# What are the other components of the weak isospin?

Sorry for a lazy question, I was just looking this up on the internet for a basic grasp of Standard Model. However, I couldn’t find anything on Wikipedia or just somehow managed to miss it on my google search. I get it that the 3rd component of the Weak Isospin is conserved and so everything is all about it, but what about the other two components? Do their behaviors have any significance at all? Does anybody care about how they change during an interaction?

Does anybody care about how they change during an interaction?

Absolutely! our very existence cares quite a bit.

Weak isospin, or $SU(2)_L$, like conventional spin, is characterized by 3 generators, which do not commute, of course. However, as in plain spin settings, choosing a reference direction, such as the third, to moor the charges of the particles in a weak multiplet, $$Q = T_3 +\tfrac{1}{2}Y_\mathrm{W} ,$$ we usually utilize the other two generators together through their quadratic Casimir, $$T_3^2+ T_2^2+T_1^2\equiv T(T+1).$$

The eigenvalue T, then, shared by all particles in a given multiplet (so T=1/2 for the electron and neutrino, T=1 for the original pre-SSB gauge-boson triplet, etc), are necessary to account for the gauge invariance of the Lagrangian.

As a result, the couplings of the various fields, even after SSB, reflect the weak isospin invariance underlying them. Thus, e.g., the coupling $$\frac{e}{\sin \theta_W \sqrt{2}} W^+_\mu (\overline{\nu_L} \gamma^\mu e_L+ \bar u \gamma^\mu d_L)$$ "knows about the other two generators" in ensuring the two lepton isodoublets have composed into an isotriplet to match the W isotriplet into an isosinglet. And likewise for the quarks.

So it is not enough that $T_3$ adds up to 0, across the reaction, here, e.g., W decay: the residual symmetries of the multiplet composition structure are also severely constrained, and you cannot have the W decaying to particle groupings conserving $T_3$ but not T.

To be sure, SSB slightly complicates (qualifies) the above statements, but, as other questions illustrate, the underlying weak isospin symmetry is evident all over the place.