I'm trying to understand how to see if a gauge symmetry would be fully or partially broken on Higgsing. Specifically I am looking at Lagrangians of the form

$$ \mathcal L = - \frac{1}{4} {F_a}_{\mu \nu} {F_a}^{\mu \nu} + (D_\mu \phi)_a^*(D^\mu \phi)_a + \mu^2 \phi_a^* \phi_a - \lambda (\phi_a^* \phi_a) $$

where the fields $\phi$ are in the fundametal representation of some group $G$ and $D_\mu = \partial_m - i {A_a}_\mu \tau^a$ making the gauge fields $A_\mu$ in the adjoint representation of $G$.

Explicit computations are given in Rabakov's book on classical fields sections 6.1 for $G=U(1)$, section 6.2 for $G=SU(2)$ and section 6.3 for $G=SU(2) \times U(1)$.

In the case of $G=U(1)$ we get one massive vector field so gauge symmetry is completely broken (as much as it breaking means in this context). For $G=SU(2)$ we get 3 massive vector fields and again gauge symmetry is complete broken. However, for $G=SU(2) \times U(1)$ there are 3 massive vector fields and one massless one. This situation corresponds to the standard model Higgs.

I follow the calculations, but am trying to understand:

(a) if there is a general way to look at the group and say how many gauge fields will acquire mass, i.e. how much symmetry will be broken

(b) if there is a way I can tweak the Lagrangian fields and couplings (but not the symmetry group) so that I can get full symmetry breaking for $G=SU(2) \times U(1)$?


Sure thing, you can change how the symmetry breaking works out by modifying the Higgs fields. For example, consider a theory with a second Higgs field $\varphi$, a complex scalar which transforms trivially under $SU(2)$ but nontrivially under $U(1)$. If this Higgs field also acquires a vev, then it breaks the $U(1)$ symmetry. The original Higgs field then breaks the $SU(2)$ symmetry, leaving none.

  • $\begingroup$ Yes of course. And the general answer is a Higgs charged under each group you wanna break. $\endgroup$ Mar 23 '18 at 7:09
  • $\begingroup$ @BorunChowdhury No, the Standard Model Higgs is charged under both $SU(2)$ and $U(1)$ but does not break both alone. $\endgroup$
    – knzhou
    Mar 23 '18 at 12:13
  • $\begingroup$ Yes it is. I meant as a counting scheme you need to have as many Higgs as symmetries you wanna break. If you don't break all then a combination of the original ones is still a symmetry. $\endgroup$ Mar 23 '18 at 13:34

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