# Linking axion-like dark matter, the XENON1T excess, inflation and the tiny
active neutrino masses

H. N. Longa,b<EMAIL_ADDRESS>D. V. Soac<EMAIL_ADDRESS>V. H.
Binhd<EMAIL_ADDRESS>A. E. Cárcamo Hernándeze<EMAIL_ADDRESS>a
Theoretical Particle Physics and Cosmology Research Group, Advanced Institute
of Materials Science, Ton Duc Thang University, Ho Chi Minh City, Vietnam
bFaculty of Applied Sciences, Ton Duc Thang University, Ho Chi Minh City,
Vietnam
c Faculty of Natural Sciences and Technology, Hanoi Metropolitan University,
98 Duong Quang Ham, Cau Giay, Hanoi, Vietnam
d Institute of Physics, VAST, 10 Dao Tan, Ba Dinh, Hanoi, Vietnam
e Departamento de Física, Universidad Técnica Federico Santa María,
Casilla 110-V, Valparaíso, Chile

(August 27, 2024)

###### Abstract

We consider a renormalizable theory, which successfully explains the number of
Standard Model (SM) fermion families and whose non-SM scalar sector includes
an axion dark matter as well as a field responsible for cosmological
inflation. In such theory, the axion gets its mass via radiative corrections
at one-loop level mediated by virtual top quark, right handed Majorana
neutrinos and SM gauge bosons. Its mass is obtained in the range $4$ keV$\div$
$40$ keV, consistent with the one predicted by XENON1T experiment, when the
right handed Majorana neutrino mass is varied from $100$ GeV up to $350$ GeV,
thus implying that the light active neutrino masses are generated from a low
scale type I seesaw mechanism. Furthermore, the theory under consideration can
also successfully accommodates the XENON1T excess provided that the PQ
symmetry is spontaneously broken at the $10^{10}$ GeV scale.

###### pacs:

11.30.Fs, 12.15.Ff, 12.60.-i

##  I I. Introduction

Currently, an axion is a very attractive subject in Particle Physics in both
theoretical and experimental aspects sikivie ; axiondm , thus providing a
motivation to consider extensions of the Standard Model that include this
particle in its field content. The axion is a CP-odd scalar field which arises
in the solution of the strong-CP problem, and originally it is always massless
particle; and in order to generate a mass for the axion, one can consider the
implementation of radiative corrections as shown in rm or gravitational
effects peigra . It is interesting to note that nowadays the axion is widely
considered as a candidate of dark matter (DM)dmaxion . The dark matter
candidate is existed only in some beyond standard models. Among the SM
extensions, the models based on the
$\mbox{SU(3)}_{C}\times\mbox{SU(3)}_{L}\times\mbox{U(1)}_{X}$ gauge symmetries
(called 3-3-1 models, for short) flt have several very interesting features,
some of them being, the natural explanation of the number of SM fermion
families, the electric charge quantization and the solution of the strong CP
problem from the PQ symmetry pal , which is automatically fulfilled in the
3-3-1 models. In one of the 3-3-1 models, there exist both interesting
features, namely the axion dark matter candidate and inflaton for Early
Universe ax1 ; ax2 . In the above mentioned papers, the axion gets mass only
from gravitational contribution.

It is worth mentioning that the CP-odd sector of the 3-3-1 model with axion
has been considered in ax1 ; ax2 ; ax3 . However, the rotation matrix that
diagonalizes the squared mass matrix for the CP odd neutral scalar fields
given in ax1 , which was obtained by Mathematica is not unitary, and thus such
mixing matrix cannot be used for further studies such coupling constants,
collider phenomenology, etc. The aim of this work is to reconsider the CP-odd
scalar sector of the aforementioned model with the inclusion of radiative
corrections mediated by virtual top quark, right handed Majorana neutrinos and
SM gauge bosons, which in turn generate the mass of the axion. Such radiative
corrections to the axion mass where not considered in ax1 . Our paper is
organized as follows: in Section II II we present brief review of the model.
Section III III is devoted to discrete and PQ symmetries needed for the axion
existence. The Higgs potential and the resulting physical scalar spectrum is
discussed in Section IV IV. In Section VV, we present radiative correction for
the axion mass. Finally, we state our conclusions in Section VI VI.

##  II II. Review of the model

The model under consideration is based on
$\mbox{SU(3)}_{C}\times\mbox{SU(3)}_{L}\times\mbox{U(1)}_{X}\times
Z_{11}\times Z_{2}$ and has the following fermion content:

$\displaystyle\psi_{aL}=\left(\nu_{a},l_{a},(\nu^{c}_{R})^{a}\right)_{L}^{T}\sim\left(1,3,-1/3\right),\hskip
8.5359ptl_{aR}\sim\left(1,1,-1\right),$ $\displaystyle
N_{aR}\sim(1,1,0),Q_{3L}=\left(u_{3},d_{3},U\right)_{L}^{T}\sim\left(3,3,1/3\right),$
$\displaystyle Q_{\alpha
L}=\left(d_{\alpha},-u_{\alpha},D_{\alpha}\right)_{L}^{T}\sim\left(3,3^{\ast},0\right),$
$\displaystyle u_{aR},U_{R}\sim\left(3,1,2/3\right),\hskip
8.5359ptd_{aR},D_{\alpha R}\sim(3,1,-1/3),$ (1)

where $\alpha=1,2$ and $a=\\{3,\alpha\\}$ are family indices. The $U$ and $D$
are exotic quarks with ordinary electric charges, whereas $N_{aR}$ are right-
handed neutrinos.

The scalar sector of the model is composed of three $SU(3)_{L}$ scalar
triplets and one $SU(3)_{L}$ singlet scalar field. They have the following
transformations under the
$\mbox{SU(3)}_{C}\times\mbox{SU(3)}_{L}\times\mbox{U(1)}_{X}$ symmetry:

$\displaystyle\chi^{T}$ $\displaystyle=$
$\displaystyle\left(\chi^{0}_{1},\chi^{-}_{2},\chi_{3}^{0}\right)\sim\left(1,3,-\frac{1}{3}\right),\,\eta^{T}=\left(\eta^{0}_{1},\eta^{-}_{2},\eta_{3}^{0}\right)$
$\displaystyle\rho^{T}$ $\displaystyle=$
$\displaystyle\left(\rho^{+}_{1},\rho_{2}^{0},\rho_{3}^{+}\right)\sim\left(1,3,\frac{2}{3}\right),\,\hskip
8.5359pt\phi\sim(1,1,0)\,.$ (2)

To generate masses for gauge bosons and fermions, the scalar fields should
acquire vacuum expectation values (VEVs). These fields can be expanded around
the minimum as follows

$\displaystyle\chi$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}\frac{1}{\sqrt{2}}(R_{\chi}^{1}+iI_{\chi}^{1})\\\
\chi^{-}\\\ \frac{1}{\sqrt{2}}(v_{\chi}+R_{\chi}^{3}+iI_{\chi}^{3})\\\
\end{array}\right),\,\eta=\left(\begin{array}[]{c}\frac{1}{\sqrt{2}}(v_{\eta}+R_{\eta}^{1}+iI_{\eta}^{1})\\\
\eta^{-}\\\ \frac{1}{\sqrt{2}}(R_{\eta}^{3}+iI_{\eta}^{3})\\\
\end{array}\right),$ (9) $\displaystyle\rho$ $\displaystyle=$
$\displaystyle\left(\begin{array}[]{c}\rho_{1}^{+}\\\
\frac{1}{\sqrt{2}}(v_{\rho}+R_{\rho}+iI_{\rho})\\\ \rho^{+}_{3}\\\
\end{array}\right),\phi=\frac{1}{\sqrt{2}}\left(v_{\phi}+R_{\phi}+iI_{\phi}\right).$
(13)

Note that since $\phi$ carries non-zero PQ charge (as shown below), it has to
be complex as shown in (13). The VEV $v_{\chi}$ is responsible for the first
stage of gauge symmetry breaking, whereas $v_{\eta},v_{\rho}$ trigger the
second stage of electroweak symmetry breaking.

##  III III. Discrete and Peccei-Quinn symmetries

In order to keep intact the physics results of the Ref.ax1 the Lagrangian of
the model must be invariant by the discrete symmetries $Z_{11}\times Z_{2}$
which are summarised in Table 1. Here we have used a notation
$\omega_{k}\equiv e^{i2\pi\frac{k}{11}},k=0,\pm 1\cdots\pm 5$.

| $Q_{\alpha L}$ | $Q_{3L}$ | $u_{aR}$ | $d_{aR}$ | $U_{3R}$ | $D_{\alpha R}$ | $\psi_{aL}$ | $l_{aR}$ | $N_{aR}$ | $\eta$ | $\chi$ | $\rho$ | $\phi$
---|---|---|---|---|---|---|---|---|---|---|---|---|---
$SU(3)_{C}$ | $\mathbf{3}$ | $\mathbf{3}$ | $\mathbf{3}$ | $\mathbf{3}$ | $\mathbf{3}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{1}$
$SU(3)_{L}$ | $\overline{\mathbf{3}}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{3}$ | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{3}$ | $\mathbf{3}$ | $\mathbf{3}$ | $\mathbf{1}$
$U(1)_{X}$ | $0$ | $\frac{1}{3}$ | $\frac{2}{3}$ | $-\frac{1}{3}$ | $\frac{2}{3}$ | $-\frac{1}{3}$ | $-\frac{1}{3}$ | $-1$ | $0$ | $-\frac{1}{3}$ | $-\frac{1}{3}$ | $\frac{2}{3}$ | $0$
$Z_{11}$ | $\omega^{-1}_{4}$ | $\omega_{0}$ | $\omega_{5}$ | $\omega_{2}$ | $\omega_{3}$ | $\omega_{4}$ | $\omega_{1}$ | $\omega_{3}$ | $\omega^{-1}_{5}$ | $\omega^{-1}_{5}$ | $\omega^{-1}_{3}$ | $\omega^{-1}_{2}$ | $\omega^{-1}_{1}$
$Z_{2}$ | $1$ | $1$ | $-1$ | $-1$ | $-1$ | $-1$ | $1$ | $-1$ | $1$ | $1$ | $-1$ | $-1$ | $1$

Table 1: $SU(3)_{C}\times SU(3)_{L}\times U(1)_{X}\times Z_{11}\times Z_{2}$
charge assignments of the particle content of the model. Here $a=1,2,3$ and
$\alpha=1,2$.

Let us discuss about the PQ symmetry. These discrete symmetries yield the
following Yukawa interactions

$\displaystyle-\mathcal{L}^{Y}=y_{1}\bar{Q}_{3L}U_{3R}\chi+\sum\limits_{\alpha,\beta=1}^{2}\left(y_{2}\right)_{\alpha\beta}\bar{Q}_{\alpha
L}D_{\beta R}\chi^{\ast}$
$\displaystyle+\sum\limits_{a=1}^{3}\left(y_{3}\right)_{3a}\bar{Q}_{3L}u_{aR}\eta+\sum\limits_{\alpha=1}^{2}\sum\limits_{a=1}^{3}\left(y_{4}\right)_{\alpha
a}\bar{Q}_{\alpha L}d_{aR}\eta^{\ast}$
$\displaystyle+\sum\limits_{a=1}^{3}\left(y_{5}\right)_{3a}\bar{Q}_{3L}d_{aR}\rho+\sum\limits_{\alpha=1}^{2}\sum\limits_{a=1}^{3}\left(y_{6}\right)_{\alpha
a}\bar{Q}_{\alpha L}u_{aR}\rho^{\ast}$
$\displaystyle+\sum\limits_{a=1}^{3}\sum\limits_{b=1}^{3}g_{ab}\bar{\psi}_{aL}e_{bR}\rho+\sum\limits_{a=1}^{3}\sum\limits_{b=1}^{3}\left(y_{\nu}^{D}\right)_{ab}\bar{\psi}_{aL}\eta
N_{bR}$
$\displaystyle+\sum\limits_{a=1}^{3}\sum\limits_{b=1}^{3}\left(y_{N}\right)_{ab}\phi\bar{N}_{aR}^{C}N_{bR}+\mbox{H.c.}.$
(14)

where unlike Ref. ax1 , the neutrino Yukawa term
$y^{(\rho)}_{ab}\left(\bar{\psi}_{aL}\right)\left(\psi_{bL}\right)\left(\rho^{*}\right)$
is not present since that such term is not invariant under the $Z_{2}$
symmetry. It is worth mentioning that the Yukawa interactions given above do
not allow for terms which interchange $\chi\leftrightarrow\eta$, since they do
not respect the $Z_{11}\times Z_{2}$ symmetry.

Assuming fermions of opposite chiralities have opposite PQ charges and
$X_{d}=X_{D}=1$ we summarise PQ charges of fermions in Table 2.

| $u_{aL}$ | $d_{aL}$ | $U_{L}$ | $D_{L}$ | $l_{a}$ | $l_{aR}$ | $\nu_{aL}$ | $\nu_{aR}$ | $N_{aR}$
---|---|---|---|---|---|---|---|---|---
$X_{PQ}$ | $-1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $1$ | $-1$ | $1$
Table 2: PQ charges of fermions in the model

It is now clear that the entire Lagrangian of the model is $U_{PQ}(1)$
invariant, providing a natural solution to the strong-CP problem.

The following remarks are in order

i) The discrete symmetry $Z_{N}$ can naturally be accommodated when it has
enough number of fields in its spectrum.
ii) If the $Z_{11}\times Z_{2}$ symmetry is imposed, then the PQ symmetry
automatically appears in the model. Therefore, the CP problem can be solved by
the dynamical properties of the axion field, which is a physical scalar field
belonging to the CP odd neutral scalar sector.
iii) From the last two terms of Eq. (14), it follows that the tiny masses for
the light active neutrinos are generated from a type I seesaw mechanism
mediated by right handed Majorana neutrinos, thus implying that the resulting
light active neutrino mass matrix has the form:

$M_{\nu}=M^{D}_{\nu}M^{-1}_{N}\left(M^{D}_{\nu}\right)^{T},\,M^{D}_{\nu}=y^{D}_{\nu}\frac{v_{\eta}}{\sqrt{2}},\,M_{N}=y_{N}\frac{v_{\phi}}{\sqrt{2}}.$
(15)

##  IV IV. Higgs potential

The model scalar potential takes the form:

$\displaystyle
V=\mu^{2}_{\phi}\phi^{*}\phi+\mu_{\chi}^{2}\chi^{\dagger}\chi+\mu_{\rho}^{2}\rho^{\dagger}\rho+\mu_{\eta}^{2}\eta^{\dagger}\eta+\lambda_{1}(\chi^{\dagger}\chi)^{2}+\lambda_{2}(\eta^{\dagger}\eta)^{2}$
$\displaystyle+\lambda_{3}(\rho^{\dagger}\rho)^{2}+\lambda_{4}(\chi^{\dagger}\chi)(\eta^{\dagger}\eta)+\lambda_{5}(\chi^{\dagger}\chi)(\rho^{\dagger}\rho)+\lambda_{6}(\eta^{\dagger}\eta)(\rho^{\dagger}\rho)$
$\displaystyle+\lambda_{7}(\chi^{\dagger}\eta)(\eta^{\dagger}\chi)+\lambda_{8}(\chi^{\dagger}\rho)(\rho^{\dagger}\chi)+\lambda_{9}(\eta^{\dagger}\rho)(\rho^{\dagger}\eta)$
$\displaystyle+\lambda_{10}(\phi^{*}\phi)^{2}+\lambda_{11}(\phi^{*}\phi)(\chi^{\dagger}\chi)+\lambda_{12}(\phi^{*}\phi)(\rho^{\dagger}\rho)$
$\displaystyle+\lambda_{13}(\phi^{*}\phi)(\eta^{\dagger}\eta)+\left(\lambda_{\phi}\epsilon^{ijk}\eta_{i}\rho_{j}\chi_{k}\phi+H.c.\right)$
(16)

The VEV $v_{\phi}$ is responsible for PQ symmetry breaking resulting in
existence of invisible axion due to very high scale around $10^{10}-10^{11}$
GeV. Then $\mbox{SU(3)}_{L}\times\mbox{U(1)}_{X}$ breaks to the SM group by
$v_{\chi}$ and two others $v_{\rho},v_{\eta}$ are needed for the usual
$\mbox{U(1)}_{Q}$ symmetry. Hence $v_{\phi}\gg v_{\chi}\gg v_{\rho},v_{\eta}$.
The constraint conditions of such scalar potential were analyzed in Ref. ax1 .

From an analysis of the scalar potential, we find that the physical CP odd
neutral scalar mass eigenstates are:

$\left(\begin{array}[]{c}G_{Z^{\prime}}\\\ A_{5}\\\ G_{Z}\\\ a\\\
\end{array}\right)=\left(\begin{array}[]{cccc}\cos\theta_{3}&-\sin\theta_{3}&0&0\\\
\sin\theta_{3}\cos\theta_{4}\cos\theta_{\phi}&\cos\theta_{3}\cos\theta_{4}\cos\theta_{\phi}&-\sin\theta_{4}\cos\theta_{\phi}&\sin\theta_{\phi}\\\
-\sin\theta_{3}\sin\theta_{4}&\sin\theta_{4}\cos\theta_{3}&\cos\theta_{4}&0\\\
-\sin\theta_{3}\sin\theta_{\phi}\cos\theta_{4}&-\sin\theta_{\phi}\cos\theta_{3}\cos\theta_{4}&-\sin\theta_{4}\sin\theta_{\phi}&\cos\theta_{\phi}\\\
\end{array}\right)\left(\begin{array}[]{c}I_{\chi}^{3}\\\ I_{\eta}^{1}\\\
I_{\rho}\\\ I_{\phi}\\\ \end{array}\right)\,,$ (17)

where the mixing angles in the CP odd scalar sector take the form:

$\displaystyle\tan\alpha$ $\displaystyle=$
$\displaystyle\frac{v_{\eta}}{v_{\rho}},\hskip
28.45274pt\tan\theta_{3}=\frac{v_{\eta}}{v_{\chi}},$ (18)
$\displaystyle\tan\theta_{\phi}$ $\displaystyle=$
$\displaystyle\frac{v_{\rho}v_{\eta}}{v_{\phi}\sqrt{v_{\rho}^{2}+v^{2}_{\eta}}}\approx\frac{v_{\rho}}{v_{\phi}}\,.\hskip
28.45274pt\tan\theta_{4}\approx\tan\alpha.$ (19)

The mixing angles in the CP odd scalar sector depend on the ratio of
$v_{\eta}$ to $v_{\rho},v_{\chi}$ and $v_{\phi}$.

By rotating the scalar fields to the physical basis, considering the limit
$v_{\chi}\gg v_{\eta},v_{\rho}$ and a scenario close to the decoupling limit
$v_{\eta}\gg v_{\rho}$, it follows that the scalar fields of the model under
consideration can be expressed as follows Hernandez:2013hea ;
CarcamoHernandez:2019iwh :

$\displaystyle\chi\simeq\left(\begin{array}[]{c}G_{X^{0}}\\\ G_{Y^{-}}\\\
\frac{1}{\sqrt{2}}\left(v_{\chi}+H_{\chi}+iG_{Z^{\prime}}\right)\\\
\end{array}\right),\,\eta\simeq\left(\begin{array}[]{c}\frac{1}{\sqrt{2}}\left(v_{\eta}+h+iG_{Z}\right)\\\
G_{W^{-}}\\\ \varphi^{0}\\\ \end{array}\right),$ (26)
$\displaystyle\rho\simeq\left(\begin{array}[]{c}H_{1}^{+}\\\
\frac{1}{\sqrt{2}}\left(v_{\rho}+h_{5}+iA_{5}\right)\\\ H_{2}^{+}\\\
\end{array}\right),\,\phi\simeq\frac{1}{\sqrt{2}}\left(v_{\phi}+\Phi+ia\right)\,.$
(30)

The scalar spectrum of the model has the following features:
1) In the charged scalar sector, one has four singly charged fields: two of
them are massless being the Goldstones bosons eaten by the longitudinal
components of the bilepton gauge bosons $W^{\pm}$ and $Y^{\pm}$, whereas the
physical charged scalar fields have masses at the TeV scale.
2) In the CP-odd scalar sector, there are six fields: two massless Goldstones
eaten by the longitudinal components of the $Z$ and $Z^{\prime}$ gauge bosons
and one massless is part of Goldstones boson associated to the longitudinal
components of the neutral bilepton $G_{X^{0}}$ gauge boson. One massive CP-odd
field has the same mass as another in CP-even sector. Hence the above pair is
a complex bilepton scalar field denoted by $\varphi^{0}$. Another field
associated with the singlet $\phi$ is identified to the axion $a$. Originally,
this axion is massless, but it becomes massive thanks to radiative orrections
at one loop level. Therefore, at tree level we have only one massive CP-odd
scalar field $A_{5}$.
3) In the CP-even scalar sector, there are six fields. One massless field is
part of $G_{X^{0}}$, another massive is associated to $\varphi^{0}$. One heavy
field associated with the imaginary part of the singlet $\phi$ is identified
with the inflaton $\Phi$ and one SM-like Higgs boson $h$ with mass 126 GeV.
Besides that, in order to trigger inflation, the inflaton $\Phi$ should
acquire a very large vacuum expectation value of around $10^{11}$ GeV. For a
detailed study of how the field $\Phi$ drives inflation in 3-3-1 models, see
for instance Ref hlpan ; ax2 . The remaining two fields are one heavy with
mass at TeV scale and another with mass at EW scale.
4) After including the axion field $a(x)$, the total Lagrangian as follows

$\displaystyle\mathcal{L}_{Total}=\mathcal{L_{SM}}+\left(\overline{\theta}+\frac{a(x)}{f_{a}}\right)\frac{g^{2}_{s}}{32\pi^{2}}G^{a\mu\nu}\tilde{G^{a}}_{\mu\nu}$
$\displaystyle+kinetic+interactions,$

where the axion decay constant $f_{a}=\frac{v_{\phi}}{\sqrt{2}}$, is related
to the magnitude of the VEV that breaks the $U(1)_{PQ}$ symmetry. Hence, the
CP violating term $G\tilde{G}$ is proportional to
$(\overline{\theta}+\frac{a(x)}{f_{a}})$, when the axion field is redefined,
$a(x)\rightarrow a(x)-\langle a(x)\rangle$, thus implying that the CP
violating term $G\tilde{G}$ is no longer present in the Lagrangian, so that
the Strong CP problem is solved.
5) Another important point is that the invisible axion of the model has a
small component of $I^{3}_{\chi}$ which couples only to exotic quarks at tree
level via the following Yukawa interaction

$\displaystyle\mathcal{L}_{(aq^{\prime}q^{\prime})}=-\frac{i}{\sqrt{2}}(\sin\theta_{3}\sin\theta_{\phi}\cos\theta_{4})\left[h^{U}\overline{U}_{L}U_{R}-h^{D}_{\alpha\beta}\overline{D}_{\alpha
L}D_{\beta R}\right]a$ $\displaystyle+H.c.$

6) From Eq. (17) we find
$I_{\rho}=-\sin\theta_{4}\cos\theta_{\phi}A_{5}+\cos\theta_{4}G_{Z}-\sin\theta_{4}\sin\theta_{\phi}a$.
Hence from Eq. (14) we obtain that the electron-positron-axion interaction has
the form:

$\displaystyle
L_{ae}=-i\frac{\tilde{g}_{11}}{\sqrt{2}}\sin\theta_{\phi}\sin\theta_{4}a\bar{e}\gamma_{5}e=-i\frac{\sqrt{2}m_{e}}{v_{\rho}}\sin\theta_{\phi}\sin\theta_{4}a\bar{e}\gamma_{5}e$
$\displaystyle\equiv-ig_{ae}a\bar{e}\gamma_{5}e\,,$ (31)

where $m_{e}=0.488$ MeV is the electron mass at the $M_{Z}$ scale.
Considering, $v_{\rho}=v_{\eta}=\frac{v}{\sqrt{2}}\approx 174$ GeV, which
corresponds to $\theta_{4}\approx\frac{\pi}{2}$, $v_{\phi}\approx 1.95\times
10^{10}$ GeV, we get the following value for the electron-positron-axion
coupling:

$g_{ae}=2.5\times 10^{-14},$ (32)

which is consistent with its corresponding experimental value
$g_{ae}=2.5\times 10^{-14}$ arising the best fit data arising from the XENON1T
excess xenon .
7) From Eqs. (14) and (17), we conclude that the interactions among axion and
fermions are, as expected, proportional to the masses of the latter.

##  V V. The axion mass

At tree level, the axion is massless. In order to generate the axion mass,
there are two methods: using the gravitational effect as done in ax2 or
relying on radiative corrections. In this paper, the latter method is used.
Before proceeding with the calculation of the loop diagrams, it is worth
mentioning that for a successful reheating after inflation ax2 and for
avoiding troubles with divergences in higher order interactions of the Higgs
in electroweak phase transition phong , the scalar couplings between axion and
scalar fields such as: $\lambda_{\phi},\lambda_{11},\lambda_{12},\lambda_{13}$
must be very small around $10^{-10}-10^{-6}$. Therefore, in the self-energy
correction of the axion, the following limits are applied: i)
$\sin\theta_{3}\simeq\frac{v_{\eta}}{v_{\chi}}\rightarrow 0$. Within these
considerations, the dominant contribution to the axion mass at one-loop level
consists of the following diagrams: ii) Loop with fermions and gauge bosons in
the internal lines. However, in the concerning to the fermionic contributions,
we only consider the leading ones which correspond to the loops with the top
quark and right-handed neutrinos $N_{aR}$ ($a=1,2,3$). iii) Tadpole diagrams
with the $W$, $Z$, $Z^{\prime}$, $X$ and $Y$ gauge bosons. Hence the axion
mass is given by

$\displaystyle-\frac{m^{2}_{a}}{2}$ $\displaystyle=$ $\displaystyle
12\frac{1}{v^{2}_{\eta}}\sin^{2}\theta_{\phi}\cos^{2}\theta_{4}\,m_{t}^{4}\left[1-\textrm{ln}\,\left(\frac{m_{t}^{2}}{m_{Z}^{2}}\right)\right]+\frac{12}{v^{2}_{\phi}}m_{N}^{4}\left[1-\textrm{ln}\left(\frac{m_{N}^{2}}{m_{Z}^{2}}\right)\right]+3\sum\limits_{V=Z,W,Z^{\prime},X,Y}g_{aaVV}m_{V}^{2}\left[1-\ln\left(\frac{m_{V}^{2}}{m_{Z}^{2}}\right)\right]$
(33) $\displaystyle+$
$\displaystyle\sum\limits_{V=Z,W,Z^{\prime},X,Y,H=h,h_{5},\cdot}4g^{2}_{aVH}\left[-m^{2}_{H}\left(\textrm{ln}\frac{m^{2}_{H}}{m^{2}_{Z}}-1\right)+m^{2}_{V}\left(-1+\frac{m_{V}^{2}+m^{2}_{H}}{m_{V}^{2}-m^{2}_{H}}\right)\textrm{
ln}\frac{m_{V}^{2}}{m^{2}_{H}}\right]$

where the terms in (33) are the contributions due to $t,N_{R}$ and gauge
bosons, respectively. In (33), the necessary quartic couplings are given by

$\displaystyle
g_{aaW^{+}W^{-}}=\frac{1}{2}g^{2}\sin^{2}\left(\theta_{\phi}\right)\left(\sin^{2}\left(\theta_{4}\right)+\cos^{2}\left(\theta_{3}\right)\cos^{2}\left(\theta_{4}\right)\right),$
$\displaystyle
g_{aaZZ}=\frac{g^{2}\sin^{2}\left(\theta_{\phi}\right)\left(\sin^{2}\left(\theta_{4}\right)+\cos^{2}\left(\theta_{3}\right)\cos^{2}\left(\theta_{4}\right)\right)}{c_{W}^{2}},$
$\displaystyle
g_{aaZ^{\prime}Z^{\prime}}=\frac{TC}{8s_{W}^{4}-14s_{W}^{2}+6},g_{aaX^{*}X}=\frac{1}{2}g^{2}\cos^{2}\left(\theta_{4}\right)\sin^{2}\left(\theta_{\phi}\right)$
$\displaystyle
g_{aaY^{+}Y^{-}}=\frac{1}{2}g^{2}\sin^{2}\left(\theta_{\phi}\right)\left(\sin^{2}\left(\theta_{4}\right)+\sin^{2}\left(\theta_{3}\right)\cos^{2}\left(\theta_{4}\right)\right)$
(34)

where

$\displaystyle
TC=2g^{2}\sin^{2}\left(\theta_{\phi}\right)\left(\sin^{2}\left(\theta_{4}\right)+\frac{1}{2}\left(5-3\cos\left(2\theta_{3}\right)\right)\cos^{2}\left(\theta_{4}\right)\right.$
$\displaystyle\left.+2\cos^{2}\left(\theta_{4}\right)s_{W}^{2}\left(\cos\left(2\theta_{3}\right)+2s_{W}^{2}-3\right)\right)$
(35)

The triple couplings are determined as follows
i) For neutral gauge bosons

$\displaystyle L_{aVR}=g_{aVR}V_{\mu}(R\partial^{\mu}a-a\partial^{\mu}R)\,,$
(36)

where $V$ stands for the neutral gauge bosons $Z$ and $Z^{\prime}$, and $R$
are the physical CP-even Higgs bosons. The calculation yields

$\displaystyle
g_{aZh_{5}}=-\frac{g}{4c_{W}}\sin\theta_{\phi}\sin\theta_{4}\,,\hskip
8.5359ptg_{aZh}=\frac{g}{4c_{W}}\sin\theta_{\phi}\cos\theta_{3}\cos\theta_{4}$
$\displaystyle
g_{aZ^{\prime}h_{5}}=\frac{g}{4c_{W}\sqrt{3-4s_{W}^{2}}}\sin\theta_{\phi}\sin\theta_{4}\,,$
$\displaystyle
g_{aZ^{\prime}h}=\frac{g}{4c_{W}\sqrt{3-4s_{W}^{2}}}\sin\theta_{\phi}\cos\theta_{3}\cos\theta_{4}\cos(2\theta_{W})\,,$
$\displaystyle
g_{aZ^{\prime}H_{\chi}}=-2\frac{g}{4c_{W}\sqrt{3-4s_{W}^{2}}}\sin\theta_{\phi}\sin\theta_{3}\cos\theta_{4}\,,$
(37)

where $h$ is the SM-like Higgs boson.
ii) For the charged gauge bosons, one has

$\displaystyle
L_{aVC}=g_{aVC}V^{+}_{\mu}(C^{-}\partial^{\mu}a-a\partial^{\mu}C^{-})+H.c.\,,\,V=W,Y;$
$\displaystyle C=H_{1},H_{2},$ $\displaystyle
L_{aXH_{\chi}}=g_{aXH_{\chi}}X^{0}_{\mu}(H_{\chi}\partial^{\mu}a-a\partial^{\mu}H_{\chi})\,,$
(38)

with

$\displaystyle
g_{aWH_{1}}=\frac{g}{2\sqrt{2}}\sin\theta_{\phi}\cos\theta_{3}\cos\theta_{4}\,,\,g_{aYH_{2}}=\frac{g}{2\sqrt{2}}\sin\theta_{\phi}\sin\theta_{4}\,,$
$\displaystyle
g_{aXH_{\chi}}=\frac{g}{2\sqrt{2}}\sin\theta_{\phi}\sin\theta_{3}\,.$ (39)

In this model, the masses of the non SM gauge bosons satisfy the relation (see
the last reference in flt ), hence in our calculation, we assume

$m_{Y}^{2}\simeq
m_{X}^{2}\simeq\frac{g^{2}}{4}v_{\chi}^{2}\simeq\frac{(3-4s_{W}^{2})}{4}m_{Z^{\prime}}^{2}$
(40)

Using data in huenpb ; Cogollo:2020afo , we take $m_{Z^{\prime}}\geq
4\,\textrm{TeV}$. Figure 1 displays the correlation between the axion and the
Majorana neutrino masses, for $0.5$ TeV$\leqslant m_{H_{5}}\leqslant$$10$ TeV,
$4$ TeV$\leqslant m_{H_{\chi}}\leqslant$$10$ TeV, $0.5$ TeV$\leqslant
m_{H_{1}^{\pm}}\leqslant$$1$ TeV, $2$ TeV$\leqslant
m_{H_{2}^{\pm}}\leqslant$$10$ TeV, $4$ TeV$\leqslant
m_{Z^{\prime}}\leqslant$$8$ TeV and $10$$\leqslant\tan\alpha\leqslant$$50$.
Furthermore, we have set $v_{\phi}=1.95\times 10^{10}$ GeV. From the Figure 1,
we find that the axion gets mass in the range $4$ keV$\div$ $40$ keV, when the
right handed Majorana neutrino mass is varied from $100$ GeV up to $350$ GeV.
Notice that the XENON1T experiment predicts a range for the axion mass xenon1t
consistent with our results. For the BBN bound where
$m_{a}\in(20\,\textrm{keV}\div 1\,\textrm{MeV})$ ghost , the mass of $N_{R}$
is in the range $270$ GeV $\div 1375$ GeV.

Figure 1: Correlation between the axion and the right handed Majorana neutrino
masses.

##  VI VI. Conclusions

The model under consideration contains three important features: Dark Matter
axion, inflation and low scale type I seesaw mechanism to generate the light
active neutrino masses. In addition, from the analysis of the model scalar
potential we find that in the CP-odd neutral scalar sector, there are six
fields: one massive CP-odd field $A_{4}$, two massless Goldstones eaten by the
longitudinal components of the $Z$ and $Z^{\prime}$ and one massless which is
part of Goldstones boson associated to the longitudinal component of the
neutral bilepton $G_{X^{0}}$ gauge boson. One massive CP-odd field has the
same mass of another one in the CP-even sector. Hence the above pair is a
complex bilepton scalar field denoted by $\varphi^{0}$. Finally, another field
associated to the neutral scalar singlet $\phi$ is identified with axion $a$
which becomes massive due to radiative corrections at one-loop level, mediated
by virtual top quark, right handed Majorana neutrinos and SM gauge bosons. Its
mass is obtained in the range $4$ keV$\div$ $40$ keV, coincided with the one
predicted by XENON1T experiment, when the right handed Majorana neutrino mass
is varied from $100$ GeV up to $350$ GeV, thus implying that the light active
neutrino masses are generated from a low scale type I seesaw mechanism as in
CarcamoHernandez:2019kjy . In the model under consideration, the couplings
between axion and fermion are proportional to the fermion masses. Such axion
is also very useful for successfully accommodating the XENON1T excess provided
that the PQ symmetry is spontaneously broken at the $10^{10}$ GeV scale.

## Acknowledgments

A.E.C.H received funding from Chilean grant Fondecyt No. 1170803.

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# Some identities on degenerate $r$-Stirling numbers via boson operators

Taekyun Kim  Department of Mathematics, Kwangwoon University, Seoul 139-701,
Republic of Korea<EMAIL_ADDRESS>and  DAE SAN KIM  Department of Mathematics,
Sogang University, Seoul 121-742, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

Broder introduced the $r$-Stirling numbers of the first kind and of the second
kind which enumerate restricted permutations and respectively restricted
partitions, the restriction being that the first $r$ elements must be in
distinct cycles and respectively in distinct subsets. Kim-Kim-Lee-Park
constructed the degenerate $r$-Stirling numbers of both kinds as degenerate
versions of them. The aim of this paper is to derive some identities and
recurrence relations for the degenerate $r$-Stirling numbers of the first kind
and of the second kind via boson operators. In particular, we obtain the
normal ordering of a degenerate integral power of the number operator
multiplied by an integral power of the creation boson operator in terms of
boson operators where the degenerate $r$-Stirling numbers of the second kind
appear as the coefficients.

###### Key words and phrases:

unsigned degenerate $r$-Stirling numbers of the first kind; degenerate
$r$-Stirling numbers of the second kind; boson operators; normal ordering

###### 2010 Mathematics Subject Classification:

11B73; 11B83

##  1\. Introduction

The $r$-Stirling numbers of the first kind and of the second kind enumerate
restricted permutations and respectively restricted partitions, the
restriction being that the first $r$ elements must be in distinct cycles and
respectively in distinct subsets (see [2]).

Carlitz initiated a study of degenerate versions of some special numbers and
polynomials (see [3]), where the degenerate Bernoulli and Euler numbers were
investigated. It is remarkable that in recent years intensive studies have
been done for degenerate version of quite a few special polynomials and
numbers and have yielded many interesting results (see [7-15]).

The normal ordering of an integral power of the number operator $a^{+}a$ in
terms of boson operators $a$ and $a^{+}$ can be written in the form

$(a^{+}a)^{k}=\sum_{l=0}^{k}S_{2}(k,l)(a^{+})^{l}a^{l},$

where $S_{2}(k,l)$ are the Stirling numbers of the second kind (see (2)). The
normal ordering of the degenerate $k$th power of the number operator $a^{+}a$,
namely $(a^{+}a)_{k,\lambda}$, in terms of boson operators $a$ and $a^{+}$ can
be written in the form

$(a^{+}a)_{k,\lambda}=\sum_{l=0}^{k}S_{2,\lambda}(k,l)(a^{+})^{l}a^{l},$

where $S_{2,\lambda}(k,l)$ are the degenerate Stirling numbers of the second
kind (see (5)) and the generalized falling factorials $(x)_{n,\lambda}$ are
given by (3).

The aim of this paper is to derive some identities and recurrence relations
for the degenerate $r$-Stirling numbers of the first kind and of the second
kind via boson operators. In particular, we obtain the normal ordering of a
degenerate integral power of the number operator multiplied by an integral
power of the creation boson operator in trems of boson operators where the
degenerate $r$-Stirling numbers of the second kind appear as the coefficients.
In addition, we derive recurrence relations for the degenerate $r$-Stirling
numbers of both kinds from certain normal orderings which generalize
aforementioned ones.

The outline of this paper is as follows. In Section 1, we recall the Stirling
numbers of both kinds, the unsigned Stirling numbers of the first kind, the
degenerate Stirling numbers of both kinds, and the unsigned degenerate
Stirling numbers of the first kind. We remind the reader of the normal
ordering of an integral power of the number operator in terms of boson
operators and its degenerate version, namely the normal ordering of a
degenerate integral power of the number operator. We also recall the
degenerate $r$-Stirling numbers of both kinds and the unsigned degenerate
$r$-Stirling numbers of the first kind. Section 2 is the main result of this
paper. Let $D=\frac{d}{dx}$. In Theorem 1, we show that the degenerate
differential operator $(xD)_{m,\lambda}x^{r}$ applied to the formal power
series $f(x)$ can be expressed as a linear combination of $x^{l+r}D^{l}f(x)$
with the coefficients given by the degenarate $r$-Stirling numbers of the
second kind. Thereby we obtain the normal ordering of
$(a^{+}a)_{n,\lambda}(a^{+})^{r}$ in terms of boson operators (see Corollary
2):

$(a^{+}a)_{n,\lambda}(a^{+})^{r}=\sum_{l=0}^{n}{n+r\brace
l+r}_{r,\lambda}(a^{+})^{l+r}a^{l},$

where ${n+r\brace l+r}_{r,\lambda}$ are the degenerate $r$-Stirling numbers of
the second kind (see (15)). In Theorem 3, we find that the differential
operator $x^{k}D^{k}$ applied to the formal power series $f(x)$ can be
represented as a linear combination of $(xD+r)_{m,\lambda}f(x)$ with the
coefficients given by the unsigned degenerate $r$-Stirling numbers of the
first kind. Hence we get the inverse relation to the normal ordering of
$(a^{+}a+r)_{n,\lambda}$ (see Theorem 4). The inverse relation to Theorem 3 is
derived in Theorem 5 in connection with the degenerate $r$-Stirling numbers of
the second kind and hence we have the normal ordering of
$(a^{+}a+r)_{n,\lambda}$ (see Theorem 6). In Theorems 7 and 9, two recurrence
relations are obtained for the degenerate $r$-Stirling numbers of the second
kind. Finally, we get a recurrence relation for the degenerate $r$-Stirling
numbers of the first kind. For the rest of this section, we recall the facts
that are needed throughout this paper.

It is well known that the Stirling numbers of the first kind are defined by

(1) $(x)_{n}=\sum_{k=0}^{n}S_{1}(n,k)x^{k},\quad(n\geq 0),\quad(\mathrm{see}\
[4-17]),$

where $(x)_{0}=1,\ (x)_{n}=x(x-1)\cdots(x-n+1),\ (n\geq 1)$.

As the inversion formula of (1), the Stirling numbers of the second kind are
given by

(2) $x^{n}=\sum_{k=0}^{n}S_{2}(n,k)(x)_{k},\quad(n\geq 0),\quad(\mathrm{see}\
[4,5,17]).$

The unsigned Stirling number of the first kind is defined by

${n\brack k}=(-1)^{n-k}S_{1}(n,k),\quad(n,k\geq 0).$

For any $\lambda\in\mathbb{R}$, the degenerate factorial sequence is defined
by

(3)
$(x)_{0,\lambda}=1,\quad(x)_{n,\lambda}=x(x-\lambda)(x-2\lambda)\cdots(x-(n-1)\lambda),\quad(n\geq
1),\quad(\mathrm{see}\ [8,10]).$

In [8], the degenerate Stirling numbers of the first kind are defined by

(4) $(x)_{n}=\sum_{k=0}^{n}S_{1,\lambda}(n,k)(x)_{k,\lambda},\quad(n\geq 0).$

In view of (2), the degenerate Stirling numbers of the second kind are given
by

(5) $(x)_{n,\lambda}=\sum_{k=0}^{n}S_{2,\lambda}(n,k)(x)_{k},\quad(n\geq
0),\quad(\mathrm{see}\ [8]).$

The unsigned degenerate Stirling numbers of the first kind are given by

(6) $\langle x\rangle_{n}=\sum_{k=0}^{n}{n\brack k}_{\lambda}\langle
x\rangle_{k,\lambda},\quad(n\geq 0),\quad(\mathrm{see}\ [13-15]),$

where

$\langle x\rangle_{n}=(-1)^{n}(-x)_{n}\quad\mathrm{and}\quad\langle
x\rangle_{n,\lambda}=(-1)^{n}(-x)_{n,\lambda},\quad(n\geq 0).$

Note that

${n\brack k}_{\lambda}=(-1)^{n-k}S_{1,\lambda}(n,k).$

We note that the boson annihilation and creation operators $a$ and $a^{+}$
satisfy the commutation relation given by

(7) $[a,a^{+}]=aa^{+}-a^{+}a=1,\quad(\mathrm{see}\ [1,6,9,11]).$

In the corresponding Fock space, we have

(8) $a^{+}|k\rangle=\sqrt{k+1}|k+1\rangle,\quad
a|k\rangle=\sqrt{k}|k-1\rangle,\quad(\mathrm{see}\ [1,12]).$

Now, the number operator $\hat{n}$ is defined by

(9) $\hat{n}|k\rangle=k|k\rangle.$

By (8), we get $a^{+}a|k\rangle=k|k\rangle$, that is $\hat{n}=a^{+}a$. Thus,
we note that

$[a,\hat{n}]=a\hat{n}-\hat{n}a=a,\quad[\hat{n},a^{+}]=\hat{n}a^{+}-a^{+}\hat{n}=a^{+}.$

The normal ordering of an integral power of the number operator
$\hat{n}=a^{+}a$ in terms of boson operators $a$ and $a^{+}$ can be written in
the form

(10) $(a^{+}a)^{k}=\sum_{l=0}^{k}S_{2}(k,l)(a^{+})^{l}a^{l},\quad(k\geq
0),\quad(\mathrm{see}\ [1,6,16]).$

Recently, the normal ordering of the degenerate $k$-th power of the number
operator $a^{+}a$, namely $(a^{+}a)_{k,\lambda}$, in terms of boson operators
$a$ and $a^{+}$ can be written in the form

(11)
$(a^{+}a)_{k,\lambda}=\sum_{l=0}^{k}S_{2,\lambda}(k,l)(a^{+})^{l}a^{l},\quad(k\geq
0),\quad(\mathrm{see}\ [12]).$

For $r\geq 0$, the unsigned $r$-Stirling numbers are defined by

(12) $\langle x+r\rangle_{n}=\sum_{k=0}^{n}{n+r\brack
k+r}_{r}x^{k},\quad(n\geq 0),\quad(\mathrm{see}\ [13-15]),$

where

$\langle x\rangle_{n}=x(x+1)\cdots(x+n-1),\quad(n\geq 1),\quad\langle
x\rangle_{0}=1.$

In view of (2), the $r$-Stirling numbers of the second kind are given by

(13) $(x+r)^{n}=\sum_{k=0}^{n}{n+r\brace k+r}_{r}(x)_{k},\quad(n\geq
0),\quad(\mathrm{see}\ [14,15]).$

Recently, the authors introduced the unsigned degenerate $r$-Stirling numbers
of the first kind, which are given by

(14) $\langle x+r\rangle_{n}=\sum_{k=0}^{n}{n+r\brack k+r}_{r,\lambda}\langle
x\rangle_{k,\lambda},\quad(n\geq 0),\quad(\mathrm{see}\ [13-15]),$

In view of (5), the degenerate $r$-Stirling numbers of the second kind are
defined by the authors, which are given by

(15) $(x+r)_{n,\lambda}=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(x)_{k},\quad(n\geq 0),\quad(\mathrm{see}\ [13-15]).$

In addition, the degenerate $r$-Stirling numbers of the first kind are given
by

(16)
$(x-r)_{n}=\sum_{k=0}^{n}S_{1,\lambda}^{(r)}(n+r,k+r)(x)_{k,\lambda},\quad(n\geq
0),\quad(\mathrm{see}\ [13,15]).$

Note that

${n+r\brack
k+r}_{r,\lambda}=(-1)^{n-k}S_{1,\lambda}^{(r)}(n+r,k+r),\quad(n\geq 0).$

In this paper, we study the degenerate $r$-Stirling numbers associated with
the number operator $\hat{n}=a^{+}a$ in terms of boson operators $a$ and
$a^{+}$ that satisfy $[a,a^{+}]=aa^{+}-a^{+}a=1$.

That is, we derive some identities involving the degenerate $r$-Stirling
numbers arising from the normal ordering of degenerate integral powers of
number operator. $\hat{n}=a^{+}a$ in terms of boson operators $a$ and $a^{+}$,
which are given by

$(a^{+}a+r)_{n,\lambda}=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(a^{+})^{k}a^{k}.$

##  2\. Some identities on degenerate $r$-Stirling numbers via boson operators

Let $f(x)$ be a formal power series in $x$ with coefficients in $\mathbb{C}$,
given by $f(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$. For $r\geq 0$, we observe that

(17) $\displaystyle(xD)_{m,\lambda}x^{r}f(x)$
$\displaystyle=(xD)(xD-\lambda)\cdots(xD-(m-1)\lambda)x^{r}\sum_{k=0}^{\infty}a_{k}x^{k}$
$\displaystyle=\sum_{k=0}^{\infty}a_{k}(k+r)_{m,\lambda}x^{k+r}=\sum_{k=0}^{\infty}a_{k}\sum_{l=0}^{m}{m+r\brace
l+r}_{r,\lambda}(k)_{l}x^{k+r}$ $\displaystyle=\sum_{l=0}^{m}{m+r\brace
l+r}_{r,\lambda}\sum_{k=0}^{\infty}a_{k}(k)_{l}x^{k}x^{r}=\sum_{l=0}^{m}{m+r\brace
l+r}_{r,\lambda}x^{l+r}f^{(l)}(x),$

where

$f^{(l)}(x)=\bigg{(}\frac{d}{dx}\bigg{)}^{l}f(x),\quad D=\frac{d}{dx}.$

From now on throughout this paper, $D$ will denote the operator
$D=\frac{d}{dx}$.

Now, by (17), we get

(18) $(xD)_{m,\lambda}x^{r}f(x)=\sum_{l=0}^{m}{m+r\brace
l+r}_{r,\lambda}\Big{(}D^{l}f(x)\Big{)}x^{l+r}.$

In particular, for $m,r\geq 0$ with $m\geq r$, we have

(19) $(xD)_{m-r,\lambda}x^{r}f(x)=\sum_{l=r}^{m}{m\brace
l}_{r,\lambda}\Big{(}D^{l-r}f(x)\Big{)}x^{l}.$

Replacing $f(x)$ by $f^{(r)}(x)$, we get

(20) $(xD)_{m-r,\lambda}x^{r}D^{r}f(x)=\sum_{l=r}^{m}{m\brace
l}_{r,\lambda}x^{l}D^{l}f(x).$

Therefore, by (18) and (20), we obtain the following theorem.

######  Theorem 1.

For $m,r\in\mathbb{Z}$ with $m,r\geq 0$, we have

$(xD)_{m,\lambda}x^{r}f(x)=\sum_{l=0}^{m}{m+r\brace
l+r}_{r,\lambda}x^{l+r}D^{l}f(x).$

In addition, for $m,r\in\mathbb{Z}$ with $m\geq r\geq 0$, we have

$(xD)_{m-r,\lambda}x^{r}D^{r}f(x)=\sum_{l=r}^{m}{m\brace
l}_{r,\lambda}x^{l}D^{l}f(x).$

Note that $Dx-xD=1$. We recall that the bosonic commutation relation
$[a,a^{+}]=aa^{+}-a^{+}a=1$ can be realized formally in a suitable space of
functions by letting $a=\frac{d}{dx}$ and $a^{+}=x$. Therefore we obtain the
following corollary.

######  Corollary 2.

For $m,r\in\mathbb{Z}$ with $m,r\geq 0$, we have

$(a^{+}a)_{m,\lambda}(a^{+})^{r}=\sum_{l=0}^{m}{m+r\brace
l+r}_{r,\lambda}(a^{+})^{l+r}a^{l}.$

In addition, for $m,r\in\mathbb{Z}$ with $m\geq r\geq 0$, we have

$(a^{+}a)_{m-r,\lambda}(a^{+})^{r}a^{r}=\sum_{l=r}^{m}{m\brace
l}_{r,\lambda}(a^{+})^{l}a^{l}.$

By (15), we get

(21) $\displaystyle\sum_{k=0}^{n+1}{n+r+1\brace k+r}_{r,\lambda}(x)_{k}$
$\displaystyle=(x+r)_{n+1,\lambda}=(x+r)_{n,\lambda}(x+r-n\lambda)$
$\displaystyle=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(x)_{k}(x-k+k+r-n\lambda)$
$\displaystyle=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(x)_{k+1}+\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(k+r-n\lambda)(x)_{k}$
$\displaystyle=\sum_{k=0}^{n+1}\bigg{(}{n+r\brace
k+r-1}_{r,\lambda}+(k+r-n\lambda){n+r\brace k+r}_{r,\lambda}\bigg{)}(x)_{k}.$

Comparing the coefficients on both sides of (21), we have

${n+r+1\brace k+r}_{r,\lambda}={n+r\brace
k+r-1}_{r,\lambda}+(k+r-n\lambda){n+r\brace k+r}_{r,\lambda},$

where $n,k\in\mathbb{Z}$ with $n+1\geq k\geq 0$.

Now, we observe that

(22) $\displaystyle x^{k}D^{k}f(x)$
$\displaystyle=\sum_{n=0}^{\infty}a_{n}(n)_{k}x^{n}$
$\displaystyle=\sum_{n=0}^{\infty}a_{n}\bigg{(}\sum_{m=0}^{k}S_{1,\lambda}^{(r)}(k+r,m+r)(n+r)_{m,\lambda}\bigg{)}x^{n}$
$\displaystyle=\sum_{m=0}^{k}S_{1,\lambda}^{(r)}(k+r,m+r)\sum_{n=0}^{\infty}a_{n}(n+r)_{m,\lambda}x^{n}$
$\displaystyle=\sum_{m=0}^{k}S_{1,\lambda}^{(r)}(k+r,m+r)\bigg{(}x\frac{d}{dx}+r\bigg{)}_{m,\lambda}f(x)$
$\displaystyle=\sum_{m=0}^{k}(-1)^{k-m}{k+r\brack
m+r}_{r,\lambda}\bigg{(}x\frac{d}{dx}+r\bigg{)}_{m,\lambda}f(x).$

Therefore, by (22), we obtain the following theorem.

######  Theorem 3.

For $k\geq 0$, we have

(23) $x^{k}D^{k}f(x)=\sum_{m=0}^{k}(-1)^{k-m}{k+r\brack
m+r}_{r,\lambda}(xD+r)_{m,\lambda}f(x).$

In (23), the normal ordering of degenerate integral power of the number
operator $\hat{n}a^{+}a$ in terms of boson operators $a^{+}$ and $a$ can be
presented in the form

(24) $\displaystyle(a^{+})^{k}a^{k}$
$\displaystyle=\sum_{m=0}^{k}(-1)^{k-m}{k+r\brack
m+r}_{r,\lambda}(a^{+}a+r)_{m,\lambda}$
$\displaystyle=\sum_{m=0}^{k}(-1)^{k-m}{k+r\brack
m+r}_{r,\lambda}(\hat{n}+r)_{m,\lambda}.$

For $k,r\in\mathbb{Z}$ with $k\geq r\geq 0$, by (24), we get

(25) $\displaystyle(a^{+})^{k-r}a^{k-r}$
$\displaystyle=\sum_{m=r}^{k}(-1)^{k-m}{k\brack
m}_{r,\lambda}(a^{+}a+r)_{m-r,\lambda}$
$\displaystyle=\sum_{m=r}^{k}(-1)^{k-m}{k\brack
m}_{r,\lambda}(\hat{n}+r)_{m-r,\lambda}.$

Therefore, by (24) and (25), we obtain the following theorem.

######  Theorem 4.

For $k,r\in\mathbb{Z}$ with $k\geq r\geq 0$, we have

$(a^{+})^{k}a^{k}=\sum_{m=0}^{k}(-1)^{k-m}{k+r\brack
m+r}_{r,\lambda}(\hat{n}+r)_{m,\lambda},$

and

$(a^{+})^{k-r}a^{k-r}=\sum_{m=r}^{k}(-1)^{k-m}{k\brack
m}_{r,\lambda}(\hat{n}+r)_{m-r,\lambda}.$

As the inversion formula of (23), we consider the following differential
equation:

(26) $\displaystyle\bigg{(}x\frac{d}{dx}+r\bigg{)}_{n,\lambda}f(x)$
$\displaystyle=\sum_{m=0}^{\infty}a_{m}\bigg{(}x\frac{d}{dx}+r\bigg{)}_{n,\lambda}x^{m}=\sum_{m=0}^{\infty}a_{m}(m+r)_{n,\lambda}x^{m}$
$\displaystyle=\sum_{m=0}^{\infty}a_{m}\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(m)_{k}x^{m}$ $\displaystyle=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}\sum_{m=0}^{\infty}a_{m}(m)_{k}x^{m}$
$\displaystyle=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}x^{k}\bigg{(}\frac{d}{dx}\bigg{)}^{k}f(x).$

Therefore, by (26), we obtain the following theorem.

######  Theorem 5.

For $n\geq 0$, we have

(27) $(xD+r)_{n,\lambda}f(x)=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}x^{k}D^{k}f(x).$

From Theorems 1 and 5, we note that

(28) $x^{r}(xD+r)_{n,\lambda}=(xD)_{n,\lambda}x^{r},\quad(n\geq 0).$

Thus, by (28), we get

(29)
$(a^{+})^{r}(a^{+}a+r)_{n,\lambda}=(a^{+}a)_{n,\lambda}(a^{+})^{r},\quad(n\geq
0).$

From (27), we have

(30) $(a^{+}a+r)_{n,\lambda}=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(a^{+})^{k}a^{k}.$

Therefore, by (29) and (30), we obtain the following theorem.

######  Theorem 6.

For $n\geq 0$, we have

$(a^{+})^{r}(a^{+}a+r)_{n,\lambda}=(a^{+}a)_{n,\lambda}(a^{+})^{r},$

and

(31) $(a^{+}a+r)_{n,\lambda}=\sum_{k=0}^{n}{n+r\brace
k+r}_{r,\lambda}(a^{+})^{k}a^{k}.$

By (31), we get

(32) $\displaystyle a^{+}(\hat{n}+r+1-\lambda)_{k,\lambda}a$
$\displaystyle=a^{+}\sum_{l=0}^{k}\binom{k}{l}(\hat{n}+r)_{l,\lambda}(1-\lambda)_{k-l,\lambda}a$
$\displaystyle=\sum_{l=0}^{k}\binom{k}{l}(1-\lambda)_{k-l,\lambda}\sum_{m=0}^{l}{l+r\brace
m+r}_{r,\lambda}(a^{+})^{m+1}a^{m+1}$
$\displaystyle=\sum_{m=0}^{k}\sum_{l=m}^{k}\binom{k}{l}(1-\lambda)_{k-l,\lambda}{l+r\brace
m+r}_{r,\lambda}(a^{+})^{m+1}a^{m+1}.$

On the other hand, by (30) and (32), we get

(33) $\displaystyle\sum_{m=0}^{k+1}{k+1+r\brace
m+r}_{r,\lambda}(a^{+})^{m}a^{m}=(\hat{n}+r)_{k+1,\lambda}=(\hat{n}+r)(\hat{n}+r-\lambda)_{k,\lambda}$
$\displaystyle=\hat{n}(\hat{n}+r-\lambda)_{k,\lambda}+r(\hat{n}+r-\lambda)_{k,\lambda}=a^{+}(\hat{n}+r+1-\lambda)_{k,\lambda}a+r(\hat{n}+r-\lambda)_{k,\lambda}$
$\displaystyle=\sum_{m=0}^{k}\sum_{l=m}^{k}\binom{k}{l}(1-\lambda)_{k-l,\lambda}{l+r\brace
m+r}_{r,\lambda}(a^{+})^{m+1}a^{m+1}+r\sum_{l=0}^{k}\binom{k}{l}(-\lambda)_{k-l,\lambda}(\hat{n}+r)_{l,\lambda}$
$\displaystyle=\sum_{m=1}^{k+1}\sum_{l=m-1}^{k}\binom{k}{l}(1-\lambda)_{k-l,\lambda}{l+r\brace
m-1+r}_{r,\lambda}(a^{+})^{m}a^{m}+r\sum_{l=0}^{k}\binom{k}{l}(-\lambda)_{k-l,\lambda}\sum_{m=0}^{l}{l+r\brace
m+r}_{r,\lambda}(a^{+})^{m}a^{m}$
$\displaystyle=\sum_{m=1}^{k+1}\sum_{l=m-1}^{k}\binom{k}{l}(1-\lambda)_{k-l,\lambda}{l+r\brace
m+r-1}_{r,\lambda}(a^{+})^{m}a^{m}+\sum_{m=0}^{k}\bigg{(}r\sum_{l=m}^{k}\binom{k}{l}(-\lambda)_{k-l,\lambda}{l+r\brace
m+r}_{r,\lambda}\bigg{)}(a^{+})^{m}a^{m}$
$\displaystyle=\sum_{m=0}^{k+1}\bigg{(}\sum_{l=m-1}^{k}\binom{k}{l}(1-\lambda)_{k-l,\lambda}{l+r\brace
m+r-1}_{r,\lambda}+r\sum_{l=m}^{k}\binom{k}{l}(-\lambda)_{k-l,\lambda}{l+r\brace
m+r}_{r,\lambda}\bigg{)}(a^{+})^{m}a^{m}.$

Therefore, by comparing the coefficients on both sides of (33), we obtain the
following theorem.

######  Theorem 7.

For $m,k\in\mathbb{Z}$ with $k\geq m\geq 0$, we have

${k+1+r\brace
m+r}_{r,\lambda}=\sum_{l=m-1}^{k}\binom{k}{l}(1-\lambda)_{k-l,\lambda}{l+r\brace
m+r-1}_{r,\lambda}+r\sum_{l=m}^{k}\binom{k}{l}(-\lambda)_{k-l,\lambda}{l+r\brace
m+r}_{r,\lambda}.$

We observe that

(34) $\displaystyle\hat{n}(\hat{n}+r-1)_{k,\lambda}$
$\displaystyle=a^{+}a(a^{+}a+r-1)_{k,\lambda}$
$\displaystyle=a^{+}a(a^{+}a+r-1)(a^{+}a+r-1-\lambda)\cdots(a^{+}a+r-1-(k-1)\lambda)$
$\displaystyle=a^{+}(aa^{+}+r-1)a(a^{+}a+r-1-\lambda)(a^{+}a+r-1-2\lambda)\cdots(a^{+}a+r-1-(k-1)\lambda)$
$\displaystyle=a^{+}(aa^{+}+r-1)(aa^{+}+r-1-\lambda)a(a^{+}a+r-1-2\lambda)\cdots(a^{+}a+r-1-(k-1)\lambda)$
$\displaystyle=\cdots$
$\displaystyle=a^{+}(aa^{+}+r-1)(aa^{+}+r-1-\lambda)(aa^{+}+r-1-2\lambda)\cdots(aa^{+}+r-1-(k-1)\lambda)a$
$\displaystyle=a^{+}(a^{+}a+r)(a^{+}a+r-\lambda)(a^{+}a+r-2\lambda)\cdots(a^{+}a+r-(k-1)\lambda)a$
$\displaystyle=a^{+}(a^{+}a+r)_{k,\lambda}a=a^{+}(\hat{n}+r)_{k,\lambda}a.$

Therefore, by (34), we obtain the following lemma.

######  Lemma 8.

For $k\geq 0$, we have

(35) $\hat{n}(\hat{n}+r-1)_{k,\lambda}=a^{+}(\hat{n}+r)_{k,\lambda}a.$

From (35) and Theorem 6, we note that

(36) $\displaystyle
a^{+}(\hat{n}+r)_{k,\lambda}a=\hat{n}(\hat{n}+r-1)_{k,\lambda}=\sum_{m=0}^{k}\binom{k}{m}(-1)_{k-m,\lambda}\hat{n}(\hat{n}+r)_{m,\lambda}$
$\displaystyle=\sum_{m=0}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}(\hat{n}+r-m\lambda+m\lambda-r)(\hat{n}+r)_{m,\lambda}$
$\displaystyle=\sum_{m=0}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}\Big{(}(\hat{n}+r)_{m+1,\lambda}+(m\lambda-r)(\hat{n}+r)_{m,\lambda}\Big{)}$
$\displaystyle=\sum_{m=0}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}(\hat{n}+r)_{m+1,\lambda}$
$\displaystyle\qquad+\lambda\sum_{m=0}^{k}\binom{k}{m+1}(-1)^{k-m-1}\langle
1\rangle_{k-m-1,\lambda}(m+1)(\hat{n}+r)_{m+1,\lambda}$
$\displaystyle\qquad-r\sum_{m=0}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}(\hat{n}+r)_{m,\lambda}$
$\displaystyle=\sum_{m=0}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}\sum_{p=0}^{m+1}{m+1+r\brace
p+r}_{r,\lambda}(a^{+})^{p}a^{p}$
$\displaystyle\qquad+\lambda\sum_{m=0}^{k}\binom{k}{m+1}(-1)^{k-m-1}\langle
1\rangle_{k-m-1,\lambda}(m+1)\sum_{p=0}^{m+1}{m+1+r\brace
p+r}_{r,\lambda}(a^{+})^{p}a^{p}$
$\displaystyle\qquad-r\sum_{m=0}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}\sum_{p=0}^{m+1}{m+r\brace
p+r}_{r,\lambda}(a^{+})^{p}a^{p}$
$\displaystyle=\sum_{p=1}^{k+1}\sum_{m=p-1}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}{m+1+r\brace p+r}_{r,\lambda}(a^{+})^{p}a^{p}$
$\displaystyle\qquad+\lambda\sum_{p=1}^{k+1}\sum_{m=p-1}^{k}\binom{k}{m+1}(-1)^{k-m-1}\langle
1\rangle_{k-m-1,\lambda}(m+1){m+1+r\brace p+r}_{r,\lambda}(a^{+})^{p}a^{p}$
$\displaystyle\qquad-r\sum_{p=1}^{r+1}\sum_{m=p-1}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}{m+r\brace p+r}_{r,\lambda}(a^{+})^{p}a^{p}$
$\displaystyle=\sum_{p=0}^{k}\bigg{\\{}\sum_{m=p}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}{m+1+r\brace p+1+r}_{r,\lambda}$
$\displaystyle\qquad+\lambda\sum_{m=p}^{k}\binom{k}{m+1}(-1)^{k-m-1}\langle
1\rangle_{k-m-1}(m+1){m+1+r\brace p+1+r}_{r,\lambda}$
$\displaystyle\qquad-r\sum_{m=p}^{k}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}{m+r\brace
p+1+r}_{r,\lambda}\bigg{\\}}(a^{+})^{p+1}a^{p+1}.$

On the other hand, by (30), we get

(37) $a^{+}(\hat{n}+r)_{k,\lambda}a=\sum_{p=0}^{k}{k+r\brace
p+r}_{r,\lambda}(a^{+})^{p+1}a^{p+1}.$

Therefore, by (36) and (37), we obtain the following theorem.

######  Theorem 9.

For $p,k\in\mathbb{Z}$ with $k\geq p\geq 0$, we have

$\displaystyle{k+r\brace
p+r}_{r,\lambda}=\sum_{m=p}^{k}\bigg{[}\binom{k}{m}(-1)^{k-m}\langle
1\rangle_{k-m,\lambda}\bigg{(}{m+1+r\brace p+1+r}_{r,\lambda}-r{m+r\brace
p+1+r}_{r,\lambda}\bigg{)}$ $\displaystyle\qquad\qquad\qquad\qquad+\lambda
k\binom{k-1}{m}\langle 1\rangle_{k-m-1}(-1)^{k-m-1}{m+1+1\brace
p+1+r}_{r,\lambda}\bigg{]}.$

From Theorem 4, we note that

(38) $\displaystyle(a^{+})^{k+1}a^{k+1}$
$\displaystyle=\sum_{l=0}^{k}(-1)^{k-l}{k+r\brack
l+r}_{r,\lambda}a^{+}(\hat{n}+r)_{l,\lambda}a$
$\displaystyle=\sum_{l=0}^{k}(-1)^{k-l}{k+r\brack
l+r}_{r,\lambda}\hat{n}(\hat{n}+r-1)_{l,\lambda}$
$\displaystyle=\sum_{l=0}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-l}\sum_{m=0}^{l}\binom{l}{m}(-1)^{l-m}\langle
1\rangle_{l-m,\lambda}(\hat{n}+r)_{m,\lambda}\hat{n}$
$\displaystyle=\sum_{l=0}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-l}\sum_{m=0}^{l}\binom{l}{m}(-1)^{l-m}\langle
1\rangle_{l-m,\lambda}(\hat{n}+r)_{m,\lambda}(\hat{n}+r-m\lambda+m\lambda-r)$
$\displaystyle=\sum_{l=0}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-l}\sum_{m=0}^{l}\binom{l}{m}(-1)^{l-m}\langle
1\rangle_{l-m,\lambda}(\hat{n}+r)_{m+1,\lambda}$
$\displaystyle\quad+\sum_{l=0}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-l}\sum_{m=0}^{l}\binom{l}{m}(-1)^{l-m}\langle
1\rangle_{l-m,\lambda}(\hat{n}+r)_{m,\lambda}(m\lambda-r)$
$\displaystyle=\sum_{m=0}^{k}\sum_{l=m}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m}\binom{l}{m}\langle
1\rangle_{l-m,\lambda}(\hat{n}+r)_{m+1,\lambda}$
$\displaystyle\quad+\sum_{m=0}^{k}\sum_{l=m}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m}\binom{l}{m}\langle
1\rangle_{l-m,\lambda}(m\lambda-r)(\hat{n}+r)_{m,\lambda}$
$\displaystyle=\sum_{m=1}^{k+1}\bigg{(}\sum_{l=m-1}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m+1}\binom{l}{m-1}\langle
1\rangle_{l-m+1,\lambda}\bigg{)}(\hat{n}+r)_{m,\lambda}$
$\displaystyle\quad+\sum_{m=0}^{k}(m\lambda-r)\sum_{l=m}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m}\binom{l}{m}\langle
1\rangle_{l-m,\lambda}(\hat{n}+r)_{m,\lambda}$
$\displaystyle=\sum_{m=0}^{k+1}\bigg{\\{}\sum_{l=m-1}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m+1}\binom{l}{m-1}\langle 1\rangle_{l-m+1,\lambda}$
$\displaystyle\qquad+(m\lambda-r)\sum_{l=m}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m}\binom{l}{m}\langle
1\rangle_{l-m,\lambda}\bigg{\\}}(\hat{n}+r)_{m,\lambda}.$

On the other hand, by (24), we get

(39) $\displaystyle(a^{+})^{k+1}a^{k+1}$
$\displaystyle=\sum_{m=0}^{k+1}(-1)^{k-m+1}{k+r+1\brack
m+r}_{r,\lambda}(a^{+}a+r)_{m,\lambda}$
$\displaystyle=\sum_{m=0}^{k+1}(-1)^{k-m+1}{k+r+1\brack
m+r}_{r,\lambda}(\hat{n}+r)_{m,\lambda}.$

Therefore, by (38) and (39), we obtain the following theorem.

######  Theorem 10.

For $m,k\in\mathbb{Z}$ with $k\geq m\geq 0$, we have

$\displaystyle(-1)^{k-m+1}{k+r+1\brack m+r}_{r,\lambda}$
$\displaystyle=\sum_{l=m-1}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m+1}\binom{l}{m-1}\langle 1\rangle_{l-m+1,\lambda}$
$\displaystyle\quad+(m\lambda-r)\sum_{l=m}^{k}{k+r\brack
l+r}_{r,\lambda}(-1)^{k-m}\binom{l}{m}\langle 1\rangle_{l-m,\lambda}.$

##  3\. conclusion

We have witnessed that various degenerate versions of some special numbers and
polynomials were studied by means of many different methods, which include
combinatorial methods, generating functions, $p$-adic analysis, umbral
calculus, probability theory, differential equations, operator theory and
analytic number theory.

The Stirling numbers of the second kind appear as the coefficients in the
normal ordering of an integral power of the number operator in terms of the
boson operators. Recently, it was shown that the degenrate Stirling numbers of
the second kind appear as the coefficients in the normal ordering of a
degenerate integral power of the number operator in terms of boson operators.
In this paper, we obtained the normal ordering of a degenerate integral power
of the number operator multiplied by an integral power of the creation boson
operator in trems of boson operators where the degenerate $r$-Stirling numbers
of the second kind appear as the coefficients. In addition, we derived
recurrence relations for the degenerate $r$-Stirling numbers of both kinds
from certain normal orderings for boson operators.

It is one of our future research projects to continue to explore various
degenerate versions of some special numbers and polynomials and to find their
applications in physics, science and engineering as well as in mathematics.

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