\label{eq:kt}$$ which diagonalizes $\mL'$. In Eq.  $i,j$ range from $1$ to $n_S$, and $K$ is a $n_S\times n_S$ orthogonal matrix. Nevertheless, to avoid spoiling the already diagonalized physical scalar mass term, it is advisable to consider the following block-diagonal form of $K$: $$\begin{aligned}
K=\begin{pmatrix}
K_M & \textbf{0}_{n_M\times n_E} \\
\textbf{0}_{n_E\times n_M} & I_{n_E\times n_E}
\end{pmatrix}\end{aligned}$$ Here $I_{n_E\times n_E}$ is the $n_E\times n_E$ identity matrix, and $K_M$ is a $n_M\times n_M$ orthogonal matrix. With this form of matrix $K$ it is made clear that the $\bar{S}_r$’s actually don’t get transformed in this step, however the $-\frac{1}{2}\xi^p\tilde{G}_p^2$ term is diagonalized by $K_M$.

It remains to find the $n_M\times n_M$ orthogonal matrix $K_M$. We note that $\mL'$ written in the form of Eq.  is highly suggestive, because it has already completed the square. Therefore it seems natural to guess that the transformation we need is simply $$\tilde{S}_p=\alpha_p \tilde{G}_p,\quad p=1,2,...,n_M\,\text{(no summation over $p$)}
\label{eq:st}$$ Here the $\alpha_p$’s are constants chosen to make the transformed fields canonically normalized. Because the $\tilde{G}_p$’s can be expressed as linear combinations of $\bar{S}_i,i=1,2,...,n_M$, Eq.  effectively leads to a transformation from $\bar{S}_i,i=1,2,...,n_M$ to $\tilde{S}_i,i=1,2,...,n_M$, from which the matrix $K_M$ can be inferred.

There is one remaining potential loophole that we 