_{ab}{\varphi }_b=0$$

   This equation must hold for any field configuration $\varphi$, not just for the vacuum.

3. Now, differentiate this identity with respect to another field component, ${\varphi }_c$:

   $$\frac{{\partial }^{2}V}{\partial {\varphi }_c\partial {\varphi }_a}(T^A{)}_{ab}{\varphi }_b+\frac{\partial V}{\partial {\varphi }_a}(T^A{)}_{ac}=0$$

4. Finally, evaluate this equation at a minimum of the potential, $\varphi ={\varphi }_0$. By definition of a minimum, the first derivatives of the potential vanish: ${\frac{\partial V}{\partial {\varphi }_a}|}_{{\varphi }_0}=0$. This causes the second term to drop out, leaving:

   $${\left(\frac{{\partial }^{2}V}{\partial {\varphi }_c\partial {\varphi }_a}\right)}_{{\varphi }_0}(T^A{)}_{ab}{\varphi }_{0b}=0$$

5. Let's interpret this equation.

   • The term $(M^2{)}_{ca}\equiv {\left(\frac{{\partial }^{2}V}{\partial {\varphi }_c\partial {\varphi }_a}\right)}_{{\varphi }_0}$ is the mass-squared matrix of the fields. Its eigenvalues are the squared masses of the physical particles.
   • The term ${\vec{v}}_A\equiv T^A{\varphi }_0$ is a vector in the field space.
     The equation has the form $M^2{\vec{v}}_A=0$. This means that for each generator $T^A$, the vector ${\vec{v}}_A$ is an eigenvector of the mass-squared matrix with an eigenvalue of zero.

6. When is this eigenvector ${\vec{v}}_A$ non-zero? It is non-zero precisely when $T^A{\varphi }_0\ne 0$. This is the definition of a broken symmetry generator: a generator is broken if its action on the vacuum state ${\varphi }_0$ is non-trivial.

Therefore, for every broken symmetry generator, we have explicitly constructed a non-zero eigenvector of the mass-squared matrix with eigenvalue zero. This proves the existence of a corresponding massless particle.

This elegant proof is one of the cornerstones of modern particle physics. To see it presented in a lecture format, you can watch the following segment.

Conclusion

Key Takeaways:

• Spontaneous Symmetry Breaking (SSB) occurs when the ground state (vacuum) of a theory lacks a symmetry present in its Lagrangian.
• SSB is signaled by a non-zero vacuum expectation value (VEV) for a field, $⟨\varphi ⟩\ne 0$.
• In theories with a continuous global symmetry, SSB leads to a continuous manifold of degenerate vacua (e.g., the "Mexican hat" potential).
• Goldstone's Theorem states that for every broken generator of a continuous global symmetry, a massless spin-0 particle, a Goldstone boson, must appear in the spectrum.
• The number of Goldstone bosons is equal to the number of broken generators, $\text{dim}(G)-\text{dim}(H)$.
• Physically, Goldstone bosons correspond to fluctuations along the flat directions of the potential's vacuum manifold.

You have now mastered a critical concept in QFT. Spontaneous symmetry breaking provides a powerful mechanism for generating specific particle spectra, and as we are about to see, it holds the key to understanding the masses of fundamental particles.

Preview of the Next Lesson:

What happens if the symmetry that is spontaneously broken is not a global symmetry (where the phase rotation $\alpha$ is constant everywhere), but a local gauge symmetry (where $\alpha (x)$ depends on spacetime)? We will see that something magical happens: the would-be Goldstone boson is "eaten" by the gauge boson, which becomes massive as a result. This is the celebrated Higgs mechanism, the cornerstone of the electroweak theory.