1
$\begingroup$

Say we have a group $G$ and a set of Higgs fields in a representation $R$ of $G$. One of the Higgs fields in $R$ gets a VEV, how can I determine the remaining subgroup after this symmetry breaking?

In the standard model things are easy, because the Higgs field that gets a VEV has nonzero Isospin and Hypercharge, but zero Electric Charge. Therefore $$SU(2)\otimes U(1)_Y \rightarrow U(1)_Q$$

There is no room for ambiguities.

Nevertheless, for some bigger group it's not sufficient to know that the corresponding Higgs field has zero charge under the corresponding subgroup $G'$ because there can be some bigger subgroup $G' \subset G''$ under which the Higgs field has zero charge, too.

This task is commonly called "finding the maximal little group" and I'm looking for a concrete way of determine this "maximal little group" if some Higgs field gets a VEV.

Any ideas would be much appreciated!

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ In general it depends how the Higgs take his vev, as the maximal (some adjective) subgroup of a given group can depend on the embedding. $\endgroup$ – Oscar Jul 2 '15 at 10:35
  • $\begingroup$ @Oscar What do you mean by "how the Higgs take his vev"? For simplicity let's say only one Higgs field $\Phi$ inside the representation $R$ gets a VEV and the embedding has been fixed. My problem is determining the subgroup $G'$ to which my starting group $G$ is broken to by this $\Phi \rightarrow VEV$. A necessary but not sufficient condition is that $\Phi$ is a singlet under $G'$ but a non-singlet under $G$. $\endgroup$ – jak Jul 2 '15 at 10:47
1
$\begingroup$

Since you start with a group $G$ which is a symmetry of the theory (ie of Lagrangian) there will be some generators for $G$. Call them $T_1,\ldots,T_N$. In the original Lagrangian, (before the Higgs gets a vev) you can do a transformation with any of the $T_i$'s and the Lagrangian will not change. After the field gets a vev, you can try the same and you will find some of the $T_i$'s still leave the Lagrangian unchanged (unbroken generators) and some of them do not (broken generators). The remaining subgroup after the symmetry breaking is simply the group generated by the unbroken generators.

I would recommend chapter 84 of Srednicki's book for more details and some examples.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.