In the case of massive bosons of weak interaction we have Higgs mechanism to save the symmetry. I do not see how with Higgs the symmetry is preserved? Because the mass term is what gives us problems so I do not see how we get rid off this with higgs. Do we need to include Higgs field into the weak lagrangian but exclude mass terms and then through gauge of the whole thing get the mass and gauge fields? I dont get this part.

  • $\begingroup$ I usually think of the Higgs mechanism as a symmetry-breaking mechanism, not a symmetry-preserving one. $\endgroup$
    – G. Smith
    Commented Dec 12, 2019 at 6:55
  • $\begingroup$ Well there is a problem with massive gauge boson and Higgs solves it. How? $\endgroup$ Commented Dec 12, 2019 at 8:46
  • $\begingroup$ So, first it says that in order to give mass we have to break gauge invariance, which is not a problem, just add the mass term and it is broken, right? I stil dont see the point of Higgs mechanism...I am totaly confused by these explanations, that is why I am asking here.. $\endgroup$ Commented Dec 12, 2019 at 13:28
  • $\begingroup$ So, what its all about is applying SSB to local gauge invariance and we get the mass term for that gauge field? So the Lagrangian is still gauge invariant just doesnt look like that? $\endgroup$ Commented Dec 12, 2019 at 13:56
  • $\begingroup$ Yes, and we choose a solution for a stable configuration. But then we apply to this gauge invariance(local) and our gauge field appeares with mass. But what is this gauge field and what does it have to do with other gauge fields? How will they get mass? We plug in Higgs field in the lagrangian for weak force and gauge fields of weak force get mass through Higgs mechanism? Is that it? $\endgroup$ Commented Dec 12, 2019 at 18:20

2 Answers 2


A naïve mass term for a vector boson looks something like

$$\sim -m^2 \mathrm{Tr} \, A^\mu A_\mu.$$

However, under a gauge transformation,

$$A_\mu \to G A_\mu G^{-1} - \frac{i}{g} (\partial_\mu G) G^{-1}$$

Where $G$ is an element of the gauge group and $g$ is the coupling constant. It’s straightforward to check that the above mass term is not invariant under this transformation.

The Higgs mechanism circumvents this problem by coupling the gauge field to a scalar field $\phi$ via the covariant derivative term

$$\sim (D_\mu \phi)^\dagger (D^\mu \phi),$$

which can be checked to be gauge invariant. Expanding this out in the unitary gauge gives a 4-point coupling

$$\sim -g^2|\phi|^2 A^2$$

(the precise details depend on the gauge group in question and the representation the scalar field transforms under). If we also add a potential for the scalar field such that $\langle \phi \rangle \ne 0$, this behaves like a mass term for the gauge field, whilst maintaining gauge invariance.

This is quite a brief overview; more details will be in pretty much any QFT textbook.

  • $\begingroup$ Yes, but this is what I was looking for so thank you. $\endgroup$ Commented Dec 13, 2019 at 20:45

(This was meant to be a quick comment written on the phone, but became too long. So I post it here.)

I think this is a way to think about it. Gauge invariance is not a physical symmetry, it's a redundancy of our description. Breaking this (local) symmetry would render the theory inconsistent and create new unphysical degrees of freedom. You can therefore not just add a mass term by hand. (Also look up Elitzur's theorem).

So what's going on? Note that there are some subtleties in what is a redundancy. Usually we choose a boundary condition of the form $A(x) \rightarrow 0$ as $|x|\rightarrow \infty$. Any local $U(1)$ transformation that preserves this is a redundancy, if you mod out with this subgroup the remaining transformations are real physical symmetries. These will correspond to global $U(1)$ transformations and these give rise to charge conservation due to Noethers theorem.

This is the symmetry of the ungauged model. Having this global symmetry, you can either have a ground state that is invariant under this symmetry or you have a bunch of these that transform into each other (they'll have the same energy as the symmetry commutes with the Hamiltonian).

Usually people talk about this in terms of classical field theory. Given any real number $\phi_0$, you can always parameterize your complex scalar field as $\phi(x) = (\phi_0 + \rho(x)) e^{i\theta(x)}$. It's just using polar coordinates and shifting the origo. Note that the amplitude fluctuations are invariant under gauge transformations, while the phase fluctuations are compact scalar field that shifts under gauge transformation. Writing your action in terms of these fields doesn't change anything, the theory is still fully gauge invariant.

Now imagine your theory has a Mexican hat potential and your equations of motion doesn't have a unique vacuum. Choose $\phi_0$ to be one of these possible vev's. In the ungauged the theory still have the symmetry, but it's broken if you only think about the vacuum. The phase fluctuations will be massless goldstone modes while the amplitude fluctuations will become massive. In the gauged model the theory is still fully gauge invariant and consistent. The main difference is that the phase fluctuations are not completely physical, you can change their values by gauge transformations. In particular, you can make them vanish by gauge fixing. The goldstone modes thus do not exist anymore. Doing this you get a mass term for the gauge field, but that doesn't break gauge invariance as you've fixed your gauge. By not fixing $\theta(x)$ and the gauge, the theory is still fully gauge invariant. So the gauge field becomes massive through the dynamical interactions with the scalar field. You can say that the Higgs mechanism let's us effectively add a mass term for the gauge field, without ruining gauge invariance.

What's really spontaneously broken is not the local gauge transformations, but the global part and thus charge conservation. Your vacuum state does not conserve particle number and charge anymore but is a coherent state of states of any number of particles. (Also look up superconductivity).

  • $\begingroup$ Thank you for your answer but this is pretty much what i heard before. And I still dont get it. I dont know why. $\endgroup$ Commented Dec 12, 2019 at 18:15

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