Page 80 - IJOCTA-15-1

E.M. Rabei et al. / IJOCTA, Vol.15, No.1, pp.71-81 (2025)
Using the method of separation of variables, let us
  express the wave function ψ β (x i ) as the product
β(D s − 1) β 2 2β (D s − 2) β
β
∂ 2β + ∂ +  ∂ + ∂  of radial and angular components:
r
r
r β r 2β θ 1 θ β θ 1
tan 1
β
 
β
β
β 2 2β (D s − 3) β ψ β (x i ) = R β (r )Π N−1 Θ iβ (θ ). (12)
i=1
+ β  ∂ θ 2 + β ∂ θ 2 
r 2β sin 2 θ 1 tan θ 2
β β Substituting this into, eq.(11) and dividing both
 
sides by ψ β (x i ) we get
β 2 1 2β (D s − 4) β
+  ∂ + ∂  + . . .
r 2β 2 θ β 2 θ β θ 3 θ β θ 3
sin 1 sin 2 tan 3
β β β
1 2β β(D s − 1) β
r
r
β 2 2β ∂ R β + r β ∂ R β
+ β β β ∂ θ N−1 R β  
r 2β sin 2 θ 1 sin 2 θ 2 . . . sin 2 θ N−2 β 2
β β β +  ∂ 2β + (D s − 2) β 
∂
 2β θ 1 θ β θ 1 Θ 1β
r Θ 1β 1
tan β
D s − N β  
+ ∂  2
θ β θ N−1  β 2β (D s − 3) β
tan N−1 + β  ∂ + β ∂  Θ 2β
β 2β 2 θ θ 2 θ θ 2
r Θ 2β sin β 1 tan β 2
2m β β  
− (V β (ˆx β ) − E )ψ β (x i ) = 0. 2
2β β 1 2β (D s − 4) β
ℏ +  ∂ + ∂  Θ 3β
β r Θ 3β 2 θ β 2 θ β θ 3 θ β θ 3
2β
(10) sin β 1 sin β 2 tan β 3
β 2 2β
4. Free particle in N-dimensional space + · · · + β β θ β ∂ θ N−1
2β
Cartesian coordinates r Θ (N−1)β sin 2 θ β 1 sin 2 θ β 2 . . . sin 2 N−2
β

In this section, we aim to particle in N- D s − N β
+ ∂  Θ (N−1)β

dimensional space polar coordinates illustrate θ β θ N−1
tan N−1
the Conformable Schr¨odinger Equation in N- β
dimensional space by solving it for Free particles. 2m β β
+ E ψ β (x i ) = 0.
For free particles V β (ˆx β ) = 0 the eq.(10) becomes: ℏ 2β
β
(13)
β(D s − 1) β
2β
∂ ψ β (x i ) + ∂ ψ β (x i ) Now, all sides of the equation must be constants.
r
r
r β
  Let us denote this constant as λ, the conformable
β 2 2β (D s − 2) β radial equation is:
+  ∂ + ∂  ψ β (x i )
r 2β θ 1 θ β θ 1
tan 1
β
 
1 2β β(D s − 1) β
β 2 2β (D s − 3) β ∂ R β + ∂ R β
r
r
+ β  ∂ θ 2 + β ∂ θ 2  ψ β (x i ) R β r β
r 2β sin 2 θ 1 tan θ 2 β 2
β β 2m β β λ r
+ E − = 0. (14)
  2β 2β
β 2 1 2β (D s − 4) β ℏ β r
+  ∂ + ∂  ψ β (x i )
r 2β 2 θ β 2 θ β θ 3 θ β θ 3
sin 1 sin 2 tan 3
β β β and the conformable angular part is:
β 2 2β
+ · · · + β β θ β ∂ θ N−1
r 2β sin 2 θ 1 sin 2 θ 2 . . . sin 2 N−2  
β β β (D s − 2)
∂ + ∂ + λ r − Θ 1β = 0,
 2β β λ 1
 
θ 1 θ β θ 1 2 θ β
D s − N β tan β 1 sin β 1
+ β ∂  ψ β (x i )

θ θ N−1 (15)
tan N−1 " #
β
2β (D s−3) β λ 2
2m β ∂ + β ∂ + λ 1 − β Θ 2β = 0.
β
+ E ψ β (x i ) = 0. θ 2 θ 2 θ 2 2 θ 2
2β tan sin
ℏ β β
β (16)
(11)
.
.
.
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