CH. V
Chapter V
Spherically Symmetric Potentials
In this chapter we are going to solve the radial equation (eq.27 in the previous chapter) for several particular
cases. Recalling the radial equation we have

 l l  1 2

 2 d 2U 


 V r U r   EU r 

2
2
2 dr


 2r

(1)
With U r   rRr  and the condition that U 0  0 for  to be finite at r  0 .
The Free Particle:
For free particle we have V(r) =0, the Schrödinger Eq. becomes
Letting
2

 2   E
2
2r
(2)
r   X xY  y  Z z 
(3)
Substituting back in Eq.(2) 
d2
X x   k x2 X x ,
2
dx
d2
Y  y   k y2Y  y  ,
2
dy
d2
Z z   k z2 Z z 
2
dz
(4)
with
k x2  k x2  k x2 
2
E
2
(5)
The solutions of Eqs.(4) gives
 
r   Ck eik r
(6)
Let us now solve the free particle problem as a central force problem:
It is known that
r , ,   Rr Y  , 
To find R(r) we have to solve the radial equation, Eq.(1). which now becomes
1
CH. V
  2 d  2 d  l l  1 2 
r


Rr   ERr 
2
2r 2 
 2r dr  dr 
2E
r  kr
2
Letting  
d 2 Rr 
d
2

(7)

2 dRr   l l  1 
 1
Rr   0
 d 
 2 
(8)
The two linearly independent solutions to Eq.(8) are called the spherical Bessel's functions, i.e.,
R   Al jl    Bl nl  
(9)
Where jl and nl are called, respectively, the lth order spherical Bessel and spherical Neumann functions.
Knowing that nl   
  it is not an acceptable solutions.
0
r , ,    jl krY  ,  
with E 
 for a free particle
2k 2
2
(10)
Comparing Eqs(6 & 10) 


m l
e ik r    C lm jl kr Y  ,  
(11)
l 0 m   l
As a special case if k is along the z-axis

2
CH. V

e ikz  e ikr cos   al jl kr Pl cos  
(12)
l 0
Multiplying by Pl  cos   and integrate 
1
e
ikr cos 
1

1
l 0
1
Pl ' cos  d cos     al jl kr  Pl ' cos  Pl cos  d cos  
Using the orthogonality relation of the Legender's functions we get
1

1
l 0
ikr cos 
 e Pl ' cos  d cos     al jl kr
2
2
 ll '  al jl kr
2l  1
2l  1
2l  1 ikr cos
al jl kr 
e
Pl ' cos  d cos  
2 1

1
(13)
Using the identity
1
jl  z   l
2j
1
e
izs
Pl s d s 
(14)
1
And the relation of Eq.(13), Eq.(12) gives

e ikz   2l  1 j l jl kr Pl cos  
(15)
l 0
This formula is especially used in scattering theory. Using the addition theorem for spherical harmonics one can
generalize Eq.(15) for any arbitrary direction of k to be

 

e ik r   2l  1 j l jl kr Pl kˆ  rˆ
l 0
Some Properties of Bessel's Functions:
jl z  

2z
J l  1 z 
2
3
CH. V
j0 z  
j1  z  
sin z
,
z
n0  z   
sin z cos z

z
z2
n1  z   
cos z
,
z
3
 3 1
j 2 z    3   sin z  cos z
z
z
z
3
 3 1
n2  z    3   cos z  sin z
z
z
z
cos z sin z

z
z2
l
 d   sin z 
jl ( z )   1 z 


 zdz   z 
l
l
l
 d   cos z 
nl ( z )    1 z 


 zdz   z 
l
l
 

a2

J 1 P 2  pq
J

J


d






0   P a    q a 
2
a
a

0
 

a3

2
jl  lP  jl  lq   2 d   jl 1  lP   pq
a 
a
2


  j z 
2
l
dz 


2l  1
The Square Potential Well:
Consider the potential
 V
V r    o
 0
r a
0
a
r
ra
The radial Eq. now reads
V0
  2 d  2 d  l l  1 2 
r


Rr   E  V0 Rr 
2
2r 2 
 2r dr  dr 
And
4
for r  a
(16)
CH. V
  2 d  2 d  l l  1 2 
r


Rr   ERr 
2
2r 2 
 2r dr  dr 
for r  a
(17)
Eq.(16) has the same form as Eq.(7) but now E is replaced by E+V0, that is the solution of Eq.(16) is again
Bessel's function
R   Al jl kr,
with
k
2 E  V0 
2
r a
(18)
For the exterior region (r > a) the solution must vanish as r  ∞. Since E<0 for bound states  Eq.(17) has
also the same solution as Eq.(7) but k now is an imaginary number. Defining
k'  i
 2E
2

R   Bl hl(1) k ' r 
ra
(19)
The boundary conditions requires that the two solutions of Eq.( 18 & 19) must match at r = a and so their
derivatives which give a complicated equation. If one make the substitution
Rr  
U r 
r
(20)
The radial equation now reads


 2 d 2U  l l  1 2

 V r U r   EU r 
2
2
2 dr
 2r

(21)
With
1
 U r U r dr  1

(22)
0
Since V(r)  0 for r>>1, Eq.(21) reduces to
d 2U 2E
 2 U r   0
dr 2

(23)
5
CH. V
If E < 0 Eq.(23) has the solution
U r   e

2 E
2
r
(24)
Also near r=0 V(r) can be neglected compared with r-2, Eq.(21) reduces to
d 2U l l  1

U r   0
dr 2
r2
l0
(25)
The series solution of Eq.(25) requires that

U r    C n r n  
(26)
n 0
Substituting back in Eq.(25) 


n 0
n 0
 C n n   n    1r n  l l  1 C n r n  0
Equating the coefficient of r to zero 
C0    1  l l  1  0

  l
or
  l 1
Now equating the coefficient of r+n to zero 
Cn n   n    1  l l  1  0
for   l 
Cn=0 except for n=0 and the same for   l  1
U r   A r l 1  B r l
But since
U 0  0
Cn n  l n  l  1  l l  1  0
 Eq.(26) now gives
l 1

(27)
B0 
The solution of Eq.(21) that satisfies the boundary conditions can be written as
U r    l 1 e   W  
  kr
(28)
Substituting Eq.(28) into Eq.(21) we get
6

CH. V
 l  1  dW V 2l l  1
d 2W
 2
 1
 
W 0
2
 
d
 
 d  E
(29)
The Coulomb Potential:
The Coulomb potential is defined as V r   
ze 2


r
r
The radial Eq. now reads

 2 d 2U  l l  1 2  

 U r   EU r 
2  dr 2  2r 2
r 
(30)
With
1
 U r U r dr  1

(31)
0
Notice that the radial equation can be though of as a consisting of a radial kinetic energy term and an effective
potential with.
Veff 
l l  1 2
2r
2


(32)
r
There is a positive contribution due to the repulsive potential for l > 0, and a negative contribution due to the
Coulomb attractive potential.
At small r, centripetal potential dominates, and large r
both terms approach zero.
Applying the series solution to Eq.(30) we will not
obtain a two-terms recurrence relation.
In order to simplify the solution we test the limiting
cases, r→ 0, and r→ ∞.
and r→ ∞ , Eq.(30) reduces to
d 2U 2E
 2 U r   0
dr 2

(33)
With the solution
U r   e

2 E
2
r
(34)
7
CH. V
Also near r=0 V(r) can be neglected compared with r-2, Eq.(29) reduces to
d 2U l l  1

U r   0
dr 2
r2
l0
(35)
The series solution of Eq.(35) requires that

U r    C n r n  
(36)
n 0
Substituting back in Eq.(35) 


n 0
n 0
 C n n   n    1r n  l l  1 C n r n  0
Equating the coefficient of r to zero 
C0    1  l l  1  0

  l
or
  l 1
Now equating the coefficient of r+n to zero 
Cn n   n    1  l l  1  0
for   l 
Cn=0 except for n=0 and the same for   l  1
U r   A r l 1  B r l
But since
U 0  0
Cn n  l n  l  1  l l  1  0
 Eq.(36) now gives
l 1

(37)
B0 
The solution of Eq.(30) that satisfies the boundary conditions can be written as
U r    l 1 e   W  
  kr
(38)
Substituting Eq.(38) into Eq.(30) we get
 l  1  dW  k 2l  1
d 2W
 2
 1
 

W 0
2
 
d
 
 d  E
8
(39)

CH. V
Letting
k
E

  0 , Eq.(39) becomes
d 2W
dW
 2l  1   
  0  2l  1W  0
2
d
d
(40)
Applying the series solution to Eq.(40) 

W     C n  n
(41)
n 0
Substituting back in Eq.(40) 




n 0
n 0
n 0
n 0
 Cn nn  1 n1  2l  1 C n n n1  2 C n n n   0  2l  1 C n  n  0
Equating the coefficient of n to zero 


C n 1 nn  1 n 1  2l  1n  1  2n   0  2l  1C n
C n 1 

2n  l  1   0
C
n  1n  2l  2 n
It is clear from Eq.(42) that
(42)
C n1
2
n


Cn
n
2n  n

n!
n 0
C n 1
2 n 1 n! 2


n  1! 2 n n
Cn

Let us now examine the series e 2   
W   behaves like e 2  for large 

 the solution given by Eq.(38) diverge at infinity and that violate the
boundary condition. We have to terminate the series after a finite number of terms say N  from Eq.(33) we let
C N 1  0

Since l & N  0,1,2,
n  N  l 1 
2N  l  1   0

0
2
0

2

 1,2,  n
l  n 1
9
 N  l 1
CH. V
0
Now as
2

n
 02
2 z 2 e 4
n 

4
4E  2
2

z 2 e 4
z 2e4
2
with a  2 is the Bohr radius.
E

2n 2
2an 2
e
It should be noted that Eq.(40) can be written as
 d 2W
dW
 l  1   
 NW  0
2
2 d
d
Letting z  2  ,
z
p  2l  1,
(43)
q  N  2l  1  n  l ,
Eq.(43) becomes
d 2W
dW
  p  1  z
 q  p W  0
2
dz
dz
(44)
Eq.(44) is the associated Laguerre differential equation with the solution
W z   Lqp p z   L2nll11 z 
(45)
The wave-function of the Coulomb potential is now written as
nlm r ,  ,   
2k 3 n  l  1! e kr L2l 1 2krY m  ,  
n l 1
l
3
2nn  l !
Properties of the associated Laguerre polynomials:
Lqp z    1
q
e  zt 1t 

t
1  t  
q 1
p 0
p
dq 0
e z z q d q pq  z


L
z

z e
pq
p! dz p
dz q


Lqp z 
zLqp1 z    p  q  1Lqp z    p  1Lqp1 z 
zLqp1 z    p  z Lqp z    p  q Lqp1 z 

p z q
q
 z e L p z L p' z  
0
 p  q !3 
p!
pp '
10
(46)
CH. V
Table 1: Wave functions and their components
n
1
2
2
2
3
3
3
3
3
3
m
0
0
0
0
1
0
1
0
0
1
0
1
2
0
2
2
11
CH. V
Plots of the radial wave-functions are shown below. It is clear from the graphs, the wave functions of H-atom
vanish at the origin except for l=0 (S states). This means that the capture of the nucleus of an atomic electron
can occur only from a level with l=0. The points at which the wave-function is zero and are called nodes. The
electron has zero probability of being located at these nodes. In general the number of nodes is n-l-1.
12
CH. V
It is of great important to draw the radial probability density as a function of r. from these graphs we conclude:
1-The radial probability distribution function reconciles our intuition that the probability of finding an electron
at the nucleus (r = 0) is zero since the electron and nucleus cannot occupy the same point in space at the same
time.
2- We also see that higher principal quantum number orbitals have higher probability of being further from the
nucleus; they are less strongly bound and can be further away
Now if one calculates the expectation value of the position operator r he found
r
100

2

But  x n e ax dx 
0
r 
3
 1   2 r
   rr ddr   R10 rR10 r dr  4    r 3e ao dr
 ao  0
0

n!
a n1
2
a 0, n is  ve integer
,
3
ao
2
Recall that the probability for finding the particle in the ground state at a distance r from the origin is given by
R10 r 
 1
r  4 
 ao
2 2
3
 2 2 r ao
 r e

Which is maximum at r=a0. In general the probability function has a maximum given by rn =n2a0.
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