SPHERICAL COORDINATES
Next we consider Laplace’s Equation in spherical coordinates. We have seen
2 2   

2
 2 
  2
  ctn  2 

r
  sin 2   2 
r r r 2 
2
Recall that last semester we found that


2
 2 
L   ctn  2 

  sin 2   2 

2
where
is the orbital angular momentum. Hence
 2 2  L2
  2

r r r 2
r
2
We now try for a solution of the forms
V  R  r  Y  ,  
Then
 d 2 R 2 dR  A 2
Y 2 
 L Y0
r dr  r 2
 dr
1  d 2 R 2 dR  1 2
r


 L Y0
R  dr 2 r dr  Y
2
As before this is only possible if each term is a constant. But we found last semester that the
eigenvalues of L2 were ℓ(ℓ+ 1) with ℓ an integer. Hence
L2 Y      1 Y 
Thus
1 2
L Y      1
Y
1  2 d 2R
dR 
r
2r


      1
R  dr 2
dr 
We try
R  r
Then
r 2    1 r 2  2r r 1      1 r 
 2        1      1
This has solutions
  ,
      1
Hence
B 

V   r    1  Y m  ,  
r 

Where
Y m
is the eigenfunction of L2 and Lz. We found last semester that these were generated from
 
 
L   ei   ictn   L x  iLy
 
 
 
 
L   e i   i ctn   L x  iLy
 
 
L  ,  z       1   z   z  1 
Recall that we started by requiring
1/2
,  z  1
L  ,  z min  0
 ,  z min L  L  ,  z min  0
But
L  L    L x  iL y  L x  iL y   L2  L 2 2  i  L y , L x   L2  L2z min  L z min
Similarly
L  ,  z max  0  L2  L2z max  L z max  0
But we also found that
L z max  L z min  n  1, 2,3,
L2  L2z min  L z min  L2   L z min  n    L z min  n   L z min  2nL z min  n 2  L z min  n
2
nn 
 L2    1 
     1
2  2   n/2
At this point ℓ would appear to be either integer or half integer. But we also had
n
i 

Y n
L z  i  i
 Ye 2

 2
But we know that
    2       
n
2
integer. Thus
  0,1, 2,;  ,    1,,    z
Henceforth, we denote ℓz by m. Then the eigenstates are denoted as
Y m
To find them we again use
L  Y   0
 
 
 e i   i ctn   Y   0
 
 
 
 
   i ctn   e i  0
 
 


In
d
 ictn i  0
d
d
d
  ctn  
  ctn 
d


  In  sin      Asin  
A
We normalize this by
A
2

2
  sin
2
 sin  d d  1
0 0


  !  1
 2A  sin 2  1  d  2A 2  2 2  1
 2  1!

2
2
0
1   2  1!
A  


 !2  1   
1/2
It is traditional to use + for even ℓ and – for odd ℓ. Thus
Y0 0 
1
 4 
1/2
1/2
Y11
 3 
    sin  ei
 8 
1/2
1  15 
Y2    sin 2  e 2i
4  2 
2
We can now generate all the other m values by applying L.
L  Y11  21/2 Y10  Y10 
1
L  Y11
2
1/2
 
1  i  1
 3 

e   i ctn   1   sin  e i
 
2
 8 
 
1/2
 3 


 16 
Proceeding in this fashion we find
e
2 
1/2
3
cos  e  i cos  e    

  4 
i
i
cos 
0
 1
1/2
 1 
Y0   
 4 
0
2
1/2
Y11
 3 
   sin  e 
 8 
Y10
 3 
 
 4 
1/2
1  15 
Y2    sin 2  e 2i
4  2 
2
1/2
Y1
1
1/2
cos 
1/2
 3 
   sin  e i
 8 
Y21
 15 
    sin  cos  ei
 8 
1/2
1
 5  3
Y20     cos 2   
2
 4   2
Y2
1
Y2
2
1/2
 15 
   sin  cos  e i
 8 
1/2
1  15 
   sin 2  e 2i
4  2 
etc.
From the orthogonality condition we have
2

  sin  dd Y 'm 'Y,m    ' mm '
0 0
We can now write a general solution of Laplace’s equation in spherical coordinates as
B 

V  r, ,      A e r    1  Y m  ,  
r 
 ,m 
Azimuthal Symmetry (m = 0)
Then
1
d d 2 

ctn
      1
d d 2 

change variables to x = cos . Then
x
1  x 
2 1/2
dx d dx d  dx d 

     1   0
d dx d dx  d dx 
d
  sin    1  x 2
dx


1/2
Let
  x   P  x 
x
 11  x 
1/2
1  x 
2
1/2

dP
x
1/2 
  11  x  

dx
1 x2



dP
 1 x2
1/2
dx


1/2

d 2 P 
      1 P
dx

2
dP
2 d P
 2x
 1 x
     1 P  0
dx
dx 2


As usual we solve this by series. Let

P   a n x n
n 0
Then

 2xna n x n1  a n n  n  1 x n2  a n n  n  1 x n      1 a n x n   0
n 0

  a n 2n  n  n  1      1  a n  2  n  2  n  1 x n  0
n 0
 n 2  n      1 
 a n 2  a n 

  n  2  n  1 
As n we get an+2 =-an. Hence at  =   x = -1 we have the ratio of successive terms
a n 2 x n 2
1
anxn
Hence series diverges. Thus it must terminate. This means ℓ is an integer as we had already
found. It is customary to choose a0 and a1 so that Pℓ(1) = 1. Then we find
P0  1,
P1  x,
P2 
P4 


1
3x 2  1 ,
2

P3 

1
5x 3  3x
2


1
35x 4  30x 2  3 
8
We then have the general solution for azimuthal symmetry

B 

V    A  r    1  P  cos  
r 
 0 
The Pℓ are, of course, orthogonal (since the Yℓm are). However, they are not orthonormal.
Instead
1
 P  x  P '  x  dx 
1
2
  , '
2  1
This is because we choose Pℓ(1) = 1 rather than
1
 P  x  P  x  dx  1
1
We can use the Pℓ to find an important expansion of
1
 
r  r'
If the angle between and  is γ we have
 
    1/2

  1/2
r  r '   r  r ' r  r '    r 2  r '2  2r  r '
  r 2  r '2  2rr 'cos  
1/2
If we rotate axis so that  lies along ̂ , the problem has azimuthal symmetry. Thus


 
B 
1
B 



r  r '    A  r   1  P  cos   


  A  r   A  r    1 
0
r  r '  0
r 
r 

 0 
Expanding the LHS gives
1    r 
  
r   0  r 


 
B 
    A  r    1 
r 
  0 

1
r 
      1 P  cos  
r  r '  0 r
This also enables us to find a “generating function” for Pℓ (recall last semester). We have
1
2
 2

 r  r '  2rr 'cos  
1/2
1
   
r  r'
1
  r 2
r 

r 1     2 x 
r 
  r 

1/2
where r< is the smaller of r and r, r> is the larger of r and r, and x = cos γ. Let t = r</r>. Then

1
1  t 2  2xt 


1/2
  t  P  x 
 0
To see how this works we expand the LHS as
1


1  2xt  t 2 


1/2
1


 
1
3
2xt  t 2  2xt  t 2
2
8
1 3  5
2xt  t 2
246


3


2
1 3  5  7
2xt  t 2
2  468


4

Then
t 2 3 2 2 3 3 3 4 1 3  5
1  t   x t  xt  t 
8x 3 t 3  12x 2 t 4  6xt 5  t 6  
2 2
2
8
246


5 
 1 3 
 3
 1  t  x   t 2    x2   t3   x  x3   
6 
 2 2 
 2
 P0  1;
P1  x ;
as above.
P2 


1 2
x 1 ;
2
P3 

1
5x 3  3x
2

We now use the generating function to find the normalization factor above
1
  t    'P  x  P '  x 
2
1  2tx  x
 , '
1
1
1

  t    '  P  x  P '  x  dx

2
 1  2tx  x
 , '
1
1



n 1  2tx  x 2
2t

1

1
n 1  t   n 1  t  1 1  t
 n
t
t 1 t

1  t3 t5 t7
2t n 1
 2  t       
  t 2t  P 2  x dx
t 
3 5 7

 n odd n
(because Pℓ are orthogonal). Now let n = 2ℓ + 1. Then
1
2t 2 
 2  1   t 2  P 2  x  dx


1
Hence
1
2
2

 P  x  dx 

2  1
1
as above.
We can also use the generating function to obtain a recursion relation for Pℓ
1  2t  t  t
1
2
2 1/2
1  2tx  t 



 1  2tx  t 2

x  t
2 1/2
1  2tx  t 


But

1
1  2tx  t 2 



 1  2tx  t 2
1/2
  t

  t  P  x 
 0
P   x  t   t  P
 1

  t 1  2xt   t 1  xt   t 1  P  0


xt
1  2tx  t 2 


1/2


t     1 P 1   2x  x  P  P 1   0
P 1    1  x  2  1 P  P 1
P 1 
 2  1 xP  P 1
 1
This is a very useful relation both for doing integrals and for finding higher order Pℓ.
Spherical Harmonic Addition Theorem
We now derive another form of the expansion of
1
 
r  r'
which is useful in doing integrals for non-azimuthal symmetries. Consider a sphere of radius R.
The potential on the surface of the sphere is
V  R, ,  
Then outside the sphere we have
B 

V  r, ,      A m r   m1  Y m  ,  
r 
 ,m 
But
V  , ,    0  A   0
Then
B m
Y ,  
 1 m 
R
 ,m
V  R, ,    

  V  R, ,  Ym' '  ,   R 2 sin  d d

B m
B m
B 'm '
m
m'
Y
,
Y
,
sin
d
d
















'
'
mm
'



 1 
 1
R  '1
 ,m R
 ,m R


 B m  R  1  V  R, ,  Ym  ,   sin  d d
Then

R  1 m
V  r, ,      1 Y  ,    V  R,  ',  ' Ym   ',  ' sin  'd 'd '
 ,m r
We now imagine two sets of axis with same origin but rotated so that

r   r, 1 , 1    r,  2 ,  2 
Then since V is a scalar, as are r and R, we must have

 Ym  1, 1   V  R, 1 ', 1 ' Ym  1 ', 1 ' sin 1 ' d1 'd1 '
 ,m

  Ym   2 ,  2   V  R,  2 ',  2 ' Ym   2 ',  2 '  sin  2 ' d 2 'd 2 '
 ,m
Since V(R,,ϕ) is arbitrary we must have

m

m
 Ym  ,   Y  1 ', 1 '   Ym  ,   Y  2 ', 2 '
m
m
(since it is a power series in 1/rℓ+1 each ℓ term must match). The (,ϕ), (,,ϕ) correspond to
and  in the two systems . If we now let  be along the ̂ 2 axis we have 2 = 0. Then only the
m = 0 terms on the RHS are 0. Thus

m
 Ym  1, 1 Y  1 ', 1 '  Y0  2  Y0  0 
m
We know that
Y0     KP  cos  
and that
1
2
 P  x  dx  2  1
2
1
while

2
 

Y0 Y0 sin  d d
0 0

 2


Y0 Y0 sin  d
0
 2K
2
1

1
1

 2  Y0  x  Y0  x  dx
1
1/2
P2
4
 2  1 
K2 1  K  
 x  dx 

2  1
 4 
1/2
 Y0
 2  1 


 4 
P

 2  1 
  Ym  ,  Ym   ',  '  
 P  cos   P 1
 4 
m
 4 
m
m
Y
,
Y
 P  cos    




   ',  ' 
 
 2  1  m
where γ is the angle between and . Thus
1
r  4  m
m
Y
,
Y




     1 
   ',  ' 
 
r  r '  ,m r  2  1 
We are now in a position to solve any Laplace’s Equation problem where the boundary
conditions are given on spheres.