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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.

Konstantin Schubert
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1 Answers1

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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 $Q$.

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.

lionelbrits
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