Tunneling in Fractional Quantum Mechanics

Home , Delta potential

arXiv:1011.1948v2 [math-ph] 15 Mar 2011 n te ra a rw.Sm xmlsae[,2 ] mn man two-parame among a by 3], driven 2, oscillator [4]. harmonic [1, noise a are of propose behavior is examples diffusive equation lous Some Langevin generalized equa grown. fractional integrodifferential the has recently, fractional areas of other study and the years recent In Introduction 1. n 1]ivligtedlaptnil oee,i 1]teFEwt delt with FSE one the correct [12] only in However, the potential. that delta out the pointed involving al. [12] one et Jeng wrong. intrinsically r 8 ,1] oee,rcnl ege l 1]hv hw htso that shown fract have the [11] that fact al. the account et into Jeng taken recently not nonlocal have However, pa FSE the the 10]. solve of 9, motion context Brownian [8, the the are of in generalization obtained natural a integrals was path is It approach, which this c paths, In flight 6]. we mechanics. [5, paper quantum to this in approach In Laskin (FSE). by Schrödinger equation introduced fractional the on 2 1 [email protected] [email protected] rcinlQatmMcais(Q)i h hoyo unu mec quantum of theory the is (FQM) Mechanics Quantum Fractional hr r oeppr nteltrtr tdigsltoso S.S FSE. of solutions studying literature the in papers some are There prto.A osqec,altoeatmt ae nlcla local on based attempts those all consequence, a As operation. re ftedrvtv (1 transmis- derivative the the of of by order 2. given limit is from energy coefficient different zero sion is the the derivative potentials, when fractional both tunneling Riesz For energy the zero of have order can behaviour we interesting very particular, a In have coeffi- coefficients transmission These and cients. reflection corresponding the we culate th potentials, solving these After for po- Schrödinger equation delta mechanics. fractional double quantum and fractional delta in through tentials tunneling the study We eatmnod ae´tc piaa-IMECC - Matemática Aplicada de Departamento .Cplsd Oliveira de Capelas E. unln nFractional in Tunneling unu Mechanics Quantum nvriaeEtda eCampinas de Estadual Universidade 38-5 apns P Brazil SP, Campinas, 13083-859 Abstract α < T 0 1 1 ≤ cos = n am a Jr. Vaz Jayme and 2). 2 ( π/α ,where ), in ple ophysics to applied tions odsusteanoma- the discuss to d α 2 r endoe Lévy over defined are sthe is oa eiaini a is derivation ional nie h S as FSE the onsider cal- e Mittag-Leffler ter hyfudi the is found they e . [7]. tes More others. y oeta was potential a m examples ome poce are pproaches ecam to claims me aisbased hanics hintegral th studied only in the case of negative energies. This has been generalized in [13], where we have solved the FSE for the delta and double delta potentials for positive and negative energies. The objective of this paper is to study the tunneling through delta and double delta potentials in the context of the FSE. As a result, we found some very interesting properties that are not observed in the usual α = 2 quantum mechanics. Probably the most interesting is the presence of tunneling through delta and double delta potentials even at zero energy. Moreover, in the case of the double delta potential, this zero energy tunneling is independent of the relation of the two delta functions. In Lin et al. [14] the problem of calculating the transmission coefficient in FQM for the double delta potential has been addressed; however, the authors have used that same local approach that Jeng et al. [11] have shown to be wrong. As expected in this case, our results differs from theirs. We organized this paper as follows. Firstly, and for the sake of completeness, we reproduce the solution of the FSE for the delta and double delta potentials, as given in [13], presenting their respective solutions in terms of Fox’s H-function. Some calculations and properties of the Fox’s H-function are given in the Appendixes. Then we study the asymptotic behaviour of those solutions, calculate the reflection and transmission coefficients, and study some of their properties. The limit α 2 for these coefficients → and the boundary conditions satisfied by the solutions are also discussed in two other Appendixes.

2. The Fractional Schr¨odinger Equation

The one-dimensional FSE is ∂ψ(x, t) i~ = D ( ~2 )α/2ψ(x, t)+ V (x)ψ(x, t), (1) ∂t α − △ 2 ~2 α/2 where 1 <α 2, Dα is a constant, = ∂x is the Laplacian, and ( ) is the Riesz fractional derivative≤ [15], that is, △ − △

+∞ ~2 α/2 1 ipx/~ α ( ) ψ(x, t)= ~ e p φ(p, t)dp, (2) − △ 2π Z−∞ | | where φ(p, t) is the Fourier transform of the wave function,

+∞ +∞ −ipx/~ 1 ipx/~ φ(p, t)= e ψ(x, t)dx, ψ(x, t)= ~ e φ(p, t)dp. (3) Z−∞ 2π Z−∞ The time-independent FSE is D ( ~2 )α/2ψ(x)+ V (x)ψ(x)= Eψ(x). (4) α − △ In the momentum representation, this equation is written as (W φ)(p) D p αφ(p)+ ∗ = Eφ(p), (5) α| | 2π~

2 where (W φ)(p) is the convolution ∗ +∞ (W φ)(p)= W (p q)φ(q)dq, (6) ∗ Z−∞ − and W (p)= [V (x)] is the Fourier transform of the potential V (x). Solutions ofF the FSE for delta and double delta potentials are given in [13] in the situations of bound and scattering states. Since we need to study the asymptotic behaviour of these solutions in order to ﬁnd the transmission coeﬃcients, and for the sake of completeness, we will reproduce the calculations of the wave functions in the case of scattering states.

2.1. FSE for Delta Potential Let us consider the case V (x)= V0δ(x), (7) where δ(x) is the Dirac delta function and V0 is a constant. Its Fourier transform is W (p)= V and the convolution (W φ)(p) is 0 ∗ (W φ)(p)= V K, (8) ∗ 0 where the constant K is +∞ K = φ(q)dq. (9) Z−∞ The FSE in the momentum representation (5) is E p α φ(p)= γK, (10) | | − Dα − where V γ = 0 . (11) 2π~Dα Since we are interested in scattering states, we will consider that E > 0 and write E = λα, (12) Dα where λ> 0. Since f(x)δ(x)= f(0)δ(x), the solution of Eq.(10) in this case is γK φ(p)= − +2π~C δ(p λ)+2π~C δ(p + λ), (13) p α λα 1 − 2 | | − where C1 and C2 are arbitrary constants and the constant 2π~ was introduced for later convenience. Using this in Eq.(9) gives that

+∞ dp K = γK +2π~C1 +2π~C2, (14) − Z p α λα −∞ | | −

3 where the integral is interpreted in the sense of Cauchy principal value, and it gives +∞ dp +∞ dq π π =2λ1−α = 2λ1−α cot , (15) Z p α λα Z qα 1 − α α −∞ | | − 0 − where we have used formula 3.241.3 (pg. 322) of [16] - see Eq.(89). The constant K is therefore 2π~(C + C )αλα−1 K = 1 2 , (16) αλα−1 2πγ cot(π/α) − and we have 2π~γ(C + C )αλα−1 1 φ(p)=2π~C δ(p λ)+2π~C δ(p + λ) 1 2 . (17) 1 − 2 − (αλα−1 2πγ cot(π/α)) ( p α λα) − | | − Next we need to calculate the inverse Fourier transform of φ(p) to obtain ψ(x), that is, 2πγα(C + C ) λx ψ(x)= C eiλx/~ + C e−iλx/~ 1 2 J , (18) 1 2 − (αλα−1 2πγ cot(π/α)) α ~ − where Jα(w) is the Cauchy principal value of the integral 1 +∞ cos wq Jα(w)= dq, (19) π Z qα 1 0 − and such that +∞ ipx/~ e 1−α dp =2πλ Jα(λx/~). (20) Z p α λα −∞ | | − The above integral is calculated in the Appendix B, and is given by Eq.(91). Using this result, and the deﬁnition of γ in Eq.(11) and λ in Eq.(12), we can write that

with H2,1 − denoting a Fox’s H-function (see Appendix A), and 2,3 · − − α 1 −1 E α π Ω = cot , (23) α U − α and α/(α−1) V0 U = 1/α . (24) α~Dα

4 2.2. FSE for Double Delta Potential Now let the potential be given by V (x)= V [δ(x + R/2)+ µδ(x R/2)], (25) 0 − with µ, V0 and R real constants. When V0 < 0 this potential can be seen as a model for + the one-dimensional limit of the molecular ion H2 [17]. The parameter R is interpreted as the internuclear distance and the coupling parameters are V0 and µV0. Its Fourier transform is ipR/2~ −ipR/2~ W (p)= V0e + V0µe (26) and for the convolution (W φ)(p)= V eipR/2~K (R)+ V µe−ipR/2~K (R), (27) ∗ 0 1 0 2 where K1(R) and K2(R) are constants given by +∞ −iRq/2~ K1(R)= K2( R)= e φ(q)dq. (28) − Z−∞ The FSE in momentum space is

α E iRp/2~ −iRp/2~ p φ(p)= γe K1(R) γµe K2(R), (29) | | − Dα − − where we used the notation introduced in Eq.(11). Since we are interested in scattering states, we have E > 0 and we write λ as in Eq.(12) and for the solution of Eq.(29) we have γeiRp/2~K (R) µγe−iRp/2~K (R) φ(p)=2π~C δ(p λ)+2π~C δ(p + λ) 1 2 . (30) 1 − 2 − p α λα − p α λα | | − | | − Using this expression for φ(p) in Eq.(28) of deﬁnition of K1(R) and K2(R) we have 1−α 1−α ~ ~ ′ (1+2πγλ Jα(0))K1(R)+ µ2πγλ Jα(λR/ )K2(R)=2π C1, (31) 1−α ~ 1−α ~ ′ 2πγλ Jα(λR/ )K1(R)+(1+ µ2πγλ Jα(0))K2(R)=2π C2, (32) where ′ −iRλ/2~ iRλ/2~ ′ iRλ/2~ −iRλ/2~ C1 = C1e + C2e , C2 = C1e + C2e . (33) In order to write the solution of these equations it is convenient to deﬁne

α−1 λα−1 1 E α ε = = , (34) 2πγ α U where U was deﬁned in Eq.(24), in such a way that we have 2π~ε K (R)= [(εµ−1 + J (0))C′ J (λR/~)C′ ], (35) 1 W α 1 − α 2 2π~ε K (R)= [(ε + J (0))C′ J (λR/~)C′ ], (36) 2 µW α 2 − α 1

5 where W =(ε + J (0))(εµ−1 + J (0)) (J (λR/~))2. (37) α α − α Using K1(R) and K2(R) in Eq.(30) gives φ(p). Then, for ψ(x), we have

iλx/~ −iλx/~ ψ(x)= C1e + C2e 1 λ x + R/2 + [(εµ−1 + J (0))C′ J (λR/~)C′ ]Φ | | 2αW α 1 − α 2 α ~ (38) 1 λ x R/2 + [(ε + J (0))C′ J (λR/~)C′ ]Φ | − | , 2αW α 2 − α 1 α ~ where we have expressed the result in terms of the function Φα deﬁned in Eq.(22).

3. Calculation of the Transmission Coeﬃcients

In order to calculate the transmission coeﬃcients, we need to know the asymptotic bahaviour of the solutions. The asymptotic behaviour of Fox’s H-function is given, if ∆ > 0, by Eq.(82) or Eq.(84) according to ∆∗ > 0or∆∗ = 0, respectively – see Eq.(81). In Φ (λ x /~) we have the diﬀerence between two Fox’s H-functions of the form α | | (1, 1), (1,µ) H2,1 wα , 2,3 (1,α), (1, 1), (1,µ)

for µ = (2+ α)/2 and µ = (2 α)/2. In both cases we have ∆ = α> 0, but ∆∗ = 0 for µ = (2+α)/2 and ∆∗ > 0 for µ−= (2 α)/2. Therefore, using Eq.(82) when µ = (2 α)/2 − − and Eq.(84) when µ =(2+ α)/2 we have, respectively, that (1, 1), (1, (2+ α)/2) 2w H2,1 wα = sin w + o(1), w , (39) 2,3 (1,α), (1, 1), (1, (2+ α)/2) α | |→∞

(1, 1), (1, (2 α)/2) H2 ,1 wα − = o(1), w , (40) 2,3 (1,α), (1, 1), (1, (2 α)/2) | |→∞ − and then λ x λ x Φ | | = 2 sin | | + o x −1 , x . (41) α ~ ~ | | | |→∞ 3.1. Transmission Coeﬃcient for the Delta Potential The behaviour of the wave function ψ(x) given by Eq.(21) for x is therefore → ±∞ λx ψ(x)= C eiλx/~ + C e−iλx/~ Ω (C + C ) sin + o x−1 , x . (42) 1 2 ± α 1 2 ~ → ±∞ or ψ(x)= Aeiλx/~ + Be−iλx/~ + o x−1 , x , (43) → −∞ ψ(x)= Ceiλx/~ + De−iλx/~ + ox−1, x + , (44) → ∞

6 where we deﬁned

A = C + i(C + C )Ω /2, B = C i(C + C )Ω /2, (45) 1 1 2 α 2 − 1 2 α C = C i(C + C )Ω /2,D = C + i(C + C )Ω /2. (46) 1 − 1 2 α 2 1 2 α Now let us consider the situation of particles coming from the left and scattered by the delta potential. In this case D = 0 (no particles coming from the right) and B = rA and C = tA, where the reflexion and transmission coefficients are given by = r 2 R T R | | and = t 2 (see, for example, [18]). The result is T | | iΩ 1 r = − α , t = , (47) 1+ iΩα 1+ iΩα and then 2 Ωα 1 = 2 , = 2 . (48) R 1 + Ωα T 1 + Ωα In Fig.(1) we show the behaviour of these coefficients for different values of α. This plot and the other ones of this paper have been done by means of numerical integration of Eq.(19) using Mathematica 7.

Figure 1: Reflection and transmission coefficients as function of E/U, as given by Eq.(48), for different values of α.

We must note that transmission coeﬃcient has a very interesting behaviour at zero energy. If we take the limit E 0 in the expression for Ω in Eq.(23) we see that → α π lim Ωα = tan , (49) E→0 − α

7 and then for the transmission coeﬃcient we have T π lim = cos2 . (50) E→0 T α This is an unexpected and very interesting eﬀect, which demands further interpretation (see Conclusions).

3.2. Transmission Coeﬃcient for the Double Delta Potential Let us introduce the following notations:

εµ−1 + J (0) ε + J (0) J (λR/~) = α , = α , = α . (51) U αW V αW X αW

The asymptotic behaviour of the wave function ψ(x) given by Eq.(38) for x is therefore → ±∞

iλx/~ −iλx/~ ′ ′ λ x + R/2 ψ(x)= C1e + C2e +(UC1 XC2) sin | ~ | − (52) λ x R/2 +(VC′ XC′ ) sin | − | + o x−1 , x , 2 − 1 ~ → ±∞ or

ψ(x)= A′eiλx/~ + B′e−iλx/~ + o x−1 , x , (53) → −∞ ψ(x)= C′eiλx/~ + D′e−iλx/~ + ox−1, x + , (54) → ∞ where we deﬁned

′ ′ A = C1 + M1, B = C2 + M2, (55) C′ = C M ,D′ = C M , (56) 1 − 1 2 − 2 and

M1 = i(ρC1 + σC2 + iτC2), (57) M = i(σC + ρC iτC ), (58) 2 − 1 2 − 1 with + λR + λR ρ = U V cos , σ = U V cos , 2 − X ~ 2 ~ − X (59) λR τ = U−V sin . 2 ~ As in the case of the delta potential, let us consider the situation of particles coming from the left and scattered by the double delta potential. In complete analogy we have D′ = 0 (no particles coming from the right) and B′ = rA′ and C′ = tA′, where the

8 reﬂexion and transmission coeﬃcients are given by = r 2 and = t 2. The R T R | | T | | result is 2(τ + iσ) (ρ2 σ2 τ 2 + 1) r = , t = − − . (60) (ρ2 σ2 τ 2 1) 2iρ −(ρ2 σ2 τ 2 1) 2iρ − − − − − − − − and and can be written as R T 4(σ2 + τ 2) (ρ2 σ2 τ 2 + 1)2 = , = − − . (61) R (ρ2 σ2 τ 2 + 1)2 + 4(σ2 + τ 2) T (ρ2 σ2 τ 2 + 1)2 + 4(σ2 + τ 2) − − − − We can simplify these expressions a little bit once we note that

sin2 (λR/~) ρ2 σ2 τ 2 = , (62) − − α2W and then 2 ∆α 1 = 2 , = 2 , (63) R 1 + ∆α T 1 + ∆α with 4α4W 2(σ2 + τ 2) ∆2 = . (64) α (α2W + sin2 (λR/~))2 In Fig.(2) we show the behaviour of the transmission coefficient for different values of α and in Fig.(3) for different values of µ.

Figure 2: Transmission coeﬃcients as function of λR/2~, as given by Eq.(63), for diﬀer- ent values of α, when 2πγ = 10 and µ = 1.

9 Figure 3: Transmission coeﬃcients as function of λR/2~, as given by Eq.(63), for diﬀer- ent values of µ, when 2πγ = 20 and α =1.8.

As in the case of the delta potential, we have a very interesting behaviour for these coeﬃcients when E 0. Firstly, using Eq.(99) in the Appendix B, we have → J (λR/~)= J (0) + A λα−1 + A λ2 + (λ3α−1), (65) α α 1 2 O where (R/~)α−1 (R/~)2 cot (3π/α) A = , A = . (66) 1 2Γ(α)cos(πα/2) 2 2α Using this, the expression Eq.(37) for W gives

W = J (0)(H(1+µ−1) 2A )λα−1 +(H2µ−1 A2)λ2(α−1) 2J (0)A λ2 + (λα+1), (67) α − 1 − 1 − α 2 O where H =1/2πγ. Then, with some calculations, we obtain that

4α2W 2(σ2 + τ 2)= B λ2(α−1) + B λα+1 + (λ2α), (68) 1 2 O with

B =(H(1+ µ−1) 2A )2, B = 2(H(1+ µ−1) 2A )(2A + J (0)(R/~)2), (69) 1 − 1 2 − − 1 2 α and α2W 2(ρ2 σ2 τ 2 +1) = B′ λ2(α−1) + B′ λ4(α−1) + B′ λα+1 + (λ2α), (70) − − 1 2 3 O

10 with ′ 2 2 −1 2 ′ 2 2 −1 2 B = α J (0)(H(1 + µ ) 2A1) , B = α (H µ A1) , 1 α − 2 − (71) B′ = 2αJ (0)(H(1 + µ−1) 2A )(2α2J (0)A (R/~)2). 3 − α − 1 α 2 − Using these results we can easily see that, for E/D = λα 0, α → 2 1 lim ∆α = 2 2 , (72) E→0 α Jα(0) and, since J (0) = (1/α)cot(π/α), that α − π lim T = cos2 . (73) E→0 α This the same result we obtained for the zero energy limit of the transmission coeﬃcient for a single delta potential. Moreover, this limit does not depends on µ, which is the parameter that relates the two delta functions in the potential in Eq.(25). Again, this is a very interesting and unexpected result.

4. Conclusions

The tunneling effect in fractional quantum mechanics has some very interesting properties which are not observed in the usual α = 2 quantum mechanics. The most interesting is the presence of tunneling through delta and double delta potentials even at zero energy. Moreover, in the case of the double delta potential, this zero energy tunneling is independent of the relation of the two delta functions. Let us give a possible explanation to these results. FQM was defined from the point of view of path integrals. As well-known, in this approach, when the sum is taken over paths of Brownian motion type, we have the standard quantum mechanics. On the other hand, in FQM the sum is taken over the paths of Lévy flights [5], which are generalizations of Brownian motion, and such that the corresponding probability distribution has infinite variance. By means of Lévy flights, there is a not negligible probability of a particle reaching faraway points in a single jump, in contrast to a random walk of Brownian motion type [19]. In FQM, an uncertainty principle still holds, but with an appropriated modification, that is, we have [6] ~ ∆x µ 1/µ ∆p µ 1/µ > , µ < α, 1 <α 2, (74) h| | i h| | i (2α)1/µ ≤ and such that in the standard quantum mechanics limit we can also have µ = α = 2. Thus, even when E = 0, the particle can have energy ∆E = ∆p µ 2/µ/2m and momentum ∆p µ 1/µ. This may not be enough for tunneling throughh| a delta| i potential in the standardh| case,| i but in the fractional one, where long jumps of Lévy flights enter the sum in the path integral, this may be responsible for the tunneling with probability cos2 (π/α). Acknowledgements: We are grateful to the referees for many valuable suggestions.

11 A. Fox’s H Function − The Fox’s H function, also known as H function or Fox’s function, was introduced in − − the literature as an integral of Mellin-Barnes type [20]. Let m, n, p and q be integer numbers. Consider the function m n Γ(b + B s) Γ(1 a A s) i i − i − i Yi=1 Yi=1 Λ(s)= q p (75) Γ(1 b B s) Γ(a + A s) − i − i i i i=Ym+1 i=Yn+1 with 1 m q and 0 n p. The coeﬃcients A and B are positive real numbers; a ≤ ≤ ≤ ≤ i i i and bi are complex parameters. The Fox’s H function, denoted by, − m,n m,n (ap, Ap) m,n (a1, A1), , (ap, Ap) Hp,q (x)= Hp,q x = Hp,q x ··· (76) (bq, Bq) (b1, B1), , (bq, Bq) ··· is deﬁned as the inverse Mellin transform, i.e.,

m,n 1 −s Hp,q (x)= Λ(s) x ds (77) 2πi ZL where Λ(s) is given by Eq.(75), and the contour L runs from L i to L+i separating − ∞ ∞ the poles of Γ(1 a A s), (i = 1,...,n) from those of Γ(b + B s), (i = 1,...,m). − i − i i i The complex parameters ai and bi are taken with the imposition that no poles in the integrand coincide. There are some interesting properties associated with the Fox’s H function. We consider here the following ones: −

P.1. Change the independent variable Let c be a positive constant. We have

m,n (ap, Ap) m,n c (ap, c Ap) Hp,q x = cHp,q x . (78) (bq, Bq) (bq, c Bq)

To show this expression one introduce a change of variable s cs in the integral of inverse Mellin transform. →

P.2. Change the ﬁrst argument Set α R. Then we can write ∈ α m,n (ap, Ap) m,n (ap + αAp, Ap) x Hp,q x = Hp,q x . (79) (bq, Bq) (bq + αBq, Bq)

To show this expression ﬁrst we introduce the change a a + αA and take s s α p → p p → − in the integral of inverse Mellin transform.

12 P.3. Lowering of Order

If the ﬁrst factor (a1, A1) is equal to the last one, (bq, Bq), we have

m,n (a1, A1), , (ap, Ap) m,n−1 (a2, A2), , (ap, Ap) Hp,q x ··· = Hp−1,q−1 x ··· . (b1, B1), , (bq−1, Bq−1)(a1, A1) (b1, B1), , (bq−1, Bq−1) ··· ··· (80)

To show this identity is suﬃcient to simplify the common arguments in the Mellin-Barnes integral.

P.4. Asymptotic Expansions The asymptotic expansions for Fox’s H-functions have been studied in [21]. Let ∆ and ∆∗ be deﬁned as

q p n p m q ∆= B A , ∆∗ = A A + B B . (81) i − i i − i i − i Xi=1 Xi=1 Xi=1 i=Xn+1 Xi=1 i=Xm+1 If ∆ > 0 and ∆∗ > 0 we have [22]

n Hm,n(x)= h x(ar −1)/Ar + o x(ar −1)/Ar , x . (82) p,q r | |→∞ Xr=1 where m n 1 Γ(bj + (1 ar)Bj/Ar) Γ(1 aj (1 ar)Aj/Ar) h = j=1 − j=1,j6=r − − − , (83) r A Qp Γ(a (1 a )A /A Q) q Γ(1 b (1 a )B /A ) r j=n+1 j − − r j r j=m+1 − j − − r j r Q Q and if ∆ > 0 and ∆∗ = 0 we have [22]

n m,n (ar −1)/Ar (ar −1)/Ar Hp,q (x)= hrx + o x Xr=1 (84) + Ax(ν+1/2)/∆ c exp[i(B + Cx1/∆)] d exp[ i(B + Cx1/∆)] 0 − 0 − + o x(ν+1/2)/|∆| , x , | |→∞ where

p m c = (2πi)m+n−p exp πi a b , 0 r − j r=Xn+1 Xj=1 p m d =( 2πi)m+n−p exp πi a b πi , 0 − − r − j r=Xn+1 Xj=1 p q 1 ∆∆ (ν+1/2)/∆ A = (2π)(p−q+1)/2∆−ν A−ar+1/2 Bbj −1/2 , 2πi∆ r j δ Yr=1 Yj=1

13 (2ν + 1)π ∆∆ 1/∆ B = , C = , 4 δ p q q p p q δ = A −Al B Bj , ν = b a + − . | l| | j| j − j 2 Yl=1 Yj=1 Xj=1 Xj=1

P.5. Series Expansion In [20] we can see that in some cases there is a series expansion for Fox’s H-function. m For example, when the poles of j=1 Γ(bj + Bjs) are simple, we can write Q m ∞ m,n (bj +ν)/Bj Hp,q (x)= hjνx , (85) Xj=1 Xν=0 where ν m n ( 1) Γ(bi Bi(bj + ν)/Bj) Γ(1 ai + Ai(bj + ν)/Bj) h = − i=1,i6=j − i=1 − . (86) jν ν!B Qq Γ(1 b + B (b + ν)/BQ) p Γ(a A (b + ν)/B ) j i=m+1 − i i j j i=n+1 i − i j j Q Q B. Calculation of the Integral in Eq.(19)

Let Jα(w) be given by 1 +∞ cos wy Jα(w)= dy. (87) π Z yα 1 0 − Taking the Mellin transform we have that 1 πz +∞ y−z w[Jα(w)](z)= Γ(z) cos dy. (88) M π 2 Z yα 1 0 − This last integral is given by formula 3.241.3 (pg.322) of [16], that is, +∞ xµ−1 π µπ dx = cot , (89) Z 1 xν ν ν 0 − where the integration is understood as the Cauchy principal value3. Therefore we have 1 π(1 z) π(1 z) [J (w)](z)= Γ(z) sin − cot − . (90) Mw α −α 2 α Using the relation 2 sin A cos B = sin (A + B)+sin (A B) and writing the sine function in terms of the product of gamma functions we can write− that 1−z 1−z 1 Γ(z)Γ α Γ 1 α w[Jα(w)](z)= − M −2α Γ (1 z) (2+α) Γ 1 (1 z) (2+α) − 2α − − 2α 1−z 1−z 1 Γ(z)Γ α Γ 1 α + − = F2(z). 2α Γ (1 z) (2−α) Γ 1 (1 z) (2−α) − 2α − − 2α 3 We remember that in the inversion of the Fourier transform the integration is to be done in the sense of the Cauchy principal value [23].

14 Taking the inverse Mellin transform and using the deﬁnition of the Fox’s H-function we have that

1 2,1 (1 1/α, 1/α), (1 (2 + α)/2α, (2+ α)/2α) Jα(w)= H w − − + −2α 2,3 (0, 1), (1 1/α, 1/α), (1 (2 + α)/2α, (2 + α)/2α) − − 1 (1 1/α, 1/α), (1 (2 α)/2α, (2 α)/2α) + H2,1 w − − − − . 2α 2,3 (0, 1), (1 1/α, 1/α), (1 (2 α)/2α, (2 α)/2α) − − − −

Using the properties given by Eqs.(78,79) and replacing w by w since Jα( w)= Jα(w) we obtain | | − 1 (1, 1), (1, (2+ α)/2) J (w)= H2,1 w α + α −2 w 2,3 | | (1,α), (1, 1), (1, (2+ α)/α) | | (91) 1 (1, 1), (1, (2 α)/2) + H2,1 w α − . 2 w 2,3 | | (1,α), (1, 1), (1, (2 α)/2) − | | Let us see what happens in the particular case α = 2. From the deﬁnition of Fox’s H-function we can see that (1, 1), (1, 0) H2,1 w 2 =0 (92) 2,3 | | (1, 2), (1, 1), (1, 0)

and that (1, 1), (1, 2) (1, 1) (0, 1) H2,1 w2 = H1,1 w2 = w2H1,1 w2 . 2,3 (1, 2), (1, 1), (1, 2) 1,2 (1, 1), (1, 2) 1,2 (0, 1), ( 1, 2) −

But [20] (0, 1) H1,1 z = E (z), (93) 1,2 − (0, 1), (1 b, a) a,b − where Ea,b(z) is the two-parameter Mittag-Leﬄer function. However, it is known [24] that sinh √z E (z)= . (94) 2,2 √z Consequently, we have

(1, 1), (1, 2) H2,1 w2 = w 2E ( w 2)= w sin w . (95) 2,3 (1, 2), (1, 1), (1, 2) | | 2,2 −| | | | | |

15 We are also interested in the expression of Jα(w) for small w. From Eq.(91) we see that we need to know the behaviour of (1, 1), (1,µ) H2,1 w α 2,3 | | (1,α), (1, 1), (1,µ)

for small w. This is given by the series expansion from Eq.(85), which gives

(1, 1), (1,µ) H2,1 z 2,3 (1,α), (1, 1), (1,µ)

Γ(1 1/α)Γ(1/α) z1/α Γ(1 2/α)Γ(2/α) z2/α = − − Γ(1 µ/α)Γ(µ/α) α − Γ(1 2µ/α)Γ(2µ/α) α − − (98) Γ(1 3/α)Γ(3/α) z3/α + − + (z4/α) Γ(1 3µ/α)Γ(3µ/α) 2α O − Γ(1 α)Γ(1) Γ(1 2α)Γ(2) Γ(1 3α)Γ(3) z3 + − z − z2 + − + (z4). Γ(1 µ)Γ(µ) − Γ(1 2µ)Γ(2µ) Γ(1 3µ)Γ(3µ) 2 O − − − Using this in Eq.(91) we arrive, after some manipulations, to 1 cot(3π/α) J (w)= J (0) + wα−1 + w2 + (w3α−1), (99) α α 2Γ(α)cos(πα/2) 2α O where 1 π J (0) = cot . (100) α −α α C. The Limit α =2

Let us calculate the transmission coefficients in the standard quantum mechanical limit and see that we recover the usual results. Firstly, let us consider the delta potential. The transmission coefficient is given by Eq.(48), so that we need to calculate Ω2 in this case, where Ωα is given by Eq.(23). The result is that E −1 mV 2 Ω = , U = 0 , (101) 2 U 2~2 where we used the definition of U in Eq.(24) and D2 =1/2m, and such that 1 = , (102) 2 ~2 T 1+(mV0 /2 E) which is the well-known result [18]. Now let us consider the transmission coefficient for the double delta potential. Let us also consider the case µ = 1 since the result in this case is well-known [25]. In order to calculate ∆2 in Eq.(63) we need Eq.(96), which by the way gives

J2(0) = 0. (103)

16 When µ = 1 we have 1 = = , τ =0, (104) U V 2W and then 64W 2σ2 ∆2 = . (105) 2 [4W + sin2(λR/~)]2 But in this case Eq.(37) together with Eq.(96) gives

4W + sin2(λR/~)=4ǫ2. (106)

From Eq.(51) and Eq.(59) we also have that

λR 1 λR 2 4W 2σ2 = ǫ cos + sin . (107) ~ 2 ~

Then for ∆2 we have 1 λR 1 λR 2 ∆2 = cos + sin . (108) 2 ǫ ~ 2ǫ2 ~ Let us change the notation a little bit in order to compare with the standard result in the literature. Let us deﬁne λR µ µ = , β = 0 . (109) 0 2~ ǫ Using this notation in the above expression for ∆ , we have for in Eq.(63) that 2 T 4 µ0 = 4 2 2 , (110) T µ0 + [βµ0 cos2µ0 +(β /2) sin 2µ0] which is the result in [25], pg 160 (where µ0 = µ).

D. The Boundary Conditions

In [13] we have discussed how the Riesz fractional derivative can be written in terms of the Riesz potentials, that is, for 0 <α< 1 we have d ( )α/2ψ(x)= ˜ 1−αψ(x), (111) −△ dxR and for 1 <α< 2 we have d2 ( )α/2ψ(x)= 2−αψ(x), (112) −△ −dx2 R

′ where α ψ(x) is the Riesz potential of ψ(x) of order α′ given by [15] R +∞ α′ 1 ψ(ξ) ψ(x)= ′ dξ, (113) R 2Γ(α′)cos(α′π/2) Z x ξ 1−α −∞ | − |

17 ′ for 0 <α′ < 1, and ˜α ψ(x) its conjugated Riesz potential given by R +∞ ˜ α′ 1 sign(x ξ)ψ(ξ) ψ(x)= − ′ dξ. (114) R 2Γ(α′) sin (α′π/2) Z x ξ 1−α −∞ | − | If we use these expressions for the Riesz fractional derivative in the FSE for the delta potential V (x)= V0δ(x) and integrate as usual from ǫ to +ǫ and take the limit ǫ 0 we obtain that − → d 2−α d 2−α V0 ψ(x) ψ(x) = α ψ(0). (115) dxR + − dxR − ~ D 0 0 α

This condition and the continuity one ψ(0−)= ψ(0+ ) are the boundary conditions to be satisﬁed by ψ(x). There is an important point to be noted here: the expression d d 2−αψ(x)= 1−(α−1)ψ(x) dxR dxR is not the Riesz fractional derivative of order α 1, which is given, for 0 < α 1 < 1, − − by d ˜ 1−(α−1)ψ(x). dxR Therefore, it is wrong to write the condition (115) as

(α−1)/2 (α−1)/2 V0 ( ) ψ(x) ( ) ψ(x) − = ψ(0), (116) −△ 0+ − −△ 0 ~αD α ′ with ( )α /2 being the Riesz fractional derivative. Maybe in another version of FQM −△ involving a fractional derivative deﬁned in another sense, it can holds one such condition, but this is not the case when it comes to the Riesz fractional derivative. In [14] the authors have used a boundary condition of the above type in their attempt to solve the problem for the double delta potential. Besides the already discussed problem with their local approach, it also seems that the use of that boundary condition is not justiﬁed. Let us consider the solution in the case of the delta potential and show that it satisfy Eq.(115). In [13] we have seen that

d i ∞ 2−αψ(x)= e−ipx/~ p α−1 sign(p)φ(p) dp. (117) ~α dxR 2π Z−∞ | | Using φ(p) given by Eq.(17) we obtain, after calculating the integrals with the delta functions, that

d λ α−1 λ α−1 2−αψ(x)=2iC eiλx/~ 2iC e−iλx/~ dxR 1 ~ − 2 ~ (118) α λ α−1 λx + (C + C )Ω Ξ , π 1 2 α ~ α ~

18 where ∞ qα−1 Ξα (w)= sin wq dq. (119) Z qα 1 0 − This integral can be calculated in a way analogous as we did in Appendix B. The result is π (0, 1), (0, (2+ α)/2) Ξ (w)= sign(w) H2,1 w α α 2 2,3 | | (0,α), (0, 1), (0, (2+ α)/2) (120) (0, 1), (0 , (2 α)/2) H2,1 w α − . − 2,3 | | (0,α), (0, 1), (0, (2 α)/2) −

In order to study the limit x 0± we need to use →

2,1 α (0, 1), (0, β) β lim H2,3 w = , (121) w→0 | | (0,α), (0, 1), (0, β) α

which can be calculated using the deﬁnition of the Fox’s H-function and the residue theorem (the RHS comes from the residue at w = 0), and which gives π lim Ξα(w)= . (122) w→0± ± 2 Using this result, we have

α−1 d 2−α α λ lim ψ(x)= (C1 + C2)Ωα . (123) x→0± dxR ± 2 ~ On the other hand, ψ(0) can be calculated using Eq.(100) in Eq.(18), which gives

α−1 α α−1 (C1 + C2)αΩαλ ~ Dα λ ψ(0) = = α(C1 + C2)Ωα , (124) 2πγ V0 ~ where we used the deﬁnition of γ in Eq.(11). We see, therefore, that Eq.(115) is satisﬁed.

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