Multiply by ψ0*and integrate:
(12.57)
The operators used in Schrödinger’s equation are called Hermetian oper-
ators and have the property:
(12.58)
and
(12.59)
Resulting in the expression for the first-order energy correction:
(12.60)
For the n 1 =1, n 2 =1 state this is:
(12.61)
dτ 1 =r 12 sinθ 1 dr 1 dθ 1 dφ 1 (12.62)
dτ 2 =r 22 sinθ 2 dr 2 dθ 2 dφ 2
The term r 12 must be rewritten in spherical coordinates, where the >or
<signs denote the larger or smaller values of r 1 or r 2 :
(12.63)
Experimentally, a value of −79.01 eV is determined for the n 1 =1 and
n 2 =1 state of helium. Substitution of these values into E^0 gives −108.8 eV.
The first-order term has a value of +34.0 eV, giving −74.8 eV, and using
up to the third order gives −78.9 eV.
Spin–orbital coupling
The electron spin influences the energies of the electronic states because
associated with the spin is a magnetic dipole moment. Likewise, the orbital
angular momentum, for the states with lgreater than 0, will possess
a dipole moment due to the angular momentum. The interaction of
the spin magnetic moment with the magnetic field arising from orbital
14
rl 12 211 11
r
r
YY
ml
l
l l
m
l
m
l
= ll()
∑∑ +
<
>
+
π
θφ*(()θφ 22
E
Z
a
(^12) eeZr a Zr a
6
0
6
2
12
(^110) / (^22)
,
= /
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟∫∫−−
π
00
2
012
4 12
e 1
πε r
ττ
⎛
⎝
⎜⎜
⎞
⎠
⎟⎟dd
EH^1010 =∫ψψτ* d
∫∫ψψτ^010 EEdd==1 0 0ψψτE1**
∫∫ ∫ψψτψψ τψψ τ^001 HH Eddd===^100 ()^100 ()∫ψψψτ001E d
∫∫ ∫ψψτψψτ ψψτψ^001 HE Hdd d−=^001 −+^010 ∫∫^0 E^10 ψτd