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.

potential shapes for unbroken and spontaneously broken symmetry)

  • 2
    $\begingroup$ The symmetry is broken whenever $\langle \phi \rangle \neq 0$. However, just setting $\langle \phi \rangle$ to some value will not be stable unless it's a minimum of the potential. $\endgroup$
    – knzhou
    May 22, 2019 at 22:01
  • $\begingroup$ @knzhou do you mean when the field value $\phi \neq 0$, if yes then any transformation of $\phi =0$ will be $\phi =0$ . So nothing seems to be changing. Or you mean expectation value of something ?? I interpreted ⟨𝜙⟩≠0 as $\phi \neq 0$, if this is incorrect can you please explain with some example why it is not the same. Thanks! $\endgroup$
    – 23rduser
    May 22, 2019 at 22:49
  • $\begingroup$ Are you thinking of the field classically or as a quantum field? I’m talking about the expectation value of a quantum field. $\endgroup$
    – knzhou
    May 22, 2019 at 22:54
  • $\begingroup$ I meant in general for both. The expectation of field would mean it's most probable value which would always be ground state. My question is about the actual value of field at any given instant. If we talk in terms of the expectation value ⟨𝜙⟩≠0.( not ground state) does it mean that the conservation of symmetry is possible only in the ground state ? $\endgroup$
    – 23rduser
    May 22, 2019 at 23:08

1 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.

  • $\begingroup$ I understand how and why goldstone is formulatred. " While being more limiting then mere non-invariance of any chosen state". I did not understand this part. let me put my question very specifically. 1) when $\phi$ changes (is transformed ) to $\phi e^{i \theta}$.then when we say it is symmetric, it means lagrangian does not change. isn't it at all values of $\phi$ (not expectation value). 2)even in ground state in the mexican hat potential the lagrangian is invariant under transformation. So, if symmetry is broken only in 2nd case what is the difference between 1st and 2nd case. $\endgroup$
    – 23rduser
    May 23, 2019 at 7:01
  • 1
    $\begingroup$ Spontaneous symmetry breaking is a property of a state, not the Lagrangian (whose invariance for any $\phi$ is assumed, otherwise there is no symmetry to be spontaneously broken). There are many, in fact most, states of a given system that are not invariant under the symmetry. In case of your complex scalar, only the point $\phi=0$ is invariant: this is a single point in the complex plane. What I tried to explain in my answer is the dramatic difference between the non-invariance of an arbitrarily chosen state, and the non-invariance of the ground state. $\endgroup$ May 23, 2019 at 8:13
  • $\begingroup$ "Spontaneous symmetry breaking is a property of a state" and "There are many, in fact most, states of a given system that are not invariant under the symmetry", 1)So if I understand correctly what you are saying is that symmetry is spontaneously broken at all points except at $\phi = 0$. ? 2)and when it is said that the symmetry is unbroken in the potential as in left figure and broken in mexican hat potential. They are talking about the invariance or ground states under s symmetry operation ? $\endgroup$
    – 23rduser
    May 23, 2019 at 9:49
  • 1
    $\begingroup$ @AkshanshSingh As to 1) I do prefer to say that all states except $\phi=0$ are non-invariant under the symmetry - but I guess we agree here except for a different wording. As to 2) yes, both potentials are invariant, but only the Mexican-hat one has spontaneous-symmetry-breaking vacua. $\endgroup$ May 23, 2019 at 13:30
  • $\begingroup$ As to 2) if Lagrangian is invariant then it's called a symmetry. as you said both potentials are invariant that I understand. "only the Mexican-hat one has spontaneous-symmetry-breaking vacua.", is this because of the changing state ($\phi$ ) under the symmetry operation? If yes then the state ($\phi$) changes everywhere not just in vacua , so there must be spontaneous-symmetry-breaking everywhere under symmetry operation not just in vacua. and this would mean that the potential on left has spontaneous symmetry breaking at all states except in vacua . $\endgroup$
    – 23rduser
    May 23, 2019 at 14:49

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