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 10 +E 20 =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
(^4012)
1
=+
⎡
⎣
⎢
⎢
⎤
⎦
⎥
πε ⎥
H
m
eZ
r
Z
r
0
2
1
2
2
2
2
(^24) 01 2
=− ()∇ +∇ − +
⎡
⎣
⎢
⎢
Z
πε
⎤⎤
⎦
⎥
⎥
−∇ − =
Z^2
2
20
2
2
02
0
22
00
24 m r^2
eZ
r
ψ rE r
πε
() ψψ() ())
−∇ − =
Z^2
1
20
1
2
01
0
11
00
24 m r^1
eZ
r
ψ rE r
πε
() ψψ() ())
−∇+∇ − +
Z^2 ⎡
1
2
2
20
12
2
(^24) 01 2
11
m
rr
e
rr
()()ψ
πε ⎣⎣
⎢
⎢
⎤
⎦
⎥
⎥
ψψ^0 ()rr 12 =E^00 ()rr 12
p
i
→∇ →rr