be $$\begin{aligned}
& \mathbb{C}^{Hd} = \nonumber \\
& \begin{pmatrix}
\frac{\sqrt{2}g}{c_W t_{2\beta}}\xi-\frac{g(7c_{2\beta}+c_{6\beta})}{3\sqrt{2}c_W s_{2\beta}^3}\xi^3 &
-\frac{g}{\sqrt{2}c_W}\xi+\frac{g(5+3c_{4\beta})}{6\sqrt{2}c_W s_{2\beta}^2}\xi^3 &
\frac{g}{2c_W}-\frac{g(5+3c_{4\beta})}{8c_W s_{2\beta}^2}\xi^2 &
\frac{g}{c_W t_{2\beta}}\xi^2 \\
\frac{2g\rho}{t_{2\beta}}\xi-\frac{g\rho(7c_{2\beta}+c_{6\beta})}{3s_{2\beta}^3}\xi^3
& -g\rho\xi+\frac{g\rho(5+3c_{4\beta})}{6s_{2\beta}^2}\xi^3
& g\kappa-\frac{g\kappa(5+3c_{4\beta})}{4s_{2\beta}^2}\xi^2
& -\frac{g}{c_W^2\sqrt{3-t_W^2}t_{2\beta}}\xi^2 \\
-g+\frac{g(5+3c_{4\beta})}{2s_{2\beta}^2}\xi^2
& -\frac{2g}{t_{2\beta}}\xi^2
& \frac{2\sqrt{2}g}{3t_{2\beta}}\xi-\frac{\sqrt{2}g(7c_{2\beta}+c_{6\beta})}{15s_{2\beta}^3}\xi^3
& 0
\end{pmatrix} \nonumber \\
& +\ord(\xi^4)\end{aligned}$$ The matrix $R$ can be computed as $$\begin{aligned}
R=\begin{pmatrix}
1+\ord(\xi^4) & -\frac{c_{2W}(1+2c_{2W})}{8c_W^5\sqrt{3-t_W^2}}\xi^2+\ord(\xi^4) & -\frac{\sqrt{2}}{3c_W t_{2\beta}}\xi^3+\ord(\xi^5) \\
\frac{c_{2W}(1+2c_{2W})}{8c_W^5\sqrt{3-t_W^2}}\xi^2+\ord(\xi^4) & 1+\ord(\xi^4) & -\frac{\sqrt{2}(1+2c_{2W})}{3c_W^2\sqrt{3-t_W^2}t_{2\beta}}\xi^3+\ord(\xi^5) \\
\frac{\sqrt{2}}{3c_W t_{2\beta}}\xi^3+\ord(\xi^5) & \frac{\sqrt{2}(1+2c_{2W})}{3c_W^2\sqrt{3-t_W^2}t_{2\beta}}\xi^3+\ord(\xi^5) & 1+\ord(\xi^6)
\end{pmatrix}\end{aligned}$$ With this precision it is feasible to obtain $c^{as}_{ZH\eta}$ and $c^{s}_{ZH\eta}$ via Eq.  and Eq.  to $\ord(\xi^3)$, the results of which are $$\begin{aligned}
c^{as}_{ZH\eta}=-\frac{g}{4\sqrt{2}c_W^3 t_{2\beta}}\xi^3+\ord(\xi^5)
\label{eq:casv}\end{aligned}$$ $$\begin{aligned}
c^{s}_{ZH\eta}=\frac{g}{\sqrt{2}c_W t_{2\beta}}\xi
+\frac{g}{24\sqrt{2}c_W s_{2\beta}}\left[\frac{8}{s_{2\beta}t_{2\beta}}+3c_{2\beta}\left(8+\frac{6}{c_W^2}-\frac{1}{c_W^4}\right)\right]\xi^3
+\ord(\xi^5)
\label{eq:csv}\end{aligned}$$ Therefore we arrive at the conclusion that the symmetric $ZH\eta$ vertex appear at $\ord(\xi)$, while the antisymmetric $ZH\eta$ vertex does not appear until $\ord(\xi^3)$. The coefficients of these two vertices are presented in Eq.  and Eq. , respectively. We note that this conclusion differs from what has been derived and used in the literature [@Kilian:2004pp; @Kilian:2006eh] for a long time.

Discussion and conclusion {#sec:conc}
=========================

In this paper we revisited the issue of deriving the mass eigenstate $ZH\eta$ vertex in the SLH. We found that the scalar kinetic terms are not canonically normalized in the usual 