Page 83 - IJOCTA-15-1

Conformable Schr¨odinger equation in D-dimensional space

2β β β γ N−2
(1 − x )D D T (x) (41) λ r = ℓ(ℓ + D − 3), ℓ = 0, 1, · · · , (47)
N−2 x N−2 x N−2 βm N−4
− 2β(γ N−2 + 1)x β D β T γ N−2 (x) λ 1 = m 0 (m 0 + D − 4), (48)
N−2 x N−2 βm N−4
m 0 = 0, 1, · · · , m 0 ≤ ℓ,
2
+ β m N−4 (m N−4 + 2γ N−2 + 1)T γ N−2 (x) = 0.
βm N−4
λ 2 = m 1 (m 1 + D − 5), (49)
m 1 = 0, 1, · · · , m 1 ≤ m 0
Taking the conformable derivative of order m N−3 ,
.
for (41), we find . . (50)
λ N−3 = m N−4 (m N−4 + D − N), (51)
m N−4 = 0, 1, · · · , m N−4 ≤ m N−5 ,
D m N−3 β [(1 − x 2β )D β D β T γ N−2 (x)] λ N−2 = m N−3 (m N−3 + D − N − 1), (52)
N−2 x N−2 x N−2 βm N−4
− 2β(γ N−2 + 1)D m N−3 β [x β D β T γ N−2 (x)] m N−3 = 0, 1, · · · , m N−4 ≤ m N−5 .
N−2 x N−2 βm N−4
2
+ β m N−4 (m N−4 + 2γ N−2 + 1)T γ N−2 (x) The generalized quantum number n, ℓ, m 1 , m 2 ,
βm N−4
..., m N−2 , clearly follows the following quantiza-
= 0. (42) 5
tion rule
after using conformable Leibniz rule, 52 we have
m N−3 ≤ m N−4 ≤ m N−5 · · · ≤ m 0 ≤ ℓ ≤ n (53)

4.1. Free particle in N-dimensional space
D m N−3 β [(1 − x 2β )D β D β T γ N−2 (x)]
N−2 x N−2 x N−2 βm N−4 Cartesian coordinates
β
β
= (1 − x 2β )D m N−3 β D D T γ N−2 (x)
N−2 x x βm N−4 The time-independent conformable Schr¨odinger
β m N−3 β β γ N−2
− 2m N−3 βx N−2 D D T (x) equation in N-dimensional space in Cartesian co-
x βm N−4
ordinates is written as
2
− m N−3 (m N−3 − 1)β D m N−3 β T γ N−2 (x) (43)
βm N−4
α N − 1
D m N−3 β [x β D β T γ N−2 (x)] (44) ∂ 2β + ∂ 2β + · · · + ∂ 2β + β ∂ β (54)
N−2 x N−2 βm N−4 x 1 x 2 x N β x N
x N
= x β D m N−3 β D β T γ N−2 (x) !
N−2 x N−2 βm N−4 2m β β
− m N−3 βD m N−3 β T γ N−2 (x) − 2β (V β (ˆx β ) − E ) ψ β (x i ) = 0.
βm N−4 ℏ
β
Substituting eqs(43)and (44) in (42), we have For free particle after using D s = α N + (N − 1)
this equation becomes
2β β β m N−3 β γ N−2 D s − N
(1 − x )D D D T (x) 2β 2β 2β β
N−2 x N−2 x N−2 βm N−4 ∂ + ∂ + · · · + ∂ + β ∂ (55)
x 1 x 2 x N β x N
β β m N−4 β γ N−2 x
− 2βx (m N−4 + γ 1 + 1)D D T (x) N
N−2 x N−2 βm N−4 !
2
+ β [m N−4 (m N−4 + 2γ N−2 + 1) + 2m β E β ψ β (x i ) = 0.
2β
− m N−3 (m N−3 + 2γ N−2 + 1)]D m N−4 β T γ N−2 (x) ℏ β
βm N−4
= 0.
By employing the separation method ψ β (x i ) =
(45)
β
Π N X iβ (x ), we derive
i=1
comparing this equation with eq.(40) we find the
eq.(35) becomes 1 2β 1 2β
∂ X 1β + ∂ X 2β + . . .
x 1 x 2
X 1β X 2β
" #
1 D s − N
m + ∂ 2β X Nβ + β ∂ β
2β N−3 m N−3 β γ N−2 β X Nβ
X β(N−2) = (1 − x N−2 ) 2 D T βm N−4 (x) X Nβ x N x N x N
(46) 2m β
+ (E β + E β + · · · + E β ) = 0.(56)
2β x 1 x 2 x N
ℏ
β
the separations constants take the values, 48 Thus, we obtain
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