Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Lecture 28, November 24, 2010
(Some material in this lecture has been adapted from
Cramer, C. J.
Essentials of
Computational Chemistry
, Wiley, Chichester:
2002; pp. 118119, 154158.)
Solved Homework
Consider the diagonal and offdiagonal kinetic energy matrix elements
T
μμ
=
μ
−
1
2
∇
2
μ
T
μ
ν
=
μ
−
1
2
∇
2
ν
Let's think about what the Laplacian operator does to the function on which it is
operating. It reports back the sum of second derivatives in all coordinate directions. That
is, it is a measure of how fast the slope of the function is changing in various directions.
If two functions
μ
and
ν
are far apart, then since good basis functions go to zero at least
exponentially fast with distance,
ν
is likely to be very flat where
μ
is large. The second
derivative of a flat function is zero. So, every point in the integration will be roughly the
amplitude of
μ
times zero, and not much will accumulate. For the diagonal element, on
the other hand, the interesting second derivatives will occur where the function has
maximum
amplitude (amongst other places) so the accumulation should be much larger.
Thus, we expect diagonal elements in this case to be larger than offdiagonal elements.
Analogous arguments can be made for the nuclear attraction integrals.
V
μμ
=
μ
−
Z
k
r
k
k
nuclei
∑
μ
V
μ
ν
=
μ
−
Z
k
r
k
k
nuclei
∑
ν
The 1/
r
operator acting on
ν
will ensure that the largest contribution to the overall
integral will come from the nucleus
k
on which basis function
ν
resides. Unless
μ
also
has significant amplitude around that nucleus, it will multiply the result by roughly zero
and the whole integral will be small. In this case, however, it is conceivable (i) that off
diagonal elements involving
Z
=100 might be larger than diagonal elements about
Z
=1
and (ii) that offdiagonal elements involving two basis functions on the
same
nuclear
center might compete with diagonal elements involving a more diffuse basis function on
that center (e.g., a 1s2s interaction on the same atom might be larger than a 5f interaction
with itself, since the latter is so spread out that its amplitude isn't very large anywhere).
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282
The 2
J
–
K
terms are somewhat more subtle to evaluate, but it will still generally
be the case that diagonal terms will generate larger magnitudes than offdiagonal terms.
We will not explore the point closely.
The HartreeFock SelfConsistent Field Procedure
To find the HF MOs we need to solve the HF secular determinant
F
11
–
ES
11
F
12
–
ES
12
F
1
N
–
ES
1
N
F
21
–
ES
21
F
22
–
ES
22
F
2
N
–
ES
2
N
F
N
1
–
ES
N
1
F
N
2
–
ES
N
2
F
NN
–
ES
NN
=
0
(281)
and find its various roots. We know that we can compute overlap integrals and that Fock
matrix elements are defined by
F
μ
ν
=
μ
–
1
2
∇
2
ν
–
Z
k
k
nuclei
∑
μ
1
r
k
ν
+
P
λσ
λσ
∑
μ
ν
λσ
(
)
–
1
2
μ
λ
νσ
(
)
(282)
So, now for the tricky part.
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 Physical chemistry, pH, Quantum Chemistry, Computational chemistry, basis functions, STOs

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