We have seen that the energy levels of a diatomic molecule in a
state may be written as

where the three terms are the energies of the electron cloud, of nuclear vibration along the internuclear axis , and rotation of the nuclei about an axis normal to .

What is the situation when the electrons have non-zero orbital angular
momentum ? In this case, if the orbital angular momentum of
the nuclei is now denoted by , the total angular momentum
is given by

Clearly

Because **L** is non-zero, the electronic potential field in which
the nuclei move is no longer really axisymmetric, and **N** is no
longer a good quantum number. However, we may approximate the main
effect of non-zero **L** on the nuclear wave functions by replacing
the centrifugal term
in the nuclear
wave equation by one in the expectation value
:

Now

Because the system is in an eigenstate of , the expectation value of a component of

The total energy of the molecule is then approximately

The quantity in square brackets depends only on the electronic state, so we rename it . All the dependence of on (the total nuclear plus electronic orbital angular momentum) is in the term , where is a constant. This looks very much like our previous result - except that now must satisfy , because must be at least as big as the component .

We now consider the rotational spectrum that should be emitted by a
diatomic molecule. Classically, we expect that radiation could be
emitted as a result of the rotation of the molecule *if* the molecule
has a net electric dipole moment, which will be the case for example
in a molecule with ionic bonding between different nuclei (e.g. CN or
OH). On the other hand, a homonuclear molecule, one with two
identical nuclei (H, N, O, etc), has no overall electric
dipole because the two ends are identical. Thus we expect that such
molecules may be inhibited from emitting dipole radiation due to
rotation.

Now look at the rotational emission problem quantum mechanically,
first for a homonuclear molecule. In the dipole approximation, we
expect that the transition amplitude between two rotational states
(having the same vibrational and electronic wave functions) will be
proportional to the matrix element of

The matrix element we are interested is that of the dipole operator between an initial state given approximately by

and a final state of the same form but with different quantum numbers and . To discover whether radiation can be emitted as a result of transitions between rotational levels, we must examine how these various terms in the matrix element change when the two nuclei are exchanged (i.e.

To summarize the effects of exchange of the nuclei:

- the dipole operator is unchanged
- a rotation state is replaced by
- the vibrational state is unchanged
- the electronic state is replaced by .

The same conclusion applies to radiation in which the vibrational
state (but not the electronic state) changes, i.e. to purely
vibrational transitions. Such radiation also requires the rotational
state to change by , so again exchange causes the sign of
the matrix element to change. We conclude that *a homonuclear
molecule in the ground electronic state does not emit purely
rotational or vibrational spectra by dipole radiation.* Such
molecules can only emit dipole radiation if the electronic state
changes; they can also emit quadrupole radiation or magnetic dipole
radiation, but this is much weaker than the suppressed dipole
radiation would be.

Heteronuclear molecules can emit a purely rotational, or a
vibrational-rotational spectrum. From the dependence of the energy of
the eigenstates on , as
, we immediately see that
that spectral lines due to rotational transitions satisfying will have frequencies (for
)

The spectral lines will be uniformly spaced in frequency, with wavelengths of the order of 0.1 mm, in the far infrared or microwave region of the spectrum.

To have spectral lines arise through the change of vibrational levels,
the matrix element

must be non-zero (note that the factor from the definition of cancels the factor from the volume element in spherical coordinates). The angular integral (over the rotational eigenfunctions) is just the one we have seen before in the one-electron problem; it is from this factor that one gets the selection rule , and we see that this rule also holds for vibrational transitions (with ). Now if

We ignore higher terms since the vibration amplitude is small. Now the matrix element that matters is given by

Using a recursion relation for the Hermite polynomials, this integral vanishes unless

(Because of the fact that the simple harmonic oscillator is only an approximation to the actual potential, transitions with can also occur, but these are usually considerably weaker than those for which .)

Now for a given pair of vibration levels and , the
transitions fall into two groups, those with (the R
branch) and those with (the P branch). The frequencies
for the R branch are given approximately by

while the P branch are at frequencies

The two branches together make up a vibrational-rotational band. The lines are now in the infra-red (typical wavelengths of a few m), and are evenly spaced in frequency (with spacing ) except for a gap wide at the position of caused by the fact that must change by . In actual spectra, the spacing of the lines is not quite uniform because the constant is actually slightly different in the two vibrational bands.

For a molecule with , transitions with (the Q branch) are also possible. If the two values are essentially equal, all the Q branch lines occur at the frequency , which instead of being absent is quite strong.

Another important effect involving vibrational and rotational levels
is Raman scattering. In this effect, a photon is scattered by the
molecule, effectively by an absorption immediately followed by an
emission to a state near the original one, so that the frequency of
the scattered photon is changed slightly. The absorption changes
by one, and the re-emission does so as well. Thus in the end, the
final emission satisfies the selection rule
. It
is found that the Raman effect does *not* require a permanent
electric dipole moment; the moment induced by the radiation field
itself is enough to make the process occur. This means that the Raman
effect provides a means of probing the vibrational-rotational levels
in molecules such as O and N which normally have no intrinsic
vibrational-rotational spectrum.

Electronic spectra arise from transitions in which the electronic
state of the molecule changes - these are the transitions most nearly
analogous to atomic transitions, and typically involve photons in the
visible and ultraviolet parts of the spectrum. At low resolution,
electronic spectra seem to be made up of series of more or less evenly
space *bands*; at higher resolution, each band is made of many
individual spectral lines.

To understand these spectra, recall that we have found that the energy
of a single level of electronic state , vibrational state , and
rotational level may be written as a sum of these three energies,
, so that the frequency of a particular
transition will be given by

For a particular pair of electronic levels , , this equation is very similar to that for vibrational-rotational spectrum, except in two respects. First, because two different electronic levels are involved, will be rather different from and (more importantly) will be different from (because different electronic states have different equilibrium positions and hence different moments of inertia); and secondly, the frequency of the transition is moved from the infrared into the visible or UV by the addition of the large term due to .

Selection rules control which electronic states can make strong transitions with each other; these are somewhat complicated and will not be treated here. There is no selection rule on the difference , since the two vibrational levels are not formed in the same electronic potential well, and values of up to 5 or 6 are not uncommon. Since the rotational energies involve the same angular functions (the 's) in both states, they continue to observe the selection rule between two states, or for states with .

For a given pair of electronic levels , , each of the bands
seen at low resolution corresponds to a particular value of . Writing the part of the energy difference due to the vibrational
levels as

we see that for a given , various values of will form an evenly spaced series of bands with spacing proportional to , while different values will produce more widely spaced sequences of such series of bands, with spacing proportional not to but to . The equation describing this structure is known as the Deslandres formula.

Each series of bands for given values of and will have
a large number of lines (*fine structure*) because of the rich
structure of rotational levels possessed by each vibrational
level. Two or three series of rotational lines will be present for
each pair of vibrational levels, corresponding to (the
P branch), (the R branch), and perhaps
(the Q branch). For transitions between two levels, the
frequency series for the P and R branches are given by

and

Here both series of lines are clearly quadratic in because . Because the coefficient of has opposite signs for the two series of rotational lines, the two series set out in opposite directions from , but as the term dominates, with the same coefficient, one of the two series reverses direction and both go off towards higher frequency (if ) or lower (if ). This phenomenon places several lines at nearly the same frequency near the point where the direction changes, and gives each series in a

In diatomic molecules we actually have four different kinds of angular
momentum that combine in different ways. These are the orbital angular
momentum of the electrons **L**, the spin angular momentum of the
electrons **S**, the nuclear rotational angular momentum **N**, and
the nuclear spin **I**, which can almost always be neglected except
for its influence on symmetries in homonuclear molecules (see
below). These may combine in a variety of fairly complicated ways.

Hund identified some of the most common ways in which the angular
momentum combines, or *couples*. To appreciate this phenomenon, it
is useful to have first studied many-electron atoms, which you have
not yet done, so we will simply summarize a sample situation.

One possible situation is Hund's case (a), when **L** couples
strongly to **R**, so that is a good quantum number
(i.e. is well-defined for each quantum state), and also couples
strongly to **S**, so that the projection of **S** on **R**,
, is also a good quantum number. Then the sum of these
two components can take on the values

and is also a good quantum number. Each pair of numbers and define a different basic state that can have a series of rotational levels defined by the total angular momentum of the molecule

Other cases arise as other kinds of coupling dominate.

Nuclei have spin due to the intrinsic spins of protons and neutrons, which like electrons have spins of . These spins combine (couple) to form the total spin of the nucleus, which may be different for different excited states. The ground state of the nucleus always has a definite spin; for example, H has spin 1/2, while O has spin 0. The nuclear spin can couple with other angular momenta, but as mentioned above this coupling has no direct effect of importance on molecular spectra.

There are however important effects of nuclear spin in homonuclear molecules due to the operation of the Pauli exclusion principle. We know that the exclusion principle requires that the total wave function of a system be antisymmetric under exchange of identical fermions, or symmetric under exchange of bosons. Since electrons have spin 1/2 and do not combine into aggregates, the wave function must always be antisymmetric under exchange of electrons, but since nuclei can act as bosons (if they have an even number of nuclei) or fermions (with an odd number of nucleons), both symmetries are possible for nuclear exchange. Let's look at one example of how this restriction affects molecular spectra of a homonuclear molecule.

We have seen that the total wavefunction (without nuclear spin) of a
homonuclear molecule (we add in the electron spin function )

has definite symmetry under exchange of nuclei: may be symmetric or antisymmetric, is always symmetric, and is multiplied by , so that alternate levels have opposite symmetry. Now suppose for example that the nuclear spin is zero, as in O, so that the spin function is symmetrical under exchange of the nuclei. Then the rest of the wave function must be symmetric under exchange too. Suppose is also symmetric; in this case only the symmetric rotation functions can occur, and thus only the values . If is antisymmetric, is restricted to the values . Thus in each case half the normal rotational levels are missing. This will be true also of excited electronic levels, and for a pair of electronic levels between which transitions are permitted, the absence of half the rotational levels will mean that half the rotational lines are missing from each band. In the almost identical molecule OO this restriction does no apply, and the missing lines are restored.

This kind of effect in homonuclear molecules is very helpful in determining the spin of the nuclei, in spite of the fact that these nuclear spins have almost no interaction with the rest of the molecule.

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