will be assumed whenever we use the index $r$), and $\nu_r$’s depend only on model parameters, not on field variables. With this labeling convention the latter $n_E$ fields in $\bar{S}_i$’s correspond to physical scalars while the remaining ones are unphysical Goldstone bosons. The matrix $P$ and the $\nu_r$’s can be made independent of the $\xi^p$’s, because in the course of diagonalizing the $-\frac{1}{2}((U^{-1})^T\bM_G^2 U^{-1})_{ij}S_i S_j$ term, the $-\frac{1}{2}\xi^p\tilde{G}_p^2$ term is left untouched.

It is helpful to recall that in Eq.  the $\tilde{G}_p$’s can be viewed as linear combinations of $\bar{S}_i$’s. In fact, because $n_E$ physical scalars must exist, the matrix $P$ can be chosen so that the $\tilde{G}_p$’s do not contain the $\bar{S}_r$’s. That is to say, the $\tilde{G}_p$’s can be expressed as linear combinations of $\bar{S}_i,i=1,2,...,n_M$. Therefore, by examining Eq.  it is obvious that in $\mL_{quad}+\mL_{gf}$ the $n_E$ physical scalars are clearly separated from the unphysical ones after the orthogonal transformation Eq. .

At this stage we need to take a closer look at the unphysical scalar mass term in Eq. , which is $$\mL'\equiv-\frac{1}{2}\xi^p\tilde{G}_p^2
\label{eq:lp}$$ Recalling that the $\tilde{G}_p$’s are linear combinations of $\bar{S}_i,i=1,2,...,n_M$, the next thing we need to do is to find an orthogonal transformation $$\tilde{S}_i=K_{ij}\bar{S}_j
