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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.

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Well, after symmetry breaking, all that remains is electromagnetic $U(1)$, so the only generator that is truly a symmetry generator is $Q$. – lionelbrits Nov 28 '13 at 22:22
@lionelbrits that's basically an answer. – innisfree Nov 28 '13 at 23:02

1 Answer 1

up vote 2 down vote accepted

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.

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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. – Konstantin Nov 29 '13 at 2:49
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 $SU(2)_L$. – Konstantin Nov 29 '13 at 3:06
$U(1)_Q$ mixes $SU(2)_L$ and $U(1)_Y$, and is the only symmetry that remains. – lionelbrits Nov 29 '13 at 11:02

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