Prescription for finite oblique parameters S and U in extensions of the SM with m_W≠m_Z cos θ_W

The present knowledge of particle physics is encapsulated in the standard model (SM) [1–3]. This is an SU(3) × SU(2) × U(1) gauge theory that describes all the fundamental particles observed until now and the way they interact with each other. The scalar sector of the SM contains just one doublet of SU(2); it is responsible for giving masses m_W and m_Z to the gauge bosons W^± and Z⁰, respectively, as well as to the fermions.

Despite being one of the most accurate theories in Science, there are phenomena that the SM cannot explain. Therefore particle physicists try to either extend or complete it. One of the ways to extend the SM is by enlarging its scalar sector [4]. The most studied extension of that sector is the addition of another SU(2) doublet, obtaining a two-Higgs-doublet model [5]. Extensions of the SM with scalar SU(2) singlets are also frequent in the literature (see for instance references [6–8]). There are also some extensions of the SM with scalar SU(2) triplets in the literature (see for instance references [9–17]).

A feature of the SM is custodial symmetry. This is a symmetry of the scalar potential of the SM that is broken both by the Yukawa interactions and by the gauge coupling of the U(1) group. Custodial symmetry leads at the tree level to the relation

$$\begin{equation}{m}_{W}={m}_{Z}{c}_{W},\end{equation} \tag{ 1 }$$

where c_W is the cosine of the Weinberg angle. The relation (1) is in good agreement with observation: experimental results give [18]

$$\begin{equation}\rho \equiv \frac{{m}_{W}^{2}}{{m}_{Z}^{2}{c}_{W}^{2}}=1.000\,38\pm 0.000\,20.\end{equation} \tag{ 2 }$$

Models in which only scalar SU(2) singlets and doublets have vacuum expectation values (VEVs) preserve the relation (1) at the tree level. That relation is broken at the loop level because the custodial symmetry is not exact, as noted above.

Peskin and Takeuchi have identified three so-called 'oblique parameters' that they named S, T, and U [19, 20] ¹ . These three observables parameterize some effects of new physics (NP), i.e. physics of extensions of the SM. At the one-loop level, vacuum polarization produces the tensors ${{\Pi}}_{V{V}^{\prime }}^{\mu \nu }(q)$, where V and V' are gauge bosons that may be either A and A (A stands for the photon), A and Z, Z and Z, or W and W, and q is the four-momentum of the gauge bosons. We write those tensors as

$$\begin{equation}{{\Pi}}_{V{V}^{\prime }}^{\mu \nu }(q)={g}^{\mu \nu }{A}_{V{V}^{\prime }}({q}^{2})+{q}^{\mu }{q}^{\nu }{B}_{V{V}^{\prime }}({q}^{2}).\end{equation} \tag{ 3 }$$

We then define $\delta {A}_{V{V}^{\prime }}({q}^{2})\equiv {\left.{A}_{V{V}^{\prime }}({q}^{2})\right\vert }_{\text{NP}}-{\left.{A}_{V{V}^{\prime }}({q}^{2})\right\vert }_{\text{SM}}$, where ${\left.{A}_{V{V}^{\prime }}({q}^{2})\right\vert }_{\text{NP}}$ is A_VV'(q²) computed in a given extension of the SM and ${\left.{A}_{V{V}^{\prime }}({q}^{2})\right\vert }_{\text{SM}}$ is A_VV'(q²) computed in the SM. The oblique parameters are defined as

$$\begin{equation}T=\frac{1}{\alpha {m}_{Z}^{2}}\left[\frac{\delta {A}_{WW}(0)}{{c}_{W}^{2}}-\delta {A}_{ZZ}(0)\right],\end{equation} \tag{ 4a }$$

$$\begin{align}\hfill S& \hfill =\frac{4{s}_{W}^{2}{c}_{W}^{2}}{\alpha }\left[{\left.\frac{\partial \delta {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}-{\left.\frac{\partial \delta {A}_{AA}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}+\frac{{c}_{W}^{2}-{s}_{W}^{2}}{{c}_{W}{s}_{W}}{\left.\frac{\partial \delta {A}_{AZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\right],\end{align} \tag{ 4b }$$

$$\begin{align}\hfill U& =\frac{4{s}_{W}^{2}}{\alpha }\left[{\left.\frac{\partial \delta {A}_{WW}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}-{c}_{W}^{2}{\left.\frac{\partial \delta {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}-{s}_{W}^{2}{\left.\frac{\partial \delta {A}_{AA}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}+2{c}_{W}{s}_{W}{\left.\frac{\partial \delta {A}_{AZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\right].\hfill \end{align} \tag{ 4c }$$

The oblique parameters have already been computed for several extensions of the SM, in particular for models with arbitrary numbers of scalar doublets and singlets [23, 24], wherein they are finite. In models where larger scalar multiplets are added to the SM and have VEVs, m_W is in general different from m_Z c_W . As we will show in this paper, this gives rise to complications when computing the oblique parameters ² . First of all, the oblique parameter T is divergent at one-loop level for models that do not have custodial symmetry [29, 30]. Furthermore, as noted above, when computing the oblique parameters one needs to subtract the functions A_VV'(q²) computed in the SM from the same functions computed in the extension of the SM. This subtraction is non-trivial, since in the SM the masses of the gauge bosons obey equation (1) while in the extension they do not obey it. This leads to divergent parameters S and U. We have found out that, in order to obtain finite (and gauge-independent) results for S and U at the one-loop level in models with m_W ≠ m_Z c_W , one needs to multiply the usual SM Feynman rules ³ for some vertices containing the SM Higgs boson by factors that become equal to 1 when m_W = m_Z c_W but are different from 1 for m_W ≠ m_Z c_W .

This paper is organized as follows. In section 2 we consider an extension of the SM with arbitrary numbers of arbitrarily large SU(2) × U(1) scalar multiplets, subject only to the constraint of the preservation of electric charge. In section 3 we illustrate the problem that arises in the computation of both S and U in that extension of the SM. In section 4 we display our proposed solution to that problem. In section 5 we restrict the model of section 2 to the case where only scalar SU(2) triplets and singlets with U(1) charges 1 or 0 are added to the SM; we give the explicit expression for S in that model. Section 6 summarizes our findings. Several appendices contain technical material that may be avoided by a hurried reader.

We consider an SU(2) × U(1) electroweak model where the scalar sector consists of an arbitrary number of multiplets M_JY labeled by their weak isospin J and weak hypercharge Y. We restrict ourselves to models where all the scalars have integer electric charges ⁴ ; therefore, if J is a (half-)integer then Y is a (half-)integer too. Furthermore, we consider only complex multiplets; if some multiplets are real, then our conclusions are still valid but there are some modifications in the intermediate steps. Finally, for the sake of simplicity we assume that there is only one M_JY for each value of the pair $\left(J,Y\right)$; our conclusions are still valid otherwise, but the notation would become clumsier.

Each multiplet M_JY has a component ${M}_{JY}^{Q}$ with electric charge Q if and only if $J-\left\vert Q-Y\right\vert \in {\mathbb{N}}_{0}$. Some multiplets M_JY may have, for some values of Q ≠ 0, both a component ${M}_{JY}^{Q}$ with electric charge Q and a component ${M}_{JY}^{-Q}$ with electric charge −Q; in our notation, ${M}_{JY}^{Q}\ne {\left({M}_{JY}^{-Q}\right)}^{\ast }$.

The covariant derivative for the SU(2) × U(1) electroweak model is, as usual ⁵ ,

$$\begin{equation}{D}_{\mu }={\partial }_{\mu }+\mathrm{i}eQ{A}_{\mu }-\mathrm{i}\frac{g}{{c}_{W}}\left({T}_{3}-Q{s}_{W}^{2}\right){Z}_{\mu }-\mathrm{i}g\left({W}_{\mu }^{+}{T}_{+}+{W}_{\mu }^{-}{T}_{-}\right),\end{equation} \tag{ 5 }$$

where e = gs_W is the unit of electric charge, the T_i (i = 1, 2, 3) are the three components of weak isospin, and the T_± are given by ${T}_{\pm }=\left({T}_{1}\pm \mathrm{i}{T}_{2}\right)/\sqrt{2}$. We use the SU(2) representation for weak isospin J:

$$\begin{equation}{\left({T}_{3}\right)}_{rc}={\delta }_{rc}\left(J+1-r\right),\end{equation} \tag{ 6a }$$

$$\begin{equation}{\left({T}_{+}\right)}_{rc}={\delta }_{r+1,c}\sqrt{\frac{r\left(2J+1-r\right)}{2}},\end{equation} \tag{ 6b }$$

$$\begin{equation}{\left({T}_{-}\right)}_{rc}={\delta }_{r-1,c}\sqrt{\frac{\left(r-1\right)\left(2J+2-r\right)}{2}},\end{equation} \tag{ 6c }$$

where r stands for the row of the matrix and c stands for the column of the matrix, with 1 ⩽ r, c ⩽ 2J + 1. We write M_JY as a column matrix with 2J + 1 rows upon which the matrices of equation (6) act. Clearly, from equation (6a), ${M}_{JY}^{Q}$ is in the row J + 1 − T₃ = J + Y + 1 − Q of that column matrix. Then, using equations (5) and (6),

$$\begin{equation}{D}_{\mu }{M}_{JY}^{Q}={\partial }_{\mu }{M}_{JY}^{Q}+\mathrm{i}eQ{A}_{\mu }{M}_{JY}^{Q}+\mathrm{i}\frac{g}{{c}_{w}}{Z}_{\mu }{M}_{JY}^{Q}\left(Y-Q{c}_{W}^{2}\right)\end{equation} \tag{ 7a }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{+}{M}_{JY}^{Q-1}\sqrt{\frac{\left(J+Y+1-Q\right)\left(J-Y+Q\right)}{2}}\end{equation} \tag{ 7b }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{-}{M}_{JY}^{Q+1}\sqrt{\frac{\left(J+Y-Q\right)\left(J-Y+Q+1\right)}{2}}.\end{equation} \tag{ 7c }$$

We define $\mathcal{A}$ to be the set of multiplets that have a component with zero electric charge. A multiplet M_JY belongs to $\mathcal{A}$ if and only if $J-\left\vert Y\right\vert \in {\mathbb{N}}_{0}$.

For every $Q\in \mathbb{N}$, we define two sets:

• ${\mathcal{R}}_{Q}$ is the set of multiplets that have a component with electric charge Q; a multiplet M_JY belongs to ${\mathcal{R}}_{Q}$ if and only if $J-\left\vert Q-Y\right\vert \in {\mathbb{N}}_{0}$.
• ${\mathcal{S}}_{Q}$ is the set of multiplets that have a component with electric charge −Q; a multiplet M_JY belongs to ${\mathcal{S}}_{Q}$ if and only if $J-\left\vert Q+Y\right\vert \in {\mathbb{N}}_{0}$.

We emphasize that both ${\mathcal{R}}_{Q}$ and ${\mathcal{S}}_{Q}$ are defined only for Q > 0. For those Q, let n_Q denote the total number of charge-Q scalars and let ${S}_{a}^{Q}$ (a = 1, ..., n_Q ) denote the physical (mass eigenstate) charge-Q scalars. For definiteness, we fix ${S}_{1}^{1}\equiv {G}^{+}$ to be the charged Goldstone boson. We define ${S}_{a}^{-Q}\equiv {\left({S}_{a}^{Q}\right)}^{\ast }$.

For every $Q\in \mathbb{N}$ and ${M}_{JY}\in {\mathcal{R}}_{Q}$, we write

$$\begin{equation}{M}_{JY}^{Q}={R}_{JY}^{Q}\left(\begin{matrix}\hfill {S}_{1}^{Q}\hfill \\ \hfill {S}_{2}^{Q}\hfill \\ \hfill {\vdots}\hfill \\ \hfill {S}_{{n}_{Q}}^{Q}\hfill \end{matrix}\right),\end{equation} \tag{ 8 }$$

where ${R}_{JY}^{Q}$ is a 1 × n_Q row matrix. For every $Q\in \mathbb{N}$ and ${M}_{JY}\in {\mathcal{S}}_{Q}$, we write

$$\begin{equation}{\left({M}_{JY}^{-Q}\right)}^{\ast }={S}_{JY}^{Q}\left(\begin{matrix}\hfill {S}_{1}^{Q}\hfill \\ \hfill {S}_{2}^{Q}\hfill \\ \hfill {\vdots}\hfill \\ \hfill {S}_{{n}_{Q}}^{Q}\hfill \end{matrix}\right),\end{equation} \tag{ 9 }$$

where ${S}_{JY}^{Q}$ is a 1 × n_Q row matrix. We form the n_Q × n_Q unitary (mixing) matrix U^Q by piling up all the row matrices ${R}_{JY}^{Q}$ and ${S}_{JY}^{Q}$ (for fixed $Q\in \mathbb{N}$) on top of each other. Since U^Q is unitary,

$$\begin{equation}\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}{\left({R}_{JY}^{Q}\right)}_{1a}{\left({R}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }+\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}{\left({S}_{JY}^{Q}\right)}_{1a}{\left({S}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }={\delta }_{a{a}^{\prime }},\end{equation} \tag{ 10a }$$

$$\begin{equation}\sum\limits _{a=1}^{{n}_{Q}}{\left({R}_{JY}^{Q}\right)}_{1a}{\left({R}_{{J}^{\prime }{Y}^{\prime }}^{Q}\right)}_{1a}^{\ast }={\delta }_{J{J}^{\prime }}{\delta }_{Y{Y}^{\prime }},\end{equation} \tag{ 10b }$$

$$\begin{equation}\sum\limits _{a=1}^{{n}_{Q}}{\left({S}_{JY}^{Q}\right)}_{1a}{\left({S}_{{J}^{\prime }{Y}^{\prime }}^{Q}\right)}_{1a}^{\ast }={\delta }_{J{J}^{\prime }}{\delta }_{Y{Y}^{\prime }},\end{equation} \tag{ 10c }$$

$$\begin{equation}\sum\limits _{a=1}^{{n}_{Q}}{\left({R}_{JY}^{Q}\right)}_{1a}{\left({S}_{{J}^{\prime }{Y}^{\prime }}^{Q}\right)}_{1a}^{\ast }=0.\end{equation} \tag{ 10d }$$

Each multiplet M_JY that belongs to $\mathcal{A}$ has VEV v_JY (which may in some cases be zero) in its component ${M}_{JY}^{0}$ with electric charge zero. We write

$$\begin{equation}{M}_{JY}^{0}={v}_{JY}+\frac{{A}_{JY}+\mathrm{i}{B}_{JY}}{\sqrt{2}}\left(\begin{matrix}\hfill {S}_{1}^{0}\hfill \\ \hfill {S}_{2}^{0}\hfill \\ \hfill {\vdots}\hfill \\ \hfill {S}_{{n}_{0}}^{0}\hfill \end{matrix}\right),\end{equation} \tag{ 11 }$$

where A_JY and B_JY are real 1 × n₀ row matrices and the real fields ${S}_{b}^{0}(b=1,\dots ,{n}_{0})$ are the physical neutral scalars. Without lack of generality, we fix ${S}_{1}^{0}\equiv {G}^{0}$ to be the neutral Goldstone boson. We form the n₀ × n₀ real orthogonal matrix V by piling up all the row matrices A_JY and B_JY on top of each other. Since V is orthogonal,

$$\begin{equation}\sum\limits _{b=1}^{{n}_{0}}{\left[{\left({A}_{JY}\right)}_{1b}\right]}^{2}=\sum\limits _{b=1}^{{n}_{0}}{\left[{\left({B}_{JY}\right)}_{1b}\right]}^{2}=1,\end{equation} \tag{ 12a }$$

$$\begin{equation}\sum\limits _{b=1}^{{n}_{0}}{\left({A}_{JY}\right)}_{1b}{\left({B}_{JY}\right)}_{1b}=0.\end{equation} \tag{ 12b }$$

Hence,

$$\begin{equation}\sum\limits _{b=1}^{{n}_{0}-1}\sum\limits _{{b}^{\prime }=b+1}^{{n}_{0}}{\left[{\left({A}_{JY}\right)}_{1b}{\left({B}_{JY}\right)}_{1{b}^{\prime }}-{\left({B}_{JY}\right)}_{1b}{\left({A}_{JY}\right)}_{1{b}^{\prime }}\right]}^{2}=1,\end{equation} \tag{ 13a }$$

$$\begin{equation}\sum\limits _{b=1}^{{n}_{0}}{\left\vert {\left({A}_{JY}+\mathrm{i}{B}_{JY}\right)}_{1b}\right\vert }^{2}=2.\end{equation} \tag{ 13b }$$

The masses of the gauge bosons are given in terms of the VEVs of the scalar fields by

$$\begin{equation}\frac{{m}_{Z}^{2}}{2}=\frac{{g}^{2}}{{c}_{W}^{2}}\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}{Y}^{2},\end{equation} \tag{ 14a }$$

$$\begin{align}\hfill {m}_{W}^{2}& ={g}^{2}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}{\left\vert {v}_{JY}\right\vert }^{2}\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}\right.\hfill \\ \hfill & \quad +\left.\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}{\left\vert {v}_{JY}\right\vert }^{2}\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}\right]\hfill \end{align} \tag{ 14b }$$

$$\begin{equation*}={g}^{2}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}\right.\end{equation*}$$

$$\begin{equation}\quad +\left.\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}\right]\end{equation} \tag{ 14c }$$

$$\begin{equation}={g}^{2}\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}\left({J}^{2}-{Y}^{2}+J\right).\end{equation} \tag{ 14d }$$

In line (14b) we have used firstly line (7b) with Q = 1 and secondly line (7c) with Q = −1. In passing from line (14b) to line (14c) we have used

$$\begin{equation}{M}_{JY}\in \mathcal{A}{\backslash}{\mathcal{R}}_{1}{\Rightarrow}J=-Y,\qquad {M}_{JY}\in \mathcal{A}{\backslash}{\mathcal{S}}_{1}{\Rightarrow}J=Y.\end{equation} \tag{ 15 }$$

One sees in equation (14) that ${m}_{Z}^{2}{c}_{W}^{2}={m}_{W}^{2}$ in general requires J² + J = 3Y² for all the nonzero v_JY . ⁶ This holds for the standard case of doublets with J = Y = 1/2, and also for neutral singlets with J = Y = 0, but it does not hold for most other choices of J and Y.

We consider the vacuum-polarization one-loop diagram with two scalars as internal particles, see figure 1. Let the Feynman rules for its vertices be ${f}_{\mu }=K{\left(p-{p}^{\prime }\right)}_{\mu }$ and ${f}_{\nu }^{\prime }={K}^{\prime }{\left(p-{p}^{\prime }\right)}_{\nu }$. Then the contribution of that diagram to

$$\begin{equation}{\left.\frac{\partial {A}_{V{V}^{\prime }}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\end{equation} \tag{ 16 }$$

is

$$\begin{equation}{\left.\frac{\partial {A}_{V{V}^{\prime }}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}=-K{K}^{\prime }\enspace \mathrm{d}\mathrm{i}\mathrm{v}+\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\end{equation} \tag{ 17 }$$

where 'div' is a divergent quantity defined in equation (68) of appendix A. That quantity is independent both of q² and of the masses of the scalars in the diagram of figure 1.

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We want to compute the divergent contributions to the quantities (16) for VV' = AA, AZ, ZZ, and WW. Those contributions originate solely in diagrams like the one in figure 1. (Other types of diagrams give either vanishing or finite contributions.) We therefore need the factors K and K' for all the diagrams, i.e. we need the Lagrangian terms that describe the interactions between one gauge boson and two scalars. Those terms are derived from equation (7) and may be found in appendix B.

Firstly consider A_AA (q²). This is generated by loops with charged scalars ${S}_{a}^{Q}$ and ${S}_{a}^{-Q}$. According to equation (69) of appendix B, they generate

$$\begin{equation}{\left.\frac{\partial {A}_{AA}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}={e}^{2}\,\mathrm{d}\mathrm{i}\mathrm{v}\sum\limits _{Q > 0}{Q}^{2}{n}_{Q}+\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}.\end{equation} \tag{ 18 }$$

Secondly consider A_AZ (q²). This is once again generated by loops with charged scalars ${S}_{a}^{Q}$ and ${S}_{a}^{-Q}$. According to equations (69) and (70) of appendix B, they generate

$$\begin{align}\hfill {\left.\frac{\partial {A}_{AZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}& =\frac{eg}{{c}_{W}}\mathrm{d}\mathrm{i}\mathrm{v}\sum\limits _{Q > 0}\left(-{c}_{W}^{2}{Q}^{2}{n}_{Q}+\!\!\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}QY\right.-\left.\!\!\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}QY\right)+\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\hfill \end{align} \tag{ 19 }$$

where we have used equations (10b) and (10c).

Thirdly consider A_ZZ (q²). This is generated either by loops with charged scalars ${S}_{a}^{Q}$ and ${S}_{{a}^{\prime }}^{-Q}$, where a' may be different from a, or by loops with neutral scalars ${S}_{b}^{0}$ and ${S}_{{b}^{\prime }}^{0}$, where b' may differ from b. The relevant equations are equations (70) and (71) of appendix B, respectively. They generate

$$\begin{align}\hfill {\left.\frac{\partial {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}& =\left.\frac{{g}^{2}}{{c}_{W}^{2}}\mathrm{d}\mathrm{i}\mathrm{v}\sum\limits _{Q > 0}\ \sum\limits _{a,{a}^{\prime }=1}^{{n}_{Q}}\right\vert -{c}_{W}^{2}Q{\delta }_{a{a}^{\prime }}+\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}Y{\left({R}_{JY}^{Q}\right)}_{1a}\hfill \\ \hfill & \quad \times {\left({R}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }-{\left.\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}Y{\left({S}_{JY}^{Q}\right)}_{1a}{\left({S}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }\right\vert }^{2}\hfill \end{align} \tag{ 20a }$$

$$\begin{align}\hfill & \quad +\frac{{g}^{2}}{{c}_{W}^{2}}\mathrm{d}\mathrm{i}\mathrm{v}\sum\limits _{b=1}^{{n}_{0}-1}\sum\limits _{{b}^{\prime }=b+1}^{{n}_{0}}\left\{\sum\limits _{{M}_{JY}\in \mathcal{A}}Y\left[{\left({A}_{JY}\right)}_{1b}{\left({B}_{JY}\right)}_{1{b}^{\prime }}\right.\right.\hfill \\ \hfill & \quad -{\left.\left.{\left({B}_{JY}\right)}_{1b}{\left({A}_{JY}\right)}_{1{b}^{\prime }}\right]\right\}}^{2}+\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s}.\hfill \end{align} \tag{ 20b }$$

This leads to

$$\begin{align}\hfill {\left.\frac{\partial {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}& =\frac{{g}^{2}}{{c}_{W}^{2}}\mathrm{d}\mathrm{i}\mathrm{v}\left[\sum\limits _{Q > 0}\left({c}_{W}^{4}{Q}^{2}{n}_{Q}-2{c}_{W}^{2}Q\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}Y+2{c}_{W}^{2}Q\right.\right.\hfill \\ \hfill & \quad \times \sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}Y\left.\left.+\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}{Y}^{2}+\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}{Y}^{2}\right)\right.\hfill \\ \hfill & \quad +\sum\limits _{{M}_{JY}\in \mathcal{A}}{Y}^{2}\left.\right]+\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\hfill \end{align} \tag{ 21 }$$

where we have used equations (10b)–(10d) and (13a).

Finally consider A_WW (q²). This is generated either by loops with one charged scalar ${S}_{a}^{Q}$ and one neutral scalar ${S}_{b}^{0}$, or by loops with two charged scalars ${S}_{a}^{\pm Q}$ and ${S}_{{a}^{\prime }}^{\mp \left(Q+1\right)}$. The relevant equations are equation (72) and (73) of appendix B, respectively. They generate

$$\begin{equation}{\left.\frac{\partial {A}_{WW}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}=\frac{{g}^{2}}{2}\mathrm{d}\mathrm{i}\mathrm{v}\left\{2\sum\limits _{{M}_{JY}\in \mathcal{A}}\left({J}^{2}-{Y}^{2}+J\right)\right.\end{equation} \tag{ 22a }$$

$$\begin{align}\hfill & \qquad +\sum\limits _{Q > 0}\left[\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}\left(J+Y-Q\right)\left(J-Y+Q+1\right)\right.\hfill \\ \hfill & \qquad +\left.\left.\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}\left(J+Y+Q+1\right)\left(J-Y-Q\right)\right]\right\}\hfill \\ \hfill & \qquad +\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\hfill \end{align} \tag{ 22b }$$

where we have used equations (10b)–(10d) and (13b), and also

$$\begin{equation}{M}_{JY}\in {\mathcal{R}}_{Q}{\backslash}{\mathcal{R}}_{Q+1}{\Rightarrow}J+Y-Q=0,\qquad {M}_{JY}\in {\mathcal{S}}_{Q}{\backslash}{\mathcal{S}}_{Q+1}{\Rightarrow}J-Y-Q=0.\end{equation} \tag{ 23 }$$

It is now clear that the combinations relevant for the parameters S and U in equations (4c) and (4d), respectively, namely

$$\begin{align}\hfill & {\left.\frac{\partial {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}-{\left.\frac{\partial {A}_{AA}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\hfill \\ \hfill & \qquad +\frac{{c}_{W}^{2}-{s}_{W}^{2}}{{c}_{W}{s}_{W}}{\left.\frac{\partial {A}_{AZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\equiv \frac{{g}^{2}}{{c}_{W}^{2}}\mathrm{d}\mathrm{i}\mathrm{v}\,\psi +\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\hfill \end{align} \tag{ 24a }$$

$$\begin{align}\hfill & {\left.\frac{\partial {A}_{WW}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}-{c}_{W}^{2}{\left.\frac{\partial {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}-{s}_{W}^{2}{\left.\frac{\partial {A}_{AA}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\hfill \\ \hfill & \qquad +2{c}_{W}{s}_{W}{\left.\frac{\partial {A}_{AZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\equiv \frac{{g}^{2}}{2}\mathrm{d}\mathrm{i}\mathrm{v}\,\theta +\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\hfill \end{align} \tag{ 24b }$$

are given by

$$\begin{equation}\psi =\sum\limits _{{M}_{JY}\in \mathcal{A}}{Y}^{2}+\sum\limits _{Q > 0}\left[\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}\left({Y}^{2}-QY\right)+\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}\left({Y}^{2}+QY\right)\right],\end{equation} \tag{ 25a }$$

$$\begin{align}\hfill \theta & =2\sum\limits _{{M}_{JY}\in \mathcal{A}}\left({J}^{2}+J-2{Y}^{2}\right)+\sum\limits _{Q > 0}\left\{-2{Q}^{2}{n}_{Q}\right.\hfill \\ \hfill & \quad +\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}\left[\left(J+Y-Q\right)\left(J-Y+Q+1\right)-2{Y}^{2}+4QY\right]\hfill \\ \hfill & \quad +\left.\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}\left[\left(J+Y+Q+1\right)\left(J-Y-Q\right)-2{Y}^{2}-4QY\right]\right\}.\hfill \end{align} \tag{ 25b }$$

It is proven in appendix C that ψ and θ are both zero, no matter what the set of M_JY (and no matter what the VEVs of their neutral components are) present in any particular SU(2) × U(1) electroweak model, i.e. either in the SM or in any extension thereof.

Since the quantities ψ and θ vanish for the NP model and, identically, also for the SM, one might think that S and U would turn out finite. However, if we subtract the SM result for ψ and θ, computed by using the usual SM Feynman rules for the Goldstone-boson vertices, from the NP result, which uses a different set of Feynman rules for those vertices, then we get a gauge-dependent result for both S and U. This happens because the gauge-dependent parts of the NP and SM results are different when equation (1) does not hold in the NP model. Indeed, as is shown in appendix D, some Goldstone-boson vertices depend on ρ, hence they are different in the SM and in the NP model. In equation (26) we display the relevant Goldstone-boson vertices and their ρ-dependent value, as derived in appendix D; we also display, for each vertex and crossed over, its value in the SM, i.e. when ρ = 1, as given in reference [31].

We thus have a problem. We may obtain ψ = θ = 0 in both the SM and its extension by using, respectively, the Feynman rules adequate for each of them, with different values of ρ (viz ρ = 1 in the SM and ρ ≠ 1 in the extension); but, since the Feynman rules for the Goldstone-boson vertices differ in the two cases, we will then end up with a gauge-dependent result. Or else we may use in both the SM and its extension the same Feynman rules, as they are written in equation (26), i.e. with the same value of ρ, and then we will get a gauge-independent result—but ψ and θ will be non-zero in the SM and, therefore, S and U, respectively, will diverge.

We propose to make S and U finite by multiplying other SM Feynman rules—not the ones in equation (26)—by factors that are equal to 1 when ρ = 1. This is possible because in the SM there is the Higgs particle H, with mass m_h , that participates in the computation of ${\left.{A}_{ZZ}({q}^{2})\right\vert }_{\text{SM}}$ through a loop with G⁰ and H, and in the computation of ${\left.{A}_{WW}({q}^{2})\right\vert }_{\text{SM}}$ through a loop with G⁺ and H. We thus propose to use, in the SM, the following Feynman rules for the vertices ZG⁰ H, ZZH, W^± G^∓ H, and W⁺ W⁻ H, respectively:

The vertices ZG⁰ H and ZZH are multiplied by the same factor $\sqrt{\aleph }$ due to gauge invariance. The same happens with the vertices W^± G^∓ H and W⁺ W⁻ H, which are both multiplied by $\sqrt{\beth }$. Obviously, in the true SM, i.e. when relation (1) holds, both ℵ and ℶ are 1.

In the computation of the SM contribution to S and U, we must consider six diagrams like the one in figure 1:

• (a)  
  A diagram with inner scalars G⁺ and H contributing to A_WW (q²).
• (b)  
  A diagram with inner scalars G⁺ and G⁰ contributing to A_WW (q²).
• (c)  
  A diagram with inner scalars G⁰ and H contributing to A_ZZ (q²).
• (d)  
  A diagram with inner scalars G⁺ and G⁻ contributing to A_ZZ (q²).
• (e)  
  A diagram with inner scalars G⁺ and G⁻ contributing to A_AA (q²).
• (f)  
  A diagram with inner scalars G⁺ and G⁻ contributing to A_AZ (q²).

A factor ℵ multiplies the diagram (c) and a factor ℶ multiplies the diagram (a).

Using equation (3) and the results for the quantities (16) computed in the SM, we get for the parameter S

$$\begin{equation}\frac{\alpha }{4{s}_{W}^{2}{c}_{W}^{2}}\ S=\frac{{g}^{2}}{{c}_{W}^{2}}\mathrm{d}\mathrm{i}\mathrm{v}\left[{\psi }_{\text{NP}}-\frac{\aleph }{4}-{c}_{W}^{4}{\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right)}^{2}+{s}_{W}^{2}{c}_{W}^{2}\right.\end{equation} \tag{ 28a }$$

$$\begin{equation}\quad +\left.{c}_{W}^{2}\left({c}_{W}^{2}-{s}_{W}^{2}\right)\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right)\right]+\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\end{equation} \tag{ 28b }$$

where the second to fifth terms inside the square brackets originate in diagrams (c)–(f), respectively. We have used equation (26b) in the third and fifth terms inside the square brackets. Since ψ_NP = 0, we obtain a finite result for S if

$$\begin{equation}\aleph =4\left[-{c}_{W}^{4}{\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right)}^{2}+{s}_{W}^{2}{c}_{W}^{2}+{c}_{W}^{2}\left({c}_{W}^{2}-{s}_{W}^{2}\right)\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right)\right].\end{equation} \tag{ 29 }$$

Thus, we must choose

$$\begin{equation}\aleph =\frac{2}{\rho }-\frac{1}{{\rho }^{2}}.\end{equation} \tag{ 30 }$$

Notice that ℵ = 1 when equation (1) holds, viz when the extension of the SM only has scalar doublets and/or singlets.

For the parameter U we get

$$\begin{equation}\frac{\alpha }{4{s}_{W}^{2}}U=\frac{{g}^{2}}{2}\mathrm{d}\mathrm{i}\mathrm{v}\left[{\theta }_{\text{NP}}-\frac{\beth }{2}-\frac{{m}_{Z}^{2}{c}_{W}^{2}}{2{m}_{W}^{2}}+\frac{\aleph }{2}+2{c}_{W}^{4}{\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right)}^{2}\right.\end{equation} \tag{ 31a }$$

$$\begin{equation}\quad +\left.2{s}_{W}^{4}+4{s}_{W}^{2}{c}_{W}^{2}\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right)\right]+\mathrm{fi}\mathrm{n}\mathrm{i}\mathrm{t}\mathrm{e}\ \mathrm{t}\mathrm{e}\mathrm{r}\mathrm{m}\mathrm{s},\end{equation} \tag{ 31b }$$

where the second to seventh terms inside the square brackets originate in the diagrams (a)–(f), respectively. We have used equation (26c) in the third term and equation (26b) in the fifth and seventh terms inside the square brackets. Since θ_NP = 0, we get a finite U if

$$\begin{equation}\beth =-\frac{{m}_{Z}^{2}{c}_{W}^{2}}{{m}_{W}^{2}}+\aleph +4{c}_{W}^{4}{\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right)}^{2}+4{s}_{W}^{4}+8{s}_{W}^{2}{c}_{W}^{2}\left(1-\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}\right).\end{equation} \tag{ 32 }$$

Using equation (30), we obtain

$$\begin{equation}\beth =4-\frac{3}{\rho }.\end{equation} \tag{ 33 }$$

Befittingly, ℶ = 1 when equation (1) holds, viz in an extension of the SM with only scalar doublets and singlets.

5.1. The model

In this section we consider an SU(2) × U(1) electroweak model where the scalar sector comprises arbitrary numbers of

• SU(2) doublets with weak hypercharge 1/2
  $$\begin{equation}{{\Phi}}_{k}=\left(\begin{matrix}\hfill {\phi }_{k}^{+}\hfill \\ \hfill {\phi }_{k}^{0}\hfill \end{matrix}\right)\quad (k=1,\dots ,{n}_{d}),\end{equation} \tag{ 34 }$$
• SU(2) triplets with weak hypercharge 1
  $$\begin{equation}{{\Xi}}_{p}=\left(\begin{matrix}\hfill {\xi }_{p}^{++}\hfill \\ \hfill {\xi }_{p}^{+}\hfill \\ \hfill {\xi }_{p}^{0}\hfill \end{matrix}\right)\quad (p=1,\dots ,{n}_{{t}_{1}}),\end{equation} \tag{ 35 }$$
• real SU(2) triplets with weak hypercharge 0
  $$\begin{equation}{{\Lambda}}_{q}=\left(\begin{matrix}\hfill {\lambda }_{q}^{+}\hfill \\ \hfill {\lambda }_{q}^{0}={{\lambda }_{q}^{0}}^{\ast }\hfill \\ \hfill -{{\lambda }_{q}^{+}}^{\ast }\hfill \end{matrix}\right)\quad (q=1,\dots ,{n}_{{t}_{0}}),\end{equation} \tag{ 36 }$$
• SU(2) singlets with weak hypercharge 1
  $$\begin{equation}{\chi }_{j}^{+}\quad (j=1,\dots ,{n}_{{s}_{1}}),\end{equation} \tag{ 37 }$$
• real SU(2) singlets with weak hypercharge 0
  $$\begin{equation}{\chi }_{l}^{0}={{\chi }_{l}^{0}}^{\ast }\quad (l=1,\dots ,{n}_{{s}_{0}}),\end{equation} \tag{ 38 }$$
• and SU(2) singlets with weak hypercharge 2
  $$\begin{equation}{\chi }_{r}^{++}\quad (r=1,\dots ,{n}_{{s}_{2}}).\end{equation} \tag{ 39 }$$

There is then a total of

${n}_{2}={n}_{{t}_{1}}+{n}_{{s}_{2}}$ complex scalar fields with electric charge 2,

${n}_{1}={n}_{d}+{n}_{{t}_{1}}+{n}_{{t}_{0}}+{n}_{{s}_{1}}$ complex scalar fields with electric charge 1,

${n}_{0}=2{n}_{d}+2{n}_{{t}_{1}}+{n}_{{t}_{0}}+{n}_{{s}_{0}}$ real scalar fields with electric charge 0.

The neutral fields are allowed to have non-zero VEVs

$$\begin{equation}\left\langle 0\left\vert {\phi }_{k}^{0}\right\vert 0\right\rangle =\frac{{v}_{k}}{\sqrt{2}},\qquad \left\langle 0\left\vert {\xi }_{p}^{0}\right\vert 0\right\rangle =\frac{{w}_{p}}{\sqrt{2}},\qquad \left\langle 0\left\vert {\lambda }_{q}^{0}\right\vert 0\right\rangle ={x}_{q},\qquad \left\langle 0\left\vert {\chi }_{l}^{0}\right\vert 0\right\rangle ={u}_{l},\end{equation} \tag{ 40 }$$

where the v_k and the w_p are in general complex and the x_q and the u_l are real. The masses of the gauge bosons W^± and Z⁰ are given in terms of the VEVs of the scalar fields as

$$\begin{equation}{m}_{Z}^{2}=\frac{{g}^{2}}{4{c}_{W}^{2}}\left({v}^{2}+4{w}^{2}\right),\qquad {m}_{W}^{2}=\frac{{g}^{2}}{4}\left({v}^{2}+2{w}^{2}+4{x}^{2}\right),\end{equation} \tag{ 41 }$$

where we have defined

$$\begin{equation}v=\sqrt{\sum\limits _{k=1}^{{n}_{d}}{\left\vert {v}_{k}\right\vert }^{2}},\qquad w=\sqrt{\sum\limits _{p=1}^{{n}_{{t}_{1}}}{\left\vert {w}_{p}\right\vert }^{2}},\qquad x=\sqrt{\sum\limits _{q=1}^{{n}_{{t}_{0}}}{x}_{q}^{2}}.\end{equation} \tag{ 42 }$$

Comparing the second equation (41) with equation (14d), note that there is an extra factor 1/2 multiplying x. This is because the field ${\lambda }_{q}^{0}$ is real, hence it has a factor 1/2 in its gauge-kinetic term.

We expand the neutral fields around their VEVs as

$$\begin{equation}{\phi }_{k}^{0}=\frac{{v}_{k}+{{\phi }_{k}^{0}}^{\prime }}{\sqrt{2}},\qquad {\xi }_{p}^{0}=\frac{{w}_{p}+{{\xi }_{p}^{0}}^{\prime }}{\sqrt{2}},\qquad {\lambda }_{q}^{0}={x}_{q}+{{\lambda }_{q}^{0}}^{\prime },\qquad {\chi }_{l}^{0}={u}_{l}+{{\chi }_{l}^{0}}^{\prime }.\end{equation} \tag{ 43 }$$

We define the mixing matrices of the scalar fields as

$$\begin{equation}{\xi }_{p}^{++}=\sum\limits _{c=1}^{{n}_{2}}{\left({T}_{2}\right)}_{pc}{S}_{c}^{++},\qquad {\chi }_{r}^{++}=\sum\limits _{c=1}^{{n}_{2}}{\left({T}_{4}\right)}_{rc}{S}_{c}^{++},\end{equation} \tag{ 44 }$$

$$\begin{align}\hfill {\phi }_{k}^{+}& =\sum\limits _{a=1}^{{n}_{1}}{\left({U}_{1}\right)}_{ka}{S}_{a}^{+},\hfill \\ \hfill {\xi }_{p}^{+}& =\sum\limits _{a=1}^{{n}_{1}}{\left({U}_{2}\right)}_{pa}{S}_{a}^{+},\hfill \\ \hfill {\lambda }_{q}^{+}& =\sum\limits _{a=1}^{{n}_{1}}{\left({U}_{3}\right)}_{qa}{S}_{a}^{+},\hfill \\ \hfill {\chi }_{j}^{+}& =\sum\limits _{a=1}^{{n}_{1}}{\left({U}_{4}\right)}_{ja}{S}_{a}^{+},\hfill \end{align} \tag{ 45 }$$

$$\begin{align}\hfill {{\phi }_{k}^{0}}^{\prime }& =\sum\limits _{b=1}^{{n}_{0}}{\left({V}_{1}\right)}_{kb}{S}_{b}^{0},\hfill \\ \hfill {{\xi }_{p}^{0}}^{\prime }& =\sum\limits _{b=1}^{{n}_{0}}{\left({V}_{2}\right)}_{pb}{S}_{b}^{0},\hfill \\ \hfill {{\lambda }_{q}^{0}}^{\prime }& =\sum\limits _{b=1}^{{n}_{0}}{\left({V}_{3}\right)}_{qb}{S}_{b}^{0},\hfill \\ \hfill {{\chi }_{l}^{0}}^{\prime }& =\sum\limits _{b=1}^{{n}_{0}}{\left({V}_{4}\right)}_{lb}{S}_{b}^{0},\hfill \end{align} \tag{ 46 }$$

where the dimensions of the matrices are

$$\begin{equation}{T}_{2}:{n}_{{t}_{1}}\times {n}_{2},\qquad \ {T}_{4}:{n}_{{s}_{2}}\times {n}_{2},\end{equation} \tag{ 47a }$$

$$\begin{equation}{U}_{1}:\ {n}_{d}\times {n}_{1},\qquad {U}_{2}:\ {n}_{{t}_{1}}\times {n}_{1},\quad {U}_{3}:\ {n}_{{t}_{0}}\times {n}_{1},\quad {U}_{4}:\ {n}_{{s}_{1}}\times {n}_{1},\end{equation} \tag{ 47b }$$

$$\begin{equation}{V}_{1}:\ {n}_{d}\times {n}_{0},\qquad {V}_{2}:\ {n}_{{t}_{1}}\times {n}_{0},\quad {V}_{3}:\ {n}_{{t}_{0}}\times {n}_{0},\quad {V}_{4}:\ {n}_{{s}_{0}}\times {n}_{0}.\end{equation} \tag{ 47c }$$

The matrices V₃ and V₄ are real while the other ones are complex. The matrix

$$\begin{equation}\tilde{T}=\left(\begin{matrix}\hfill {T}_{2}\hfill \\ \hfill {T}_{4}\hfill \end{matrix}\right)\end{equation} \tag{ 48 }$$

is n₂ × n₂ unitary; it diagonalizes the Hermitian mass matrix of the scalars with charge 2. The matrix

$$\begin{equation}\tilde{U}=\left(\begin{matrix}\hfill {U}_{1}\hfill \\ \hfill {U}_{2}\hfill \\ \hfill {U}_{3}\hfill \\ \hfill {U}_{4}\hfill \end{matrix}\right)\end{equation} \tag{ 49 }$$

is n₁ × n₁ unitary; it diagonalizes the Hermitian mass matrix of the scalars with charge 1. The matrix

$$\begin{equation}\tilde{V}=\left(\begin{matrix}\hfill \mathrm{R}\mathrm{e}\,{V}_{1}\hfill \\ \hfill \mathrm{I}\mathrm{m}\,{V}_{1}\hfill \\ \hfill \mathrm{R}\mathrm{e}\,{V}_{2}\hfill \\ \hfill \mathrm{I}\mathrm{m}\,{V}_{2}\hfill \\ \hfill {V}_{3}\hfill \\ \hfill {V}_{4}\hfill \end{matrix}\right)\end{equation} \tag{ 50 }$$

is n₀ × n₀ real orthogonal; it diagonalizes the symmetric mass matrix of the real components of the neutral scalar fields.

One may obtain the form of the Goldstone bosons by applying to the vacuum state the generators of the gauge group that are spontaneously broken. Using this method, we obtain

$$\begin{align}\hfill {\left({U}_{1}\right)}_{k1}& =\frac{{v}_{k}}{\sqrt{{v}^{2}+2{w}^{2}+4{x}^{2}}},\ \hfill \\ \hfill {\left({U}_{2}\right)}_{p1}& =\frac{\sqrt{2}{w}_{p}}{\sqrt{{v}^{2}+2{w}^{2}+4{x}^{2}}},\hfill \\ \hfill {\left({U}_{3}\right)}_{q1}& =\frac{2{x}_{q}}{\sqrt{{v}^{2}+2{w}^{2}+4{x}^{2}}},\hfill \end{align} \tag{ 51 }$$

$$\begin{equation}{\left({V}_{1}\right)}_{k1}=\frac{\mathrm{i}{v}_{k}}{\sqrt{{v}^{2}+4{w}^{2}}},\qquad {\left({V}_{2}\right)}_{p1}=\frac{2\mathrm{i}{w}_{p}}{\sqrt{{v}^{2}+4{w}^{2}}},\end{equation} \tag{ 52 }$$

while the first columns of U₄, V₃, and V₄ are identically zero.

5.2. The formula for S

There are two kinds of diagrams that produce the parameter S: the ones like in figure 1 and those like in figure 2. Diagrams of the type of figure 1 produce a function $K\left(I,J\right)$, where I and J are the squared-masses of the two scalars in the loop. Diagrams of the type of figure 2 produce a function $K\left(I,J\right)-6J\tilde{K}\left(I,J\right)$, where I is the squared-mass of the scalar and J is the squared-mass of the gauge boson in the loop. The functions K and $\tilde{K}$ are defined in equation (67) of appendix A.

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Let M_c denote the mass of the scalar ${S}_{c}^{++}$, m_a denote the mass of the scalar ${S}_{a}^{+}$, and μ_b denote the mass of ${S}_{b}^{0}$. Using the Feynman diagrams from appendix E and the prescription described in section 2, we have obtained for the oblique parameter S the following result:

$$\begin{align}\hfill S& =\frac{1}{12\pi }\left\{4\sum\limits _{c,{c}^{\prime }=1}^{{n}_{2}}{\left\vert 2{s}_{W}^{2}{\delta }_{c{c}^{\prime }}-{\left({T}_{2}^{{\dagger}}{T}_{2}\right)}_{c{c}^{\prime }}\right\vert }^{2}\left[K\left({M}_{c}^{2},{M}_{{c}^{\prime }}^{2}\right)+\frac{\mathrm{ln}\,{M}_{c}^{2}+\mathrm{ln}\,{M}_{{c}^{\prime }}^{2}}{2}\right]\right.\hfill \\ \hfill & \quad +8\sum\limits _{c=1}^{{n}_{2}}\left\{\left({c}_{W}^{2}-{s}_{W}^{2}\right)\left[2{s}_{W}^{2}-{\left({T}_{2}^{{\dagger}}{T}_{2}\right)}_{cc}\right]-2{c}_{W}^{2}{s}_{W}^{2}\right\}\mathrm{ln}\,{M}_{c}^{2}\hfill \\ \hfill & \quad +\sum\limits _{a,{a}^{\prime }=2}^{{n}_{1}}{\left\vert 2{s}_{W}^{2}{\delta }_{a{a}^{\prime }}-{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{a{a}^{\prime }}-2{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{a{a}^{\prime }}\right\vert }^{2}\left[K\left({m}_{a}^{2},{m}_{{a}^{\prime }}^{2}\right)+\frac{\mathrm{ln}\,{m}_{a}^{2}+\mathrm{ln}\,{m}_{{a}^{\prime }}^{2}}{2}\right]\hfill \\ \hfill & \quad +2\sum\limits _{a=2}^{{n}_{1}}\left\{\left({c}_{W}^{2}-{s}_{W}^{2}\right)\left[2{s}_{W}^{2}-{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{aa}-2{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{aa}\right]-2{c}_{W}^{2}{s}_{W}^{2}\right\}\mathrm{ln}\,{m}_{a}^{2}\hfill \\ \hfill & \quad +\sum\limits _{b=2}^{{n}_{0}-1}\sum\limits _{{b}^{\prime }=b+1}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{b{b}^{\prime }}+2\,\mathrm{I}\mathrm{m}{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{b{b}^{\prime }}\right]}^{2}\left[K\left({\mu }_{b}^{2},{\mu }_{{b}^{\prime }}^{2}\right)+\frac{\mathrm{ln}\,{\mu }_{b}^{2}+\mathrm{ln}\,{\mu }_{{b}^{\prime }}^{2}}{2}\right]\hfill \\ \hfill & \quad +2\,\sum\limits _{a=2}^{{n}_{1}}{\left\vert {\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{a1}-{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{a1}\right\vert }^{2}\left[K\left({m}_{a}^{2},{m}_{W}^{2}\right)+\frac{\mathrm{ln}\,{m}_{a}^{2}+\mathrm{ln}\,{m}_{W}^{2}}{2}\right.\hfill \\ \hfill & \quad -\left.6{m}_{W}^{2}\tilde{K}\left({m}_{a}^{2},{m}_{W}^{2}\right)\right]\hfill \\ \hfill & \quad +\sum\limits _{b=2}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{1b}+2\,\mathrm{I}\mathrm{m}{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{1b}\right]}^{2}\left[K\left({\mu }_{b}^{2},{m}_{Z}^{2}\right)+\frac{\mathrm{ln}\,{\mu }_{b}^{2}+\mathrm{ln}\,{m}_{Z}^{2}}{2}\right.\hfill \\ \hfill & \quad -\left.6{m}_{Z}^{2}\tilde{K}\left({\mu }_{b}^{2},{m}_{Z}^{2}\right)\right]\hfill \\ \hfill & \quad -\left.\left(\frac{2}{\rho }-\frac{1}{{\rho }^{2}}\right)\left[K\left({m}_{h}^{2},{m}_{Z}^{2}\right)+\frac{\mathrm{ln}\,{m}_{h}^{2}+\mathrm{ln}\,{m}_{Z}^{2}}{2}-6{m}_{Z}^{2}\tilde{K}\left({m}_{h}^{2},{m}_{Z}^{2}\right)\right]\right\}.\hfill \end{align} \tag{ 53 }$$

One may use the unitarity of the matrices $\tilde{T}$ and $\tilde{U}$ and the orthogonality of $\tilde{V}$ to write the following relations:

$$\begin{align}\hfill & \sum\limits _{a,{a}^{\prime }=2}^{{n}_{1}}{\left\vert 2{s}_{W}^{2}{\delta }_{a{a}^{\prime }}-{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{a{a}^{\prime }}-2{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{a{a}^{\prime }}\right\vert }^{2}\,\mathrm{ln}\,{m}_{a}^{2}\hfill \\ \hfill & \qquad =\sum\limits _{a=2}^{{n}_{1}}\left[4{s}_{W}^{4}+\left(1-4{s}_{W}^{2}\right){\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{aa}+4\left(1-2{s}_{W}^{2}\right){\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{aa}\right]\mathrm{ln}\,{m}_{a}^{2}\hfill \\ \hfill & \qquad \quad -\sum\limits _{a=2}^{{n}_{1}}{\left\vert {\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{a1}-{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{a1}\right\vert }^{2}\mathrm{ln}\,{m}_{a}^{2},\hfill \end{align} \tag{ 54 }$$

$$\begin{align}\hfill & \frac{1}{2}\,\sum\limits _{b=2}^{{n}_{0}-1}\sum\limits _{{b}^{\prime }=b+1}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{b{b}^{\prime }}+2\,\mathrm{I}\mathrm{m}{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{b{b}^{\prime }}\right]}^{2}\left(\mathrm{ln}\,{\mu }_{b}^{2}+\mathrm{ln}\,{\mu }_{{b}^{\prime }}^{2}\right)\hfill \\ \hfill & \qquad =\frac{1}{2}\sum\limits _{b=2}^{{n}_{0}}\left[{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{bb}+4{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{bb}\right]\mathrm{ln}\,{\mu }_{b}^{2}\hfill \\ \hfill & \qquad \quad -\frac{1}{2}\,\sum\limits _{b=2}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{1b}+2\,\mathrm{I}\mathrm{m}{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{1b}\right]}^{2}\mathrm{ln}\,{\mu }_{b}^{2},\hfill \end{align} \tag{ 55 }$$

$$\begin{equation}\sum\limits _{b=2}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{1b}+2\,\mathrm{I}\mathrm{m}{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{1b}\right]}^{2}={\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{11}+4{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{11},\end{equation} \tag{ 56 }$$

$$\begin{equation}\sum\limits _{a=2}^{{n}_{1}}{\left\vert {\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{a1}-{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{a1}\right\vert }^{2}={\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{11}+{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{11}-{\left(1-\frac{1}{\rho }\right)}^{2},\end{equation} \tag{ 57 }$$

$$\begin{align}\hfill \sum\limits _{c,{c}^{\prime }=1}^{{n}_{2}}{\left\vert 2{s}_{W}^{2}{\delta }_{c{c}^{\prime }}-{\left({T}_{2}^{{\dagger}}{T}_{2}\right)}_{c{c}^{\prime }}\right\vert }^{2}\,\mathrm{ln}\,{M}_{c}^{2}& =\sum\limits _{c=1}^{{n}_{2}}\left[4{s}_{W}^{4}+\left(1-4{s}_{W}^{2}\right){\left({T}_{2}^{{\dagger}}{T}_{2}\right)}_{cc}\right]\hfill \\ \hfill & \quad \times \mathrm{ln}\,{M}_{c}^{2}.\hfill \end{align} \tag{ 58 }$$

We use equations (54)–(58) to simplify equation (53), obtaining

$$\begin{equation}S=\frac{1}{12\pi }\left\{4\,\sum\limits _{c,{c}^{\prime }=1}^{{n}_{2}}{\left\vert 2{s}_{W}^{2}{\delta }_{c{c}^{\prime }}-{\left({T}_{2}^{{\dagger}}{T}_{2}\right)}_{c{c}^{\prime }}\right\vert }^{2}K\left({M}_{c}^{2},{M}_{{c}^{\prime }}^{2}\right)\right.\end{equation} \tag{ 59a }$$

$$\begin{equation}\quad +\sum\limits _{a,{a}^{\prime }=2}^{{n}_{1}}{\left\vert 2{s}_{W}^{2}{\delta }_{a{a}^{\prime }}-{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{a{a}^{\prime }}-2{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{a{a}^{\prime }}\right\vert }^{2}K\left({m}_{a}^{2},{m}_{{a}^{\prime }}^{2}\right)\end{equation} \tag{ 59b }$$

$$\begin{equation}\quad +\sum\limits _{b=2}^{{n}_{0}-1}\sum\limits _{{b}^{\prime }=b+1}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{b{b}^{\prime }}+2\,\mathrm{I}\mathrm{m}{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{b{b}^{\prime }}\right]}^{2}K\left({\mu }_{b}^{2},\,{\mu }_{{b}^{\prime }}^{2}\right)\end{equation} \tag{ 59c }$$

$$\begin{equation}\quad +2\,\sum\limits _{a=2}^{{n}_{1}}{\left\vert {\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{a1}-{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{a1}\right\vert }^{2}\left[K\left({m}_{a}^{2},\,{m}_{W}^{2}\right)-6{m}_{W}^{2}\tilde{K}\left({m}_{a}^{2},{m}_{W}^{2}\right)\right]\end{equation} \tag{ 59d }$$

$$\begin{equation}\quad +\sum\limits _{b=2}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{1b}+2\,\mathrm{I}\mathrm{m}{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{1b}\right]}^{2}\left[K\left({\mu }_{b}^{2},\,{m}_{Z}^{2}\right)-6{m}_{Z}^{2}\tilde{K}\left({\mu }_{b}^{2},{m}_{Z}^{2}\right)\right]\end{equation} \tag{ 59e }$$

$$\begin{equation}\quad -\left(\frac{2}{\rho }-\frac{1}{{\rho }^{2}}\right)\left[K\left({m}_{h}^{2},{m}_{Z}^{2}\right)-6{m}_{Z}^{2}\tilde{K}\left({m}_{h}^{2},{m}_{Z}^{2}\right)\right]\end{equation} \tag{ 59f }$$

$$\begin{equation}\quad -4\,\sum\limits _{c=1}^{{n}_{2}}{\left({T}_{2}^{{\dagger}}{T}_{2}\right)}_{cc}\,\mathrm{ln}\,{M}_{c}^{2}-\sum\limits _{a=2}^{{n}_{1}}{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{aa}\,\mathrm{ln}\,{m}_{a}^{2}\end{equation} \tag{ 59g }$$

$$\begin{equation}\quad +\sum\limits _{b=2}^{{n}_{0}}\left[{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{bb}+4{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{bb}\right]\frac{\mathrm{ln}\,{\mu }_{b}^{2}}{2}\end{equation} \tag{ 59h }$$

$$\begin{equation}\quad +\left[{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{11}+4{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{11}\right]\frac{\mathrm{ln}\,{m}_{Z}^{2}}{2}\end{equation} \tag{ 59i }$$

$$\begin{equation}\quad +\left[{\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{11}+{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{11}-{\left(1-\frac{1}{\rho }\right)}^{2}\right]\mathrm{ln}\,{m}_{W}^{2}\end{equation} \tag{ 59j }$$

$$\begin{equation}\quad -\left.\left(\frac{2}{\rho }-\frac{1}{{\rho }^{2}}\right)\frac{\mathrm{ln}\,{m}_{h}^{2}+\mathrm{ln}\,{m}_{Z}^{2}}{2}\right\}.\end{equation} \tag{ 59k }$$

The coefficients of the logarithms in lines (59g)–(59k) add up to zero as they should:

$$\begin{equation}-4\,\sum\limits _{c=1}^{{n}_{2}}{\left({T}_{2}^{{\dagger}}{T}_{2}\right)}_{cc}-\sum\limits _{a=2}^{{n}_{1}}{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{aa}+\frac{1}{2}\ \sum\limits _{b=1}^{{n}_{0}}\left[{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{bb}+4{\left({V}_{2}^{{\dagger}}{V}_{2}\right)}_{bb}\right]\end{equation} \tag{ 60a }$$

$$\begin{equation}\quad +{\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{11}+{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{11}-{\left(1-\frac{1}{\rho }\right)}^{2}-\left(\frac{2}{\rho }-\frac{1}{{\rho }^{2}}\right)\end{equation} \tag{ 60b }$$

$$\begin{align}\hfill & \quad =-4\,\mathrm{t}\mathrm{r}\left({T}_{2}{T}_{2}^{{\dagger}}\right)-\mathrm{t}\mathrm{r}\left({U}_{1}{U}_{1}^{{\dagger}}\right)+{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{11}+\frac{1}{2}\,\mathrm{t}\mathrm{r}\left({V}_{1}{V}_{1}^{{\dagger}}\right)+2\,\mathrm{t}\mathrm{r}\left({V}_{2}{V}_{2}^{{\dagger}}\right)\hfill \\ \hfill & \qquad +{\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{11}+{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{11}-1\hfill \end{align} \tag{ 60c }$$

$$\begin{equation}\quad =-4{n}_{{t}_{1}}-{n}_{d}+{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{11}+{n}_{d}+4{n}_{{t}_{1}}+{\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{11}+{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{11}-1\end{equation} \tag{ 60d }$$

$$\begin{equation}\quad ={\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{11}+{\left({U}_{2}^{{\dagger}}{U}_{2}\right)}_{11}+{\left({U}_{3}^{{\dagger}}{U}_{3}\right)}_{11}-1\end{equation} \tag{ 60e }$$

$$\begin{equation}\quad =0,\end{equation} \tag{ 60f }$$

where in the last step we have used equation (51).

Equation (59) for S generalizes the result given in reference [24] for the case without any scalar SU(2) triplets and without charge-2 singlets. In that case the matrices T₂, T₄, U₂, U₃, V₂, and V₃ and zero. Reference [24] then gave

$$\begin{equation}S=\frac{1}{24\pi }\left\{\sum\limits _{a=2}^{{n}_{1}}{\left[2{s}_{W}^{2}-{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{aa}\right]}^{2}G\left({m}_{a}^{2},{m}_{a}^{2},{m}_{Z}^{2}\right)\right.\end{equation} \tag{ 61a }$$

$$\begin{equation}\quad +2\,\sum\limits _{a=2}^{{n}_{1}-1}\sum\limits _{{a}^{\prime }=a+1}^{{n}_{1}}{\left\vert {\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{a{a}^{\prime }}\right\vert }^{2}G\left({m}_{a}^{2},{m}_{{a}^{\prime }}^{2},{m}_{Z}^{2}\right)\end{equation} \tag{ 61b }$$

$$\begin{equation}\quad +\sum\limits _{b=2}^{{n}_{0}-1}\sum\limits _{{b}^{\prime }=b+1}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{b{b}^{\prime }}\right]}^{2}G\left({\mu }_{b}^{2},{\mu }_{{b}^{\prime }}^{2},{m}_{Z}^{2}\right)\end{equation} \tag{ 61c }$$

$$\begin{equation}\quad +\sum\limits _{b=2}^{{n}_{0}}{\left[\mathrm{I}\mathrm{m}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{1b}\right]}^{2}\hat{G}\left({\mu }_{b}^{2},{m}_{Z}^{2}\right)-\hat{G}\left({m}_{h}^{2},{m}_{Z}^{2}\right)\end{equation} \tag{ 61d }$$

$$\begin{equation}\quad -\left.\mathrm{ln}\,{m}_{h}^{2}-2\,\sum\limits _{a=2}^{{n}_{1}}{\left({U}_{1}^{{\dagger}}{U}_{1}\right)}_{aa}\,\mathrm{ln}\,{m}_{a}^{2}+\sum\limits _{b=2}^{{n}_{0}}{\left({V}_{1}^{{\dagger}}{V}_{1}\right)}_{bb}\,\mathrm{ln}\,{\mu }_{b}^{2}\right\}.\end{equation} \tag{ 61e }$$

It must be stressed that reference [24] used a different definition of S [33], wherein

$$\begin{equation}{\left.\frac{\partial \,\delta {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}\end{equation} \tag{ 62 }$$

in equation (4c) was substituted by

$$\begin{equation}\frac{\delta {A}_{ZZ}\left({m}_{Z}^{2}\right)-\delta {A}_{ZZ}\left(0\right)}{{m}_{Z}^{2}}.\end{equation} \tag{ 63 }$$

This caused the appearance of the functions $G\left(I,J,{m}_{Z}^{2}\right)$ and $\hat{G}\left(J,{m}_{Z}^{2}\right)$ in equation (61). In order to make the connection with the present work, one must use

$$\begin{equation}G\left(I,J,{m}_{Z}^{2}\right)=2K\left(I,J\right)+O\left({m}_{Z}^{2}\right),\end{equation} \tag{ 64a }$$

$$\begin{equation}\hat{G}\left(J,{m}_{Z}^{2}\right)=2\left[K\left(J,{m}_{Z}^{2}\right)-6{m}_{Z}^{2}\tilde{K}\left(J,{m}_{Z}^{2}\right)\right]+O\left({m}_{Z}^{2}\right).\end{equation} \tag{ 64b }$$

In this paper we have discovered a problem that arises in the computation of the oblique parameters S and U whenever the model of NP at hand does not obey relation (1), because it contains scalars with nonzero VEVs in multiplets of SU(2) larger than doublets. Namely, the Feynman rules for some Goldstone-boson vertices depend on ρ—see equation (26)—and this causes a mismatch between the NP model, wherein ρ ≠ 1, and the SM, wherein ρ = 1. Depending on how one (mis)handles this mismatch, S and U may turn out either gauge-dependent or divergent. We have proceeded to suggest a solution for this problem through a redefinition of the SM Feynman rules for some vertices containing the Higgs boson H. Namely, we have inserted by hand into those Feynman rules factors $\sqrt{\aleph }$ and $\sqrt{\beth }$—see equation (27)—that are equal to 1 when relation (1) holds. One easily finds out that, if ℵ and ℶ take the values in equations (30) and (33), respectively, then the problem is solved.

We must emphasize that the presence of the extra factors ℵ and ℶ in the Feynman rules of the SM does not alter any precision calculation made in the context of the SM, because in the SM ρ = 1 and when this relation is verified ℵ = ℶ = 1, such that the Feynman rules of the SM remain exactly the same. Our prescription may only be relevant when one is working in a NP model wherein ρ ≠ 1 and one wants to compare some loop result within the NP model with the SM result for the same quantity, viz to subtract one from the other.

An example of quantities that one might try to compute in a NP model by using our prescription is the couplings ${g}_{b}^{L}$ and ${g}_{b}^{R}$ that parametrize the $Zb\bar{b}$ vertex [34]. In order to compute those two quantities—that do not vanish at the tree level—at the one-loop level, either in the SM or in a NP model, one needs to fully renormalize the theory [25]. This involves the computation of very many Feynman diagrams and is clearly beyond the scope of the present paper; we consider carrying out this calculation in specific NP models in subsequent paper(s).

We have utilized our insight to explicitly compute the value of S in a NP model containing arbitrary numbers of triplets of SU(2) (and also extra doublets and singlets) with weak hypercharges either 0 or 1.

In the future, it may be interesting to study whether the changes to the Feynman rules that we have suggested may somehow be justified, whether they cure the problems in the computation of S and U also at the two-loop level, and whether they may cure other possible problems elsewhere.

This work was supported by the Portuguese Foundation for Science and Technology through projects CERN/FIS-PAR/0004/2019, CERN/FIS-PAR/0008/2019, PTDC/FIS-PAR/29436/2017, UIDB/00777/2020, and UIDP/00777/2020. FA acknowledges a fellowship from Project CERN/FIS-PAR/0008/2019.

No new data were created or analysed in this study.

We define the function

$$\begin{equation}D\left(Q,I,J,x\right)=Qx\left(x-1\right)+I\left(1-x\right)+Jx.\end{equation} \tag{ 65 }$$

We have used the following results:

$$\begin{align}\hfill & {\int }_{0}^{1}\mathrm{d}x{\left.\frac{\partial }{\partial Q}\left\{\frac{4}{d}{\mu }^{4-d}\int \frac{{\mathrm{d}}^{d}k}{{\left(2\pi \right)}^{d}}\frac{{k}^{2}}{{\left[{k}^{2}-D\left(Q,I,J,x\right)+\mathrm{i}{\epsilon}\right]}^{2}}\right\}\right\vert }_{Q=0}\hfill \\ \hfill & \qquad =\mathrm{i}\left[\text{div}+\frac{K\left(I,J\right)}{48{\pi }^{2}}\right.\left.\,+\frac{\mathrm{ln}\,I+\mathrm{ln}\,J}{96{\pi }^{2}}\right],\hfill \end{align} \tag{ 66a }$$

$$\begin{align}\hfill & {\int }_{0}^{1}\mathrm{d}x{\left.\frac{\partial }{\partial Q}\left\{{\mu }^{4-d}\int \frac{{\mathrm{d}}^{d}k}{{\left(2\pi \right)}^{d}}\ \frac{1}{{\left[{k}^{2}-D\left(Q,I,J,x\right)+\mathrm{i}{\epsilon}\right]}^{2}}\right\}\right\vert }_{Q=0}\hfill \\ \hfill & \qquad =\frac{\mathrm{i}}{32{\pi }^{2}}\ \tilde{K}\left(I,J\right),\hfill \end{align} \tag{ 66b }$$

where

$$\begin{equation}K\left(I,J\right)=\left\{\begin{array}{lcl}-\frac{5}{6}+\frac{2IJ}{{\left(I-J\right)}^{2}}+\frac{{I}^{3}+{J}^{3}-3IJ\left(I+J\right)}{2{\left(I-J\right)}^{3}}\mathrm{ln}\frac{I}{J}\quad \hfill & \hfill {\Leftarrow}\hfill & I\ne J,\hfill \\ 0\quad \hfill & \hfill {\Leftarrow}\hfill & I=J,\hfill \end{array}\right.\end{equation} \tag{ 67a }$$

$$\begin{equation}\tilde{K}\left(I,J\right)=\left\{\begin{array}{lcl}\frac{I+J}{{\left(I-J\right)}^{2}}-\frac{2IJ}{{\left(I-J\right)}^{3}}\mathrm{ln}\frac{I}{J}\quad \hfill & \hfill {\Leftarrow}\hfill & I\ne J,\hfill \\ \frac{1}{3I}\quad \hfill & \hfill {\Leftarrow}\hfill & I=J,\hfill \end{array}\right.\end{equation} \tag{ 67b }$$

and

$$\begin{equation}\mathrm{d}\mathrm{i}\mathrm{v}=\frac{-1}{48{\pi }^{2}}\left[\frac{2}{4-d}-\gamma +\mathrm{ln}\left(4\pi {\mu }^{2}\right)\right].\end{equation} \tag{ 68 }$$

In equations (66) and (68), μ is an arbitrary mass scale. In equation (68), γ is Euler's constant. The quantity named 'div' diverges in the limit d → 4.

In this appendix we derive some consequences of equation (7).

The interaction terms of the photon with the charge-Q scalars (Q > 0) are given by

$$\begin{align}\hfill & {\mathcal{L}}_{A{S}^{Q}{S}^{-Q}}\hfill \\ \hfill & \qquad =\mathrm{i}eQ{A}^{\mu }\left\{\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}\left[{M}_{JY}^{Q}\,{\partial }_{\mu }{\left({M}_{JY}^{Q}\right)}^{\ast }-{\left({M}_{JY}^{Q}\right)}^{\ast }{\partial }_{\mu }{M}_{JY}^{Q}\right]\right.\hfill \end{align} \tag{ 69a }$$

$$\begin{equation}\qquad \quad +\left.\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}\left[{\left({M}_{JY}^{-Q}\right)}^{\ast }{\partial }_{\mu }{M}_{JY}^{-Q}-{M}_{JY}^{-Q}\,{\partial }_{\mu }{\left({M}_{JY}^{-Q}\right)}^{\ast }\right]\right\}\end{equation} \tag{ 69b }$$

$$\begin{equation}\qquad =\mathrm{i}eQ{A}^{\mu }\sum\limits _{a,{a}^{\prime }=1}^{{n}_{Q}}\left({S}_{a}^{Q}\,{\partial }_{\mu }{S}_{{a}^{\prime }}^{-Q}-{S}_{{a}^{\prime }}^{-Q}\,{\partial }_{\mu }{S}_{a}^{Q}\right)\end{equation} \tag{ 69c }$$

$$\begin{equation}\qquad \quad \times \left[\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}{\left({R}_{JY}^{Q}\right)}_{1a}{\left({R}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }+\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}{\left({S}_{JY}^{Q}\right)}_{1a}{\left({S}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }\right]\end{equation} \tag{ 69d }$$

$$\begin{equation}\qquad =\mathrm{i}eQ{A}^{\mu }\sum\limits _{a=1}^{{n}_{Q}}\left({S}_{a}^{Q}\,{\partial }_{\mu }{S}_{a}^{-Q}-{S}_{a}^{-Q}\,{\partial }_{\mu }{S}_{a}^{Q}\right),\end{equation} \tag{ 69e }$$

where we have used equation (10a).

The interaction terms of the Z with two charged scalars are given by

$$\begin{align}\hfill {\mathcal{L}}_{Z{S}^{Q}{S}^{-Q}}=& \mathrm{i}\frac{g}{{c}_{W}}{Z}^{\mu }\left\{\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}\left(Y-Q{c}_{W}^{2}\right)\left[{M}_{JY}^{Q}{\partial }_{\mu }{\left({M}_{JY}^{Q}\right)}^{\ast }\!-{\left({M}_{JY}^{Q}\right)}^{\ast }{\partial }_{\mu }{M}_{JY}^{Q}\right]\right.\hfill \end{align} \tag{ 70a }$$

$$\begin{equation}\left.\quad -\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}\left(Y+Q{c}_{W}^{2}\right)\left[{\left({M}_{JY}^{-Q}\right)}^{\ast }{\partial }_{\mu }{M}_{JY}^{-Q}-{M}_{JY}^{-Q}\,{\partial }_{\mu }{\left({M}_{JY}^{-Q}\right)}^{\ast }\right]\right\}\end{equation} \tag{ 70b }$$

$$\begin{equation}\quad =-\mathrm{i}g{c}_{W}Q{Z}^{\mu }\sum\limits _{a=1}^{{n}_{Q}}\left({S}_{a}^{Q}\,{\partial }_{\mu }{S}_{a}^{-Q}-{S}_{a}^{-Q}{\partial }_{\mu }{S}_{a}^{Q}\right)\end{equation} \tag{ 70c }$$

$$\begin{equation}\qquad +\,\mathrm{i}\frac{g}{{c}_{W}}{Z}^{\mu }\sum\limits _{a,{a}^{\prime }=1}^{{n}_{Q}}\left({S}_{a}^{Q}\,{\partial }_{\mu }{S}_{{a}^{\prime }}^{-Q}-{S}_{{a}^{\prime }}^{-Q}{\partial }_{\mu }{S}_{a}^{Q}\right)\end{equation} \tag{ 70d }$$

$$\begin{equation}\qquad \times \left[\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}Y{\left({R}_{JY}^{Q}\right)}_{1a}{\left({R}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }-\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}Y{\left({S}_{JY}^{Q}\right)}_{1a}\!{\left({S}_{JY}^{Q}\right)}_{1{a}^{\prime }}^{\ast }\right].\end{equation} \tag{ 70e }$$

The interaction terms of the Z with two neutral scalars are given by

$$\begin{equation}{\mathcal{L}}_{Z{S}^{0}{S}^{0}}=\mathrm{i}\frac{g}{{c}_{W}}{Z}^{\mu }\!\!\sum\limits _{{M}_{JY}\in \mathcal{A}}Y\left[{M}_{JY}^{0}{\partial }_{\mu }{\left({M}_{JY}^{0}\right)}^{\ast }-{\left({M}_{JY}^{0}\right)}^{\ast }{\partial }_{\mu }{M}_{JY}^{0}\right]\end{equation} \tag{ 71a }$$

$$\begin{equation}=\mathrm{i}\frac{g}{2{c}_{W}}{Z}^{\mu }\sum\limits _{b,{b}^{\prime }=1}^{{n}_{0}}\left({S}_{b}^{0}{\partial }_{\mu }{S}_{{b}^{\prime }}^{0}-{S}_{{b}^{\prime }}^{0}{\partial }_{\mu }{S}_{b}^{0}\right)\end{equation} \tag{ 71b }$$

$$\begin{equation}\quad \times \sum\limits _{{M}_{JY}\in \mathcal{A}}Y\left[{\left({A}_{JY}\right)}_{1b}+\mathrm{i}{\left({B}_{JY}\right)}_{1b}\right]\left[{\left({A}_{JY}\right)}_{1{b}^{\prime }}-\mathrm{i}{\left({B}_{JY}\right)}_{1{b}^{\prime }}\right]\end{equation} \tag{ 71c }$$

$$\begin{equation}=\frac{g}{{c}_{W}}{Z}^{\mu }\sum\limits _{b=1}^{{n}_{0}-1}\sum\limits _{{b}^{\prime }=b+1}^{{n}_{0}}\left({S}_{b}^{0}{\partial }_{\mu }{S}_{{b}^{\prime }}^{0}-{S}_{{b}^{\prime }}^{0}{\partial }_{\mu }{S}_{b}^{0}\right)\end{equation} \tag{ 71d }$$

$$\begin{equation}\quad \times \sum\limits _{{M}_{JY}\in \mathcal{A}}Y\left[{\left({A}_{JY}\right)}_{1b}{\left({B}_{JY}\right)}_{1{b}^{\prime }}-{\left({A}_{JY}\right)}_{1{b}^{\prime }}{\left({B}_{JY}\right)}_{1b}\right].\end{equation} \tag{ 71e }$$

The interaction terms of the W with one singly-charged scalar and one neutral scalar are given by

$$\begin{equation}{\mathcal{L}}_{{W}^{\pm }{S}^{\mp }{S}^{0}}=\mathrm{i}g\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\end{equation} \tag{ 72a }$$

$$\begin{equation}\quad \times \left\{{W}^{\mu +}\left[{\left({M}_{JY}^{0}\right)}^{\ast }\,{\partial }_{\mu }{M}_{JY}^{-1}-{M}_{JY}^{-1}\,{\partial }_{\mu }{\left({M}_{JY}^{0}\right)}^{\ast }\right]\right.\end{equation} \tag{ 72b }$$

$$\begin{equation}\quad +\left.{W}^{\mu -}\left[{\left({M}_{JY}^{-1}\right)}^{\ast }\,{\partial }_{\mu }{M}_{JY}^{0}-{M}_{JY}^{0}\,{\partial }_{\mu }{\left({M}_{JY}^{-1}\right)}^{\ast }\right]\right\}\end{equation} \tag{ 72c }$$

$$\begin{equation}\quad +\mathrm{i}g\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\end{equation} \tag{ 72d }$$

$$\begin{equation}\quad \times \left\{{W}^{\mu +}\left[{\left({M}_{JY}^{1}\right)}^{\ast }{\partial }_{\mu }{M}_{JY}^{0}-{M}_{JY}^{0}\,{\partial }_{\mu }{\left({M}_{JY}^{1}\right)}^{\ast }\right]\right.\end{equation} \tag{ 72e }$$

$$\begin{equation}\quad +\left.{W}^{\mu -}\left[{\left({M}_{JY}^{0}\right)}^{\ast }{\partial }_{\mu }{M}_{JY}^{1}-{M}_{JY}^{1}\,{\partial }_{\mu }{\left({M}_{JY}^{0}\right)}^{\ast }\right]\right\}\end{equation} \tag{ 72f }$$

$$\begin{equation}=\mathrm{i}\frac{g}{2}\sum\limits _{{M}_{JY}\in \mathcal{A}}\,\sum\limits _{a=1}^{{n}_{1}}\,\sum\limits _{b=1}^{{n}_{0}}\,\sqrt{\left(J+Y+1\right)\left(J-Y\right)}\end{equation} \tag{ 72g }$$

$$\begin{equation}\quad \times \left[{W}^{\mu +}{\left({A}_{JY}-\mathrm{i}{B}_{JY}\right)}_{1b}{\left({S}_{JY}^{1}\right)}_{1a}^{\ast }\left({S}_{b}^{0}\,{\partial }_{\mu }{S}_{a}^{-}-{S}_{a}^{-}\,{\partial }_{\mu }{S}_{b}^{0}\right)\right.\end{equation} \tag{ 72h }$$

$$\begin{equation}\quad +\left.{W}^{\mu -}{\left({A}_{JY}+\mathrm{i}{B}_{JY}\right)}_{1b}{\left({S}_{JY}^{1}\right)}_{1a}\left({S}_{a}^{+}{\partial }_{\mu }{S}_{b}^{0}-{S}_{b}^{0}{\partial }_{\mu }{S}_{a}^{+}\right)\right]\end{equation} \tag{ 72i }$$

$$\begin{equation}\quad +\mathrm{i}\frac{g}{2}\sum\limits _{{M}_{JY}\in \mathcal{A}}\sum\limits _{a=1}^{{n}_{1}}\sum\limits _{b=1}^{{n}_{0}}\,\sqrt{\left(J+Y\right)\left(J-Y+1\right)}\end{equation} \tag{ 72j }$$

$$\begin{equation}\quad \times \left[{W}^{\mu +}{\left({A}_{JY}+\mathrm{i}{B}_{JY}\right)}_{1b}{\left({R}_{JY}^{1}\right)}_{1a}^{\ast }\left({S}_{a}^{-}{\partial }_{\mu }{S}_{b}^{0}-{S}_{b}^{0}{\partial }_{\mu }{S}_{a}^{-}\right)\right.\end{equation} \tag{ 72k }$$

$$\begin{equation}\quad +\left.{W}^{\mu -}{\left({A}_{JY}-\mathrm{i}{B}_{JY}\right)}_{1b}{\left({R}_{JY}^{1}\right)}_{1a}\left({S}_{b}^{0}{\partial }_{\mu }{S}_{a}^{+}-{S}_{a}^{+}{\partial }_{\mu }{S}_{b}^{0}\right)\right],\end{equation} \tag{ 72l }$$

where we have used equation (15).

The interaction terms of the W with two charged scalars are given by

$$\begin{equation}\mathcal{L}=\cdots +\mathrm{i}\frac{g}{\sqrt{2}}\,\sum\limits _{a=1}^{{n}_{Q}}\sum\limits _{{a}^{\prime }=1}^{{n}_{Q+1}}\,\left\{{W}_{\mu }^{+}\left({S}_{a}^{Q}{\partial }^{\mu }{S}_{{a}^{\prime }}^{-Q-1}-{S}_{{a}^{\prime }}^{-Q-1}{\partial }^{\mu }{S}_{a}^{Q}\right)\right.\end{equation} \tag{ 73a }$$

$$\begin{equation}\quad \times \left[\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}\cap {\mathcal{S}}_{Q+1}}\!\sqrt{\left(J+Y+Q+1\right)\left(J-Y-Q\right)}{\left({S}_{JY}^{Q}\right)}_{1a}{\left({S}_{JY}^{Q+1}\right)}_{1{a}^{\prime }}^{\ast }\right.\end{equation} \tag{ 73b }$$

$$\begin{equation}\left.\quad -\!\!\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}\cap {\mathcal{R}}_{Q+1}}\!\sqrt{\left(J+Y-Q\right)\left(J-Y+Q+1\right)}{\left({R}_{JY}^{Q}\right)}_{1a}{\left({R}_{JY}^{Q+1}\right)}_{1{a}^{\prime }}^{\ast }\right]\end{equation} \tag{ 73c }$$

$$\begin{equation}\quad +{W}_{\mu }^{-}\left({S}_{{a}^{\prime }}^{Q+1}{\partial }_{\mu }{S}_{a}^{-Q}-{S}_{a}^{-Q}\,{\partial }_{\mu }{S}_{{a}^{\prime }}^{Q+1}\right)\end{equation} \tag{ 73d }$$

$$\begin{equation}\quad \times \left[\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}\cap {\mathcal{S}}_{Q+1}}\!\sqrt{\left(J+Y+Q+1\right)\left(J-Y-Q\right)}{\left({S}_{JY}^{Q}\right)}_{1a}^{\ast }{\left({S}_{JY}^{Q+1}\right)}_{1{a}^{\prime }}\right.\end{equation} \tag{ 73e }$$

$$\begin{align}\hfill & \quad -\!\!\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}\cap {\mathcal{R}}_{Q+1}}\!\sqrt{\left(J+Y-Q\right)\left(J-Y+Q+1\right)}\left.\left.{\left({R}_{JY}^{Q}\right)}_{1a}^{\ast }{\left({R}_{JY}^{Q+1}\right)}_{1{a}^{\prime }}\right]\right\}.\hfill \end{align} \tag{ 73f }$$

We show in this appendix that the quantities ψ and θ in equation (25) are equal to 0. We shall use the sums

$$\begin{equation}1+2+3+\cdots +n=\frac{n\left(n+1\right)}{2},\end{equation} \tag{ 74a }$$

$$\begin{equation}1+4+9+\cdots +{n}^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6},\end{equation} \tag{ 74b }$$

which are valid for any positive integer n.

We shall separately consider the following types of multiplets M_JY :

• (a)  
  Y < −J. A multiplet of this type belongs neither to $\mathcal{A}$ nor to ${\mathcal{R}}_{Q}$ for any Q > 0. It belongs to ${\mathcal{S}}_{Q}$ for Q = −Y − J, ..., −Y + J.
• (b)  
  Y = −J. A multiplet of this type belongs to $\mathcal{A}$. It also belongs to ${\mathcal{S}}_{Q}$ for Q = 1, 2, ..., −2Y. It does not belong to ${\mathcal{R}}_{Q}$ for any value of Q > 0.
• (c)  
  −J < Y < J. A multiplet of this type belongs to $\mathcal{A}$, to ${\mathcal{R}}_{Q}$ for Q = 1, ..., J + Y, and to ${\mathcal{S}}_{Q}$ for Q = 1, ..., J − Y.
• (d)  
  Y = J. A multiplet of this type belongs to $\mathcal{A}$ and also to ${\mathcal{R}}_{Q}$ for Q = 1, 2, ..., 2Y. It does not belong to ${\mathcal{S}}_{Q}$ for any value of Q > 0.
• (e)  
  Y > J. A multiplet of this type belongs neither to $\mathcal{A}$ nor to ${\mathcal{S}}_{Q}$ for any Q > 0. It belongs to ${\mathcal{R}}_{Q}$ for Q = Y − J, ..., Y + J.

Firstly consider

$$\begin{align}\hfill \psi & =\sum\limits _{{M}_{JY}\in \mathcal{A}}{Y}^{2}+\sum\limits _{Q > 0}\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}{Y}^{2}-\sum\limits _{Q > 0}\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}QY\hfill \\ \hfill & \quad +\sum\limits _{Q > 0}\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}{Y}^{2}+\sum\limits _{Q > 0}\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}QY.\hfill \end{align} \tag{ 75 }$$

Take a multiplet of type (a). It contributes to ψ as

$$\begin{align}\hfill & \left(0\right)+\left(0\right)-\left(0\right)+\left[\left(-Y+J\right)-\left(-Y-J\right)+1\right]{Y}^{2}\hfill \\ \hfill & \qquad \quad +\left[\left(-Y-J\right)+\cdots +\left(-Y+J\right)\right]Y\hfill \end{align} \tag{ 76a }$$

$$\begin{equation}\qquad =\left(2J+1\right){Y}^{2}+\frac{\left(-Y+J\right)\left(-Y+J+1\right)-\left(-Y-J-1\right)\left(-Y-J\right)}{2}\ Y\end{equation} \tag{ 76b }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 76c }$$

Take a multiplet of type (b). It gives the following contribution to ψ:

$$\begin{equation}\left({Y}^{2}\right)+\left(0\right)-\left(0\right)+\left(-2Y\right){Y}^{2}+\left[1+2+\cdots +\left(-2Y\right)\right]Y\end{equation} \tag{ 77a }$$

$$\begin{equation}\qquad =\left(-2Y+1\right){Y}^{2}+\frac{\left(-2Y\right)\left(-2Y+1\right)}{2}\ Y\end{equation} \tag{ 77b }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 77c }$$

Take a multiplet of type (c). Its contribution to ψ is

$$\begin{equation}{Y}^{2}+\left(J+Y\right){Y}^{2}-\left[1+\cdots +\left(J+Y\right)\right]Y\end{equation} \tag{ 78a }$$

$$\begin{equation}\qquad \quad +\left(J-Y\right){Y}^{2}+\left[1+\cdots +\left(J-Y\right)\right]Y\end{equation} \tag{ 78b }$$

$$\begin{equation}\qquad =\left(2J+1\right){Y}^{2}+\frac{\left(J-Y\right)\left(J-Y+1\right)-\left(J+Y\right)\left(J+Y+1\right)}{2}\ Y\end{equation} \tag{ 78c }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 78d }$$

Take a multiplet of type (d). It contributes to ψ as

$$\begin{equation}\left({Y}^{2}\right)+\left(2Y\right){Y}^{2}-\left(1+2+\cdots +2Y\right)Y+\left(0\right)+\left(0\right)\end{equation} \tag{ 79a }$$

$$\begin{equation}\qquad =\left(2Y+1\right){Y}^{2}-\frac{\left(2Y\right)\left(2Y+1\right)}{2}\ Y\end{equation} \tag{ 79b }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 79c }$$

Take a multiplet of type (e). Its contribution to ψ is

$$\begin{align}\hfill & \left(0\right)+\left[\left(Y+J\right)-\left(Y-J\right)+1\right]{Y}^{2}\hfill \\ \hfill & \qquad \ -\left[\left(Y-J\right)+\cdots +\left(Y+J\right)\right]Y+\left(0\right)+\left(0\right)\hfill \end{align} \tag{ 80a }$$

$$\begin{equation}\qquad =\left(2J+1\right){Y}^{2}-\frac{\left(Y+J\right)\left(Y+J+1\right)-\left(Y-J-1\right)\left(Y-J\right)}{2}\ Y\end{equation} \tag{ 80b }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 80c }$$

We conclude that no multiplet produces a nonzero contribution to ψ, hence ψ is zero.

We next consider

$$\begin{equation}\theta =2\sum\limits _{{M}_{JY}\in \mathcal{A}}\left({J}^{2}+J-2{Y}^{2}\right)\end{equation} \tag{ 81 }$$

$$\begin{equation}\quad +\sum\limits _{Q > 0}\,\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{Q}}\left[\left({J}^{2}+J-3{Y}^{2}+Y\right)+\left(6Y-1\right)Q-3{Q}^{2}\right]\end{equation} \tag{ 82 }$$

$$\begin{equation}\quad +\sum\limits _{Q > 0}\,\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{Q}}\left[\left({J}^{2}+J-3{Y}^{2}-Y\right)-\left(6Y+1\right)Q-3{Q}^{2}\right].\end{equation} \tag{ 83 }$$

Take a multiplet of type (a). It belongs to neither $\mathcal{A}$ nor ${\mathcal{R}}_{Q}$; it belongs to ${\mathcal{S}}_{Q}$ from Q = −Y − J to Q = −Y + J. Its contribution to θ reads

$$\begin{equation}\left({J}^{2}+J-3{Y}^{2}-Y\right)\left[\left(-Y+J\right)-\left(-Y-J\right)+1\right]\end{equation} \tag{ 84a }$$

$$\begin{equation}-\left(6Y+1\right)\left[\left(-Y-J\right)+\cdots +\left(-Y+J\right)\right]-3\left[{\left(-Y-J\right)}^{2}+\cdots +{\left(-Y+J\right)}^{2}\right]\end{equation} \tag{ 84b }$$

$$\begin{equation}=\left({J}^{2}+J-3{Y}^{2}-Y\right)\left(2J+1\right)\end{equation} \tag{ 84c }$$

$$\begin{equation}\quad -\left(6Y+1\right)\frac{\left(-Y+J\right)\left(-Y+J+1\right)-\left(-Y-J-1\right)\left(-Y-J\right)}{2}\end{equation} \tag{ 84d }$$

$$\begin{align}\hfill & -\frac{\left(-Y+J\right)\left(-Y+J+1\right)\left(-2Y+2J+1\right)-\left(-Y-J-1\right)\left(-Y-J\right)\left(-2Y-2J-1\right)}{2}\hfill \\ \hfill & \hfill \end{align} \tag{ 84e }$$

$$\begin{equation}=\left({J}^{2}+J-3{Y}^{2}-Y\right)\left(2J+1\right)\end{equation} \tag{ 84f }$$

$$\begin{equation}\quad -\left(6Y+1\right)\frac{{\left(-Y+J\right)}^{2}-Y+J-{\left(-Y-J\right)}^{2}-Y-J}{2}\end{equation} \tag{ 84g }$$

$$\begin{equation}\quad -\frac{2{\left(-Y+J\right)}^{3}+3{\left(-Y+J\right)}^{2}-Y+J-2{\left(-Y-J\right)}^{3}+3{\left(-Y-J\right)}^{2}+Y+J}{2}\end{equation} \tag{ 84h }$$

$$\begin{equation}=0.\end{equation} \tag{ 84i }$$

Take a multiplet of type (b), which has J = −Y. Its contribution to θ is

$$\begin{equation}2\left({Y}^{2}-Y-2{Y}^{2}\right)+\left({Y}^{2}-Y-3{Y}^{2}-Y\right)\left(-2Y\right)\end{equation} \tag{ 85a }$$

$$\begin{equation}\qquad \quad -\left(6Y+1\right)\left[1+2+\cdots +\left(-2Y\right)\right]-3\left[1+4+\cdots +{\left(-2Y\right)}^{2}\right]\end{equation} \tag{ 85b }$$

$$\begin{align}\hfill & \qquad =-2Y-2{Y}^{2}+4{Y}^{2}+4{Y}^{3}+\left(6Y+1\right)Y\left(-2Y+1\right)\hfill \\ \hfill & \qquad \quad +Y\left(-2Y+1\right)\left(-4Y+1\right)\hfill \end{align} \tag{ 85c }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 85d }$$

Now consider a multiplet of type (c). It produces the following contribution to θ:

$$\begin{align}\hfill & 2\left({J}^{2}+J-2{Y}^{2}\right)+\left({J}^{2}+J-3{Y}^{2}+Y\right)\left(J+Y\right)\hfill \\ \hfill & \qquad \quad +\left(6Y-1\right)\frac{\left(J+Y\right)\left(J+Y+1\right)}{2}\hfill \end{align} \tag{ 86a }$$

$$\begin{equation}\qquad \quad -\frac{\left(J+Y\right)\left(J+Y+1\right)\left(2J+2Y+1\right)}{2}+\left({J}^{2}+J-3{Y}^{2}-Y\right)\left(J-Y\right)\end{equation} \tag{ 86b }$$

$$\begin{equation}\qquad \quad -\left(6Y+1\right)\frac{\left(J-Y\right)\left(J-Y+1\right)}{2}-\frac{\left(J-Y\right)\left(J-Y+1\right)\left(2J-2Y+1\right)}{2}\end{equation} \tag{ 86c }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 86d }$$

Take a multiplet of type (d), i.e. with J = Y. It contributes to θ as

$$\begin{equation}2\left({Y}^{2}+Y-2{Y}^{2}\right)+\left({Y}^{2}+Y-3{Y}^{2}+Y\right)\left(2Y\right)\end{equation} \tag{ 87a }$$

$$\begin{equation}\qquad \quad +\left(6Y-1\right)\left[1+2+\cdots +\left(2Y\right)\right]-3\left[1+4+\cdots +{\left(2Y\right)}^{2}\right]\end{equation} \tag{ 87b }$$

$$\begin{equation}\qquad =2Y-2{Y}^{2}+4{Y}^{2}-4{Y}^{3}+\left(6Y-1\right)Y\left(2Y+1\right)-Y\left(2Y+1\right)\left(4Y+1\right)\end{equation} \tag{ 87c }$$

$$\begin{equation}\qquad =0.\end{equation} \tag{ 87d }$$

Take a multiplet of type (e), that has only ${\mathcal{R}}_{Q}$ from Q = Y − J to Q = Y + J. It contributes to θ as

$$\begin{equation}\left({J}^{2}+J-3{Y}^{2}+Y\right)\left[\left(Y+J\right)-\left(Y-J\right)+1\right]\end{equation} \tag{ 88a }$$

$$\begin{equation}\quad +\left(6Y-1\right)\left[\left(Y-J\right)+\cdots +\left(Y+J\right)\right]-3\left[{\left(Y-J\right)}^{2}+\cdots +{\left(Y+J\right)}^{2}\right]\end{equation} \tag{ 88b }$$

$$\begin{equation}=\left({J}^{2}+J-3{Y}^{2}+Y\right)\left(2J+1\right)\end{equation} \tag{ 88c }$$

$$\begin{equation}\quad +\left(6Y-1\right)\frac{\left(Y+J\right)\left(Y+J+1\right)-\left(Y-J-1\right)\left(Y-J\right)}{2}\end{equation} \tag{ 88d }$$

$$\begin{equation}\quad -\frac{\left(Y+J\right)\left(Y+J+1\right)\left(2Y+2J+1\right)-\left(Y-J-1\right)\left(Y-J\right)\left(2Y-2J-1\right)}{2}\end{equation} \tag{ 88e }$$

$$\begin{equation}=\left({J}^{2}+J-3{Y}^{2}+Y\right)\left(2J+1\right)\end{equation} \tag{ 88f }$$

$$\begin{equation}\quad +\left(6Y-1\right)\frac{{\left(Y+J\right)}^{2}+Y+J-{\left(Y-J\right)}^{2}+Y-J}{2}\end{equation} \tag{ 88g }$$

$$\begin{equation}\quad -\frac{2{\left(Y+J\right)}^{3}+3{\left(Y+J\right)}^{2}+Y+J-2{\left(Y-J\right)}^{3}+3{\left(Y-J\right)}^{2}-Y+J}{2}\end{equation} \tag{ 88h }$$

$$\begin{equation}=0.\end{equation} \tag{ 88i }$$

Thus, all the multiplets produce a null contribution to θ, which means that θ = 0.

The terms that describe the mixing of the gauge bosons with the scalars allow one to obtain the form of the Goldstone bosons. For the mixing of the W boson with the charged scalars we get

$$\begin{equation}{\mathcal{L}}_{{W}^{\pm }{G}^{\mp }}=\mathrm{i}g{W}_{\mu }^{-}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}{v}_{JY}^{\ast }\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\ {\partial }^{\mu }{M}_{JY}^{1}\right.\end{equation} \tag{ 89a }$$

$$\begin{equation}\quad -\left.\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}{v}_{JY}\,\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\ {\partial }^{\mu }{\left({M}_{JY}^{-1}\right)}^{\ast }\right]+\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 89b }$$

We identify the quantity in equation (89) as being equal to $\mathrm{i}{m}_{W}{W}_{\mu }^{-}{\partial }^{\mu }{G}^{+}+\mathrm{h}.\mathrm{c}.$ [31]. This implicitly defines the phase of

$$\begin{equation}{G}^{+}=\frac{g}{{m}_{W}}\left[\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{1}}{v}_{JY}^{\ast }\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\ {M}_{JY}^{1}\right.\end{equation} \tag{ 90a }$$

$$\begin{equation}\quad -\left.\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{1}}{v}_{JY}\,\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\ {\left({M}_{JY}^{-1}\right)}^{\ast }\right],\end{equation} \tag{ 90b }$$

where we have used

$$\begin{equation}{M}_{JY}\in {\mathcal{R}}_{1}{\backslash}\mathcal{A}{\Rightarrow}J-Y+1=0,\qquad {M}_{JY}\in {\mathcal{S}}_{1}{\backslash}\mathcal{A}{\Rightarrow}J+Y+1=0.\end{equation} \tag{ 91 }$$

It follows from equation (90) that the row matrices defined in equations (8) and (9) have their first matrix elements given by

$$\begin{align}\hfill {\left({R}_{JY}^{1}\right)}_{11}& =\frac{g{v}_{JY}}{{m}_{W}}\,\sqrt{\frac{{J}^{2}-{Y}^{2}+J+Y}{2}},\hfill \\ \hfill {\left({S}_{JY}^{1}\right)}_{11}& =-\frac{g{v}_{JY}^{\ast }}{{m}_{W}}\,\sqrt{\frac{{J}^{2}-{Y}^{2}+J-Y}{2}}.\hfill \end{align} \tag{ 92 }$$

For the mixing of the Z boson with the neutral scalars we get

$$\begin{equation}{\mathcal{L}}_{Z{G}^{0}}=\mathrm{i}\frac{g}{{c}_{W}}{Z}_{\mu }\sum\limits _{{M}_{JY}\in \mathcal{A}}Y{v}_{JY}\,{\partial }^{\mu }{\left({M}_{JY}^{0}\right)}^{\ast }+\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 93 }$$

We identify this as being equal to m_Z Z_μ ∂^μ G⁰ + h.c. [31]. This implicitly defines the sign of

$$\begin{equation}{G}^{0}=\frac{\mathrm{i}g}{{c}_{W}{m}_{Z}}\sum\limits _{{M}_{JY}\in \mathcal{A}}Y\left[{v}_{JY}{\left({M}_{JY}^{0}\right)}^{\ast }-{v}_{JY}^{\ast }{M}_{JY}^{0}\right].\end{equation} \tag{ 94 }$$

Therefore, the row matrices defined in equation (11) have their first matrix elements given by

$$\begin{equation}{\left({A}_{JY}\right)}_{11}=\frac{-\sqrt{2}g}{{c}_{W}{m}_{Z}}\ Y\,\mathrm{I}\mathrm{m}\,{v}_{JY},\qquad {\left({B}_{JY}\right)}_{11}=\frac{\sqrt{2}g}{{c}_{W}{m}_{Z}}\ Y\,\mathrm{R}\mathrm{e}\,{v}_{JY}.\end{equation} \tag{ 95 }$$

We shall next demonstrate the Feynman rules in equation (26). Using the result of equation (70), the ZG⁺ G⁻ interaction terms are

$$\begin{align}\hfill & {\mathcal{L}}_{Z{G}^{+}{G}^{-}}\hfill \\ \hfill & \qquad ={Z}^{\mu }\left({G}^{+}{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{+}\right)\left\{-\mathrm{i}g{c}_{W}+\mathrm{i}\frac{g}{{c}_{W}}\left[\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{1}}Y{\left\vert {\left({R}_{JY}^{1}\right)}_{11}\right\vert }^{2}\!\!\right.\right.\hfill \\ \hfill & \left.\left.\qquad \quad -\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{1}}Y{\left\vert {\left({S}_{JY}^{1}\right)}_{11}\right\vert }^{2}\right]\right\}\hfill \end{align} \tag{ 96a }$$

$$\begin{align}\hfill & \qquad ={Z}^{\mu }\left({G}^{+}\,{\partial }_{\mu }{G}^{-}-{G}^{-}\,{\partial }_{\mu }{G}^{+}\right)\left\{-\mathrm{i}g{c}_{W}\right.\hfill \\ \hfill & \qquad \quad +\mathrm{i}\frac{{g}^{3}}{2{m}_{W}^{2}{c}_{W}}\left[\sum\limits _{{M}_{JY}\in {\mathcal{R}}_{1}}{\left\vert {v}_{JY}\right\vert }^{2}Y\left({J}^{2}-{Y}^{2}+J+Y\right)\right.\hfill \\ \hfill & \qquad \quad -\left.\left.\sum\limits _{{M}_{JY}\in {\mathcal{S}}_{1}}{\left\vert {v}_{JY}\right\vert }^{2}Y\left({J}^{2}-{Y}^{2}+J-Y\right)\right]\right\}\hfill \end{align} \tag{ 96b }$$

$$\begin{align}\hfill & \qquad ={Z}^{\mu }\left({G}^{+}\,{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{+}\right)\left\{-\mathrm{i}g{c}_{W}\right.\hfill \\ \hfill & \qquad \quad +\mathrm{i}\frac{{g}^{3}}{2{m}_{W}^{2}{c}_{W}}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}Y\left({J}^{2}-{Y}^{2}+J+Y\right)\right.\hfill \\ \hfill & \qquad \quad -\left.\left.\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}Y\left({J}^{2}-{Y}^{2}+J-Y\right)\right]\right\}\hfill \end{align} \tag{ 96c }$$

$$\begin{equation}\qquad ={Z}^{\mu }\left({G}^{+}{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{+}\right)\left(-\mathrm{i}g{c}_{W}+\mathrm{i}\frac{{g}^{3}}{{m}_{W}^{2}{c}_{W}}\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}{Y}^{2}\right)\end{equation} \tag{ 96d }$$

$$\begin{equation}\qquad =\mathrm{i}g{c}_{W}\left(\frac{{m}_{Z}^{2}}{2{m}_{W}^{2}}-1\right){Z}^{\mu }\left({G}^{+}{\partial }_{\mu }{G}^{-}-{G}^{-}\,{\partial }_{\mu }{G}^{+}\right),\end{equation} \tag{ 96e }$$

where we have used firstly equation (92), secondly equations (15) and (91), and thirdly equation (14a). Equation (96e) proves that the Feynman rule in equation (26b) is correct for a model with an arbitrary scalar content.

Using the result of equation (72), the W^± G^∓ G⁰ interaction terms are

$$\begin{align}\hfill {\mathcal{L}}_{{W}^{\pm }{G}^{\mp }{G}^{0}}& =\mathrm{i}\frac{g}{2}\,\sum\limits _{{M}_{JY}\in \mathcal{A}}\,\sqrt{\left(J+Y+1\right)\left(J-Y\right)}\hfill \end{align} \tag{ 97a }$$

$$\begin{equation}\qquad \times \left[{W}^{\mu +}{\left({A}_{JY}-\mathrm{i}{B}_{JY}\right)}_{11}{\left({S}_{JY}^{1}\right)}_{11}^{\ast }\left({G}^{0}{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{0}\right)\right.\end{equation} \tag{ 97b }$$

$$\begin{equation}\qquad +\left.{W}^{\mu -}{\left({A}_{JY}+\mathrm{i}{B}_{JY}\right)}_{11}{\left({S}_{JY}^{1}\right)}_{11}\left({G}^{+}{\partial }_{\mu }{G}^{0}-{G}^{0}{\partial }_{\mu }{G}^{+}\right)\right]\end{equation} \tag{ 97c }$$

$$\begin{equation}\qquad +\mathrm{i}\frac{g}{2}\,\sum\limits _{{M}_{JY}\in \mathcal{A}}\sqrt{\left(J+Y\right)\left(J-Y+1\right)}\end{equation} \tag{ 97d }$$

$$\begin{equation}\qquad \times \left[{W}^{\mu +}{\left({A}_{JY}+\mathrm{i}{B}_{JY}\right)}_{11}{\left({R}_{JY}^{1}\right)}_{11}^{\ast }\left({G}^{-}{\partial }_{\mu }{G}^{0}-{G}^{0}{\partial }_{\mu }{G}^{-}\right)\right.\end{equation} \tag{ 97e }$$

$$\begin{equation}\qquad +\left.{W}^{\mu -}{\left({A}_{JY}-\mathrm{i}{B}_{JY}\right)}_{11}{\left({R}_{JY}^{1}\right)}_{11}\left({G}^{0}{\partial }_{\mu }{G}^{+}-{G}^{+}{\partial }_{\mu }{G}^{0}\right)\right]\end{equation} \tag{ 97f }$$

$$\begin{equation}\quad =-\frac{{g}^{3}}{2{m}_{Z}{m}_{W}{c}_{W}}\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}Y\left(J+Y+1\right)\left(J-Y\right)\end{equation} \tag{ 97g }$$

$$\begin{equation}\qquad \times \left[{W}^{\mu +}\left({G}^{0}\,{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{0}\right)-{W}^{\mu -}\left({G}^{+}{\partial }_{\mu }{G}^{0}-{G}^{0}{\partial }_{\mu }{G}^{+}\right)\right]\end{equation} \tag{ 97h }$$

$$\begin{equation}\qquad +\frac{{g}^{3}}{2{m}_{Z}{m}_{W}{c}_{W}}\,\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}Y\left(J+Y\right)\left(J-Y+1\right)\end{equation} \tag{ 97i }$$

$$\begin{equation}\qquad \times \left[{W}^{\mu +}\left({G}^{0}\,{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{0}\right)-{W}^{\mu -}\left({G}^{+}{\partial }_{\mu }{G}^{0}-{G}^{0}{\partial }_{\mu }{G}^{+}\right)\right]\end{equation} \tag{ 97j }$$

$$\begin{equation}\quad =\frac{{g}^{3}}{{m}_{Z}{m}_{W}{c}_{W}}\,\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}{Y}^{2}\end{equation} \tag{ 97k }$$

$$\begin{equation}\qquad \times \left[{W}^{\mu +}\left({G}^{0}\,{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{0}\right)+{W}^{\mu -}\left({G}^{0}\,{\partial }_{\mu }{G}^{+}-{G}^{+}{\partial }_{\mu }{G}^{0}\right)\right]\end{equation} \tag{ 97l }$$

$$\begin{equation}\quad =\frac{g{m}_{Z}{c}_{W}}{2{m}_{W}}\left[{W}^{\mu +}\left({G}^{0}{\partial }_{\mu }{G}^{-}-{G}^{-}{\partial }_{\mu }{G}^{0}\right)\!+{W}^{\mu -}\left({G}^{0}{\partial }_{\mu }{G}^{+}-{G}^{+}{\partial }_{\mu }{G}^{0}\right)\right]\!\!.\end{equation} \tag{ 97m }$$

This proves that the Feynman rule in equation (26c) is correct for a model with an arbitrary scalar content.

From equation (7),

$$\begin{equation}{D}_{\mu }{M}_{JY}^{-1}={\partial }_{\mu }{M}_{JY}^{-1}-\mathrm{i}e{A}_{\mu }{M}_{JY}^{-1}+\mathrm{i}\frac{g}{{c}_{W}}{Z}_{\mu }{M}_{JY}^{-1}\left(Y+{c}_{W}^{2}\right)\end{equation} \tag{ 98a }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{+}{M}_{JY}^{-2}\sqrt{\frac{\left(J+Y+2\right)\left(J-Y-1\right)}{2}}\end{equation} \tag{ 98b }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{-}{M}_{JY}^{0}\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}},\end{equation} \tag{ 98c }$$

$$\begin{equation}{D}_{\mu }{M}_{JY}^{0}={\partial }_{\mu }{M}_{JY}^{0}+\mathrm{i}\frac{g}{{c}_{W}}{Z}_{\mu }{M}_{JY}^{0}Y\end{equation} \tag{ 98d }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{+}{M}_{JY}^{-1}\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\end{equation} \tag{ 98e }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{-}{M}_{JY}^{1}\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}},\end{equation} \tag{ 98f }$$

$$\begin{equation}{D}_{\mu }{M}_{JY}^{1}={\partial }_{\mu }{M}_{JY}^{1}+\mathrm{i}e{A}_{\mu }{M}_{JY}^{1}+\mathrm{i}\frac{g}{{c}_{W}}{Z}_{\mu }{M}_{JY}^{1}\left(Y-{c}_{W}^{2}\right)\end{equation} \tag{ 98g }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{+}{M}_{JY}^{0}\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\end{equation} \tag{ 98h }$$

$$\begin{equation}\quad -\mathrm{i}g{W}_{\mu }^{-}{M}_{JY}^{2}\sqrt{\frac{\left(J+Y-1\right)\left(J-Y+2\right)}{2}}.\end{equation} \tag{ 98i }$$

Therefore, the interaction terms of the Z⁰ with W^± and a charged scalar are given by

$$\begin{equation}{\mathcal{L}}_{Z{W}^{\pm }{S}^{\mp }}=-\frac{{g}^{2}}{{c}_{W}}{Z}^{\mu }{W}_{\mu }^{+}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}\left(Y+{c}_{W}^{2}\right)\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\ {v}_{JY}^{\ast }{M}_{JY}^{-1}\right.\end{equation} \tag{ 99a }$$

$$\begin{equation}\left.\quad +\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}Y\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\ {v}_{JY}^{\ast }{M}_{JY}^{-1}\right]\end{equation} \tag{ 99b }$$

$$\begin{equation}\quad -\frac{{g}^{2}}{{c}_{W}}{Z}_{\mu }{W}_{\mu }^{-}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}Y\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\ {v}_{JY}^{\ast }{M}_{JY}^{1}\right.\end{equation} \tag{ 99c }$$

$$\begin{equation}\left.\quad +\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}\left(Y-{c}_{W}^{2}\right)\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\ {v}_{JY}^{\ast }{M}_{JY}^{1}\right]+\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 99d }$$

$$\begin{equation}=-\frac{{g}^{2}}{{c}_{W}}{Z}^{\mu }{W}_{\mu }^{+}\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}\left(2Y+{c}_{W}^{2}\right)\sqrt{\frac{\left(J+Y+1\right)\left(J-Y\right)}{2}}\ {v}_{JY}^{\ast }{M}_{JY}^{-1}\end{equation} \tag{ 99e }$$

$$\begin{equation}\quad -\frac{{g}^{2}}{{c}_{W}}{Z}_{\mu }{W}_{\mu }^{-}\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}\left(2Y-{c}_{W}^{2}\right)\sqrt{\frac{\left(J+Y\right)\left(J-Y+1\right)}{2}}\ {v}_{JY}^{\ast }{M}_{JY}^{1}\end{equation} \tag{ 99f }$$

$$\begin{equation}\quad +\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 99g }$$

$$\begin{equation}=\frac{g{m}_{W}}{{c}_{W}}{Z}^{\mu }{W}_{\mu }^{+}\!\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}\left(2Y+{c}_{W}^{2}\right){\left({S}_{JY}\right)}_{11}\,\sum\limits _{a=1}^{{n}_{1}}{\left({S}_{JY}^{1}\right)}_{1a}^{\ast }{S}_{a}^{-}\end{equation} \tag{ 99h }$$

$$\begin{equation}\quad -\frac{g{m}_{W}}{{c}_{W}}{Z}_{\mu }{W}_{\mu }^{-}\!\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}\left(2Y-{c}_{W}^{2}\right){\left({R}_{JY}\right)}_{11}^{\ast }\,\sum\limits _{a=1}^{{n}_{1}}{\left({R}_{JY}^{1}\right)}_{1a}{S}_{a}^{+}\end{equation} \tag{ 99i }$$

$$\begin{equation}\quad +\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 99j }$$

$$\begin{equation}=\frac{g{m}_{W}}{{c}_{W}}{Z}_{\mu }{W}^{\mu -}\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}\left(2Y+{c}_{W}^{2}\right){\left({S}_{JY}^{1}\right)}_{11}^{\ast }\sum\limits _{a=1}^{{n}_{1}}{\left({S}_{JY}^{1}\right)}_{1a}{S}_{a}^{+}\end{equation} \tag{ 99k }$$

$$\begin{equation}\quad -\frac{g{m}_{W}}{{c}_{W}}{Z}_{\mu }{W}^{\mu -}\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}\left(2Y-{c}_{W}^{2}\right){\left({R}_{JY}^{1}\right)}_{11}^{\ast }\,\sum\limits _{a=1}^{{n}_{1}}{\left({R}_{JY}^{1}\right)}_{1a}{S}_{a}^{+}\end{equation} \tag{ 99l }$$

$$\begin{equation}\quad +\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 99m }$$

Therefore, the ZW^± G^∓ interaction terms are

$$\begin{equation}{\mathcal{L}}_{Z{W}^{\pm }{G}^{\mp }}\!=\!\frac{g{m}_{W}}{{c}_{W}}{Z}_{\mu }\left[\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}{\left\vert {\left({S}_{JY}^{1}\right)}_{11}\right\vert }^{2}\left(2Y+{c}_{W}^{2}\right)\right.\end{equation} \tag{ 100a }$$

$$\begin{equation}\left.\quad -\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}{\left\vert {\left({R}_{JY}^{1}\right)}_{11}\right\vert }^{2}\left(2Y-{c}_{W}^{2}\right)\right]\left({W}^{\mu -}{G}^{+}+{W}^{\mu +}{G}^{-}\right)\end{equation} \tag{ 100b }$$

$$\begin{equation}=\frac{{g}^{3}}{2{m}_{W}{c}_{W}}{Z}_{\mu }{W}^{\mu -}{G}^{+}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{S}}_{1}}{\left\vert {v}_{JY}\right\vert }^{2}\left({J}^{2}-{Y}^{2}+J-Y\right)\left(2Y+{c}_{W}^{2}\right)\right.\end{equation} \tag{ 100c }$$

$$\begin{equation}\left.\quad -\sum\limits _{{M}_{JY}\in \mathcal{A}\cap {\mathcal{R}}_{1}}{\left\vert {v}_{JY}\right\vert }^{2}\left({J}^{2}-{Y}^{2}+J+Y\right)\left(2Y-{c}_{W}^{2}\right)\right]+\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 100d }$$

$$\begin{equation}=\frac{{g}^{3}}{2{m}_{W}{c}_{W}}{Z}_{\mu }{W}^{\mu -}{G}^{+}\left[\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}\left({J}^{2}-{Y}^{2}+J-Y\right)\left(2Y+{c}_{W}^{2}\right)\right.\end{equation} \tag{ 100e }$$

$$\begin{equation}\left.\quad -\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}\left({J}^{2}-{Y}^{2}+J+Y\right)\left(2Y-{c}_{W}^{2}\right)\right]+\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 100f }$$

$$\begin{equation}=\frac{{g}^{3}}{{m}_{W}{c}_{W}}{Z}_{\mu }{W}^{\mu -}{G}^{+}\sum\limits _{{M}_{JY}\in \mathcal{A}}{\left\vert {v}_{JY}\right\vert }^{2}\left[-2{Y}^{2}+{c}_{W}^{2}\left({J}^{2}-{Y}^{2}+J\right)\right]+\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 100g }$$

$$\begin{equation}=\frac{g}{{m}_{W}{c}_{W}}{Z}_{\mu }{W}^{\mu -}{G}^{+}\left(-{c}_{W}^{2}{m}_{Z}^{2}+{c}_{W}^{2}{m}_{W}^{2}\right)+\mathrm{h}.\mathrm{c}.\end{equation} \tag{ 100h }$$

This proves the Feynman rule in equation (26a) for a model with an arbitrary scalar content.

In order to compute the oblique parameter S we need to compute the vacuum-polarization tensors ${{\Pi}}_{V{V}^{\prime }}^{\mu \nu }$, where V and V' may be either A and A (A is the photon), A and Z, or Z and Z.

The diagrams that contribute to ${\left.\frac{\partial \,\delta {A}_{ZZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}$ at the one-loop level are ⁷

The diagrams that contribute to ${\left.\frac{\partial \,\delta {A}_{AA}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}$ at one-loop level are

The diagrams that contribute to ${\left.\frac{\partial \,\delta {A}_{AZ}({q}^{2})}{\partial {q}^{2}}\right\vert }_{{q}^{2}=0}$ at one-loop level are