Biomedical Engineering Reference
In-Depth Information
1 with all remaining terms zero. If at least one of the spinorbitals are different, say
U
=
U
,
then the complete overlap integral is zero. Thus
D
1
D
2
dτ
n
D
2
0 otherwise
This gives a Slater determinant normalizing factor of 1/
√
n
!
if
D
1
=
=
!
.
The rules for sums of one- and two-electron operators can be found in more advanced
texts such as Eyring
et al.
(1944); all we need to note are the following results.
1. If two or more spinorbitals are zero, then the expectation value of a sum of one-electron
operators is zero.
2. If three or more spinorbitals are different, then the expectation value of a sum of two-
electron operators is zero.
14.9 Hartree Model
We now return to the problem of the helium atom. We have established that the electronic
wavefunction would be exactly a product of hydrogenic orbitals in the absence of electron
repulsion. We have also seen that neglect of electron repulsion leads to impossibly poor
agreement with experiment. The orbital model is an extremely attractive one, so the question
is how can we allow for electron repulsion in some average way, whilst retaining the orbital
picture? Douglas R. Hartree's solution to the problem was to allow each electron to come
under the influence of an
average
potential due to the other electron and the nucleus.
Suppose for the sake of argument that electrons 1 and 2 occupy orbitals ψ
A
and ψ
B
(which
might be the same, but are to be determined). We could have a guess that the ψ's might be
hydrogenic, to get the calculation started.
According to the familiar Born interpretation, ψ
B
dτ is a probability and so we can regard
the second electron as a charge distribution, with density
e
ψ
B
. Electron 1 therefore sees
a potential due to the nucleus and the smeared-out electron 2, as shown in Figure 14.3.
−
electron 1
r
12
r
1
τ
d
Ze
electron 2
Figure 14.3
The Hartree model
