The simplest molecule possible is the* *H_{2}^{+} molecular
ion. It consists of two protons and one electron. Assume that you could arbitrarily fix
the distance between the protons. When the distance between the protons goes to infinity
then they do no longer interact. The ground state energy of the system will be -13.6 eV,
when the electron forms a hydrogen atom with one of the protons. When the distance between
the protons is on the order of nuclear dimensions then the electron will see a nucleus of
charge *Z*=2. The binding energy of the electron in the ground state will be -54.4
eV, the binding energy of the electron in the He^{+} ion. The electron will
be more strongly bound, but the protons will also repel each other strongly, so that the
total energy of the system will be positive. As the distance between the protons is
varied, the total energy of the ground state reaches a minimum for a certain value of the
distance. This explains the stability of the H_{2}^{+} molecular
ion.

In order to treat the problem exactly, one must solve the eigenvalue equation of the
three particle system. However, it is possible to simplify the problem by using the **Born-Oppenheimer approximation**.
The mass of the electron is
much smaller than the mass of the protons, therefore the electron moves much more quickly
than the protons. The electronic wavefunction adapts itself nearly instantaneously to any
change in the inter-nuclear distance *r*. At any inter-nuclear distance *r* the
Hamiltonian describing the nuclear motion therefore contains the terms associated with the
kinetic energy of the nuclei, the nucleus-nucleus interaction, and the total energy of the
electron at that inter-nuclear distance. The nuclei move in an effective potential

.

To find *E _{electron}(r)* the eigenvalue equation of the Hamiltonian of
the electron, subject to the attraction of two fixed protons, must be solved.
The distance

For a general diatomic molecule we want to solve the problem of two nuclei of mass *m*_{1}
and* m*_{2} in a potential *V _{eff}(r)*. There exist vibrational
and rotational degrees of freedom. These degrees of freedom are coupled. When the molecule
vibrates, its moment of inertia changes because the inter-nuclear distance

Let us first consider only **small amplitude vibrations**.

The eigenvalue equation of the Hamiltonian for the relative motion of the two nuclei is

,

where *m* is the reduced mass, .

If the angular momentum of the nuclei is zero, then the eigenvalue equation reduces to
the radial equation with *l*=0,

.

Let us expand *V(r)* about the equilibrium position *r=r _{e}*.

.

For small *r’=r-r _{e}* we have

,

or

.

This is the eigenvalue equation of a one-dimensional harmonic oscillator Hamiltonian. The eigenvalues are

and we therefore have for the eigenenergies of the vibrational motion of the two nuclei

.

Typical vibrational frequencies are on the order of 10^{12} to 10^{14}
Hz and fall in the infrared.

The ground state wavefunction of the harmonic oscillator has a finite spread on the order of . (See notes.) If , then the moment of inertia is nearly constant and to a good approximation the vibrational and rotational degrees of freedom are decoupled.

**Heteropolar molecules**, composed of two different atoms, generally have a
permanent dipole moment * p(r)*, because the electrons are attracted more
towards one of the atoms. We may expand

**Homopolar molecules**, consisting of two identical atoms, are inactive in the
infrared. The Hamiltonian for the nuclear motion is not perturbed by the oscillating
electric field, because the charge distribution has no permanent dipole moment.
However
the Hamiltonian for the electronic motion is perturbed, and therefore the molecule can
acquire an induced dipole moment. If the frequency of the oscillating electric field lies
in the optical range transitions between electronic states can be induced.
Photons of
energy can be absorbed and reemitted.
This phenomenon is called Rayleigh scattering. If during the scattering a transition
between occurs,
then the energy of the scattered photon will be .
We therefore can observe scattered photons of three different
frequencies.

( Rayleigh line ),

( Raman-Stokes line ),

( Raman-anti Stokes line ).

This can also be observed with heteropolar molecules.

Let us now study the **rotational motion** of
molecules. We neglect vibrations and assume that the distance *r* between the nuclei
remains constant. We then have a **rigid rotator**. (See notes.) We have

.

The eigenfunctions of *H* are the eigenfunctions of *L ^{2}*.
These
are the spherical harmonics, . The
eigenvalues are . It is customary to
set , and write .
The separation
between adjacent levels is

,

it increases linear with *l*. Each energy eigenvalue is (*2l+1*) fold
degenerate.

If a heteropolar molecule with a dipole moment * p* is placed in an
oscillating electric field , the
probability that the field induces a transition from an eigenstate of the rotational
Hamiltonian

With , the matrix element is proportional to

.

With

we have

.

The selection rules are .
For
polarization of the electric field along the *x*- and *y*-axis we find .
The pure rotational absorption and
emission spectrum is therefore composed of a series of equidistant lines. The frequency
separation between two successive lines is *2B*. Pure rotational spectra fall in the
very far infrared or the microwave region.

Homopolar molecules which have no permanent dipole moment are inactive in the microwave region. However, like the vibrations, the rotation of the molecule can be observed via the Raman effect in the inelastic scattering of light by the molecule.

In the Born-Oppenheimer approximation we assume that the potential energy of
interaction between the two nuclei depends only on the inter-nuclear separation *r.*
The relative motion of the two nuclei can therefore be described in terms of the motion of
a fictitious particle of reduced mass *m* moving in a
central potential. The stationary state wavefunctions of this particle are of the form , where satisfies

.

We have .

For *l=0* we have already solved the problem for small amplitude oscillations and
obtained the purely vibrational energy levels of the molecule, .

For we approximate .
We assume that *l* is sufficiently
small such that . Variations in the centrifugal
potential are of the order and are
neglected just as anharmonic terms in *V(r)* are neglected. Then satisfies the equation

,

which again is the eigenvalue equation of a harmonic oscillator. We therefore have

.

The radial functions do not depend on *l*, since does not depend on* l*.
The wave functions of the stationary states therefore are , with being a harmonic
oscillator eigenfunction. The first two vibrational levels with their rotational structure
are shown in the figure below.

Infrared absorption and emission by heteropolar molecules

If a heteropolar molecule with a dipole moment * p* is placed in an
oscillating electric field , the
probability that the field induces a transition from an eigenstate of the
vibrational-rotational Hamiltonian

.

The selection rules for emission and absorption of infrared radiation therefore are .

The set of lines with constitutes a purely rotational spectrum.
For let n*’*
be the larger of the two vibrational quantum numbers. Then we can separate the
vibrational-rotational lines into two groups, .

For the frequency of the photon is given by

.

We observe the lines in the R branch of the figure below.

For the frequency of the photon is given by

.

We observe the lines in the P branch. In each branch the distance between adjacent
lines is *2B*. The central interval separating the two branches is of width *4B*.
There is no line at the pure vibrational frequency . The pure vibrational spectrum consisting of a single line does not exist.
But with an instrument with poor resolution we may not be able to resolve the rotational
structure and treat the band shown in the figure as a single line.

Links:

- Microwave and Infrared Spectra from Web Databases
- Vibrational-Rotational Spectrum Simulator
- Molecular Vibrations (in Spanish)