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62 Foundations
symmetric with something else that is totally symmetric must give a result that is totally
symmetric. On the other hand, if either the upper or lower state vibrational wavefunction is
non-totally symmetric, then the integrand will be non-totally symmetric.
5
At this point it will be helpful to consider a specific example, CO
2
. This has the three
normal modes shown schematically in Figure 5.5. The symmetric C
−O stretch v
1
, has
a totally symmetric vibrational wavefunction for all values of the vibrational quantum
number. However, the two non-totally symmetric modes, the degenerate bend, v
2
, and the
antisymmetric stretch, v
3
, are different. For these modes, the vibrational wavefunction has a
totally symmetric component for v even (0, 2, 4,...)butmust be non-totally symmetric for
v odd (1, 3, 5, . . .). We can now work out the symmetries of the integrands in the Franck–
Condon factors and hence selection rules for all possible upper and lower state vibrational
quantum numbers in an electronic transition.
For v
1
,any value of v
1
is possible, although as for diatomic molecules there will be a
propensity for certain values. Suppose the equilibrium C
−O bond lengths are substantially
larger or smaller in the upper electronic state than in the lower. This will be the equivalent
of the case II scenario in diatomics, and a long vibrational progression in v
1
would be
expected for this electronic transition. On the other hand, if the equilibrium bond lengths
are virtually the same in the two electronic states, then this corresponds to the case I limit
and v
1
= 0 transitions will dominate, i.e. no significant vibrational progression will be
observed.
For modes v
2
and v
3
, because they involve non-totally symmetric normal coordinates,
only even quantum number changes v = 0, ±2, ±4, etc., are allowed. In fact a little more
thought will show that v = 0 transitions will dominate for these modes. For example,
unless one bond becomes longer than the other in the excited electronic state, then there is
no change in equilibrium structure in the direction of normal coordinate v
3
. This is equivalent
to a case I Franck–Condon situation applying for this mode. Similarly, if the molecule is
linear in both electronic states then there is no propensity for v
2
= 0 transitions.
This illustrates a general and important point that will be met in many examples later,
namely that the vibrational structure in electronic spectra is normally dominated by modes
with totally symmetric normal coordinates. Furthermore, the propensity for formation of a
progression in a particular mode will depend on whether there is a change in equilibrium
structure in the direction of that coordinate. If there is a substantial structural change in the
direction of only one coordinate, then only this mode will show any significant activity in
the spectrum. Thus one may have, and often finds, very simple vibrational structure arising
in the spectrum of a relatively complicated molecule.
5
A useful analogy is to liken the direct product of representations with products of the numbers +1 and −1, where
+1 represents totally symmetric and −1 represents a non-totally symmetric representation. We can therefore
instantly see that the direct product of two totally symmetric representations will give a totally symmetric result
since (+1) ×(+1) =+1. The direct product of two (identical) non-totally symmetric representations will also give
a totally symmetric result, since (−1) × (−1) =+1. On the other hand, the direct product of a totally symmetric
and non-totally symmetric representation (or vice versa) will give a non-totally symmetric representation, since
(+1) × (−1) =−1.