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 θand 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)
11
2
2
sinθ^2 (,) sin sin
δ
δφ
θφ
θ
δ
δθ
θ
δ
δθ
Y +
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟YYllY(,)θφ=− +( 1 ) (,)θφ
−
+
−
−+
ZZ^22
2
2
0
2
24 m
r
Rr
drRr
dr
er
rE
()
[()]
πε
22
2
10
m
ll()+=
Λ^2
1
(,) (,)
(,)
()
θφ θφ
θφ
Y
Y
=− +ll
−+
−
−
⎛
⎝
⎜⎜
Z^22
2
2
0
2
24 m
r
Rr
rRr
r
er
Er
()
δ[()]
δπε
⎞⎞
⎠
⎟⎟−
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟=
(,)
(,) (,)
Z^22
2
1
mY
Y
θφ
Λ θφ θφ 00
−+
Z^22
2
2
2
1
m
r
Rr
rRr
rY
Y
()
[()]
(,)
(,)(,
δ
δθφ
Λφθφ θφ
πε
)
⎧
⎨
⎪
⎩⎪
⎫
⎬
⎪
⎭⎪
+
−
=
er
Er
2
0
2
4
−+
Z^22
22
2
2
11
m
Y
r
rRr
r
Rr
r
(,) Y
[()]
θφ () ( , )
δ
δ
Λ θφ((,)θφ ()(,) (
πε
θφ
⎧
⎨
⎪
⎩⎪
⎫
⎬
⎪
⎭⎪
+
−
=
e
r
RrYER
2
(^40)
rrY)(,)θφ