Imagine that you have some model with an enlarged scalar potential, such that there is, for instance, a quartic coupling $\kappa$ between the Higgs charged component and three other scalars, which do not get a vev.
After electroweak symmetry breaking in the Feynman-t’ Hooft gauge ($\xi=1$), one should have then a similar quartic coupling $\kappa$ between the three scalars and the Goldstone boson associated to $W$ with mass $m_W$.
However, in the unitary gauge ($\xi \rightarrow \infty$), the Goldstone boson gets “eaten” by the $W$ and the coupling $\kappa$ disappears, as the mass of the Goldstone is infinity in this gauge.
So my question is, where does the information of the coupling $\kappa$ go in the unitary gauge? Is it some how hidden in the momenta structure that appears in the $W$ propagator in the unitary gauge?
For clarification I give an example: Consider adding an $SU(2)_L$ doublet with hypercharge $3/2$, $\eta \equiv (\eta^{++},\eta^+)$. Along with an $SU(2)_L$ singlet with hypercharge $-1$, $S \equiv S^-$.
For this particle content the full scalar potential is, \begin{eqnarray} \mathcal{V} = \mathcal{V}_{SM} &+& m_{\eta} \eta^\dagger \eta + m_S S^* S \\ &+& ( \, \mu_2 \, \eta^\dagger H S + \kappa \, H \eta S S + \text{h.c.} \, ) \\ &+& \lambda_{\eta} \, (\eta^\dagger\eta)^2 + \lambda_{\eta S} \,(S^* S)^2 \\ &+& \lambda_{H\eta,1} \, (H^\dagger H)(\eta^\dagger\eta) + \lambda_{H\eta,3} \, (H^\dagger\eta^\dagger)(H\eta) + \lambda_{HS} \, (H^\dagger H)(S^* S) + \lambda_{\eta S} \, (\eta^\dagger\eta)(S^* S) \, . \end{eqnarray}
Regarding the couling $\kappa \, H \eta S S$,
\begin{equation} H \eta S S \supset H^+ \eta^+ S^- S^- \, , \end{equation} which is the coupling I am talking about in the main text of the question.