Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2009
Laura Gagliardi
Lecture 27, November 16, 2009
(Some material in this lecture has been adapted from
Cramer, C. J.
Essentials of
Computational Chemistry
, Wiley, Chichester:
2002; pp. 116118.)
Solved Homework
In the groundstate He atom at the first step, each electron is assumed to occupy a
hydrogenic 1s orbital. Recall that, with
Z
= 2 for He, this orbital is defined as
!
100
r
,
"
,
#
(
)
=
8
$
%
&
’
(
1/2
e
)
2
r
The square of this wave function defines the charge cloud that creates the repulsive
potential. Thus, at any point
r
0
the total repulsion
E
felt by a point charge
q
at point
r
0
will be
E
r
0
(
)
=
q
1
r
"
r
0
#
100
r
,
$
,
%
(
)
0
2
&
’
0
&
’
0
(
’
2
r
2
dr
sin
$
d
$
d
%
where somewhat sloppy notation has been used to try to keep things simpler. The
distance between point
r
0
and any point in the integration depends on all three polar
coordinates (not just
r
), so the integral is not necessarily easy to solve.
What is easier to notice is that, since the electron we are interested in is itself
also
delocalized (and not a point charge
q
), the total repulsion is a
double
integral, where we
integrate over the space of the first electron the repulsion felt by the fractional charge at
any point defined by the equation above that involves integrating over the space of the
second
electron. The result is just our old friend the Coulomb integral
E
=
!
100
1
( )
2
""
1
r
12
!
100
2
( )
2
dr
1
dr
2
where each
dr
now stands for a complete spherical polar differential volume element. The
solution to this integral is fairly painful, but can be achieved through some heroic
coordinate transformations. We’ll forego the pleasure.
Since 1s orbitals have their maximum amplitudes at the nucleus, electron 1 will be
most repelled by electron 2 near the nucleus, and less so further away. So, electron 1 will
tend to localize its density further from the nucleus than would be true were electron 2
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272
not to be there (which is what
ψ
100
assumes, since it is a oneelectron orbital, not a two
electron orbital). We've seen this previously in discussion of the variational process
where we treated the He atomic number as a parameter and determined it to be 1.69—this
drop from 2 causes the electron to localize further away.
Of course, each electron does this to the other, so in the next step, the repulsion
near the nucleus felt by each will be reduced (because each electron is now spread out
further), and after step 2 the electron will contract back in a bit. Now, at step 3, since
charge has reconcentrated near the nucleus, the electronic orbitals will expand out a little
bit more, and this will continue in smaller and smaller incremental steps until we declare
convergence to have been achieved based on the increment being sufficiently small.
If the original second electron had been in a 2s orbital instead of a 1s, it would
have repelled electron 1 considerably less, since it would have much smaller amplitude
near the nucleus. The original electron 1 would not have expanded as far outward at step
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