Pauli Sum Rules After Higgs Boson Discovery

1. Introduction

In the frame of General Relativity (GR), any energy density influences the universe’s expansion. This also concerns the zero-point energy density of the quantum fields. Sum rules [1] have been suggested to exclude this influence by mutual compensation of the contributions from fermions and bosons in each term of the vacuum energy density. This was done before discussions about supersymmetry and before the common acceptance of the renormalization procedure. After the development of a renormalization technique, Pauli sum rules become less accurate, because one could exclude vacuum energy density by some renormalization [2,3,4]. Nevertheless, sum rules in a Pauli version remain a subject of discussion [5,6,7,8]. After the suggestion of the Higgs particle, vacuum fluctuations are widely discussed in relation [9,10,11,12,13,14,15,16,17,18] to the Higgs mechanism of particle mass generation. In particular, Veltman’s sum rule was introduced [19,20,21,22,23].

Here, we move away from the GR paradigm and work within the framework of a five-vector theory of gravity (FVT) [24,25], which allows an arbitrary choice of the reference energy density level. In other words, the Hamiltonian constraint in this theory is not exactly zero but satisfied up to some constant. Besides, we do not use a renormalization procedure. That is, in this paper, we believe that the energy density of zero-point fluctuations actually exists. The purpose of the present work is not to construct a unique ultraviolet-complete extension of the standard model. Rather, we ask a narrower question: what constraints on additional bosonic degrees of freedom follow if the two Pauli-type mass moments retained in the present FVT/Higgs framework are imposed literally? The resulting sum rules constrain moments of a spectrum, not the spectrum uniquely. Therefore, all statements below about a minimal number of new states or characteristic masses are to be understood as conditional on the explicitly stated ansatz. This distinction is essential when comparing algebraic sum-rule analysis with realistic model building.

2. Pauli Sum Rules

Pauli sum rules [1] suggested to exclude the influence of the zero point fluctuations on the universe expansion include four relations:

$$\begin{array}{c}\hfill \sum _{boson}i=\sum _{ferm}i,\end{array}$$

(1)

$$\begin{array}{c}\hfill \sum _{boson}{m}_i^2=\sum _{ferm}{m}_i^2,\end{array}$$

(2)

$$\begin{array}{c}\hfill \sum _{boson}{m}_i^4=\sum _{ferm}{m}_i^4,\end{array}$$

(3)

$$\begin{array}{c}\hfill \sum _{boson}{m}_i^{4}ln{m}_i=\sum _{ferm}{m}_i^{4}ln{m}_i,\end{array}$$

(4)

where index i implies summation over all bosonic and fermionic degrees of freedom, including kind of particle, spin, color, etc. In the expanding flat universe with the metric

$$d{s}^2=g_{\mu \nu }d{x}^{\nu }d{x}^{\mu }=a^2\left(\eta \right)\left(d{\eta }^2-d{\mathit{x}}^2\right),$$

(5)

vacuum energy of a single massive boson field is written as

$$\begin{array}{c}\hfill {\rho }_v={\displaystyle \frac{1}{4{\pi }^2{a}^4}}{\int }_0^{k_{max}}{k}^2\sqrt{k^2+a^2{m}^2}dk\approx {\displaystyle \frac{1}{16{\pi }^2}}({\displaystyle \frac{k_{max}^4}{a^4}}+\\ \hfill {\displaystyle \frac{m^2{k}_{max}^2}{a^2}}+{\displaystyle \frac{m^4}{8}}\left(1+2ln\left({\displaystyle \frac{m^2{a}^2}{4k_{max}^2}}\right)\right)),\end{array}$$

(6)

where ultraviolet cut off $k_{max}\sim M_p$ of comoving momentums at a level of a Planck mass $M_p$ is implied [25]. For a single massive fermion with spin $1/2$, there will be a minus sign and a multiplier 2 in front of the expression (6). Here the factor 2 refers to a two-component spin-$1/2$ field; for a charged Dirac fermion the factor is 4, as used below for the top-quark contribution. The terms in the expression (6) have different dependence on the scale factor a. First term corresponds to the energy density of “invisible” radiation $\rho \sim M_p^4/a^4$. To cancel these terms, supersymmetry is needed, which corresponds to an equal number of bosonic and fermionic degrees of freedom. There is no experimental evidence for supersymmetric particles, and here we assume the absence of equality of bosonic and fermionic degrees of freedom. In FVT gravity, uniform energy density corresponding to the first term in (6) does not influence the universe expansion and we retain the second (2) and third (3) Pauli sum rules. The fourth logarithmic rule (4) will not be used below, because the Higgs-based treatment developed in Sec. III does not yield it as an independent consistency condition.

For clarity, the sums are over physical degrees of freedom. Equivalently, one may write the supertrace [8] schematically

$$\mathrm{Str}f\left(m^2\right)=\sum _i{(-1)}^{2s_i}{d}_{i}f\left(m_i^2\right)=\sum _{\mathrm{bosons}}{d}_{i}f\left(m_i^2\right)-\sum _{\mathrm{fermions}}{d}_{i}f\left(m_i^2\right),$$

(7)

where

$$d_i=\left\{\begin{array}{cc}1,\hfill & \mathrm{for}\mathrm{a}\mathrm{real}\mathrm{scalar}\mathrm{field},\hfill \\ 2,\hfill & \mathrm{for}\mathrm{a}\mathrm{complex}\mathrm{scalar}\mathrm{field},\hfill \\ 3,\hfill & \mathrm{for}\mathrm{a}\mathrm{massive}\mathrm{vector}\mathrm{boson},\hfill \\ 4,\hfill & \mathrm{for}\mathrm{a}\mathrm{charged}\mathrm{Dirac}\mathrm{fermion},\hfill \end{array}\right..$$

denotes the physical degeneracy. Thus, a real scalar counts as one degree of freedom, a massive vector as three, and a Dirac fermion as four before color or other multiplicities are included. In Sec. IV, the integers $N_A$ and $N_B$ denote such bosonic degrees of freedom, not necessarily the number of distinct particle species. Keeping this bookkeeping explicit prevents a spectral interpretation from being conflated with the algebraic weights that enter the sum rules.

3. Vacuum Energy Density and Higgs Mechanism of Particle Mass Generation

After the discovery of the Higgs boson, it is desirable to account explicitly for the mechanism of particle mass generation. The standard model of particle physics is based on the gauge group

$$SU{\left(3\right)}_C\times SU{\left(2\right)}_L\times U{\left(1\right)}_Y,$$

and the mechanism of mass generation can be described in terms of a single complex scalar Higgs doublet. The Higgs doublet contains four real scalar degrees of freedom. After spontaneous electroweak symmetry breaking, three of them become the longitudinal polarization states of the massive $W^{\pm }$ and Z gauge bosons. At the same time, the remaining degree of freedom corresponds to the physical Higgs boson. Let us write a schematic action functional of the standard model in the form of

$$\begin{array}{c}\hfill S=\int ({\displaystyle \frac{1}{2}}{\partial }_{\mu }\varphi g^{\mu \nu }{\partial }_{\nu }\varphi +V\left(\varphi \right)-{\displaystyle \frac{1}{2}}{W}_{\mu \nu }^{+}{W}^{\mu \nu }+g_W^2{\varphi }^2{W}_{\mu }^{+}{W}^{\mu }+\\ \hfill -{\displaystyle \frac{1}{4}}{Z}_{\mu \nu }{Z}^{\mu \nu }+{\displaystyle \frac{g_Z^2}{2}}{\varphi }^2{Z}^{\mu }{Z}_{\mu }+i\overline{\psi }{\gamma }^{\mu }{D}_{\mu }\psi -g_f\varphi \overline{\psi }\psi )\sqrt{g}{d}^{4}x,\end{array}$$

(8)

where the Higgs field $\mathbf{\Phi }$ is taken in the unitary gauge $\mathbf{\Phi }=\{0,\varphi /\sqrt{2}\}$, $g=|det{g}_{\mu \nu }|$, $g_{\mu \nu }$ is space-time metric tensor, $W_{\mu \nu }={\partial }_{\mu }{W}_{\nu }-{\partial }_{\nu }{W}_{\mu }$, and $Z_{\mu \nu }={\partial }_{\mu }{Z}_{\nu }-{\partial }_{\nu }{Z}_{\mu }$. As one could see, the quantity $g_i\varphi$ plays the role of “effective mass” $m_i$ for the vector boson and fermion fields. We will use this fact in our heuristic model to calculate the vacuum energy density. Let us begin with the only field $\varphi$ described in the metric (5) by the Lagrangian density:

$$\begin{array}{c}\hfill \mathcal{L}=\left({\displaystyle \frac{1}{2}}{\partial }_{\mu }\varphi g^{\mu \nu }{\partial }_{\nu }\varphi +{\displaystyle \frac{1}{2}}{\mu }^2{\varphi }^2-{\displaystyle \frac{1}{4}}\lambda {\varphi }^4\right)\sqrt{g}=\end{array}$$

(9)

$$\begin{array}{c}\hfill ={\displaystyle \frac{a^2}{2}}\left({\varphi }^{\prime 2}-{(\nabla \varphi )}^2\right)+a^4\left({\displaystyle \frac{1}{2}}{\mu }^2{\varphi }^2-{\displaystyle \frac{1}{4}}\lambda {\varphi }^4\right),\end{array}$$

(10)

where $\mu$ (the dimension of mass) and $\lambda$ (dimensionless) are the parameters of the Higgs potential $V\left(\varphi \right)$. Hamiltonian density, corresponding to (10) takes the form

$$\mathcal{H}={\varphi }^{\prime }{\displaystyle \frac{\partial \mathcal{L}}{\partial {\varphi }^{\prime }}}-\mathcal{L}={\displaystyle \frac{a^2}{2}}\left({\varphi }^{\prime 2}+{(\nabla \varphi )}^2\right)+a^4\left({\displaystyle \frac{1}{4}}\lambda {\varphi }^4-{\displaystyle \frac{1}{2}}{\mu }^2{\varphi }^2\right).$$

(11)

The problem consists of quantizing the scalar field and calculating the mean value $\langle 0\left|\widehat{\mathcal{H}}\right|0\rangle$ of the energy density in the vacuum state. An approximate method is used here [26] to obtain the vacuum state. Let us explain it with an example of a single nonlinear oscillator. To find the approximate function of the vacuum state, the wave function of the vacuum state of a harmonic oscillator is taken, but its frequency is considered as a free parameter. Then, the energy of the nonlinear oscillator in this state is calculated, and a minimum over frequency is found. The condition of a minimum allows for determining this frequency. In a quantum field, we have a number of coupled nonlinear oscillators and need to find the minimum of the energy density over all frequencies.

We will consider a relatively slow universe expansion and write the operator of the scalar field in the form [27]

$$\widehat{\varphi }={\varphi }_0\left(\eta \right)+\sum _{\mathit{k}}{\displaystyle \frac{1}{a\sqrt{2{\omega }_{\mathit{k}}}}}\left({\widehat{\mathrm{a}}}_{\mathit{k}}{e}^{i\mathit{k}\mathbf{r}-i\int {\omega }_{\mathit{k}}\left(\eta \right)d\eta }+{\widehat{\mathrm{a}}}_{\mathit{k}}^{+}{e}^{-i\mathit{k}\mathbf{r}+i\int {\omega }_{\mathit{k}}\left(\eta \right)d\eta }\right),$$

(12)

where the creation and annihilation operators satisfy the relation $[{\widehat{\mathrm{a}}}_{\mathit{k}},{\widehat{\mathrm{a}}}_{\mathit{k}}^{+}]=1$, ${\varphi }_0$ is the condensate of the scalar field, and ${\omega }_{\mathit{k}}$ are the frequencies of the field oscillators to be determined. Let us determine a state $|0\rangle$, such that ${\widehat{\mathrm{a}}}_{\mathit{k}}|0\rangle =0$. This state is not the vacuum state, before frequencies ${\omega }_{\mathit{k}}$ is determined by minimization, and only after minimization of the energy density, this state is some approximation of the vacuum state. Calculation of the mean energy density according to (11), (12) over state $|0\rangle$ gives

$$a^4{\rho }_v=\langle 0\left|\widehat{\mathcal{H}}\right|0\rangle ={\displaystyle \frac{1}{4}}\left(\sum _{\mathit{k}}{\omega }_k+{\displaystyle \frac{k^2-a^2({\mu }^2-3\lambda {\varphi }_0^2)}{{\omega }_k}}\right)+{\displaystyle \frac{3}{16}}\lambda {\left(\sum _{\mathit{k}}{\displaystyle \frac{1}{{\omega }_k}}\right)}^2+{\displaystyle \frac{\lambda a^4}{4}}{\varphi }_0^4-{\displaystyle \frac{{\mu }^2{a}^4}{2}}{\varphi }_0^2,$$

(13)

where it is taken into account that

$$\begin{array}{c}\hfill \langle 0\left|{\widehat{\varphi }}^2\right|0\rangle ={\varphi }_0^2+{\displaystyle \frac{1}{2a^2}}S,\langle 0\left|{\widehat{\varphi }}^4\right|0\rangle ={\varphi }_0^4+{\displaystyle \frac{3}{a^2}}{\varphi }_0^{2}S+{\displaystyle \frac{3}{4a^4}}{S}^2,\\ \hfill S=\sum _{\mathit{k}}{\displaystyle \frac{1}{{\omega }_k}}.\end{array}$$

(14)

Let us find the instantaneous minimum of the expression (13) by equating the derivatives to zero: ${\displaystyle \frac{\partial {\rho }_v}{\partial {\omega }_k}}=0$, ${\displaystyle \frac{\partial {\rho }_v}{\partial {\varphi }_0}}=0$. As a result, the following equations arise

$$\begin{array}{c}\hfill {\omega }_k^2-k^2-3\lambda a^2{\varphi }_0^2+a^2{\mu }^2-{\displaystyle \frac{3\lambda S}{2}}=0,\end{array}$$

(15)

$$\begin{array}{c}\hfill {\varphi }_0^2={\displaystyle \frac{2a^2{\mu }^2-3S\lambda }{2a^2\lambda }}.\end{array}$$

(16)

Substituting (16) into (15) allows obtaining the dispersion equation for frequency:

$${\omega }_k^2=k^2+2a^2{\mu }^2-3S\lambda .$$

(17)

The last term in (17) represents the quantum contribution to the Higgs boson mass

$$m_H^2=2{\mu }^2-3S\lambda /a^2.$$

(18)

The quantity S is calculated explicitly:

$$\begin{array}{c}\hfill S\left(m_H\right)=\sum _{\mathit{k}}{\displaystyle \frac{1}{\sqrt{k^2+a^2{m}_H^2}}}={\displaystyle \frac{1}{2{\pi }^2}}{\int }_0^{k_{max}}{\displaystyle \frac{k^{2}dk}{\sqrt{k^2+a^2{m}_H^2}}}=\\ \hfill {\displaystyle \frac{1}{8{\pi }^2}}\left(2k_{max}\sqrt{k_{max}^2+a^2{m}_H^2}+2a^2{m}_H^{2}ln{\displaystyle \frac{a{m}_H}{k_{max}+\sqrt{k_{max}^2+a^2{m}_H^2}}}\right)\approx \\ \hfill {\displaystyle \frac{1}{32{\pi }^2}}\left(8k_{max}^2+4a^2{m}_H^2\left(1+2ln{\displaystyle \frac{a{m}_H}{2k_{max}}}\right)-{\displaystyle \frac{3a^4{m}_H^4}{k_{max}^2}}\right).\end{array}$$

(19)

Further, we have to take into account other particles of the standard model. In our heuristic model, we proceed only from the fact that the coupling of the particles with the Higgs field is equivalent to some “effective mass” of a particle $m_i\left(\varphi \right)=g_i\varphi$, where $g_i$ is the interaction constant. Thus, we calculate the vacuum energy density with this effective mass to obtain an additional effective potential

$$\begin{array}{c}\hfill a^4{V}_{i,\mathrm{eff}}\left(\varphi \right)=a^4{\rho }_{vi}={\displaystyle \frac{1}{4{\pi }^2}}{\int }_0^{k_{max}}{k}^2\sqrt{k^2+a^2{m}_i^2\left(\varphi \right)}dk\approx {\displaystyle \frac{1}{16{\pi }^2}}(k_{max}^4+\\ \hfill a^2{m}_i^2\left(\varphi \right)k_{max}^2+{\displaystyle \frac{a^4{m}_i^4\left(\varphi \right)}{8}}\left(1+4ln\left({\displaystyle \frac{m_i\left(\varphi \right)a}{2k_{max}}}\right)\right)).\end{array}$$

(20)

After adding this quantity to expression (11) one could see, that the potential term has the same form, but with the effective interaction constants

$$\begin{array}{c}\hfill \tilde{\lambda }(a,\varphi )=\lambda +{\displaystyle \frac{(1+4ln\left(\frac{a\varphi }{2k_{max}}\right))}{32{\pi }^2}}\left(\sum _{i\ne H}{g}_{\mathrm{bos}i}^4-g_{\mathrm{ferm}i}^4\right)+\end{array}$$

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{8{\pi }^2}}\left(\sum _{i\ne H}{g}_{\mathrm{bos}i}^{4}ln{g}_{\mathrm{bos}i}-g_{\mathrm{ferm}i}^{4}ln{g}_{\mathrm{ferm}i}\right),\end{array}$$

(21)

$$\begin{array}{c}\hfill {\tilde{\mu }}^2\left(a\right)={\mu }^2-{\displaystyle \frac{k_{max}^2}{8{\pi }^2{a}^2}}\left(\sum _{i\ne H}{g}_{\mathrm{bos}i}^2-g_{\mathrm{ferm}i}^2\right),\end{array}$$

(22)

where the sum over fermions and bosons includes all the particles except the Higgs boson. According to (21), (22), the Higgs boson mass (18) is redefined as

$$m_H^2=2{\tilde{\mu }}^2\left(a\right)-{\displaystyle \frac{3S\tilde{\lambda }}{a^2}},$$

(23)

and, with S given by (19), we come to:

$$m_H^2=2{\tilde{\mu }}^2-{\displaystyle \frac{3\tilde{\lambda }}{32{\pi }^2}}\left({\displaystyle \frac{8k_{max}^2}{a^2}}+4m_H^2\left(1+2ln{\displaystyle \frac{a{m}_H}{2k_{max}}}\right)-{\displaystyle \frac{3a^2{m}_H^4}{k_{max}^2}}\right).$$

(24)

Sum rules arise from (24) in demand that the Higgs boson mass does not depend on the scale factor and does not contain large Planck mass quantities. The quantity $2{\tilde{\mu }}^2$ in (24) contains large quantity $\sim k_{max}^2\sim M_p^2$ according to equation (22) and, besides, depends on $\sim a^{-2}$. To compensate for these effects in the Higgs mass, one needs firstly to exclude dependence on the scale factor in $\tilde{\lambda }$ by the requirement

$$\sum _{i\ne H}{g}_{\mathrm{bos}i}^4-g_{\mathrm{ferm}i}^4=0.$$

(25)

In contrast to the conventional Pauli rule, summation does not include the Higgs boson. Under this condition $\tilde{\lambda }$ does not depend on the scale factor now, we could exclude the dependence on $a^{-2}$ in (24) by the demand

$${\displaystyle \frac{1}{3}}\left(\sum _{i\ne H}{g}_{\mathrm{bos}i}^2-g_{\mathrm{ferm}i}^2\right)=-\tilde{\lambda }.$$

(26)

The factor $1/3$ originates from the coefficient of the $k_{max}^2/a^2$ term in Eq. (24). Taking into account that the condensate field ${\varphi }_0$ is given by (16), but with the tilded quantities, the Higgs boson mass could be rewritten as

$$m_H^2=2{\varphi }_0^2\tilde{\lambda }.$$

(27)

Multiplying (26) by ${\varphi }_0^2$ leads to

$${\displaystyle \frac{1}{3}}\left(\sum _{i\ne H}{m}_{\mathrm{bos}i}^2-m_{\mathrm{ferm}i}^2\right)=-m_H^2/2,$$

(28)

instead of the original Pauli sum rule (2). With this placement, expanding Eq. (28) over the heavy Standard-Model content (top quark, W, and Z) reproduces Eq. (32) below. Here we assume heuristically (more rigorously, we have to obtain dispersion equations for every particle) that $m_i\equiv g_i{\varphi }_0$. Analogously, multiplication (25) by ${\varphi }_0^4$ gives

$$\sum _{i\ne H}{m}_{\mathrm{bos}i}^4-m_{\mathrm{ferm}i}^4=0.$$

(29)

This sum rule replaces the Pauli sum rule (3). The sum rules (28), (29) arise as conditions to exclude dependence of the Higgs boson mass on a large quantity $k_{max}\sim M_p$ and on the scale factor a.

It is important to state the mathematical scope of these two relations. If the unknown sector contains bosonic degrees of freedom with $x_j=m_j^2$ and degeneracies $d_j$, then Eqs. (28) and (29) fix only two moments, schematically ${\sum }_j{d}_j{x}_j=C_2$ and ${\sum }_j{d}_j{x}_j^2=C_4$, after the known standard model contributions are subtracted. For more than two independent mass levels, these moment constraints do not reconstruct a unique spectrum. Consequently, a lower bound on the number of new degrees of freedom or a statement about individual masses requires an additional ansatz concerning spins, degeneracies, the presence or absence of new fermions and scalars. In Sec. IV we make this ansatz explicit and presenting numerical examples.

However, we are not able to exclude the dependence on $lna$ in (24) completely here, although this dependence is relatively weak, because $|lna|\ll |ln{\displaystyle \frac{m_H}{k_m}}|$ for not very small a. Possibly, the potential with quadratic and quartic terms only is too oversimplified and, at least, the term ${\varphi }^{4}ln\varphi$ has to be included initially, but this strongly complicates calculations. It should also be noted that the last term in (24) is strongly suppressed, because it contains $m_H^2/M_p^2$ if $k_{max}\sim M_p$ [25].

Thus we have the general equations (13)-(18), but have to imply tilded quantities $\tilde{\mu }$ and $\tilde{\lambda }$ there, because the contribution of the other particles has been taken into account. Let us calculate the energy density using (13). Expressing ${\tilde{\mu }}^2$ from (24), ${\varphi }_0^2$ from (27) and substituting them to (13) gives

$$\begin{array}{c}\hfill a^4{\rho }_v={\displaystyle \frac{k_{max}^4}{16{\pi }^2}}\left(1-{\displaystyle \frac{3\tilde{\lambda }}{16{\pi }^2}}\right)-{\displaystyle \frac{a^2{k_{max}}^2{m}_H^2\left(6\tilde{\lambda }ln\left(\frac{a{m}_H}{4k_{max}}\right)+3\tilde{\lambda }+8{\pi }^2\right)}{256{\pi }^4}}-\\ \hfill {\displaystyle \frac{a^4{m}_H^4}{1024{\pi }^4}}(12\tilde{\lambda }ln\left({\displaystyle \frac{2m_H}{k_{max}}}\right)ln\left({\displaystyle \frac{8m_H}{a^2{k}_{max}}}\right)+12\tilde{\lambda }{ln}^2\left({\displaystyle \frac{2}{a}}\right)+\\ \hfill 12\tilde{\lambda }ln\left({\displaystyle \frac{a{m}_H}{4k_{max}}}\right)+64{\pi }^{2}ln\left({\displaystyle \frac{a{k}_{max}}{2m_H}}\right)+3\tilde{\lambda }+40{\pi }^2+{\displaystyle \frac{64{\pi }^4}{\tilde{\lambda }}}).\end{array}$$

(30)

This expression for the energy density does not produce any sum rules. First term in (30) does not contribute to the universe expansion in a framework of FVT gravity [24,25]. Next term $\sim a^2{k_{max}}^2{m}_H^2$ turns to zero when

$$\tilde{\lambda }=-{\displaystyle \frac{8{\pi }^2}{3\left(2ln\left(\frac{a}{4}\right)+2ln\left(\frac{m_H}{k_{max}}\right)+1\right)}}.$$

(31)

The value of $\tilde{\lambda }$ can be chosen arbitrarily, as the value of $\lambda$ from (21) is not determined by any observations, so no sum rules arise from (31). However, this $\tilde{\lambda }$ depends on $lna$ in contrast to equation (21) under equation (25). This is a flaw of our heuristic theory, which, it seems, can be trusted up to a logarithm a. It should be noted that we use a comoving momentum cutoff of the order of the Planck mass here. Sometimes, a physical momentum cut off $p_{max}=k_{max}/a$ is used [28], which is more appropriate for the renormalization procedure.

4. New Bosons

4.1. Scope and Assumptions of the Minimal Bosonic Solutions

The Higgs boson has a mass $m_H\simeq 125\mathrm{GeV}$, while the heaviest Standard-Model states relevant for the present mass moments are the top quark, $m_t=173.2\mathrm{GeV}$, the charged vector boson, $m_W=80.4\mathrm{GeV}$, and the neutral vector boson, $m_Z=91.2\mathrm{GeV}$. In the numerical estimate, lighter Standard-Model masses are neglected. This approximation is particularly well motivated for the fourth moment, but it should be kept in mind when interpreting sharp integer thresholds.

To obtain transparent benchmark solutions of Eqs. (28) and (29), we now adopt a deliberately restricted ansatz: (i) the unknown contribution is purely bosonic; (ii) no additional fermions are introduced; (iii) the new bosonic degrees of freedom are grouped into two mass levels $m_A$ and $m_B$ with non-negative integer weights $N_A$ and $N_B$; and (iv) these weights count physical degrees of freedom according to Eq. (7). The resulting statements about the minimal number of states are therefore conditional on this ansatz and are not general theorems about arbitrary BSM spectra.

The two modified sum rules then give

$$\begin{array}{c}\hfill {\displaystyle \frac{3}{2}}{m}_H^2+N_A{m}_A^2+N_B{m}_B^2+6m_W^2+3m_Z^2=12m_t^2,\end{array}$$

(32)

$$\begin{array}{c}\hfill N_A{m}_A^4+N_B{m}_B^4+6m_W^4+3m_Z^4=12m_t^4.\end{array}$$

(33)

Within this two-mass, boson-only ansatz, real positive solutions exist only when $N_A+N_B\ge 8$. This threshold is not merely a numerical accident of the two-level parametrization: it follows from the Cauchy–Schwarz inequality. Writing the right-hand sides of Eqs. (32) and (33) after subtracting the known Standard-Model contributions as ${\sum }_j{N}_j{m}_j^2=C_2$ and ${\sum }_j{N}_j{m}_j^4=C_4$, with $N_j\ge 0$ and $m_j^2\ge 0$, one has

$$C_2^2={\left(\sum _j{N}_j{m}_j^2\right)}^2\le \left(\sum _j{N}_j\right)\left(\sum _j{N}_j{m}_j^4\right)=(N_A+N_B)C_4,$$

so that $N_A+N_B\ge C_2^2/C_4$. With the masses adopted here, $C_2^2/C_4\simeq 7.2$, hence $N_A+N_B\ge 8$. Hence, the robust statement from the present calculation is that at least eight additional bosonic degrees of freedom are required in this specific minimal ansatz. It does not imply eight distinct particle species, nor does it imply that every new mass must exceed $m_H$. Indeed, for $N_A=7$ and $N_B=1$, the two branches are $m_A=173\mathrm{GeV}$, $m_B=253\mathrm{GeV}$ and $m_A=196\mathrm{GeV}$, $m_B=63\mathrm{GeV}$. The second branch explicitly demonstrates why a universal claim that all new bosons are heavier than the Higgs would be too strong.

For comparison, the original Pauli sum rules (2), (3):

$$\begin{array}{c}\hfill m_H^2+N_A{m}_A^2+N_B{m}_B^2+6m_W^2+3m_Z^2=12m_t^2,\end{array}$$

(34)

$$\begin{array}{c}\hfill m_H^4+N_A{m}_A^4+N_B{m}_B^4+6m_W^4+3m_Z^4=12m_t^4,\end{array}$$

(35)

which have two solutions at $N_A=7$, $N_B=1$: $m_A=179GeV$, $m_B=236GeV$ or $m_A=195GeV$, $m_B=121GeV$.

4.2. Scalar-Sector Illustrations

A natural question is how the required additional bosonic degrees of freedom might be realized in extensions of the standard model [29,30]. A two-Higgs-doublet model adds four physical scalar states relative to the standard model [31]; therefore, by itself, it does not saturate the present $N_A+N_B\ge 8$ benchmark. Likewise, the six real degrees of freedom of an additional complex scalar triplet in the minimal type-II seesaw construction [32] do not by themselves saturate this restricted benchmark. These observations should not be read as exclusions of the corresponding quantum field theories: additional particles can alter the moment balance.

With two additional scalar doublets, a three-Higgs-doublet model contains eight additional physical scalar degrees of freedom relative to the standard model. It can therefore realize the minimal counting used here. As an illustrative two-mass grouping, $N_A=N_B=4$ in Eqs. (32) and (33) gives $m_A\simeq 151\mathrm{GeV}$ and $m_B\simeq 213\mathrm{GeV}$. This establishes compatibility with the two mass-moment constraints at the level of the present ansatz; it is not, by itself, a complete phenomenological validation of a specific 3HDM potential, vacuum structure, or collider benchmark [33].

4.3. $SU{\left(3\right)}_F$-Based Vector Realizations: Local Family Symmetry and a Hidden-Sector Interpretation

The sum-rule analysis motivates additional bosonic degrees of freedom, but it does not select a unique gauge completion. The $SU\left(3\right)$-based constructions below should therefore be viewed as illustrative realizations rather than unique completions.

Flavor symmetry of quark and lepton generations can be described by the $SU{\left(3\right)}_F$ group, which has eight generators ${\mathit{t}}_a$. To introduce this symmetry locally, eight vector fields are required, with a covariant derivative

$$D_{\mu }={\partial }_{\mu }+i{g}_F{\mathit{t}}_a{\mathcal{A}}_{\mu }^a.$$

(36)

Family gauge bosons generically induce flavor-changing currents specifying the smallness of $g_F$ according to the experimental restrictions. Sometimes, $SU{\left(3\right)}_F$ local symmetry is used to explain neutrino masses and hierarchy of quark masses [34,35]. For this aim two additional Higgs field, Yukawa couplings of these fields to quark and leptons and additional heavy fermions are introduced [34]. These studies find symmetry-breaking scales of order $300\mathrm{TeV}$ and lightest family-gauge-boson masses of order $100\mathrm{TeV}$. Here, our only aim is to complete the Standard Model relative sum rules. Thus, it is sufficient to introduce one additional complex scalar triplet ${\mathsf{\Phi }}_F$ with a nonzero vacuum expectation value, which realizes the breaking pattern $SU{\left(3\right)}_F\to SU{\left(2\right)}_F$. Therefore, five gauge bosons become massive, three remain massless, five of the six real scalar degrees of freedom are eaten as Goldstone modes, and one physical scalar remains. In the present sum-rule bookkeeping, this corresponds to $N_a=15$ for the five massive vectors and $N_b=1$ for the extra scalar. The illustrative solution of Eqs. (32) and (33) then gives approximately $m_a\simeq 110\mathrm{GeV}$ and $m_b\simeq 300\mathrm{GeV}$.

A distinction must be made between a genuine local family symmetry $SU{\left(3\right)}_F$ acting on the known quark and lepton generations and a hidden non-Abelian symmetry $SU{\left(3\right)}_X$ under which Standard-Model fields are neutral, and communication with the visible sector can proceed through a Higgs portal. A minimal scalar potential illustrating this possibility is

$$\begin{array}{cc}\hfill V({\mathsf{\Phi }}_H,{\mathsf{\Phi }}_X)=& -{\mu }_H^2{\mathsf{\Phi }}_H^{†}{\mathsf{\Phi }}_H+{\lambda }_H{\left({\mathsf{\Phi }}_H^{†}{\mathsf{\Phi }}_H\right)}^2-{\mu }_X^2{\mathsf{\Phi }}_X^{†}{\mathsf{\Phi }}_X+{\lambda }_X{\left({\mathsf{\Phi }}_X^{†}{\mathsf{\Phi }}_X\right)}^2\hfill \\ & +{\lambda }_{HX}\left({\mathsf{\Phi }}_H^{†}{\mathsf{\Phi }}_H\right)\left({\mathsf{\Phi }}_X^{†}{\mathsf{\Phi }}_X\right),\hfill \end{array}$$

(37)

where ${\mathsf{\Phi }}_X$ generates the hidden-vector masses and ${\lambda }_{HX}$ mixes the hidden mode with the observed Higgs sector. In this interpretation, the five massive vectors are candidates for vector dark matter and the three massless gauge bosons are dark-radiation-like degrees of freedom. More complete hidden-vector constructions, including fully broken $SU{\left(3\right)}_X$ models with a richer scalar sector are known [36,37]. Certainly, any Higgs-portal realization must also satisfy existing Higgs signal-strength and invisible-decay constraints [38,39].

4.4. Electroweak-Higgs-Linked Vector Masses as an Effective Ansatz

Let us finally consider a deliberately phenomenological possibility in which vector masses are linked to the electroweak Higgs field through terms of the form

$$\Delta {\mathcal{L}}_{\mathrm{eff}}=\sum _a{\displaystyle \frac{g_a^2}{2}}{\varphi }^2{\mathcal{A}}_{\mu }^a{\mathcal{A}}^{a\mu }.$$

(38)

For a genuine local $SU{\left(3\right)}_F$ gauge symmetry the Standard-Model Higgs does not automatically carry family charge, thus equation (38) violates this symmetry explicitly.

Within the present vector-only two-mass ansatz, no solution with fewer than three massive vector bosons is found, because $N_A+N_B\le 6$. As a benchmark with three massive vectors, take $N_A=6$ (two vectors of equal mass, $2\times 3$ degrees of freedom) and $N_B=3$ (one vector with a different mass). Eqs. (32) and (33) give two branches,

$$(m_a,m_b)=(140,227)\mathrm{GeV},(m_a,m_b)=(202,95)\mathrm{GeV},$$

In the first branch, all three massive vectors lie above the Higgs mass, while five additional vectors are taken massless. A heavy-vector decay into Higgs boson and two massless vector (“Higgs from nothing”) are suppressed by need of ${\mathcal{A}}^a{\mathcal{A}}^b{\mathcal{A}}^b$ vertex, that is by smallness of $g_F$.

The second branch imply an invisible on-shell decay of the $125\mathrm{GeV}$ Higgs. The diagonal interaction in Eq. (38) would generate an $h{\mathcal{A}}^a{\mathcal{A}}^a$ vertex, for which the two-body channel $h\to {\mathcal{A}}^b{\mathcal{A}}^b$ requires $2m_b<m_H$; this condition is not satisfied for $m_b=95\mathrm{GeV}$. A decay into one massive and two massless vector would again require an additional vertex ${\mathcal{A}}^a{\mathcal{A}}^b{\mathcal{A}}^b$. Thus, Higgs decay “into nothing” is suppressed.

However, let us take $N_a=15$ (five vectors of equal mass) and $N_b=6$ (two vector with another mass). One vector remains massless. The unique solution is

$$m_a=40\mathrm{GeV},m_b=203\mathrm{GeV},$$

which allows standard model Higgs decay into two invisible vector $40.3\mathrm{GeV}$ particle. Thus, unsuppressed decay of Higgs “into nothing” is still possible. Concerning another values of $N_a$ and $N_b$, for instance, $N_a=18$ (six vectors of equal mass) and $N_b=6$ (two vector with another mass) give

$$m_a=37\mathrm{GeV},m_b=204\mathrm{GeV},$$

whereas $N_a=15$ (five vectors of equal mass), $N_b=9$ (three vector with another mass) or $N_a=12$ (four vectors of equal mass), $N_b=9$ (three vector with another mass) give no solution at all.

5. Conclusion

Within the FVT/Higgs framework developed here, the conventional Pauli mass-moment relations are replaced by Eqs. (28) and (29), in which the physical Higgs contribution is treated explicitly. These two relations constrain the second and fourth moments of the particle-mass spectrum, although do not determine a unique BSM spectrum.

When the unknown contribution is restricted to additional bosonic states grouped into two mass levels, with no new fermions and with the heaviest Standard-Model states retained in the numerical balance, real positive solutions require $N_A+N_B\ge 8$. The resulting statement is therefore conditional but nontrivial: within this minimal ansatz, at least eight additional bosonic degrees of freedom are required. It should not be reformulated as a universal prediction of eight distinct particles or as a lower bound $m_i>m_H$ on every new mass. A three-Higgs-doublet model provides a successful illustrative scalar realization of the required counting.

The $SU\left(3\right)$ examples further illustrate sum-rule bookkeeping and a model building. The benchmark with seven or eight massive vectors predicts unsuppressed Higgs decay “into nothing”.