4
$\begingroup$

I was wondering what is the group theoretic way to find the resulting charges of matter fields after a scalar field is given a vev.

In the case of the EW symmetry breaking, one can directly read the charges from the Lagrangian by setting the Higgs field $H=v+h'$ and going in the unitary gauge.

Given a gauge group $\mathcal{G}$, a set of field with their charges under that group. What is the way to find the charges if I give a vev to $n$ fields under the remnant group $\mathcal{G}_\text{br.}$. This is a priori totally unrelated to any Lagrangian and should have a purely group theoretic answer.

A simple example would be $\mathcal{G}=U(1)^k$ with $m$ fields. If I give a vev to $n$ of them, we'll have $U(1)^k\to U(1)^{k-n}$ (assuming the $n$ fields have linearly independent charges). My problem is that I can't find how to get the charges.

I would also be interested in the non-abelian case, and with not only scalar fields but other spin in the spectrum. Any references would also be very welcome!

$\endgroup$
7
  • 2
    $\begingroup$ What do you call the charge for non-abelian groups? The value of the quadratic Casimir? $\endgroup$
    – ACuriousMind
    Sep 5, 2014 at 16:00
  • 1
    $\begingroup$ @ACuriousMind, something like eigenvalues of Cartan operators? e.g. weak isospin? $\endgroup$
    – innisfree
    Sep 5, 2014 at 18:24
  • $\begingroup$ In that case by "charge", I meant representation for non-abelian groups, eg for SU(2) and EW breaking, the Higgs is a 2 of SU(2) (fundamental), the neutrino is a 1 (singlet) of SU(2) etc... $\endgroup$
    – Bulkilol
    Sep 5, 2014 at 20:27
  • $\begingroup$ The charge assignment doesn't change, does it? The quark charges remain the same before and after EWSB. I mean, the charge is a well definite operator $Q=T^3+Y$ so that if you know the representation you know how it acts. All you need to know in general is which charge remain unbroken and that's all determined by the representations that take vevs. $\endgroup$
    – TwoBs
    Sep 6, 2014 at 7:19
  • 1
    $\begingroup$ @Trimok well, if you consider a more extreme breaking as in SU(5) GUT. You have 10 and 5 representation of SU(5) which give rise to an SU(5)xSU(2)xU(1) after breaking. In the case of the $U(1)^n\to U(1)^{n-k}$ case, I could in principle choose to break not one of the U(1), but a linear combinaison of all of them. In that case, I still have trouble to see how I can group theoretically find the remaining charges. $\endgroup$
    – Bulkilol
    Sep 8, 2014 at 9:45

1 Answer 1

0
$\begingroup$

As @TwoBs and @Trimok mentioned, in the case of the breaking $U(1)^n\to U(1)^{n-k}$, the charges don't change. This is however true only in a basis the broken fields are diagonals (only charge under one U(1).

As an example, consider $U(1)^3$ and the following three fields with their charges:

\begin{aligned} \Phi_1:& (1,1,0)\\ \Phi_2:& (1,-1,0)\\ \Phi_3:& (1,1,1) \end{aligned}

We can change the basis of the three $U(1)$'s such that

\begin{aligned} \Phi'_1:& (a_1,0,0)\\ \Phi'_2:& (0,a_2,0)\\ \Phi'_3:& (b_1,b_2,c_3) \end{aligned} Then if $\left<\Phi_1\right>\neq0\neq\left<\Phi_2\right>$, we'll be left with a theory where with have only one field $\Phi'_3$ with charge $U(1)$:

\begin{equation}Q_{U(1)}(\Phi_3)=c_3.\end{equation}

The generalisation to a higher number of $U(1)$'s and fields is obvious.

$\endgroup$

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.