Born-Oppenheimer Approximation
Atoms and molecules consist of heavy nuclei and light electrons. Consider (for simplicity)
a diatomic molecule (e.g. HCl). Clamp/freeze the nuclei in space, a distance r
0
apart.
Under this condition, solve the Schroedinger equation for the stationary electronic states.
This solution gives us the ground electronic state energy E
elect
0
(r
0
), the first excited electronic
state energy E
elect
1
(r
0
), etc. Now repeat this procedure for a distance r
1
, r
2
, . . .. “connect
the dots” to obtain the potential energy surfaces. Consider being at r
0
with clamped/frozen
nuclei, and being in the ground electronic state. Unclamp/unfreeze the nuclei, and suppose
r begins to increase and is now r = r
1
. What is the energy of the electrons (which interact,
of course, with the nuclei)? The Born-Oppenheimer, or adiabatic, approximation says that
at r = r
1
, the electrons behave as if the nuclei were always at r = r
1
(frozen). In other words,
the electrons respond instantaneously (adapt immediately) to changes in r. The electrons
“shadow” the nuclei. Since the electrons “track” the nuclei, we only need to figure out how
the nuclei behave, and then we will know where the electrons are (in the Born-Oppenheimer
picture). E
elect
also includes the potential energy between the (clamped) nuclei. Hence E
elect
is the potential energy function under which the nuclei move. Whence to find the stationary
nuclear states and energies, we should solve
N
X
i=1
−
~
2
2m
i
∇
2
i
ψ
| {z }
kinetic energy
+ E
elect
ψ
| {z }
potential energy
= E
|{z}
eigenvalues
ψ
|{z}
eigenfunctions
for N nuclei (1)
For a diatomic this equation becomes a central force problem. Why? because E
elect
=
E
elect
(r) where r = internuclear separation. The energy E for a central force problem is
obtained by solving the radial equation after changing to spherical polar coordinates. This
looks like (using the reduced mass 1/µ = 1/M
1
+ 1/M
2
)
−
~
2
2µ
1
r
2
d
dr
r
2
R
r
− `(` + 1)R
+ E
elect
(r)R = ER (2)
To proceed, define F (r) = rR(r). Then
dR
dr
=
d
dr
F
r
=
1
r
dF
dr
−
F
r
2
(3)
and
d
dr
r
2
dR
dr
=
d
dr
r
dF
dr
− F
= r
d
2
F
dr
2
+
dF
dr
−
dF
dr
= r
d
2
F
dr
2
(4)
1