# Localization of Massive and Massless Fermions on Two-Field Branes

Authors: Ali Farokhtabar; Ali Tofighi
Journal: Advances in High Energy Physics
 (2017)

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## Abstract

We study fermion localization and resonances on a special type of brane-world model supporting brane splitting. In such models one can construct multiwall branes which cause considerable simplification in the field equations. We use a polynomial superpotential to construct this brane. The suitable Yukawa coupling between the background scalar field and the localized fermion is determined. The massive fermion resonance spectrum is obtained. The number of resonances is increased for higher values of Yukawa coupling.


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## Body

## 1. Introduction

Brane-world scenarios have attracted considerable attention in the literature during the last two decades because these models can address some important issues in theoretical physics problems such as hierarchy [1–3] and cosmological constant problem [4, 5]. The branes in Randall and Sundrum (RS) models are fixed in some points along extra dimension and have a δ function form [1, 6]. This brane-world model is very ideal and its formation has no dynamical mechanism. But for realistic models thickness of brane should be considered. By now several thick brane construction mechanisms have been developed such as thick branes generated from pure gravity [7–11], fermion self-interaction branes [12, 13], and thick brane scenarios with the gravity coupled to five-dimensional scalar fields [14–21]. In the last scenario, the scalar field configuration is usually a kink. It is found that a single kink becomes unstable when it moves in a discrete lattice with a large velocity while multikink solutions remain stable [22]. This phenomenon is associated with interaction between kink and radiation, and the resonances were observed experimentally [23, 24]. Furthermore, in cosmology we encounter models in which our universe is the result of continuous collision of branes and nucleation and therefore splitting of branes is a fundamental scenario in these models [25–28].Therefore, the universal aspects of thick brane splitting in warped bulk are important. Such branes are constructed from a complexϕ4 scalar field potential [29] or from a real ϕ6 scalar field potential [30]. These branes can be constructed from deformation of ϕ4 scalar field potential as well [31–33].Recently, Dutra and coworkers proposed a new model of thick brane in which multibrane scenario arises from scalar field models generating usual kink solutions [34]. It suggests a special type of brane splitting. In this method superpotential function and warp factor will decompose in a special form and field equations will be simplified significantly. In this work we deal with this thick brane model which arises from polynomial superpotential.The localization of spin1/2 fermions on the brane is very interesting and important. Usually, in five dimensions fermion does not have a normalizable zero mode without scalar-fermion coupling [35–43]. In five dimensions, with a Yukawa scalar-fermion coupling, there may exist a massless bound state and a continuous gapless spectrum of massive Kaluza-Klein (KK) states [39], while, in some of other brane models, there exist some discrete KK states and a continuous gapless mass spectrum [44].This paper is organized as follows. In the next section, we present the brane model that is constructed in [34]. In Section 3 we investigate localization of the zero mode of the fermion field on the brane which is derived from a polynomial potential. In Section 4 we study localization of massive fermionic modes. Finally, in the last section we present our conclusions.


## 2. The Model

We consider the following action in which two scalar fields are coupled to gravity in 5 dimensions:(1)S=∫d4xdr-g14R-12∂Mϕ1∂Mϕ1-12∂Mϕ2∂Mϕ2-Vϕ1,ϕ2,where g=det⁡(gMN), M,N=0,1,2,3,5. The coordinates on the brane are represented by xμ (μ=0,1,2,3) while the coordinate in the bulk is shown by x5=r. The line element is written as(2)ds2=gMNdxMdxN=e2Arημνdxμdxν+dr2,where ημν is the usual Minkowski metric with diag⁡(-,+,+,+) and e2A(r) is called the warp factor. For this brane-world scenario, the equations of motion are obtained as(3)ϕi′′+4A′ϕi′=dVdϕi,i=1,2,A′′+23ϕ1′2+ϕ2′2=0,A′2+13Vϕ1,ϕ2=16ϕ1′2+ϕ2′2,and the potential V(ϕ1,ϕ2) can be written in terms of a superpotential W(ϕ1,ϕ2) as(4)Vϕ1,ϕ2=12∑i=12∂W∂ϕi2-43W2.Therefore, equations of motion can be reduced to the following first-order equations:(5)dϕidr=∂W∂ϕi,dAdr=-23W,i=1,2.But(6)Wϕ1,ϕ2=W1ϕ1+W2ϕ2,Ar=A1r+A2r.The first-order equations in (5) are converted to(7)dϕidr=∂Wiϕi∂ϕi,dAidr=-23Wiϕi,i=1,2,for polynomial superpotential(8)Wiϕi=λiϕi-ϕi33,i=1,2;ϕi and Ai are given by(9)ϕi=tanh⁡λir-ri,i=1,2,Ar=19∑i=12tanh2⁡λir~-ri-tanh2⁡λir-ri-49ln⁡∏i=12sech⁡λir~-risech⁡λir-ri,where ri is an integration constant, representing the center of the kink. For solitonic solutions we consider λi>0 and r~ is defined as the average value of coordinates of center of the kinks(10)r~=12r1+r2.In order for this model to support brane splitting mechanism, we consider two symmetric kink solutions. Therefore, λ1=λ2=λ and r1=-r2=a. So r~=0 and we can write (9) as(11)ϕ1=tanh⁡λr-a,ϕ2=tanh⁡λr+a,Ar=192tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a-49ln⁡cosh⁡λr-acosh⁡λr+acosh2⁡λa.In Figure1 we show the warp factor for different values of a. For a=0 we have a single brane that warp factor has a sharp peak. For a>0 a plateau is formed in the interior of the brane where the energy density vanishes. This is attributed to the presence of a new phase inside the brane. The plateau region in the brane grows when a increases.Figure 1
Plots of warp factore2A(r) for a=0 (solid line), a=3.0 (dashed line), and a=5.0 (dotted line) and we put λ=1.0.In the next section, we investigate localization of zero mode of the fermion on the brane.


## 3. Localization of Zero Mode

In order to get the mass-independent potential for KK modes of fermion, we perform the following coordinate transformation:(12)dz=e-Adr;by this transformation we get a conformally flat metric(13)ds2=e2Aημνdxμdxν+dz2.In five dimensions, fermions are four component spinors and their Dirac structure can be described by ΓM=eMM¯ΓM¯ with eMM¯ being the vielbein and {ΓM,ΓN}=2gMN. ΓM¯ are the gamma matrices in 5D flat space-time. The vielbein is given by(14)eMM¯=eAe^μν¯00eA,ΓM=e-A(e^μν¯γν¯,γ5)=e-A(γμ,γ5), where γμ=e^μν¯γν¯, γν¯ and γ5 are the usual flat gamma matrices in the 4D Dirac representation. The action of a spin 1/2 fermion coupled to the background scalar fields and gravity in five dimensions is(15)Sf=∫d5x-gΨ¯ΓMDMΨ-ηΨ¯Fϕ1,ϕ2Ψ,where the sign of coupling constant η of the spinor Ψ to scalars is arbitrary and represents a coupling to either kink or antikink domain wall. For definiteness, we shall consider only the case of a kink coupling and thus assume that η>0. F(ϕ1,ϕ2) is a function of ϕ1 and ϕ2 determining the form of coupling between the fermion and scalars. The covariant derivative is defined as DM=∂M+ωM with the spin connection ωM defined as ωM=1/4ωMM¯N¯ΓM¯ΓN¯ and ωMM¯N¯ is(16)ωMM¯N¯=12eNM-∂MeNN--∂NeMN--12eNN-∂MeNM--∂NeMM--12ePM-eQN-∂PeQR--∂QePR-eMR-.The nonvanishing components of ωM are(17)ωμ=12A′γμγ5+ω^μ,where the prime denotes the derivative with respect to z and ω^μ=1/4ω¯μμ¯ν¯Γμ¯Γν¯ is the spin connection derived from the metric g^μν(x)=e^μμ¯(x)e^νν¯(x)ημ¯ν¯. Here we neglect the back-reaction from the fermion on the brane solution and the scalar field is considered to be unchanged. Thus, by using the metric (13), the equation of motion for fermion is(18)γμ∂μ+ω^μ+γ5∂z+2∂zA-ηeAFϕ1,ϕ2Ψ=0,where γμ(∂μ+ω^μ) is the Dirac operator on the brane.For solving this equation, we separate variables with KK and chiral decomposition [44, 45](19)Ψx,z=e-2A∑nψLnxfLnz+ψRnxfRnz.The 4D left-handed and right-handed fermions satisfy the Dirac equations(20)γμ∂μ+ω^μψLn=mnψRn,γμ∂μ+ω^μψRn=mnψLn,while the KK modes satisfy(21)∂z+2∂zA+ηeAFϕ1,ϕ2fLnz=mnfRnz,∂z+2∂zA-ηeAFϕ1,ϕ2fRnz=-mnfLnz,with the following orthonormality condition(22)∫-∞∞dzfLnfLm=∫-∞∞dzfRnfRm=δnm,∫-∞∞dzfLnfRm=0,and the Schrödinger-like equations are obtained:(23)-∂z2+VLzfLn=mn2fLn,(24)-∂z2+VRzfRn=mn2fRn,where the effective potentials are given by(25)VLz=η2e2AF2ϕ1,ϕ2-η∂zeAFϕ1,ϕ2,VRz=η2e2AF2ϕ1,ϕ2+η∂zeAFϕ1,ϕ2.Because of the complexity of A(r), we can not use relation (12) to obtain analytical form of z(r). Therefore we apply numerical method to get pair of (r,z):(26)∂zA=eAr∂rA,∂zF=eAr∂rF.By these relations, we can rewrite the potentials as a function of r:(27)VLzr=ηe2AηF2-∂rF-F∂rAr,VRzr=ηe2AηF2+∂rF+F∂rAr.By substituting mn=0 in (21), the left-handed and right-handed zero modes of fermions are obtained as(28)fL0∝exp⁡-η∫0zdz′eAz′Fϕ1,ϕ2,fR0∝exp⁡η∫0zdz′eAz′Fϕ1,ϕ2. The normalization condition for localization of zero mode of left-handed fermions on the brane is(29)∫-∞∞dzexp⁡-2η∫0zdz′e-Az′Fϕ1z′,ϕ2z′<∞,and in r coordinate, we have(30)∫-∞∞drexp⁡-Ar-2η∫0rdr′Fϕ1r′,ϕ2r′<∞.For right-handed fermion this condition is achieved by replacing η→-η. For studying localization of zero mode of fermion on the brane, we should determine the suitable form of F(ϕ1,ϕ2). The procedure for determining F(ϕ1,ϕ2) in this paper is similar to the methods used in [46, 47]. In the following subsections we try to determine this function by using the normalization condition.


### 3.1.F=ϕ1ϕ2

The integrand in (30) can be expressed as(31)I0≡exp⁡-192tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a-2ηr×cosh⁡λr-acosh⁡λr+acosh2⁡λa4/91+e-2λr-a1+e-2λr+a-2η/λ.Since η>0, when r→-∞ we have(32)I0⟶exp⁡-89-2ηr⟶∞;therefore, the integral in (30) is divergent. So the left-handed fermion zero mode can not be localized on the brane. On the other hand, with η→-η, we can see that the zero mode of right-handed fermions can not be localized on the brane too.


### 3.2.F=ϕ1-ϕ2

The integrand in (30) is written as(33)I1≡exp⁡-192tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa4/9cosh⁡λr-acosh⁡λr+a-2η/λ,from which we have(34)I1⟶exp⁡±89r⟶∞,whenr⟶±∞.Hence, left-handed zero mode can not be localized on the brane in this case. Moreover, because I1 is independent of η, therefore, we can conclude that right-handed fermion can not be localized on the brane.


### 3.3.F=ϕ1+ϕ2

The integrand in (30) is expressed as(35)I2≡exp⁡-192tanh2λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa4/9-2η/λ;therefore, we have(36)I2⟶exp⁡±89r-4ηrλ,whenr⟶±∞.So the normalization condition for localization of left-handed fermion zero mode is(37)η>2λ9.By changing η→-η one can find the zero mode of the right-handed fermion; however this mode can not be localized on the brane.The effective potentials for left-handed and right-handed KK fermion have the form(38)VL=ηexp⁡292tanh2⁡λa-tanh2λr-a-tanh2λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa-8/9ηtanh⁡λr-a+tanh⁡λr+a2-λsech2⁡λr-a+sech2⁡λr+a+23tanh⁡λr-a+tanh⁡λr+a×tanh⁡λr-a+tanh⁡λr+a-tanh3⁡λr-a+tanh3⁡λr+a3,VR=VLη→-η.The values of potential in r or z=0 and r or z→±∞ are(39)VL0=-VR0=-2ηλsech2⁡λa,VL±∞=VR±∞=0.It can be seen that the asymptotic behaviors of two potentials are the same when r→±∞, but opposite at the origin, z=0. This reveals that only one of the massless left or right chiral fermions can be localized on the brane. The shapes of effective potentials are shown in Figure 2. The form of VL(z) is volcano type and therefore, there is no mass gap between the zero mode and KK excitation modes. On the other hand, VR(z) is always positive at the brane location. We know that this type of potential can not trap any bound state of right-handed fermion and there is no zero mode of right-handed fermion. This is consistent with our pervious knowledge that only one chirality of massless fermion can exist.Figure 2
The shapes of potentials: (a)VL(z) and (b) VR(z), a=0 (solid line), a=3 (dashed line), and a=5 (dotted line).
(a)
(b)The zero mode of left-handed fermion is written as(40)fL0z∝exp⁡-η∫0zdz′eAz′ϕ1z′+ϕ2z′=exp⁡-η∫0rdr′ϕ1r′+ϕ2r′=cosh⁡λr-acosh⁡λr+a-2η/λ.Figure 3 shows the form of fermion zero modes on the brane. One can see that the width of the function is increased with splitting.Figure 3
Fermion zero modes localized on the brane:a=1 (solid line), a=3 (dashed line), and a=5 (dotted line).


### 3.4. The CaseF=ϕ1+βϕ2

The integrand in (30) is expressed as(41)I3≡exp⁡-192tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λa4/9-2η/λcosh⁡λr+acosh⁡λa4/9-β2η/λ.Therefore, we have(42)I3⟶exp±89r-2β+1ηλr,whenr⟶±∞.So the normalization condition becomes(43)η>4λ9β+1,β≠-1.For β=-1 the relation (42) is reduced to (34). Hence, left-handed zero mode can not be localized on the brane. For β=0 we have η>4λ/9. This means that coupling of fermion to every subbrane can localize zero mode on the brane. For the case β=1 the normalization condition is reduced to (37). The zero mode of left-handed fermion is turned out to be(44)fL0z∝exp⁡-η∫0rdr′ϕ1r′+βϕ2r′=coshλr-a-2η/λcosh⁡λr+a-2βη/λ.The explicit forms of the potentials are(45)VL=ηexp⁡292tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa-8/9ηtanh⁡λr-a+βtanh⁡λr+a2-λsech2⁡λr-a+β2sech2⁡λr+a+23tanh⁡λr-a+βtanh⁡λr+a×tanh⁡λr-a+tanh⁡λr+a-tanh3⁡λr-a+tanh3⁡λr+a3,VR=VLη→-η.The asymptotic behaviors of the potentials are(46)VL0=ηηβ-12-ηβ-12+λβ2+1sech2⁡λa,VR0=ηηβ-12-ηβ-12-λβ2+1sech2⁡λa,VL±∞=VR±∞=0.It is realized that VL(0) can be negative or positive, while, for η>0 and λ>0, VR(0) is always greater than VL(0). This means that localization of zero mode of the left-handed fermion is more supported.


## 3.1.F=ϕ1ϕ2

The integrand in (30) can be expressed as(31)I0≡exp⁡-192tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a-2ηr×cosh⁡λr-acosh⁡λr+acosh2⁡λa4/91+e-2λr-a1+e-2λr+a-2η/λ.Since η>0, when r→-∞ we have(32)I0⟶exp⁡-89-2ηr⟶∞;therefore, the integral in (30) is divergent. So the left-handed fermion zero mode can not be localized on the brane. On the other hand, with η→-η, we can see that the zero mode of right-handed fermions can not be localized on the brane too.


## 3.2.F=ϕ1-ϕ2

The integrand in (30) is written as(33)I1≡exp⁡-192tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa4/9cosh⁡λr-acosh⁡λr+a-2η/λ,from which we have(34)I1⟶exp⁡±89r⟶∞,whenr⟶±∞.Hence, left-handed zero mode can not be localized on the brane in this case. Moreover, because I1 is independent of η, therefore, we can conclude that right-handed fermion can not be localized on the brane.


## 3.3.F=ϕ1+ϕ2

The integrand in (30) is expressed as(35)I2≡exp⁡-192tanh2λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa4/9-2η/λ;therefore, we have(36)I2⟶exp⁡±89r-4ηrλ,whenr⟶±∞.So the normalization condition for localization of left-handed fermion zero mode is(37)η>2λ9.By changing η→-η one can find the zero mode of the right-handed fermion; however this mode can not be localized on the brane.The effective potentials for left-handed and right-handed KK fermion have the form(38)VL=ηexp⁡292tanh2⁡λa-tanh2λr-a-tanh2λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa-8/9ηtanh⁡λr-a+tanh⁡λr+a2-λsech2⁡λr-a+sech2⁡λr+a+23tanh⁡λr-a+tanh⁡λr+a×tanh⁡λr-a+tanh⁡λr+a-tanh3⁡λr-a+tanh3⁡λr+a3,VR=VLη→-η.The values of potential in r or z=0 and r or z→±∞ are(39)VL0=-VR0=-2ηλsech2⁡λa,VL±∞=VR±∞=0.It can be seen that the asymptotic behaviors of two potentials are the same when r→±∞, but opposite at the origin, z=0. This reveals that only one of the massless left or right chiral fermions can be localized on the brane. The shapes of effective potentials are shown in Figure 2. The form of VL(z) is volcano type and therefore, there is no mass gap between the zero mode and KK excitation modes. On the other hand, VR(z) is always positive at the brane location. We know that this type of potential can not trap any bound state of right-handed fermion and there is no zero mode of right-handed fermion. This is consistent with our pervious knowledge that only one chirality of massless fermion can exist.Figure 2
The shapes of potentials: (a)VL(z) and (b) VR(z), a=0 (solid line), a=3 (dashed line), and a=5 (dotted line).
(a)
(b)The zero mode of left-handed fermion is written as(40)fL0z∝exp⁡-η∫0zdz′eAz′ϕ1z′+ϕ2z′=exp⁡-η∫0rdr′ϕ1r′+ϕ2r′=cosh⁡λr-acosh⁡λr+a-2η/λ.Figure 3 shows the form of fermion zero modes on the brane. One can see that the width of the function is increased with splitting.Figure 3
Fermion zero modes localized on the brane:a=1 (solid line), a=3 (dashed line), and a=5 (dotted line).


## 3.4. The CaseF=ϕ1+βϕ2

The integrand in (30) is expressed as(41)I3≡exp⁡-192tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λa4/9-2η/λcosh⁡λr+acosh⁡λa4/9-β2η/λ.Therefore, we have(42)I3⟶exp±89r-2β+1ηλr,whenr⟶±∞.So the normalization condition becomes(43)η>4λ9β+1,β≠-1.For β=-1 the relation (42) is reduced to (34). Hence, left-handed zero mode can not be localized on the brane. For β=0 we have η>4λ/9. This means that coupling of fermion to every subbrane can localize zero mode on the brane. For the case β=1 the normalization condition is reduced to (37). The zero mode of left-handed fermion is turned out to be(44)fL0z∝exp⁡-η∫0rdr′ϕ1r′+βϕ2r′=coshλr-a-2η/λcosh⁡λr+a-2βη/λ.The explicit forms of the potentials are(45)VL=ηexp⁡292tanh2⁡λa-tanh2⁡λr-a-tanh2⁡λr+a×cosh⁡λr-acosh⁡λr+acosh2⁡λa-8/9ηtanh⁡λr-a+βtanh⁡λr+a2-λsech2⁡λr-a+β2sech2⁡λr+a+23tanh⁡λr-a+βtanh⁡λr+a×tanh⁡λr-a+tanh⁡λr+a-tanh3⁡λr-a+tanh3⁡λr+a3,VR=VLη→-η.The asymptotic behaviors of the potentials are(46)VL0=ηηβ-12-ηβ-12+λβ2+1sech2⁡λa,VR0=ηηβ-12-ηβ-12-λβ2+1sech2⁡λa,VL±∞=VR±∞=0.It is realized that VL(0) can be negative or positive, while, for η>0 and λ>0, VR(0) is always greater than VL(0). This means that localization of zero mode of the left-handed fermion is more supported.


## 4. Resonances of Massive Modes

We can rewrite (23) and (24) as(47)Q†QfLn=-∂z+ηFeA∂z+ηFeAfLn=m2fLn,QQ†fRn=∂z+ηFeA-∂z+ηFeAfRn=m2fRn. From these equations, we can see that the tachyonic modes in spectrum are excluded. With converting the equations of motion for fermion to Schrödinger-like equations, we can present a quantum mechanical interpretation for fLn and fRn. By studying resonant modes we are able to obtain information about the coupling between massive modes and the brane.In order to derive KK modes from (23) and (24) we apply relative probability method [48–50]. Since (23) and (24) are Schrödinger-like, we can interpret normalized |fL,R(z)|2 as the probability of finding massive KK modes on the brane. But the massive modes can not be normalized because they are oscillating when far away from brane along extra dimension. Therefore, the relative probability function is defined as [48] (48)PLn,Rnm=∫-zbzbdzfLn,Rnz2∫-zmaxzmaxdzfLn,Rnz2,where 2zb is brane thickness approximately and zmax=10zb here. fLn,Rn(z) is solution of (23) or (24) with two boundary conditions:(49)fLn,Rn0=1,fLn,Rn′0=0,for even parity, and(50)fLn,Rn0=0,fLn,Rn′0=1,for odd parity. If m2≫VL,Rmax, fL,R will be approximately a plane wave with PL,R(m)=zb/zmax=0.1. For simplicity we have chosen β=1. Figures 4 and 5 show plots of the relative probability for different values of a and η for left-handed and right-handed fermions, respectively. The masses of resonances are presented in Table 1. The fermion resonant wave functions for η=1, λ=1, and a=5.0 are shown in Figure 6.Table 1
Masses of resonances for different values ofη and a.
(a)
Left
η
=
0.5
η
=
1.0
η
=
2.0
a
=
1.0
Absent
1.185465
1.719258
a
=
3.0
0.453441
0.562819;0.994057
0.689911;1.232968;1.664278;2.034932
a
=
5.0
0.288833;0.567862
0.337121;0.646626;0.924378;1.198549
0.377447;0.731511;1.058468;1.359395;1.631559;1.884065;2.146727
(b)
Right
η
=
0.5
η
=
1.0
η
=
2.0
a
=
1.0
Absent
1.110830
1.711422
a
=
3.0
0.442279
0.562819;0.994057
0.689911;1.232968;1.662400;2.027853
a
=
5.0
0.283917;0.561504
0.337121;0.645689;0.921857;1.190822
0.377447; 0.731480; 1.058468; 1.359395;1.631559; 1.882145; 2.139101Figure 4
Plots of relative probability for left-handed fermion resonances witha=1.0 (left column), a=3.0 (middle column), and a=5.0 (right column). Coupling constants are η=0.5 (first raw), η=1.0 (second raw), and η=2.0 (third raw). Solid lines and dashed lines correspond to even and odd parity, respectively. In all cases λ=1.0.Figure 5
Plots of relative probability for right-handed fermion resonances witha=1.0 (left column), a=3.0 (middle column), and a=5.0 (right column). Coupling constants are η=0.5 (first raw), η=1.0 (second raw), and η=2.0 (third raw). Solid lines and dashed lines correspond to even and odd parity, respectively. In all cases λ=1.0.Figure 6
Plots of the left-handed and right-handed fermion resonances withη=1, λ=1, and a=5.From the figures, we can see that the spectra of massive KK modes of left-handed and right-handed fermions are almost the same which reveals that a Dirac fermion is composed from left-handed and right-handed bound KK modes. Furthermore for fixed value ofa the larger the value of coupling parameter, the larger the number of resonances. Also for fixed value of η, the larger value of a leads to larger number of peaks. The first peak is the most narrow and the resonances will become broader with increasing m. This means that first resonance has larger lifetime and the lifetime declines with increasing the mass of peak.We can also see from figures that there are successively even and odd parity wave functions for left and right chiral modes with the same values ofm2. In other words, the zero mode is beginning of left chirality with even parity spectra; therefore the two first resonances (if exist) with the same m2 are odd parity left chiral mode and even parity right chiral mode. Next the two second resonances (if exist) with the same m2 are even parity left chiral mode and odd parity right chiral mode.From Figures4 and 5 we can estimate the lifetime of every resonance as the reciprocal of width of peak at half of the height. With this definition we can find visually that heavier resonances are broader and they have shorter lifetimes. For evaluating this fact, we determine approximately the lifetimes of resonances of right-handed fermions for λ=1.0, η=1.0, and a=5.0 for instance. From Figure 5 we can see that in this case we have 4 resonances. The lifetimes of resonances are τ1=242.1894 for m1=0.327121, τ2=41.3873 for m2=0.645689, τ3=12.4500 for m3=0.924378, and τ4=4.5196 for m4=1.190822. Therefore we can see that the lighter resonances have longer lifetimes.


## 5. Conclusions

In this paper we have studied the issue of the localization of fermion field on the double wall brane. This brane includes two scalar fields coupled minimally to brane. For observing whether the zero mode can be localized on this brane or not, we use a Yukawa coupling between fermion and background scalar fields. By investigating normalization condition, we can see that fermion couples with summation of kinks. We can see that by this coupling the zero mode of left-handed fermion can be localized on the brane under the special condition. This condition provides relation between coupling constant andλ parameter. Also we found that the zero mode of right-handed fermion can not be localized on the brane.The massive mode resonances were investigated numerically. From the volcano shape of effective potential for left-handed fermions, it results that the spectrum is continuous and there is no gap between zero mode and KK excitation modes. Also larger values of coupling constant and the distance of subbranes support more resonances in the spectrum. In one spectrum, heavier resonances have broader peaks rather than lighter ones. This means the lighter fermions couple stronger to the brane rather than heavier KK modes. Because of very narrow peak, we may not see light resonances. Fortunately because of supersymmetric feature of Schrödinger-like equation, we have successively even and odd parity for left or right chirality modes with the same mass in the spectrum. This helps us to search a small region when a resonance peak is not seen. Therefore numeric procedure becomes simple and fast. The lifetime of a resonance is proportional to inverse of peak width at half maximum. Hence generally, light resonances have longer lifetimes.



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*Source: 3926286-2017-03-26.xml*