Applied Mathematics, 2011, 2, 999-1004
doi:10.4236/am.2011.28138 Published Online August 2011 (http://www.SciRP.org/journal/am)
Determine the Eigen Function of Schrodinger Equation
with Non-Central Potential by Using NU Method
Hamdollah Salehi
Department of Physi cs , Shahid Chamran University, Ahvaz, Iran
Received April 11, 2011; revised June 21, 2011; accepted June 28, 2011
Abstract
So far, Schrodinger equation with central potential has been solved in different methods but solving this
equation with non-central potentials is less dealt with. Solving such equations are way more difficult and
complicated and a certain and limited number of non-central potentials can be solved. In this paper, we in-
troduce one of the solvable kinds of such potentials and we will use NU method for solving Schrodinger
equation and then by using this method we have calculated particular figures of its energy and function.
Keywords: Schrodinger Equation (SE), Non-Central Potentials, NU Method, Central Potential
1. Introduction
One of the important tasks of quantum mechanic is find-
ing accurate answers of Schrodinger eq uation with a cer-
tain potential. It is obvious that finding exact answers of
SE by the usual and traditional methods is impossible,
except certain cases such as a system with Qualeny po-
tential or a coordinating oscillator. Thus, it is inevitable
to use methods to help us solve this problem. Among the
cases where we have to refuse ordinary methods and
seek new methods is solving SE with non-central poten-
tials. Such potentials are of high importance in quantum
chemistry and nuclear physics. Recently, a lot of studies
are being done about such potentials. Accordingly, dif-
ferent methods are used to solve SE with non-central
potentials among which we can name symmetrical cloud,
SUSY, SIP idea [1,2], route integral [3] and Factorial
method [4].
There is also another method known as NU (Niki-
forov-Uvarov) which gives a clear instruction for ob-
taining exact answers of certain states ,Eigen value of
energy and the related functions based on Orthogonal
polynomials [5]. NU method is based on reducing a sec-
ond degree differential equation of SE into an equation
of hyper geometric type [6-8]. Based on this, in this pa-
per we will try to solve SE with a suitable potential
without any limits. Therefore, we consider a potential as
follows to meet all our need s in order to solve SE by NU
method:
 
2
22
cot
,D
AB
Vr rrr

  
After choosing a suitable change of variable


s
sr,
the transformed equ a tion will be as follows:



    
20
nn n
ss
ss
ss


 s

 
 (1)
In which

and

 are polynomials of maximum
second degree and

 is a polynomial of maximum first
degree. By considering wave function

n
s
 as:


nnn

s
sy s   (2)
Equation (1) will be redu ced to the following equation
which is of hyper geometric type [5];





 
0
nnn
syssys ys 
 

 (3)
That in equations:
 

s
ss
s



 (4)




2

s
ss   
,0

 (5)

is a parameter which is defined as follows[5]:



1, 0,1,2,
2
n
nn
ns sn  



  (6)

1000 H. SALEHI
The polynomial ()
s

 shows that its first derivative
must be negative. We should notice that

and n

are
obtained from a particular answer of



n
y
sys
which is a polynomial of n degree. Furthermore, state-
ment

n
y
s,wave function of Equation (2), is a function
of hyper geometric type which is obtained from the fol-
lowing Rodriguez equation [5].
  

d
d
nn
n
nn
n
yss s
s





(7)
In which n is the normalization constant and B


s

is a weight function which has to meet the following
condition [5,6].
 
 
d
d
s
s
s
ss



,
 
s
ss   (8)
Function

s
 and parameter

are defined as
follow:
    
2
22
ss ss
s
sKs
 



 
 




(9)

K
s 
  (10)
Since

s

4ac
has to be a polynomial of maximum
first degree, the statements under the radical in Equation
(9) have to be sorted in the form of a first degree poly-
nomial and this is possible when its determiner,
, is zero. In this case, an equation is ob-
tained for K and after solving the equation, the obtained
figures for K are placed in Equation (9) and by compar-
ing with Equations (6) and (10) we will calculate Eigen
value of energy.
2
b 
2. Schrodinger Equation with Central and
Non-Central Potentials
We consider time-independent SE as follows:
   
2
2
20
rr
mEV
  
r
 (11)
Wave function


r
n
elaborates certain states and
their related energy levels, n, for a particle in a poten-
tial field. First, central and non-central potentials
E
 
2
22
cot
,rABD VVrrrr

 
are considered in spiracle coordinates and put them in the
above equation. ()
2
2mh J


 

2
2
2
222 2
2
22
1
11 1
sin
sin sin
cot 0
r
r
rr r
rr
r
ABD
Er
rrr


















 
 






 


(12)
By considering the whole wave function as:




,,rrRr


    (13)
And replacing in Equation (12), we can write the
equation separately as follows:




22
22
dd
21 0
d d
Rr RrErArBR r
rr rr





 (14)




222
22
dd
cotcot 0
d dsin
mD

 
 
 

 


(15)


22
2
d0
dm


 
(16)
In which m2 and


1ll 

 are the separation fix
amounts. The answer for Equation (16) is as follows [9].

1e,0, 1, 2,,
2
im
mml





 (17)
Equations (14) and (15) are radial and angular equa-
tions respectively which we are going to solve by NU
method.
3. Solving Radial Equation and Calculating
Eigen Values of Energy by Using NU
Method
For solving radial part of SE, by considering 2 EE



in Equation (14) we will have:




222
22
dd
21 0
d d
Rr RrrArBRr
rr rr



 

(18)
Now if we compare the above equation with Equation
(1), the general form of equation in NU method, we will
have:


2r 

,


rr 

,


22
rrArB





 (19)
Therefore, according to the definition of


r  in

H. SALEHI
1001
Equation (9), we can write


r  as follows:
 
22
114441
22
rrKArB
   


(20)
From the above equation, considering that under the
radical there should be a of a first degree polynomial.
 

11
24
24


B
B





1
22
11 1
22
rB
r
rB




 
 



  





We will determine K and we will have:


1/2
1
1/2
2
41
41
KA
KA
  
  

(21)
To continue, we choose the suitable amount of


r 
which can meet the condition 0

. So:
 
 


1/
1/
2
2
41
41
rr
rr












2
1
2
2
11
(i)4 1
22
11
(ii)41
22
B
B
KA B
KA B



 
 



  




 
 
(22)
Now, for (i), we can write


r 
24rr
from Equation (5) as
follows:

B

     (23)
And from Equations (6) and (10) we will calculaten

and

respectively
2
nn

 

(24)
1
41 

4  

21BA


 
 

(25)
So:
 
1/2
,21 1
nnA B
 

 


(26)
And eventually, we can obtain Eigen value of energy
for (i) from the above equation:

1
2
412
nA
Bn



 
 

1

2
(27)
In the same way, for (ii), Equation (22), by repeating
the above process we can obtain Eigen value of energy.
Therefore, from Equation (5), we write as:

r 

24rrB    (28)
And from Equations (6) and (10) we calculate n

and

:
2
nn

  (29)

1
2
41AB   1



 





 (30)
n




1
2
241nA B 

1

  



(31)
From the above equations, Eigen value of energy is
obtained:
 
1
2
412
nA
Bn




1


 





(32)
We considered 2 E


 and from (27) and (32), we
have Eigen value of energy as:
 
2
2
1
2
4121
nA
E
Bn 




  






(33)
4. Calculating Eigen Functions Related to
Radial Share of Wave Function
In order to obtain Eigen radial functions, by using Equa-
tion (4) we have:


  



d, lnd
d
s
s
s
ss
s
ss




 (34)
Thus, for (i) and (ii) in Equation (22) we ha ve :
 

11
24 1
22
11
2
22
rrB
r





 



 
(35)
in which


4B  1

 and considering


rr 

We will have:


(1)
2
(1)
2
)e
)e
r
r
irr
iir r















(36)
On the other hand, from Equation (8) we have:




  

dln( )d
d
ws tsts
ws wss
ssS sS

 (37)
From Equation (37), as we have

2rr



 in
which 1




, we will have:


2
er rr





 (38)

1002 H. SALEHI
So we can write the weight function




r
rr

  as
follows:

2
er rr




 (39)
And to continue, from Equation (7), we will write

n
y
r as follows:


2
2
de
ed
nn
n
nrn
B
yr r
rr






r
(40)
From Rodriguez sequence polynomial, we will have
Laguerres dependent fu nction s as [9]:


ed
ee
!d
xk n
kx
nn
x
Lx x
nx



nk
(41)
By comparing the two functions, (39) and (40) and by
considering k

 and 2rx


we will have:
 
!1 2
n
nn n
yr BnLr



 (42)
In the same way for (ii) in which
from Equation (37) we will have:

21rr   

(1)2
er rr




 (43)




2
er
r
rr
r





 (44)
And finally, from Equation (7), we can write


n
y
r
as follows:


2
2
de
ed
nn
nrn
Bn
yr r
rr






r
(45)
By comparing Equations (41) and (45) and consider-
ing k

 and 2
x
r

 we will have:
(ii)

!1 2
n
nn n
yr BnLr


 (46)
Therefore, from Equations (2), (35), (36), (42), (46)
the whole radial wave function can be written as:


(1)
2
(1)
2
(i)1!e2
(ii)1 !e2
nr
n
nr
nn
Rr BnrLr
Rr BnrLr




n











(47)
Normalization coefficient is determined by

2
2
0
drRrr



1 by taking orthogonal condition of
associated Laguerre polynomials and we will have:

 

 
2
2
2
)!!21
2
)!!2
n
n
r
iB nn n
r
ii Bnn n

















1
(48)
5. Solving Angular Equation and
Calculating Eigen Values
With a variable change as cosx

 we will
tion (15) as the following transformation: have Equa-





22 2
Dx 

2
22
d
d() 2
d d1
1
x
xx
x xx


2
210
1xm x
x

 
(49)
By comparing Equations (4 9) with (1) we wi ll have:




2
x
x 


,


2
1
x
x  





22
x
Dx m      (50)
By using (9), we write

x

as:
 

22
x
DKx mK     (51)
We determine K on condition that there
gree polynomial under the radical: is a first de-

2
2
mD
xmD



x









(52)
1
2
2
K
D
K
m





0



And for we have the right choice as:


1/2
2
x
mDx    (53)
2
2
K
m   (54)
m EqAlso, frouation (5), we can obtain


x

e will hav Then we:




1/2
2
21
x
Dm x    (55)
According to Equations (6) and (10), we will calculate
n

and

respectively:


1
22
21 1
nnmDnn



 


(56)

1/2
22
mDm  ,n

  (57)
Finally we will obtain

from the above equation:


1/2
2
321nnnDmm 
2

 (58)

H. SALEHI
1003
6. Calculation Eigen Functionof A
Equation
To fiby
using Equation (4) we will have:
s ngular
nd Eigen function related to angular equation,




ln d
s
s
s
s

  (59)
Thus, for

x
x

  in which

1/2
2
Dm   and
considering that


2
1
x
x   we will have:


/2
2
1xx

 
er hand, fr o(8) we have:
(60)
On the othm Equation




ln d
s
s
s
s


  (61)
As
 
21
x
x

   , so:


2
1
x
x

  (62)
Therefore, we can write the weight function
 

r
rr

  as follows:



2
1
x
x

 
it
as:
(63)
And eventually, from Equation (7), we wre)(xyn

 
2
2
d1
d
1
nn
n
nn
B
yx x
x
x







 (64)
acobi’sFrom Rodriguez sequence, we will have J
polynomials as [10,11]:

 




,111
2!
d11
dnzz
z
 

We will rewrite Equat
nn
nnn
Pzz z
n




 




ion (64) as:
(65)
 
 
21
2!1 11
2!
d
111
d
n
n
nn n
nnn
n
yxBnxx
n
xxx
x






 





) and (64) and c
sidering
(64)
Now, by comparing Equations (65on-
,



 we will have:
And finally angular wave function will be obtained by
us
 


2,
2!1 1
n
n
nn n
yxBnxPx
 


 (65)
ing Equations (2), (60) and (65):






 



,x

222
2!111
nnn
n
n
nn
y
Bnxx P








This way, we calculated Eigen of functions and Eigen
value of energy. According to this wave function, we can
determine static characteristics of this system. We can
also use this method for solving SE with other non- cen-
tral potentials.
7.
rtain non-central potential by using the
rse, we should note that this method,
ethods for solving SE, does not func-
on with any non-central potential and we can get a
0375-9601(00)00252-8
xxx
(66)
Conclusions
In this paper we showed that Eigen value of energy and
its Eigen dependent functions can be obtained for a sys-
tem under a ce
NU method. Of cou
too, like previous m
ti
suitable result out of this method only with certain types
of potentials which meet the requirements of the method.
And in this paper, by considering all the requirements,
we have introduced the potential in mind and gotten the
results mentioned in the article.
8. References
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als and Spherical Harmonics Using Super-
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