Chemistry 342
Spring, 2005
Spectroscopy.
For the past several lectures, we have been concerned exclusively with solutions of the
electronic Schrödinger equation,
Ĥel Ψel
 Eel Ψel
where “el” means an operator, a function, or a constant that depends upon the electronic
coordinates. The eigenfunctions Ψel are what we call “orbitals”; their magnitudes
squared (|Ψel|2) give us information about the distributions of electrons in molecules, and
about chemical bond formation. The corresponding eigenvalues Eel tell us what the
energies of the electrons are, and how these energies depend upon the geometry of the
molecule. Knowing these dependencies gives us information about the preferred
structure of the molecule.
We now turn our attention to a variety of spectroscopic techniques that may be used to
determine what that structure is. Knowing this, of course, will also tell us about the
preferred arrangement of the electrons in the molecule.
A few, more formal, preliminaries…
The Born-Oppenheimer approximation.
The first of these is the B/O approximation. Let us begin by writing down the molecular
Schrödinger equation, in which j represents the nuclear coordinates and i the electronic
ones; then (4π ε0 = 1!)

  2


 2

j

i
1 2 2
j 
mj
2m e

j
Z je 2
rij

 i2  
i

i
j

i'

j
Z j Z'je 2
rjj'
e 2 
Ψ  EΨ
rii' 
Here, both Ψ and E depend upon both electronic and nuclear coordinates, since the
Hamiltonian ({…}) does. Clearly, we cannot solve this equation exactly for anything
other than the simplest system (e.g., H2+). To deal with larger molecules, at least
approximately, we have to involve several approximations. The most important of these
is the B/O approx…
Chemistry 342
Spring, 2005
Essentially, we say that because electrons are light “particles”, they move much more
rapidly than do the heavier nuclei, so we can separate their motions. If the motions can
be separated, then we can write the total wavefunction Ψ as a product,
Ψ = Ψel Ψnuclei
Then, we can fix the nuclei at some position, and solve the electronic Schrödinger
equation,
(A)
Ĥ el Ψel = Eel Ψel
Where Ĥ el is the electronic Hamiltonian
Ĥ el

 2
2m e


e2
rii'

i
i'
 i2  
i
j

j
Z jj' e 2
rjj'

i

j
Z je 2
rij
This is what we have been doing for the past few weeks (Oh!). Eel then tells us what the
total electronic energy is, at some particular geometry. Then we can vary that geometry,
and by solving the SEqn. again, determine how Eel varies as a function of that geometry.
(Recall H2, H2O, etc.). For a diatomic like H2, this might look like
Vel
rAB
The ‘nuclear’ Schrödinger equation.
Having solved the electronic part of the problem, we now turn our attention to the
“nuclear” part, or what is left over from the molecular Schrödinger equation. This is
Ĥnucl Ψnucl = Enucl Ψnucl
where
Ĥ nucl

 2
2

j
1 2
 j  Vel (rjj' )
mj
Chemistry 342
Spring, 2005
So V(rjj'), which we obtained by solving the electronic SEqn., plays the role of the
potential energy in the nuclear SEqn.; it provides the ‘field’ in which the nuclei move.
Now Ĥnucl must be expressed in the coordinate system of the molecule. For a diatomic
having reduced mass μ, this is
z

A
μ 
mA mB
mA  mB
x

o
y
B
origin of c/s at
center of mass.
In this c/s, Ĥnucl becomes…
  2    2  
1  
 
R

 sin θ 

2 
R  sin θ θ 
θ 
 2μR  R 


1
2

V
(
R
)
 Ψnucl (R , θ, φ)  E nuclΨnucl (R , θ, φ).
sin 2 θ φ 2

Here, we have set V(R) = Eel. Notice that this equation has exactly the same form as the
SEqn. for the H atom (!), except that me → μ and -Ze2/r → V(r). Thus, by analogy, we
can expect that the ‘nuclear’ SEqn. can also be solved by the separation of variables
technique…
Ψnucl (R, θ, φ) = Ψv(R) Ψr(θ, φ).
The Ψr (θ, φ) are wavefunctions that depend only on θ and φ; these are eigenfunctions of
the angular momentum operator squared…
Ĵ 2 Ψr (θ, φ) = J (J+1) ħ2 Ψr (θ, φ)
(B)
where
Ĵ 2
 1


1
2 

   2 
sin θ

θ sin 2 θ φ 2 
 sin θ θ
is an operator representing (the square of) the rotational angular momentum of our
molecule. Thus, the Ψr (θ, φ) are the same functions as the spherical harmonics of the
rigid rotor problem
Chemistry 342
Spring, 2005
Ψr (θ, φ)  PJk (cos θ) eiKφ
If we now substitute these results back into the nuclear SEqn., we obtain
   2

 2μR 2

  2 
Ĵ 2 

  V(R ) Ψv (R ) Ψr (θ, φ) 
R

 R
R  2 


E Ψv (R ) Ψr (θ, φ).
However, since
Ĵ 2 Ψr (θ, φ) = J (J+1) ħ2 Ψr (θ, φ)
and since all other terms are independent of θ and φ, Ψr (θ, φ) cancels on both sides of
the previous equation, leaving us with a single equation in a single variable…


   2   2  Ĵ 2 

R


V
(
R
)

 Ψv (R )  E Ψv (R )
2 
2 
R  


 2μR  R

If we define Erot = Ĵ 2 / 2 μ R 2 , and subtract it from each side of the equation, we have
  2   2  

R

  V(R ) Ψv (R )  (E  E rot )Ψv (R )  E vib Ψv (R ).

2
R 
 2μR  R

(C)
This is the vibrational SEqn. It will turn out, at least approximately, that this eqn. is
analogous to that of the (1D) harmonic oscillator, easily solved.
So, to summarize,

solve (A) to obtain Ψel, Eel

solve (B) to obtain Ψrot, Erot

solve (C) to obtain Ψvib, Evib
then
Ψtot = Ψel · Ψnucl = Ψel · Ψvib · Ψrot
Etot = Eel + Enucl = Eel + Evib + Erot.