Determine the Eigen Function of Schrodinger Equation with Non Central Potential by Using NU Method

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 Physics, Shahid Chamran University, Ahvaz, Iran E-mail: [email protected]

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 Schrodfind-inger equation 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 needs in order to solve SE by NU method:

 

2 22

 

cot , A B D

V r

r r r



   

After choosing a suitable change of variabless r

 

, the transformed equation will be as follows:

 

 

_{   }

2

 

_{ }

 

0

n n n

s s

s s

s s

 

 

  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:

 

  

n s n s yn s



  (2)

Equation (1) will be reduced to the following equation which is of hyper geometric type [5];

       

s y_n s s y_n s y_n

 

s 0

      (3)

That in equations:

 

   

_{ }

s

s s

s 

 

 

 (4)

 

s

 

s 2



s



    ,  0 (5)

is a parameter which is defined as follows[5]:

  

1

  

, 0,1, 2, 2

n

n n

n s s n

The polynomial  ( )s shows that its first derivative must be negative. We should notice that  and n are obtained from a particular answer of y s

 

 yn

 

s

which is a polynomial of n degree. Furthermore, state-ment yn

 

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 n

n n

n n

n

y s s s

s  

 





(7)

In which Bn is the normalization constant and 

 

s is a weight function which has to meet the following condition [5,6].

 

 

_{   }

d d

s

s s

s s



 



 , 

 

s 

   

s  s (8)

Function 

 

s and parameter  are defined as follow:

 

   

   

2

 

 

2 2

s s s s

s     s K s

           

 

 



(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 obob-tained 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

2

0

r m E V r



   

  r (11)

Wave function_n

 

r elaborates certain states and their related energy levels, n, for a particle in a poten-tial field. First, central and non-central potenpoten-tials

E

 

 

2 22

cot ,

r A B D

V V r

r r r

 

   

are considered in spiracle coordinates and put them in the above equation. (2mh2J)

 

 

 

 

2 2

2

2 2 2 2

2

2 2

1

1 1 1

sin

sin sin

cot

0 r

r

r r

r

r r

r

A B D

E r

r r r



 



 

 

  

 



 

 _  _



      

  _  _  _ 

   

 

 

_    _ 

 

(12)

By considering the whole wave function as:

 

r



r, , 



R r

    





       (13)

And replacing in Equation (12), we can write the equation separately as follows:

 

 

_{ }

2

2

2 2

d 2d 1

0 d

d

R r R r

Er Ar B R r

r r

r  r      (14)

 

 

_{ }

2 2

2

2 2

d d

cot cot 0

d

d sin

m D

 

   



 

   

 _   _ 

 

(15)

 

_{ }

2

2 2

d

0

d m



 



   (16)

In which m2 and  l l



1



are the separation fix amounts. The answer for Equation (16) is as follows [9].

 

1

e , 0, 1, 2, , 2

im

m m l



 

    

   (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 E  E 2 in Equation (14) we will have:

 

 

_{ }

2

2 2

2 2

d 2d 1

0 d

d

R r R r

r Ar B R r

r r

r  r      (18) Now if we compare the above equation with Equation (1), the general form of equation in NU method, we will have:

 

r 2

  ,

 

r r,

 

2 2



r r Ar B



      (19)

Equation (9), we can write 

 

r as follows:

 

1 1 2 2









4 4  4 1

2 2

r r K A r B

      _   _

(20) From the above equation, considering that under the radical there should be a of a first degree polynomial.

 









1 1 2 4 2 4   B B        1 2 2 1 1 1 2 2 r B r r B      _{      }         _{      }    

We will determine K and we will have:









1/ 2 1 1/ 2 2 4 1 4 1 K A K A      

   _  _ (21)

To continue, we choose the suitable amount of 

 

r

which can meet the condition   0. So:

 





 













1/ 1/ 2 2 4 1 4 1 r r r r                     2 1 2 2 1 1

(i) 4 1

2 2 1 1

(ii) 4 1

2 2

B

B

K A B

K A B

   _{       }         _{       }            (22)

Now, for (i), we can write 

 

r

2 4

r r

from Equation (5) as follows:

 



B



      (23)

And from Equations (6) and (10) we will calculaten and  respectively

2 n n   





(24) 1

4  1

 4  





2 ₁ B A   _   _  _

  (25)

So:









1/ 2

, 2 1 1

n n A B

     _    _

  (26)

And eventually, we can obtain Eigen value of energy for (i) from the above equation:





1 2

4 1 2

n A B n             

 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 

 

r as:

 

r 2 r 4



B



       (28)

And from Equations (6) and (10) we calculate n and :

2

n n

   (29)





12

4 1

A B

  __   _ 1_

  (30)

n

 





1 2

2n A  4 B  1 1

    __   _  _

  (31)

From the above equations, Eigen value of energy is obtained:





12





4 1 2

n A B n     1              (32)

We considered 2

E  and from (27) and (32), we have Eigen value of energy as:









2

2 1

2

4 1 2 1

n

A E

B  n

               (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

, ln d d

s s

s s s

s s s

 

 _{ }

 

 

_

(34)

Thus, for (i) and (ii) in Equation (22) we have:

 









1 1

2 4 1

2 2 1 1

2 2 2

r r B

r           _ _   _     (35)

in which  4



B 



1 and considering 

 

r r

We will have:

 

 

( 1) 2 ( 1) 2 ) e ) e r r

i r r

ii r r

        _        _  (36)

On the other hand, from Equation (8) we have:

 

 

   





 

 

d

ln ( ) d d

w s t s t s

w s w s s

s  s S  



s S (37)

From Equation (37), as we have 

 

r  2r in which    1, we will have:

 

2

e r

r r 

 _   

So we can write the weight function

 

 

 

r r r   

 as

follows:

 

2

e r

r r  

 _  

(39) And to continue, from Equation (7), we will write

 

n

y r as follows:

 



2

2 d e e d n n n

n r n

B

y r r

r r

 

 

   

 r



(40)

From Rodriguez sequence polynomial, we will have Laguerres dependent functions as [9]:

 

e d



e e ! d

x k n

k x

n n

x

L x x

n x



 

 n k



(41)

By comparing the two functions, (39) and (40) and by considering k  and 2rx we will have:

 

!

 

1n



2



n n n

y r B n  L r (42) In the same way for (ii) in which

from Equation (37) we will have:

 

r 2 r 1

     

 

_{r r}( 1)_e2r

 _  

(43)

 

 

_{ }

2

e r r r r r      

  (44)

And finally, from Equation (7), we can write yn

 

r

as follows:

 



2

2 d e e d n n

n r n

Bn

y r r

r r

 

 

  

 r



(45)

By comparing Equations (41) and (45) and consider-ing k and x2r we will have:

(ii) yn

 

r B nn !

 

1nLn



2 r



 _

  (46) Therefore, from Equations (2), (35), (36), (42), (46) the whole radial wave function can be written as:

 

 





 

 





( 1) 2 ( 1) 2

(i) 1 ! e 2

(ii) 1 ! e 2

n r

n

n r

n n

R r B n r L r

R r B n r L r

      n     _{ }         _{ }  (47)

Normalization coefficient is determined by

 

2

2 0

d r R r r





 

 



1 by taking orthogonal condition of associated Laguerre polynomials and we will have:







 









 



2 2 2 )

! ! 2 1

2 )

! ! 2

n

n

r i B

n n n

r ii B

n n n

             _              1 (48)

5. Solving Angular Equation and

Calculating Eigen Values

With a variable change as xcos we will tion (15) as the following transformation:

have

Equa- 









2



2 Dx2

 

2

2 2

d d ( ) 2

d d 1 1 x x x x x x     2 2 1 0 1

x m x

x

      (49)

By comparing Equations (49) with (1) we will have:

 



 

x 2x

   ,

 



2



1

x x

  

 





2



2



x D x m

       (50)

By using (9), we write 

 

x as:

 

_x



_{D K x}



2



_m2 _K



         (51)

We determine K on condition that there gree polynomial under the radical:

is a first

de- 

2 2 m D x m D  _{ }  x         (52) 1 2 2 K D K m       0  

And for we have the right choice as:

 





1/ 2

2

x m D x

    (53)

2 2

K   m (54) m Eq

Also, fro uation (5), we can obtain 

 

x

e will hav

Then

w e:

 





₂



1/ 2



2 1

x D m x

     (55)

According to Equations (6) and (10), we will calculate n

 and  respectively:



2



12





2 1 1

n n m D n n

      

  (56)



 



1/ 2

2 2

m D m

    ,  n (57)

Finally we will obtain  from the above equation:



 







1/ 2

2

3 2 1

n n n D m m

 _{     } 2

6. Calculation Eigen Function of A

Equation

To fi by

using Equation (4) we will have:

s

ngular

nd Eigen function related to angular equation,

 





 

_{ }

ln s s ds s

 





_

(59)

Thus, for

 

x  x in which



2



1/ 2

D m

   and

considering that 

 

x  



1 x2



we will have:

 





/ 2

2

1

x x 

  

er hand, fro (8) we have: (60) On the oth m Equation

 





 

_{ }

ln s s ds

s





 

_

(61)

As 

 

x  2 1







x , so:

 



2



1

x x 

   (62)

Therefore, we can write the weight function

 

r

 

_{ }

r

r

 



 as follows:



 



2

1

x x 

  

it as:

(63) And eventually, from Equation (7), we wr e

y

_n

(

x

)

 



2





2



d 1 d 1

n

n n

n n

B

y x x

x x

 



 

 _  _

 

 (64)

acobi’s From Rodriguez sequence, we will have J polynomials as [10,11]:

 

_{ }

_

_{ }

_





 

_

_

 

, 1

1 1

2 ! d

1 1

d n z z

z

 _   _

We will rewrite Equat

n n

n

n n

P z z z

n

 

 

 

  

 

 

 

  

 

ion (64) as:

(65)

 

  













 



2 1

2 ! 1 1 1

2 ! d

1 1 1

d n n

n n n

n

n n

n

y x B n x x

n

x x x

x

 

 



  

 

_   _ 

 

  _   _

) and (64) and c sidering

(64)

Now, by comparing Equations (65 on-,

    we will have:

And finally angular wave function will be obtained by us

 

  



2  , 

_{ }

2n ! 1n 1

n n n

y x _B n  _ x P  x (65)

ing Equations (2), (60) and (65):

 

   

  







 , 

_{ }

x

 

2 2 ₂

2 ! 1 1 1

n n n

n n

n n

y

B n x x 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 nfunc-on-central potential and we can get a

0375-9601(00)00252-8

x x x



(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

[1] B. Gonu and I. Zorba, “Supersymmeric Solutions of Non- central Potentials,” Physics Letters A, Vol. 269, No. 2-3, 2000, pp. 83-88. doi:10.1016/S

] R. Dutt, A. Gangopadhyaya and U. P Sukhatme, “Non-als and Spherical Harmonics Using Shape Invariance,” American Journal of

reen Functions [2

central Potenti symmetry and

Physics, Vol. 65, 1997, pp. 400-403.

[3] Y. Aharonov and D. Bohn, “On the Measurement of Ve-locity of Relaivistic Particles,” Il Nuovo Cimento (1955- 1965), Vol. 5, No. 3, 1957, pp. 429-439.

[4] B. M. Mandel, “Path Integral Solution of Noncentral Potential,” International Journal of Modern Physics A, Vol. 15, No. 9, 2000, pp. 1225-1238.

[5] A. F. Nikiforov and V. B. Uvarov, “Special Functions of Mathematical Physics Birkhauser,” Basel, 1988.

[6] H. F. Jones and R. J. Rivers, “Which G

does the Path Integral for Quasi-Hermitian Hamiltonians Represent?” Physics Letters A, Vol. 373, No. 37, 2009, pp. 3304-3308.

doi:10.1016/j.physleta.2009.07.034

[7] S. Sakoda, “Exactncess in the Path Integral of Coulomb Potential in One Space Dimension,” Modern Physics Let-ters A, Vol. 23, No. 36, 2008, pp. 3057-3076.

doi:10.1142/S0217732308028491

[8] F. Cooper, J. Ginocchio

between Suprsymmetry and Solvabl

Physi-cal Review D, Vol. 36, No. 8, 1987, pp. 2458-2473.

doi:10.1103/PhysRevD.36.2458

[9] R. Dutt, A. Khare and U. P. Sukhatme, “Is the Order Supersymmetric WKB App

Lowest roximation Exact for All Shape Invariant Potentials,” American Journal of Physics, Vol. 56, No. 2, 1988, pp. 163-168.

doi:10.1119/1.15697

[10] H. Katsura, H. Aoki and J. Math, “Exact Supersymmetry in the Relativistic Hydrogen Atom in General

Superchange and Ger

Dimension elized Johnson-Lippman Operator,” Physics, Vol. 47, 2006, pp. 032301.