240
PART 2
QUANTUM MECHANICS AND SPECTROSCOPY
Substitution of this into the Schrödinger equation gives:
(db12.6)
Now, multiply by:
r
2
/[R(r)Y(θ,φ)]
(db12.7)
The result is:
(db12.8)
The terms with the same variables are grouped, giving:
(db12.9)
The term in the first bracket depends only upon the variable r, whereas the second depends
only upon θ and φ, so they must both be constants for the sum to always be a constant.
Let us define the separation constant as such that:
(db12.10)
Then the radial equation becomes:
(db12.11)
Angular solution
The angular part of the equation is:
Λ
2
(θ,φ)Y(θ,φ) = −l(l + 1)Y(θ,φ)
(db12.12)
We need to substitute the definition for Λ:
(db12.13)
1
1
2
2
2
sin
( , )
sin
sin
θ
δ
δφ
θ φ
θ
δ
δθ
θ
δ
δθ
Y
+
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ Y Y
ll
Y
( , )
(
) ( , )
θ φ
θ φ
= − + 1
−
+
−
−
+
Z
Z
2
2
2
2
0
2
2
4
m
r
R r
d rR r
dr
e r r E
( )
[ ( )]
πε
2 2
2
1 0
m
l l
(
)
+ =
Λ
2
1
( , ) ( , )
( , )
(
)
θ φ θ φ
θ φ
Y
Y
l l
= − +
−
+
−
−
⎛
⎝
⎜ ⎜
Z
2
2
2
2
0
2
2
4
m
r
R r
rR r
r
e r Er
( )
[ ( )]
δ
δ
π ε
⎞ ⎞
⎠
⎟ ⎟ −
⎛
⎝
⎜ ⎜
⎞
⎠
⎟ ⎟ =
( , )
( , ) ( , )
Z
2
2
2
1
m Y
Y
θ φ
θ φ θ φ
Λ
0 0
−
+
Z
2
2
2
2
2
1
m
r
R r
rR r
r
Y
Y
( )
[ ( )]
( , )
( , ) ( ,
δ
δ
θφ
θ φ θ
Λ
φ φ
πε
)
⎧
⎨
⎪
⎩ ⎪
⎫
⎬
⎪
⎭ ⎪
+
−
=
e r Er
2
0
2
4
−
+
Z
2
2
2
2
2
2
1
1
m
Y
r
rR r
r
R r
r
Y
( , )
[ ( )]
( )
( , )
θ φ
δ
δ
θ φ
Λ
( ( , )
( ) ( , )
(
θ φ
πε
θ φ
⎧
⎨
⎪
⎩ ⎪
⎫
⎬
⎪
⎭ ⎪
+
−
=
e
r
R r Y
ER
2
0
4
r r Y
) ( , )
θ φ
9781405124362_4_012.qxd 4/30/08 20:24 Page 240