# $\rho$-parameters for several Higgs multiplets

I'm studying the Higgs mechanism of the spontaneus symmetry breaking in the SM.

The expression for the $$\rho$$-parameter is $$\rho=\frac{M^2_W}{{\cos^2\theta_w}M^2_Z},$$ that, in the case of a $$SU(2)$$ doublet with hypercharge $$Y/2=1/2$$: $$\phi=\frac{1}{\sqrt{2}}(0\;\;\;v+h(x))^T$$, is $$\rho=1$$.

I did a straightforward calculation: evaluating $$|(D_\mu\phi)|^2$$, where $$D_\mu=\partial_\mu+igT^aW^a_\mu+i\frac{g'}{2}B_\mu,$$

in which $$T^a$$ are the generators of $$SU(2)$$ in doublet representation.

However, I found that in general, for several Higgs multiplets, with $$Y_i$$, $$T_i$$ and VEV $$v_i$$, the parameter becomes

$$\rho=\frac{\sum_iv_i^2[T_i(T_i+1)-Y_i^2]}{2\sum_iv_i^2Y_i^2}.$$

Searching the web I found this, in which I understand the new way to write the $$W^a$$, but I don't really understand the way to write the multiplets and how to get to the masses of $$W$$ and $$Z$$.

Any hints?

Considering several Higgs multiplets with $$T_i$$ and $$Y_i$$, whose neutral ($$Q_i=0$$) states acquire $$v_i$$, we can write the kinetic part of the Lagrangian as $$(D_\mu\phi_i)(D^\mu\phi_i)=\left|i\partial_\mu-\frac{g}{\sqrt{2}}(T_i^+W_\mu^++T_i^-W_\mu^-)\phi_i-\left(gT^3_iW_\mu^3+\frac{g}{2}Y_iB_\mu\right)\phi_i\right|^{\;2},$$ where $$T^{\pm}=T^1\pm iT^2$$ and $$\phi_i$$ is a multiplet.
Using the $$W^3,B\rightarrow Z, A$$ transformation: $$\sum_i |D_\mu\phi_i|^2=\\\sum_i\left|i\partial_\mu\phi_i-\frac{g}{\sqrt{2}}(T_i^+W_\mu^++T_i^-W_\mu^-)\phi_i-\frac{g}{\cos\theta_W}(T_i^3-\sin^2\theta_WQ_i)Z_\mu\phi_i-ig\sin\theta_WQ_iA_\mu\phi_i\right|^{\,\,2}$$
($$g'\cos\theta_W=g\sin\theta_W=e$$).
Then, recalling that $$Q_i=0$$ and $$T_i^3=-Y_i$$, we can find $$M_W$$ and $$M_Z$$:
$$M_W^2=\sum_i\frac{g^2}{2}\phi^+_i(T_i^+T_i^-+T_i^-T_i^+)\phi_i=\sum_i\frac{g^2v_i^2}{2}[T_i(T_i+1)-(T_i^3)^2]=\sum_i\frac{g^2v_i^2}{2}[T_i(T_i+1)-Y_i^2]$$ and $$M_Z^2=\sum_i\frac{g^2}{\cos^2\theta_W}(T_i^3)^2v_i^2=\sum_i\frac{g^2}{\cos^2\theta_W}v_i^2Y_i^2.$$
Finally: $$\rho=\frac{\sum_iv_i^2[T_i(T_i+1)-Y_i^2]}{2\sum_iv_i^2Y_i^2}.$$