1
|~r 1 −~r 2 |=
1
√
r 12 +r^22 − 2 r 1 r 2 cosθDo thedΩ 1 integral and prepare the other.
∆Egs=4 π
π^2e^2(
Z
a 0) 6 ∫∞
0r 12 dr 1 e−^2 Zr^1 /a^0∫∞
0r^22 dr 2 e−^2 Zr^2 /a^0∫
dφ 2 dcosθ 21
√
r^21 +r^22 − 2 r 1 r 2 cosθ 2The angular integrals are not hard to do.
∆Egs =4 π
π^2e^2(
Z
a 0) 6 ∫∞
0r 12 dr 1 e−^2 Zr^1 /a^0∫∞
0r^22 dr 2 e−^2 Zr^2 /a^02 π[
− 2
2 r 1 r 2√
r^21 +r 22 − 2 r 1 r 2 cosθ 2] 1
− 1∆Egs =4 π
π^2e^2(
Z
a 0) 6 ∫∞
0r 12 dr 1 e−^2 Zr^1 /a^0∫∞
0r^22 dr 2 e−^2 Zr^2 /a^02 π
r 1 r 2
[
−√
r^21 +r^22 − 2 r 1 r 2 +√
r^21 +r^22 + 2r 1 r 2]
∆Egs =4 π
π^2e^2(
Z
a 0) 6 ∫∞
0r 12 dr 1 e−^2 Zr^1 /a^0∫∞
0r^22 dr 2 e−^2 Zr^2 /a^02 π
r 1 r 2[−|r 1 −r 2 |+ (r 1 +r 2 )]∆Egs = 8e^2(
Z
a 0) 6 ∫∞
0r 1 dr 1 e−^2 Zr^1 /a^0∫∞
0r 2 dr 2 e−^2 Zr^2 /a^0 (r 1 +r 2 −|r 1 −r 2 |)We can do the integral forr 2 < r 1 and simplify the expression. Because of the symmetry between
r 1 andr 2 the rest of the integral just doubles the result.
∆Egs = 16e^2(
Z
a 0) 6 ∫∞
0r 1 dr 1 e−^2 Zr^1 /a^0∫r^10r 2 dr 2 e−^2 Zr^2 /a^0 (2r 2 )∆Egs = e^2Z
a 0∫∞
0x 1 dx 1 e−x^1∫x^10x^22 dx 2 e−x^2=
Ze^2
a 0∫∞
0x 1 dx 1 e−x^1
−x^21 e−x^1 +∫x^102 x 2 dx 2 e−x^2
=
Ze^2
a 0∫∞
0x 1 dx 1 e−x^1
−x^21 e−x^1 − 2 x 1 e−x^1 + 2∫x^10e−x^2 dx 2
=
Ze^2
a 0∫∞
0x 1 dx 1 e−x^1{
−x^21 e−x^1 − 2 x 1 e−x^1 − 2(
e−x^1 − 1)}
= −
Ze^2
a 0∫∞
0[(
x^31 + 2x^21 + 2x 1)
e−^2 x^1 − 2 x 1 e−x^1]
dx 1