Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Lecture 20, November 01, 2010
Solved Homework
We determined that the two coefficients in our twogaussian wave function were
c
1
= 0.3221 and
c
2
= 0.7621. We also determined that
ψ
1
s
r
,
θ
,
φ
;
c
1
,
α
1
,
c
2
,
α
2
(
)
−
1
2
∇
2
ψ
1
s
r
,
θ
,
φ
;
c
1
,
α
1
,
c
2
,
α
2
(
)
=
1.5
c
1
2
+
0.6435
c
1
c
2
+
0.3
c
2
2
Using the normalized coefficient values, we have <
T
> = 0.4873. Since <
H
> was
−
0.4819,
<
V
> must be
−
0.9692. We could also simply plug the coefficients into
ψ
1
s
r
,
θ
,
φ
;
c
1
,
α
1
,
c
2
,
α
2
(
)
−
1
r
ψ
1
s
r
,
θ
,
φ
;
c
1
,
α
1
,
c
2
,
α
2
(
)
=
−
1.5958
c
1
2
−
1.5908
c
1
c
2
−
0.7136
c
2
2
which is also (as it must be)
−
0.9692. If we take
−
2<
T
> we have –0.9746. So, although
the virial theorem is
almost
satisfied, it is not quite. The error is less than 1%.
Exact wave functions
must
satisfy the virial theorem, but that is
not
true of
approximate wave functions. As such, one may examine the virial ratio to assess the
quality of an approximate wave function. As one gets closer to the exact wave function,
one should get closer to a ratio of –2.
Antisymmetry
Consider a quantum mechanical system consisting of two indistinguishable
particles, e.g., two electrons. If we wanted to compute the probability that we would find
electron 1 in some volume of space characterized by
r
a
≤
r
≤
r
b
,
θ
a
≤
θ
≤
θ
b
, and
φ
a
≤
φ
≤
φ
b
, and at the same time find electron 2 in some volume of space characterized by
r
c
≤
r
≤
r
d
,
θ
c
≤
θ
≤
θ
d
, and
φ
c
≤
φ
≤
φ
d
, we would compute this probability as
P V
1
1
( )
,
V
2
2
( )
[
]
=
Ψ ψ
1
1
( )
,
ψ
2
2
( )
[
]
φ
c
φ
d
∫
θ
c
θ
d
∫
r
c
r
d
∫
φ
a
φ
b
∫
θ
a
θ
b
∫
r
a
r
b
∫
2
r
1
2
dr
1
sin
θ
1
d
θ
1
d
φ
1
r
2
2
dr
2
sin
θ
2
d
θ
2
d
φ
2
(201)
where the cumbersome notation is meant to emphasize that the probability has to do with
electron 1 being in volume 1 and electron 2 being in volume 2, that
Ψ
depends (in an
unspecified way) on two individual electron wave functions
ψ
, the first of which is
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202
occupied by electron 1 and the second by electron 2, and that the first set of integration
coordinates corresponds to electron 1 and the second set to electron 2.
Now, since the electrons are indistinguishable, we are not permitted to really label
them. Put differently, the probability of finding electron 2 in volume 1 and electron 1 in
volume 2 must be the same as the probability we’ve already discussed up to this point
(that is, there is a single probability of finding “an” electron in volume one and “another”
electron in volume 2). We would write the second probability just mentioned as
P V
1
2
( )
,
V
2
1
( )
[
]
=
Ψ ψ
2
1
( )
,
ψ
1
2
( )
[
]
φ
c
φ
d
∫
θ
c
θ
d
∫
r
c
r
d
∫
φ
a
φ
b
∫
θ
a
θ
b
∫
r
a
r
b
∫
2
r
1
2
dr
1
sin
θ
1
d
θ
1
d
φ
1
r
2
2
dr
2
sin
θ
2
d
θ
2
d
φ
2
(202)
where the change is simply that we have swapped the coordinates that we used to use for
electron 1 to now be associated with electron 2 and vice versa. Given that the
probabilities in eqs. 201 and 202 are equal, and given that both integrations are over the
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