Can the Higgs field feel a dark force?

Abstract

We argue that if an electroweak Higgs field possesses a dark gauge charge responsible for dark matter stability, the W-boson mass deviation is properly induced, besides appropriately generated neutrino masses. We examine a simple model in which the usual Higgs doublet plays the role but dark matter candidates are somewhat input by ad hoc. We look for a realistic model that fully realizes such observation, thereby neutrino mass and dark matter are naturally supplied by a dark non-abelian gauge symmetry.

Appendices

Appendix A: SU(2) tensor products

Consider two representations $|x\rangle$ and $|y\rangle$ determined in ket-bases $|a \rangle$ for $a=1,2,3,\cdots ,n$ and $|\alpha \rangle$ for $\alpha =1,2,3,\cdots , m$, such that $|x\rangle = x_a |a\rangle$ and $|y\rangle = y_\alpha |\alpha \rangle$, respectively. We can label each representation according to its dimension and entries as $|x\rangle = {\underline{n}}(1,2,3,\cdots ,n)$ and $|y\rangle = {\underline{m}}(1,2,3,\cdots ,m)$. Their tensor product is $|x\rangle |y\rangle =x_a y_\alpha |a\rangle |\alpha \rangle = {\underline{n}}\otimes {\underline{m}}(\cdots ,a\alpha ,\cdots )$. Notice that in usual notation, ${\underline{n}}$ [${\underline{m}}$] corresponds to the spin-$j=(n-1)/2$ [$j'=(m-1)/2$] representation and the index $a=1,2,3,\cdots ,n$ [$\alpha =1,2,3,\cdots ,m$] marks weights $j,j-1,\cdots ,-(j-1),-j$ [$j',j'-1,\cdots ,-(j'-1),-j'$]—the $T_3$ eigenvalues—whose corresponding eigenstates forming the ket-basis, respectively. The tensor product is just ${\underline{n}}\otimes {\underline{m}}=\textrm{spin}$-$j\otimes \textrm{spin}$-$j'$.

With the aid of Clebsch-Gordan coefficients (cf. [18]), decomposition rules of tensor products into irreducible representations are straightforwardly derived,

$$\begin{aligned} {\underline{2}}\otimes {\underline{2}} &= {\underline{1}}\left( \frac{12-21}{\sqrt{2}}\right) \oplus {\underline{3}}\left( 11,\frac{12+21}{\sqrt{2}},22\right) ,\nonumber \\ {\underline{3}}\otimes {\underline{3}}&={\underline{1}}\left( \frac{13-22+31}{\sqrt{3}}\right) \oplus {\underline{3}}\left( \frac{12-21}{\sqrt{2}},\frac{13-31}{\sqrt{2}},\frac{23-32}{\sqrt{2}}\right) \nonumber \\ & \quad \oplus {\underline{5}}\left( 11,\frac{12+21}{\sqrt{2}},\frac{13+2.22+31}{\sqrt{6}},\frac{23+32}{\sqrt{2}},33\right) ,\nonumber \\ {\underline{2}}\otimes {\underline{3}} &= {\underline{2}}\left( \frac{\sqrt{2}.21-12}{\sqrt{3}},\frac{22-\sqrt{2}.13}{\sqrt{3}}\right) \oplus {\underline{4}}\left( 11,\frac{21+\sqrt{2}.12}{\sqrt{3}},\frac{\sqrt{2}.22+13}{\sqrt{3}},23\right) ,\nonumber \\ {\underline{4}}\otimes {\underline{4}}&={\underline{1}}\left( \frac{14-23+32-41}{2}\right) \nonumber \\ & \quad \oplus {\underline{3}}\left( \frac{\sqrt{3}.13-2.22+\sqrt{3}.31}{\sqrt{10}},\frac{3.14-23-32+3.41}{2\sqrt{5}},\frac{\sqrt{3}.24-2.33+\sqrt{3}.42}{\sqrt{10}}\right) \nonumber \\ & \quad \oplus {\underline{5}}\left( \frac{12-21}{\sqrt{2}},\frac{13-31}{\sqrt{2}},\frac{14+23-32-41}{2},\frac{24-42}{\sqrt{2}},\frac{34-43}{\sqrt{2}}\right) \nonumber \\ & \quad \oplus {\underline{7}}\left( 11,\frac{12+21}{\sqrt{2}},\frac{13+\sqrt{3}.22+31}{\sqrt{5}},\frac{14+3.23+3.32+41}{2\sqrt{5}},\frac{24+\sqrt{3}.33+42}{\sqrt{5}},\frac{34+43}{\sqrt{2}},44\right) ,\nonumber \\ {\underline{2}}\otimes {\underline{4}} &= {\underline{3}}\left( \frac{-12+\sqrt{3}.21}{2},\frac{-13+22}{\sqrt{2}},\frac{-\sqrt{3}.14+23}{2}\right) \nonumber \\ & \quad \oplus {\underline{5}}\left( 11,\frac{\sqrt{3}.12+21}{2},\frac{13+22}{\sqrt{2}},\frac{14+\sqrt{3}.23}{2},24\right) ,\nonumber \\ {\underline{3}}\otimes {\underline{4}} &={\underline{2}}\left( \frac{\sqrt{3}.31-\sqrt{2}.22+13}{\sqrt{6}},\frac{32-\sqrt{2}.23+\sqrt{3}.14}{\sqrt{6}}\right) \nonumber \\ & \quad \oplus {\underline{4}} \left( \frac{\sqrt{3}.21-\sqrt{2}.12}{\sqrt{5}},\frac{\sqrt{6}.31+22-2\sqrt{2}.13}{\sqrt{15}},\frac{2\sqrt{2}.32-23-\sqrt{6}.14}{\sqrt{15}},\frac{\sqrt{2}.33-\sqrt{3}.24}{\sqrt{5}}\right) \nonumber \\ & \quad \oplus {\underline{6}}\left( 11,\frac{\sqrt{2}.21+\sqrt{3}.12}{\sqrt{5}},\frac{31+\sqrt{6}.22+\sqrt{3}.13}{\sqrt{10}},\frac{\sqrt{3}.32+\sqrt{6}.23+14}{\sqrt{10}},\frac{\sqrt{3}.33+\sqrt{2}.24}{\sqrt{5}},34\right) ,\nonumber \\ {\underline{5}}\otimes {\underline{5}}&= {\underline{1}}\left( \frac{15-24+33-42+51}{\sqrt{5}}\right) \nonumber \\ & \quad \oplus {\underline{3}}\left( \frac{\sqrt{2}.14-\sqrt{3}.23+\sqrt{3}.32-\sqrt{2}.41}{\sqrt{10}},\frac{2.15-24+42-2.51}{\sqrt{10}},\right. \nonumber \\ & \quad \left. \frac{\sqrt{2}.25-\sqrt{3}.34+\sqrt{3}.43-\sqrt{2}.52}{\sqrt{10}}\right) \nonumber \\ & \quad \oplus {\underline{5}}\left( \frac{\sqrt{2}.13-\sqrt{3}.22+\sqrt{2}.31}{\sqrt{7}},\frac{\sqrt{6}.14-23-32+\sqrt{6}.41}{\sqrt{14}},\frac{2.15+24-2.33+42+2.51}{\sqrt{14}},\right. \nonumber \\ & \quad \left. \frac{\sqrt{6}.25-34-43+\sqrt{6}.52}{\sqrt{14}},\frac{\sqrt{2}.35-\sqrt{3}.44+\sqrt{2}.53}{\sqrt{7}}\right) \nonumber \\ & \quad \oplus {\underline{7}}\left( \frac{12-21}{\sqrt{2}},\frac{13-31}{\sqrt{2}},\frac{\sqrt{3}.14+\sqrt{2}.23-\sqrt{2}.32-\sqrt{3}.41}{\sqrt{10}},\frac{15+2.24-2.42-51}{\sqrt{10}},\right. \nonumber \\ & \quad \left. \frac{\sqrt{3}.25+\sqrt{2}.34-\sqrt{2}.43-\sqrt{3}.52}{\sqrt{10}},\frac{35-53}{\sqrt{2}},\frac{45-54}{\sqrt{2}}\right) \nonumber \\ &\quad {\underline{9}}\left( 11,\frac{12+21}{\sqrt{2}},\frac{\sqrt{3}.13+2\sqrt{2}.22+\sqrt{3}.31}{\sqrt{14}},\frac{14+\sqrt{6}.23+\sqrt{6}.32+41}{\sqrt{14}},\right. \nonumber \\ & \quad \left. \frac{15+4.24+6.33+4.42+51}{\sqrt{70}},\frac{25+\sqrt{6}.34+\sqrt{6}.43+52}{\sqrt{14}},\right. \nonumber \\ & \quad \left. \frac{\sqrt{3}.35+2\sqrt{2}.44+\sqrt{3}.53}{\sqrt{14}},\frac{45+54}{\sqrt{2}},55\right) ,\nonumber \end{aligned}$$

and so forth for ${\underline{2}}\otimes {\underline{5}}$, ${\underline{3}}\otimes {\underline{5}}$, and ${\underline{4}}\otimes {\underline{5}}$. To be concrete, for instance $x=(x_1,x_2)$ and $y=(y_1,y_2)$, we have $xy=(xy)_{{\underline{1}}}\oplus (xy)_{{\underline{3}}}$, where

$$\begin{aligned} (xy)_{{\underline{1}}}= & (x_1y_2-x_2y_1)/\sqrt{2}, \end{aligned}$$

(A1)

$$\begin{aligned} (xy)_{{\underline{3}}}= & (x_1 y_1, (x_1y_2+x_2y_1)/\sqrt{2},x_2y_2). \end{aligned}$$

(A2)

It is noted that since all SU(2) representations are real, we need not necessarily consider their conjugated representations. If a conjugated representation exists, by contrast, we can transform it to the normal one, say ${\underline{2}}=\epsilon {\underline{2}}^*$, ${\underline{3}}=\epsilon ' {\underline{3}}^*$, and so forth, where

$$\begin{aligned} \epsilon =\begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix},\hspace{0.5cm}\epsilon '=\begin{pmatrix}0 & 0 & 1\\ 0 & -1 & 0\\ 1 & 0 & 0\end{pmatrix}, \end{aligned}$$

(A3)

and so forth, and the above rules apply.

The overall factors on resultant irreducible representations, e.g., $1/\sqrt{2}$ in $(xy)_{{\underline{1}}}$, which are field normalization (exactly arising from orthonormalized bases) coefficients can be conveniently omitted or not. This work uses the full forms given above for the Yukawa Lagrangian and scalar potential, since the overall factors do not make scene as possibly absorbed into the mass and coupling parameters. However, for the kinetic term, including its gauge interaction, such overall factors are suppressed, in order to keep the canonical form.

Appendix B: Lagrangian for the model with $n=m=5$

For this model, the Lagrangian relevant to the new fields, including all scalars, takes the form,

$$\begin{aligned} {\mathcal {L}}&\supset {\bar{\xi }} i \gamma ^\mu {\mathcal {D}}_\mu \xi +\sum _S ({\mathcal {D}}^{\mu } S)^\dagger ({\mathcal {D}}_{\mu } S) -(1/4) A'_{j\mu \nu } A^{\prime \mu\nu}_j\nonumber \\ & +\left[ h {\bar{l}}_{L} \Phi \xi -(1/2) (M+f \varphi )\xi \xi +H.c.\right] -V(S), \end{aligned}$$

(B1)

where $S=\{H,\Phi ,\varphi \}$, ${\mathcal {D}}_{\mu }=\partial _\mu + ig T_j A_{j\mu }+i g_B Y B_\mu + i g' T'_j A'_{j\mu }$ is the covariant derivative as coupled to $SU(2)_L\otimes U(1)_Y\otimes SU(2)'_L$, and $A'_{j\mu \nu }=\partial _\mu A'_{j\nu }-\partial _\nu A'_{j\mu }-g'\epsilon _{jkl} A'_{k\mu }A'_{l\nu }$ is $SU(2)'_L$ field strength. Notice that the adjoint dark gauge boson takes the form $A'=T'_j A'_j\sim (A^{\prime +},A^{\prime 0},A^{\prime -})$ where the last one is given in the basis of $T'_3$ eigenstates, i.e., arranged in the $T'_3$ weight order, with $A^{\prime \pm} \equiv (A'_1\mp i A'_2)/\sqrt{2}$ and $A^{\prime 0}\equiv A'_3$. Here, the superscripts $^{\pm ,0}$ label only $T'_3$ values (i.e., weights), not electric charge; indeed, their electric charge is zero.

The scalar potential is

$$\begin{aligned} V(S)&= \mu ^2_1{\tilde{H}}H+\mu ^2_2{\tilde{\Phi }}\Phi +\mu ^2_3 {\tilde{\varphi }}\varphi \nonumber \\ &\quad +[\mu _4 {\tilde{\Phi }}\Phi \varphi +\mu _5 H\Phi \varphi +H.c.]\nonumber \\ &\quad +\lambda _1({\tilde{H}}H)^2+\lambda _2({\tilde{\Phi }}\Phi )^2+\lambda _3({\tilde{\varphi }}\varphi )^2\nonumber \\ &\quad +\lambda _4 ({\tilde{H}} H)({\tilde{\Phi }} \Phi )+(\lambda _5{\tilde{H}}H+\lambda _6{\tilde{\Phi }}\Phi )({\tilde{\varphi }}\varphi )\nonumber \\ &\quad +\lambda _7 ({\tilde{H}}\Phi )({\tilde{\Phi }}H)+\lambda _8({\tilde{\Phi }}\varphi )({\tilde{\varphi }}\Phi )\nonumber \\ &\quad +[\lambda _9(H\Phi )+H.c.]{\tilde{\varphi }}\varphi +\lambda _{10}({\tilde{H}}{\tilde{\Phi }})(H\Phi )\nonumber \\ &\quad +[\lambda _{11} (H\Phi )(H\Phi )+H.c.], \end{aligned}$$

(B2)

where ${\tilde{H}}=\epsilon H^*$ and ${\tilde{\Phi }}=\epsilon \Phi ^* \epsilon '$, with $\epsilon ,\epsilon '$ supplied in Appendix A. The soft terms $\mu _{4,5}$ and the couplings $\lambda _{9,11}$ are generally complex. However, they can be considered to be real for the following computation, since otherwise their phases can be removed by redefining the relevant fields. Additionally, the combinations of types $\varphi \varphi$ and ${\tilde{\varphi }}{\tilde{\varphi }}$ are possible, in addition to the canonical form ${\tilde{\varphi }}\varphi$. However, they violate a phase transformation $e^{i x}$ unlike ${\tilde{\varphi }}\varphi$. Furthermore, we might have many/alternative possibilities for constructing an invariant tensor product, e.g.

$$\begin{aligned} \lambda _6({\tilde{\Phi }}\Phi )({\tilde{\varphi }}\varphi )\rightarrow & \lambda ^{(1)}_6({\tilde{\Phi }}\Phi )_{{\underline{1}}}({\tilde{\varphi }}\varphi )_{{\underline{1}}}+\lambda ^{(3)}_6({\tilde{\Phi }}\Phi )_{{\underline{3}}}({\tilde{\varphi }}\varphi )_{{\underline{3}}}+\lambda ^{(5)}_6({\tilde{\Phi }}\Phi )_{{\underline{5}}}({\tilde{\varphi }}\varphi )_{{\underline{5}}}\nonumber \\ & +\lambda ^{(7)}_6({\tilde{\Phi }}\Phi )_{{\underline{7}}}({\tilde{\varphi }}\varphi )_{{\underline{7}}}+\lambda ^{(9)}_6({\tilde{\Phi }}\Phi )_{{\underline{9}}}({\tilde{\varphi }}\varphi )_{{\underline{9}}}.\end{aligned}$$

(B3)

Obviously, not all of the tensor combinations are independent and that the physics with minimal couplings by smallest dimension decompositions is the most relevant. In other words, the non-minimal couplings if viable will be not interpreted for the above potential.

To let the potential be bounded from below and achieve the relevant vacuum structure for scalar fields, the potential parameters must obey

$$\begin{aligned} \lambda _{1,2,3}>0,\hspace{0.5cm}\mu ^2_{1,2,3}<0. \end{aligned}$$

(B4)

Here, the conditions for $\lambda _{1,2,3}$ are determined if $V(S)>0$ for $S=H,\Phi ,\varphi$ separately tending to infinity. Additionally, the supplemental conditions for $V(S)>0$ when two of the scalar fields simultaneously tending to infinity are

$$\begin{aligned} & \lambda _4+(\lambda _7+\lambda _{10}+2\lambda _{11})\Theta (-\lambda _7-\lambda _{10}-2\lambda _{11})>-2\sqrt{\lambda _1\lambda _2}, \end{aligned}$$

(B5)

$$\begin{aligned} & \lambda _6+\lambda _8\Theta (-\lambda _8)>-2\sqrt{\lambda _2\lambda _3},\hspace{0.5cm}\lambda _5>-2\sqrt{\lambda _1\lambda _3}, \end{aligned}$$

(B6)

where $\Theta (x)$ is the Heaviside step function. Note that the conditions for $V(S)>0$ when three of the scalar fields simultaneously tending to infinity would supply extra conditions. Additionally, the conditions for physical scalar masses squared to be positive may be also presented. All such constraints are skipped for brevity.

At the minimum of the potential energy, we obtain the condition

$$\begin{aligned} 0&=\, \mu _1^2+\lambda _1 v^2-\frac{\lambda _{10}}{2\sqrt{5}}\left( u_1^2+u_3^3+u_5^2\right) +\frac{\lambda _{11}}{\sqrt{5}}\left( u_3^2+2u_1u_5\right) \nonumber \\ &\quad -\frac{\lambda _{4}}{2\sqrt{5}}\left( u_1^2+u_3^3+u_5^2\right) -\frac{\lambda _5 }{2\sqrt{5}}\Lambda ^2-\frac{\lambda _9 u_3 }{\sqrt{70}v}\Lambda ^2 -\frac{u_5}{\sqrt{10}v} \mu _5 \Lambda , \end{aligned}$$

(B7)

$$\begin{aligned} 0&=\, \mu _2^2+\frac{\lambda _2}{\sqrt{5}}\left( u_1^2+u_3^2+u_5^2 \right) -\frac{\lambda _4}{2}v^2+\frac{\lambda _6}{2\sqrt{5}}\Lambda ^2-\frac{\lambda _{10}}{2}v^2\nonumber \\ &\quad +\lambda _{11}\frac{u_5}{u_1}v^2+\sqrt{\frac{5 }{7}}\frac{u_3 }{u_1}\mu _4\Lambda ,\end{aligned}$$

(B8)

$$\begin{aligned} 0&=\, \mu _2^2+\frac{\lambda _2}{\sqrt{5}}\left( u_1^2+u_3^2+u_5^2 \right) -\frac{\lambda _4}{2}v^2+\frac{\lambda _6}{2\sqrt{5}}\Lambda ^2 -\frac{v}{\sqrt{14}u_3}\lambda _9 \Lambda ^2\nonumber \\ &\quad -\frac{\lambda _{10}}{2}v^2+\lambda _{11}v^2 +\sqrt{\frac{5}{7}}\frac{u_1+u_5}{u_3} \mu _4 \Lambda , \end{aligned}$$

(B9)

$$\begin{aligned} 0&=\, \mu _2^2+\frac{\lambda _2}{\sqrt{5}}\left( u_1^2+u_3^2+u_5^2 \right) -\frac{\lambda _4}{2}v^2+\frac{\lambda _6}{2\sqrt{5}}\Lambda ^2 -\frac{\lambda _{10}}{2}v^2\nonumber \\ &\quad +\lambda _{11}\frac{u_1}{u_5}v^2+\sqrt{\frac{5}{7}}\frac{u_3}{u_5}\mu _4 \Lambda -\frac{v}{\sqrt{2}u_5}\mu _5 \Lambda ,\end{aligned}$$

(B10)

$$\begin{aligned} 0&=\, \mu _3^2 +\frac{\lambda _3}{\sqrt{5}}\Lambda ^2-\frac{\lambda _5}{2}v^2 +\frac{\lambda _6}{2\sqrt{5}}\left( u_1^2+u_3^2+u_5^2\right) -\sqrt{\frac{2}{7}} \lambda _9 u_3 v\nonumber \\ &\quad + \sqrt{\frac{5}{7}} \frac{u_3}{\Lambda }\left( u_1+u_5 \right) \mu _4- \frac{u_5}{\sqrt{2}\Lambda } \mu _5 v. \end{aligned}$$

(B11)

The five equations always give a solution of $(\Lambda ,v,u_{1,3,5})$ in which $\Lambda$ is governed by $|\mu _3|$ scale, while $v,u_{1,3,5}$ are by $|\mu _{1,2}|$ scales, with appropriately-adjusting scalar self-couplings and $\mu _{4,5}$. Although we do not deal with the issues of $v,u_{1,3,5}\ll \Lambda$ and $m_H\ll m_{\Phi }$ in detail, these hierarchies are typically only $1\ \textrm{TeV}/100\ \textrm{GeV}\sim 10$, a necessary fine-tuning between the mass parameters as well as the dimensionless couplings easily supply an expected solution.

To proceed further, we define $\varphi _1=\left( \Lambda +S_{1\varphi }+iA_{1\varphi }\right) /\sqrt{2}$, $\varphi _a=\left( S_{a\varphi }+iA_{a\varphi }\right) /\sqrt{2}$, for $a=2,3,4,5$, $\Phi ^0_b=\left( u_b+S_b+iA_b \right) /\sqrt{2}$, for $b=1,3,5$, and $\Phi ^0_c=\left( S_c+iA_c \right) /\sqrt{2}$, for $c=2,4$. Additionally, we conveniently denote $H_1\equiv S_{1\varphi }$ for using throughout the text.

The dark scalars $(S_2, S_4)$ mix via a mass-squared matrix as follows:

$$\begin{aligned} -\frac{1}{2} \begin{pmatrix} S_2&S_4 \end{pmatrix} \begin{pmatrix} m_{S_2 S_2}^2 & m_{S_2 S_4}^2 \\ m_{S_4 S_2}^2 & m_{S_{4}S_4}^2\end{pmatrix} \begin{pmatrix} S_2 \\ S_4 \end{pmatrix}, \end{aligned}$$

(B12)

where

$$\begin{aligned} m_{S_{2}S_2}^2=m^2_{S_{4}S_4}= & \frac{u^2_3-u_1u_5-u_5^2}{\left( u_1^2+u_1 u_5-u_3^2 \right) } \frac{\lambda _{11}v^2}{2\sqrt{5}}-\frac{1}{2\sqrt{70}}\frac{\lambda _9 u_3v\Lambda ^2}{\left( u_1^2+u_1 u_5-u_3^2 \right) }, \end{aligned}$$

(B13)

$$\begin{aligned} m_{S_2 S_4}^2=m^2_{S_4 S_2}= & -\frac{\lambda _{11}}{2\sqrt{5}}v^2-\sqrt{\frac{3}{40}}\frac{\lambda _{11}\left( u_1+u_5 \right) u_3v^2}{u_1^2-u_3^2+u_1u_5}+\sqrt{\frac{3}{35}}\frac{\lambda _9 u_1v\Lambda ^2}{4\left( u_1^2-u_3^2+u_1u_5\right) }. \end{aligned}$$

(B14)

We obtain the physical eigenstates

$$\begin{aligned} S_{24}=(S_2-S_{4})/\sqrt{2},\hspace{0.5cm}S'_{24}=(S_2+S_{4})/\sqrt{2},\end{aligned}$$

(B15)

with respect to the mass eigenvalues,

$$\begin{aligned} m_{S_{24}}^2=m_{S_2 S_2}^2-m_{S_2 S_4}^2, \hspace{0.5cm}m_{S'_{24}}^2 =m_{S_{2}S_2}^2+m_{S_{2}S_{4}}^2. \end{aligned}$$

(B16)

It is clear that the scalar masses are completely separated and proportional to $\Lambda$ scale. The lightest dark scalar is $S_{24}$ whose mass is below or above a TeV, depending on the magnitude of $\lambda _{9}$.

The relevant couplings among the lightest dark scalar, the usual Higgs boson, and the new Higgs fields are computed, collected in Table 3.

Table 3 Couplings of $S_{24}$ with various Higgs fields as well as those of the usual and new Higgs fields, where note that the couplings of $S_{24}$ with CP-odd scalar components vanish