#  Flavor Changing Neutral Current processes and family discrimination in
3-3-1 models

Vinícius Oliveiraa<EMAIL_ADDRESS>C. A. de S. Piresa
<EMAIL_ADDRESS>aDepartamento de Física, Universidade Federal da
Paraíba, Caixa Postal 5008, 58051-970, João Pessoa, PB, Brazil

###### Abstract

In 3-3-1 models anomaly cancellation requires that one of the three families
of quarks comes in different $\text{SU}(3)_{\text{L}}$ representation.
Consequently each version of the 3-3-1 models dismembers in other three
variants obeying family discrimination. The main consequence of such
arrangement of families is the arising of processes violating flavor at tree
level by means of neutral currents mediated by gauge and scalar fields. In
this work we focus on the 3-3-1 model with right-handed neutrinos and
calculated, within each variant, the contributions of the pseudoscalar to the
$K^{0}-\bar{K}^{0}$ mixing process and confront our findings with the current
experimental prediction. As result, the constraints on the mass of the
pseudoscalar obtained in each case allow us distinguish one variants from the
others. We give an example for an illustrative parametrization of the quark
mixing matrices and, as result, we found that the variant in which the first
family of quarks transform as triplet is the most interesting one from the
point of view of collider phenomenology.

##  I Introduction

The 3-3-1 model with right-handed neutrinos Singer  _et al._ (1980); Foot  _et
al._ (1994); Montero  _et al._ (1993) (331RHN for short) is a
phenomenologically viable version of models based on the
$\text{SU}(3)_{\text{C}}\times\text{SU}(3)_{\text{L}}\times\text{U}(1)_{\text{N}}$
gauge symmetry111For other variants, see: Frampton (1992); Pisano and Pleitez
(1992); Ozer (1996); Sanchez  _et al._ (2001).. In 3-3-1 models anomaly is not
canceled within each generation Pisano (1996); Frampton (1992). It is canceled
among three generations Ng (1994); Liu and Ng (1994); Pisano (1996). Then
fermions come in equal numbers of triplet and anti-triplet representations of
$\text{SU}(3)_{\text{L}}$. Leptons are treated democratically with the three
families coming in triplet representation. Consequently two families of quarks
must come in anti-triplet representation. Theory is not able to discriminate
which families transform differently. There are three possibilities with each
one leading to different phenomenological aspects. Which of these three
possibilities describe nature is the question we refer to as the problem of
family discrimination. It is this problem that we attack in this paper.

The original scalar content of the 331RHN involves three triplet of scalars
Foot  _et al._ (1994); Montero  _et al._ (1993). After spontaneous breaking of
the symmetry, the spectrum of scalars poses one pseudoscalar de S. Pires and
Rodrigues da Silva (2007). We then calculate within each variant of the model
the contributions of the pseudoscalar to the $K^{0}-\bar{K}^{0}$ transition
and obtain constraints on its mass. We find that such constraints provide an
efficient way of distinguishing one variant from the others and solve the
problem of family discrimination.

This paper is organized in the following way: in section II we summarize the
essence of the model; in section III we discuss the possibles choices for
quarks anti-triplet representation; in section IV we derive the mass
difference terms for the neutral mesons; in section V we present the numerical
results; finally we summarize and draw our conclusions in section VI.

##  II Preliminaries

In the 331RHN the third component of the leptonic triplets is a right-handed
neutrino, $f_{a}=(\nu_{a_{L}}\,,\,e_{a_{L}}\,,\,\nu^{C}_{a_{R}})^{T}$, where
$a=1,2,3$. Its gauge sector involves nine gauge bosons Long (1996), namely the
four standard ones, $W^{\pm}$, $Z^{0}$, and $\gamma$. The other five are
$W^{\prime\pm}$, $U^{0}$, $U^{0\dagger}$, and $Z^{\prime}$.

The original scalar sector of the model is composed by three triplet of
scalars

$\displaystyle\eta=\left(\begin{array}[]{c}\eta^{0}\\\ \eta^{-}\\\
\eta^{\prime 0}\end{array}\right),\,\rho=\left(\begin{array}[]{c}\rho^{+}\\\
\rho^{0}\\\
\rho^{\prime+}\end{array}\right),\,\chi=\left(\begin{array}[]{c}\chi^{0}\\\
\chi^{-}\\\ \chi^{\prime 0}\end{array}\right),$ (10)

with $\eta$ and $\chi$ transforming as $(1\,,\,3\,,\,-1/3)$ and $\rho$ as
$(1\,,\,3\,,\,2/3)$.

The most economical potential of the model is obtained by demanding lepton
number is conserved at tree level Pal (1995); de S. Pires and Rodrigues da
Silva (2007). In this case we have:

$\displaystyle V(\eta,\rho,\chi)$ $\displaystyle=$
$\displaystyle\mu_{\chi}^{2}\chi^{2}+\mu_{\eta}^{2}\eta^{2}+\mu_{\rho}^{2}\rho^{2}+\lambda_{1}\chi^{4}+\lambda_{2}\eta^{4}+\lambda_{3}\rho^{4}+$
(11)
$\displaystyle\lambda_{4}(\chi^{\dagger}\chi)(\eta^{\dagger}\eta)+\lambda_{5}(\chi^{\dagger}\chi)(\rho^{\dagger}\rho)+\lambda_{6}(\eta^{\dagger}\eta)(\rho^{\dagger}\rho)+$
$\displaystyle\lambda_{7}(\chi^{\dagger}\eta)(\eta^{\dagger}\chi)+\lambda_{8}(\chi^{\dagger}\rho)(\rho^{\dagger}\chi)+\lambda_{9}(\eta^{\dagger}\rho)(\rho^{\dagger}\eta)$
$\displaystyle-\frac{f}{\sqrt{2}}\epsilon^{ijk}\eta_{i}\rho_{j}\chi_{k}+\mbox{H.c.}\,.$

We restrict our approach to the case where only $\eta^{0}\,\,,\rho^{0}$, and
$\chi^{\prime 0}$ develop VEV. We then shift the fields in the usual way

$\eta^{0}\,\,,\rho^{0}\,\,,\chi^{\prime
0}=\frac{1}{\sqrt{2}}(v_{\eta\,,\rho\,,\chi^{\prime}}+R_{{}_{\eta\,,\rho\,,\chi^{\prime}}}+iI_{{}_{\eta\,,\rho\,,\chi^{\prime}}})\,.$
(12)

The potential above provides the following set of minimal constraint equations

$\displaystyle\mu^{2}_{\chi}+\lambda_{1}v^{2}_{\chi^{\prime}}+\frac{\lambda_{4}}{2}v^{2}_{\eta}+\frac{\lambda_{5}}{2}v^{2}_{\rho}-\frac{f}{2}\frac{v_{\eta}v_{\rho}}{v_{\chi^{\prime}}}=0,$
$\displaystyle\mu^{2}_{\eta}+\lambda_{2}v^{2}_{\eta}+\frac{\lambda_{4}}{2}v^{2}_{\chi^{\prime}}+\frac{\lambda_{6}}{2}v^{2}_{\rho}-\frac{f}{2}\frac{v_{\chi^{\prime}}v_{\rho}}{v_{\eta}}=0,$
$\displaystyle\mu^{2}_{\rho}+\lambda_{3}v^{2}_{\rho}+\frac{\lambda_{5}}{2}v^{2}_{\chi^{\prime}}+\frac{\lambda_{6}}{2}v^{2}_{\eta}-\frac{f}{2}\frac{v_{\eta}v_{\chi^{\prime}}}{v_{\rho}}=0.$
(13)

After spontaneous breaking of the symmetry the scalar sector get composed by
three CP-even scalars, $h$, $H_{1}$, and $H_{2}$ where the first recovers the
features of the standard Higgs and the other two are heavy scalars with mass
proportional to $v_{\chi^{\prime}}$ de S. Pires and Rodrigues da Silva (2007).
There are also two heavy singly charged scalars and one pseudoscalar $A$. Our
interest here lies in $A$. The justification comes below.

Considering the basis $(I_{\chi^{\prime}},I_{\eta},I_{\rho})$, the mass matrix
that involves the CP-odd scalars are

$M_{I}=\begin{pmatrix}fv_{\eta}v_{\rho}/4v_{\chi^{\prime}}&fv_{\rho}/4&fv_{\eta}/4\\\
fv_{\rho}/4&fv_{\chi^{\prime}}v_{\rho}/4v_{\eta}&fv_{\chi^{\prime}}/4\\\
fv_{\eta}/4&fv_{\chi^{\prime}}/4&fv_{\chi^{\prime}}v_{\eta}/4v_{\rho}\end{pmatrix}.$
(14)

Besides $M_{I}$ is a complex $3\times 3$ matrix, its texture presents a high
level of symmetry such that its diagonalization leads to two null eigenvalues.
In order to simplify the level of complexity, we also assume throughout this
work that $v_{\chi^{\prime}}\gg v_{\eta},v_{\rho}$ and that
$v_{\eta}=v_{\rho}=v$. In this case the eigenvectors $P\sim I_{\chi^{\prime}}$
and $G\approx\frac{1}{\sqrt{2}}I_{\eta}-\frac{1}{\sqrt{2}}I_{\rho}$ represent
the goldstone bosons eaten by the gauge bosons $Z^{\prime}$ and $Z$ Cao and
Zhang (2016); de S. Pires and Rodrigues da Silva (2007). The third eigenvector
is massive and we call $A$. Its mass is given by
$m^{2}_{A}=\frac{f}{2}v_{\chi^{\prime}}$ and its mixing with the eigenstates
of symmetry is $A=\frac{1}{\sqrt{2}}I_{\eta}+\frac{1}{\sqrt{2}}I_{\rho}$.
Observe that $m_{A}$ depends directly on $f$ which is a free parameter of the
model. The LHC lower bound on $v_{\chi^{\prime}}$ found in Refs. Coutinho  _et
al._ (2013); Alves  _et al._ (2022) requires $v_{\chi^{\prime}}>10^{4}$ GeV.
Then, whatever be the value of $v_{\chi^{\prime}}$, the parameter $f$ may
lower or rise $m_{A}$. Once we do not know how to constraint $f$, then $A$ may
be the lightest 3-3-1 particle that mediate Flavor Changing Neutral Current
(FCNC) processes as $K^{0}-\bar{K}^{0}$ mixing. If this is the case, it may
give the main contribution to $\Delta m_{K}$222For previous studies of FCNC in
3-3-1 models, see: Ng (1994); Liu and Ng (1994); Gomez Dumm  _et al._ (1994);
Long and Van (1999); Rodriguez and Sher (2004); Cabarcas  _et al._ (2008);
Promberger  _et al._ (2007); Benavides  _et al._ (2009); Cabarcas  _et al._
(2010, 2012); Jaramillo and Sanchez (2011); Cogollo  _et al._ (2012); Machado
_et al._ (2013); Okada  _et al._ (2016); Queiroz  _et al._ (2016); Huitu and
Koivunen (2019); Nguyen Tuan  _et al._ (2021).. This is the reason why we
focus on $A$ to investigate the role of family discrimination in flavor
changing neutral processes. In other words, we investigate what are the bounds
on $m_{A}$ put by meson mixing data according to family discrimination. In
what follow we obtain the Yukawa interactions composed by $A$ and the quarks
that contribute to meson mixing transitions for the three case of interest
here.

##  III Possibles versions

###  III.1 Case I

In this case the first two families of quarks transform as anti-triplet while
the third one transforms as triplet by $\text{SU}(3)_{\text{L}}$

$\displaystyle Q_{i_{L}}=\left(\begin{array}[]{c}d_{i}\\\ -u_{i}\\\
d^{\prime}_{i}\end{array}\right)_{L}\sim(3\,,\,\bar{3}\,,\,0)\,,u_{iR}\,\sim(3,1,2/3),\,\,\,$
(18)
$\displaystyle\,\,d_{iR}\,\sim(3,1,-1/3)\,,\,\,\,\,d^{\prime}_{iR}\,\sim(3,1,-1/3),$
$\displaystyle Q_{3L}=\left(\begin{array}[]{c}u_{3}\\\ d_{3}\\\
u^{\prime}_{3}\end{array}\right)_{L}\sim(3\,,\,3\,,\,1/3),u_{3R}\,\sim(3,1,2/3),$
(22)
$\displaystyle\,\,d_{3R}\,\sim(3,1,-1/3)\,,\,u^{\prime}_{3R}\,\sim(3,1,2/3),$
(23)

where the index $i=1,2$ is restricted to only two generations. The primed
quarks are new heavy quarks with the usual $(+\frac{2}{3},-\frac{1}{3})$
electric charges.

In this case the simplest Yukawa interactions that generate the correct mass
for all standard quarks are composed by the terms333For the most general
Yukawa interactions involving terms that violate lepton number, see: Doff  _et
al._ (2006).,

$-{\cal L}_{Y}\supset
g^{1}_{ia}\bar{Q}_{i_{L}}\eta^{*}d_{a_{R}}+h^{1}_{3a}\bar{Q}_{3_{L}}\eta
u_{a_{R}}+g^{1}_{3a}\bar{Q}_{3_{L}}\rho
d_{a_{R}}+h^{1}_{ia}\bar{Q}_{i_{L}}\rho^{*}u_{a_{R}}+\mbox{H.c.}\,,$ (24)

where $a=1,2,3$.

Let us consider the standard up quarks. For the basis
$u=(u_{1}\,,\,u_{2}\,,\,u_{3})$ we get the following mass matrix

$M_{u}=\frac{v}{\sqrt{2}}\begin{pmatrix}-h_{11}&-h_{12}&-h_{13}\\\
-h_{21}&-h_{22}&-h_{23}\\\ h_{31}&h_{32}&h_{33}\end{pmatrix},$ (25)

diagonalizing this matrix by a bi-unitary transformation

$V_{L}^{u\dagger}M_{u}V_{R}^{u}=\begin{pmatrix}m_{u}&0&0\\\ 0&m_{c}&0\\\
0&0&m_{t}\end{pmatrix},$ (26)

we get the masses of the up quarks. The relation among the basis is given by

$\hat{u}_{L,R}=V^{\dagger u}_{L,R}u_{L,R}\,,$ (27)

with $\hat{u}=(u\,\,,\,\,c\,\,,\,\,t)^{T}$.

For the down quarks we have the mass matrix

$M_{d}=\frac{v}{\sqrt{2}}\begin{pmatrix}g_{11}&g_{12}&g_{13}\\\
g_{21}&g_{22}&g_{23}\\\ g_{31}&g_{32}&g_{33}\end{pmatrix},$ (28)

diagonalizing this matrix by a bi-unitary transformation

$V_{L}^{d\dagger}M_{d}V_{R}^{d}=\begin{pmatrix}m_{d}&0&0\\\ 0&m_{s}&0\\\
0&0&m_{b}\end{pmatrix},$ (29)

we get the masses of the down quarks. The relation among the basis is given by

$\hat{d}_{L,R}=V^{\dagger d}_{L,R}d_{L,R}$ (30)

with $\hat{d}=(d\,\,,\,\,s\,\,,\,\,b)^{T}$. These mixing matrices
$V_{L}^{d,u}$ are related by the constraint
$V^{u}_{L}V^{d\dagger}_{L}=V_{\text{CKM}}$, where Workman (2022)

$|V_{\text{CKM}}|=\begin{pmatrix}0.97435&0.22500&0.00369\\\
0.22486&0.97349&0.04182\\\ 0.00857&0.04110&0.999118\end{pmatrix},$ (31)

is the CKM matrix.

Note that $V_{R}^{d,u}$ play no role at all. Then we assume $V_{R}^{d,u}=I$.
In this case Eqs.(25), (26), (28), and (31) give

$\displaystyle
g^{1}_{ia}=\sqrt{2}(V^{d}_{L})_{ia}\frac{(m_{down})_{a}}{v}\,,\,g^{1}_{3a}=\sqrt{2}(V^{d}_{L})_{3a}\frac{(m_{down})_{a}}{v}\,,$
$\displaystyle
h^{1}_{ia}=-\sqrt{2}(V^{u}_{L})_{ia}\frac{(m_{up})_{a}}{v}\,,\,h^{1}_{3a}=\sqrt{2}(V^{u}_{L})_{3a}\frac{(m_{up})_{a}}{v},$
(32)

where $i=1,2$, $a=1,2,3$, $(m_{down})_{a}=m_{d},m_{s},m_{b}$, and
$(m_{up})_{a}=m_{u},m_{c},m_{t}$.

After all this, in the end of the day we obtain the following Yukawa
interactions among $A$ and the physical standard quarks

$\displaystyle{\cal L}_{Y}^{A}$ $\displaystyle=$ $\displaystyle
iA\bar{\hat{u}}_{b_{L}}\left(\frac{1}{\sqrt{2}}(V^{u}_{L})_{3a}(V^{u}_{L})_{b3}\frac{(m_{up})_{a}}{v}-\frac{1}{\sqrt{2}}(V^{u}_{L})_{ia}(V^{u}_{L})_{bi}\frac{(m_{up})_{a}}{v}\right)\hat{u}_{a_{R}}$
(33) $\displaystyle+$ $\displaystyle
iA\bar{\hat{d}}_{b_{L}}\left(-\frac{1}{\sqrt{2}}(V^{d}_{L})_{ia}(V^{d}_{L})_{bi}\frac{(m_{dowm})_{a}}{v}+\frac{1}{\sqrt{2}}(V^{d}_{L})_{3a}(V^{d}_{L})_{b3}\frac{(m_{dowm})_{a}}{v}\right)\hat{d}_{a_{R}}+\text{H.c.}\,,$

with the subscript $b=1,2,3$.

###  III.2 Case II

In this case the first family transforms as triplet and the second and third
as anti-triplet, which means

$\displaystyle Q_{1_{L}}=\left(\begin{array}[]{c}u_{1}\\\ d_{1}\\\
u^{\prime}_{1}\end{array}\right)_{L}\sim(3\,,\,3\,,\,1/3)\,,u_{1R}\,\sim(3,1,2/3),\,\,\,$
(37)
$\displaystyle\,\,d_{1R}\,\sim(3,1,-1/3)\,,\,\,\,\,u^{\prime}_{iR}\,\sim(3,1,2/3),$
$\displaystyle Q_{iL}=\left(\begin{array}[]{c}d_{i}\\\ -u_{i}\\\
d^{\prime}_{i}\end{array}\right)_{L}\sim(3\,,\,\bar{3}\,,\,0),u_{3R}\,\sim(3,1,2/3),$
(41)
$\displaystyle\,\,d_{3R}\,\sim(3,1,-1/3)\,,\,d^{\prime}_{3R}\,\sim(3,1,-1/3),$
(42)

where the index $i=2,3$ is restricted to only two generations.

The minimal set of Yukawa interactions that leads to the correct quark masses
involves the terms,

$\displaystyle-$ $\displaystyle{\cal L}_{Y}\supset
g^{2}_{1a}\bar{Q}_{1_{L}}\rho
d_{a_{R}}+g^{2}_{ia}\bar{Q}_{i_{L}}\eta^{*}d_{a_{R}}$ (43)
$\displaystyle+h^{2}_{1a}\bar{Q}_{1_{L}}\eta
u_{a_{R}}+h^{2}_{ia}\bar{Q}_{i_{L}}\rho^{*}u_{a_{R}}+\mbox{H.c.}\,.$

Following all the previous procedures done in Case I, we obtain the following
Yukawa interactions among $A$ and the standard quarks

$\displaystyle{\cal L}_{Y}^{A}$ $\displaystyle=$ $\displaystyle
iA\bar{\hat{u}}_{b_{L}}\left(\frac{1}{\sqrt{2}}(V^{u}_{L})_{1a}(V^{u}_{L})_{b1}\frac{(m_{up})_{a}}{v}-\frac{1}{\sqrt{2}}(V^{u}_{L})_{ia}(V^{u}_{L})_{bi}\frac{(m_{up})_{a}}{v}\right)\hat{u}_{a_{R}}$
(44) $\displaystyle+$ $\displaystyle
iA\bar{\hat{d}}_{b_{L}}\left(\frac{1}{\sqrt{2}}(V^{d}_{L})_{1a}(V^{d}_{L})_{b1}\frac{(m_{dowm})_{a}}{v}-\frac{1}{\sqrt{2}}(V^{d}_{L})_{ia}(V^{d}_{L})_{bi}\frac{(m_{dowm})_{a}}{v}\right)\hat{d}_{a_{R}}+\mbox{H.c.}\,,$

with the subscript $b=1,2,3$.

###  III.3 Case III

In this case the second family transforms as triplet while the first and third
transform as anti-triplet

$\displaystyle Q_{2_{L}}=\left(\begin{array}[]{c}u_{2}\\\ d_{2}\\\
u^{\prime}_{2}\end{array}\right)_{L}\sim(3\,,\,3\,,\,1/3)\,,u_{2R}\,\sim(3,1,2/3),\,\,\,$
(48)
$\displaystyle\,\,d_{2R}\,\sim(3,1,-1/3)\,,\,\,\,\,u^{\prime}_{2R}\,\sim(3,1,2/3),$
$\displaystyle Q_{iL}=\left(\begin{array}[]{c}d_{i}\\\ -u_{i}\\\
d^{\prime}_{i}\end{array}\right)_{L}\sim(3\,,\,\bar{3}\,,\,0),u_{iR}\,\sim(3,1,2/3),$
(52)
$\displaystyle\,\,d_{iR}\,\sim(3,1,-1/3)\,,\,d^{\prime}_{iR}\,\sim(3,1,-1/3),$
(53)

where $i=1,3$.

In this case the Yukawa interactions among the scalars and the standard quarks
are

$\displaystyle-$ $\displaystyle{\cal L}_{Y}\supset
g^{3}_{2a}\bar{Q}_{2_{L}}\rho
d_{a_{R}}+g^{3}_{ia}\bar{Q}_{i_{L}}\eta^{*}d_{a_{R}}$ (54)
$\displaystyle+h^{3}_{2a}\bar{Q}_{2_{L}}\eta
u_{a_{R}}+h^{3}_{ia}\bar{Q}_{i_{L}}\rho^{*}u_{a_{R}}+\mbox{H.c.},$

where $i=1,3$.

Following the procedure of Case I the interactions of the pseudoscalar $A$
with the standard quarks are

$\displaystyle{\cal L}_{Y}^{A}$ $\displaystyle=$ $\displaystyle
iA\bar{\hat{d}}_{b_{L}}\left(\frac{1}{\sqrt{2}}(V^{d}_{L})_{2a}(V^{d}_{L})_{b2}\frac{(m_{down})_{a}}{v}-\frac{1}{\sqrt{2}}(V^{d}_{L})_{ia}(V^{d}_{L})_{b}i\frac{(m_{down})_{a}}{v}\right)\hat{d}_{a_{R}}$
(55)
$\displaystyle+iA\bar{\hat{u}}_{b_{L}}\left((V^{u}_{L})_{2a}(V^{u}_{L})_{b2}\frac{(m_{up})_{a}}{v}-\frac{1}{\sqrt{2}}(V^{u}_{L})_{ia}(V^{u}_{L})_{bi}\frac{(m_{up})_{a}}{v}\right)\hat{u}_{a_{R}}+\mbox{H.c.}\,,$

with the subscript $b=1,2,3$.

Observe that the interactions in Eqs. (33), (44) and (55) lead to processes
that violate flavor mediated by $A$ as, for example, $K^{0}-\bar{K}^{0}$
oscillation. We call the attention to the fact that each case has its proper
set of Yukawa couplings but we used only one set of $V^{u,d}_{L}$ to
diagonalize the different set of mass matrices that appear in each case. This
is possible because each mixing matrix may diagonalize more than one mass
matrix when this mass matrix is Dirac type.

##  IV FCNC constraint on $\mathbf{m_{A}}$ from $K^{0}-\bar{K}^{0}$ transition

We saw in the previous section that $A$ mediates processes that violate flavor
at tree level. Those processes may be, for example,
$K^{0}(s\bar{d})-\bar{K}^{0}(\bar{s}d)$,
$D^{0}(c\bar{u})-\bar{D}^{0}(\bar{c}u)$, and
$B^{0}(d\bar{b})-\bar{B}^{0}(\bar{d}b)$ transitions. For our proposal here it
is just necessary to investigate the impact of the contribution of $A$ to
$K^{0}(s\bar{d})-\bar{K}^{0}(\bar{s}d)$ in the three cases discussed in the
previous sections. On requiring that such contribution recover the
experimental results, we obtain a lower bound on $m_{A}$. We will see that
such constraint vary for each case of family discrimination. For attaining our
proposal we must obtain the effective lagrangian responsible by the
transition.

From the lagrangian in Eqs.(33), (44), and (55), following the procedure in
Ref. Gabbiani  _et al._ (1996), we obtain the following effective lagrangian
that leads to the $K^{0}-\bar{K}^{0}$ transition,

$\displaystyle{\cal
L}^{K}_{eff}=-\frac{1}{m^{2}_{A}}[\bar{d}(C^{R}_{K}P_{R}-C^{L}_{K}P_{L})s]^{2},$
(56)

where $P_{R,L}$ are the left-handed and right-handed projections while the
coefficients $C^{L,R}_{K}$ are given below.

With this lagrangian in hand, and following the procedure of Ref.Gabbiani  _et
al._ (1996), we obtain the following general expression to the mixing
parameter $\Delta m_{K}$,

$\Delta
m_{K}=\frac{2m_{K}f_{K}^{2}}{m^{2}_{A}}\left[\frac{5}{24}Re\left((C^{L}_{K})^{2}+(C^{R}_{K})^{2}\right)\left(\frac{m_{K}}{m_{s}+m_{d}}\right)^{2}+2Re(C^{L}_{K}C^{R}_{K})\left[\frac{1}{24}+\frac{1}{4}\left(\frac{m_{K}}{m_{s}+m_{d}}\right)^{2}\right]\right]\,,$
(57)

where $f_{K}$ is the decay constant of $K^{0}$.

As we show below ,family discrimination will manifest clearly in the
expressions for the coefficients $C^{L,R}$ because each case leads to a
specific behavior of $C^{L,R}$ with the elements of $V^{u,d}_{L}$.

  * •

Case I

$\displaystyle
C^{R}_{K}=\frac{1}{\sqrt{2}}(V^{d}_{L})_{32}(V^{d}_{L})_{13}\frac{m_{s}}{v}-(V^{d}_{L})_{i2}(V^{d}_{L})_{1i}\frac{m_{s}}{v}\,,$
$\displaystyle
C^{L}_{K}=\frac{1}{\sqrt{2}}(V^{d}_{L})^{*}_{13}(V^{d}_{L})^{*}_{32}\frac{m_{d}}{v}-\frac{1}{\sqrt{2}}(V^{d}_{L})^{*}_{1i}(V^{d}_{L})^{*}_{i2}\frac{m_{d}}{v}\,,$
(58)

where $i=1,2$.

  * •

Case II

$\displaystyle
C^{R}_{K}=\frac{1}{\sqrt{2}}(V^{d}_{L})_{12}(V^{d}_{L})_{11}\frac{m_{s}}{v}-\frac{1}{\sqrt{2}}(V^{d}_{L})_{i2}(V^{d}_{L})_{1i}\frac{m_{s}}{v}\,,$
$\displaystyle
C^{L}_{K}=\frac{1}{\sqrt{2}}(V^{d}_{L})^{*}_{11}(V^{d}_{L})^{*}_{12}\frac{m_{d}}{v}-\frac{1}{\sqrt{2}}(V^{d}_{L})^{*}_{1i}(V^{d}_{L})^{*}_{i2}\frac{m_{d}}{v}\,,$
(59)

where $i=2,3$.

  * •

Case III

$\displaystyle
C^{R}_{K}=\frac{1}{\sqrt{2}}(V^{d}_{L})_{22}(V^{d}_{L})_{12}\frac{m_{s}}{v}-\frac{1}{\sqrt{2}}(V^{d}_{L})_{i2}(V^{d}_{L})_{1i}\frac{m_{s}}{v}\,,$
$\displaystyle
C^{L}_{K}=\frac{1}{\sqrt{2}}(V^{d}_{L})^{*}_{12}(V^{d}_{L})^{*}_{22}\frac{m_{d}}{v}-\frac{1}{\sqrt{2}}(V^{d}_{L})^{*}_{1i}(V^{d}_{L})^{*}_{i2}\frac{m_{d}}{v}\,,$
(60)

where $i=1,3$.

Observe that the process is very sensitive to family discrimination because
$\Delta m_{K}$ varies with $\left(V^{u,d}_{L}\right)^{4}$.

We now have in hand all the ingredients necessary for we obtain the
contributions of $A$ to the $K^{0}-\bar{K}^{0}$ meson transition for the three
cases and check how sensitive is this transition regarding family
discrimination.

##  V Numerical results

The coefficients $C^{L,R}_{K}$ depend on the mixing matrices $V^{u}_{L}$ and
$V^{d}_{L}$ whose entries are free parameters that obey the constraint
$V^{u}_{L}V^{d\dagger}_{L}=V_{\text{CKM}}$. Remember that the idea here is to
check if family discrimination has impact on these kind of process. As usual,
we must choose a parametrization to $V^{u,d}_{L}$. Here, as illustrative
examples, we use the following one:

Figure 1: Numerical results for case I (continuous curve), case II (dashed
curve) and case II (dashed dotted curve).

$V_{L}^{d}=\begin{pmatrix}-0.40573&0.73916&-0.53108\\\
0.073685&-0.11229&-0.98774\\\ 0.88355&0.46336&-0.06802\end{pmatrix}\,,$ (61)
$V_{L}^{u}=\begin{pmatrix}-0.45440&0.82278&-0.34139\\\
0.13857&-0.31329&-0.93949\\\ 0.87996&0.47421&-0.02834\end{pmatrix}\,.$ (62)

This parametrization reproduces the experimental values of the elements of CKM
matrix. Thus it is in the game. Concerning the VEV’s, we take
$v_{\chi^{\prime}}=10^{4}$ GeV and $v^{2}_{\eta}+v^{2}_{\rho}=(246\mbox{
GeV})^{2}$. The other input parameters are find in Table 1. These VEV’s
guarantee that the mass of $Z^{\prime}$ satisfies the LHC constraint Coutinho
_et al._ (2013); Alves  _et al._ (2022). In what concern $H_{1}$ and $H_{2}$
their masses are proportional to $v_{\chi^{\prime}}$ then they are heavy
particle de S. Pires and Rodrigues da Silva (2007). By the fact that
$m^{2}_{A}=\frac{1}{2}fv_{\chi^{\prime}}$, and as $f$ is a free parameter,
this pseudoscalar may be light enough and, then, give the dominant
contribution to $K^{0}-\bar{K}^{0}$ mixing. This justify why we are not
worried about the contributions of $Z^{\prime}$, $H_{1}$ and $H_{2}$ to such
transition.

Input Parameters
---
Quark masses |  $K^{0}$
$m_{u}=2.16\ \text{MeV}$ |  $m_{K}=497.611\ \text{MeV}$
$m_{c}=1.27\ \text{GeV}$ |  $f_{K}=156\ \text{MeV}$
$m_{s}=93.4\ \text{MeV}$ |  $\Delta m_{K}=\left(3.484\pm 0.006\right)\times 10^{-12}\ \text{MeV}$
$m_{b}=4.18\ \text{GeV}$ |
$m_{d}=4.67\ \text{MeV}$ |
Table 1: Input parameters Workman (2022).

Before go to the results, there are two things we would like to comment.
Firstly, we do not dispose of a mechanism or process that provides the correct
parametrization of $V^{u,d}_{L}$. The only constraint we dispose is that they
recover CKM. So there are many others proposals of parametrization to these
matrices. Our result here depends strongly on the parametrization of the
mixing matrices $V^{u,d}_{L}$. However, we can not believe that any other
possible parametrization would change considerably our results.

As second point, the current experimental value for the mass difference
$\Delta m_{K}$ is Amhis  _et al._ (2021); Yao  _et al._ (2006); Chen  _et al._
(2021); Aaij  _et al._ (2013)

$\displaystyle\Delta m_{K}=\left(3.484\pm 0.006\right)\times 10^{-12}\text{
MeV}.$

The discrepancy among experimental and theoretical (standard model $(\Delta
m)_{\text{SM}}$) predictions define the window to new physics. It happens that
the standard model prediction involves a considerable amount of uncertainties
due to the QCD corrections Wang (2019, 2022). But even with the uncertainties
the standard model prediction is in good accordance with experimental result
Buchalla  _et al._ (1996); Di Luzio  _et al._ (2019); De Bruyn  _et al._
(2022). In view of this, we adopted a conservative perspective and took the
experimental errors as the window to new physics. In other word, we demand
that the 3-3-1 contribution be no larger than the experimental error,

$\displaystyle\Delta m_{K}=0.006\times 10^{-12}\mbox{ MeV}.$ (63)

Doing this we assure that the dominant contribution of the 331RHN to the
$K^{0}-\bar{K}^{0}$ mixing obey whatever will be the window to new physics
when the uncertainties of the standard model prediction reduce to acceptable
values Wang (2022). It is based on these arguments that we displayed our
findings in FIG.1. According to our illustrative parametrization, the case III
suffers the most severe bound 444It is curious because this case never was
developed in literature, while case II, where the first family of quarks
transform as triplet, received the less severe bounds and case I receives
intermediate bound but much more severe that case II and well less severe than
case III. Then, family discrimination indicates that the 331RHN with the first
family of quarks transforming as triplet is the case energetically favoured to
be probed at the LHC.

According to our results, if LHC detects a pseudoscalar with mass at the
electroweak scale, the whole 3-3-1 model is discarded. But if LHC finds a
pseudoscalar around $2$ TeV, the case I and case III that are discarded. In
summary, as flavor changing processes are a natural outcome of the 3-3-1
models, then, it is this kind of process that has the potential to confirm or
discard one of the three cases or discard even the whole model. So the study
of flavor changing processes is very important for the status of the 3-3-1
model.

Note that once $m^{2}_{A}=\frac{1}{2}fv_{\chi^{\prime}}$, then for the case I
we get $f\gtrsim 12$ TeV and in case II we get $f\gtrsim 578$ GeV while in
case III we get $f\gtrsim 145$ TeV. So, the contribution of $A$ to
$K^{0}-\bar{K}^{0}$ also constraint the parameter $f$.

##  VI Conclusions

In this work we investigated flavor changing neutral current in order to see
the effects of family replication in the 331RHN. The process base was
$K^{0}-\bar{K}^{0}$. We took into account the LHC bound on the $Z^{\prime}$
and assumed that the only pseudoscalar of the model gives the dominant
contribution to the process. As verified family discrimination dismembers the
331RHN into three variants and for each variant studied we obtained different
bounds on $m_{A}$. Family discrimination manifests strongly in flavor changing
neutral current processes. Moreover, the constraint we got on the mass of $A$
translates directly into constraints on the parameter $f$. As far as we know,
this is the first work in the literature that constrain the parameter $f$.
According to our results, the variant in which the third family of quarks
transform as triplet puts the strongest constraint on the mass of $A$. So this
variant is poorly favoured phenomenologically. The variant in which the second
family of quarks transform as triplet is also poorly favoured, but the variant
in which the first family of quarks transform as triplet is mildly constrained
which means that such case is the most interesting one from the point of view
of collider phenomenology. Finally, for we know exactly what variant of the
331RHN may be realized in nature, we need to conjugate information from flavor
changing neutral currents and collider phenomenology.

## Acknowledgments

C.A.S.P was supported by the CNPq research grants No. 311936/2021-0 and V.O
was supported by CAPES.

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# Braiding and fusion of Majorana fermions in minimal Kitaev spin liquid
on a single hexagon with $5$ qubits

Motohiko Ezawa  Department of Applied Physics, University of Tokyo, Hongo
7-3-1, 113-8656, Japan

###### Abstract

We investigate the minimal Kitaev spin liquid on a single hexagon with three
Ising-type exchange interactions proportional to $K_{x}$, $K_{y}$ and $K_{z}$.
In the limit $K_{z}=0$, we find 32-fold zero-energy states, leading to 10 free
Majorana fermions, and hence, 5 qubits are constructed. These qubits are
protected by particle-hole symmetry even for $K_{z}\neq 0$. Braiding of these
Majorana fermions is possible by temporally controlling a spin-correlation
Hamiltonian. In addition, the fusion is possible by measuring the spin
correlation. By switching on the Heisenberg interaction together with magnetic
field, only one zero-energy state persists, which can be used as an
initialization of qubits. Furthermore, it is shown that $3L+2$ qubits are
constructed on the Kitaev spin liquid model on connected $L$ hexagons. All the
processes of initialization, operation and readout of qubits are executable in
terms of spin operators.

Introduction: A quantum computer is expected to be a most promising next
generation computer[1, 2, 3], which can store $2^{N}$ information in $N$ qubit
systems. A topological quantum computation based on Majorana fermions[4, 5, 6,
7, 8, 9, 10, 11] is attractive, where $N$ qubits are constructed from $2N$
Majorana fermions. Majorana fermions are theoretically proposed to emerge in
fractional quantum Hall effects[12, 13, 9, 14], topological
superconductors[15, 16, 17, 18] and Kitaev spin liquids[19, 20, 21, 22]. The
Kitaev topological superconductor model is the simplest fermionic model that
hosts Majorana fermions[6]. Recently, the Minimal Kitaev model consisting of
only two sites is realized in double quantum dots[23, 24]. In the view point
of quantum computation, it is desirable to construct a model hosting many
Majorana fermions with the use of smaller number of sites.

The Kitaev spin liquid is one of the prominent exactly solvable models on the
honeycomb lattice realizing spin liquid with the emergence of Majorana zero
modes[19]. It is defined on the honeycomb lattice, where there are three
Ising-type exchange interactions ($\varpropto K_{x},K_{y},K_{z}$) depending on
the directions of bonds as shown in Fig.1(a). By representing the spin
operator by a combination of Majorana fermion operators, the system turns into
a free Majorana fermion model on the honeycomb lattice[19]. The theoretical
proposal on the Kitaev spin liquid based on perovskite materials[20] evokes an
intensive researches[25, 26, 27]. Experimental signature of Majorana fermions
is observed by measuring a half quantization of thermal conductivity[27, 28].
Brading of Majorana states at vortices in the Kitaev spin liquid is
theoretically proposed[29]. In addition to perovskite material realization,
there are several proposals on the realization of the Kitaev spin liquid model
in artificial systems such as qubits[30, 31], trapped ions[32], cold atoms[33,
34] and quantum dots[35]. One of the merit of these systems is that it is
possible to construct an extremely small-size system and control model
parameters temporally. The simplest system is a single hexagon. It is an
interesting problem to study whether the Kitaev spin liquid model hosts
Majorana fermions on a single hexagon.

In this paper, we investigate the minimal Kitaev spin liquid on a single
hexagon. When $K_{z}=0$, there are 32=25 zero-energy states, which leads to 5
qubits. These qubits are protected by particle-hole symmetry even for
$K_{z}\neq 0$. Then, we construct braiding operators in terms of spin
operators. A readout process of qubits is performed by the fusion protocol
which measures the local spin correlation. An initialization is executed by
switching on the Heisenberg interaction and magnetic field term, where only
one zero-energy state is present.

Figure 1: Kitaev spin liquid model on a single hexagon. (a) Illustration of
the Kitaev spin liquid model on a single hexagon, where there are three-types
of Ising interactions. (b) Energy spectrum with $K_{z}=0$. There are 32-fold
degenerate zero-energy states. (c) Energy spectrum as a function of $K_{z}/K$.
There are 24-fold degenerate zero-energy states and 8 nonzero energy states
near the zero energy. (d) Energy spectrum with $K_{z}=K/4$. We have set
$K_{x}=K_{y}=K$. The horizontal axes in (b) and (d) are eigenindices in the
increasing order of the eigenenergy.

Minimal Kitaev spin liquid model: We study the Kitaev spin liquid model in an
artificial system. Especially, we analyze the spin $1/2$ system on the single
hexagon based on the Kitaev model,

$\hat{H}_{\text{K}}=\sum_{\alpha=1}^{2}\hat{H}_{\text{K}\alpha}+\hat{H}_{\text{K}z},$
(1)

with

$\displaystyle\hat{H}_{\text{K}\alpha}$
$\displaystyle=-K_{x}\sigma_{3\alpha-2}^{x}\sigma_{3\alpha-1}^{x}-K_{y}\sigma_{3\alpha-1}^{y}\sigma_{3\alpha}^{y},$
(2) $\displaystyle\hat{H}_{\text{K}z}$
$\displaystyle=-K_{z}\left(\sigma_{3}^{z}\sigma_{4}^{z}+\sigma_{6}^{z}\sigma_{1}^{z}\right),$
(3)

where $\sigma_{i}^{\gamma}$ is the Pauli matrix at the site $i$ with
$\gamma=x,y,z$. The Ising-type exchange interaction
$K_{\gamma}\sigma_{i}^{\gamma}\sigma_{j}^{\gamma}$ is anisotropic depending on
the direction of the link $\gamma$ as illustrated in Fig.1(a). It contains $6$
spins, and hence, there are $2^{6}$ states in total.

The Hamiltonian for the Kitaev quantum spin liquid is rewritten in terms of
Majorana fermions by way of the Jordan Wigner transformation[36, 37, 38, 39,
40, 41]. We number the site from 1 to 6 as shown in Fig.1(a). There are
relations between the spin operators and the fermion operators:
$\sigma_{j}^{-}=\Omega_{j}c_{j}$, $\sigma_{j}^{+}=\Omega_{j}c_{j}^{\dagger}$,
$\sigma_{j}^{z}=c_{j}^{\dagger}c_{j}-1/2$ with
$\Omega_{j}\equiv\prod\limits_{\ell=1}^{j-1}\exp[i\pi
c_{\ell}^{\dagger}c_{\ell}]$ and
$\sigma_{j}^{+}\equiv\frac{1}{2}(\sigma_{j}^{x}+i\sigma_{j}^{y})$ and
$\sigma_{j}^{-}\equiv\frac{1}{2}(\sigma_{j}^{x}-i\sigma_{j}^{y})$. Here,
$c_{j}$ and $c_{j}^{\dagger}$ satisfy the anti-commutation relations,
$\\{c_{i},c_{\ell}\\}=\\{c_{i}^{\dagger},c_{\ell}^{\dagger}\\}=0$,
$\\{c_{i},c_{\ell}^{\dagger}\\}=\delta_{i\ell}$. Furthermore, we introduce
Majorana operators as $\gamma_{2j}^{A}=c_{2j}+c_{2j}^{\dagger}$,
$\gamma_{2j}^{B}=(c_{2j}-c_{2j}^{\dagger})/i$,
$\gamma_{2j+1}^{A}=(c_{2j+1}-c_{2j+1}^{\dagger})/i$,
$\gamma_{2j+1}^{B}=c_{2j+1}+c_{2j+1}^{\dagger}$ for $j=1,2,3$. Then, the
Hamiltonians (2) and (3) read

$\displaystyle\hat{H}_{\text{K}\alpha}$
$\displaystyle=-i\left(K_{x}\gamma_{3\alpha-2}^{A}\gamma_{3\alpha-1}^{A}-K_{y}\gamma_{3\alpha-1}^{A}\gamma_{3\alpha}^{A}\right),$
(4) $\displaystyle\hat{H}_{\text{K}z}$
$\displaystyle=-\frac{K_{z}}{4}(\gamma_{3}^{A}\gamma_{3}^{B}\gamma_{4}^{A}\gamma_{4}^{B}+\gamma_{6}^{A}\gamma_{6}^{B}\gamma_{1}^{A}\gamma_{1}^{B}),$
(5)

in the Majorana form.

Particle-hole symmetry: The zero-energy states of the Hamiltonian with
particle-hole symmetry are Majorana fermions[15, 16, 17, 18]. We discuss
particle-hole symmetry in the Kitaev spin liquid. Particle-hole symmetry acts
as $P^{-1}\gamma_{j}^{A}P=\gamma_{j}^{A}$ and
$P^{-1}\gamma_{j}^{B}P=\gamma_{j}^{B}$ in terms of Majorana fermion operators,
or $P^{-1}c_{j}P=c_{j}^{\dagger}$ and $P^{-1}c_{j}^{\dagger}P=c_{j}$ in terms
of fermion operators. In terms of the spin operators, it acts as

$\displaystyle P^{-1}\sigma_{j}^{x}P$
$\displaystyle=\left(-1\right)^{j-1}\sigma_{j}^{x},\qquad
P^{-1}\sigma_{j}^{y}P=\left(-1\right)^{j}\sigma_{j}^{y},$ $\displaystyle
P^{-1}\sigma_{j}^{z}P$ $\displaystyle=-\sigma_{j}^{z},$ (6)

where we have used the relation
$P^{-1}\Omega_{j}P=\left(-1\right)^{j-1}\Omega_{j}$.

Under the particle-hole symmetry transformation, the Hamiltonian (1) is mapped
to
$P^{-1}\hat{H}_{\text{K}}\left(K_{x},K_{y},K_{z}\right)P=-\hat{H}_{\text{K}}\left(K_{x},K_{y},-K_{z}\right)$.
Hence, the Hamiltonian has particle-hole symmetry for $K_{z}=0$. We later show
that particle-hole symmetry is present even for $K_{z}\neq 0$ in the present
model, as is consistant with Fig.1(d).

Minimal Kitaev spin chain models: We analyze the minimal Kitaev spin liquid
model where $K_{z}$ is much smaller than $K_{x}$ and $K_{y}$. We first
consider the limit $K_{z}=0$ and later include the effect due to $K_{z}\neq
0$.

When we set $K_{z}=0$ in Hamiltonian (1), it is decomposed[8, 37, 41] into two
independent Kitaev spin chain models $\hat{H}_{\text{K}\alpha}$. There are
$2^{3}$ states because there are 3 spins for each $\alpha$. By exactly
diagonalizing $\hat{H}_{\text{K}\alpha}$ for each $\alpha$, we find that there
are 4-fold degenerate states with
$E_{\text{K1}}=E_{\text{K2}}=\pm\sqrt{K_{x}^{2}+K_{y}^{2}}$. Note that there
are no zero-energy states in each minimal Kitaev spin chain model.

However, the combined system has $8\times 8=64$ states made of 32 zero-energy
states and 32 nonzero-energy states irrespective of $K_{x}$ and $K_{y}$. The
Hamiltonian (4) is rewritten in the form

$\sum_{\alpha=1}^{2}\hat{H}_{\text{K}\alpha}=\sqrt{K_{x}^{2}+K_{y}^{2}}\left(i\gamma_{2}^{A}\bar{\gamma}_{1}^{A}-i\gamma_{5}^{A}\bar{\gamma}_{4}^{A}\right),$
(7)

where we have defined new Majorana operators

$\bar{\gamma}_{1}^{A}=\frac{K_{x}\gamma_{1}^{A}+K_{y}\gamma_{3}^{A}}{\sqrt{K_{x}^{2}+K_{y}^{2}}},\qquad\
\bar{\gamma}_{4}^{A}=\frac{K_{x}\gamma_{4}^{A}+K_{y}\gamma_{6}^{A}}{\sqrt{K_{x}^{2}+K_{y}^{2}}}.$
(8)

Since the Hamiltonian (7) does not contain Majorana operators
$\gamma_{3}^{A}$, $\gamma_{6}^{A}$, $\gamma_{j}^{B}$ with $j=1,2,\cdots,6$, we
have
$\left[\hat{H}_{\text{K}},\gamma_{3}^{A}\right]=\left[\hat{H}_{\text{K}},\gamma_{6}^{A}\right]=0$
and $\left[\hat{H}_{\text{K}},\gamma_{j}^{B}\right]=0$. Hence, there are 8
free Majorana fermions, from which we construct 4 fermion operators as
$f_{1}^{A}\equiv\left(\gamma_{3}^{A}-i\gamma_{6}^{A}\right)/2$ and
$f_{j}^{B}=\left(\gamma_{2j-1}^{B}-i\gamma_{2j}^{B}\right)/2$ with $j=1,2,3$.
In addition, we introduce 2 fermion operators
$f_{2}^{A}\equiv\left(\bar{\gamma}_{1}^{A}-i\gamma_{2}^{A}\right)/2$ and
$f_{3}^{A}\equiv\left(\bar{\gamma}_{4}^{A}-i\gamma_{5}^{A}\right)/2$. The
Hamiltonian (7) is rewritten in terms of these fermion operators as

$\sum_{\alpha=1}^{2}\hat{H}_{\text{K}\alpha}=2\sqrt{K_{x}^{2}+K_{y}^{2}}\left(\hat{n}_{2}^{A}-\hat{n}_{3}^{A}\right),$
(9)

where we have defined the number operators $\hat{n}_{2}^{A}\equiv
f_{2}^{A\dagger}f_{2}^{A}=\left(i\gamma_{2}^{A}\bar{\gamma}_{1}^{A}+1\right)/2$
and $\hat{n}_{3}^{A}\equiv
f_{3}^{A\dagger}f_{3}^{A}=\left(i\gamma_{5}^{A}\bar{\gamma}_{4}^{A}+1\right)/2$.
In the similar way, we define $\hat{n}_{j}^{A}=f_{j}^{A\dagger}f_{j}^{A}$ and
$\hat{n}_{j}^{B}=f_{j}^{B\dagger}f_{j}^{B}$ with $j=1\sim 3$. We consider the
Hilbert space where $\hat{n}_{j}^{A}$ and $\hat{n}_{j}^{B}$ take the
eigenvalues $0$ and $1$. We take $f_{2}^{A}$ to be a free fermion. Then,
$f_{3}^{A}$ is determined by the zero-energy condition of the Hamiltonian (9).
As a result, the Kitaev spin liquid model on the single hexagon contains 10
free Majorana fermions, or 5 qubits defined by
$\left|n_{2}^{A}n_{1}^{A}n_{3}^{B}n_{2}^{B}n_{1}^{B}\right\rangle$.

Braiding: The basic operation on qubits is braiding defined[5] by
$\mathcal{B}_{\alpha\beta}=\exp\left[\frac{\pi}{4}\gamma_{\beta}\gamma_{\alpha}\right]$.
It is generalized[11] to an arbitrary angle such that
$\mathcal{B}_{\alpha\beta}\left(\theta\right)=\exp\left[\theta\gamma_{\beta}\gamma_{\alpha}\right]$.
The unitary dynamics under the Hamiltonian $H$ reads
$U=\exp\left[-iHt/\hbar\right]$. The generalized brading is executed by
setting $-iHt/\hbar=\theta\gamma_{\beta}\gamma_{\alpha}$. Generalized braiding
operators for $B$ Majorana fermions are rewritten in terms of spin operators
as
$\exp\left[\theta\gamma_{2j-1}^{B}\gamma_{2j}^{B}\right]=\exp\left[i\theta\sigma_{2j-1}^{y}\sigma_{2j}^{y}\right]$
for $j=1,2,3$, and
$\exp\left[\theta\gamma_{2j}^{B}\gamma_{2j+1}^{B}\right]=\exp\left[i\theta\sigma_{2j}^{x}\sigma_{2j+1}^{x}\right]$
for $j=1,2$. Generalized braiding operators for $A$ Majorana fermions are
rewritten in terms of spin operators as

$\displaystyle\exp\left[\theta\bar{\gamma}_{3j-2}^{A}\gamma_{3j-1}^{A}\right]$
$\displaystyle=\exp\left[\frac{\theta}{\sqrt{K_{x}^{2}+K_{y}^{2}}}\left(K_{x}\gamma_{3j-2}^{A}\gamma_{3j-1}^{A}-K_{y}\gamma_{3j-1}^{A}\gamma_{3j}^{A}\right)\right]$
$\displaystyle=\exp\left[\frac{i\theta}{\sqrt{K_{x}^{2}+K_{y}^{2}}}\left(K_{x}\sigma_{3j-2}^{x}\sigma_{3j-1}^{x}-K_{y}\sigma_{3j-1}^{y}\sigma_{3j}^{y}\right)\right]$
(10)

for $j=1,2$. Generalized braiding operators consisting of $A$ and $B$ Majorana
fermions are rewritten in terms of spin operators as

$\displaystyle\exp\left[\theta\gamma_{2j-1}^{A}\gamma_{2j-1}^{B}\right]$
$\displaystyle=\exp\left[-2i\theta\sigma_{2j-1}^{z}\right],$ (11)
$\displaystyle\exp\left[\theta\gamma_{2j}^{A}\gamma_{2j}^{B}\right]$
$\displaystyle=\exp\left[-2i\theta\sigma_{2j}^{z}\right].$ (12)

Hence, it is possible to execute braiding by temporally controlling the spin
Hamiltonian.

Fusion: In order to readout the information of qubits based on Majorana
fermions, the fusion protocol is used, where the fermion number constructed
from Majorana fermions are observed. The fusion is a pair annihilation process
of two Majorana fermions, which results in a single fermion ($n_{j}=1$) or a
vacuum ($n_{j}=0$), and hence, the qubit $n_{j}$ can be readout. The fermion
numbers of Majorana fermions are expressed in terms of spin operators as

$\displaystyle\hat{n}_{j}^{B}$ $\displaystyle\equiv
f_{j}^{B\dagger}f_{j}^{B}=-i\gamma_{2j-1}^{B}\gamma_{2j}^{B}/2=-\sigma_{2j-1}^{y}\sigma_{2j}^{y},\quad
j=1,2,3,$ $\displaystyle\hat{n}_{2}^{A}$ $\displaystyle\equiv
f_{2}^{A\dagger}f_{2}^{A}=-i\bar{\gamma}_{1}^{A}\gamma_{2}^{A}=-i\frac{K_{x}\gamma_{1}^{A}+K_{y}\gamma_{3}^{A}}{\sqrt{K_{x}^{2}+K_{y}^{2}}}\gamma_{2}^{A}$
$\displaystyle=-\frac{K_{x}}{\sqrt{K_{x}^{2}+K_{y}^{2}}}\sigma_{1}^{x}\sigma_{2}^{x}+\frac{K_{y}}{\sqrt{K_{x}^{2}+K_{y}^{2}}}\sigma_{2}^{y}\sigma_{3}^{y}.$
(13)

Hence, the fusion is executed by measuring the local spin correlation
$\sigma_{j}^{x}\sigma_{j+1}^{x}$ and $\sigma_{j}^{y}\sigma_{j+1}^{y}$. On the
other hand, it is difficult to readout $\hat{n}_{1}^{A}$ because the number
operator $\hat{n}_{1}^{A}=-i\gamma_{3}^{A}\gamma_{6}^{A}$ cannot be
represented by a local spin correlation operator.

Nonzero $K_{z}$: We next consider the realistic case with $K_{z}\neq 0$. There
are conserved quantities[36, 37, 38, 39, 40, 41] known as the $Z_{2}$ gauge
fields in the Hamiltonian (1). They are real variables $\hat{u}_{34}\equiv
i\gamma_{3}^{B}\gamma_{4}^{B}$ and $\hat{u}_{61}\equiv
i\gamma_{6}^{B}\gamma_{1}^{B}$, satisfying
$\left[\hat{H}_{\text{K}},\hat{u}_{34}\right]=\left[\hat{H}_{\text{K}},\hat{u}_{61}\right]=0$
and $\hat{u}_{34}^{2}=\hat{u}_{61}^{2}=1$. The Hilbert space is decomposed
into the subspaces, where $\hat{u}_{34}$ and $\hat{u}_{61}$ take eigenvalues
$\pm 1$. In these subspaces, because the Hamiltonian (5) becomes in terms of
Majorana fermions as in

$\hat{H}_{\text{K}z}=-i\frac{K_{z}}{4}(\hat{u}_{34}\gamma_{3}^{A}\gamma_{4}^{A}+\hat{u}_{61}\gamma_{6}^{A}\gamma_{1}^{A}),$
(14)

particle-hole symmetry is present. Hence, the Majorana states are particle-
hole symmetry protected even for $K_{z}\neq 0$. Furthermore, it is possible to
diagonalize exactly the Hamiltonian.

We show the energy spectrum as a function of $K_{z}/K$ with $K_{x}=K_{y}=K$ in
Fig.1(c). There are 24 zero-energy states, 8 states with
$E=\pm\sqrt{K_{x}^{2}+K_{y}^{2}+K_{z}^{2}}$ and other 24 states as shown in
Fig.1(d). The energy spectrum is symmetric with respect to $E=0$ as shown in
Fig.1(c). For $\left|K_{z}/K\right|\ll 1$, 32 states are almost degenerate and
it is possible to use 5 qubits even for $K_{z}\neq 0$.

Initialization: In quantum computation, it is necessary to prepare one unique
quantum state as an initial state. For this purpose, we introduce the
Heisenberg interaction[42, 43, 44] together with the magnetic field $B_{z}$
along $z$ direction at the initial stage,

$\hat{H}_{J}=J\sum_{\left\langle
i,j\right\rangle}\sum_{\gamma=x,y,z}\sigma_{i}^{\gamma}\sigma_{j}^{\gamma}-B_{z}\sum_{j=1}^{6}\sigma_{j}^{z},$
(15)

where we have set $K_{x}=K_{y}=K_{z}\equiv K$. The Hamiltonian
$\hat{H}=\hat{H}_{\text{K}}+\hat{H}_{J}$ with Eq.(1) is analytically
diagonalizable for the zero-energy state, and we find it given by

$\displaystyle\left|\psi_{0}\right\rangle\propto$
$\displaystyle\left|010010\right\rangle+\left|010101\right\rangle+\left|011001\right\rangle+\left|110001\right\rangle$
$\displaystyle-(\left|001100\right\rangle+\left|100100\right\rangle+\left|101000\right\rangle+\left|101011\right\rangle).$
(16)

The state $\left|\psi_{0}\right\rangle$ is used for the initialization process
of the qubits.

Figure 2: Kitaev spin liquid model on double hexagons and triple hexagons. (a)
Illustration of the Kitaev spin liquid model on double hexagons. Its energy
spectrum with (b) $K_{z}=0$ and (c) $K_{z}=K/8$. (d) Illustration for the
Kitaev spin liquid model on triple hexagons. Its energy spectrum with (e)
$K_{z}=0$ and (f) $K_{z}=K/8$. Insets in (c) and (f) show the enlarged figures
around the zero-energy states. We have set $K_{x}=K_{y}=K$.

Kitaev spin liquid on connected hexagons: It is possible to generalize the
Kitaev spin liquid model on a single hexagon to that on connected $L$
hexagons, where there are $4L+2$ spins. It is illustrated in the case of $L=2$
and $3$ in Fig.2(a) and (d). The Hamiltonian reads

$\displaystyle\hat{H}_{\text{K}}$
$\displaystyle=-\sum_{j=1}^{L}(K_{x}\sigma_{2j-1}^{x}\sigma_{2j}^{x}+K_{y}\sigma_{2j}^{y}\sigma_{2j+1}^{y})$
$\displaystyle-\sum_{j=1}^{L}(K_{x}\sigma_{2j-1+2L+1}^{x}\sigma_{2j+2L+1}^{x}+K_{y}\sigma_{2j+2L+1}^{y}\sigma_{2j+1+2L+1}^{y})$
$\displaystyle-K_{z}\sum_{j=1}^{L+1}\sigma_{2j-1}^{z}\sigma_{4L+4-2j}^{z}.$
(17)

It is rewritten in terms of Majorana fermions as

$\displaystyle\hat{H}_{\text{K}}=$
$\displaystyle-i\sum_{j=1}^{L}\left(K_{x}\gamma_{2j-1}^{A}\gamma_{2j}^{A}-K_{y}\gamma_{2j}^{A}\gamma_{2j+1}^{A}\right)$
$\displaystyle-i\sum_{j=1}^{L}(K_{x}\gamma_{2j-1+2L+1}^{A}\gamma_{2j+2L+1}^{A}$
$\displaystyle\qquad\qquad-K_{y}\gamma_{2j+2L+1}^{A}\gamma_{2j+1+2L+1}^{A})$
$\displaystyle-i\sum_{j=1}^{L+1}\frac{K_{z}}{4}u_{2j-1,4L+4-2j}\gamma_{2j-1}^{A}\gamma_{4L+4-2j}^{A},$
(18)

where the $\mathbb{Z}_{2}$ gauge fields are given by

$u_{2j-1,4L+4-2j}\equiv i\gamma_{2j-1}^{B}\gamma_{4L+4-2j}^{B},\quad
j=1,\cdots,L+1.$ (19)

We first consider the case $K_{z}=0$, where the system of hexagons is
decomposed into two chains. The system is particle-hole symmetric. The energy
spectrum is shown in Fig.2(b) and (e) for the case of $L=2$ and $3$. There are
$2^{3L+2}$-fold degenerate zero-energy states. It is understood as follows.
There are $2L+1$ free $B$ Majorana fermions because $\gamma_{j}^{B}$ does not
appear in the Hamiltonian of each chain. On the other hand, there is one free
$A$ Majorana fermion according to the Lieb theorem dictating the number of the
zero-energy states in the bipartite system (18). Accordingly, the number of
one type of sites is $L+1$, while that of the other type of sites is $L$.
Hence, the difference is $1$ in each chain, implying the presence of one free
$A$ Majorana fermion in Hamiltonian (18). Hence, the number of free Majorana
fermions is $2L+2$ in total, which results in the $2^{L+1}$-fold degeneracy in
the energy spectrum. In addition, the diagonalization of the quadratic
Hamiltonian (18) for $\gamma_{j}^{A}$ gives different eigenenergies each
other. Hence, one Kitaev spin chain model with length $L$ has $2^{L}$
different states with $2^{L+1}$-fold degeneracy. They produces
$2^{3L+2}=2^{L}\times 2^{L+1}\times 2^{L+1}$-fold degenerate zero energy
states in total. There emerge $2^{3L+2}$ zero-energy states, and hence, $3L+2$
qubits are constructed.

We next consider the case $K_{z}\neq 0$. The energy spectrum is shown in
Fig.2(c) and (f) for the case of $L=2$ and $3$, where particle-hole symmetry
holds manifestly. This can be shown with the aid of the $Z_{2}$ gauge fields
(19) as in the case of the single hexagon. Hence, the Majorana states are
particle-hole symmetry protected even for $K_{z}\neq 0$.

Conclusion: In this paper, we have shown that the Kitaev spin liquid model on
a single hexagon acts as a 5-qubit system. It is possible to prepare a unique
initial state by introducing the Heisenberg interaction together with magnetic
field and to execute braiding by controlling the local spin correlation
Hamiltonian, while qubits are readout with the use of the fusion protocol by
observing local spin correlators. In addition, an arbitrary number of qubits
is constructed by using connected hexagons. Our results are more efficient
comparing previous results on the emergence of Majorana fermions at in the
Kitaev spin liquid model.

This work is supported by CREST, JST (Grants No. JPMJCR20T2) and Grants-in-Aid
for Scientific Research from MEXT KAKENHI (Grant No. 23H00171).

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