International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 DOI: 10.5923/j.ijtmp.20150505.03
Some Solutions to the Fractional and Relativistic Schrödinger Equations
Yuchuan Wei
Department of Radiation Oncology, Wake Forest University, NC
Abstract Laskin introduced the fractional quantum mechanics and several common problems were solved in a piecewise fashion. However, Jeng et al pointed out that it was meaningless to solve a nonlocal equation in a piecewise fashion and that all the solutions in publication were wrong except the solution for the delta function potential, which was obtained in the momentum space. Jeng’s critique results in a crisis of fractional quantum mechanics, that is, in mathematics it is quite difficult to find solutions to the fractional Schrödinger equation and in physics there is no realization for the fractional quantum mechanics. In order to eliminate this crisis, this paper reports some analytic solutions to the fractional Schrödinger equation without using piecewise method, and introduces the relativistic Schrödinger equation as a realization of the fractional quantum mechanics. These two sister equations should be studied at the same time. Keywords Fractional quantum mechanics, Fractional Schrödinger equation, Relativistic quantum mechanics, Relativistic Schrödinger equation
themselves provided a solution for the one dimensional 1. Introduction harmonic oscillator potential for the case α =1 [4]. Readers have been looking forward to some other solutions In 2000, Laskin introduced the fractional quantum to the fractional Schrödinger equation since the simple mechanics [1-3]. As the first example he solved the infinite solutions for the infinite square well potential were square well problem in a piecewise fashion [3]. Since then, disproved. Jeng et al [4] also showed their concern that there the piecewise method has been widely used in this field. In was no realization of the fractional quantum mechanics. 2010, however, Jeng, et al [4] criticized that it was Jeng’s critique resulted in a crisis within fractional meaningless to solve a nonlocal equation in a piecewise quantum mechanics. In mathematics it is not easy to find a fashion and they demonstrated thatit was impossible for the solution to the fractional Schrödinger equation, and in ground state function to satisfy the fractional Schrödinger physics it is not easy to find realizations for the fractional equation near the boundary inside the well. In a series of quantum mechanics. In order to eliminate this crisis, this papers [5-8], Bayin insisted that he explicitly completed the paper reports some solutions to the fractional Schrödinger calculation in Jeng’s paper and the wave functions did satisfy equation without using the piecewise method, and introduces the fractional Schrödinger equation inside the well. Hawkins the relativistic Schrödinger equation [18-21], as a realization and Schwarz [9] claimed that Bayin’s calculation contained of the fractional quantum mechanics. Several solutions for serious mistakes. Luchko [10] provided some evidence that the solution did not satisfy the equation outside the well. On the relativistic quantum mechanics are also presented. the other hand, Dong [11] re-obtained the Laskin’s solution by solving the fractional Schrödinger equation with the path 2. The Relativistic Schrödinger integral method. It is not easyfor readers to judge their mathematical argument [12, 13], but weagree with Jeng that Equation: A Realization of the the piecewise method to solve the equation is wrong, since Fractional Schrödinger Equation recently we explicitly and inarguably showed that the In this section we will list the standard, fractional, and Laskin’s functions did not satisfy the fractional Schrödinger relativistic Schrödinger equations in one- and equation with α =1 anywhere on the x-axis [14]. three-dimensional spaces, and explain why we claim the According to Jeng, et al [4], only the solution for the delta relativistic Schrödinger equation is an approximate function potential [15-17] was acceptable and they realization for the fractional Schrödinger equation.
* Corresponding author: 2.1. The Schrödinger Equation [email protected] (Yuchuan Wei) Published online at http://journal.sapub.org/ijtmp In the standard quantum mechanics [22, 23], the Copyright © 2015 Scientific & Academic Publishing. All Rights Reserved time-independent Schrödinger equation is
88 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations
HEψψ(rr) = ( ) , (1) where ψ (r) is a wave function defined in the 3 dimensional Euclideanspace R3, E is an energy, and r is a vector in the 3 dimensional space. The Hamiltonian operator H= TV + (r) (2) is the summation of the kinetic energy operator and the potential energy operator. The standard kinetic energy operator is 22 p 2 T = =−∇, (3) 22mm =−∇ where p i is the momentum operator. As usual, m is the mass of a particle and is the reduced Plank constant. The one dimensional time-independent Schrödinger equation is ℏ 22d −ψ( x) += Vx() ψψ( x) E( x) , (4) 2m dx2 where the wave function ψ ( x) and the potential Vx() are defined on the x-axis.
2.2. The Fractional Schrödinger Equation and Its Scaling Property In 2000 [1-3], Laskin generalized the classical kinetic energy and momentum relation (3) to
α α α 2||p 22 /2 Tαα= D||p =χ mc= Dα −∇ , (5) α mc ( ) 2 α where α is the fractional parameter, the coefficient ≡ χ , χ is a positive number dependent on Dα α mc/( mc ) α α , and c is the speed of the light. Originally Laskin [1-3] ever required the fractional parameter 12<≤α , but in this paper we allow 0 <α <∞, as in [4, 9], with an emphasis on the simplest nonlocal case α =1 .
In the case α = 2 , taking χ =1/2, the fractional kinetic energy is the same as the classical kinetic energy 2 2 2 221 ||pp T22= D||p = mc = = T . (6) 22mc m In the case α =1 , taking χ =1, the fractional kinetic energy is the approximate kinetic energy in the extremely 1 relativistic case, ||p T= D||pp = mc2 = c ||. (7) 11 mc The definition of the fractional kinetic energy operator is
1 α p 3 Tααψϕ()r= D (pp ) | | exp(id⋅ r ) p ∫R3 , (8) (2π )3 where 1 p ϕψ()p= (r )exp(−⋅id rr ) 3 ∫R3 (9) (2π )3 is called the wavefunction in the momentum space [22, 23]. The fractional Schrödinger equation is
International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 89
Hαψψ(rr) = E ( ) (10)
Hαα= TV + ()r . (11) When α = 2 , the fractional Schrödinger equation recovers the standard Schrödinger equation; when α =1 , the fractional Schrödinger equation is
H1ψψ(rr) = E ( ) . (12) For α =1 , the following scaling property is straightforward.
Scaling property. If a wave function ψ (r) and an energy E is a solution of the fractional Schrödinger equation for the potential V (r)
DE1 |p |ψ ( r ) +=V ( rr) ψψ ( ) () r, (13) then ψ (λr) and λE is a solution of the fractional Schrödinger equation for the potential λV (λr) , i.e.
DE1 ||()prψλ += λλV ( r ) ψ ()λ r λ ψλ() r. (14)
In this paper λ is an arbitrary positive number.
The proof is trivial. From (13) we have 1 D |p |ψλ ( r ) +=V ()λ rrψ (λ ) E ψλ() r. (15) λ 1 2 For a potential satisfying λVV(λrr) = ( ) , such as (1) the coulomb V(r) = − Ze/ r with e the charge of an electron and Z the order number of an atom, or (2) the radial delta function potential V(r) = − Vr0δ () with the constant
V0 > 0 , the scaling property can be described simply as follows.
Scaling property. For a potential V (r) with a property VV(rr) = λ (λ ) , if a wave function ψ (r) and an energy
E is a solution of the fractional Schrödinger equation H1ψψ(rr) = E ( ) , then the wave function ψ (λr) and the energy λE is also a solution.
The one dimensional fractional Schrödinger equation is
Hαψψ( xEx) = ( ) (16)
Hαα= T + Vx(). (17) When α =1 , we have
H1ψψ( xEx) = ( ) (18) d H=+= T Vx() D | p | += Vx () D H + Vx(). (19) 11 1 1dx In this paper, the bold H denotes the Hilbert transform [24] while a normal H denotes the Hamiltonian operator. From the definition of the fractional kinetic energy operator, we have
90 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations
1 ∞∞ T1ψψ( x) = D11| p | ( x) = D| p | ψ( x ') exp(− ipx '/ ) dx ' exp( ipx / ) dp 2π ∫∫−∞ −∞ D ∞∞ = 1 |k | ψ ( x ') exp(− ikxdx ') ' exp( ikxdk ) 2π ∫∫−∞ −∞ D ∞∞ =1 (−−i sgn( k ))( ik ) ψ ( x ') exp( ikx ') dx ' exp( ikx ) dk 2π ∫∫−∞ −∞ D d ∞ ∞ =−−1 (i sgn( k ))ψ ( x ') exp( ikx ')dx' exp( ikx ) dk (20) 2π dx ∫−∞ ∫−∞ dd = D Hψψ( x) = DH( x) 11dx dx dd11 = DxDx11ψψ( )** = ( ) dxππ x dx x −1 = Dx1ψ ( )* π x2 2 where * denotes the convolution, and 1/ x and (1/xx )'= − 1/ are generalized functions [25]. We point out that 22 the generalized function −1/ (2π x ) is the well-known ideal ramp filter, which plays an important role in the theory and applications of Computed Tomography [26, 27]. We will discuss the relationship between fractional quantum mechanics and the computed tomography in another paper. The Dirac delta potential VV( x) = − 0δ ()x with V0 > 0 satisfies V ( xx) = λV (λ ) . Based on the scaling property, if a wave function ψ ( x) and an energy E is a solution of the fractional Schrödinger equation H1ψψ= E with a delta potential well, then the wave function ψ (λx) and the energy λE is also a solution. See Problems 3, 7, & 9.
2.3. The Relativistic Schrödinger Equation According to the special relativity, the revised kinetic energy is [18]
22 24 Tr =p c + mc , (21) where the subscript r means special relativity. For the case of low speed, the relativistic kinetic energy is approximately the summation of the rest energy and the classical kinetic energy (α = 2 ) 2 22p T≈ mc += mc + T , (22) r 2m 2 and for the case of extremely high speed, where the rest energy can be neglected, the relativistic kinetic energy is the fractional kinetic energy with α =1
Tr ≈=||p cT1 . (23)
Generally speaking, if the speed of a particle increases from low to high, the relativistic kinetic energy Tr will approximately correspond to a fractional kinetic energy Tα , whose parameter α changes from 2 to 1. Therefore the relativistic kinetic energy is an approximate realization of the fractional kinetic energy. The definition of the relativistic kinetic energy operator is
22 2 24 Trψψ()rr= − c ∇+ mc () 1 p ,(24) =ϕ()pp22c +⋅ mc 24exp( ir ) d3 p ∫R3 (2π )3
International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 91
where ϕ()p is the wave function in the 3D momentum space. The relativistic Schrödinger equation is
Hrψψ(rr) = E ( ) (25)
Hrr= TV + . (26) Accordingly, the 1D relativistic Schrödingere quationis
Hrψψ( xx) = E ( ) (27)
2 22∂ 24 Hrr=T +=V −+ c m c + Vx() .(28) ∂x2 As two sisters, the relativistic and fractional Schrödinger equations should be studied at the same time.
3. Solutions to the Fractional Schrödinger Equation We will study the one dimensional problems first and then the 3D problems.
3.1. One Dimensional Problems Problem 1. (Thefree particle.) For Vx()= 0, the fractional Schrödinger equation H ψψ= E has the solutions α ψ ( x) = exp(ikx ) (29) α ED= α ( | k |) (30) with −∞ α ED= α () k (32) with k ≥ 0 . Proof. According to the definition of the fractional kinetic energyoperator [1-3], we obviously have 2 2α /2 Hααααψψ( xDTxD) =( ) =( −∇ ) exp(ikx ) , (33) α = Dα ( | k |) ψψ( xEx) = ( ) with −∞ Hααcos( kx) = E sin( kx) , (35) Hα 1= 0. Generally, if Vx()= V0 with V0 a constant, the eigen-functions do not change but the new eigen-energies become α ED= α ( | k |) + V0 . (36) 92 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations Problem 2. (A periodic potential.) For the potential α b Vx( ) = Dα ( k0 ) , (37) b+ cos( kx0 ) where ka0 = π / , a is a positive real number (as throughout this paper), b is a real number, and 0 <α <∞, the fractional Schrödinger equation Hαψψ= E has a solution ψ ( x) = b + cos( kx0 ) (38) α ED= α () k0 . (39) Proof. Since αα |p | cos( kx0 ) = ( k 00 ) cos( kx ) (40) |p |α 1 = 0 We have α Dpα ||ψ α =Dα | p |( b + cos( kx0 )) α =0 + Dα ( k00 ) cos( kx ) αα (41) =Dkαα()0( b +− cos() kxDkb 00) () ααb =Dαα( k0 )( b +− cos( kx 00 )) D ( k ) (b+cos( kx0 )) b+ cos( kx0 ) =EVψψ − . This completes the proof. Further, we can calculate the average of the kinetic and potential energies of the particle. Since aa2 ψψ* ( x) ( x) dx=( b + cos( k x )) dx ∫∫−−aa0 (42) =2ab( 22 + 1 / 2) = a (2 b + 1) the normalized function is 1 Ψ=( x) (b +cos( kx0 )) . (43) ab(22 + 1) The averages of the kinetic and the potential energy are a * αα1 International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 93 For a Dirac delta function potential Vx()= − V0δ () x with V0 > 0 , the fractional Schrödinger equation H1ψψ= E has a solution ψδ( xx) = ( ), E= -∞ (47) in the sense of a certain limit. Proof. The factional Schrödinger equation Dp10| |ψ( xV) - δψ (x)( x) = E ψ( x) (48) can be rewritten in the momentum space as [22] ∞ V0 (49) DE1 |p |ϕ( pp) - ϕϕ( )dp = ( p) 2π ∫−∞ ∞ V0 Dp1 ||ϕϕ( pp) + || E ( ) = ϕ( p)dp, (50) 2π ∫−∞ where ϕ ( p) is the wavefunction in the momentum space. We first change the integral limit in the above equation to a finite positive number p0 , and then let p0 →∞. Thus we have + p V0 0 (51) Dp1 | |ϕψ( pp) + | E |( ) = ∫ ϕ( p)dp 2π − p0 V 1 + p ϕϕ( pp) = 0 0 ( )dp (52) ∫− p 2π Dp1 | |||+ E 0 Taking an integral of the two sides, we have +pV ++ pp1 0ϕϕ( p)dp = 0 00dp ( p)dp (53) ∫−p ∫∫ −− pp 02π 00Dp1 | |||+ E If + p0 ϕ ( p)dp ≠ 0 (54) ∫− p0 we have V + p0 1 1 = 0 dp (55) ∫− p 2π 0 Dp1 | |||+ E V0 1 = ln( 1 + pD01 / | E |) (56) π D1 Dp E = − 10 (57) exp(π DV10 / )− 1 11 ϕ ( p) . (58) |Dp11 |||++ E | Dp / E |1 Here means that the wave functions at both sides are equal except for a constant coefficient. Obviously, a constant coefficient is not important for an eigen-wavefunction. Therefore we have 94 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations lim E = −∞ , (59) p0→∞ limϕ ( p) =1. (60) p0→∞ Accordingly, in real space we have limψδ( x) → (x ) . (61) p0→∞ We can simply write the solution as ψδ( xx) = ( ), E= -∞ , (62) which completes the proof. Let us compare this solution with the solution in the standard quantum mechanics. This solution indicates that the particle falls inside the potential well completely, while the solution to the standard Schrödinger equation with a delta potential well indicates that the particle appears outside the well mainly. Therefore we see that Laskin’s particles are easier to be trapped than Schrödinger’s particles. The second difference is that the delta potential well problem has a unique bound state in the standard quantum mechanics but has more than one bound states in the fractional quantum mechanics (α =1 ), as we will see soon. Problem 4. (The linear potential.) For a linear potential V= Fx with F > 0 , the solutions to the fractional Schrödinger equation H1ψψ= E are π ξξ2 π 2 ψ1(x) =− C(ξ /2 ) cos +−S(ξ /2 ) sin (63) 8 48 4 where the functions C() and S() are Fresnel integrals, and 2F E ξ =x − (64) D1 F with ER∈ . Proof. The fractional Hamilton is H1 =D1 | p| + Fx (65) In the momentum representation [22] d H1 =D1 |p | + iF (66) dp The Fractional Schrödinger equation is d D|1 p |()ϕ p+= iF ϕϕ () p E () p (67) dp 1 dED ϕ()p= i1 | pi | − . (68) ϕ()p dp F F Its solution is DE ϕ(p )= A exp( i pp | |− i p ) (69) 2FF with A an arbitrary real number. In the real space, the wavefunction is International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 95 1 ∞ ψϕ(x) =∫ (p )exp( ipx / ) dp 2π −∞ A ∞ iD iE ix = ∫ exp( 1 pp||−+ p p)dp 2π −∞ 2FF 2A ∞ D Ex = ∫ cos( 1 p2 −+ pp)dp 2π 0 2FF 2A ∞ D 1 E = ∫ cos( 1 p2 +−() xp)dp 2π 0 2FF 22AF ∞ 2 1 2FE = cos(u + ()x − u)du ∫0 2π DD11 F F ∞ = 2Aucos(2 +ξu )du ∫0 D1π (70) ∞ = C cos(u2 +ξξu +−22/4 ξ /4)du ∫0 ∞ = Ccos ( u +ξξ/2)22− /4 du ∫0 ∞ =++ C cos( u ξ/2)2 cos ξ 2 /4+ sin(u ξξ /2)22 sin /4 du ∫0 ξξ22∞∞ = C cos cos(u++ξ / 2)22 du + C sin sin(uξ / 2) du 4 ∫∫0 4 0 ξ 22∞∞ξ = C cos cosu22 du + C sin sin u du 4 ∫∫ξξ/2 4 /2 π ξξ22π = C − C(ξ /22/) cos +−C S(ξ ) sin 8 48 4 where the coefficient F CA= 2 . (71) D1π This completes the proof. Furthermore, since C(+∞ ) = S(+ ∞ )=ππ / 8, C(−∞ ) = S(- ∞ )=- / 8 , (72) the limit behavior of the wavefunction is 0 x → +∞ ψ (x) = ξ 2 π (73) C π sin( + ) x → −∞ 44 Problem 5. (A periodic potential.) The periodic function X(x) defined by X( x )=− | x | + a / 2 x ∈− [ aa , ] (74) Xx(+ 2 a ) = Xx ( ) x ∈ ( −∞ , ∞ ) (75) is called a triangular wave, where a>0 is a real number, whose properties have been studied carefully in electronics [28-30]. 96 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations For the potential 2 D1 π x Vx( ) = ln tg (76) π Xx() 2 a the fractional Schrödinger equation H1ψψ= E has the solution ψ ( x) = Xx( ) (77) E = 0 . (78) Proof. We have H11ψ () x= D | pXx | () d = DH ( Xx()) 1 dx π = D1H(− sgn(sinx )) a (79) 2 π x = −D1 ln tg π 2a = −VxXx() () = −Vx()ψ (x ). Therefore we have Hx1ψψ()+=Vx() () x 0, (80) which completes the proof. In the above proof, we used a formula π 2 π x H(sgn(sin x)) = ln tg (81) a π 2a which can be seen in book [23] (Equation 6.14, page 292, vol. 1). Problem 5’. (The Dirac comb.) The Schrödinger equation Hψψ= E with the Dirac comb potential ∞ 22 V( x) =−−∑ δ ()x na (82) ma n=−∞ has a solution ψ ( x) = Xx( ) (83) E = 0 . (84) Proof. Since 22 2 2∞ ddπ n −X x = sgn(sinx )= ( −− 1)δ (x na ) , (85) 2 ( ) ∑ 22m dx m dx a m n=−∞ and International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 97 ∞ 22 VxXx() ( ) =−− Xx( ) ∑ δ () x na ma n=−∞ (86) ∞∞ 222 =−−∑∑δδ()x na X (na) =−−−(1)(),n x na ma nn=−∞ m =−∞ we have 22d −Xx( ) += VXx( ) 0 , (87) 2m dx2 which completes the proof. We include this well-known result in standard quantum mechanics here so that the reader can compare the standard and the fractional Schrödinger equations conveniently. Problem 6. For the potential 2a Vx( ) = − D1 , (88) xa22+ with a>0, the fractional Schrödinger equation H1ψψ= E has two and only two bound states a ψ1( x) = , E11= − Da / , (89) xa22+ x ψ 22( xE) = , = 0. (90) xa22+ Proof. We need a well-known Hilbert transform pair [24] ax H = (91) xa22++ xa 22 xa− H = . (92) xa22++ xa 22 Taking the derivative of the above two equations, we have da 12a2 H =−+ (93) dx 22 22 2 xa++ xa ( xa22+ ) d x2 ax H = (94) dx xa22++() xa 222 Multiplying by the constant D1 , we have d aD1 a2 aa DD11H =−+ (95) dx xa22+a xa 22 + xaxa 2222 ++ d x2 ax DD11H = (96) dx xa22+ xaxa 2222 ++ Therefore we have Dp1| |ψ 1 () x= E 11 ψψ () x − V 1 () x (97) Dp1| |ψψ 2 () x= E 22 − V () x (98) 98 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations where E11=−= D / aE , 2 0. (99) From the shapes of the wave functions [24], we know that ψ1 ( x) is the ground state, and ψ 2 ( x) is the first excited state. Since the excited energy E2 = 0 , we know that there are no more excited states. The particle has only two bound states, the ground state and the excited state. This completes the proof. Let us further calculate the averages of the kinetic and potential energies in this elegant problem. The normalized functions are 2aa Ψ=1 ( x) (100) π xa22+ 2ax Ψ=2 ( x) (101) π xa22+ Since the wavefunction Ψ1 is even and Ψ2 is odd, the two states obviously are orthogonal, i.e. ∞∞ ΨΨ**( x) ( x) dx =ΨΨ( x) ( x) dx =0 . (102) ∫∫−∞ 12−∞ 21 In the ground state, the averages of the kinetic and potential energies are 31DD Vx( ) = −2 D1πδ () x (105) the fractional Schrödinger equation H1ψψ( xEx) = ( ) (106) has two solutions ψ11( xx) = πδ ( ), E = −∞ (107) 1 ψ ( x) =, E = 0 (108) 22x Proof. In the above example, let a → 0 , and notice that a lim = πδ (),x a→0 xa22+ (109) x 1 lim = . a→0 xa22+ x This completes the proof. The two solutions can also be written as ψδ11( xx) =( ), E = −∞ (110) 1 ψ ( x) =, E = 0 (111) 22π x International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 99 This result is consistent with the solution for Problem 3 and the scaling property discussed in Sec. II. Problem 8. For the potential 4a Vx( ) = − D1 (112) xa22+ with a>0, the fractional Schrödinger equation H1ψψ( xEx) = ( ) has a bound state 2ax2 D ψ ( x) = , E= − 1 (113) ()xa2+ 22 a Proof. Taking the derivative of the two sides of Equation (95), we have ′′ ′ daD1 a4 a a DD11H=−+ (114) dx 22 a 22 22 xa++ xa ( xa22+ ) xa + aa′′ a ′ D1 | p | = E - Vx ( ) (115) xa22++ xa 22 xa 22 + This completes the proof. Notice that the potential in this problem is just 2 times the potential in Problem5, but their solutions are completely different. Problem 9. For the potential Vx( ) = −4 D1πδ () x (116) the fractional Schrödinger equation H1ψψ( xEx) = ( ) (117) has a bound state ψ( xx) = πδ '( ), E = −∞ . (118) Proof. By letting a → 0 in Problem8, this statement follows immediately. 3.2. Three Dimensional Problems Problem 10. (The Free particle.) For V ()r = 0, the solutions for the fractional Schrödinger equation Hαψψαα= E are ψα (rr) =exp(ik ⋅ ) (119) α Eαα= Dk() . (120) with k any three dimensional vector. An alternative form of the eigen-functions is m ψα (r) = jll( kr ) Y (,θϕ ) (121) m where j is the spherical Bessel function of order l, Y (,θϕ )is the spherical harmonic function of degree l and order m l l [22], (,r θϕ , ) is the spherical coordinate system, and k ≡||k is the length of the vector k. Proof. For V ()r = 0, the standard Schrödinger equation 100 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations 2 2 Hψψ(r) =−∇( rr) =EV ( ) (122) 2m has the solutions ψ (r) =exp(ik ⋅r ) (123) 22k E = . (124) 2m From the relationship between the fractional Hamiltonian and the standard Hamiltonian αα2 /2 α/2 α/2 Hαα= T = D α|pp | = D α ( ) = D α (2 mT )= Dα (2 mH ) , (125) we see that the fractional Schrödinger equation HEααψψ()rr= αα () (126) has the solutions ψα (rr) =exp(ik ⋅ ) (127) α Eαα= Dk() . (128) In the spherical coordinate system, the classical Schrödinger equation 2 2 Hψψ(r) =−∇( rr) =EV ( ) (129) 2m has solutions m ψ(r) = jll( kr ) Y (,θϕ ). (130) Since mmα /2 HjkrYααll( ) (,θϕ )= D (2 mH ) jkrYll ( ) (,θϕ ) (131) 2 2αα /2 mm = Dαα( k ) jkrYll ( ) (,θϕ )= DkjkrY ( )ll ( ) (,θϕ ) the solutions to the fractional Schrödinger equation has an alternative form m ψα (r) = jll( kr ) Y (,θϕ ), (132) α Eαα= Dk() . (133) Problem 11. The function 11p 3 Yr() = exp(i⋅rp ) d (134) 3 ∫R3 α (2π )Dα |p | and E=0 is a solution of the fractional Schrödinger equation H ψψ= E with the potential α Vr()= −δ ()/()r Yr, (135) 2 where δ()r = δδδ ()xyz () ()= δ ()/(2 r π r ) is the Dirac’s delata function in the 3D space. Proof. It is easy to verify that α Dα |p | Yr ()+ VrYr () () =−=δδ ()rr () 0. (136) This completes the proof. There are two special cases, where the wave functions and the potential energy can be given explicitly: International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 101 (1) When α = 2 we have 12m p = ⋅ 3 Yr() ∫ 3 exp(irp ) d (2π )32R p (137) 1 2m21 mm 1 = ⋅=3 = ∫ 3 exp(idkr ) k (2π )3R 22k 242ππrr 2 and 2 1 Vr()= −=δ ()/r Yr( ) − δ (r) . (138) m2 r 2 1 Therefore we say that for the central potential Vr( ) = − δ (r) , the Schrödinger equation Hψψ= E has a m2 r m 1 solution ψ ()r = and E=0. 2 2π r (2) When α =1 , we have 11p 11 Yr() = exp(i⋅=rp ) d33exp(ikrk⋅ ) d 33∫∫RR33 (2ππ ) DD11||pk(2 ) || , (139) 4π 1 1 11 = exp(idkr⋅= ) 3 k 3∫R3 22 DD11(2ππ ) 4||π k 2 r and Vr()=−=δ ()/()r Yr −Dr1δ (). (140) Therefore, for the central potential V (r) = −D1δ (r) , the fractional Schrödinger equation H1ψψ= E has a 11 solution ψ ()r = and E=0. 22 D1 2π r Problem 12. (The harmonic oscillator potential.) For a harmonic oscillator potential 2 Vk()rr= , (141) the fractional Schrödinger equation H1ψψ= E has solutions(in the momentum representation) 1 ϕϕ(p )= (p ) = Ai( κ pr − ) (142) p n 2 2 1/3 = 1 . (143) Enn(2 kD 1 )|| r 2 1/3 where Ai(x) is the Airy function, rn is its n-th zero point, and κ ≡ (2Dk1 / ( )) . Proof. The fractional Hamiltonian is 2 HD11=||pr + k. (144) In the momentum representation, the Hamiltonian operator and its eigen-equation are 22 Hk11=− ∇+p D||p (145) 22 −∇k pϕ()p + DE | pp | ϕϕ () = () p (146) 102 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations 2 In the spherical coordinate system, the Laplace operator ∇p is expressed as 22 2 11∂ ∂∂ 1 ∂ ∇=p p + (sinθ ) + . (147) p ∂∂pp2 2sinθ∂∂θθp22sin θϕ 2 For a s-state wave function, ψψ()p = ()p , the eigen-equation becomes 2 2 1 ∂ −k ppDppEpψ() +=1 ψψ () (). (148) p ∂p2 Let up()= pψ () p. Then we have 2 2 d −k up() += Dpup1 () Eup (). (149) dp2 The solution of this equation under the condition uu(0)= ( ∞= ) 0 is up( )= Ai(κ p − rn ) (150) 2 2 1/3 = 1 (151) Enn(2 kD 1 ) | r |, 2 1/3 where Ai is the Airy function, rn is its n-th zero point, and κ ≡ (2Dk1 / ( )) [4]. This completes the proof. Problem 13. (The Coulomb potential.) 2 For the Coulomb potential V() r= − Ze / r with Z>0, and e is the charge of the electron, the fractional Schrödinger 2 equation has a solution ψ (r )= 1/r with E=0 when Ze= 2/ D1 π . Proof. In fact we have 1 11 = ⋅ 3 DD1 ||pp1 || ∫ 3 exp(idkr ) k 4π r (2π )32R k DD11 =11|k | exp(id kr⋅= ) 33 k exp(idkr⋅ ) k 32∫∫RR333 (2ππ ) k (2 ) k (152) DD44ππ11 =11⋅=3 ∫ 3 exp(ikr )d k (2π )33R 4π k (2π ) r2 D 1 = 1 2π 22r and 1Ze2 11 D Vr() =−=−. (153) 44ππrrrr. 2π 22 Therefore we have 1 1 D1 ||p +V = 0 4π r 4π r (154) D1 |pr |ψ ()+Vψ () r= 0. This completes the proof. International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 103 4. Solutions to the Relativistic Schrödinger Equation Again, let us study the one dimensional problems first and then the 3D problems. 4.1. One Dimensional Problems Problem i. (The free particle.) For V(x)=0, the solution for the relativistic Schrödinger equation Hrψψ= E is ψ ( x) = exp( ikx ) (155) 2 24 E=() kc + m c . (156) with −∞ 2 24 E=() kc + m c . (159) with k ≥ 0 . Further, when V(x)=V0 with V0 a constant, the wavefunctions do not change but the new energy levels become 2 24 E=() kc ++ m c V0 (160) Problem ii. (A periodic potential.) For the potential b =2 +− 24 2 V( x) ( () k0 c m c mc ) (161) b+ cos( kx0 ) where ka0 = π / , a is a length, and b is a real number, the relativistic Schrödinger equation Hrψψ= E has a solution ψ ( x) = b + cos( kx0 ) (162) 2 24 E=() kc0 + mc . (163) Proof. From the definition of the square root operator [18-20], 22 24 222 24 pc+=+∈ mcexp( ikx ) kc mcexp( ikxkR ) , (164) We get that 22 24 222 24 pc+=+ mccos kx00 kc mccos( kx 0 ) (165) 22 24 2 p c+= m c 1 mc 1. (166) Therefore we have 104 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations Trrψ 22 24 =++pc mc( bcos( kx0 )) 2 24 2 =++(k00 c ) m c( cos( k x )) mc b (167) =2 + 24 + −2 +− 24 2 ()kc0 mc( b cos()() kx00) ( kc mc mc) b b =2 + 24 + −2 +− 24 2 + ()kc0 mc( bcos( kx00 )) ( ( kc ) mc mc ) (bcos( kx0 )) b+ cos( kx0 ) =EVψψ − . This completes the proof. Further, we can calculate the average of the kinetic and potential energies. The normalized function is 1 Ψ=( x) (b +cos( kx0 )) . (168) ab(22 + 1) The averages of the kinetic and the potential energies are a 21b2 * 2 2 24. (169) ∞ 22 24 V0 p c+= mcϕ( p) - ϕϕ( pp)dp Er ( ) (174) 2π ∫−∞ ∞ 22 24 V0 p c +=mcϕϕ( pp) +| Er |( ) ϕ( p)dp (175) 2π ∫−∞ We first change the integral limit to a positive number p0 > mc , and then let p0 → +∞ . Thus we have + 22 24 V0 p0 pc+= mcϕψ( p) +|Er |( pp) ∫ ϕ( )dp (176) 2π − p0 International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 105 V 1 + p0 ϕϕ( pp) = 0 ∫ ( )dp (177) 2π 22 24 − p0 pc++ mc|| Er Taking integral of the two sides, we have ++ppV 1 + p 00ϕϕ( pp)dp = 0 dp ⋅ 0( )dp (178) ∫∫−− ∫ − pp002π pc22++ mc 24 ||E p0 r + p0 If ϕ ( p)dp ≠ 0 , we have ∫− p0 VVpp11 1 = 00∫∫00dp = dp . (179) 2ππ− p0 22 24 0 22 24 pc++ mc|| Errpc++ mc|| E If the above integration is calculated only on a subinterval [,][0,]mc p00⊂ p , we have V p 1 0 0 dp <1. (180) π ∫mc 22 24 pc++ mc|| Er Since p>⇒> mc p2 mc 22 ⇒ pc 22 > mc 24 ⇒2 pc22 >+ pc 22 mc 24 ⇒>2 pc p22 c + m 24 c (181) 22 24 ⇒2pc +> || Err p c + m c +|| E 11 ⇒< , 2pc+ || E 22 24 r pc++ mc|| Er we have V p0 1 0 dp <1 (182) ∫mc π 2pc+ || Er V pE+ | | /( 2 c ) 00ln r <1 (183) π 2c mc+ | Er | /( 2 c ) Further we have 2 p00 c− mcexp( 2π c / V )) |Er | > 2 . (184) exp( 2π cV /0 )− 1 We see |Er |→∞ , as p0 →∞, (185) and hence the bound state energy E r = −∞ . (186) The wave function 11 ϕ ( p) (187) 22 24 22 24 pc++ mc|| Err pc + mc/||1 E + 106 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations From Equation (185), we have ϕ ( p) →1 as p0 →∞. (188) Accordingly, in the real space we have ψδ( x) →(xp ) as 0 →∞. (189) Again, a particle with a relativistic kinetic energy is easier to be trapped than a particle with a Newtonian kinetic energy. Problem iv. (The linear potential.) For a linear potential V= Fx with F > 0 , the solution to the relativistic Schrödinger equation Hrψψ= E is ∞ ψ (x) = cos(bu u22++ 1 b ln( u + u + 1) −uξ )du (190) ∫0 E mc m23c ξ =−=(xb ), (191) FF2 with ER∈ . Proof. The relativistic Hamilton is 22 24 Hr = pc+ mc + Fx . (192) In the momentum representation, d H = pc22+ mc 24+ iF . (193) r dp The relativistic Schrödinger equation is d pc22+= mc 24ϕ() p+ iF ϕϕ () p E () p (194) dp This equation can be solved easily 1 d i iE ϕ()p= pc22 +− m 24c . (195) ϕ()p dp F F ii lnϕ (p ) = pc22 +− mc 24 Edp . (196) ∫ F F ic lnϕ (p ) =p2 +− m 22 c E/ c dp F ∫ ( ) 2 22 (197) ic p++ p mc iE =p p2 ++ mc 22 mc 22ln −p 2 FFmc 2 22 ic im23 c p++ p mc iE ϕ(p )= A exp pp2++ m 22c ln −p , (198) 22F F mc F where A is an arbitrary constant. Via Fourier transform, we can get the wavefunction in the real space International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 107 1 ∞ ψϕ1(x) =∫ (p )exp( ipx / ) dp 2π −∞ A∞ ic im23 c p++ p2 mc 22 iE ipx = ∫ exp( p p2++ m 22 c ln −+p )dp 2π −∞ 22FF mc F 2A∞ c mc23 p ++p2 mc 22 Ep = ∫ cos( p p2++ m 22 c ln +(x − ) )dp (199) 2π 0 22FFmc F 2Amc∞ mc23 mc23 E umc = ∫ cos( uu22++1 ln(u + u++ 1) (x − ) )du 2π 0 22FF F ∞ =Ccos( bu u22++ 1 b ln( u + u + 1) +uξ ) du ∫0 2Amc E mc C=, p = umc, ξ = (x − ), (200) 2π F Further, we point out that ψ (x)→0 as x → +∞. (201) It is because the value of ψ (x) is equal to the sum of an alternating series and the absolute values of the terms in the series become smaller when x → +∞ . Specifically, let us consider the integral ∞ I(ξ )= cosG ( u ,ξ ) du (202) ∫0 with 22 G(u,ξξ )= (bu u ++ 1 b ln( u +u + 1) +u . (203) For a fixed ξ > 0 , suppose that un with n=0,1,2,3 satisfy that Gu( ,ξ )= n ππ + /2 (204) n cosGu(n,ξ )0= . (205) We have I(ξ ) =I012 ++ II ++ In + (206) With u I (ξ )= 0 cosG ( u ,ξ ) du 0 ∫0 u1 I1 (ξ )= ∫ cosG ( u ,ξ ) du u0 (207) un In (ξ )= cosG ( u ,ξ ) du ∫un−1 Since the terms In alternately change their sign, and |II12 ||< | << | In | <. (208) The series of I(ξ ) converges for any given ξ > 0 . As ξ →∞, the interval between any two adjacent points, uunn+1 − , becomes closer to each other, every term 108 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations In ()ξ → 0, and hence their alternating summation I()ξ → 0. In other words, for any fixed ξ →∞, we have ψ (x)→0 as x → +∞. (209) It is relatively complicated to discuss the limit behavior of the wavefunction as x → −∞ , so we omit it temporarily. Interested readers can observe its behavior intuitively on a graph. 4.2. Three Dimensional Problems Problem v. (The free particle.) For V ()r = 0, the solutions to the relativistic Schrödinger equation HErrψψ()rr= rr () are ψ = ⋅ (210) r (rr) exp(ik ) 22 2 24 Er = ck + mc (211) with k a three dimensional vector. An alternative form of the eigen-functions is m ψr(r) = j ll( kr ) Y (,θϕ ). (212) Proof. For V ()r = 0, the standard Schrödinger equation 2 2 Hψψ(r) =−∇( rr) =EV ( ) (213) 2m has the solutions ψ (r) =exp(ik ⋅r ) (214) 22k E = (215) 2m where k =||k is the length of the vector k. From the relationship between the relativistic Hamiltonian and the standard Hamiltonian 22 24 2 24 2 24 Hr =+++p c mc =2 mcT mc =2 mcH mc (216) we know that the relativistic Schrödinger equation HErrψψ()rr= rr () (217) has the solutions ψ r (rr) =exp(ik ⋅ ) (218) 22 2 24 Er = ck + mc . (219) Obviously, the wavefunction can also be expressed in the spherical coordinate system. Problem vi. The function 11p Yr() = exp(i⋅rp ) d3 3 ∫ 3 (220) (2π ) R p22c+ mc 24 and E=0 is a solution of the relativistic Schrödinger equation HErψψ()rr= ()with the potential Vr()= −δ ()/()r Yr. (221) Proof. It is easy to verify that International Journal of Theoretical and Mathematical Physics 2015, 5(5): 87-111 109 22 24 pc+ mcYr() + V ()rr Y () =−=δδ () r () r 0. (222) This completes the proof. Problem vii. (The harmonic oscillator potential) 2 For a harmonic oscillator potential Vk()rr= , the s-state energies for the relativistic Schrödinger equation, En , satisfy 2 2 1/3 2 11<< −ω + (223) (22k c ) | rnn | E (2 n ) mc where n =1, 2, 3, , ω ≡ 2/km, and rn is the n-th zero point of the Airy function Ai(x). Proof. In the momentum space, we have 22 24 2 22 24 2 2 H=p c ++=+−∇ mc k rp c mc k (224) r p The Schrödinger equation is 2 2 22 24 −∇k pϕ()pp +c + mc ϕϕ() p = Er () p. (225) Up to some constants, this equation is mathematically the same as the Schrödinger equation in the real space with a square root potential 2 2 22 −∇ψ()r +kr + r ψψ() rr = E (), (226) 2m 0 with k and r0 are positive numbers. We also know that in standard quantum mechanics the energy levels become higher if the potential becomes higher. Now let us return to the current problem. Since p2 ||ppc<22 c + m 24 c <+ mc 2 , (227) 2m 2 we see that the relativistic kinetic energy is smaller than the classical kinetic energy plus the rest energy mc , but greater than the energy levels of the fractional energy with α =1 and Dc1 = . Specifically, for s-state, we have 1 (1 k2 c 2 ) 1/3 | r |<< E mc2 +(2 n − ) ω (228) 2 nn 2 n =1, 2, 3, . Here we used the result of Problem 12 and the energy formula for the classical harmonic oscillator [22]. Problem viii. (The Coulomb potential.) 2 For the Coulomb potential V= − er/ , wheree is the charge of the electron, the energy eigen value of the relativistic Schrödinger equation HErψψ()rr= ()is 21 2 1 n 34 64 1 56 Enl = mc 1 −α − −+α α δ0l +O(α ) (229) 2n24 2n l +1/ 2. 4 15π n3 2 where α=e / c is the fine structure coefficient, n is the principle quantum number, l is the angular momentum quantum number. Only in this problem and its solution, α is not fractional parameter. Proof. The relativistic Hamiltonian is 110 Yuchuan Wei: Some Solutions to the Fractional and Relativistic Schrödinger Equations 2 22 24 e H=p c + mc - (230) r The Schrödinger equation 2 22 24 e Hψ()rp= c+= mcψ() r - ψψ () rE () r (231) r has no analytic solutions by now. Since the relativistic effect is very small, we can use the perturbation method based on the classical Hamiltonian 22 2 p e H= mc + - (232) 0 2mr The classical Schrödinger equation 22 0 20p 0e 0 0 Hψ()r=+= mc ψ () r ψ () r - ψψ () rrE () (233) 0 2mr 0 has the well-known wave function ψ nlm (r) and energy levels 02 1 2 E1n =mc − α . (234) 2n2 According to the perturbation theory [22], the first order approximation of the energy is e2 E=< H >=< p22 c + mc 24 > − < > nl r 1 (235) =ψψ0* 22 +− 24 0 3 2 2 ∫ 3 nlm ()ppc mcnlm ()p d p α mc R n2 21 2 1 n 34 64 1 56 = mc 1−α − −+α α δ0l +O(α ) 2n24 2n l +1/ 2 4 15π n3 where 1 pr⋅ φψ0()p= 03( rr)exp(−id ) nlm ∫R3 nlm (236) (2π )3 is the wavefunction in the momentum space[23]. The details realization of the fractional Schrödinger equation. Several of the calculation can be found in [19]. solutions to the relativistic Schrödinger equation are also 5 presented. The standard, fractional and relativistic The new energy levels contain a valuable α term, Schrödinger equation should be studied together. which is 41% of the observed Lamb shift [19]. We are trying to find the exact solutions for the relativistic Schrödinger We wish that the winter of the fractional quantum equation with a Coulomb potential to see whether we can mechanics could go away and its spring could come soon. explain the Lamb shift better in the framework of quantum mechanics. ACKNOWLEDGEMENTS The research on the relativistic Schrödinger equation was 5. Conclusions supported by Gansu Industry University (currently called Jeng’s critique resulted in a crisis of fractional quantum Lanzhou University of Technology) during 1989-1991, with mechanics, that is, the fractional Schrödinger equation was a project title ‘On the solvability of the square root equation difficult to solve in mathematics and had no realization in the in the relativistic quantum mechanics’. real world. 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