1
– --2
Z
( x, y, z ) – --- = E ( X , Y , Z )
r
Where the system of atomic units is adopted. The solution for this equation is:
2
Z
E n = – --------2
2n
with energy given in Hartree. Note that the solutions should be scaled with /me. Usually it is
solved in spherical coordinates because the problem can be separated in spherical polar coordinates. A problem of an 1/r potential can also be sperated in parabolic coordinates, defined as:
x =
cos
= r+z
y =
sin
= r–z
1
z = --- ( – )
2
y
= atan -x
and:
r =
1
2
2
2
x + y + z = --- ( + )
2
This gives in matrix form:
dx
dy =
dz
1
--- --- cos
2
1
--- --- cos
2
1
--- --- sin
2
1
--2
1
--- --- sin
2
1
– --2
–
sin
d
d
d
cos
0
Then:
+
2
2
2
2
ds = dx + dy + dz = ------------ d
4
2
+
+ ------------ d
4
2
+
d
2
and:
1
dx dy dz = --- ( + ) d d d
4
It can be shown that the Laplace operator is equal to:
4 d
= -----------+ d
1
– --2
4 d
-----------+ d
d
d
d
d
4 d
+ -----------+ d
4 d
+ -----------+ d
d
d
2
d
d
1 d
+ ------ 2
d
2
1 d
+ ------ 2
d
2Z
– ----------------( + )
= E
Define now a solution of this equation as:
( , , ) =
with m 0.
Insertion and deviding by 4
1
– --2
d
d
4 1 d
------------ -----+
1 d
im
2e
1
1(
)
2(
im
yields:
d
d
4 1 d
+ ------------ -----+
2 d
1
)e
2
1 2
2Z
+ ------m – ----------------- = E
( + )
Multiply by 1--4- ( + ) :
d
d
1 d
-----1 d
1
d
d
1 d
+ -----2 d
( + ) 2
1
+ -----------------m + Z = – E --- ( + )
4
2
2
and separate the m2 term as well as the Coulomb term in two parts with Z 1 + Z 2 = Z :
d 2
1 d d 1
1 d
1 2 1 2
1
1
-----+ -----+ Z 1 + Z 2 – ------ m – ------m = – --- E + --- E
d
d
4
4
2
2
1 d
2 d
These are in fact two identical differential equations, one for
d
d
d
d
1
The behaviour of the wave function
for large
e
1
– --2
+ Z1
1
1
1 2
– ------ m
4
1
1(
1
= – --- E
2
) and one for
1
in the limiting cases is:
1
--- m
2
for small
A solution is taken of the form:
1
with x =
and
=
= e
1
– --2
1
--- m
2
f 1( )
– 2E this leads to an equation:
2
x
d f1
dx
2
+ (m + 1 – x)
df1
Z
m+1
+ -----1- – ------------- f 1 = 0
dx
2
Again a Laguerre-type equation with solutions:
m
f 1 = L n1 + m ( x )
with a quantisation condition, for each equation:
Z
1
n 1 = -----1- – --- ( m + 1 )
2
Z
1
n 2 = -----2- – --- ( m + 1 )
2
2(
):
Also n1 must be a non-negative integer if is to be real and remains finite. So:
Z
Z
Z
--- = -----1- + -----2- = n 1 + n 2 + ( m + 1 ) = n
This gives the energy quantization:
1
E = – --2
2
1Z
= – --- -----22n
in hartree units. The normalized eiegenfunctions are:
1⁄2
n1 n2 m
1⁄2
im
( n 1! ) ( n 2! )
e
= ------------- ------------------------------------------------------------n ( n 1 + m )! 3 ⁄ 2 ( n 2 + m )! 3 ⁄ 2
2
m+3⁄2
e
1
– --- ( + )
2
(
1
--- m
2
m
) L n1 + m (
m
)L n2 + m (
These eigenfunctions are symmetrical with respect to the plane z=0.
For n 1 > n 2 the density is on the positive side of z, for n 1 < n 2 at the negative side.
)