# Strength of the naked singularity in critical collapse

Jun-Qi Guo<EMAIL_ADDRESS>School of Physics and Technology, University
of Jinan, Jinan 250022, Shandong, China  Lin Zhang<EMAIL_ADDRESS>College of Mathematics and Statistics, Chongqing University, Chongqing 401331,
China  Yuewen Chen<EMAIL_ADDRESS>Institute of Applied Mathematics,
Morningside Center of Mathematics and LESC, Institute of Computational
Mathematics, Academy of Mathematics and System Science, Chinese Academy of
Sciences, University of Chinese Academy of Sciences, Beijing 100190, China.
Pankaj S. Joshi<EMAIL_ADDRESS>International Center for Cosmology,
Charusat University, Anand 388421, Gujarat, India  Hongsheng Zhang
<EMAIL_ADDRESS>School of Physics and Technology, University of Jinan,
Jinan 250022, Shandong, China

(27th August 2024)

###### Abstract

The critical collapse of a scalar field is a threshold solution of black hole
formation, in which a naked singularity arises. We study here the curvature
strength of this singularity using a numerical ansatz. The behavior of the
Jacobi volume forms is examined along a non-spacelike geodesic in the limit of
approach to the singularity. These are seen to be vanishing, thus showing that
all physical objects will be crushed to zero size near the singularity.
Consequently, although the critical collapse is considered to be a fine-tuning
problem, the naked singularity forming is gravitationally strong. This implies
that the spacetime cannot be extended beyond the singularity, thus making the
singularity genuine and physically interesting. These results imply that the
nature of critical collapse may need to be examined and explored further.

##  I Introduction

It is widely believed that the vicinities of spacetime singularities are an
ideal place to decipher quantum gravity. However, the Weak Cosmic Censorship
Conjecture, a long debated issue in general relativity, states that any
spacetime singularity should be necessarily hidden inside an event horizon
Penrose_1969 . If this conjecture is valid, it would be challenging to extract
information on quantum gravity from the neighborhoods of singularities
directly. On the other hand, if realistic singularities, such as those forming
in gravitational collapse of physically reasonable matter fields, were not
always covered by an event horizon or gravity, the task of receiving quantum
gravity signals from ultra-strong gravity regions in Universe would become
easier.

Since the proposal of the Weak Cosmic Censorship Conjecture, several
counterexamples to this conjecture have been found, including collapse of dust
Eardley_1979 ; Christodoulou_1984 ; Joshi_1993 , perfect fluids Ori_1987 ;
Ori_1990 ; Joshi_1992  and scalar field Christodoulou_1994 ; Brady:1994aq ;
Zhang:2015rsa . The instability of naked singularities in scalar field
collapse was analyzed in Refs. Christodoulou_1999 ; Liu_2018 . Motivated by
the question posed by Christodoulou, namely whether the smallest black hole
forms in scalar field collapse would have finite or infinitesimal mass,
Choptuik simulated gravitational collapse of a scalar field Choptuik:1992jv ;
Choptuik:1993 . When the scalar field is weak, the field will implode and then
disperse, while a black hole will form when the scalar field is strong enough.
By fine-tuning the strength of the scalar field, a critical solution was
obtained. The main results of critical collapse analysis include discrete
self-similarity, universality and mass-scaling law. The global structure of
critical collapse was investigated in Refs. Gundlach:1995kd ; Gundlach:1996eg
; MartinGarcia:2003gz , and a real analytic solution to critical collapse was
proved to exist in Ref. Reiterer:2012mz . In critical collapse, the scalar
field carries away its energy and an infinitesimal naked singularity is left
behind Scheel:2014hfa , which is a counterexample to the original censorship
conjecture. Due to this discovery, the cosmic censorship was restated such
that it only applies to collapse with generic initial data. Consequently, the
naked singularity produced in critical collapse was suggested to become less
worrisome. For reviews on critical collapse, see Refs. Choptuik:1997mq ;
Gundlach:2007gc ; Choptuik:2015mma .

The gravitational strength is an important characteristic in describing the
nature of spacetime singularities. Tipler defined a singularity to be
gravitationally strong if an object approaching the singularity is crushed to
a zero volume Tipler_1977 . In mathematical language, a singularity is called
gravitationally strong if the volume (area) element, defined by three (two)
independent vorticity-free Jacobi fields via the exterior product along a
timelike (null) geodesic from or toward the singularity goes to zero as the
singularity is approached. The spacetime cannot be extended beyond a strong
singularity, while the extension may be implemented for a weak one. The
strength of some singularities has been investigated, including the shell-
crossing and shell-focusing singularities in dust collapse Newman_1986 ;
Nolan:1999tw ; Waugh_1988 ; Joshi_1993 , naked singularities in perfect-fluid
collapse Lake_1988 ; Waugh_1989 ; Ori_1990 ; Joshi_1992 , singularities in the
Schwarzschild, Friedmann-Robertson-Walker and Kasner solutions and in black
hole formation by scalar collapse Burko:1997xa , central singularity in
continuous self-similar critical collapse of a scalar field Nolan:1999tw ;
Roberts , and mass-inflation singularities near the Cauchy horizons of
Reissner-Nordström and Kerr black holes Ori_1991 ; Ori_1992 . For a review on
the strength of singularities, see Ref. Krolak:1999wk .

Considering the basic role that the critical collapse might play in general
relativity, in this Letter, we analyze the strength of the naked singularity
in this model. We find that, although the critical collapse may not occur
naturally or could be termed as fine-tuned, the naked singularity forming in
this model is gravitationally strong. In this sense, the nature of critical
collapse may need to be re-examined and explored further for its possible
physical implications. Throughout the Letter, we set $G=c=1$.

##  II Methodology

We simulate critical collapse of a massless scalar field $\phi$ in spherical
symmetry and investigate the strength of the naked singularity in the ansatz,

$\text{d}s^{2}=e^{-2\sigma(x,t)}(-\text{d}t^{2}+\text{d}x^{2})+{r^{2}(x,t)}\text{d}\Omega^{2}.$
(1)

The equations of motion are the following,

$-r_{,tt}+r_{,xx}-\frac{2m}{r^{2}}e^{-2\sigma}=0,$ (2)
$-\sigma_{,tt}+\sigma_{,xx}-\frac{2m}{r^{3}}e^{-2\sigma}+4\pi(\phi_{,t}^{2}-\phi_{,x}^{2})=0,$
(3) $-\phi_{,tt}+\phi_{,xx}+\frac{2}{r}(-r_{,t}\phi_{,t}+r_{,x}\phi_{,x})=0,$
(4)

where the $(_{,t})$ and $(_{,x})$ denote derivatives with respect to the
coordinates $t$ and $x$, respectively. $m$ is the Misner-Sharp mass. Details
on numerics are depicted in Ref. Guo_2018 . The simulation is implemented with
second-order finite difference method. The initial profile for the scalar
field is set as $\phi(x,0)=a\cdot\exp\left(-{x^{2}}/{\delta}\right)$. The
parameter $a$ is precisely tuned to obtain the numerical solution to critical
collapse. Compared to Ref. Guo_2018 , in this work we use $m_{,x}$, instead of
$m_{,t}$, to obtain $m$, and mesh refinement technique is applied
Garfinkle:1994jb .

We explore the curvature strength of the singularity by the aid of one test
particle’s trajectory, which is a radial timelike geodesic approaching the
singularity. The geodesic equations are

$\ddot{x}-\sigma_{,x}(\dot{t}^{2}+\dot{x}^{2})-2\sigma_{,t}\dot{x}\dot{t}=0,$
(5)
$\ddot{t}-\sigma_{,t}(\dot{t}^{2}+\dot{x}^{2})-2\sigma_{,x}\dot{x}\dot{t}=0,$
(6) $e^{-2\sigma}(\dot{t}^{2}-\dot{x}^{2})=1,$ (7)

where the dot $(\dot{})$ denotes derivative with respect to the proper time
$\tau$ for the test particle along its geodesic path, and $\tau$ is set to be
zero at the center and positive for $r>0$. We solve the geodesic equations
with the initial data being fine tuned such that the geodesic terminates at
the singularity.

Let $|p|$, $|q|$ and $|b|$ denote the norms of the Jacobi fields, in $\theta$,
$\varphi$ and radial directions, respectively. Here $p$ and $q$ satisfy the
same evolution equation. From the geodesic deviation equation, the evolution
equations for $p$ and $b$ can be obtained Nolan:1999tw

$r\ddot{p}+2\dot{r}\dot{p}=0,$ (8)
$\ddot{b}+e^{2\sigma}(\sigma_{,tt}-\sigma_{,xx})b=0.$ (9)

For simplicity, we set $p=q$. Then the norm of the volume element expanded by
the Jacobi fields is Nolan:1999tw

$||V||=|b|p^{2}r^{2}.$ (10)

---
Figure 1: $\phi$ vs. $-\ln{T}$ at the center. $T$ is the proper time for a
central observer.

Based on the condition by Clarke and Krolak Clarke_1985 , in non-vacuum
collapse, a singularity is strong in the sense of Tipler if along the non-
spacelike geodesic, the integral

$\displaystyle{\int_{c}^{0}\int_{c}^{\tau_{1}}R_{\alpha\beta}T^{\alpha}T^{\beta}}(\tau)\text{d}\tau\text{d}\tau_{1}$
(11)

diverges, where $R_{\alpha\beta}$ is the Ricci tensor, $c$ is any fixed
positive constant, and $T^{\alpha}(\equiv dx^{\alpha}/d\tau)$ is the tangent
vector to the geodesic. When the matter is a scalar field, using
$R_{\alpha\beta}=8{\pi}\phi_{,\alpha}\phi_{,\beta}$ and setting
$\lambda\equiv-\ln\tau$, we have
$R_{\alpha\beta}T^{\alpha}T^{\beta}=8\pi(\phi_{,\tau})^{2}$ and

$\displaystyle\begin{split}&\int_{c}^{0}\int_{c}^{\tau_{1}}(\phi_{,\tau})^{2}\text{d}\tau\text{d}\tau_{1}\\\
=&\int_{-\ln c}^{+\infty}\left(\int_{-\ln
c}^{\lambda_{1}}e^{\lambda}\cdot(\phi_{,\lambda})^{2}\text{d}\lambda\right)\cdot
e^{-\lambda_{1}}\text{d}\lambda_{1}\\\ =&\int_{-\ln
c}^{+\infty}(\phi_{,\lambda})^{2}\text{d}\lambda-\lim\limits_{\lambda_{1}\to+\infty}\frac{\int_{-\ln
c}^{\lambda_{1}}e^{\lambda}\cdot(\phi_{,\lambda})^{2}\text{d}\lambda}{e^{\lambda_{1}}}.\end{split}$

If $|\phi_{,\lambda}|$ is bounded by some uniformity constant, then the second
term in the last line is also bounded by some finite constant. Hence, the
divergence of (11) is equivalent to that the integral

$\int_{-\ln c}^{+\infty}(\phi_{,\lambda})^{2}\text{d}\lambda$ (12)

diverges.

In this work, we first obtained the values for $\sigma_{,x}$, $\sigma_{,t}$
and $r$ etc in Eqs. (5)-(9) and $\phi$ on the grid $(x_{i},t_{i})$ in the
simulation. Then, interpolating their values on the geodesic, we solved Eqs.
(5)-(9) with the 4th-order Runge-Kutta method, and computed the quantity
$||V||$ in (10) and the integral in (12).

---
Figure 2: Upper panel: one timelike geodesic terminating at the naked
singularity. Middle and lower panels: the Ricci scalar $R$ and Kretschmann
scalar $K$ along the geodesic.
---
Figure 3: Results on the strength of the singularity. Upper panel: the norm of
the volume element $||V||(=|b|p^{2}r^{2})$ along the geodesic approaching the
singularity. Middle and lower panels: $\phi$ and
$8\pi(\phi_{,\lambda})^{2}[=\tau^{2}R_{\alpha\beta}T^{\alpha}T^{\beta}]$ vs.
$\lambda(\equiv-\ln\tau)$ along the geodesic.

##  III Results and discussions

We first simulated critical collapse and obtained the numerical values for the
spacetime and scalar field. The numerical results for $\phi$ versus $\ln{T}$
at the center are shown in Fig. 1. $T$ is the proper time for a central
observer, $T\equiv\int_{0}^{t}e^{-\sigma(0,\tilde{t})}d\tilde{t}$, where $t$
is set to be zero (positive) as (before) the singularity is approached. The
results are similar to those in Refs. Hamade:1995ce ; Choptuik:2003ac ;
Baumgarte:2018 , where the double-null coordinates, polar-slicing coordinates
and Baumgarte-Shapiro-Shibata-Nakamura (BSSN) formalism were used,
respectively. Solving Eqs. (5) and (6) with proper initial data for the test
particle, we obtained one timelike geodesic terminating at the naked
singularity. This is verified by the divergence of the Ricci scalar $R$ and
Kretschmann scalar $K$ along the geodesic. See Fig. 2.

Via integrations of Eqs. (8) and (9), the norm of the volume element $||V||$
by the Jacobi fields was obtained. As shown in the upper panel of Fig. 3,
$||V||$ vanishes as the center is approached. Therefore, according to the
Tipler’s definition, the naked singularity is gravitationally strong. The
results for
$8\pi(\phi_{,\lambda})^{2}[=\tau^{2}R_{\alpha\beta}T^{\alpha}T^{\beta}]$
versus $\lambda(\equiv-\ln\tau)$, plotted in the lower panel of Fig. 3, show
that along the geodesic $|\phi_{,\lambda}|$ is bounded. Then the integral
$\int_{-\ln c}^{+\infty}(\phi_{,\lambda})^{2}\text{d}\lambda$ diverges. So by
this means one also concludes that the singularity is strong.

It may be instructive to compare the results of the discrete self-similar
critical collapse (Choptuik solution) with those of the continuous one
(Roberts solution). In the Roberts solution, the spacetime at the center is
not conformally flat Guo_2018 . Moreover, in the $v<0$ region, the mass is
negative. To avoid the negative mass problem, the spacetime of this region is
replaced by a flat spacetime Oshiro:1994hd ; Brady:1993np . The central
singularity is weak Nolan:1999tw . It was commented that this singularity is
actually a collapsed cone singularity rather than a naked singularity
Christodoulou_1994 ; Wald:1997wa .

Since it was found that a naked singularity arises in Choptuik critical
collapse, it has been argued that the initial data are not generic and so this
model is not a serious conterexample to the Weak Comic Censorship Conjecture.
Now it is observed here that, although critical collapse is a fine-tuning
problem, the naked singularity forming in this process is gravitationally
strong. The spacetime therefore cannot be extended beyond this singularity. So
the physical nature of critical collapse may need to be examined further in
light of this phenomenon.

## Acknowledgments

The authors are grateful to Xinliang An, Yun-Kau Lau, Junbin Li, Daoyan Wang
and Xiaoning Wu for useful discussions. JQG thanks the Morningside Center of
Mathematics, the Academy of Mathematics and Systems Science of the Chinese
Academy of Sciences for hospitality. This work is supported by Shandong
Province Natural Science Foundation under grant No.ZR2019MA068.

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# A Flavor Change Study based on Dyson-Schwinger Equation

Xue-ao Chao  Department of Physics and State Key Laboratory of Nuclear Physics
and Technology
Peking University, Beijing 100871, China  Yu-xin Liu<EMAIL_ADDRESS>Department of Physics and State Key Laboratory of Nuclear Physics and
Technology
Peking University, Beijing 100871, China  Collaborative Innovation Center of
Quantum Matter, Beijing 100871, China  Center for High Energy Physics, Peking
University, Beijing 100871, China

###### Abstract

We study the flavor change effects using the Dyson-Schwinger (DS) equation in
a multi-flavor system. By taking the Electroweak interaction as perturbation
into conditions, the
$\text{SU(4)}_{\text{L}}\rightarrow\text{SU(2)}_{\text{L}}\otimes\text{SU(2)}_{\text{L}}\rightarrow\text{SO(2)}_{\text{L}}\otimes\text{SO(2)}_{\text{L}}$
symmetry breaking chain is studied. Under this symmetry breaking pattern
fermion masses are split, and we can identify the fermions with different
masses as different generations. Quark mass spectrum is then given. Meanwhile,
there are a total of fifteen Goldstones but only four of them are independent.
The Goldstones have electric charges $0$, $0$, $+1$, $-1$, respectively. One
of them becomes pseudo-Nambu-Goldstone boson (pNGB) and gains a mass due to
the Electroweak interaction perturbation. It can be identified as the Higgs
boson. The other three Goldstones maintain massless and will be eaten by gauge
bosons to give $W^{\pm}$ and $Z^{0}$ masses via Schwinger mechanism. Thus, the
Goldstones can take the role of the Higgs boson.

##  I INTRODUCTION

The Standard Model (SM) has been proved a great success. Nevertheless, it is
not all the stories. There are many phenomena beyond the explanation given by
the SM, for instance, Dark Matter, Dark Energy, and Neutrino Masses. Moreover,
even within the SM some problems are still puzzling scientists deeply, such as
the Fine-Tuning Problem, Hierarchy Problem, Mass Splitting, and various
problems related to flavors.

Concerning flavor questions, there are four main issues (The significance of
these issues is present in the fact that 12 of 19 parameters in the SM are
related to them. They are $m_{i}$ and $\Theta_{k}$) [1]:
① Replication: Why are there three generations of fermions? Each fermion
particle seems to have three different versions, differentiated only by mass.
The three generations are literally copy-paste of the first generation. And
curiously, it is pasted three times. Why three generations? What’s the
necessity of three generations? There is no doubt this is an important issue.
The fact of three generations of fermions is vital to the cosmic structure
even to the existence of life [2].
② Fermions mass hierarchy: Why so significant the mass differences between the
three generations are? The tau lepton is roughly 3600 times more massive than
the electron, and the top quark is nearly 100,000 times heavier than the up
quark. Although we do not yet know the exact mass of neutrinos, related
experiments have shown that they have different masses [3, 4, 5], and there
seems to be a hierarchical structure between them. The mass spectrum is
approximately exponential [6]. The fermions mass hierarchy means that higher
generation fermions will decay to the first generation quickly. All the matter
in our daily life consists of the first generation particles, and those are
protons, neutrons, and electrons. The higher generation fermions only appear
in high energy environments, such as cosmic rays and higher energy particle
experiments. It seems to go back to our first question, what is the necessity
for the existence of higher generation quarks?
③ Mixing hierarchy: the Cabibbo-Kobayashi-Maskawa (CKM) matrix elements have
hierarchies, $|V_{ii}|\sim 1>|V_{12}|>|V_{23}|>|V_{13}|$.
④ The origin of CKM and its evaluations.

In this article, we mainly focus on the first two issues, trying to answer the
question of why the mass of three generation quarks split. On one hand, there
is not only a horizontal hierarchical structure between the three generations
of fermions but also a vertical hierarchical structure between them. The
horizontal hierarchies can be approximately expressed as $m_{u}/m_{c}\sim
m_{c}/m_{t}\sim\lambda^{4}$ and $m_{d}/m_{s}\sim m_{s}/m_{b}\sim\lambda^{2}$,
where $\lambda\approx 0.22$. If we look vertically, we can also roughly see a
hierarchical structure among the masses of quarks, charged leptons, and
neutrinos. For instance, in the first generation fermions, up and down quark
have a mass of $m_{u}\sim m_{d}\sim 5\text{MeV}$, the electron has a mass of
$m_{e}\sim 0.5\text{MeV}$, while electric neutrino has a mass of
$m_{1}<15\text{eV}$. Here we are talking about current mass rather than the
constituent one, which has the same origin as leptons, i.e., the Higgs
mechanism. On the other hand, we notice that quarks participate in all strong
interactions, electromagnetic interactions, and weak interactions. It has the
strongest interaction. Electron participates in electromagnetic interaction
and weak interaction, and its interaction is weaker. The electron neutrino
only participates in the weakest weak interaction. The hierarchical structure
of these interactions combined with the hierarchical structure of their mass
suggests that their mass is likely to have a dynamic origin. In other words,
the Higgs mechanism may have a dynamic mechanism, and the Higgs boson should
be composite (see, e.g., Refs. [7, 8, 9, 10]). In order to ensure the
hierarchical structure of quarks, charged leptons, and neutrinos that match
the strengths of the corresponding interactions, the Higgs boson needs to
participate in all the three kind interactions, so it can only be composed of
quarks. Based on the above arguments, we need a theory that describes
consistently these two hierarchical structures. So we start by trying to take
quantum chromodynamics (QCD) to explain the mass splittings of quarks. This
article is the first step in this direction.

As mentioned before, the three generation fermions with the same charge are
identical if there are no mass differences. The fact that mass does not affect
interaction type suggests naturally that they can be treated as the same
particles on the interaction level. At least they can be replaced each other
in interactions. This implies that there may be a symmetry among the three
generations which is usually called horizontal symmetry [1, 11]. Different
generation fermions acquire different masses in the symmetry breaking
progress. An up quark can be transformed to other particles in some progress,
and the opposite progress occurs too. Just imagine when a particle (or a group
of particles) transforms to an up quark, it may not distinguish between an up
quark or a charm quark because they look the same. As a result, an up quark
(first generation) can transform to a charm quark (another generation).
Although a pure type of this flavor change progress can happen dynamically, it
is forbidden kinematically to consider necessary quantum number conservation.
In other words, a flavor change progress can not be external legs in Feynman
diagram, and corresponding particles can not be regarded as stable particles
[12]. Even so, flavor change progress still possibly has multi-faceted impacts
on many issues, for instance, CKM and mass splitting. Flavor change progress
and its renormalization procedure have been discussed in many works (see,
e.g., Refs. [12, 13, 14]).

DS equation is an important tool to study dynamical symmetry breaking [15]. We
will give a model from an $\text{SU(2)}_{\text{L}}$ horizontal symmetry, which
will be dynamically broken to $\text{SO(2)}_{\text{L}}$ (We ignore the right-
handed symmetries in this article.). By the tool of DS equation, we will see
that there can be dynamical flavor change progress that generates different
masses for different generations. Three generations are also indicated.

The remainder of this article is organized as follows: In Sec. II, we describe
the theoretical framework. In Sec. III, we discuss the structure of the
solutions and the numerical results. The mass spectrum for different
generation quarks is obtained, and we will show that why there are three
generation fermions. In Sec. IV, we will discuss the goldstone physics of our
model. Sec. V gives a summary and some brief remarks.

##  II THEORETICAL FRAMEWORK

###  II.1 Lagrangian and Symmetry

We define two types hyperquarks first, the up-type hyperquarks and the down-
type hyperquarks.

$\displaystyle up-type:$ $\displaystyle\left(u_{1}\quad u_{2}\right),$ (1)
$\displaystyle down-type:$ $\displaystyle\left(d_{1}\quad d_{2}\right).$

We consider only QCD now first; the Lagrangian density is given as:

$L_{0}=\bar{u}_{1}i\not{\partial}u_{1}+\bar{u}_{2}i\not{\partial}u_{2}+\bar{d}_{1}i\not{\partial}d_{1}+\bar{d}_{2}i\not{\partial}d_{2}+L^{\prime}_{QCD},$
(2)

where $L^{\prime}_{QCD}$ contains the remaining terms in QCD. Obviously, it
has $\text{SU(4)}_{\text{L}}$ invariance. We will add the
$\text{SU(2)}_{\text{L}}\otimes\text{U(1)}_{\text{Y}}$ Electro-Weak terms as
perturbation into the Lagrangian later. Apparently,
$L_{0}+\text{U(1)}_{\text{Y}}$ terms are also invariant in the
$\text{SU(4)}_{\text{L}}$ transformation. The symmetry breaking is due to the
$\text{SU(2)}_{L}$ terms completely. We can rearrange the
$\text{SU(2)}_{\text{L}}\otimes\text{U(1)}_{\text{Y}}$ terms as
$\text{SU(2)}_{\text{W}}\otimes\text{U(1)}_{\text{EM}}$. The
$\text{U(1)}_{\text{EM}}$ terms break the $\text{SU(4)}_{L}$ symmetry to
$\text{SU(2)}_{\text{H}}\otimes\text{SU(2)}_{\text{H}}$ because of electric
charge conservation. The subscript H suggests that the
$\text{SU(2)}_{\text{L}}$ is horizontal symmetries under the family
$\left\\{u_{1},u_{2}\right\\}$ or $\left\\{d_{1},d_{2}\right\\}$.

We focus on the up sector first. In order to study a flavor change problem, we
adopt a formulation that is able to describe the flavor change. The action of
one of the family mentioned above after renormalization reads:

$\displaystyle S$ $\displaystyle=$
$\displaystyle\int\mathrm{d}^{4}x\big{[}Z_{2}^{11}\bar{u}_{1}\left(i\not{\partial}-Z_{m}^{11}m_{11}\right)u_{1}+Z_{2}^{22}\bar{u}_{2}\left(i\not{\partial}-Z_{m}^{22}m_{22}\right)u_{2}+Z_{1}g\bar{u}_{1}\gamma^{\mu}A^{a}_{\mu}T_{a}u_{1}+Z_{1}g\bar{u}_{2}\gamma^{\mu}A^{a}_{\mu}T_{a}u_{2}$
(3) $\displaystyle-
Z_{3}\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{Z_{3}}{2\xi}\left(\partial_{\mu}A^{\mu}_{a}\right)^{2}$
$\displaystyle+Z_{\lambda}^{12}\bar{u}_{1}\lambda_{12}\left(i\not{\partial}-Z_{m}^{12}m_{12}\right)u_{2}+Z_{\lambda}^{21}\bar{u}_{2}\lambda_{21}\left(i\not{\partial}-Z_{m}^{21}m_{21}\right)u_{1}$
$\displaystyle+\frac{Z_{\lambda}^{12}Z_{1}}{Z_{2}}g\lambda_{12}\bar{u}_{1}\gamma^{\mu}A^{a}_{\mu}T_{a}u_{2}+\frac{Z_{\lambda}^{21}Z_{1}}{Z_{2}}g\lambda_{21}\bar{u}_{2}\gamma^{\mu}A^{a}_{\mu}T_{a}u_{1}\big{]},$

where $T_{a}$ are the eight Gell-Mann matrices in color space. The terms in
the first two lines are invariant under $\text{SU(2)}_{\text{L}}$
transformation. We introduced flavor changed terms in the last two lines,
which break the symmetry. We will let $\lambda_{ij}\rightarrow 0$ at the end
of calculations. $Z_{1}^{ij}$,$Z_{2}^{ij}$,$Z_{3}^{ij}$ are vertex
renormalization constant, quark field, and gauge boson field renormalization
constant, respectively. $Z_{\lambda}^{ij}$ is necessary for the
renormalization of $\lambda$. Notice that we define
$g=\frac{Z_{2}^{ii}\sqrt{Z_{3}}}{Z_{1}^{ii}}g^{0}$,
$\lambda_{ij}=\frac{\sqrt{Z_{2}^{11}Z_{2}^{22}}}{Z_{\lambda}^{ij}}\lambda^{0}_{ij}$,
where $g^{0}$, $\lambda^{0}_{ij}$ are the corresponding bare parameters, and
the indexes take $\\{i=1,j=2\\}$ and $\\{i=2,j=1\\}$. By define a flavor
doublet $\psi\equiv\left(\begin{matrix}u_{1}\\\ u_{2}\\\ \end{matrix}\right)$,
$\bar{\psi}\equiv\psi^{\dagger}\Gamma^{0}=\left(\begin{matrix}\bar{u}_{1}&\bar{u}_{2}\\\
\end{matrix}\right)$, where
$\Gamma^{\mu}\equiv\left(\begin{matrix}\gamma^{\mu}&0\\\ 0&\gamma^{\mu}\\\
\end{matrix}\right)$, we can write the action in a compact form:

$\displaystyle S$ $\displaystyle=$
$\displaystyle\int\mathrm{d}^{4}x\big{[}\bar{\psi}Z_{2}\left(i\not{\partial}-M\right)\psi+\bar{\psi}gZ_{1}\Gamma^{\mu}A_{\mu}\psi$
(4)
$\displaystyle-\frac{Z_{3}}{4}F_{\mu\nu}F^{\mu\nu}-\frac{Z_{3}}{2\xi}\left(\partial_{\mu}A^{\mu}\right)^{2}\big{]},$

where

$\displaystyle Z_{1}$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}Z_{1}&\frac{Z_{1}Z_{\lambda}^{12}\lambda_{12}}{Z_{2}}\\\
\frac{Z_{1}Z_{\lambda}^{21}\lambda_{21}}{Z_{2}}&Z_{1}\\\ \end{matrix}\right),$
$\displaystyle Z_{2}\left(i\not{\partial}-M\right)$ $\displaystyle=$
$\displaystyle\left(\begin{matrix}Z_{2}^{11}\left(i\not{\partial}-Z_{m}^{11}m_{11}\right)&Z_{\lambda}^{12}\lambda_{12}\left(i\not{\partial}-Z_{m}^{12}m_{12}\right)\\\
Z_{\lambda}^{21}\lambda_{21}\left(i\not{\partial}-Z_{m}^{21}m_{21}\right)&Z_{2}^{22}\left(i\not{\partial}-Z_{m}^{22}m_{22}\right)\\\
\end{matrix}\right).$

Obviously, the Lagrangian has an $\text{SU(2)}_{\text{L}}$ horizontal symmetry
if $\lambda_{ij}=0$ and $m_{11}=m_{22}$ keep finite. In the following, we take
$m_{11}=m_{22}$ indeed and let $\lambda_{ij}\rightarrow 0$ in the end. Then it
will be shown that the horizontal symmetry is broken dynamically.

For convenient numerical calculations, we switch to Euclidean space and write
down the partition function as:

$\displaystyle Z_{E}\\!=\\!$
$\displaystyle\int\\!\\!D(\bar{\psi},\psi,A_{\mu})\exp\bigg{\\{}\\!\\!-\\!\\!\int\\!\mathrm{d}^{4}x\Big{[}\bar{\psi}Z_{2}\\!\left(\not{\partial}+M\right)\\!\psi$
(5)
$\displaystyle+igA_{\mu}\bar{\psi}Z_{1}\Gamma^{\mu}\psi+\\!\frac{Z_{3}}{4}F_{\mu\nu}F^{\mu\nu}\\!\\!+\\!\frac{Z_{3}}{2\xi}\\!\left(\partial_{\mu}A^{\mu}\right)^{2}\\!\Big{]}$
$\displaystyle+\\!\\!\int\\!\mathrm{d}^{4}x\left[\bar{\psi}\eta+\bar{\eta}\psi+A_{\mu}J^{\mu}\right]\bigg{\\}}.$

###  II.2 DS equations

Following Ref. [15], the DS equation for quarks is:

$\displaystyle S^{-1}\\!\left(p\right)$ $\displaystyle=$ $\displaystyle
Z_{2}\left(i\not{p}+M\right)$
$\displaystyle+i\\!\\!\int\\!\\!\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}gZ_{1}\Gamma^{\mu}S\left(k\right)\Lambda_{\varepsilon}D_{\mu}^{\varepsilon}\left(p\\!-\\!k\right),$

where $S$, $D$, and $\Lambda$ stand for the propagators of quarks, gauge
bosons, and the complete vertex, respectively. It can be represented
graphically as:

$\displaystyle\leavevmode\hbox to0pt{\vbox
to0pt{\pgfpicture\makeatletter\hbox{\hskip 0.0pt\lower 0.0pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynhand \vertex(p1) at (0,0);
\vertex[dot](p2) at (1,0) {}; \vertex[particle](p0) at (1,-0.5)
{$S_{0}^{-1}$}; \vertex(p3) at (2,0); \propag[fermion] (p1) to (p2);
\propag[fermion] (p2) to (p3); \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}^{-1}$
$\displaystyle=$ $\displaystyle\leavevmode\hbox to0pt{\vbox
to0pt{\pgfpicture\makeatletter\hbox{\hskip 0.0pt\lower 0.0pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynhand \vertex(p1) at (0,0);
\vertex[grayblob](p2) at (1,0) {}; \vertex[particle](p0) at (1,-0.5)
{$S^{-1}$}; \vertex(p3) at (2,0); \propag[fermion] (p1) to (p2);
\propag[fermion] (p2) to (p3); \pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}^{-1}$
$\displaystyle+$

where $S_{0}$, $S$, and $\Sigma$ stand for the bare quark propagator, the
complete quark propagator, and the self-energy of quarks, respectively. The
gauge boson propagator is modeled by:

$g^{2}D_{\mu\nu}\left(p\right)=D_{\mu\nu}^{free}p^{2}G\left(p\right),$ (7)

where free gauge boson propagator is

$D_{\mu\nu}^{free}=\frac{1}{p^{2}}\left(g_{\mu\nu}-\frac{p_{\mu}p_{\nu}}{p^{2}}\right).$
(8)

In this article, Landau gauge is in use. We take the bare vertex approximation
and absorb $g$ into the interaction model, thus

$\Lambda^{\mu}=\left(\begin{matrix}\gamma^{\mu}&0\\\ 0&\gamma^{\mu}\\\
\end{matrix}\right).$ (9)

As mentioned before, we are using a formulation that can describe the flavor
change progress. The basic idea of our method to study flavor change is to
regard the term that describes flavor change as a renormalization flow. The
realistic version of Eq. (3) is obtained by taking $\lambda_{ij}=0$. We view
this as the definition of the theory at some energy scale $\xi^{2}$. At that
scale, we can not distinguish between $S_{1}=S_{0}$ and
$S_{2}=S_{0}+\lambda(\mu^{2})S_{\lambda}$, where $S_{0}$ represents the first
two lines in Eq. (3) while $S_{\lambda}$ represents the other terms and the
flavor change term satisfies

$\lambda(\mu^{2}=\xi^{2})=0.$ (10)

Thus, it is reasonable to assume that they describe the same theory elsewhere.
When we consider the effective theory at any other scale, the structure of
$S_{2}$ will change significantly due to radiative corrections, and $\lambda$
will generate a non-zero value. That is just the flavor change effect we
desired. The renormalization group can be used as a dynamical tool to include
all of the effects of the full theory. Eq. (10) defines the renormalization
group boundary condition for the full theory and can be viewed as our
renormalization condition. For more related discussions, see, eg., Refs. [16,
17, 18, 19].

At the renormalization point $\xi^{2}$, the propagator becomes free, and the
interaction vertex is bare. Here is our renormalization condition:

$\displaystyle
S\left(\xi^{2}\right)=\left(\begin{matrix}\frac{1}{i\not{\xi}+m_{11}}&0\\\
0&\frac{1}{i\not{\xi}+m_{22}}\\\ \end{matrix}\right).$ (11)

We can parameterize the quark propagator as:

$S\\!=\\!\left(\\!\begin{matrix}-i\not{p}V_{1}(p^{2})\\!+\\!S_{1}(p^{2})&-i\not{p}T_{1}(p^{2})\\!+\\!R_{1}(p^{2})\\\
-i\not{p}T_{2}(p^{2})\\!+\\!R_{2}(p^{2})&-i\not{p}V_{2}(p^{2})\\!+\\!S_{2}(p^{2})\\\
\end{matrix}\right).$ (12)

Substituting these to Eq. (II.2) and carrying out many sterile derivations, we
obtain eight nonlinear integral equations. Because these equations are too
long, we only list the first equation explicitly here. The complete set of the
equations are listed in appendix A.

$\displaystyle 1$ $\displaystyle=$
$\displaystyle\Big{[}Z_{2}^{11}Z_{m}^{11}m_{11}S_{1}\left(p^{2}\right)+Z_{2}^{11}p^{2}V_{1}\left(p^{2}\right)$
(13)
$\displaystyle+Z_{m}^{12}m_{12}R_{2}\left(p^{2}\right)+Z_{\lambda}^{12}\lambda_{12}p^{2}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}3\Big{[}S_{1}\left(k^{2}\right)S_{1}\left(p^{2}\right)$
$\displaystyle+R_{1}\left(k^{2}\right)R_{2}\left(p^{2}\right)\Big{]}+\Big{[}V_{1}\left(k^{2}\right)V_{1}\left(p^{2}\right)$
$\displaystyle+T_{1}\left(k^{2}\right)T_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}}$

All the others are structurally similar to this one. By the reason of
simplicity and solubility, we take the point interaction model first, where
$G\left(p\right)=G$ is a constant. Then due to the rotational symmetry of the
system, we can write the 4-dimensional integration as

$\int\frac{\mathrm{d}^{4}k}{(2\pi)^{4}}=\frac{1}{8\pi^{3}}\int_{0}^{\Lambda^{2}}\mathrm{d}k^{2}k^{2}\int_{0}^{\pi}\mathrm{d}\theta\sin^{2}\theta,$
(14)

where $\Lambda$ is the regularization momentum truncation in the numerical
calculations. Looking closely at these formulas, we find that there is only
one type of term involving angle in these integrations. That is the term like

$2pk\cos\theta+\frac{\left(p^{2}\\!+\\!k^{2}\right)pk\cos\theta\\!-\\!2p^{2}k^{2}}{p^{2}\\!+\\!k^{2}\\!-\\!2pk\cos\theta}\equiv\\!f\left(p,k,\theta\right).$

Thus, using the integration identity

$\int_{0}^{\pi}\mathrm{d}\theta\sin^{2}\theta
f\left(p,k,\theta\right)=-\frac{3\pi}{4}p^{2}+\frac{\pi}{4}\frac{p^{4}}{k^{2}},$
(15)

we can integrate the angle integration out first. We define

$\left\\{\begin{aligned}
\frac{G}{\left(8\pi\right)^{3}}\int\mathrm{d}k^{2}k^{2}V_{1}\left(k^{2}\right)&=CV_{11},\\\
\frac{G}{\left(8\pi\right)^{3}}\int\mathrm{d}k^{2}k^{2}\frac{1}{k^{2}}V_{1}\left(k^{2}\right)&=CV_{12},\\\
\frac{G}{\left(8\pi\right)^{3}}\int\mathrm{d}k^{2}k^{2}S_{1}\left(k^{2}\right)&=CS_{11},\\\
\frac{G}{\left(8\pi\right)^{3}}\int\mathrm{d}k^{2}k^{2}\frac{1}{k^{2}}S_{1}\left(k^{2}\right)&=CS_{12},\\\
\end{aligned}\right.$ (16)

and likewise for $V_{2}$, $S_{2}$, $T_{1}$, $R_{1}$, $T_{2}$, $R_{2}$. We have
then sixteen parameters: $CV_{ij}$, $CS_{ij}$, $CT_{ij}$, and $CR_{ij}$, which
are functionals of $V_{i}$, $S_{i}$, $T_{i}$, or $R_{i}$, and are independent
of $p$ and $k$. As can be seen, functions of $p$ and $k$ in these equations
are decoupled so that by defining the above functionals, we can collect the
integrations to these parameters. The Eq. (13) now reads

$\displaystyle 1=$
$\displaystyle\Big{[}Z_{2}^{11}Z_{m}^{11}m_{11}S_{1}\left(p^{2}\right)+Z_{2}^{11}p^{2}V_{1}\left(p^{2}\right)$
(17)
$\displaystyle+Z_{m}^{12}m_{12}R_{2}\left(p^{2}\right)+Z_{\lambda}^{12}\lambda_{12}p^{2}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\frac{3\pi}{2}CS_{11}S_{1}(p^{2})+\frac{3\pi}{2}CR_{11}R_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CV_{11}V_{1}(p^{2})+\frac{\pi}{4}p^{4}CV_{12}V_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CT_{11}T_{2}(p^{2})+\frac{\pi}{4}p^{4}CT_{12}T_{2}(p^{2}),$

and similarly for others (see appendix A). With the renormalization conditions
defined at renormalization point $\xi^{2}$ by Eqs (11), we have

$\left\\{\begin{aligned}
V_{1}\left(\xi^{2}\right)&=\frac{1}{\xi^{2}+m_{11}^{2}},\\\
V_{2}\left(\xi^{2}\right)&=\frac{1}{\xi^{2}+m_{22}^{2}},\\\
T_{1}\left(\xi^{2}\right)&=0,\\\ T_{2}\left(\xi^{2}\right)&=0,\\\
\end{aligned}\qquad\begin{aligned}
S_{1}\left(\xi^{2}\right)&=\frac{m_{11}}{\xi^{2}+m_{11}^{2}},\\\
S_{2}\left(\xi^{2}\right)&=\frac{m_{22}}{\xi^{2}+m_{22}^{2}},\\\
R_{1}\left(\xi^{2}\right)&=0,\\\ R_{2}\left(\xi^{2}\right)&=0.\\\
\end{aligned}\right.$ (18)

Using the above formulas, we can solve the renormalization constants, and then
put them into the above formulas, and finally, we get

$\left\\{\begin{aligned}
1&\\!=\\!m_{ii}S_{i}\\!\left(p^{2}\right)\\!+\\!\left(\\!1\\!-\\!\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)\\!CV_{i2}\\!\right)p^{2}V_{i}\\!\left(p^{2}\right)\\\
&\qquad\qquad\qquad-\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)p^{2}CT_{i2}T_{j}\\!\left(p^{2}\right),\\\
0&\\!=\\!m_{ii}R_{i}\\!\left(p^{2}\right)\\!+\\!\left(\\!1\\!-\\!\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)\\!\right)\\!p^{2}CV_{i2}T_{i}\\!\left(p^{2}\right)\\\
&\qquad\qquad\qquad-\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)p^{2}CT_{i2}V_{j}\\!\left(p^{2}\right),\\\
0&\\!=\\!-m_{ii}V_{i}\\!\left(p^{2}\right)\\!+\\!\left(\\!1\\!-\\!\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)\\!\right)\\!CV_{i2}S_{i}\\!\left(p^{2}\right)\\\
&\qquad\qquad\qquad-\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)CT_{i2}R_{j}\\!\left(p^{2}\right),\\\
0&\\!=\\!-m_{ii}T_{i}\\!\left(p^{2}\right)\\!+\\!\left(\\!1\\!-\\!\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)\\!\right)\\!CV_{i2}R_{i}\\!\left(p^{2}\right)\\\
&\qquad\qquad\qquad-\frac{\pi}{4}\left(\xi^{2}\\!-\\!p^{2}\right)CT_{i2}S_{j}\\!\left(p^{2}\right),\\\
\end{aligned}\right.$ (19)

for $\left\\{i=1,j=2\right\\}$ and $\left\\{i=2,j=1\right\\}$. We have taken
$\lambda=0$ and $m_{12}=m_{21}=0$ in the previous derivations. It is found
that only $\left\\{CV_{12},CV_{22},CT_{12},CT_{22}\right\\}$ appear in the
final equations. Luckily, these equations are linear for
$\left\\{V_{i}\left(p^{2}\right),S_{i}\left(p^{2}\right),T_{i}\left(p^{2}\right),R_{i}\left(p^{2}\right)\right\\}$,
and $\left\\{CV_{12},CV_{22},CT_{12},CT_{22}\right\\}$ are just constants
which are independent of momentum so that we can solve them analytically. Even
so, they are still very tedious. So we just abbreviate them as:

$\left\\{\begin{aligned}
&V_{i}\left(p^{2};CV_{12},CV_{22},CT_{12},CT_{22}\right),\\\
&S_{i}\left(p^{2};CV_{12},CV_{22},CT_{12},CT_{22}\right),\\\
&T_{i}\left(p^{2};CV_{12},CV_{22},CT_{12},CT_{22}\right),\\\
&R_{i}\left(p^{2};CV_{12},CV_{22},CT_{12},CT_{22}\right).\\\
\end{aligned}\right.$ (20)

If we substitute the functions in Eq. (20) back to the definition Eq. (16), we
can determine the four variables numerically. Once this is done, we have
$\left\\{CV_{12},CV_{22},CT_{12},CT_{22}\right\\}$ then
$\left\\{V_{i},S_{i},T_{i},R_{i}\right\\}$ and all.

Due to the Hermiticity of the Lagrangian, we have

$S^{-1}_{ij}(p)=\gamma^{0\dagger}S_{ij}^{\dagger}(p)\gamma^{0},$ (21)

which has been given in Ref. [14]. Thus the solutions must satisfy
$T_{1}\left(p^{2}\right)=T_{2}\left(p^{2}\right)$ and
$R_{1}\left(p^{2}\right)=R_{2}\left(p^{2}\right)$. We take these relations as
conditions in the following discussion. Therefore, we have totally three
unknown variables $\left\\{CV_{12},CV_{22},CT_{2}\equiv
CT_{12}=CT_{22}\right\\}$, which can be organized in three dimensional
Cartesian coordinate.

##  III NUMERICAL RESULT

###  III.1 Structure of the solutions

As mentioned above, one solution can be parameterized by three variables,
namely $\\{CV_{12},CV_{22},CT_{2}\\}$. We can present the three variables in a
three-dimensional space in which every solution is presented by a point. As
shown in Fig. 1, there are infinite solutions as expected because the original
Lagrangian has a SU(2) symmetry. These solutions form exactly some ellipses.
Apparently, they can be classified into three groups, and each one is an
ellipse with eccentricity $e=\frac{\sqrt{2}}{2}$. Here is the reason why we
take the point interaction model. In this model, we can solve the DS equations
semi-analytically. The more important is that there are only three variables
to parameterize the solutions in this model. So, it gives us an easy way to
find the structure of the solutions. We can also choose other more realistic
interaction models, certainly. However, in that case, we will be only able to
solve the equations numerically, and it will be difficult to show the
structure of the solutions without any advanced knowledge of it.

Figure 1:  Obtained structure of the gap solutions. These solutions are solved
under the parameters as $G=1000\text{GeV}^{-2}$, $m_{11}=m_{22}=50\text{GeV}$,
$\xi^{2}=1000\text{GV}^{2}$. There are two isolated (red) points on both sides
of each ellipse which are the solutions that have no mixing between two flavor
hyperquarks. There exist totally three such points. These points do not break
the SU(2) symmetry.

On the one hand, a SO(2) rotation of $\theta^{\prime}$ in the flavor space
$\psi=\left(\begin{matrix}u_{1}\\\ u_{2}\\\ \end{matrix}\right)$ can be
defined by

$\left(\begin{matrix}u^{\prime}_{1}\\\ u^{\prime}_{2}\\\
\end{matrix}\right)=\left(\begin{matrix}\cos{\theta^{\prime}}&-\sin{\theta^{\prime}}\\\
\sin{\theta^{\prime}}&\cos{\theta^{\prime}}\\\
\end{matrix}\right)\left(\begin{matrix}u_{1}\\\ u_{2}\\\ \end{matrix}\right).$
(22)

It is not hard to find that it induces a rotation $\theta$ defined by

$\displaystyle\qquad\qquad\qquad\qquad\left(\begin{matrix}CV_{12}&CT_{2}\\\
CT_{2}&CV_{22}\\\ \end{matrix}\right)\rightarrow$ (23)
$\displaystyle\left(\begin{matrix}\cos\theta&-\sin\theta\\\
\sin\theta&\cos\theta\\\
\end{matrix}\right)\left(\begin{matrix}CV_{12}&CT_{2}\\\ CT_{2}&CV_{22}\\\
\end{matrix}\right)\left(\begin{matrix}\cos\theta&\sin\theta\\\
-\sin\theta&\cos\theta\\\ \end{matrix}\right)$

in the parameter space. The rotation $\theta$ is also equivalent to the one in
the ellipse, as showing in Fig. 2. The relation between the two rotations is

$\tan{2\theta^{\prime}}=\frac{\tan{\theta}}{\sqrt{2}}.$ (24)

This suggests that the system still has a SO(2) symmetry. The other two
transformation directions in the SU(2) group drive the transformed point out
of the ellipse. However, the point $CV_{12}=CV_{22}$ ($\theta=0$ on the
ellipse) is a fixed point of SU(2) translation. So we can say that the SU(2)
symmetry is broken to SO(2) symmetry.

Figure 2:  Schematic diagram of the solutions in one ellipse. A rotation
$\theta$ in this ellipse can be regarded as a SO(2) transformation. The plane
of the ellipse has an angle of $\frac{\pi}{4}$ with the $\text{CV}_{12}$ axis
and $\text{CV}_{22}$ axis. Note that the origin of the coordinate system in
the figure is not necessary at $(0,0,0)$.The point with $\theta=0$
($\text{CV}_{12}=\text{CV}_{22}$) is a fixed point of SU(2). The points with
$\theta=\frac{\pi}{2}$ and $\theta=\frac{3\pi}{2}$ correspond to the
diagonalized points.

On the other hand, we can diagonalize the propagator or the mass matrix by a
rotation defined in Eq. (22) or Eq. (23) to obtain the physical states. The
fact that the eccentricity of each ellipse is always $\frac{\sqrt{2}}{2}$
guarantees whatever point we choose in the ellipse, after the diagonalization,
we get always the same result. This means each point in the same ellipse
corresponding to the same physical state. Therefore, they are degenerate, and
every different ellipse maps to a different physical state one by one. In
other words, there is a one-to-one correspondence between the ellipses and the
physical states. The diagonalization procedure is actually equivalent to
rotate the point on the ellipse to the major axis point. If we interpret the
different excited states of hyperquarks as different particles, each ellipse
corresponds to one particle. We can then regard the states in the three
ellipses as the up family quarks $\\{u,c,t\\}$ with the same quantum numbers
except for masses.

###  III.2 Mass splitting and spectrum

As mentioned above, by diagonalizing the propagator matrix, we can get the
mass spectrum of up family quarks. So do the down family quarks. The solution
can be parameterized by an ellipse equation

$\displaystyle\left(\\!CV_{12}-\frac{c_{0}}{2}\\!\right)^{2}\\!\\!\\!+\\!\left(\\!CV_{22}-\frac{c_{0}}{2}\\!\right)^{2}\\!\\!+2CT_{2}^{2}$
$\displaystyle=a_{0}^{2},\qquad$ (25a) $\displaystyle CV_{12}+CV_{22}$
$\displaystyle=c_{0},$ (25b)

as shown in Fig. 3. The $a_{0}$ determines the size of the ellipse, and the
$c_{0}$ determines its position. We only need to obtain at least two points in
the ellipse, for figuring out $a_{0}$ and $c_{0}$.

Figure 3:  Schematic diagram of the ellipse. The coordinates in the
$(CV_{12},CV_{22},CT_{2})$ space of characteristic points on the ellipse are
as follows. A: $\left(c_{0},c_{0},0\right)$, B:
$\left(c_{0},c_{0},\frac{a_{0}}{\sqrt{2}}\right)$, C:
$\left(CV_{12},CV_{22},CT_{2}\right)$, D:
$\left(\frac{c_{0}-\sqrt{2}a_{0}}{2},\frac{c_{0}+\sqrt{2}a_{0}}{2},0\right)$,
E:
$\left(\frac{c_{0}+\sqrt{2}a_{0}}{2},\frac{c_{0}-\sqrt{2}a_{0}}{2},0\right)$.

The parameter matrix $\left(\begin{matrix}CV_{12}&CT_{2}\\\ CT_{2}&CV_{22}\\\
\end{matrix}\right)$ can be diagonalized to

$\left(\begin{matrix}\frac{c_{0}+\sqrt{2}a_{0}}{2}&0\\\
0&\frac{c_{0}-\sqrt{2}a_{0}}{2}\\\ \end{matrix}\right).$ (26)

This corresponds to a diagonalized propagator

$S=\left(\begin{matrix}i\not{p}V_{1}(p^{2})+S_{1}(p^{2})&0\\\
0&i\not{p}V_{2}(p^{2})+S_{2}(p^{2})\\\ \end{matrix}\right),$ (27)

where

$\left\\{\begin{aligned}
&V_{1}(p^{2})=\frac{1+\frac{\pi}{4}\frac{c_{0}+\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})}{m_{11}^{2}+p^{2}\left(1+\frac{\pi}{4}\frac{c_{0}+\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})\right)^{2}},\\\
&S_{1}(p^{2})=\frac{m_{11}}{m_{11}^{2}+p^{2}\left(1+\frac{\pi}{4}\frac{c_{0}+\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})\right)},\\\
&M_{1}(p^{2})=\frac{S_{1}(p^{2})}{V_{1}(p^{2})}=\frac{m_{11}}{1+\frac{\pi}{4}\frac{c_{0}+\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})},\end{aligned}\right.$
(28) $\left\\{\begin{aligned}
&V_{2}(p^{2})=\frac{1+\frac{\pi}{4}\frac{c_{0}-\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})}{m_{22}^{2}+p^{2}\left(1+\frac{\pi}{4}\frac{c_{0}-\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})\right)^{2}},\\\
&S_{2}(p^{2})=\frac{m_{22}}{m_{22}^{2}+p^{2}\left(1+\frac{\pi}{4}\frac{c_{0}-\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})\right)},\\\
&M_{2}(p^{2})=\frac{S_{1}(p^{2})}{V_{1}(p^{2})}=\frac{m_{22}}{1+\frac{\pi}{4}\frac{c_{0}-\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})}.\end{aligned}\right.$
(29)

The off-diagonal terms which describes the flavor mixed effects become zero
because $T_{1}=R_{1}=T_{2}=R_{2}=0$ after diagonalization.

There are three groups of different solutions in total and the relationships
between them are quite interesting and subtle. If we look at the top view of
it, these ellipses are always from the top left to the bottom right, inscribed
to the same square, as shown in Fig. 4. Each solution corresponds to an
ellipse in the figure and can be diagonalized to obtain two different mass
eigenvalues. Therefore, we can obtain six mass eigenvalues totally, but they
are repeated in pairs, so we only obtain three different results finally.
According to Ref. [6], one may interpret either the different solutions of a
gap equation or the different propagator poles of one solution as particles of
different generations. Thus, we can obtain the three generation quark masses
from the diagonalized propagator.

Figure 4:  Top view of the solutions. Each ellipse corresponds to two
diagonalized states, but they are repeated in pairs. The diagonalized states
are at the ellipses’ major axis point, but they can also be related
numerically to the red points in this figure. There are three different
diagonalized states corresponding to physical states shown by red points.

When $\xi^{2}=4.52\times 10^{4}\text{GeV}^{2}$, $G=2.88\times
10^{-6}\text{GeV}^{-2}$, $m_{11}=m_{22}=2.15\text{MeV}$, the three
diagonalized solutions (corresponding to the three red points in Fig. 4) are
$(2.812\times 10^{-5},2.812\times 10^{-5},0)$, $(2.817\times
10^{-5},2.817\times 10^{-5},0)$, and $(3.223\times 10^{-7},3.223\times
10^{-7},0)$ in unit $\text{GeV}^{-2}$. Corresponding to one ellipse in the
figure, after the diagonalization, the propagator structure functions and the
mass functions under this group of parameters are shown in Fig. 5. From these
mass functions, we can obtain the constituent mass and the current mass. We
take the mass value at 2GeV as the current mass here. However, it is worth
noting that the definition of the current masses of heavy quarks is slightly
different in the next section. We get then $m_{u}=2.17\text{MeV}$,
$m_{c}=1.27\text{GeV}$, and $m_{t}=171\text{GeV}$. As we see, the mass
function is large but limited in the infrared region and tends to zero in the
ultraviolet region. Besides, their masses have split as we expected.

Figure 5:  The diagonalized propagator structure functions. The parameters are
taken as: $\xi^{2}=4.52\times 10^{4}\text{GeV}^{2}$, $G=2.88\times
10^{-6}\text{GeV}^{-2}$, $m_{11}=m_{22}=2.15\text{MeV}$. The two color lines
correspond to the two different solutions obtained by diagonalization. (a):
The vector structure functions $V_{i}(p^{2})$. (b): The scalar structure
functions $S_{i}(p^{2})$. (c): The mass functions $M_{i}(p^{2})$.

###  III.3 Numerical results with a more realistic interaction model

While the point interaction model described above clearly shows the structure
of the solutions, it has some problems. First of all, the mass
$m_{11}=m_{22}\neq 0$ can not be set to zero because of infrared divergence.
Secondly, the mass function may be divergent at some scale and becomes
negative. As we can see, the denominator of the mass function is
$1+\frac{\pi}{4}\frac{c_{0}-\sqrt{2}a_{0}}{2}(p^{2}-\xi^{2})$, so if
$c_{0}-\sqrt{2}a_{0}>\frac{8}{\pi}\xi^{2}$, the mass function will diverge at
$p^{2}=\xi^{2}-\frac{8}{\pi\left(c_{0}-\sqrt{2}a_{0}\right)}$ and becomes
negative below it. Lastly, higher generation quarks always have more masses
while lower generation quarks have fewer masses at the same momentum scale.
Clearly, the mass functions shown in Fig. 5 do not have correct running
behaviors.

These problems arise from the inappropriate interaction model. In our point
interaction model, the interaction function $G(p)$ is constant while the
dominant strong interaction is asymptotically free actually. So we should take
a more realistic interaction model. Different interaction models are embodied
in different $G(p^{2})$. We take

$G\left(p^{2}\right)=\frac{4}{3}G_{1}\left(p^{2}\right)+G_{2}\left(p^{2}\right),$
(30)

where $\frac{4}{3}$ arises from the sum of color matrices and

$G_{1}\left(p^{2}\right)=\frac{8\pi^{2}}{\omega^{4}}De^{-\frac{p^{2}}{\omega^{2}}}$
(31)

is the infrared constant model (usually simply say QC model) [20] in order to
describe the strong interaction, while

$G_{2}\left(p^{2}\right)=\frac{e^{2}}{p^{2}+\mu^{2}},$ (32)

in order to model the other remainder interactions including electromagnetic
interaction. Because we are using a more complex model, we are only able to
carry out pure numerical calculations now. Thanks to the previous point
interaction model, we have understood the basic structure of the solutions,
which makes the numerical calculation less difficult. Using the same method as
before, we can deal with the current numerical problems only needing a simple
expansion. It is proved that the above discussions on the basic structures of
the solutions are also valid under the current model. Under a group of typical
parameters, an example of the obtained mass functions are shown in Fig. 6. The
best fitted results are shown in Table 1.

Figure 6:  The obtained mass function in terms of the $p^{2}$. The parameters $e^{2}=21$, $\mu^{2}=1000\text{GeV}^{2}$, $\xi^{2}=10^{6}\text{GeV}^{2}$, $D=0.9\text{GeV}^{2}$, $\omega=0.4\text{GeV}$, $m_{11}=m_{22}=0\text{GeV}$ are taken. (a): The mass function of the lightest quark, $m_{1}$ or $m_{u}$. (b): The mass function of the intermediate mass quark, $m_{2}$ or $m_{c}$. (c): The mass function of the heaviest quark, $m_{3}$ or $m_{t}$. Table 1:  Parameter $e^{2}$ dependence of the best fitted constituent mass spectrum. See the footnote for the definition of quality in parentheses. $\xi^{2}=10^{6}\text{GeV}^{2}$, $\mu^{2}=1000\text{GeV}^{2}$, $D=0.9\text{GeV}^{2}$, $\omega=0.4\text{GeV}$, $m_{11}=m_{22}=0\text{GeV}$. Dimentional quantities are in unit GeV. $e^{2}$ | $m_{1}$ | $m_{2}$ | $m_{3}$
---|---|---|---
$33$ |  $0.48$111The current mass is not obtained because of the negative value at 2GeV; refer to (a) in Fig. 6. |  $0.95(0.77)$222Mass at the pole. |  $166(137)$22footnotemark: 2
$20$ |  $0.51(0.00080)$333Mass at 2GeV. |  $0.63(0.016)$33footnotemark: 3 |  $14(13)$22footnotemark: 2

We define the value of the mass function at the zero point as the constituent
mass of the quark. The current masses of heavy quarks and light quarks adopt
different definitions (see the footnotes in the table).

It can be seen that the masses of different generation quarks have obvious
split, and the higher generation the quarks are, the more massive the quarks
are. The constituent masses of all flavor quarks are greater than their
current masses. The constituent masses of heavy flavor quarks are almost the
same as their current masses, while the constituent masses of light flavor
quarks are significantly different from their current masses.

##  IV Goldstone Bosons

Now let us take the down-type sector into account and discuss Goldstone
physics. The $\text{SU(2)}_{\text{L}}$ symmetry of the down-type hyperquarks
is also dynamically broken to $\text{SO(2)}_{\text{L}}$. We assume that the
$\text{SU(4)}_{\text{L}}$ symmetry is also dynamically (or spontaneously)
broken to $\text{SU(2)}_{\text{L}}\otimes\text{SU(2)}_{\text{L}}$ at some
higher scale due to something more fundamental. The $U(1)_{EM}$ terms may be
one of the candidate reasons. As we said at the beginning of the second
section, the $\text{U(1)}_{\text{EM}}$ terms break the
$\text{SU(4)}_{\text{L}}$ symmetry to
$\text{SU(2)}_{\text{H}}\otimes\text{SU(2)}_{\text{H}}$ because of electric
charge conservation. The $\text{U(1)}_{\text{EM}}$ terms are the mixing of the
$\text{U(1)}_{\text{Y}}$ and $\text{SU(2)}_{\text{L}}$. Given that the
$\text{U(1)}_{\text{Y}}$ terms are invariant under the $\text{SU(4)}_{L}$
symmetry and we regard the $\text{SU(2)}_{\text{L}}$ terms as perturbative, it
appears that $\text{SU(4)}_{L}$ is spontaneously broken because the original
Lagrangian is $\text{SU(4)}_{L}$ invariant. It remains to be shown whether
this argument makes sense or not. However, we will not address to that
question here. The symmetry breaking pattern is
$\text{SU(4)}_{\text{L}}\rightarrow\text{SO(2)}_{\text{L}}\otimes\text{SO(2)}_{\text{L}}$
now. So there will be $15-2=13$ Goldstone bosons totally. We follow the
approach provided in Ref. [21] to analyze these Goldstones. Let us briefly
review the approach here first.

Introducing the symmetry breaking term to the Lagrangian as

$L_{SB}=\lambda\bar{\psi}M\psi,$ (33)

where $M\in g=\\{T_{a}\\}$, and the $T_{a}$ are the basis of the Lie algebra
of the symmetry group, one can define the order parameter as

$\Delta=\left\langle\bar{\psi}M\psi\right\rangle.$ (34)

In the limit $\lambda\rightarrow 0$, if the order parameter $\Delta$ is not
zero, then the corresponding symmetry generated by $T_{a}$ with $[T_{a},M]\neq
0$ is broken. The corresponding susceptibility is defined as

$\chi_{\Delta}=\lim\limits_{\lambda\rightarrow 0}\frac{\Delta}{\lambda}.$ (35)

Then, the divergence of $\chi_{\Delta}$ suggests a spontaneous symmetry
breaking. The Goldstones can be characterized by

$\Pi_{a}=i\bar{\psi}\left[T_{a},M\right]\psi.$ (36)

Here, we choose the matrix $M=T_{1}+T_{13}$, where $T_{i}$, $i=1,\cdots,15$,
are the generators of SU(4). We define

$\displaystyle\Delta_{1}=$ $\displaystyle\langle\bar{\psi}T_{1}\psi\rangle,$
(37) $\displaystyle\Delta_{2}=$
$\displaystyle\langle\bar{\psi}T_{13}\psi\rangle,$ (38)

then, it can be shown that

$\sum_{c}f_{ac1}^{2}\Delta_{1}+\sum_{c}f_{ac13}^{2}\Delta_{2}=\lambda
D_{aa}(\omega=0,\textbf{q}=0),$ (39)

where $D_{ij}$ stand for propagators of Goldstones. This formula is an
extended multi-flavor version of what Takashi Yanagisawa obtained there. See
appendix B for the proof. It can be seen that the divergence of susceptibility
means $\Pi_{a}$ has a pole at $\omega=0$ and $q=0$, so $\Pi_{a}$ is indeed
massless. The symmetry breaking term is then

$L_{SB}=\lambda\\!\left(\begin{matrix}\bar{u}_{1}&\bar{u}_{2}&\bar{d}_{1}&\bar{d}_{2}\end{matrix}\right)_{\\!R}\\!\\!\left(\begin{matrix}0&1&0&0\\\
1&0&0&0\\\ 0&0&0&1\\\ 0&0&1&0\\\
\end{matrix}\right)\\!\\!\left(\begin{matrix}u_{1}\\\ u_{2}\\\ d_{1}\\\
d_{2}\\\ \end{matrix}\right)_{\\!\\!L},$ (40)

and the order parameter is

$\Delta=\left\langle\bar{\psi}M\psi\right\rangle=\left\langle\bar{u}_{2_{R}}u_{1_{L}}+\bar{d}_{2_{R}}d_{1_{L}}\right\rangle+h.c.$
(41)

When there is flavor change between $\\{u_{1},u_{2}\\}$ and
$\\{d_{1},d_{2}\\}$, the symmetry is broken. The corresponding symmetry
breaking pattern is desired as
$\text{SU(4)}_{L}\rightarrow\text{SO(2)}_{L}\otimes\text{SO(2)}_{L}$.

According to Eq. (36), the Goldstones can be written exactly as (For brevity,
we omit the subscripts L and R in the following formula, but keep that in
mind.):

$\displaystyle\pi_{1}=0,$ (42)
$\displaystyle\pi_{2}=2\left(\bar{u}_{1}u_{1}-\bar{u}_{2}u_{2}\right),$
$\displaystyle\pi_{3}=2i\left(\bar{u}_{1}u_{2}-\bar{u}_{2}u_{1}\right),$
$\displaystyle\pi_{4}=i\left(\bar{u}_{1}d_{2}-\bar{u}_{2}d_{1}+\bar{d}_{1}u_{2}-\bar{d}_{2}u_{1}\right),$
$\displaystyle\pi_{5}=\left(\bar{u}_{1}d_{2}-\bar{u}_{2}d_{1}-\bar{d}_{1}u_{2}+\bar{d}_{2}u_{1}\right),$
$\displaystyle\pi_{6}=i\left(\bar{u}_{2}d_{2}-\bar{u}_{1}d_{1}+\bar{d}_{1}u_{1}-\bar{d}_{2}u_{2}\right),$
$\displaystyle\pi_{7}=\left(\bar{u}_{2}d_{2}-\bar{u}_{1}d_{1}-\bar{d}_{1}u_{1}+\bar{d}_{2}u_{2}\right),$
$\displaystyle\pi_{8}=\frac{2i}{\sqrt{3}}\left(\bar{d}_{2}d_{1}-\bar{d}_{1}d_{2}\right),$
$\displaystyle\pi_{9}=i\left(\bar{u}_{1}d_{1}-\bar{u}_{2}d_{2}+\bar{d}_{2}u_{2}-\bar{d}_{1}u_{1}\right),$
$\displaystyle\pi_{10}=\left(\bar{u}_{1}d_{1}-\bar{u}_{2}d_{2}-\bar{d}_{2}u_{2}+\bar{d}_{1}u_{1}\right),$
$\displaystyle\pi_{11}=i\left(\bar{u}_{2}d_{1}-\bar{u}_{1}d_{2}+\bar{d}_{2}u_{1}-\bar{d}_{1}u_{2}\right),$
$\displaystyle\pi_{12}=\left(\bar{u}_{2}d_{1}-\bar{u}_{1}d_{2}-\bar{d}_{2}u_{1}+\bar{d}_{1}u_{2}\right),$
$\displaystyle\pi_{13}=0,$
$\displaystyle\pi_{14}=2\left(\bar{d}_{1}d_{1}-\bar{d}_{2}d_{2}\right),$
$\displaystyle\pi_{15}=\left(\sqrt{\frac{3}{2}}+\sqrt{\frac{1}{6}}\right)i\left(\bar{d}_{1}d_{2}-\bar{d}_{2}d_{1}\right).$

As we can see, they are not all completely independent. There are only eight
independent Goldstones. After appropriate linear combinations, we can identify
them as (ignoring the normalization constant):

$\begin{array}[]{ll}\Pi_{1}=\pi_{2},&\qquad\Pi_{5}=\pi_{3},\\\
\Pi_{2}=\pi_{14},&\qquad\Pi_{6}=\pi_{15},\\\
\Pi_{3}=\pi_{7}-i\pi_{6},&\qquad\Pi_{7}=i\pi_{5}+\pi_{4},\\\
\Pi_{4}=\pi_{7}+i\pi_{6},&\qquad\Pi_{8}=i\pi_{5}-\pi_{4}.\\\ \end{array}$ (43)

It is obvious that $\Pi_{3}$ conjugate with $\Pi_{4}$ and $\Pi_{7}$ conjugate
with $\Pi_{8}$ (ignoring the unimportant overall phase).

Under the preserved transformation $M=T_{1}+T_{13}$, the hyperquarks transform
as:

$\displaystyle\left(\begin{matrix}u_{1}\\\ u_{2}\\\
\end{matrix}\right)_{L}\rightarrow\left(\begin{matrix}\cos{\theta_{1}}&i\sin{\theta_{1}}\\\
i\sin{\theta_{1}}&\cos{\theta_{1}}\\\
\end{matrix}\right)\left(\begin{matrix}u_{1}\\\ u_{2}\\\
\end{matrix}\right)_{L},$ (44) $\displaystyle\left(\begin{matrix}d_{1}\\\
d_{2}\\\
\end{matrix}\right)_{L}\rightarrow\left(\begin{matrix}\cos{\theta_{2}}&i\sin{\theta_{2}}\\\
i\sin{\theta_{2}}&\cos{\theta_{2}}\\\
\end{matrix}\right)\left(\begin{matrix}d_{1}\\\ d_{2}\\\
\end{matrix}\right)_{L}.$

So, the Goldstones transform as

$\displaystyle\Pi_{5}$
$\displaystyle=i\left(\bar{u}_{1_{R}}u_{2_{L}}-\bar{u}_{2_{R}}u_{1_{L}}\right)$
(45) $\displaystyle=\left(\begin{matrix}\bar{u}_{1}&\bar{u}_{2}\\\
\end{matrix}\right)_{R}\left(\begin{matrix}0&i\\\ -i&0\\\
\end{matrix}\right)\left(\begin{matrix}u_{1}\\\ u_{2}\\\
\end{matrix}\right)_{L}$
$\displaystyle\rightarrow\left(\begin{matrix}\bar{u}_{1}&\bar{u}_{2}\\\
\end{matrix}\right)_{R}\left(\begin{matrix}c&-is\\\ -is&c\\\
\end{matrix}\right)\left(\begin{matrix}0&i\\\ -i&0\\\ \end{matrix}\right)$
$\displaystyle\qquad\qquad\qquad\quad\times\left(\begin{matrix}c&is\\\ is&c\\\
\end{matrix}\right)\left(\begin{matrix}u_{1}\\\ u_{2}\\\
\end{matrix}\right)_{\\!\\!L}$
$\displaystyle=\left(\begin{matrix}\bar{u}_{1}&\bar{u}_{2}\\\
\end{matrix}\right)_{R}\left(\begin{matrix}-\sin{2\theta}&i\cos{2\theta}\\\
-i\cos{2\theta}&\sin{2\theta}\\\
\end{matrix}\right)\left(\begin{matrix}u_{1}\\\ u_{2}\\\
\end{matrix}\right)_{\\!\\!L}.$

Because of

$\displaystyle\Pi_{1}$
$\displaystyle=\left(\bar{u}_{1_{R}}u_{1_{L}}-\bar{u}_{2_{R}}u_{2_{L}}\right)$
(46) $\displaystyle=\left(\begin{matrix}\bar{u}_{1}&\bar{u}_{2}\\\
\end{matrix}\right)_{R}\left(\begin{matrix}1&0\\\ 0&-1\\\
\end{matrix}\right)\left(\begin{matrix}u_{1}\\\ u_{2}\\\
\end{matrix}\right)_{\\!\\!L},$

we see that $\Pi_{5}$ is transformed to $\Pi_{1}$ under the preserved symmetry
group $\text{SO(2)}_{L}\otimes\text{SO(2)}_{L}$ by $\theta=-\frac{\pi}{4}$.
Similarly, $\Pi_{6}$ is transformed to $\Pi_{2}$, $\Pi_{7}$ to $\Pi_{3}$, and
$\Pi_{8}$ to $\Pi_{4}$. We can thus conclude that each two Goldstones are
degenerate. We have, in turn, only four Goldstones to deal with. They are:

$\displaystyle\Pi_{1}$
$\displaystyle=\bar{u}_{1_{R}}u_{1_{L}}-\bar{u}_{2_{R}}u_{2_{L}},$ (47)
$\displaystyle\Pi_{2}$
$\displaystyle=\bar{d}_{1_{R}}d_{1_{L}}-\bar{d}_{2_{R}}d_{2_{L}},$
$\displaystyle\Pi_{3}$
$\displaystyle=\bar{u}_{2_{R}}d_{2_{L}}-\bar{u}_{1_{R}}d_{1_{L}},$
$\displaystyle\Pi_{4}$
$\displaystyle=\bar{d}_{2_{R}}u_{2_{L}}-\bar{d}_{1_{R}}u_{1_{L}}.$

When we consider the $\text{SU(2)}_{\text{W}}$ terms, the $\text{SU(4)}_{L}$
symmetry becomes approximate so that some of the Goldstones become pNGBs.
Notice that $\Pi_{1}$ and $\Pi_{2}$ are due to the horizontal
$\text{SU(2)}_{L}$ symmetry, which is exactly broken by the
$\text{SU(2)}_{\text{W}}$ terms, so one mixing state of them must become pNGB
and have a small mass while the other one is still massless. On the contrary,
$\Pi_{3}$ and $\Pi_{4}$ are due to the vertical $\text{SU(2)}_{L}$ symmetry,
which is not broken; therefore, they are still massless. Since the up-type
quarks have electric charge $+\frac{2}{3}$ and down-type quarks have electric
charge $-\frac{1}{3}$, $\Pi_{1}$ and $\Pi_{2}$ have charge $0$ while $\Pi_{3}$
has charge $-1$ and $\Pi_{4}$ has charge $+1$.

According to the well-known Schwinger mechanism [22, 23, 24, 25, 26, 27],
gauge bosons can absorb a massless boson to gain their masses. They eat the
corresponding bosons as their longitudinal component. This can be
schematically represented as below ($\alpha$ and $\beta$ represent the
relative phase between $\Pi_{1}$ and $\Pi_{2}$):

$Z^{0}$$Z^{0}$$\alpha\Pi_{1}+\beta\Pi_{2}$ $W^{-}$$W^{-}$$\Pi_{3}$
$W^{+}$$W^{+}$$\Pi_{4}$

By calculating the four-points Green functions $G(\bar{f},f;\bar{f},f)$, we
can get the amplitude of two quarks changing to the Goldstones because these
Goldstones give the contributions around poles. Recalling the Lagrangian in
Eq. (1), in the limit $\lambda_{ij}\rightarrow 0$ and $m_{ij}=0$, all fermions
and gauge bosons are symmetric. So we conclude that their amplitudes
translating to Goldstones are identical, and so do $F$’s, which have mass
dimension and are defined by

$p^{\mu}F=\leavevmode\hbox to0pt{\vbox
to0pt{\pgfpicture\makeatletter\hbox{\hskip 0.0pt\lower 0.0pt\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{
}\definecolor{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@rgb@stroke{0}{0}{0}\pgfsys@invoke{
}\pgfsys@color@rgb@fill{0}{0}{0}\pgfsys@invoke{
}\pgfsys@setlinewidth{0.4pt}\pgfsys@invoke{ }\nullfont\hbox
to0.0pt{\pgfsys@beginscope\pgfsys@invoke{ } \feynhand \vertex(p1) at (0,0);
\vertex(p2) at (1,0); \vertex(p3) at (2.5,0); \vertex(p4) at (3.5,0);
\vertex(p5) at (5,0); \vertex(p6) at (6,0); \propag[bos] (p1) to (p2);
\propag[fermion] (p2) to [half left](p3); \propag[fermion] (p3) to [half
left](p2); \vertex(p31) at (2.5,0.07); \vertex(p32) at (2.5,-0.07);
\vertex(p41) at (3.5,0.07); \vertex(p42) at (3.5,-0.07); \propag[fermion]
(p31) to [edge label = $\Pi$] (p41); \propag[fermion] (p42) to (p32);
\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope{}{}{}\hss}\pgfsys@discardpath\pgfsys@invoke{\lxSVG@closescope
}\pgfsys@endscope\hss}}\lxSVG@closescope\endpgfpicture}}$ (48)

Notice that $Z^{0}_{\mu}=\frac{1}{\sqrt{g^{2}+g^{\prime
2}}}\left(gA^{3}_{\mu}-g^{\prime}B_{\mu}\right)$, and
$A_{\mu}=\frac{1}{\sqrt{g^{2}+g^{\prime
2}}}\left(g^{\prime}A^{3}_{\mu}+gB_{\mu}\right)$ as usual, where $A_{\mu}^{a}$
and $B_{\mu}$ are, respectively, the SU(2) and U(1) gauge bosons. In order to
maintain $A_{\mu}$ massless, we should have $\left\langle
A_{\mu}|\Pi\right\rangle=0$. If we write the mixing state of $\Pi_{1}$ and
$\Pi_{2}$ which keeps massless and is eaten by $Z_{0}$ as $\pi_{z}$, while the
other one as $h$, we have

$\displaystyle\left\langle A_{\mu}^{3}|\pi_{z}\right\rangle=gF,$ (49)
$\displaystyle\left\langle Z_{\mu}^{0}|\pi_{z}\right\rangle=-g^{\prime}F.$

By combining these aspects, the mass of $Z^{0}$ is given as

$\displaystyle m_{Z}^{2}$ $\displaystyle=$
$\displaystyle\frac{g}{\sqrt{g^{2}+g^{\prime
2}}}\left(g\right)\left(g\right)\frac{g}{\sqrt{g^{2}+g^{\prime 2}}}F^{2}$ (50)
$\displaystyle+\frac{g}{\sqrt{g^{2}+g^{\prime
2}}}\left(g\right)\left(-g^{\prime}\right)\frac{-g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}F^{2}$ $\displaystyle+\frac{-g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}\left(-g^{\prime}\right)\left(g\right)\frac{g}{\sqrt{g^{2}+g^{\prime
2}}}F^{2}$ $\displaystyle+\frac{-g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}\left(-g^{\prime}\right)\left(-g^{\prime}\right)\frac{-g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}F^{2}$ $\displaystyle=$ $\displaystyle(g^{2}+g^{\prime 2})F^{2}.$

Similarly, the mass of $A_{\mu}$ is given by

$\displaystyle m_{A}^{2}$ $\displaystyle=$
$\displaystyle\frac{g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}\left(g\right)\left(g\right)\frac{g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}F^{2}$ (51) $\displaystyle+\frac{g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}\left(g\right)\left(-g^{\prime}\right)\frac{g}{\sqrt{g^{2}+g^{\prime
2}}}F^{2}$ $\displaystyle+\frac{g}{\sqrt{g^{2}+g^{\prime
2}}}\left(-g^{\prime}\right)\left(g\right)\frac{g^{\prime}}{\sqrt{g^{2}+g^{\prime
2}}}F^{2}$ $\displaystyle+\frac{g}{\sqrt{g^{2}+g^{\prime
2}}}\left(-g^{\prime}\right)\left(-g^{\prime}\right)\frac{g}{\sqrt{g^{2}+g^{\prime
2}}}F^{2}$ $\displaystyle=$ $\displaystyle 0$

as expected and

$m_{W}^{2}=g^{2}F^{2}.$ (52)

Because $\cos\theta_{w}=\frac{g}{\sqrt{g^{2}+g^{\prime 2}}}$, we see that
$m_{W}=m_{Z}\cos\theta_{w}$.

Because the other boson $h$ is a mixing of $\Pi_{1}$ and $\Pi_{2}$, it
contains all components of $u_{1}$, $u_{2}$, $d_{1}$, and $d_{2}$. Therefore,
$h$ has all coupling with fermions, with weak gauge bosons, and with itself.
We can identify it with the Higgs boson in the Glashow-Weinberg-Salam model.

##  V SUMMARY

We studied the flavor change effects in a multi-flavor system. With the
extended multi-flavor DS equations, we find that the flavor symmetry SU(2) is
dynamically broken to SO(2). Combining the up-type quark sector and the down-
type quark sector and take the Electroweak interaction as perturbation into
consideration, we get the
$\text{SU(4)}_{L}\rightarrow\text{SU(2)}_{L}\otimes\text{SU(2)}_{L}\rightarrow\text{SO(2)}_{L}\otimes\text{SO(2)}_{L}$
symmetry breaking chain. This would produce massless Goldstones, which can
replace the role of the Higgs. Besides, due to the flavor mixing effects, the
fermion mass eigenstates will misalign with the flavor eigenstates. By
diagonalizing the mass matrix, the quark states are redefined, and the fermion
masses are split apparently. We can interpret the new fermion mass eigenstates
with distinct masses as different generations.

We started with a simple point interaction model. By calculating this model
analytically, we obtained the structure of their solutions. We find that there
are always three sets of closely related solutions, which can be classified
into three interrelated ellipses. These solutions generate three distinct
masses, which can be identified as $m_{u}$, $m_{c}$, and $m_{t}$. So does the
down quark sector. After that, we studied a more realistic interaction model.
Although this model can only be solved numerically, we find the structures of
solutions obtained before still hold. The masses of different generation
quarks split obviously, and the constituent masses are always greater than
their current masses. The higher generation quarks have almost the same
constituent masses as the current masses, while for the low generation quarks,
the constituent masses are much more significant due to dynamical effects.

Only consider the left-handed part, the symmetry breaking produces fifteen
Goldstones, but only four of them are independent. Two of them have electric
charge $0$, and the other two have electric charge $+1$, $-1$, respectively.
Among these Goldstones, one is pNGB and has a mass which could be identified
as the Higgs boson. The other three Goldstones keep massless and will be eaten
by gauge bosons to give $W^{\pm}$ and $Z_{0}$ masses via the Schwinger
mechanism. Besides, the famous equation $m_{W}=m_{Z}\cos\theta_{w}$ is
maintained.

There is still much work to be done. The $m_{Z}$, $m_{W}$ and $m_{H}$ should
be calculated exactly, and the Higgs couplings should be checked carefully.
The related work is under progress.

###### Acknowledgements.

The work was supported by the National Natural Science Foundation of China
under Contracts No. 11435001 and No. 11775041. Helpful discussions with
Professors Shou-hua Zhu and Qing-hong Cao are acknowledged with great thanks.

## Appendix A

The eight nonlinear integration equations for $\\{V_{i},S_{i},T_{i},R_{i}\\}$
are as follows:

$\displaystyle 1$ $\displaystyle=$
$\displaystyle\Big{[}Z_{2}^{11}Z_{m}^{11}m_{11}S_{1}\left(p^{2}\right)+Z_{2}^{11}p^{2}V_{1}\left(p^{2}\right)$
(53)
$\displaystyle+Z_{m}^{12}m_{12}R_{2}\left(p^{2}\right)+Z_{\lambda}^{12}\lambda_{12}p^{2}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}3\Big{[}S_{1}\left(k^{2}\right)S_{1}\left(p^{2}\right)$
$\displaystyle+R_{1}\left(k^{2}\right)R_{2}\left(p^{2}\right)\Big{]}+\Big{[}V_{1}\left(k^{2}\right)V_{1}\left(p^{2}\right)$
$\displaystyle+T_{1}\left(k^{2}\right)T_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}},$ $\displaystyle 1$
$\displaystyle=$
$\displaystyle\Big{[}Z_{m}^{21}m_{21}R_{1}\left(p^{2}\right)+Z_{\lambda}^{21}\lambda_{21}p^{2}T_{1}\left(p^{2}\right)$
(54)
$\displaystyle+Z_{2}^{22}Z_{m}^{22}m_{22}S_{2}\left(p^{2}\right)+Z_{2}^{22}p^{2}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}3\Big{[}R_{2}\left(k^{2}\right)R_{1}\left(p^{2}\right)$
$\displaystyle+S_{2}\left(k^{2}\right)S_{2}\left(p^{2}\right)\Big{]}+\Big{[}T_{2}\left(k^{2}\right)T_{1}\left(p^{2}\right)$
$\displaystyle+V_{2}\left(k^{2}\right)V_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}},$ $\displaystyle 0$
$\displaystyle=$
$\displaystyle\Big{[}Z_{2}^{11}Z_{m}^{11}m_{11}R_{1}\left(p^{2}\right)+Z_{2}^{11}p^{2}T_{1}\left(p^{2}\right)$
(55)
$\displaystyle+Z_{m}^{12}m_{12}S_{2}\left(p^{2}\right)+Z_{\lambda}^{12}\lambda_{12}p^{2}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}3\Big{[}S_{1}\left(k^{2}\right)R_{1}\left(p^{2}\right)$
$\displaystyle+R_{1}\left(k^{2}\right)S_{2}\left(p^{2}\right)\Big{]}+\Big{[}V_{1}\left(k^{2}\right)T_{1}\left(p^{2}\right)$
$\displaystyle+T_{1}\left(k^{2}\right)V_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}},$ $\displaystyle 0$
$\displaystyle=$
$\displaystyle\Big{[}Z_{m}^{21}m_{21}S_{1}\left(p^{2}\right)+Z_{\lambda}^{21}\lambda_{21}p^{2}V_{1}\left(p^{2}\right)$
(56)
$\displaystyle+Z_{2}^{22}Z_{m}^{22}m_{22}R_{2}\left(p^{2}\right)+Z_{2}^{22}p^{2}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}3\Big{[}R_{2}\left(k^{2}\right)S_{1}\left(p^{2}\right)$
$\displaystyle+S_{2}\left(k^{2}\right)R_{2}\left(p^{2}\right)\Big{]}+\Big{[}T_{2}\left(k^{2}\right)V_{1}\left(p^{2}\right)$
$\displaystyle+V_{2}\left(k^{2}\right)T_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}},$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle
p^{2}\Big{[}Z_{2}^{11}S_{1}\left(p^{2}\right)-Z_{2}^{11}Z_{m}^{11}m_{11}V_{1}\left(p^{2}\right)$
(57)
$\displaystyle+Z_{\lambda}^{12}\lambda_{12}R_{2}\left(p^{2}\right)-Z_{m}^{12}m_{12}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}\\!\\!-\\!3p^{2}\Big{[}S_{1}\left(k^{2}\right)V_{1}\left(p^{2}\right)$
$\displaystyle+R_{1}\left(k^{2}\right)T_{2}\left(p^{2}\right)\Big{]}+\Big{[}V_{1}\left(k^{2}\right)S_{1}\left(p^{2}\right)$
$\displaystyle+T_{1}\left(k^{2}\right)R_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}},$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle
p^{2}\Big{[}Z_{\lambda}^{21}\lambda_{21}R_{1}\left(p^{2}\right)-Z_{m}^{21}m_{21}T_{1}\left(p^{2}\right)$
(58)
$\displaystyle+Z_{2}^{22}S_{2}\left(p^{2}\right)-Z_{2}^{22}Z_{m}^{22}m_{22}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}\\!\\!-\\!3p^{2}\Big{[}R_{2}\left(k^{2}\right)T_{1}\left(p^{2}\right)$
$\displaystyle+S_{2}\left(k^{2}\right)V_{2}\left(p^{2}\right)\Big{]}+\Big{[}T_{2}\left(k^{2}\right)R_{1}\left(p^{2}\right)$
$\displaystyle+V_{2}\left(k^{2}\right)S_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}},$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle
p^{2}\Big{[}Z_{2}^{11}R_{1}\left(p^{2}\right)-Z_{2}^{11}Z_{m}^{11}m_{11}T_{1}\left(p^{2}\right)$
(59)
$\displaystyle+Z_{\lambda}^{12}\lambda_{12}S_{2}\left(p^{2}\right)-Z_{m}^{12}m_{12}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}\\!\\!-\\!3p^{2}\Big{[}S_{1}\left(k^{2}\right)T_{1}\left(p^{2}\right)$
$\displaystyle+R_{1}\left(k^{2}\right)V_{2}\left(p^{2}\right)\Big{]}+\Big{[}V_{1}\left(k^{2}\right)R_{1}\left(p^{2}\right)$
$\displaystyle+T_{1}\left(k^{2}\right)S_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}},$ $\displaystyle 0$
$\displaystyle=$ $\displaystyle
p^{2}\Big{[}Z_{\lambda}^{21}\lambda_{21}S_{1}\left(p^{2}\right)-Z_{m}^{21}m_{21}V_{1}\left(p^{2}\right)$
(60)
$\displaystyle+Z_{2}^{22}R_{2}\left(p^{2}\right)-Z_{2}^{22}Z_{m}^{22}m_{22}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\int\frac{\mathrm{d}^{4}k}{\left(2\pi\right)^{4}}G(p-k)\bigg{\\{}\\!\\!-\\!3p^{2}\Big{[}R_{2}\left(k^{2}\right)V_{1}\left(p^{2}\right)$
$\displaystyle+S_{2}\left(k^{2}\right)T_{2}\left(p^{2}\right)\Big{]}+\Big{[}T_{2}\left(k^{2}\right)S_{1}\left(p^{2}\right)$
$\displaystyle+V_{2}\left(k^{2}\right)R_{2}\left(p^{2}\right)\Big{]}\Big{[}2p\cdot
k$ $\displaystyle+\frac{\left(p^{2}+k^{2}\right)p\cdot
k-2p^{2}k^{2}}{\left(p-k\right)^{2}}\Big{]}\bigg{\\}}.$

Making use of $\\{CV_{ij},CS_{ij},CT_{ij},CR_{ij}\\}$ ($i=1,2$), we obtain the
simplified equations as follows:

$\displaystyle 1$ $\displaystyle=$
$\displaystyle\Big{[}Z_{2}^{11}Z_{m}^{11}m_{11}S_{1}\left(p^{2}\right)+Z_{2}^{11}p^{2}V_{1}\left(p^{2}\right)$
(61)
$\displaystyle+Z_{m}^{12}m_{12}R_{2}\left(p^{2}\right)+Z_{\lambda}^{12}\lambda_{12}p^{2}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\frac{3\pi}{2}CS_{11}S_{1}(p^{2})+\frac{3\pi}{2}CR_{11}R_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CV_{11}V_{1}(p^{2})+\frac{\pi}{4}p^{4}CV_{12}V_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CT_{11}T_{2}(p^{2})+\frac{\pi}{4}p^{4}CT_{12}T_{2}(p^{2}),\qquad\quad$
$\displaystyle 1$ $\displaystyle=$
$\displaystyle\Big{[}Z_{m}^{21}m_{21}R_{1}\left(p^{2}\right)+Z_{\lambda}^{21}\lambda_{21}p^{2}T_{1}\left(p^{2}\right)$
(62)
$\displaystyle+Z_{2}^{22}Z_{m}^{22}m_{22}S_{2}\left(p^{2}\right)+Z_{2}^{22}p^{2}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\frac{3\pi}{2}CR_{21}R_{1}(p^{2})+\frac{3\pi}{2}CS_{21}S_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CT_{21}T_{1}(p^{2})+\frac{\pi}{4}p^{4}CT_{22}T_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CV_{21}V_{2}(p^{2})+\frac{\pi}{4}p^{4}CV_{22}V_{2}(p^{2}),$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Big{[}Z_{2}^{11}Z_{m}^{11}m_{11}R_{1}\left(p^{2}\right)+Z_{2}^{11}p^{2}T_{1}\left(p^{2}\right)$
(63)
$\displaystyle+Z_{m}^{12}m_{12}S_{2}\left(p^{2}\right)+Z_{\lambda}^{12}\lambda_{12}p^{2}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\frac{3\pi}{2}CS_{11}R_{1}(p^{2})+\frac{3\pi}{2}CR_{11}S_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CV_{11}T_{1}(p^{2})+\frac{\pi}{4}p^{4}CV_{12}T_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CT_{11}V_{2}(p^{2})+\frac{\pi}{4}p^{4}CT_{12}V_{2}(p^{2}),$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Big{[}Z_{m}^{21}m_{21}S_{1}\left(p^{2}\right)+Z_{\lambda}^{21}\lambda_{21}p^{2}V_{1}\left(p^{2}\right)$
(64)
$\displaystyle+Z_{2}^{22}Z_{m}^{22}m_{22}R_{2}\left(p^{2}\right)+Z_{2}^{22}p^{2}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle+\frac{3\pi}{2}CR_{21}S_{1}(p^{2})+\frac{3\pi}{2}CS_{21}R_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CT_{21}V_{1}(p^{2})+\frac{\pi}{4}p^{4}CT_{22}V_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}p^{2}CV_{21}T_{2}(p^{2})+\frac{\pi}{4}p^{4}CV_{22}T_{2}(p^{2}),$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Big{[}Z_{2}^{11}S_{1}\left(p^{2}\right)-Z_{2}^{11}Z_{m}^{11}m_{11}V_{1}\left(p^{2}\right)$
(65)
$\displaystyle+Z_{\lambda}^{12}\lambda_{12}R_{2}\left(p^{2}\right)-Z_{m}^{12}m_{12}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle-\frac{3\pi}{2}CS_{11}V_{1}(p^{2})-\frac{3\pi}{2}CR_{11}T_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CV_{11}S_{1}(p^{2})+\frac{\pi}{4}p^{2}CV_{12}S_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CT_{11}R_{2}(p^{2})+\frac{\pi}{4}p^{2}CT_{12}R_{2}(p^{2}),$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Big{[}Z_{\lambda}^{21}\lambda_{21}R_{1}\left(p^{2}\right)-Z_{m}^{21}m_{21}T_{1}\left(p^{2}\right)$
(66)
$\displaystyle+Z_{2}^{22}S_{2}\left(p^{2}\right)-Z_{2}^{22}Z_{m}^{22}m_{22}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle-\frac{3\pi}{2}CR_{21}T_{1}(p^{2})-\frac{3\pi}{2}CS_{21}V_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CT_{21}R_{1}(p^{2})+\frac{\pi}{4}p^{2}CT_{22}R_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CV_{21}S_{2}(p^{2})+\frac{\pi}{4}p^{2}CV_{22}S_{2}(p^{2}),\qquad\qquad$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Big{[}Z_{2}^{11}R_{1}\left(p^{2}\right)-Z_{2}^{11}Z_{m}^{11}m_{11}T_{1}\left(p^{2}\right)$
(67)
$\displaystyle+Z_{\lambda}^{12}\lambda_{12}S_{2}\left(p^{2}\right)-Z_{m}^{12}m_{12}V_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle-\frac{3\pi}{2}CS_{11}T_{1}(p^{2})-\frac{3\pi}{2}CR_{11}V_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CV_{11}R_{1}(p^{2})+\frac{\pi}{4}p^{2}CV_{12}R_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CT_{11}S_{2}(p^{2})+\frac{\pi}{4}p^{2}CT_{12}S_{2}(p^{2}),$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\Big{[}Z_{\lambda}^{21}\lambda_{21}S_{1}\left(p^{2}\right)-Z_{m}^{21}m_{21}V_{1}\left(p^{2}\right)$
(68)
$\displaystyle+Z_{2}^{22}R_{2}\left(p^{2}\right)-Z_{2}^{22}Z_{m}^{22}m_{22}T_{2}\left(p^{2}\right)\Big{]}$
$\displaystyle-\frac{3\pi}{2}CR_{21}V_{1}(p^{2})-\frac{3\pi}{2}CS_{21}T_{2}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CT_{21}S_{1}(p^{2})+\frac{\pi}{4}p^{2}CT_{22}S_{1}(p^{2})$
$\displaystyle-\frac{3\pi}{4}CV_{21}R_{2}(p^{2})+\frac{\pi}{4}p^{2}CV_{22}R_{2}(p^{2}).\qquad\quad$

## Appendix B

We write the generators of $SU(4)$ as $\\{T_{a}\\}$, which satisfies

$\textrm{Tr}T_{a}T_{b}=2\delta_{ab}.$ (69)

The structure constants of the Lie algebra are

$\left[T_{a},T_{b}\right]=\sum_{c}if_{abc}T_{c}.$ (70)

Moreover, we have

$\left[Q_{a},\psi\right]=-T_{a}\psi$ (71)

where

$Q_{a}=\int\mathrm{d}\textbf{r}\psi^{\dagger}T_{a}\psi$ (72)

is the charge corresponding to $T_{a}$. We set M to be the superposition of
two bases: $M=T_{m}=T_{1}+T_{13}$.

We begin with

$\displaystyle e^{i\theta Q_{a}}\bar{\psi}e^{-i\theta
Q_{a}}\left[M,T_{a}\right]e^{i\theta Q_{a}}\psi e^{-i\theta Q_{a}}$
$\displaystyle=\left(\bar{\psi}+i\theta\bar{\psi}T_{a}\right)\left[M,T_{a}\right]\left(\psi-i\theta
T_{a}\psi\right)$
$\displaystyle=\bar{\psi}\left[M,T_{a}\right]\psi+i\theta\bar{\psi}\left[T_{a},\left[M,T_{a}\right]\right]\psi+O(\theta^{2})$
$\displaystyle=-i\sum_{c}f_{amc}\bar{\psi}T_{c}\psi-i\theta\sum_{cd}f_{acm}f_{acd}\bar{\psi}T_{d}\psi$
$\displaystyle\quad+O\left(\theta^{2}\right).$ (73)

Notice that from here, $m$ is an index needed to sum for $m=1$ and $m=13$, for
instance, $f_{amc}=f_{a1c}+f_{a13c}$. We assume that
$\langle\bar{\psi}M\psi\rangle=\langle\bar{\psi}T_{m}\psi\rangle\neq 0$, that
is, $\langle\bar{\psi}T_{1}\psi\rangle\neq 0$,
$\langle\bar{\psi}T_{13}\psi\rangle\neq 0$, but
$\langle\bar{\psi}T_{d}\psi\rangle=0$ when $d\neq 1$ or $d\neq 13$. Then, we
define

$\Delta=\Delta_{1}+\Delta_{2},$ (74)

where

$\displaystyle\Delta_{1}=$ $\displaystyle\langle\bar{\psi}T_{1}\psi\rangle,$
(75) $\displaystyle\Delta_{2}=$
$\displaystyle\langle\bar{\psi}T_{13}\psi\rangle.$ (76)

Notice that $f_{1a13}=0$, we have

$\displaystyle\sum_{c}f_{amc}\langle\bar{\psi}T_{c}\psi\rangle=$
$\displaystyle
f_{am1}\langle\bar{\psi}T_{1}\psi\rangle+f_{am13}\langle\bar{\psi}T_{13}\psi\rangle$
$\displaystyle=$ $\displaystyle
f_{a,1,1}\langle\bar{\psi}T_{1}\psi\rangle+f_{a,13,1}\langle\bar{\psi}T_{1}\psi\rangle$
$\displaystyle+f_{a,1,13}\langle\bar{\psi}T_{13}\psi\rangle+f_{a,13,13}\langle\bar{\psi}T_{13}\psi\rangle$
$\displaystyle=$ $\displaystyle 0.$ (77)

Also, because of $\sum_{c}f_{ac1}f_{ac13}=0$, we have

$\displaystyle\sum_{cd}f_{acm}f_{acd}\bar{\psi}T_{d}\psi$ $\displaystyle=$
$\displaystyle\sum_{cd}f_{ac1}f_{acd}\bar{\psi}T_{d}\psi+\sum_{cd}f_{ac13}f_{acd}\bar{\psi}T_{d}\psi$
$\displaystyle=$
$\displaystyle\sum_{c}f_{ac1}f_{ac1}\bar{\psi}T_{1}\psi+\sum_{c}f_{ac1}f_{ac13}\bar{\psi}T_{13}\psi$
$\displaystyle+\sum_{c}f_{ac13}f_{ac1}\bar{\psi}T_{1}\psi+\sum_{c}f_{ac13}f_{ac13}\bar{\psi}T_{13}\psi$
$\displaystyle=$
$\displaystyle\sum_{c}f_{ac1}^{2}\bar{\psi}T_{1}\psi+\sum_{c}f_{ac13}^{2}\bar{\psi}T_{13}\psi$
$\displaystyle=$
$\displaystyle\sum_{c}f_{ac1}^{2}\Delta_{1}+\sum_{c}f_{ac13}^{2}\Delta_{2}.$
(78)

These result in

$\displaystyle\sum_{c}f_{ac1}^{2}\Delta_{1}+\sum_{c}f_{ac13}^{2}\Delta_{2}$
$\displaystyle=\lim_{\theta\to 0}\frac{-1}{i\theta}e^{i\theta
Q_{q}}\bar{\psi}e^{-i\theta Q_{a}}\left[M,T_{a}\right]e^{i\theta Q_{a}}\psi
e^{-i\theta Q_{a}}$ $\displaystyle=\lim_{\theta\to
0}\frac{-1}{\theta}e^{i\theta Q_{a}}\Pi_{a}e^{-i\theta Q_{a}};$ (79)

in the third line, we have made use of $e^{-i\theta
Q_{a}}\left[M,T_{a}\right]e^{i\theta Q_{a}}=\left[M,T_{a}\right]$ because
$Q_{a}$ is an operator (not matrix).

We add the symmetry breaking term $H_{\lambda}$ to the original Hamiltonian
$H_{0}$ and write the total Hamiltonian as

$H_{T}=H_{0}+H_{\lambda},$ (80)

where

$H_{\lambda}=-\lambda\int\mathrm{d}\mathbf{r}\bar{\psi}T_{1}\psi-\lambda\int\mathrm{d}\mathbf{r}\bar{\psi}T_{13}\psi.$
(81)

Furthermore, we write $\phi$ as the ground state of the total Hamiltonian
$H_{T}$, that is $H_{T}\phi=E_{0}\phi$. We define

$\displaystyle A_{a}$ $\displaystyle\equiv\langle\phi|e^{i\theta
Q_{a}}\Pi_{a}(\textbf{r})e^{-i\theta Q_{a}}|\phi\rangle$
$\displaystyle=\langle\phi|e^{iH_{T}t}e^{i\theta
Q_{a}}\Pi_{a}(\textbf{r})e^{-i\theta Q_{a}}e^{-iH_{T}t}|\phi\rangle.$

Following Ref. [21], we get the same result:

$\displaystyle A_{a}=$ $\displaystyle\langle\phi|\Pi_{a}|\phi\rangle$ (83)
$\displaystyle-\theta\lambda\int\mathrm{d}\textbf{r}^{\prime}\int_{-\infty}^{\infty}\mathrm{d}t^{\prime}D_{aa}^{R}\left(t-t^{\prime},\textbf{r}-\textbf{r}^{\prime}\right)$
$\displaystyle+O(H_{I}^{2}),$

where the retarded Green’s function is defined as

$D_{aa}^{R}(t-t^{\prime},\textbf{r}-\textbf{r}^{\prime})\\!=\\!-i\theta(t-t^{\prime})\langle\left[\Pi_{a}(\textbf{r},t),\Pi_{a}(\textbf{r}^{\prime}\\!,t^{\prime})\right]\rangle,$
(84)

and

$H_{I}=-\theta\lambda\int\mathrm{d}\textbf{r}\Pi_{a}(\textbf{r},t).$ (85)

Substituting Eq. (83) and $\langle\phi|\Pi_{a}|\phi\rangle=0$ to Eq. (B), and
taking the Fourier transformation and analytic continuation [28], we have

$\sum_{c}f_{ac1}^{2}\Delta_{1}+\sum_{c}f_{ac13}^{2}\Delta_{2}=\lambda
D_{aa}(\omega=0,\textbf{q}=0).$ (86)

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