Biomedical Engineering Reference
In-Depth Information
We treat the nuclei separately; if we want to know about vibrational and rotational motion
we have to solve the relevant nuclear vibrational Schrödinger equation and the relevant
nuclear rotational Schrödinger equation. This is not the same as doing a MM calculation
for the nuclei: there are proper quantum mechanical vibrational and rotational Schrödinger
equations as we have seen in Chapter 11.
The separation of electronic and nuclear motions allows us to specify a molecular geo-
metry; the electrons experience an electrostatic potential energy due to the nuclei, which
are fixed at positions in space for the purpose of calculating the electronic wavefunction.
Just as for MM, molecular geometries can be found by investigating stationary points on
the potential energy surface.
The electronicwavefunction is given as the solution of an electronic Schrödinger equation
∂
2
∂
x
2
∂
x
2
ψ
e
(
R
A
,
R
B
,
r
)
+
U
ψ
e
(
R
A
,
R
B
,
r
)
=
ε
e
ψ
e
(
R
A
,
R
B
,
r
)
h
2
8π
2
m
e
∂
2
∂
x
2
∂
2
(15.1)
−
+
+
where the electrostatic potential energy is
e
2
4πε
0
−
1
1
U
=
R
A
|
+
|
r
−
|
r
−
R
B
|
We solve the electronic equation and the total energy is given by adding on the fixed nuclear
repulsion
e
2
4πε
0
|
R
A
−
ε
tot
=
ε
e
+
(15.2)
R
B
|
The hydrogen molecule ion is unique amongst molecules in that we can solve the elec-
tronic Schrödinger equation exactly (by numerical methods) to any required accuracy,
within the Born-Oppenheimer approximation. The first step is to make a change of variable
to so-called
elliptic coordinates
that are defined with reference to Figure 15.2:
r
A
+
r
B
R
AB
r
A
−
r
B
R
AB
μ
=
ν
=
φ
In this system of coordinates the electronic Schrödinger equation is
∂
∂μ
μ
2
1
1
∂
∂μ
ν
2
∂
∂ν
h
2
8π
2
m
e
4
R
AB
(μ
2
∂
∂ν
−
−
+
−
−
ν
2
)
ψ
e
+
μ
2
−
ν
2
∂
2
∂φ
2
+
U
ψ
e
=
ε
e
ψ
e
(15.3)
(μ
2
−
1)(1
−
ν
2
)
and we look for a separation of variables solution
ψ
e
(μ, ν, φ)
=
M
(μ)
N
(ν) Φ (φ)
Since φ only enters the equation in the ∂
2
/∂φ
2
term, it is at once apparent that we can factor
off the φ term. If we call the separation parameter
−
λ
2
we have
d
2
Φ
dφ
2
=−
λ
2
Φ
and so
1
√
2π
exp (jλφ)
Φ (φ)
=
(15.4)
