# What is dynamical symmetry breaking, exactly?

As far as I understand, a symmetry can be broken explicitly (either by manually putting symmetry-violating terms in the Lagrangian or via an anomaly) or spontaneously. I want to focus on the second kind.

For a spontaneously broken symmetry (SSB), the Lagrangian is invariant under the symmetry transformation, but the ground state is not. So just by looking at the ground state, you don't see the symmetry, hence the alternative name hidden symmetry.

How are the terms "Dynamical", "Higgs" and "Goldstone" related to each other in the context of spontaneous symmetry breaking?

Some of my thoughts:

• My nuclear physics lecture distinguishes global SSB ($$\to$$Goldstone, massless bosons) and local SSB ($$\to$$Higgs, bosons get "eaten"), no mention of "dynamical".

• My QCD lecture states that SSB always leads to Goldstone bosons, and that there are two mechanisms how to achieve SSB: Higgs (bosons get "eaten") and dynamical breaking.

• Wikipedia states that the Higgs mechanism and dynamical symmetry breaking are two ways of describing local SSB, whereas the Goldstone theorem applies for global SSB. They also list dynamical symmetry breaking for global symmetries if the breaking is due to quantum corrections (at the level of the effective action).

• The Stanford Encyclopedia of Philosophy also distinguishes global SSB ($$\to$$Goldstone) and local SSB ($$\to$$Higgs, dynamical). In particular, they state that dynamical symmetry breaking means that the Higgs field is "phenomenological rather than fundamental", i.e. that the Higgs field are actually bound states from within the theory.

• Oct 6, 2018 at 16:34

As correctly stated in the following answer by flippiefanus , dynamical symmetry breaking is identical to spontaneous symmetry breaking except that in the case of dynamical symmetry breaking a composite noninvariant field operator acquires a vacuum expectation value while in the spontaneous symmetry breaking case an elementary noninvariant field operator acquires a vacuum expectation value. Please see, for example, the following review by Higashijima (at the bottom of page 2).

Apart from this difference, these two cases are completely identical: In both cases, the Goldstone theorem applies; the rules for the number of Nambu-Goldstone bosons and their representations are the same. Both cases above refer to global symmetry breaking.

The Higgs mechanism differs from both cases. First, although many textbooks introduce the Higgs mechanism in classical theory as spontaneous symmetry breaking (of the global symmetry) in systems with local symmetry, this is not the only valid description. Landsman describes the two approaches in the case of the Abelian Higgs model: $$\mathcal{L} = -\frac{1}{4} F_A^2 + \frac{1}{2} D_{\mu}^A\phi D_{\mu A}\phi – V(|\phi|)$$ By performing a redefinition of the fields: $$\begin{pmatrix}\phi_1 \\\phi_1\end{pmatrix} = e^{i \theta \sigma_x}\begin{pmatrix}\rho \\0\end{pmatrix}$$ $$A_{\mu} = B_{\mu} + \partial_{\mu} \theta$$ By substituting this parametrization into the Lagrangian, the $$\theta$$ dependence vanishes completely, and we are left with: $$\mathcal{L} = -\frac{1}{4} F_B^2 + \frac{1}{2} \partial_{\mu}\rho \partial_{\mu}\rho +\frac{1}{2}\rho^2 B_{\mu}B^{\mu} – V(\rho)$$ This Lagrangian (which is gauge fixed as both $$\rho$$ and $$B$$ are invariant under the gauge transformation) describes a real scalar field and a massive gauge boson in the case when the scalar field acquires a vacuum expectation value.

Landsman also describes the conventional picture where the Nambu-Goldstone boson gets eaten by the gauge field. The question, which picture is the right one in quantum theory is not settled. The difference is that in the conventional picture, the global rigid symmetry gets spontaneously broken, while in the second picture it does not.

The conventional picture seemingly contradicts Elitzur's theorem and the fact that local gauge symmetry cannot be broken. This is the reason why some authors prefer the second picture over the conventional picture, please see the following lecture notes, on the grounds of Elitzur's theorem. However, as Landsman shows on pages 426-428, it is possible to still implement the first picture on a gauge fixed Lagrangian for which Elitzur's theorem is not valid. The only loophole remaining in the conventional picture is that gauge fixing does not get rid of all gauge redundancy.

• I needed some time to think about this. Do I understand correctly that there are two ways to describe the Higgs mechanism? One would be to decide on a gauge, as you wrote in your answer. This eliminates one dependence in the Lagrangian while rendering the $B$ massive. My problem is, I previously encountered this as "Gauge field eats NG boson". So could you please elaborate on the difference of the two ways? What am I missing? Oct 12, 2018 at 11:11
• For the Higgs mechanism to take place in classical field theory, it is not necessary that a non-singlet field acquires a vacuum expectation value. It is enough that a gauge invariant quantity such as $|\phi|^2$ acquires a vacuum expectation value. Thus, the Higgs mechanism can be obtained with or without (global) spontaneous symmetry breaking. There are no rigorous results, if this picture remains valid in the quantum domain. Oct 14, 2018 at 15:53
• cont. The difficulties lie in the fact that both the Higgs mechanism and spontaneous symmetry breaking are nonperturbative phenomena and to deal with them we need to either remove all gauge freedom and work with gauge invariant fields; or prove gauge invariance nonperturbatively at the quantum level. Both tasks are formidable and there are no rigorous results for that. Oct 14, 2018 at 15:53
• cont. Please see another derivation of the Higgs mechanism without spontaneous symmetrt breaking: arxiv.org/abs/1606.06194 Oct 14, 2018 at 16:06