Whenever I deal with spontaneous symmetry breaking (SSB), I encounter some confusions. This is my flow of logic and related questions:
We begin with UV gauge group. Find matter contents and their representation under the group.
Build a Lagrangian which is both Lorentz and gauge invariant. The Lagrangian consists of gauge and matter kinetic terms. Gauge-matter interaction comes from covariant derivative.
Introduce appropriate scalars. Add potential and possible Yukawa interactions.
Break original gauge group by giving scalars VEV. Some gauge boson get heavy(relative to SM scale)masses and disappear from spectrum.
Now this is what I am confused with. When we start to deal with broken phase, do we have to use the original Lagrangian but heavy part eliminated? or newly constructed Lagrangian with respect to broken gauge group?
Let me give you a specific example. Let us consider EW symmetry breaking. The kinetic term of the Original SM Lagrangian and Yukawa interaction are given as $$\mathcal{L} \supset i\bar{L}D_{\mu}\gamma^{\mu} L + y_{ij} \bar{L}He_R$$ where $L$ is $SU(2)_L$ doublet, $e_R$ is charged lepton singlet and $H$ is $SU(2)$ fundamental scalar. This scalar will get VEV $<H> = (0,v)^T$. This breaks $SU(2)_L \times U(1)_Y \rightarrow U(1)_{EM}$. I think this does not show that $L$ kinetic term will be broken into separate $e_L$ and $\nu_L$ kinetic term.
