# Higgs mechanism and spontaneous symmetry breaking

I'm really confused about spontaneous symmetry breaking and the Higgs mechanism. I've searched for other answers but I couldn't still resolve all my doubts. I also understand that I'm asking a lot of questions, so I don't expect an answer to all of them. From what I understand, spontaneous symmetry breaking for a quantum field theory with a global continuous symmetry group $$G$$ is effectively the request that the VEV satisfies a constraint

$$$$\langle\phi\rangle|_{J=0}\neq 0.$$$$ This translates into a constraint on the vacuum of the theory. Provided what I said until know is right (which might well not be the case), I have several questions:

• Textbooks often use the property that when the VEV is assumed to be constant throughout space, the effective action can be assumed to coincide with the effective potential. Why can the VEV assumed to be constant throughout space? I can understand if one talks about a macroscopic system, but this fact is used to rewrite the higgs field as his constant VEV plus some perturbation, and thus this impacts perturbative calculations.

• In absence of symmetry breaking $$$$\langle\phi\rangle|_{J=0}= 0.$$$$ On the other hand $$$$\frac{\delta\Gamma}{\delta\langle\phi\rangle}\Bigg|_{J=0}=0.$$$$ The second equation admits non vanishing solutions in general, so there seems to be a contradiction between the two conditions.

• In order to do perturbative calculations, one can go in the unitarity gauge expands the Higgs field around its VEV as $$H=\langle h\rangle+\delta h$$, where $$\delta h$$ is a longitudinal fluctuation around the VEV. This implies that $$$$\langle \delta h\rangle=0$$$$ which, to me, seems a self-consistency condition to be checked when doing perturbative calculations, requiring $$\delta h$$ to not have tadpoles (otherwise $$\delta h$$ could shift you to another vacuum). How is this constraint implemented when doing perturbative calculations? Is it automatically satisfied?

• The Higgs mechanism provides masses to vector bosons, but it does not completely break gauge symmetry (rather, it breaks the global part of it). This is obvious from the fact that gauge symmetry is often used to "eliminate" Goldstone bosons. In fact, I've read on some other posts that at the quantum level, gauge symmetry is not a considered a symmetry (in the sense that it is not a symmetry of observables), but rather that it is some structural redundancy that physical theories have. Indeed, gauge symmetry breaking is a name more appropriate to the act of adding a gauge fixing term to the lagrangian. I have also read that it is a do-nothing transformation, in the sense that it acts trivially on states. Can someone show me an heuristic calculation showing that a gauge transformation is a do-nothing transformation (by this I mean define some gauge transformation on states and show that the state which is obtained is exactly the one we started from)?

• For the second bullet, why do you think there is a contradiction? One is a statement about $\phi$ and the other about $\Gamma$ (namely, it's $\phi$ dependence). – Aaron Feb 14 at 19:38
• – SRS Feb 14 at 23:10
• @Aaron If it it seen as an equation for the VEV, it allows the VEV to take a non-zero value. I feel like my mistake is confusing a possible VEV solution with the VEV as calculated in a non broken theory. – Tanatofobico Feb 15 at 11:32

Textbooks often use the property that when the VEV is assumed to be constant throughout space, the effective action can be assumed to coincide with the effective potential. Why can the VEV assumed to be constant throughout space?

That can't be assumed in all cases - there are certainly interesting field theories where the VEV depends on space. But in applications to particle physics, we often note that the ground state of our universe does not appear to break translation invariance, and therefore neither can the VEV of a local quantum field in the Lagrangian for our universe. Or in explicit equations, where $$\hat{T}$$ is the translation operator, if we assume $$\hat{T}|\mathrm{GS}\rangle = |\mathrm{GS}\rangle$$, then $$\langle \mathrm{GS} | \phi(x) | \mathrm{GS} \rangle = \langle \mathrm{GS} |\hat{T}^{\dagger} \phi(x) \hat{T}| \mathrm{GS} \rangle = \langle \mathrm{GS} | \phi(x+a) | \mathrm{GS} \rangle,$$ so the VEV is clearly independent of this translation. But in applications to systems with translational symmetry breaking (like in a crystal), this won't hold.

I don't understand what contradiction you're referring to in your second bullet.

How is this constraint implemented when doing perturbative calculations? Is it automatically satisfied?

It isn't in general, so you'll need to enforce it at each order in perturbation theory. One way to proceed is to calculate the VEV $$\langle H \rangle$$ for the interacting theory to the order you're working at, and then the resulting field $$\delta h = H - \langle H \rangle$$ will have zero VEV automatically. I'm sure there are plenty of equivalent ways to enforce this. See Appendix A of this paper for one explicit example of this procedure: https://arxiv.org/abs/1108.5207

Can someone show me an heuristic calculation showing that a gauge transformation is a do-nothing transformation (by this I mean define some gauge transformation on states and show that the state which is obtained is exactly the one we started from)?

There's not much I can add to some of the previous well-written answers here on Stack Exchange, such as:
Gauge symmetry is not a symmetry?
What role does “spontaneously symmetry breaking” played in the “Higgs Mechanism”?
Understanding Elitzur's theorem from Polyakov's simple argument?