# The Extension of the Massless Fermion Behaviour in the Cosmic String Spacetime

Authors: Bengü Çağatay; Özlem Yeşiltaş; Anıl L. Aygün
Journal: Advances in High Energy Physics
 (2020)

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## Abstract

In this work, we have obtained the solutions of a massless fermion which is under the external magnetic field around a cosmic string for specific three potential models using supersymmetric quantum mechanics. The constant magnetic field, energy-dependent potentials, and position-dependent mass models are investigated for the Dirac Hamiltonians, and an extension of these three potential models and their solutions is also obtained. The energy spectrum and potential graphs for each case are discussed for theα deficit angle.


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## Body

## 1. Introduction

The massless Dirac character of the low energy electrons moving has attracted much interest in physics due to the graphene’s important electronic properties [1]. There are a series of studies on the interaction of graphene electrons in perpendicular magnetic fields which have been carried out in order to find a way for confining the charges [2, 3]. Graphene and its derivatives (nanotubes, fullerenes) have become a mine of novel technologies and studying various aspects of physics through its many subfields as well as cosmological models on honeycomb branes [4]. Moreover, topological defects which are disorders lead to important effects on the electronic properties of low dimensional systems. A cosmic string and topological defect relationship is studied where the topological defects are located at arbitrary positions on the graphene plane [5]. In line with the efforts to combine general relativity theory with quantum mechanics, these studies are of great interest [6–8]. Cosmic strings were introduced by Kibble in 1976 [9], and the geometry of the massless cosmic strings is examined in the large scale limit of the model given in [10]. In relativistic quantum mechanics, Lie algebraic approaches [11], a hydrogen atom in the background of an infinitely thin cosmic string [12], scalar particle dynamics in gravity’s rainbow through the space-time of a cosmic string [13], and N=2 supersymmetric approach to cosmic string dynamics [14] are some recent studies about the topic. For more details about the technique of SUSY, the readers may look at the book [15]. The origin of supersymmetric (SUSY) quantum mechanics is based on the very early development of quantum mechanics by Dirac who found a method to factorize the harmonic oscillator and Schrödinger who noted symmetries in the solutions to his equation [16]. After then, the context of SUSY QM was first studied by Witten [17] and Cooper and Freedman [18]. This theory, which still attracts much attention today, has also application in optics [19] and biophysics [20]; it has a very important place in both relativistic [21] and nonrelativistic quantum mechanics [22, 23]. This work points the dynamics of a massless fermion out when it is under the influence of a magnetic field. Using the fundamental aspects of SUSY quantum mechanics, three different potential models are discussed for the different values of a deficit angle. Moreover, supersymmetric extension of these models provides generated unknown and complex potential models. The paper is organized as follows: low dimensional Dirac equation is written in the curved spacetime using cosmic string line element and aspects of SUSY quantum mechanics given in Section 2. Section 3 is devoted to all potential models which are radial, hyperbolic, and nonlinear ones, and their solutions are given. Their energy and potential function graphs are shown. Section 4 includes the extended new quantum mechanical potential models and their solutions. Energy and potential graphs are also shown. We conclude our results in Section 5.


## 2. Dirac Hamiltonian for a Cosmic String in the Gravitational Background

A massless fermion dynamics can be represented by the Dirac equation-Weyl equation. Specially, we assume that the particle occurs in the external electromagnetic field in a cosmic string spacetime. Then, the particle is described by(1)iσaeaμ∇μ+ieAμψ=0.Here,σa are the Pauli matrices, and ∇μ=∂μ+Γμ is the covariant derivative. The cosmic string spacetime is represented by the line element which is
(2)ds2=−dt2+dr2+α2r2dϕ2+dz2,where −∞<t,z<∞, r≥0, and 0≤ϕ≤2π. The parameter α=1−4m~/c2 is the angular deficit changing in the interval 0,1, where m~ is the linear mass density of the cosmic string. Here, eμa are called as the tetrad fields which connect the Riemannian metric tensor to the flat spacetime metric tensor as
(3)gμν=eaμebνηab.The tetrads and spinor connections are given, respectively, as(4)eμax=10000cosϕ−αrsinϕ00sinϕαrcosϕ00001,(5)Γμ=i4ωμabΣab,where
(6)Σab=i2γa,γb,where γa is the standard Dirac matrices defined in the Minkowski spacetime:
(7)γ0=β=100−1,(8)γi=βαi=0σi−σi0,(9)Σi=σi00σi.We use the unitsℏ=c=1; the line element of the stationary cosmic string spacetime is written as
(10)ds2=−dt2+dr2+α2r2dϕ2+dz2,with −∞<t,z<∞ and r≥0 and 0≤ϕ≤2π. Here, the parameter α is the angular deficit changing in the interval 0,1. Considering the components of the metric tensor gμν,
(11)g11=−1,(12)g22=1(13)g33=α2r2(14)g33=1,and the Christoffel symbols
(15)Γijμ=12gμk∂gik∂qj+∂gjk∂qi−∂gij∂qk,the spin connection components can be calculated using the tetrad components and the Christoffel symbols,
(16)ωμba=ηaceνcebσΓσμν−ηacebν∂μeνc,which leads to
(17)ωϕab=0000001−α00−1−α000000.Substituting (17), (7), (11), and (4) in (1), we can obtain [24]
(18)iγ0∂Ψ∂t+iγ1cosϕ∂∂r−sinϕαr∂∂ϕ+isinϕαrωϕ212γ3−isinϕαreAϕΨ+iγ2sinϕ∂∂r+cosϕαr∂∂ϕ−icosϕαrωϕ212γ3+icosϕαreAϕΨ+iγ3∂Ψ∂z=0.Let us use the following form for the spinor as(19)ψ=exp−iEt+imϕ+ikzφ1rφ2r.Putting (19) in (18) gives
(20)E−kie−iϕ∂∂r+mαr+1−α2αr+eαAϕrrieiϕ∂∂r−mαr+1−α2αr−eαAϕrrE−kφ1rφ2r=0.Then, we get a couple of differential equations:(21)d2dr2−α−1αrddr+E2−k2−1+2m−3α−1+2m+α4α2r2+eα−2mAϕrr2α2−e2Aϕ2rr2α2−eAϕ′rαrφ1r=0,(22)d2dr2−α−1αrddr+E2−k2−1+2m−α1−2m−3α4α2r2−eα+2mAϕrr2α2−e2Aϕ2rr2α2+eA′ϕrαrφ2=0.Next, we transform the system given above into the form(23)−d2dr2+Wr2+W′rχ1r=εχ1r,(24)−d2dr2+Wr2−W′rχ2r=εχ2r,where ε=E2−k2 and
(25)χ1,2r=r1/21/α−1φ1,2r,Wr=eAϕr+mαrare used. Here, (21) are transformed into (23) with the same energy which also shows that the system is supersymmetric. Let us call two effective Hamiltonians for (23) as H1 and H2, respectively:
(26)H1=−d2dr2+V1r=L−L+,H2=−d2dr2+V2r=L+L−,and the intertwining operators are defined as
(27)L±=∓ddr+Wr.It is noted that (27) can be used to intertwine the system as [25]
(28)H1L−=L−H2,H2L+=L+H1.Furthermore, one can observe that(29)L−χ2=εχ1,L+χ1=εχ2.Let us discuss the exactly solvable potential models forH1 and H2 in the next section.


## 3. Potential Models

Now we can arrange some differentAϕr vector potential models which give rise to effective Hamiltonians. We can argue that the discrete spectrum of the Hamiltonian H2 is ε2,n and L−χ2,0=0; then [26],
(30)ε1,n−1=ε2,n,χ1,n−1r=1ε2,nL−χ2,nr,n=1,2,⋯.


### 3.1. Constant Magnetic Field

The constant magnetic field vectorB→=0,0,a0, where a0 is a real parameter, is perpendicular to the plane. Then, vector potential component Aϕr can be taken as Aϕr=a0r, and Wr becomes
(31)Wr=ea0α+mαr.This superpotential is introduced in [15] where the singular potential models are also discussed. Here, we have considered the one-dimensional system. Hence, V1,2r functions become
(32)V1r=a02e2α2+mm−αα2r2+2a0emα2r,(33)V2r=a02e2α2+mm+αα2r2+2a0emα2r.This system shows thata0 should be negative because of the Coulomb’s potential; let a0 be a0=−ν,ν>0. The solutions of the system (32) which is known as pseudoharmonic potential are already known [27]:
(34)En=±k2−ν2e2α2−a02e2m2α4n+1/2+mm−α/α2+1/42,n=0,1,2,⋯,(35)X1,n=Nnr1/2+μexp−∈rF11−n,2μ+1,2∈r,where μ=mm−α/α2+1/4, ε=ν2e2/α2−E, and F11−n,2μ+1,2∈r are the confluent hypergeometric functions. Let us look at the energy spectrum and the conditions for getting real energy levels. The inside of the square root in (34) must be positive for the real energy spectrum, and
(36)∣k∣>ν2e2α2+a02e2m2α4n+1/2+mm−α/α2+1/42,m>1.The normalization constant can be written as [27]
(37)Nn=Γ2μ+1n!2n+2μ+1Γn+2μ+1.Let us see how the energy and potential functions change withα.As is shown from Figure1, we can obtain physical potentials for the values of α which takes 0<α<1. Energy graph in Figure 2 shows that for the values of α≥0.1831 (green curve), we can get an increasing energy graph. For the greater values of the number n, the lower bound for the α increases.Figure 1
The graph of (32) and (40). V1r graph versus the position for the different values α. α=0.5 for orange, α=0.5 for blue, α=0.8 for green curves, and the red dashed curve stands for the vector potential.Figure 2
The graph of energy eigenvalues with respect toα (34). The green curve is drawn for k=10, n=1, and a0=2, and the yellow curve is shown for k=8, n=2, and a0=0.2.


### 3.2. Potential Models: Energy-Dependent Vector and Scalar Potentials

The energy-dependent potentials are one of the some modified problems of the quantum mechanics [28]. The condition on the density distribution is [28]
(38)ρnEn,r=ψnEn,x21−∂VEn,r∂En.As it is seen from the above equation, a positivity condition is(39)1−∂VEn,r∂En>0.The details of the related energy-dependent potentials in supersymmetry can be found in [29]. In this case, we take the vector potential Ar as a complex function
(40)Ar=C1+Srr−me,where C1 is a constant and Sr is the complex function. Then, (23) becomes
(41)χ1′′r+E2−k2−C12e2α2−2C1e2Srα2−e2Sr2α2−eαS′rχ1r=0,(42)χ2′′r+E2−k2−C12e2α2−2C1e2Srα2−e2Sr2α2+eαS′rχ2r=0,where we can give the superpotential Wr as
(43)Wr=C1eα+eSrαand let Sr be
(44)Sr=A1sechr+B1tanhr−C1,where A1,B1,C1 are the constants. We can get the partner potentials as
(45)V1r=B12e2α2+eA12e+B1−B1e+αα2sech2r+A1e2B1e−αα2sechrtanhr,(46)V2r=B12e2α2+eA12e−B1B1e+αα2sech2r+A1e2B1e+αα2sechrtanhr.Here, (45) is known as Scarf II potentials (see [30]). Comparing our system (41), (43), and (44) with the linear energy-dependence results in [31] can give a solvable model. We can use the parameter A1 as energy-dependent potential in our calculations as
(47)A1=α2ei−2E2−B12e2α2.Then, we match our system with the one where linear energy dependency can be found in page 9 in [31] and write the partner potentials as energy-dependent potentials as
(48)V1r=B12e2α2+E2+B1eα−14−2B12e2α2−iE2−B12e2α2sech2r+α−2B1e2α−i+2E2−B12e2α2sechrtanhr,(49)V2r=B12e2α2+E2−B1eα−14−2B12e2α2−iE2−B12e2α2sech2r+α+2B1e2αi−2E2−B12e2α2sechrtanhr.Then, one can find the energy eigenvalues as(50)En=±k2−n+B1eα2+B12e2α2,n=0,1,2,⋯.The condition for the real energies is(51)∣k∣>n+B1eα2−B12e2α2.And the solutions become(52)χ1,n=1−isinhr1/4−1/2n+1+k2−β2+4iβ1+isinhr1/4−1/2n+1+k2−β2−4iβPn−1/22+2n+2k2−β2+2iβ2,−1/2−2n−2k2−β2+2iβ2isinhr.In Figure3, we have obtained an effective potential graph for the real potential while we can get a potential barrier from the imaginary part. Figure 4 shows the component of the vector potential whose real component is corresponding to another effective potential curve. In Figure 5, we can see that for the greater values of n, the lowest bound for the α again increases as energy increases.Figure 3
The real and imaginary parts of the potential in (48). Green and blue curves are corresponding to the real and imaginary parts, respectively, where α=0.2.Figure 4
The real (green) and imaginary (blue) parts of the vector potential in (40).Figure 5
Energy eigenvalues versus theα values for (50).


### 3.3. Potential Models: Position-Dependent Mass Model

The idea of a position-dependent-effective-mass (PDEM) quantum Hamiltonians gains more attention because of the physical applications in graded crystals, quantum dots, and liquid crystals. The hermiticity and the uniqueness of the problem are studied in the region of mathematical physics. For more details, see [32–34]. Let us make a point transformation in (41) given by
(53)r=∫zdtft.Then, we can get
(54)fz2χ1′′z+fzf′zχ1′z+E2−k2−C12e2α2−2C1e2Szα2−e2Sz2α2+eαfzS′zχ1z=0.Let us choosefz=1+λz2 and Sz=iC1/1+λz2−c1; then, (54) turns into
(55)1+λz2χz′′+λzχz′+E2−k2+C12e2α21+λz2+iC1eλzα1+λz2χz=0.The solutions of the type of equations like (55) are already known. We can write the complete solutions as [35]
(56)En=±k2+nλ2C1e−αλα−nλ,(57)χnz=Nnin1+λz2−s/2e−rtan−1zλPnir−s−1/2,−ir−s−1/2izλ,n=0,1,2,⋯<s,where
(58)Nn=λn!s−nΓs−ir−n+1/2Γs+ir−n+1/2π2−2sΓ2s−n+1.s=2C1e−αλ/2α3/2,r=−i/2, and Pna,by are the corresponding Jacobi polynomials. Now, let us examine the SUSY of this model. First, we need to make a point transformation to (55) which is χ1z=1/1+4λ2z21/4Yz. We get
(59)Y′′z+14z2+3−4iAz+4A2z24z21+z2λ2+−1+iAz+z2ε2z21+z2λYz=0,where A=C1e/α. We propose a super potential which is given by
(60)Wz=C2z21+z2λ+C3.Then, for (59), the unknown parameters of the superpotential can be obtained as
(61)C1=−352iαεe,C2=435ε3,C3=35ε,λ=−4ε2.Figure6 shows that the energy becomes zero when α=0.5 and it is decreasing for the greater values of this angular deficit parameter. In Figure 7, the effective potential graph is shown where the potential well becomes more distinct as α increases.Figure 6
Energy eigenvalues versusα for (56). k=1, C1=1, n=1, and λ=5 for the green curve; k=1, C1=1, n=1, and λ=1 for the red curve; and k=1, C1=1, n=1, and λ=0.1 for the dashed curve.Figure 7
The effective potential graph in (55). α=0.5 for the dashed curve, α=0.9 for the red curve, and α=0.1 for the magenta curve.


## 3.1. Constant Magnetic Field

The constant magnetic field vectorB→=0,0,a0, where a0 is a real parameter, is perpendicular to the plane. Then, vector potential component Aϕr can be taken as Aϕr=a0r, and Wr becomes
(31)Wr=ea0α+mαr.This superpotential is introduced in [15] where the singular potential models are also discussed. Here, we have considered the one-dimensional system. Hence, V1,2r functions become
(32)V1r=a02e2α2+mm−αα2r2+2a0emα2r,(33)V2r=a02e2α2+mm+αα2r2+2a0emα2r.This system shows thata0 should be negative because of the Coulomb’s potential; let a0 be a0=−ν,ν>0. The solutions of the system (32) which is known as pseudoharmonic potential are already known [27]:
(34)En=±k2−ν2e2α2−a02e2m2α4n+1/2+mm−α/α2+1/42,n=0,1,2,⋯,(35)X1,n=Nnr1/2+μexp−∈rF11−n,2μ+1,2∈r,where μ=mm−α/α2+1/4, ε=ν2e2/α2−E, and F11−n,2μ+1,2∈r are the confluent hypergeometric functions. Let us look at the energy spectrum and the conditions for getting real energy levels. The inside of the square root in (34) must be positive for the real energy spectrum, and
(36)∣k∣>ν2e2α2+a02e2m2α4n+1/2+mm−α/α2+1/42,m>1.The normalization constant can be written as [27]
(37)Nn=Γ2μ+1n!2n+2μ+1Γn+2μ+1.Let us see how the energy and potential functions change withα.As is shown from Figure1, we can obtain physical potentials for the values of α which takes 0<α<1. Energy graph in Figure 2 shows that for the values of α≥0.1831 (green curve), we can get an increasing energy graph. For the greater values of the number n, the lower bound for the α increases.Figure 1
The graph of (32) and (40). V1r graph versus the position for the different values α. α=0.5 for orange, α=0.5 for blue, α=0.8 for green curves, and the red dashed curve stands for the vector potential.Figure 2
The graph of energy eigenvalues with respect toα (34). The green curve is drawn for k=10, n=1, and a0=2, and the yellow curve is shown for k=8, n=2, and a0=0.2.


## 3.2. Potential Models: Energy-Dependent Vector and Scalar Potentials

The energy-dependent potentials are one of the some modified problems of the quantum mechanics [28]. The condition on the density distribution is [28]
(38)ρnEn,r=ψnEn,x21−∂VEn,r∂En.As it is seen from the above equation, a positivity condition is(39)1−∂VEn,r∂En>0.The details of the related energy-dependent potentials in supersymmetry can be found in [29]. In this case, we take the vector potential Ar as a complex function
(40)Ar=C1+Srr−me,where C1 is a constant and Sr is the complex function. Then, (23) becomes
(41)χ1′′r+E2−k2−C12e2α2−2C1e2Srα2−e2Sr2α2−eαS′rχ1r=0,(42)χ2′′r+E2−k2−C12e2α2−2C1e2Srα2−e2Sr2α2+eαS′rχ2r=0,where we can give the superpotential Wr as
(43)Wr=C1eα+eSrαand let Sr be
(44)Sr=A1sechr+B1tanhr−C1,where A1,B1,C1 are the constants. We can get the partner potentials as
(45)V1r=B12e2α2+eA12e+B1−B1e+αα2sech2r+A1e2B1e−αα2sechrtanhr,(46)V2r=B12e2α2+eA12e−B1B1e+αα2sech2r+A1e2B1e+αα2sechrtanhr.Here, (45) is known as Scarf II potentials (see [30]). Comparing our system (41), (43), and (44) with the linear energy-dependence results in [31] can give a solvable model. We can use the parameter A1 as energy-dependent potential in our calculations as
(47)A1=α2ei−2E2−B12e2α2.Then, we match our system with the one where linear energy dependency can be found in page 9 in [31] and write the partner potentials as energy-dependent potentials as
(48)V1r=B12e2α2+E2+B1eα−14−2B12e2α2−iE2−B12e2α2sech2r+α−2B1e2α−i+2E2−B12e2α2sechrtanhr,(49)V2r=B12e2α2+E2−B1eα−14−2B12e2α2−iE2−B12e2α2sech2r+α+2B1e2αi−2E2−B12e2α2sechrtanhr.Then, one can find the energy eigenvalues as(50)En=±k2−n+B1eα2+B12e2α2,n=0,1,2,⋯.The condition for the real energies is(51)∣k∣>n+B1eα2−B12e2α2.And the solutions become(52)χ1,n=1−isinhr1/4−1/2n+1+k2−β2+4iβ1+isinhr1/4−1/2n+1+k2−β2−4iβPn−1/22+2n+2k2−β2+2iβ2,−1/2−2n−2k2−β2+2iβ2isinhr.In Figure3, we have obtained an effective potential graph for the real potential while we can get a potential barrier from the imaginary part. Figure 4 shows the component of the vector potential whose real component is corresponding to another effective potential curve. In Figure 5, we can see that for the greater values of n, the lowest bound for the α again increases as energy increases.Figure 3
The real and imaginary parts of the potential in (48). Green and blue curves are corresponding to the real and imaginary parts, respectively, where α=0.2.Figure 4
The real (green) and imaginary (blue) parts of the vector potential in (40).Figure 5
Energy eigenvalues versus theα values for (50).


## 3.3. Potential Models: Position-Dependent Mass Model

The idea of a position-dependent-effective-mass (PDEM) quantum Hamiltonians gains more attention because of the physical applications in graded crystals, quantum dots, and liquid crystals. The hermiticity and the uniqueness of the problem are studied in the region of mathematical physics. For more details, see [32–34]. Let us make a point transformation in (41) given by
(53)r=∫zdtft.Then, we can get
(54)fz2χ1′′z+fzf′zχ1′z+E2−k2−C12e2α2−2C1e2Szα2−e2Sz2α2+eαfzS′zχ1z=0.Let us choosefz=1+λz2 and Sz=iC1/1+λz2−c1; then, (54) turns into
(55)1+λz2χz′′+λzχz′+E2−k2+C12e2α21+λz2+iC1eλzα1+λz2χz=0.The solutions of the type of equations like (55) are already known. We can write the complete solutions as [35]
(56)En=±k2+nλ2C1e−αλα−nλ,(57)χnz=Nnin1+λz2−s/2e−rtan−1zλPnir−s−1/2,−ir−s−1/2izλ,n=0,1,2,⋯<s,where
(58)Nn=λn!s−nΓs−ir−n+1/2Γs+ir−n+1/2π2−2sΓ2s−n+1.s=2C1e−αλ/2α3/2,r=−i/2, and Pna,by are the corresponding Jacobi polynomials. Now, let us examine the SUSY of this model. First, we need to make a point transformation to (55) which is χ1z=1/1+4λ2z21/4Yz. We get
(59)Y′′z+14z2+3−4iAz+4A2z24z21+z2λ2+−1+iAz+z2ε2z21+z2λYz=0,where A=C1e/α. We propose a super potential which is given by
(60)Wz=C2z21+z2λ+C3.Then, for (59), the unknown parameters of the superpotential can be obtained as
(61)C1=−352iαεe,C2=435ε3,C3=35ε,λ=−4ε2.Figure6 shows that the energy becomes zero when α=0.5 and it is decreasing for the greater values of this angular deficit parameter. In Figure 7, the effective potential graph is shown where the potential well becomes more distinct as α increases.Figure 6
Energy eigenvalues versusα for (56). k=1, C1=1, n=1, and λ=5 for the green curve; k=1, C1=1, n=1, and λ=1 for the red curve; and k=1, C1=1, n=1, and λ=0.1 for the dashed curve.Figure 7
The effective potential graph in (55). α=0.5 for the dashed curve, α=0.9 for the red curve, and α=0.1 for the magenta curve.


## 4. Extended Potential Models

UsingV1 functions given in the previous section, we derive more general partner potentials for each model here.


### 4.1. Extended Potential Models: Constant Magnetic Field

Let us start with the choice of the superpotential. Here, we want to generate more general potential family of the system given in (32). Now, Wr is
(62)Wr=Ar+Br+frgr,where fr,gr are unknown functions. Then, partner potentials can be obtained as
(63)V1r=A+B1+Br2+2ABr+fr2gr2+2Afrgr+2Bfrrgr−f′rgr+frg′rgr2,(64)V2r=A+B−1+Br2+2ABr+fr2gr2+2Afrgr+2Bfrrgr+f′rgr−frg′rgr2.Here, we match (63) with (32). If we equate the rational terms to zero in (63) as below
(65)fr2gr2+2Afrgr+2Bfrrgr−f′rgr+frg′rgr2=0,then, we obtain
(66)fr=exp2Arr2BgrC1+r1+2BEn−2Ar,where n=−2B and the exponential integral function is given by
(67)Enz=∫1∞exp−rttndt.Then, the partner potentials become(68)V1r=A2+BB+1r2+2ABr,(69)V2r=A2+BB−1r2+2ABr+2e2Arr2B−1e2Arr1+2B+2C1B+Ar+2r2B+1B+ArEn−2ArC1+r2B+1En−2Ar2.Comparing (68) and (32) gives the unknown constants A and B as
(70)A=−a0eα,B=−mα.Then, one can find the solution of (69) which is χ2,n as
(71)X2,n=e−a2rra1εc2nF111−n,1+c1,c2r+A+B−a1r+a2+e2Arr2BC1+r2B+1E−2B−2ARF11−n,c1,c2r,where
(72)a1=12+μ,μ=mm−α1α2,a2=ε,c1=2μ+1,c2=2ε.We note that (69) shares the same energy level given by (34). In Figure 8, one can see that V2r fits an effective potential with a well which is the partner of a Coulombic potential.Figure 8
V1r,V2r graphs in (68) and (69). α=0.3 and a0=−0.2.


### 4.2. Extended Potential Models: Extended Energy-Dependent Vector and Scalar Potentials

Let us choose the superpotentialWr as
(73)Wr=ea1+ib1sechrα+c1fr.So, partner potentials are obtained as(74)V1r=c12fr2−c1f′r+2a1+ib1c1efrαsechr+a1+ib12e2sech2rα2+a1+ib1esechrtanhrα,(75)V2r=c12fr2+c1f′r+2a1+ib1c1efrαsechr+a1+ib12e2sech2rα2−a1+ib1esechrtanhrα.In (73), we can equate the terms to zero which are given below
(76)c12fr2−c1f′r+2a1+ib1c1efrαsechr=0.Then, we find(77)fr=exp4ea1+ib1tan−1tanhr/2/α2a1e+2ib1e+iα2a1e+2ib1e+iα−2c1αexp22a1e+2ib1e+iαtan−1tanhr/2/αF21−2ia1e+2b1e+α/2α,1,3/2+b1e−ia1e/α,−e4itan−1tanhr/2.Once can give the first partner potential as(78)V1r=a1+ib1eα2sech2r+a1+ib1eαsechrtanhr,and comparing the coefficients of the hyperbolic functions in (74) with those in (48) gives us
(79)b1=ia1−A1,A1=α2ei−2E2−1,B1=αe.Then, we can find theV2r(80)V2r=a¯eα2sech2r−a¯esechrtanrα−c1e4a¯etan−1tanhr/2/α2a1e+2b1e+α2a1e+i2b1e+αhrsechr1+e4itan−1tanhr/2αC12ia1e+α+2iαc1e22a¯+iα/αF21−2ia1e+2b1e+α/2α,1,3/2+e−ia1+b1/α,−e4itan−1tanhr/22,where
(81)hr=2C1αe22a¯e+2iαtan−1tanhr/2/α+1+e4itan−1tanhr/24a¯eC1+αc1e4a¯etan−1tanhr/2/αcoshr−8a¯c1e22a¯+3iαtan−1tanhr/2/α+e2a¯+iαtan−1tanhr/2/αeαF21−2ia1e+2b1e+α2α,1,32+e−ia1+b1α,−e4itan−1tanhr/2,and we use a¯=a1+ib1; the constants λ1,2 are given by
(82)λ1=−1+n+k2−β2+iβ,λ2=−−n−k2−β2+iβ.One can find a higher potential barrier asα decreases in Figure 9 for the real part of V1, and for the real part of the partner potential V2, it shows a potential getting deeper as α decreases in Figure 10.Figure 9
V1r graph in (78). α=0.3 for the blue curve and α=0.9 for the magenta curves. a1=0.3 and b1=0.2.Figure 10
V2r in (80), α=0.3 for the black curve and α=0.9 for the green curve. a1=0.3 and b1=0.2.


### 4.3. Extended Potential Models: Extended Position-Dependent Mass Model

The exact solutions of the position-dependent effective mass Schrödinger equation for the nonlinear isotonic oscillator are found in [36]. In this case, we search for a new family of the extension of the nonlinear isotonic oscillator type potentials within the approach mentioned in (55). Now, we take Wz which has a form
(83)Wz=C2z21+λz2+C3+Gz,where Gz is the unknown function which can be found by the following equations:
(84)Wz2−W′z=C32+C22z4+2C2z3λ1+z2λ2+2C2C3z2−2C2z1+z2λ+2C3G1z+2C2z2G1z1+z2λ+G1z2−G1′z,(85)2C3G1z+2C2z2G1z1+z2λ+G1z2−G1′z=0.Here, we note that(86)V1z=ε23−815εz51−4ε2z22.From (85), Gz can be obtained as
(87)Gz=2ε10+153/5tanh−12εz−210+15e3/5tanh−12εzεz+210+15εC1+hz,where
(88)hz=e3/5tanh−12∈z−10+15F213/52,1,1+3/52,−e2tanh−12∈z+15e3/5tanh−12∈zF213/52,1,2+3/52,−e2tanh−12∈z.Now, we can find theV2z as
(89)V2z=ε21533+815εz1−4ε2z22+c1z+c2z,where
(90)c1z=60023+415e23/5tanh−12εz10+15−2εC1+e3/5tanh−12εz2εz+2F11,3/2/2,1+3/2/2,1+2/−1+2εz−15e10+15tanh−12εz/5λz,c2z=1203+215e3/5tanh−12εz−1+4ε2z210+15−2εC1+e3/5tanh−12εz2εz+2F11,3/2/2,1+3/2/2,1+2/−1+2εz−μz.and λz=F211,1+3/5/2+2+3/5/2,1+2/−1+2∈z and μz=15e3/5tanh−12∈z2F11,1+3/2,2+3/2,1+2/−1+2∈z.Figure11 shows that the partner potentials V1z and V2z behave like Vz=z3 which goes to infinity as z→±∞. We have seen that the potential pictures can be obtained independently from α.Figure 11
The graph of the partner potentials which areV1z in (86) (blue) and V2z in (89) (green).


## 4.1. Extended Potential Models: Constant Magnetic Field

Let us start with the choice of the superpotential. Here, we want to generate more general potential family of the system given in (32). Now, Wr is
(62)Wr=Ar+Br+frgr,where fr,gr are unknown functions. Then, partner potentials can be obtained as
(63)V1r=A+B1+Br2+2ABr+fr2gr2+2Afrgr+2Bfrrgr−f′rgr+frg′rgr2,(64)V2r=A+B−1+Br2+2ABr+fr2gr2+2Afrgr+2Bfrrgr+f′rgr−frg′rgr2.Here, we match (63) with (32). If we equate the rational terms to zero in (63) as below
(65)fr2gr2+2Afrgr+2Bfrrgr−f′rgr+frg′rgr2=0,then, we obtain
(66)fr=exp2Arr2BgrC1+r1+2BEn−2Ar,where n=−2B and the exponential integral function is given by
(67)Enz=∫1∞exp−rttndt.Then, the partner potentials become(68)V1r=A2+BB+1r2+2ABr,(69)V2r=A2+BB−1r2+2ABr+2e2Arr2B−1e2Arr1+2B+2C1B+Ar+2r2B+1B+ArEn−2ArC1+r2B+1En−2Ar2.Comparing (68) and (32) gives the unknown constants A and B as
(70)A=−a0eα,B=−mα.Then, one can find the solution of (69) which is χ2,n as
(71)X2,n=e−a2rra1εc2nF111−n,1+c1,c2r+A+B−a1r+a2+e2Arr2BC1+r2B+1E−2B−2ARF11−n,c1,c2r,where
(72)a1=12+μ,μ=mm−α1α2,a2=ε,c1=2μ+1,c2=2ε.We note that (69) shares the same energy level given by (34). In Figure 8, one can see that V2r fits an effective potential with a well which is the partner of a Coulombic potential.Figure 8
V1r,V2r graphs in (68) and (69). α=0.3 and a0=−0.2.


## 4.2. Extended Potential Models: Extended Energy-Dependent Vector and Scalar Potentials

Let us choose the superpotentialWr as
(73)Wr=ea1+ib1sechrα+c1fr.So, partner potentials are obtained as(74)V1r=c12fr2−c1f′r+2a1+ib1c1efrαsechr+a1+ib12e2sech2rα2+a1+ib1esechrtanhrα,(75)V2r=c12fr2+c1f′r+2a1+ib1c1efrαsechr+a1+ib12e2sech2rα2−a1+ib1esechrtanhrα.In (73), we can equate the terms to zero which are given below
(76)c12fr2−c1f′r+2a1+ib1c1efrαsechr=0.Then, we find(77)fr=exp4ea1+ib1tan−1tanhr/2/α2a1e+2ib1e+iα2a1e+2ib1e+iα−2c1αexp22a1e+2ib1e+iαtan−1tanhr/2/αF21−2ia1e+2b1e+α/2α,1,3/2+b1e−ia1e/α,−e4itan−1tanhr/2.Once can give the first partner potential as(78)V1r=a1+ib1eα2sech2r+a1+ib1eαsechrtanhr,and comparing the coefficients of the hyperbolic functions in (74) with those in (48) gives us
(79)b1=ia1−A1,A1=α2ei−2E2−1,B1=αe.Then, we can find theV2r(80)V2r=a¯eα2sech2r−a¯esechrtanrα−c1e4a¯etan−1tanhr/2/α2a1e+2b1e+α2a1e+i2b1e+αhrsechr1+e4itan−1tanhr/2αC12ia1e+α+2iαc1e22a¯+iα/αF21−2ia1e+2b1e+α/2α,1,3/2+e−ia1+b1/α,−e4itan−1tanhr/22,where
(81)hr=2C1αe22a¯e+2iαtan−1tanhr/2/α+1+e4itan−1tanhr/24a¯eC1+αc1e4a¯etan−1tanhr/2/αcoshr−8a¯c1e22a¯+3iαtan−1tanhr/2/α+e2a¯+iαtan−1tanhr/2/αeαF21−2ia1e+2b1e+α2α,1,32+e−ia1+b1α,−e4itan−1tanhr/2,and we use a¯=a1+ib1; the constants λ1,2 are given by
(82)λ1=−1+n+k2−β2+iβ,λ2=−−n−k2−β2+iβ.One can find a higher potential barrier asα decreases in Figure 9 for the real part of V1, and for the real part of the partner potential V2, it shows a potential getting deeper as α decreases in Figure 10.Figure 9
V1r graph in (78). α=0.3 for the blue curve and α=0.9 for the magenta curves. a1=0.3 and b1=0.2.Figure 10
V2r in (80), α=0.3 for the black curve and α=0.9 for the green curve. a1=0.3 and b1=0.2.


## 4.3. Extended Potential Models: Extended Position-Dependent Mass Model

The exact solutions of the position-dependent effective mass Schrödinger equation for the nonlinear isotonic oscillator are found in [36]. In this case, we search for a new family of the extension of the nonlinear isotonic oscillator type potentials within the approach mentioned in (55). Now, we take Wz which has a form
(83)Wz=C2z21+λz2+C3+Gz,where Gz is the unknown function which can be found by the following equations:
(84)Wz2−W′z=C32+C22z4+2C2z3λ1+z2λ2+2C2C3z2−2C2z1+z2λ+2C3G1z+2C2z2G1z1+z2λ+G1z2−G1′z,(85)2C3G1z+2C2z2G1z1+z2λ+G1z2−G1′z=0.Here, we note that(86)V1z=ε23−815εz51−4ε2z22.From (85), Gz can be obtained as
(87)Gz=2ε10+153/5tanh−12εz−210+15e3/5tanh−12εzεz+210+15εC1+hz,where
(88)hz=e3/5tanh−12∈z−10+15F213/52,1,1+3/52,−e2tanh−12∈z+15e3/5tanh−12∈zF213/52,1,2+3/52,−e2tanh−12∈z.Now, we can find theV2z as
(89)V2z=ε21533+815εz1−4ε2z22+c1z+c2z,where
(90)c1z=60023+415e23/5tanh−12εz10+15−2εC1+e3/5tanh−12εz2εz+2F11,3/2/2,1+3/2/2,1+2/−1+2εz−15e10+15tanh−12εz/5λz,c2z=1203+215e3/5tanh−12εz−1+4ε2z210+15−2εC1+e3/5tanh−12εz2εz+2F11,3/2/2,1+3/2/2,1+2/−1+2εz−μz.and λz=F211,1+3/5/2+2+3/5/2,1+2/−1+2∈z and μz=15e3/5tanh−12∈z2F11,1+3/2,2+3/2,1+2/−1+2∈z.Figure11 shows that the partner potentials V1z and V2z behave like Vz=z3 which goes to infinity as z→±∞. We have seen that the potential pictures can be obtained independently from α.Figure 11
The graph of the partner potentials which areV1z in (86) (blue) and V2z in (89) (green).


## 5. Conclusions

Using the fundamental concepts of SUSY QM, we have obtained physical solutions for the extended constant magnetic field which leads to a Coulomb problem, energy-dependent hyperbolic potential and nonlinear isotonic potential, which is argued as the position-dependent mass model for a fermion near cosmic string spacetime. It is observed that the restricted values of the angular deficitα give reasonable behaviours of the potentials for our models.



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*Source: 5037471-2020-09-03.xml*