Page 84 - IJOCTA-15-1

E.M. Rabei et al. / IJOCTA, Vol.15, No.1, pp.71-81 (2025)

2β 2 1−D s+N
∂ X 1β + K X 1β = 0, (57) where q = 2 . To ensure the solution of
x 1 x 1
2m β the Schr¨odinger equation is finite and regular, the
β
K 2 = E . constant B must be set to zero.
x 1 2β x 1
ℏ
β
2β 2
βq βq
∂ X 2β + K X 2β = 0, (58) X Nβ = x k ), (67)
x 2 x 2 AJ βq (x N k x N
N x N
2m β β
2
K = E .
x 2 2β x 2
ℏ
β The total wave function for free particles in N-
. . . dimensional space is
#
h D s − N
β
∂ 2β + β ∂ β + K 2 X Nβ = 0, ψ β (x i ) = Π N X iβ (x )
x N β x N x N i=1
x
N β β β
(59) = X 1β (x )X 2β (x ) . . . X Nβ (x )
2m β β
2
K = E βq
x N 2β x N = Cx J βq (x N k x N ) (68)
N
ℏ
β β !
x
The solution to these equations are N−1 j
× Π j=1 exp ik x j · β
β !
x 1 x β β β β
X 1β = A x 1 exp ik x 1 (60) · j x 1 x 2 x N . The
β where k x j β = k x 1 β +k x 2 β +· · ·+k x N β
total energy for free particles in N-dimensional
β !
x 2 space is given by
X 1β = A x 2 exp ik x 2 (61)
β
. E β = E β + E β + · · · + E β
. x 1 x 2 x N
.
2β
β ! ℏ
x β 2 2 2
X (N−1)β = A x N−1 exp ik x N−1 N−1 .(62) = β [k x 1 + k x 2 + · · · + k x N ]. (69)
β 2m
1
For the solution of eq.(59), after multiplying by For 3D, where N = D s = 3 → q = , the wave
2
2β function takes form
x , we have
N
!
β
2β 2β β β 2 2β β 1 x
x ∂ X Nβ + βax ∂ X Nβ + k x X Nβ = 0, ψ β (x i ) = Cx 2 J 1
N x N N x N x N N 3 β 1 (x 3 k x 3 ) exp ik x 1
(63) 2 β
!
β
x
2
We then introduce the change of variables Let × exp ik x 2 β
β ρ β β β 2β 2 2β
∂
x = → ∂ x N = k x N ρ → ∂ x N = k ∂ ρ .
N k x x N
N q β
This yields Note that Ref, 53 J ) = 2 x 3 .
β 1 (x 3 k x 3 πx β sin k x 3 β
2
Thus, we can rewrite the wave functions:
β β
2β 2β
2β
ρ ∂ X Nβ + βaρ ∂ X Nβ + ρ X Nβ = 0. (64)
ρ ρ
! !
r β β
2 x 3 x 1
Here a = (D s −N). This equation is resembles to ψ β (x i ) = π sin k x 3 β exp ik x 1 β
eq.(21) when λ r = 0. Hence, the solution to this
!
equation is x β 2
× exp ik x 2
β
2 [AJ βq (ρ) + BJ −βq (ρ)] ,
X Nβ = ρ β 1−a (65)
5. Conclusions
where q = 1−a . After substituting a = (D s − N) We used the separation of the variable method to
2
β ρ β solve a time-dependent conformable Schr¨odinger
andx = , the solution becomes:
N k x equation of order 0 < β ≤ 1, in fractional space
N
domains of space dimension, 0 < D s ≤ 3. This
βq βq
N
X Nβ = x k x N [AJ βq (x N k x N ) + BJ −βq (x N k x N )] , allows us to obtain the total energy for a free par-
(66) ticle in fractional space dimensions for the general
78