Misconceptions in spontaneous symmetry breaking

Question

Spontaneous symmetry breaking occurs when we have a potential like a mexican hat as shown in figure (right) and is unbroken for the potential shape as shown in left figure. Under the Symmetry transformation $\phi \rightarrow \phi e^{i\theta}$. Where $\phi$ is a complex scalar field.

1. My question is that the symmetry is broken only in the ground state ($|\phi| = \phi_{o}$) or is it broken everywhere (for any value of $\phi$) except at $\phi$ = 0. If it is broken only in ground state and is intact at all other values of $\phi$ what is the difference between the $\phi = \phi_{o}$ and any other arbitrary value of $\phi$.

2. Similarly for the unbroken case, is the symmetry intact at all values of $\phi$ or only at $\phi$ = 0.

Answer 1

Let me for concreteness use the same language of classical field theory that you used in your question. It seems like your question is at least partially about the definition of what we call spontaneous symmetry breaking. Is this applied only to the ground state, or does it make sense to apply the same reasoning to any other state?

In fact, we already have a name for the situation where a given state, that is a given value of the field $\phi$, does or does not change under a symmetry transformation: we say that $\phi$ is or is not invariant under the symmetry.

Spontaneous symmetry breaking is a more specific concept, and indeed it is usually formulated as the requirement of non-invariance of the ground state $\langle\phi\rangle$. While being more limiting then mere non-invariance of any chosen state, this property has striking consequences. It implies degeneracy of the ground state, or in other words the existence of flat directions of the potential, which in quantum field theory translates into the existence of gapless excitations: the Nambu-Goldstone bosons. Furthermore, it implies that excited states are classified by irreducible representations of the unbroken subgroup (not the whole symmetry group) and that the Nambu-Goldstone bosons as a rule interact weakly at low energies.

The above said, it sometimes makes perfect sense to apply the same definition to other states than the ground state. For instance, an equilibrium state with nonzero density of a given conserved charge (typically the particle number), which is usually obtained by minimizing the grand-canonical Hamiltonian, can be equally well described by a time-dependent solution of the classical equations of motion, following from the microscopic (canonical) Hamiltonian. By looking at fluctuations of such a time-dependent background, one can generalize the Goldstone theorem to many-body statistical systems. Interestingly, one thus finds that such fluctuations retain most of the attributes of Nambu-Goldstone bosons, except that they need not be gapless: see 1204.1570 and 1303.1527 for more details.

Answer 2

For completeness let us write down the potential

$$V(\phi)=-\mu^2\phi^2+\lambda \phi^4$$

In what follows we will use the term "invariant", which means: "does not change".

The potential (as well as the Lagrangian) is invariant under the symmetry transformation $$\phi \rightarrow \phi e^{i\theta}$$

This symmetry of the potential is intact in the broken and in the unbroken phase for all values of $\phi$.

You can see this directly from the graphs: the symmetry transformation is a rotation of $\phi$ around the origin and both graphs are symmetric under the rotation around the z-axis, which goes through the origin of the complex $\phi$-plane, which I will call just origin in what follows.

In general a rotation of $\phi$ around any point $\phi_1$ can be written as

$$\phi = \phi_1 + \phi_2\rightarrow \phi_1 + \phi_2 e^{i\theta}$$

which will turn into the transformation above when identifying $\phi_1=0$ with the origin. For $\phi_1 \neq 0$ the potential is not invariant under the transformation in the broken as well as in the unbroken phase. Also this can be seen directly from both graphs as they have only the z-axis as symmetry axis.

The point is, that we are interested in fluctuations around the ground state. In the unbroken phase the ground state is at the minimum of the potential, so it is identical to the origin. Since the potential is symmetric under rotations around the z-axis, the potential is symmetric under rotations of $\phi$ around the ground state.

In the broken phase the ground state is one of the $\phi$-values of the minimum-circle. In particular this is not the origin. Since we have already stated that the potential is not symmetric under rotations of $\phi$ around any other $\phi_1$ than the origin, the potential is not symmetric under rotations of $\phi$ around the ground state.

To summarize: The potential is invariant under rotations of $\phi$ around the ground state in the unbroken phase, but it is not invariant under rotations of $\phi$ around the ground state in the broken phase. I think this is meant when we say: the ground state breaks the symmetry.

Note that the above symmetry statements for the potential are also true for the Lagrangian

$$\mathcal{L} = \partial^\mu \phi \partial_\mu \phi - V(\phi)$$