Consider a complex scalar field with Lagrangian $$\mathcal{L} = (\partial_{\mu} \bar{\phi})(\partial^{\mu} \phi) - V(\phi)$$ with potential $$V(\phi) = \frac{1}{4}\lambda(\bar{\phi}\phi - \eta^2)^2$$ The model is invariant under global $U(1)$ phase transformations. The minima of the potential lie on the circle $|\phi| = \eta$, and so the vacuum is characterized by a non-zero expectation value: $$\langle 0|\phi|0\rangle = \eta e^{i\theta}.$$

Now, here is where my confusion lies. The $U(1)$ phase transformation would change the phase of the ground state into $\theta + \alpha$ for some constant $\alpha$. If the symmetry were still manifest, then we would not have found this and instead returned to $\theta$ alone; therefore, the symmetry is broken. However, the broken symmetry vacua with different values of $\theta$ are all equivalent. So, what would it matter if considered $\theta + \alpha$ as opposed to $\theta$ as surely the two represent equivalent vacua? If this is the case, then why is the phase transformation not a symmetry of the vacuum, if it works only to move me to an equivalent configuration? What am I missing?

  • 1
    $\begingroup$ The vacua with different $\theta$ values are not equivalent. If they were there would be no goldstone mode associated with oscillations between the different vacua! $\endgroup$
    – fewfew4
    Jul 23, 2020 at 14:34

2 Answers 2


In general, spontaneous symmetry breaking is the phenomenon in which a stable state of a system (for example the ground state or a thermal equilibrium state) is not symmetric under a symmetry of its Hamiltonian, Lagrangian, or action. Note the "stable" word, it is important: it means that if such a state is perturbed, then it oscillates around its non-perturbed configuration.

The vacua configurations are equivalent from the energetic point of view, but are not the same configuration. Since these configurations are required to be stable, they do not "mix", i.e. it is not easy to pass from one to another, they can just oscillate (this is related to the "Goldstone modes").

Another example that may be easier to visualize: the Hamiltonian of a bunch of particles is (typically) invariant under translations, but the ground state may be a crystal. In this case the ground state is left with only a subgroup of the continuous translations, i.e. the discrete translations that are allowed by the crystal structure.

Moreover, all crystals with the same structure are equivalent and differ by a continuous translation (i.e. they can be superimposed). The point is that, when the system is realized in nature, only one configuration is "randomly selected", and this configuration does not have the full symmetry: the thermal or quantum small fluctuations acting on a real system crossing a critical point decide the system's fate and determine which branch of a bifurcation is taken.


Even though this question has been successfully answered already, I just wanted to emphasize some points about spontaneous symmetry breaking.

When a symmetry is 'spontaneously broken', it is not true that it is no longer a symmetry of the theory, as is so commonly implied in textbooks. Indeed, the broken symmetry is still represented (anti)unitarily on states.

The important difference between broken and unbroken scenarios is the spectrum of states. When a symmetry is unbroken, there is a single vacuum that is invariant with one tower of states given by exciting the vacuum.

When a symmetry is broken, there are many towers of states, each associated with a different vacuum that corresponds to a different 'orientation' (in your case a different $\theta$). If we find ourselves on one tower, and apply a broken symmetry transformation, we jump to a different tower.

The symmetry is called broken because, as Quillo said, when the theory is realized in nature, one tower of states is chosen. We don't see the other towers and so there's no way to directly observe the symmetry (of course we can do so indirectly through goldstone bosons).


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