CHAPTER 12
THE HYDROGEN ATOM
263
This is first solved at zeroth order; that is, neglecting the interactions
between electrons. These solutions are denoted by the superscript zero:
(12.48)
Assume that the wavefunction can be separated as the product of two
contributions:
ψ(r 1 r 2 ) = ψ(r 1 )ψ(r 2 )
(12.49)
This yields two separate equations:
(12.50)
(12.51)
where
E 1
0
+ E 2
0
= E
0
(12.52)
The first-order correction for energy is determined by modifying
Schrödinger’s equation to include the interaction. First, rewrite
Schrödinger’s equation using:
(12.53)
(H
0
+ H
1
)(ψ
0
+ ψ
1
) = (E
0
+ E
1
)(ψ
0
+ ψ
1
)
(12.54)
Ignoring the two second-order terms yields:
H
1
ψ
0
+ H
0
ψ
1
= E
1
ψ
0
+ E
0
ψ
1
(12.55)
H
0
ψ
1
− E
0
ψ
1
= −H
1
ψ
0
+ E
1
ψ
0
(12.56)
H
e
r
1
2
0
1 2
4
1
= +
⎡
⎣
⎢
⎢
⎤
⎦
⎥
⎥
πε
H
m
e
Z
r
Z
r
0
2
1
2
2
2
2
0
1
2
2
4
(
)
= −
∇ + ∇ −
+
⎡
⎣
⎢
⎢
Z
πε
⎤ ⎤
⎦
⎥
⎥
−
∇
−
=
Z
2
2
2 0
2
2
0 2
0
2
2
0 0
2
2
4
m
r
e Z
r
r
E
r
ψ
πε
ψ
ψ
( )
( )
( ) )
−
∇
−
=
Z
2
1
2 0
1
2
0 1
0
1
1
0 0
1
2
4
m
r
e Z
r
r
E
r
ψ
πε
ψ
ψ
( )
( )
( )
)
−
∇ + ∇
−
+
⎡
Z
2
1
2
2
2
0
1 2
2
0
1
2
2
4
1 1
m
rr
e
r
r
(
) ( )
ψ
πε ⎣ ⎣
⎢
⎢
⎤
⎦
⎥
⎥
=
ψ
ψ
0
1 2
0 0
1 2
( )
( )
rr
E
rr
p
i
r
r
→ ∇
→
Z
9781405124362_4_012.qxd 4/30/08 20:25 Page 263