Two-dimensional Dirac fermion in presence of an asymmetric vector potential
TTwo-dimensional Dirac fermion in presence of an asymmetric vectorpotential
A. Ishkhanyan a,b , V. Jakubsk´y c a Russian-Armenian University, Yerevan, 0051 Armenia b Institute of Physics and Technology, National Research Tomsk Polytechnic University,Tomsk, 634050 Russian Federation c Nuclear Physics Institute of the CAS, ˇReˇz, 25068, Czech RepublicE-mails: [email protected], [email protected]
September 3, 2018
Abstract
We introduce the new, exactly solvable model of the two-dimensional Dirac fermion in presence ofan asymmetric, P¨oschl-Teller-like vector potential. Utilizing the translation invariance of the system,the effective one-dimensional stationary equation is brought into the form of the Heun equation andits fundamental solutions are found as an irreducible combination of two Gauss hypergeometric func-tions. The energy spectrum and the scattering is studied in dependence on the conserved longitudinalmomentum as well as on the strength of the coupling.
Experimental isolation of graphene [1] ignited intensive study of this new material. Many of its remarkableproperties, e.g. unconventional quantum Hall effect, minimal conductivity, or light absorption emerge dueto the fact that the low-energy excitations of electrons in the lattice behave like mass-less, two-dimensionalDirac fermions [2], [3], [4], [5], [6]. In the recent years, there have been discovered other materials thatshare this property, hinted by existence of Dirac cones in their band structure. Let us mention silicene,germanene, dichalcogenides [7], [8], [9], or artificially prepared materials where the low-energy quasi-particles are described by two-dimensional Dirac equation, see [10] for review. These systems were calledDirac materials or Dirac matter in the literature [11], [12].Unlike standard semi-conductors, electron transport in Dirac materials cannot be controlled very wellby an electrostatic barrier; the charge carriers can tunnel it without backscattering. This phenomenonwas understood as the manifestation of Klein tunneling [13]. Alternative ways to confinement and controlof the charge carriers were proposed. The magnetic field proved to be one of the feasible options [14].Symmetries of the stationary equation can be used for separation of variables, giving rise to an effec-tively one-dimensional problem. The stationary Dirac equation is reduced to an effective one-dimensionalSchr¨odinger equation with a fixed value of the (conserved) longitudinal momentum. The models withtranslational symmetry in one direction, where either the magnetic field or the vector potential are piece-wise constant, were discussed e.g. in [14], [15], [16], [17], and [18]. The exactly solvable systems withsmooth magnetic field analyzed in the literature were usually related to the known, exactly solvable mod-els of the nonrelativistic quantum mechanics [19], [20], [21], or constructed from them with the use ofsupersymmetric techniques [22], [23], [24], [25], [26], [27], [28].In the current article, we will also utilize translational symmetry to reduce the considered systeminto the effectively one-dimensional setting. The corresponding one-dimensional Schr¨odinger equationrepresents the new exactly solvable model presented here for the first time. We discuss the spectral and1 a r X i v : . [ h e p - t h ] A ug cattering properties of both the effective one-dimensional and the initial two-dimensional relativistic sys-tem in dependence on the conserved longitudinal momentum and the strength of the coupling parameter.The article is organized as follows: in the next section, we introduce the model with asymmetric stepvector potential, we derive effectively one-dimensional stationary equation and find its explicit solutionsin terms of Gauss hypergeometric functions. In the Section 3, discrete energies of the effective one-dimensional system are discussed whereas in the Section 4, we focus on the scattering properties. Thelast section is left for discussion. Let us introduce the following Hamiltonian H D = − i (cid:126) σ ∂ x − i (cid:126) σ ∂ y + V (cid:113) e xσ σ , (1)where the coupling constant V is fixed to be real and positive number and σ , σ and σ are Paulimatrices. We use the units where (cid:126) = 1 and we fix σ = 1. The vector potential represents an asymmetricstep. Indeed, defining the reflection operator as P = P σ , where P xP = − x and P yP = − y , thepotential term breaks manifestly the reflection symmetry of H D . As (1) commutes with the generator oftranslations ˆ p y = − i∂ y , it possesses translational symmetry along y -axis.Making the partial Fourier transform in the y coordinate, we can rewrite H D in the form of directintegral H D = (cid:90) ⊕ R H D [ k y ] , where the effectively one-dimensional Hamiltonian H D [ k y ] is acting in x -coordinate with k y being fixed.Its explicit form reads H D [ k y ] = − iσ ∂ x + W ( x ) σ , (2)where W ( x ) = k y + V z ( x ) , z ( x ) = (cid:112) e x . (3)We fix the domain of H D [ k y ] as the Sobolev space W , ( R , C ), the space of C -valued square integrablefunctions with square integrable first derivatives. As the potential term W ( x ) is smooth and bounded, itfollows from the Kato-Rellich theorem that the operator H D [ k y ] is self-adjoint on this domain [29]. Oncethe spectrum of H D [ k y ] is found, it is straightforward to reconstruct the spectrum σ ( H D ) of H D . In ourcase, it can be understood as the union of σ ( H D [ k y ]), see [30] for more details.It is convenient to inspect the stationary equation for the squared Dirac Hamiltonian H D [ k y ] . Theoperator reads explicitly H D [ k y ] = (cid:18) H + S [ k y ] 00 H − S [ k y ] (cid:19) , H (cid:15)S [ k y ] = − ∂ x + W + (cid:15) ∂ x W, (cid:15) = ± . (4)Let us focus on the spectral properties of H S [ k y ] ≡ H + S [ k y ] as those of H D [ k y ] can be deduced in astraightforward manner afterwards. The explicit form of the associated nonrelativistic stationary equationreads (cid:18) − ∂ x + k y + V (2 k y − z ( x ) + V z ( x ) + V z ( x ) (cid:19) ψ E, ± = Eψ E, ± . (5)It is the key finding of this article that this equation is exactly solvable for any E . To our best knowledge,the explicit solution of (5) has not been discussed yet. The potential term in (5) is generalization of the The spectrum and eigenstates of H D for negative values of V can be obtained directly with the use of the operator P y σ where P y f ( x, y ) = f ( x, − y ). H S [ k y ] form an irreducible combination of two hypergeometric functions. To this instance,the potential belongs to the solvable potentials that have been explored recently [31], [32], [33], [34].We prefer to leave the technical details of the solution of (5) to the Appendix and present here theobtained eigenstates of H S [ k y ]. They can be written as ψ E, ± = ( z + 1) ± α ( z − α u ± ( z ) , (6)where u ± ( z ) = F (cid:18) ± α + α − α − , ± α + α + α , ± α , z + 12 (cid:19) + ( − V ∓ α + α + ( ± α + α + α ) z ) ± α × F (cid:18) ± α + α − α , ± α + α + α + 1 , ± α , z + 12 (cid:19) . (7)The quantities α , α and α are defined as α = 12 (cid:113) − E + ( k y − V ) , α = 12 (cid:113) − E + ( k y + V ) , α = (cid:113) − E + k y . (8)Let us notice that the argument z +12 in (7) falls out of the radius of convergence of the hypergeometricfunctions. In the Section 3.1, where the asymptotic behavior of the wave function (6) is discussed, thefunctions (7) will be transformed into the expressions with guaranteed convergence.Let us consider the eigenstates Ψ √ E, ± of the associated Dirac Hamiltonian H D [ k y ]. We haveΨ √ E, ± = (cid:32) ψ E, ±− i∂ x + iW √ E ψ E, ± (cid:33) , Ψ −√ E, ± = σ Ψ √ E, ± , (9)and ( H D [ k y ] ∓ √ E )Ψ ±√ E, ± = 0 . (10)The spectrum of H D [ k y ] is symmetric with respect to zero. This was to be expected as we deal with achiral Hamiltonian; it anticommutes with σ , { σ , H D [ k y ] } = 0. The zero modes of H D [ k y ] can be foundby direct calculation asΨ , − = (cid:18) exp (cid:18) − (cid:90) x W ( s ) ds (cid:19) , (cid:19) T , Ψ , + = (cid:18) , exp (cid:18)(cid:90) x W ( s ) ds (cid:19) , (cid:19) T . (11)Their square integrability depends on the asymptotic behavior of W ( x ) that varies in dependence on k y .We will discuss this point in Section 3.2 in more detail. We focus on the non-relativistic system described by the Hamiltonian H S [ k y ]. We find its bound statesby the asymptotic analysis of the wave functions (6) in dependence on the coupling constant V andon the values of k y . As we mentioned above, the spectral properties of the effective one-dimensionaloperator H D [ k y ] and the initial two-dimensional Dirac Hamiltonian H D can be deduced afterwards instraightforward manner. 3 .1 Asymptotics of the wave functions As the potential term of H S [ k y ] is smooth and asymptotically constant, W ( x ) → (cid:40) k y , x → + ∞ ,k y + V , x → −∞ , W ( x ) + ∂ x W ( x ) → (cid:40) k y , x → ∞ , ( k y + V ) , x → −∞ , (12)the wave functions have to coincide asymptotically with a linear combination of the free particle solutions.In order to find the coefficients of the linear combination, we use the following formulas that turn thehypergeometric functions into the form convergent at x → ±∞ : F ( a, b, c, z ) = Γ( c )Γ( b − a )Γ( b )Γ( c − a ) ( − z ) − a F (cid:18) a, − c + a, − b + a, z (cid:19) (13)+ Γ( c )Γ( a − b )Γ( a )Γ( c − b ) ( − z ) − b F (cid:18) b, − c + b, − a + b, z (cid:19) , (14) F ( a, b, c, z ) = Γ( c )Γ( c − a − b )Γ( c − b )Γ( c − a ) z − a F (cid:18) a, − c + a, b + a − c, − z (cid:19) (15)+ Γ( c )Γ( a + b − c )Γ( a )Γ( b ) ( − c − a − b ( − z ) c − a − b z a − c × F (cid:18) c − a, − a, c − a − b, − z (cid:19) . (16)Then we find that the functions (6) behave asymptotically as ψ E, ± ∼ a R ± e − √ E + k y + b R ± e √ E + k y , x → + ∞ , (17) ψ E, ± ∼ a L ± e − √ E +( k y + V ) + b L ± e √ E +( k y + V ) , x → −∞ , (18)where the upper indices L, R denote the asymptotics at x → ±∞ , and the lower indices ± distinguishthe two independent solutions. The coefficients can be written explicitly as a R + = ( − ) − α − α − α (1 + 2 α )Γ( − − α )Γ(2 α )Γ( α − α − α )Γ( α + α − α ) ,b R + = ( − ) α − α − α Γ(2 α )Γ(2 α )( α − α − α V )Γ(1 + α + α − α )Γ(1 + α + α + α ) ,a L + = ( − ) − α − α + α Γ(2 α )Γ(2 α )( α + 2 α − V )Γ( α + α − α )Γ( α + α + α + 1) ,b L + = 2 − α − α Γ(2 α )Γ( − α )( α − α − V )Γ( α − α − α )Γ( α − α + α + 1) ,a R − = a R + | α →− α , a L − = a L + | α →− α ,b R − = b R + | α →− α , b L − = b L + | α →− α . (19)It is convenient to fix the combination that coincides with the single exponential function at x → + ∞ , F ≡ b R − λ ψ E, + − b R + λ ψ E, − ∼ e − √ k y − Ex , x → + ∞ . (20)Here, we denoted λ = ( − ) − α ( α − α + α V )4 α α . At x → −∞ , the function F behaves in the following manner F ∼ C e − √ ( k y + V ) − Ex + C e √ ( k y + V ) − Ex , x → −∞ , (21)where C and C are complex numbers given in terms of α , α , α and V .4he relations (20) and (21) suggest that the function F will vanish exponentially for large | x | providedthat (cid:113) k y − E > , (cid:113) ( k y + V ) − E > , C = 0 . (22)The first two relations fix the range of allowed energies to the interval whose upper bound depends onthe values of k y , E ∈ [0 , E max ) , E max = (cid:40) k y , k y ≥ − V , k y (cid:54) = 0 , ( k y + V ) , k y < − V , k y (cid:54) = − V . (23)Here, the lower bound follows from the fact that E has to be non-negative as it should correspond to theeigenvalue of the operator that is related to the square of the self-adjoint Hamiltonian, see (4) and (5).When k y = 0 or k y = − V , the first two conditions in (22) cannot be fulfilled for E ≥
0. To analyze thethird condition, it is convenient to bring the coefficient C into the following simplified form C = 2 α − α α Γ(2 α )Γ(2 α )(2 α + α − V )Γ(1 − α + α + α )Γ(1 + α + α + α ) . (24)The parameters α , α and α are positive by definition (8) and by the requirement (22). It implies thatthe terms Γ(2 α ) and Γ(2 α ) are nonvanishing. Therefore, the coefficient C can vanish as long as either2 α + α − V = 0 or 1 − α + α + α is equal to a non-positive integer. In the latter case, the Γ functionin the denominator would have singularity and would make the coefficient C vanishing. Let us considerthese cases in more detail. First, let us consider the case where C vanishes due to 2 α + α − V = 0. Substituting into the equationfrom (8), we get α + 2 α − V = (cid:113) k y − E − V + (cid:113) ( k y + V ) − E = 0 . (25)The the expression on the left-hand side is monotonically decreasing function of E . Hence, the equationabove has a solution if and only if the left-hand side has different signs at E = 0 and E = E max , or it isidentically zero at E = 0, see (23). Straightforward analysis shows that this is the case for − V < k y < . (26)It is consistent with the the discussion below (10); one can see from the explicit formulas for the zero modes(11) that they are square integrable provided that the vector potential W ( x ) changes sign asymptotically.Taking into account (12), we can see that this happens exactly for the values of k y specified in (26). Thesquare integrable zero modes acquire the following explicit form where k y is supposed to satisfy (26),Ψ = ( ψ , T , H D [ k y ]Ψ = 0 , ψ = e k y x (cid:32) − √ e x √ e x (cid:33) V , H S [ k y ] ψ = 0 . (27) Now, let us focus on the cases where C = 0 due to the singularity of the Γ function in the denominatorof (24). As α , α and α are positive numbers, the only term that can be negative is I ( E ) ≡ α − α + α = (cid:113) k y − E − (cid:113) ( k y − V ) − E + 12 (cid:113) ( k y + V ) − E. (28)The values of E where I ( E ) acquires negative integer values will correspond to the bound state energiesof H S [ k y ]. Let us inspect the behavior of I ( E ) for different values of k y . We will discuss the cases ofpositive and negative k y separately. 5 .3.1 k y ≥ I ( E ) is strictly positive by finding its positive lower bound. We have − (cid:113) ( k y − V ) − E ≥ − (cid:113) k y + V − E, (cid:113) ( k y + V ) − E ≥ (cid:113) k y + V − E. (29)Then, taking into account (22), we can write I ( E ) ≥ (cid:113) k y − E − (cid:113) k y + V − E + 12 (cid:113) k y + V − E = (cid:113) k y − E > . (30)It implies that I ( E ) is always positive. Therefore, the coefficient C does not vanish for any value of E . We conclude that the Hamiltonian H S [ k y ], and, consequently, H D [ k y ] have no bound states for any k y ≥ k y < I ( E ) is a decreasing function of E for k y <
0. Indeed, its derivative,2 ∂ E I ( E ) = − (cid:113) k y − E − (cid:112) ( | k y | − V ) − E + 12 (cid:112) ( | k y | + V ) − E , (31)can be bounded from above by a negative function. It can be obtained by using the following estimatesin (31), (cid:113) ( | k y | + V ) − E ≥ (cid:113) k y + V − E, (cid:113) ( | k y | − V ) − E ≤ (cid:113) k y + V − E. (32)It yields ∂ E I ≤ − (cid:113) k y − E < . (33)Accordingly, I ( E ) is decreasing function of E on the interval [0 , E max ). It acquires its maximum at E = 0,max E ∈ [0 ,E max ) I ( E ) = I (0) = (cid:40) , for − V < k y < , | k y | − V , for k y < − V , (34)whereas its infimum is obtained at E = E max , see (23),inf E ∈ [0 ,E max ) I ( E ) = I ( E max ) = (cid:112) | k y | V − V − (cid:112) | k y | V , k y ∈ ( −∞ , − V ] , V (cid:18)(cid:113) − | k y | V − (cid:113) | k y | V (cid:19) , k y ∈ ( − V , . (35)When k y ≤ − V , we can write I ( E max ) = V (cid:115) | k y | V − − (cid:115) | k y | V ≥ V (cid:115) | k y | V + V V − − (cid:115) | k y | V = 0 . (36)Therefore, I ( E ) is strictly positive and H S [ k y ] has no bound states for k y ≤ − V .For k y ∈ ( − V , H S [ k y ] is given by the absolute value of the integerpart of (35), | [ I ( E max )] | , where we denoted the integer part of by [ . ]. As the bound states of H D [ k y ] canbe obtained from those of H S [ k y ] via (9), we can conclude that the number of bound states n of H D [ k y ]as a function of k y can be expressed as n = 2 | [ I ( E max )] | + 1 . (37)6 (cid:45) (cid:45) (cid:45) k y (cid:45) (cid:45) (cid:45) (cid:45) k y (cid:45) (cid:45) Figure 1: Spectrum of the Schr¨odinger operator H S [ k y ] (left) and of the Dirac operator H D [ k y ] (right)in dependence on the value of k y . We fixed V = 5 .
6. In accordance with the formula for the maximumnumber of bound states n max = 2 (cid:104) V √ (cid:105) + 1, the Hamiltonian H D [ k y ] has seven bound states for k y = − V .Here, the multiplicative factor 2 reflects presence of negative energies while the summand 1 stands for thesingle zero mode. As the function of k y , E max acquires its largest value for k y = − V . Substituting thisvalue to (35) and inserting the result into (37), we get that Hamiltonian H D [ k y ] has maximum numberof bound states n max = 2 (cid:104) V √ (cid:105) + 1.We solved the equation I ( E ) = 0 numerically and reconstructed the spectrum of H S [ k y ] and H D [ k y ]in dependence on k y , see Fig. 1. The probability densities of the bound states associated with the discreteenergies of H S [ k y ] and H D [ k y ] are presented in Fig. 2. Scattering properties of H D [ k y ] can be unambiguously determined from those of H S [ k y ] with the use of(9). Henceforth, let us analyze the reflection and transmission coefficients of the latter Hamiltonian.First, we consider the wave function propagating from the left, i.e. we require it to have the followingasymptotic form, ψ l = (cid:40) e x √ ( k y + V ) − E + R l e − x √ ( k y + V ) − E , x → −∞ , ˜ T l e x √ k y − E , x → ∞ . (38)The incoming wave exists for E > ( k y + V ) , and the outgoing (scattering) wave for E > k y . When( k y + V ) < k y and E ∈ (( k y + V ) , k y ), then the transmitted wave is exponentially vanishing and totalreflection takes place. Using (17), we can find the coefficients R l and ˜ T l in terms of (19), R l = a R − a L + − a R + a L − a R − b L + − a R + b L − , ˜ T l = a R − b R + − a R + b R − a R − b L + − a R + b L − . (39)The quantity R l coincides with the reflection coefficient. However, ˜ T l is proportional but not equal tothe transmission coefficient in general. It is caused by the fact that the potential converges to differentvalues at x → ±∞ , see e.g. [35]. With the use of the continuity equation, we get the following relationfor the transmission coefficient T l , |R l | + |T l | = 1 for T l = (cid:118)(cid:117)(cid:117)(cid:116) (cid:113) E − k y (cid:112) E − ( k y + V ) ˜ T l . (40)7 a (cid:76) (cid:45) (cid:45) x Ρ D (cid:72) (cid:89) (cid:76) (cid:72) b (cid:76) (cid:45) (cid:45) x Ρ S (cid:72) Ψ (cid:76) Ρ D (cid:72) (cid:89) (cid:76)(cid:72) c (cid:76) (cid:45) (cid:45) x Ρ S (cid:72) Ψ (cid:76) Ρ D (cid:72) (cid:89) (cid:76) (cid:72) d (cid:76) (cid:45) x Ρ S (cid:72) Ψ (cid:76) Ρ D (cid:72) (cid:89) (cid:76) Figure 2: Probability densities of (normalized) bound states ψ j and Ψ j , j = 0 , , ,
3, of the effectiveSchr¨odinger and Dirac Hamiltonians H S [ k y ] and H D [ k y ], respectively. Here we used ρ S ( f ) = f f and ρ D ( F ) = F † F . We fixed V = 5 . k y = − V . We did not plot probability densities for eigenstates ofnegative energies as there holds ρ D ( σ Ψ j ) = Ψ .Now, let us consider the wave function traveling from the right. It behaves asymptotically as ψ r = (cid:40) e − x √ k y − E + R r e x √ k y − E , x → ∞ , ˜ T r e − x √ ( k y + V ) − E , x → −∞ . (41)This time, the incoming (scattering) wave exists for E > k y while the outgoing wave for E > ( k y + V ) .When ( k y + V ) > k y and E ∈ ( k y , ( k y + V ) ), there are no solutions for the outgoing wave in the formof scattering states. We can find the coefficients as R r = b L − b R + − b L + b R − b L − a R + − b L + a R − , ˜ T r = b L − a L + − b L + a L − b L − a R + − b L + a R − . Like in the previous case, R r represents the reflection coefficient. We can find transmission coefficient T r such that |R r | + |T r | = 1 for T r = (cid:118)(cid:117)(cid:117)(cid:116) (cid:112) E − ( k y + V ) (cid:113) E − k y ˜ T r . (42)The formulas (40) and (42) for T l and T r imply that the transmission occurs only for E and k y satisfying E − k y > E − ( k y + V ) >
0. Fixing the energy of the incoming wave, the two inequalities hold fora limited range of k y , k y ∈ ( −√ E, − V + √ E ) , √ E > V , (43)where the restriction of √ E on the right-hand side ensures that the interval is non-empty. When √ E ≤ V or k y does not fall into the specified interval (43), then the incident waves are totally reflected; thebarrier is filtering the wave functions dependently on their energy and the value of the longitudinalmomentum k y . It also implies that there are angles behind the barrier in which the wave function cannot8 Π (cid:45) Π Φ (cid:84) l (cid:72) Φ (cid:76) Figure 3: Transmission amplitude T l for fixed energy E and k y = √ E sin φ , φ ∈ ( − π/ , π/ V = 4 and E = 4 .
01 (dotted black curve), E = 4 . E = 5 (solid black curve).be scattered. Consider the transmitted wave function e ik y y ψ l ( x ) that propagates to the right, see (38).We get E = k x + k y , where k x = (cid:113) E − k y . We can parametrize k y = √ E sin φ , where φ ∈ ( − π/ , π/ φ ∈ (cid:16) − π/ , arcsin (cid:16) − V √ E (cid:17)(cid:17) , see Fig. 4. Let usmention that the tunneling dependent on the wave-vector through magnetic barriers was discussed for thenon-relativistic particles e.g. in [36] and for the effectively one-dimensional Dirac system with symmetricvector potential in [19]. In the current article, we have presented the new exactly solvable model of two-dimensional Dirac fermionsin the presence of an asymptotically vanishing, asymmetric, magnetic barrier (1). The translation sym-metry allowed us to work with effectively one-dimensional stationary equation, whose solutions were givenin terms of an irreducible combination of two hypergeometric functions, see (6) and (7).The solution was facilitated by the fact that the Schr¨odinger equation for the upper component of thespinor can be reduced to the Heun equation and solved analytically. We elaborated its solution in detailin the Appendix. In this manner, the model extends the family of exactly solvable systems discoveredrecently, [31], [32], [33], [34]. It is also worth noticing in this context that there were analyzed quasi-exactlysolvable system whose solutions were found to be written in terms of the confluent Heun functions [37].The exactly solvable Schr¨odinger equation (5) is generalization of the one discussed in [31] whosepotential term was V + V √ x . Careful comparison reveals one striking feature whose deeper meaningis yet to be understood; the latter potential of the Schr¨odinger operator coincides with the vector potentialof our
Dirac
Hamiltonian, see (3) here and (1) in [31].We focused on the spectral properties of the two-dimensional Dirac Hamiltonian (1). Enjoying thetranslation invariance of the system, its spectrum could be obtained from the analysis of the effectiveone-dimensional operator (2). We found that the energies of the two-dimensional Dirac fermion stronglydepend on the value of the longitudinal momentum k y . When k y is positive, the effective one-dimensionalHamiltonian has no bound states, whereas it can have finite number of bound states for negative valuesof k y . The discrete energy levels of the one-dimensional Hamiltonian form spectral bands of the two-dimensional energy operator, see Fig. 1. As it was discussed in [38], there can exist wave packets in thesystem that are dispersionless in perpendicular direction on the barrier.Our system is similar to the P¨osch-Teller model with the (vector) potential V = tanh x . However,9he systems differ in their symmetries; the Hamiltonian of the P¨oschl-Teller system commutes with thereflection operator P = σ P , while our system breaks this symmetry explicitly. The difference is mani-fested when the spectra of the two systems are compared. Energy of the Dirac particle in presence ofP¨oschl-Teller potential is even function of k y , whereas the spectrum of our system is asymmetric, compareFig. 3 in [19] and Fig. 1 here.Supersymmetric techniques can be used to generate the new exactly solvable systems from the onepresented here. The (confluent) Darboux-Crum transformation was used for construction of the newsolvable models e.g. from the harmonic oscillator, Rosen-Morse model, Coulomb potential or the freeparticle [23], [24], [27]. The new systems obtained in this manner from either H S [ k y ] or H D [ k y ] wouldhave the same spectrum as up to a finite number of discrete energy levels, provided that regularity ofthe new potential is guaranteed. As much as the application of the SUSY transformation on our modelrepresents interesting research direction, we find it going beyond the scope of the present paper. Acknowledgements
The research by Artur Ishkhanyan has been supported by the State Committee of Science of the Republicof Armenia (project 18RF-139), the Armenian National Science and Education Fund (ANSEF grantPS-4986), the Russian-Armenian (Slavonic) University at the expense of the Ministry of Education andScience of the Russian Federation, as well as by the project ”Leading research universities of Russia”(grant FTI 24 2016 of the Tomsk Polytechnic University). V´ıt Jakubsk´y was supported by GA ˇCR grantno. 15-07674Y.
Appendix. Solution of the associated Schr¨odinger equation (5)
The transformation of the dependent and independent variables ψ = ϕ ( z ) u ( z ) , z = z ( x ) (44)reduces the one-dimensional Schr¨odinger equation d ψdx + 2 m (cid:126) ( E − V ( x )) ψ = 0 (45)to the equation u zz + (cid:18) ϕ z ϕ + ρ z ρ (cid:19) u z + (cid:18) ϕ zz ϕ + ρ z ρ ϕ z ϕ + 2 m (cid:126) E − V ( z ) ρ (cid:19) u = 0 , (46)where ρ = dz/dx .Consider a potential given as V ( z ) = V + V z + V z + V z (47)with z = (cid:112) e x/σ . (48)For this coordinate transformation we have ρ = dzdx = ( z + 1)( z − σz . (49)It is then readily checked by direct substitution that putting ϕ = ( z + 1) α ( z − α (50)10educes equation (45) to the general Heun equation d udz + (cid:18) γz + δz − εz + 1 (cid:19) dudz + α β z − qz ( z − z + 1) u = 0 , (51)where the involved parameters are given as( γ, δ, ε ) = ( − , α , α ) , (52) αβ = ( α + α ) + 2 mσ (cid:126) ( E − V ) , (53) q = − α + α − mσ (cid:126) V . (54) α = ± (cid:114) mσ (cid:126) ( − E + V − V + V − V ) , (55) α = ± (cid:114) mσ (cid:126) ( − E + V + V + V + V ) . (56)Now we attempt to solve the Heun equation (51) in terms of the hypergeometric functions.A helpful observation here is that γ = −
1. This means that the characteristic exponents (0 , − γ ) = (0 ,
2) of the singularity z = 0 differ by an integer, hence, the Frobenius solution of the Heunequation generally involves a logarithmic term. However, there is a case when the singularity becomes apparent (simple), that is, when the logarithmic term disappears and the solution becomes analytic inthe singularity. The condition for this is known to be (see, e.g., [39], [40]) q + q ( ε − a ( δ − aαβ = 0 , (57)where a is the location of the third regular singularity of the Heun equation ( a = − u = (cid:80) ∞ n =0 c n z n with c (cid:54) = 0, and successively calculating the coefficients c n after substitution of the seriesinto the Heun equation. In calculating c a division by zero will occur, unless the accessory parameter q satisfies equation (57), in which case the equation for c is identically satisfied.It is known that for a root of equation (57) the Heun equation is solved in terms of the Gausshypergeometric functions as [39] u = F (cid:18) α, β ; ε − a − za − (cid:19) + q + a ( δ − ε − · F (cid:18) α, β ; ε ; a − za − (cid:19) , (58)With α, β determined from (53) together with the Fuchsian condition 1 + α + β = γ + δ + ε : α, β = α + α ± (cid:114) mσ (cid:126) ( − E + V ) , (59)and a = −
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