In the standard model (omitting the QCD part), we start off with the set of generators

$T_1$, $T_2$, $T_3$, $Y$

for the four-parametric gauge group $SU(2)_L \times U(1)_Y$.

We then define a new generator $Q= T_3+Y$ and make the transition to the four-parametric gauge group $SU(2)_? \times U(1)_Q$.

What are, aside from $Q$, the new generators for this "new" gauge group?

$?$ , $?$ , $?$ , $Q$

Do we still use the $T$'s we used in $SU(2)_L$? That means the left factor in the group product is still the same as before the symmetry breaking?

My motivation for asking is the observation that in $SU(2)_L \times U(1)_Y$, the four generators are orthogonal and a basis for the space of all complex self-adjoint matrices.

The set of $T_1$, $T_2$, $T_3$, $Q$, while still a basis, is however not orthogonal, since

$( T_3| Q )$=$(T_3|T_3+Y)$=$(T_3|T_3) \neq 0$

It would seem that we would probably want to preserve that orthogonality property and thus not use $T_3$ as a generator after symmetry breaking.

  • 2
    $\begingroup$ Well, after symmetry breaking, all that remains is electromagnetic $U(1)$, so the only generator that is truly a symmetry generator is $Q$. $\endgroup$ Nov 28, 2013 at 22:22
  • $\begingroup$ @lionelbrits that's basically an answer. $\endgroup$
    – innisfree
    Nov 28, 2013 at 23:02

1 Answer 1


Well, after symmetry breaking, all that remains is electromagnetic $U(1)$, so the only generator that is truly a symmetry generator is $Q$.

The fermions couple to the "Higgs" via the Yukawa coupling:

$\mathcal{L}_y = -y_e^{ij} \bar L_{L,i} \Phi e_{R,j} - y_u^{ij} \bar Q_{L,i} \tilde{\Phi} u_{R,j} - y_d^{ij} \bar Q_{L,i} \Phi d_{R,j} + h.c.\,$

which mixes left and right handed fermions. Here $L$ is the left-handed doublet $(e_L, \nu_L)$, and $e_R$ is the right-handed singlet. Because both $L$ and $\Phi$ transform under $SU(2)_L$, there is a symmetry. After symmetry breaking,

$\mathcal{L}_m = -\frac{y_e^{ij} v}{\sqrt{2}} \bar e_{L,i} e_{R,j} -\frac{y_u^{ij} v}{\sqrt{2}} \bar u_{L,i} u_{R,j} -\frac{y_d^{ij} v}{\sqrt{2}} \bar d_{L,i} d_{R,j} + h.c.$

where $v$ is the Higg's vev. This is not invariant under $SU(2)_L$.

The same thing happens with the gauge bosons that become massive, although there the interaction term comes from the covariant derivative acting on $\Phi$.

Finally, the potential for $\Phi$, (the Mexican hat) is symmetric under SU(2), but the vacuum is not, because for the vacuum state, $\langle 0 | \Phi | 0\rangle = (0,v/\sqrt{2})$, which is not invariant.

  • $\begingroup$ Well, if you write down $L_m$ like this, without weak spinors, $SU(2)_L$ isn't really defined on the elements any more. However, if you replace the left handed leptons and the higgs vev with the corresponding spinors, you are back at $L_y$ with the higgs field in vaccum state, which is again invariant. Maybe I am messing up semantics here. I am starting to wonder if my initial question made any sense. $\endgroup$ Nov 29, 2013 at 2:49
  • $\begingroup$ As for the original question: Looking at $L_y$, can we agree that this term is invariant under $SU(2)_L \times U(1)_Y$ if and only if it is invariant under $SU(2)_L \times U(1)_Q$? Those groups are at least very similar, if not the same. However, it seems to me that when using $Q$ as a generator in $SU(2)_L \times U(1)_Y$, it seems sensible to also replace at least $T_3$ in the set of generators for $SU(2)_L$. $\endgroup$ Nov 29, 2013 at 3:06
  • 1
    $\begingroup$ $U(1)_Q$ mixes $SU(2)_L$ and $U(1)_Y$, and is the only symmetry that remains. $\endgroup$ Nov 29, 2013 at 11:02
  • $\begingroup$ I like this answer, +1 but I am confused by one thing: "after symmetry breaking, all that remains is electromagnetic $U(1)$, so the only generator that is truly a symmetry generator is $Q$." However, it is known that gauge symmetry is never broken and that "symmetry breaking" in this context means "non zero Higgs VEV", see physics.stackexchange.com/a/439384/226902 ...how to reconcile this answer with the linked one? $\endgroup$
    – Quillo
    Jan 13, 2023 at 15:02
  • $\begingroup$ I've been out of this for some time, but here goes. At low energies, at low orders of perturbation theory, we can't see the other symmetries, because of the infinite energy required to shift the v.e.v. Stated another way, the theory still has the full gauge symmetry, but the state we are in doesn't. A hydrogen atom has among other things an SO(3) symmetry, but in a magnetic field the wave function does not. We say the symmetry has been broken by the B field, although of course rotational invariance is still part of the theory. $\endgroup$ Jan 14, 2023 at 16:21

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