Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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.

share|improve this question
1  
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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